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0
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Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$
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Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.
Note that
\begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*}
so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$
It follows that the equation of $R(x)$ is \[R(x)=-\frac12x+c\] for some constant $c,$ and we wish to find $R(0)=c.$
We substitute $x=20$ into this equation to get $106=-\frac12\cdot20+c,$ from which $c=\boxed{116}.$
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116
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1
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1
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Three spheres with radii $11$, $13$, and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$, $B$, and $C$, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$. Find $AC^2$.
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This solution refers to the Diagram section.
We let $\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$. We take a cross-section that contains $A$ and $B$, which contains these two spheres but not the third, as shown below:
Because the plane cuts out congruent circles, they have the same radius and from the given information, $AB = \sqrt{560}$. Since $ABO_BO_A$ is a trapezoid, we can drop an altitude from $O_A$ to $BO_B$ to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is $\sqrt{560}$ and let the distance from $O_B$ to $D$ be $x$. Then we have $x^2 = 576-560 \implies x = 4$.
We have $AO_A = BD$ because of the rectangle, so $\sqrt{11^2-r^2} = \sqrt{13^2-r^2}-4$.
Squaring, we have $121-r^2 = 169-r^2 + 16 - 8 \cdot \sqrt{169-r^2}$.
Subtracting, we get $8 \cdot \sqrt{169-r^2} = 64 \implies \sqrt{169-r^2} = 8 \implies 169-r^2 = 64 \implies r^2 = 105$.
We also notice that since we had $\sqrt{169-r^2} = 8$ means that $BO_B = 8$ and since we know that $x = 4$, $AO_A = 4$.
We take a cross-section that contains $A$ and $C$, which contains these two spheres but not the third, as shown below:
We have $CO_C = \sqrt{19^2-r^2} = \sqrt{361 - 105} = \sqrt{256} = 16$. Since $AO_A = 4$, we have $EO_C = 16-4 = 12$. Using Pythagorean theorem, $O_AE = \sqrt{30^2 - 12^2} = \sqrt{900-144} = \sqrt{756}$. Therefore, $O_AE^2 = AC^2 = \boxed{756}$.
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756
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_10
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2
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Let $ABCD$ be a parallelogram with $\angle BAD < 90^\circ.$ A circle tangent to sides $\overline{DA},$ $\overline{AB},$ and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ,$ as shown. Suppose that $AP=3,$ $PQ=9,$ and $QC=16.$ Then the area of $ABCD$ can be expressed in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$
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Let's redraw the diagram, but extend some helpful lines.
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \cdot (3+9) = 36$. Then $AT_2 = AT_3 = \sqrt{36} = 6$. Similarly, the power of $C = 16 \cdot (16+9) = 400$ and $CT_1 = \sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and label the diagram accordingly.
Notice that because $BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x$. Let $O$ be the center of the circle. Since $OT_1$ and $OT_2$ intersect $BC$ and $AD$, respectively, at right angles, we have $T_2T_1CD$ is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from $D$ to $BC$ and $C$ to $AD$, and both are equal to $2r$. Since $T_1E = T_2D$, $20 - CE = 14+x \implies CE = 6-x$. Since $CE = DF, DF = 6-x$ and $AF = 6+14+x+6-x = 26$. We can now use Pythagorean theorem on $\triangle ACF$; we have $26^2 + (2r)^2 = (3+9+16)^2 \implies 4r^2 = 784-676 \implies 4r^2 = 108 \implies 2r = 6\sqrt{3}$ and $r^2 = 27$.
We know that $CD = 6+x$ because $ABCD$ is a parallelogram. Using Pythagorean theorem on $\triangle CDF$, $(6+x)^2 = (6-x)^2 + 108 \implies (6+x)^2-(6-x)^2 = 108 \implies 12 \cdot 2x = 108 \implies 2x = 9 \implies x = \frac{9}{2}$. Therefore, base $BC = 20 + \frac{9}{2} = \frac{49}{2}$. Thus the area of the parallelogram is the base times the height, which is $\frac{49}{2} \cdot 6\sqrt{3} = 147\sqrt{3}$ and the answer is $\boxed{150}$
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150
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_11
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3
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For any finite set $X$, let $| X |$ denote the number of elements in $X$. Define
\[S_n = \sum | A \cap B | ,\]
where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$.
For example, $S_2 = 4$ because the sum is taken over the pairs of subsets
\[(A, B) \in \left\{ (\emptyset, \emptyset) , ( \{1\} , \{1\} ), ( \{1\} , \{2\} ) , ( \{2\} , \{1\} ) , ( \{2\} , \{2\} ) , ( \{1 , 2\} , \{1 , 2\} ) \right\} ,\]
giving $S_2 = 0 + 1 + 0 + 0 + 1 + 2 = 4$.
Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p + q$ is divided by
1000.
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Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$. The problem states that for $n=2, S_n$ is $4$. Let's try it out for $n=3$.
Let's perform casework on the number of elements in $A, B$.
$\textbf{Case 1:} |A| = |B| = 1$
In this case, the only possible equivalencies will be if they are the exact same element, which happens $3$ times.
$\textbf{Case 2:} |A| = |B| = 2$
In this case, if they share both elements, which happens $3$ times, we will get $2$ for each time, and if they share only one element, which also happens $6$ times, we will get $1$ for each time, for a total of $12$ for this case.
$\textbf{Case 3:} |A| = |B| = 3$
In this case, the only possible scenario is that they both are the set $\{1,2,3\}$, and we have $3$ for this case.
In total, $S_3 = 18$.
Now notice, the number of intersections by each element $1 \ldots 3$, or in general, $1 \ldots n$ is equal for each element because of symmetry - each element when $n=3$ adds $6$ to the answer. Notice that $6 = \binom{4}{2}$ - let's prove that $S_n = n \cdot \binom{2n-2}{n-1}$ (note that you can assume this and answer the problem if you're running short on time in the real test).
Let's analyze the element $k$ - to find a general solution, we must count the number of these subsets that $k$ appears in. For $k$ to be in both $A$ and $B$, we need both sets to contain $k$ and another subset of $1$ through $n$ not including $k$. ($A = \{k\} \cup A'| A' \subset \{1,2,\ldots,n\} \land A' \not \subset \{k\}$ and
$B = \{k\} \cup B'| B' \subset \{1,2,\ldots,n\} \land B' \not \subset \{k\}$)
For any $0\leq l \leq n-1$ that is the size of both $A'$ and $B'$, the number of ways to choose the subsets $A'$ and $B'$ is $\binom{n-1}{l}$ for both subsets, so the total number of ways to choose the subsets are $\binom{n-1}{l}^2$.
Now we sum this over all possible $l$'s to find the total number of ways to form sets $A$ and $B$ that contain $k$. This is equal to $\sum_{l=0}^{n-1} \binom{n-1}{l}^2$. This is a simplification of Vandermonde's identity, which states that $\sum_{k=0}^{r} \binom{m}{k} \cdot \binom{n}{r-k} = \binom{m+n}{r}$. Here, $m$, $n$ and $r$ are all $n-1$, so this sum is equal to $\binom{2n-2}{n-1}$. Finally, since we are iterating over all $k$'s for $n$ values of $k$, we have $S_n = n \cdot \binom{2n-2}{n-1}$, proving our claim.
We now plug in $S_n$ to the expression we want to find. This turns out to be $\frac{2022 \cdot \binom{4042}{2021}}{2021 \cdot \binom{4040}{2020}}$. Expanding produces $\frac{2022 \cdot 4042!\cdot 2020! \cdot 2020!}{2021 \cdot 4040! \cdot 2021! \cdot 2021!}$.
After cancellation, we have \[\frac{2022 \cdot 4042 \cdot 4041}{2021 \cdot 2021 \cdot 2021} \implies \frac{4044\cdot 4041}{2021 \cdot 2021}\]
$4044$ and $4041$ don't have any common factors with $2021$, so we're done with the simplification. We want to find $4044 \cdot 4041 + 2021^2 \pmod{1000} \equiv 44 \cdot 41 + 21^2 \pmod{1000} \equiv 1804+441 \pmod{1000} \equiv 2245 \pmod{1000} \equiv \boxed{245}$
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245
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_12
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4
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Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a,$ $b,$ $c,$ or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, both $4$ and $410$ are counted among the distinct numerators for numbers in $S$ because $0.\overline{3636} = \frac{4}{11}$ and $0.\overline{1230} = \frac{410}{3333}.$ Find the remainder when $N$ is divided by $1000.$
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$0.\overline{abcd}=\frac{abcd}{9999} = \frac{x}{y}$, $9999=9\times 11\times 101$.
Then we need to find the number of positive integers $x$ that (with one of more $y$ such that $y|9999$) can meet the requirement $1 \leq {x}\cdot\frac{9999}{y} \leq 9999$.
Make cases by factors of $x$. (A venn diagram of cases would be nice here.)
Case $A$:
$3 \nmid x$ and $11 \nmid x$ and $101 \nmid x$, aka $\gcd (9999, x)=1$.
Euler's totient function counts these:
\[\varphi \left(3^2 \cdot 11 \cdot 101 \right) = ((3-1)\cdot 3)(11-1)(101-1)= \bf{6000}\] values (but it's enough to note that it's a multiple of 1000 and thus does not contribute to the final answer)
Note: You don't need to know this formula. The remaining cases essentially re-derive the same computation for other factors of $9999$. This case isn't actually different.
The remaining cases have $3$ (or $9$), $11$, and/or $101$ as factors of $abcd$, which cancel out part of $9999$.
Note: Take care about when to use $3$ vs $9$.
Case $B$: $3|x$, but $11 \nmid x$ and $101 \nmid x$.
Then $abcd=9x$ to leave 3 uncancelled, and $x=3p$,
so $x \leq \frac{9999}{9} = 1111$, giving:
$x \in 3 \cdot \{1, \dots \left\lfloor \frac{1111}{3}\right\rfloor\}$,
$x \notin (3\cdot 11) \cdot \{1 \dots \left\lfloor \frac{1111}{3\cdot 11}\right\rfloor\}$,
$x \notin (3 \cdot 101) \cdot \{1 \dots \left\lfloor \frac{1111}{3 \cdot 101}\right\rfloor\}$,
for a subtotal of $\left\lfloor \frac{1111}{3}\right\rfloor - (\left\lfloor\frac{1111}{3 \cdot 11}\right\rfloor + \left\lfloor\frac{1111}{3 \cdot 101}\right\rfloor ) = 370 - (33+3) = \bf{334}$ values.
Case $C$: $11|x$, but $3 \nmid x$ and $101 \nmid x$.
Much like previous case, $abcd$ is $11x$, so $x \leq \frac{9999}{11} = 909$,
giving $\left\lfloor \frac{909}{11}\right\rfloor - \left(\left\lfloor\frac{909}{11 \cdot 3}\right\rfloor + \left\lfloor\frac{909}{11 \cdot 101}\right\rfloor \right) = 82 - (27 + 0) = \bf{55}$ values.
Case $D$: $3|x$ and $11|x$ (so $33|x$), but $101 \nmid x$.
Here, $abcd$ is $99x$, so $x \leq \frac{9999}{99} = 101$,
giving $\left\lfloor \frac{101}{33}\right\rfloor - \left\lfloor \frac{101}{33 \cdot 101}\right\rfloor = 3-0 = \bf{3}$ values.
Case $E$: $101|x$.
Here, $abcd$ is $101x$, so $x \leq \frac{9999}{101} = 99$,
giving $\left\lfloor \frac{99}{101}\right\rfloor = \bf{0}$ values, so we don't need to account for multiples of $3$ and $11$.
To sum up, the answer is \[6000+334+55+3+0\equiv\boxed{392} \pmod{1000}.\]
Clarification
In this context, when the solution says, "Then $abcd=9x$ to leave 3 uncancelled, and $x=3p$," it is a bit vague. The best way to clarify this is by this exact example - what is really meant is we need to divide by 9 first to achieve 1111, which has no multiple of 3; thus, given that the fraction x/y is the simplest form, x can be a multiple of 3.
Similar explanations can be said when the solution divides 9999 by 11, 101, and uses that divided result in the PIE calculation rather than 9999.
mathboy282
\[\text{To begin, we notice that all repeating decimals of the form }0.\overline{abcd}\text{ where }a,b,c,d\text{ are digits can be expressed of the form }\frac{\overline{abcd}}{9999}\text{.}\]
\[\text{However, when }\overline{abcd}\mid 9999\text{, the fraction is not in lowest terms.}\]
\[\text{Since }9999 = 3^2 \cdot 11 \cdot 101\text{, } x\mid 9999\iff x\mid 3\lor x\mid 11\lor x\mid 101\text{.}\]
\[\text{(For those of you who have no idea what that meant, it means every divisor of 9999 is a divisor of at least one of the following: )}\]
\[(3)\]
\[(11)\]
\[(101)\]
\[\text{(Also, I'm not going to give you explanations for the other logic equations.)}\]
\[\text{Let's say that the fraction in lowest terms is }\frac{x}{y}\text{.}\]
\[\text{If }x\mid 101\text{, then }99\mid y\text{ but that can't be, since }0\text{ is the only multiple of }101\text{ below }99\text{.}\]
\[\exists! f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))\implies f=3\lor f=11 (1)\]
\[\text{If (1) is true, then we have two cases. If it isn't, we also have two cases.}\]
\[\textbf{\textit{Case 1: }}f=3\]
\[y=1111\land x=3z\implies 1\leq z\leq 370\]
\[370-33-3^{[1]}=334\]
\[\textbf{\textit{Case 2: }}f=11\]
\[y=909\land x=11z\implies 1\leq z\leq 82\]
\[82-27=55\]
\[\textbf{\textit{Case 3: }}\neg\exists! f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))\land \exists f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))=\]
\[\exists f_1(f_1\in\mathbb{N}\land f_1\neq 1\land\exists g_1(g_1\nleq 0 \land x \mid f_1^{g_1})\land\exists f_2(f_2\neq f_1\land f_2\in\mathbb{N}\land f_2\neq 1\land\exists g_2(g_2\nleq 0 \land x \mid f_2^{g_2}))\implies f_1=3\land f_2=11\lor f_1=11\land f_2=3\]
\[y=101\land x=33z\implies 1\leq z\leq 3\]
\[\textbf{\textit{Case 4: }}\neg\exists f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))\]
\[\Phi (9999)=6000\]
\[\textbf{\textit{Grand Finale}}\]
\[\text{Adding the outcomes, }N=6000+334+55+3=6392\equiv\boxed{392}\text{ (mod 1000).}\]
\[\textit{[1] This is to make sure that 3 is the \textbf{only} factor of x}\]
Note
\[\text{When I tried to write LaTeX, AoPS kept putting the LaTeX on a new line so I gave up and put most of it in LaTeX instead.}\]
\[\text{Some of the text in this section is just normal.}\]
\[\text{Example:}\]
Normal text \[\text{This is some LaTeX.}\] More normal text
\[\text{If any of you can fix this issue, please do so.}\]
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392
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_13
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5
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Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^\circ.$ Find the perimeter of $\triangle ABC.$
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Steven Chen (www.professorchenedu.com)
We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \cdot 73^2$. It can be shown that $3|a$ and $3|b$, so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for.
Notice that $r^2+r+1=(r-\omega)(r-{\omega}^2)$, where $\omega=e^{i\frac{2\pi}{3}}$. Thus,
\[x^2+xy+y^2 = y^2((x/y)^2+(x/y)+1) = y^2 (\frac{x}{y}-\omega)(\frac{x}{y}-{\omega}^2) = (x-y\omega)(x-y{\omega}^2).\]
Note that $8^2+1^2+8 \cdot 1=73$. Thus, $(8-\omega)(8-{\omega}^2)=73$. Squaring both sides yields
\begin{align} (8-\omega)^2(8-{\omega}^2)^2&=73^2\\ (63-17\omega)(63-17{\omega}^2)&=73^2. \end{align}
Thus, by $(2)$, $(63, 17)$ is a solution to $x^2+xy+y^2=73^2$. This implies that $a=189$ and $b=51$, so our final answer is $189+51+219=\boxed{459}$.
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459
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_14
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6
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Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations:
\begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*}
Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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First, let define a triangle with side lengths $\sqrt{2x}$, $\sqrt{2z}$, and $l$, with altitude from $l$'s equal to $\sqrt{xz}$. $l = \sqrt{2x - xz} + \sqrt{2z - xz}$, the left side of one equation in the problem.
Let $\theta$ be angle opposite the side with length $\sqrt{2x}$. Then the altitude has length $\sqrt{2z} \cdot \sin(\theta) = \sqrt{xz}$ and thus $\sin(\theta) = \sqrt{\frac{x}{2}}$, so $x=2\sin^2(\theta)$ and the side length $\sqrt{2x}$ is equal to $2\sin(\theta)$.
We can symmetrically apply this to the two other equations/triangles.
By law of sines, we have $\frac{2\sin(\theta)}{\sin(\theta)} = 2R$, with $R=1$ as the circumradius, same for all 3 triangles.
The circumcircle's central angle to a side is $2 \arcsin(l/2)$, so the 3 triangles' $l=1, \sqrt{2}, \sqrt{3}$, have angles $120^{\circ}, 90^{\circ}, 60^{\circ}$, respectively.
This means that by half angle arcs, we see that we have in some order, $x=2\sin^2(\alpha)$, $y=2\sin^2(\beta)$, and $z=2\sin^2(\gamma)$ (not necessarily this order, but here it does not matter due to symmetry), satisfying that $\alpha+\beta=180^{\circ}-\frac{120^{\circ}}{2}$, $\beta+\gamma=180^{\circ}-\frac{90^{\circ}}{2}$, and $\gamma+\alpha=180^{\circ}-\frac{60^{\circ}}{2}$. Solving, we get $\alpha=\frac{135^{\circ}}{2}$, $\beta=\frac{105^{\circ}}{2}$, and $\gamma=\frac{165^{\circ}}{2}$.
We notice that \[[(1-x)(1-y)(1-z)]^2=[\sin(2\alpha)\sin(2\beta)\sin(2\gamma)]^2=[\sin(135^{\circ})\sin(105^{\circ})\sin(165^{\circ})]^2\] \[=\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \left(\frac{\sqrt{2}}{8}\right)^2=\frac{1}{32} \to \boxed{033}. \blacksquare\]
- kevinmathz
(This eventually whittles down to the same concept as Solution 1)
Note that in each equation in this system, it is possible to factor $\sqrt{x}$, $\sqrt{y}$, or $\sqrt{z}$ from each term (on the left sides), since each of $x$, $y$, and $z$ are positive real numbers. After factoring out accordingly from each terms one of $\sqrt{x}$, $\sqrt{y}$, or $\sqrt{z}$, the system should look like this:
\begin{align*} \sqrt{x}\cdot\sqrt{2-y} + \sqrt{y}\cdot\sqrt{2-x} &= 1 \\ \sqrt{y}\cdot\sqrt{2-z} + \sqrt{z}\cdot\sqrt{2-y} &= \sqrt2 \\ \sqrt{z}\cdot\sqrt{2-x} + \sqrt{x}\cdot\sqrt{2-z} &= \sqrt3. \end{align*}
This should give off tons of trigonometry vibes. To make the connection clear, $x = 2\cos^2 \alpha$, $y = 2\cos^2 \beta$, and $z = 2\cos^2 \theta$ is a helpful substitution:
\begin{align*} \sqrt{2\cos^2 \alpha}\cdot\sqrt{2-2\cos^2 \beta} + \sqrt{2\cos^2 \beta}\cdot\sqrt{2-2\cos^2 \alpha} &= 1 \\ \sqrt{2\cos^2 \beta}\cdot\sqrt{2-2\cos^2 \theta} + \sqrt{2\cos^2 \theta}\cdot\sqrt{2-2\cos^2 \beta} &= \sqrt2 \\ \sqrt{2\cos^2 \theta}\cdot\sqrt{2-2\cos^2 \alpha} + \sqrt{2\cos^2 \alpha}\cdot\sqrt{2-2\cos^2 \theta} &= \sqrt3. \end{align*}
From each equation $\sqrt{2}^2$ can be factored out, and when every equation is divided by 2, we get:
\begin{align*} \sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \beta} + \sqrt{\cos^2 \beta}\cdot\sqrt{1-\cos^2 \alpha} &= \frac{1}{2} \\ \sqrt{\cos^2 \beta}\cdot\sqrt{1-\cos^2 \theta} + \sqrt{\cos^2 \theta}\cdot\sqrt{1-\cos^2 \beta} &= \frac{\sqrt2}{2} \\ \sqrt{\cos^2 \theta}\cdot\sqrt{1-\cos^2 \alpha} + \sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \theta} &= \frac{\sqrt3}{2}. \end{align*}
which simplifies to (using the Pythagorean identity $\sin^2 \phi + \cos^2 \phi = 1 \; \forall \; \phi \in \mathbb{C}$):
\begin{align*} \cos \alpha\cdot\sin \beta + \cos \beta\cdot\sin \alpha &= \frac{1}{2} \\ \cos \beta\cdot\sin \theta + \cos \theta\cdot\sin \beta &= \frac{\sqrt2}{2} \\ \cos \theta\cdot\sin \alpha + \cos \alpha\cdot\sin \theta &= \frac{\sqrt3}{2}. \end{align*}
which further simplifies to (using sine addition formula $\sin(a + b) = \sin a \cos b + \cos a \sin b$):
\begin{align*} \sin(\alpha + \beta) &= \frac{1}{2} \\ \sin(\beta + \theta) &= \frac{\sqrt2}{2} \\ \sin(\alpha + \theta) &= \frac{\sqrt3}{2}. \end{align*}
Taking the inverse sine ($0\leq\theta\frac{\pi}{2}$) of each equation yields a simple system:
\begin{align*} \alpha + \beta &= \frac{\pi}{6} \\ \beta + \theta &= \frac{\pi}{4} \\ \alpha + \theta &= \frac{\pi}{3} \end{align*}
giving solutions:
\begin{align*} \alpha &= \frac{\pi}{8} \\ \beta &= \frac{\pi}{24} \\ \theta &= \frac{5\pi}{24} \end{align*}
Since these unknowns are directly related to our original unknowns, there are consequent solutions for those:
\begin{align*} x &= 2\cos^2\left(\frac{\pi}{8}\right) \\ y &= 2\cos^2\left(\frac{\pi}{24}\right) \\ z &= 2\cos^2\left(\frac{5\pi}{24}\right) \end{align*}
When plugging into the expression $\left[ (1-x)(1-y)(1-z) \right]^2$, noting that $-\cos 2\phi = 1 - 2\cos^2 \phi\; \forall \; \phi \in \mathbb{C}$ helps to simplify this expression into:
\begin{align*} \left[ (-1)^3\left(\cos \left(2\cdot\frac{\pi}{8}\right)\cos \left(2\cdot\frac{\pi}{24}\right)\cos \left(2\cdot\frac{5\pi}{24}\right)\right)\right]^2 \\ = \left[ (-1)\left(\cos \left(\frac{\pi}{4}\right)\cos \left(\frac{\pi}{12}\right)\cos \left(\frac{5\pi}{12}\right)\right)\right]^2 \end{align*}
Now, all the cosines in here are fairly standard:
\begin{align*} \cos \frac{\pi}{4} &= \frac{\sqrt{2}}{2} \\ \cos \frac{\pi}{12} &=\frac{\sqrt{6} + \sqrt{2}}{4} & (= \cos{\frac{\frac{\pi}{6}}{2}} ) \\ \cos \frac{5\pi}{12} &= \frac{\sqrt{6} - \sqrt{2}}{4} & (= \cos\left({\frac{\pi}{6} + \frac{\pi}{4}} \right) ) \end{align*}
With some final calculations:
\begin{align*} &(-1)^2\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 \\ =& \left(\frac{1}{2}\right) \left(\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)\right)^2 \\ =&\frac{1}{2} \frac{4^2}{16^2} = \frac{1}{32} \end{align*}
This is our answer in simplest form $\frac{m}{n}$, so $m + n = 1 + 32 = \boxed{033}$.
|
033
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_15
|
7
|
Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.
|
We are given that \[100a + 10b + c = 81b + 9c + a,\] which rearranges to \[99a = 71b + 8c.\]
Taking both sides modulo $71,$ we have
\begin{align*} 28a &\equiv 8c \pmod{71} \\ 7a &\equiv 2c \pmod{71}. \end{align*}
The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$
Therefore, the requested three-digit positive integer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$
|
227
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2
|
8
|
In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$.
|
We have the following diagram:
Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$, and $YZ$ be the height of $\triangle AZB$. As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$. We can apply this logic to triangles $BCQ$ and $XCQ$ as well, giving us $BC = CX = 333$. Since $CD = 650$, $XW = DW + CX - CD = 16$.
Additionally, we can see that $\triangle XZW$ is similar to $\triangle PQZ$ and $\triangle AZB$. We know that $\frac{XW}{AB} = \frac{16}{500}$. So, we can say that the height of the triangle $AZB$ is $500u$ while the height of the triangle $XZW$ is $16u$. After that, we can figure out the distance from $Y$ to $PQ: \frac{500+16}{2} = 258u$ and the height of triangle $PZQ: 500-258 = 242u$.
Finally, since the ratio between the height of $PZQ$ to the height of $AZB$ is $242:500$ and $AB$ is $500$, $PQ = \boxed{242}.$
|
242
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3
|
9
|
Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$
|
We rewrite $w$ and $z$ in polar form:
\begin{align*} w &= e^{i\cdot\frac{\pi}{6}}, \\ z &= e^{i\cdot\frac{2\pi}{3}}. \end{align*}
The equation $i \cdot w^r = z^s$ becomes
\begin{align*} e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\ e^{i\left(\frac{\pi}{2}+\frac{\pi}{6}r\right)} &= e^{i\left(\frac{2\pi}{3}s\right)} \\ \frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\ 3+r &= 4s+12k \\ 3+r &= 4(s+3k). \end{align*}
for some integer $k.$
Since $4\leq 3+r\leq 103$ and $4\mid 3+r,$ we conclude that
\begin{align*} 3+r &\in \{4,8,12,\ldots,100\}, \\ s+3k &\in \{1,2,3,\ldots,25\}. \end{align*}
Note that the values for $s+3k$ and the values for $r$ have one-to-one correspondence.
We apply casework to the values for $s+3k:$
$s+3k\equiv0\pmod{3}$
There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\equiv0\pmod{3},$ so there are $33$ values for $s.$
There are $8\cdot33=264$ ordered pairs $(r,s)$ in this case.
$s+3k\equiv1\pmod{3}$
There are $9$ values for $s+3k,$ so there are $9$ values for $r.$ It follows that $s\equiv1\pmod{3},$ so there are $34$ values for $s.$
There are $9\cdot34=306$ ordered pairs $(r,s)$ in this case.
$s+3k\equiv2\pmod{3}$
There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\equiv2\pmod{3},$ so there are $33$ values for $s.$
There are $8\cdot33=264$ ordered pairs $(r,s)$ in this case.
Together, the answer is $264+306+264=\boxed{834}.$
|
834
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_4
|
10
|
A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$.
|
Define $m$ as the number of minutes they swim for.
Let their meeting point be $A$. Melanie is swimming against the current, so she must aim upstream from point $A$, to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must aim for a point $B$ that is $14m$ meters upstream from point $A$. Similarly, Sherry is swimming downstream for $m$ minutes, so she must also aim at point $B$ to compensate for the flow of the current.
If Melanie and Sherry were to both aim at point $B$ in a currentless river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are $80m$ and $60m$ meters. We can draw out this new scenario, with the dimensions that we have:
(While it is indeed true that the triangle above with side lengths $60m$, $80m$ and $D$ is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.)
By the Pythagorean Theorem, we have
\begin{align*} 264^{2} + \left( \frac{D}{2} - 14m \right) ^{2} &= 3600m^{2} \\ 264^{2} + \left( \frac{D}{2} + 14m \right) ^{2} &= 6400m^{2}. \end{align*}
Subtracting the first equation from the second gives us $28Dm = 2800m^{2}$, so $D = 100m$. Substituting this into our first equation, we have that
\begin{align*}264^{2} + 36^{2} m^{2} &= 60^{2}m^{2} \\ 264^{2} &= 96 \cdot 24 \cdot m^{2} \\ 11^{2} &= 4 \cdot m^{2} \\ m &= \frac{11}{2}. \end{align*}
So $D = 100m = \boxed{550}$.
|
550
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_5
|
11
|
Find the number of ordered pairs of integers $(a, b)$ such that the sequence\[3, 4, 5, a, b, 30, 40, 50\]is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
|
Since $3,4,5,a$ and $3,4,5,b$ cannot be an arithmetic progression, $a$ or $b$ can never be $6$. Since $b, 30, 40, 50$ and $a, 30, 40, 50$ cannot be an arithmetic progression, $a$ and $b$ can never be $20$. Since $a < b$, there are ${24 - 2 \choose 2} = 231$ ways to choose $a$ and $b$ with these two restrictions in mind.
However, there are still specific invalid cases counted in these $231$ pairs $(a,b)$. Since
\[3,5,a,b\]
cannot form an arithmetic progression, $\underline{(a,b) \neq (7,9)}$.
\[a,b,30,50\]
cannot be an arithmetic progression, so $(a,b) \neq (-10,10)$; however, since this pair was not counted in our $231$, we do not need to subtract it off.
\[3,a,b,30\]
cannot form an arithmetic progression, so $\underline{(a,b) \neq (12,21)}$.
\[4, a, b, 40\]
cannot form an arithmetic progression, so $\underline{(a,b) \neq (16,28)}$.
\[5, a,b, 50\]
cannot form an arithmetic progression, $(a,b) \neq 20, 35$; however, since this pair was not counted in our $231$ (since we disallowed $a$ or $b$ to be $20$), we do not to subtract it off.
Also, the sequences $(3,a,b,40)$, $(3,a,b,50)$, $(4,a,b,30)$, $(4,a,b,50)$, $(5,a,b,30)$ and $(5,a,b,40)$ will never be arithmetic, since that would require $a$ and $b$ to be non-integers.
So, we need to subtract off $3$ progressions from the $231$ we counted, to get our final answer of $\boxed{228}$.
|
228
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_6
|
12
|
Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{1}{7\cdot8\cdot9}.$
If we minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f = 1.$ Note that $a \cdot b \cdot c \cdot d \cdot e \cdot f = (a \cdot b \cdot c) \cdot (a \cdot b \cdot c - 1) \geq 6! = 720,$ so $a \cdot b \cdot c \geq 28.$ It follows that $a \cdot b \cdot c$ and $d \cdot e \cdot f$ are consecutive composites with prime factors no other than $2,3,5,$ and $7.$ The smallest values for $a \cdot b \cdot c$ and $d \cdot e \cdot f$ are $36$ and $35,$ respectively. So, we have $\{a,b,c\} = \{2,3,6\}, \{d,e,f\} = \{1,5,7\},$ and $\{g,h,i\} = \{4,8,9\},$ from which $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} = \frac{1}{288}.$
If we do not minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f > 1.$ Note that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{2}{7\cdot8\cdot9} > \frac{1}{288}.$
Together, we conclude that the minimum possible positive value of $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}$ is $\frac{1}{288}.$ Therefore, the answer is $1+288=\boxed{289}.$
|
289
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_7
|
13
|
Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$
|
We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\omega_A$ is the incircle of $\triangle AB'C'$.
[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C,dashed); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*(-1,0),linewidth(4)); dot("$C$",C,1.5*(1,0),linewidth(4)); dot("$B'$",B1,1.5*dir(B1),linewidth(4)); dot("$C'$",C1,1.5*dir(C1),linewidth(4)); dot("$O$",W,1.5*dir(90),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); [/asy]
Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$. We can use inradius or equilateral triangle properties to get the inradius of this triangle is $12$ (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is $12$.
Let $O=\omega$ be the center of the largest circle. We will set up a coordinate system with $O$ as the origin. The center of $\omega_A$ will be at $(0,-6)$ because it is directly beneath $O$ and is the length of the larger radius minus the smaller radius, or $18-12 = 6$. By rotating this point $120^{\circ}$ around $O$, we get the center of $\omega_B$. This means that the magnitude of vector $\overrightarrow{O\omega_B}$ is $6$ and is at a $30$ degree angle from the horizontal. Therefore, the coordinates of this point are $(3\sqrt{3},3)$ and by symmetry the coordinates of the center of $\omega_C$ is $(-3\sqrt{3},3)$.
The upper left and right circles intersect at two points, the lower of which is $X$. The equations of these two circles are:
\begin{align*} (x+3\sqrt3)^2 + (y-3)^2 &= 12^2, \\ (x-3\sqrt3)^2 + (y-3)^2 &= 12^2. \end{align*}
We solve this system by subtracting to get $x = 0$. Plugging back in to the first equation, we have $(3\sqrt{3})^2 + (y-3)^2 = 144 \implies (y-3)^2 = 117 \implies y-3 = \pm \sqrt{117} \implies y = 3 \pm \sqrt{117}$. Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-\sqrt{117})$.
We can solve the problem two ways from here. We can find $Y$ by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find $OX$ as they lie on the same vertical, $\angle XOY$ is $120$ degrees so we can make use of $30-60-90$ triangles, and $OX = OY$ because $O$ is the center of triangle $XYZ$. We can draw the diagram as such:
[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C^^W--X^^W--Y^^W--midpoint(X--Y),dashed); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*(-1,0),linewidth(4)); dot("$C$",C,1.5*(1,0),linewidth(4)); dot("$B'$",B1,1.5*dir(B1),linewidth(4)); dot("$C'$",C1,1.5*dir(C1),linewidth(4)); dot("$O$",W,1.5*dir(90),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); [/asy]
Note that $OX = OY = \sqrt{117} - 3$. It follows that
\begin{align*} XY &= 2 \cdot \frac{OX\cdot\sqrt{3}}{2} \\ &= OX \cdot \sqrt{3} \\ &= (\sqrt{117}-3) \cdot \sqrt{3} \\ &= \sqrt{351}-\sqrt{27}. \end{align*}
Finally, the answer is $351+27 = \boxed{378}$.
|
378
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8
|
14
|
Ellina has twelve blocks, two each of red ($\textbf{R}$), blue ($\textbf{B}$), yellow ($\textbf{Y}$), green ($\textbf{G}$), orange ($\textbf{O}$), and purple ($\textbf{P}$). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement
\[\textbf{R B B Y G G Y R O P P O}\]
is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
Consider this position chart: \[\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\]
Since there has to be an even number of spaces between each pair of the same color, spots $1$, $3$, $5$, $7$, $9$, and $11$ contain some permutation of all $6$ colored balls. Likewise, so do the even spots, so the number of even configurations is $6! \cdot 6!$ (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of $\frac{12!}{(2!)^6}$ possible arrangements, so the probability is: \[\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^4}{7\cdot11\cdot3} = \frac{16}{231},\]
which is in simplest form. So, $m + n = 16 + 231 = \boxed{247}$.
|
247
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_9
|
15
|
Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.
|
Let $x$ be the number of people at the party before the bus arrives. We know that $x\equiv 0\pmod {12}$, as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \equiv 0 \pmod{25}$, as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 50 \equiv 0 \pmod{25}$ can be reduced to $x \equiv 0 \pmod{25}$, and since we are looking for the minimum amount of people, $x$ is $300$. That means there are $350$ people at the party after the bus arrives, and thus there are $350 \cdot \frac{11}{25} = \boxed{154}$ adults at the party.
|
154
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_1
|
16
|
Find the remainder when\[\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}\]is divided by $1000$.
|
We first write the expression as a summation.
\begin{align*} \sum_{i=3}^{40} \binom{\binom{i}{2}}{2} & = \sum_{i=3}^{40} \binom{\frac{i \left( i - 1 \right)}{2}}{2} \\ & = \sum_{i=3}^{40} \frac{\frac{i \left( i - 1 \right)}{2} \left( \frac{i \left( i - 1 \right)}{2}- 1 \right)}{2} \\ & = \frac{1}{8} \sum_{i=3}^{40} i \left( i - 1 \right) \left( i \left( i - 1 \right) - 2 \right) \\ & = \frac{1}{8} \sum_{i=3}^{40} i(i - 1)(i^2-i-2) \\ & = \frac{1}{8} \sum_{i=3}^{40} i(i-1)(i+1)(i-2) \\ & = \frac{1}{8}\sum_{i=3}^{40} (i-2)(i-1)i(i+1) \\ & = \frac{1}{40}\sum_{i=3}^{40}[(i-2)(i-1)i(i+1)(i+2)-(i-3)(i-2)(i-1)i(i+1)]* \\ & = \frac{38\cdot39\cdot40\cdot41\cdot42-0}{40}\\ & = 38 \cdot 39 \cdot 41 \cdot 42 \\ & = \left( 40 - 2 \right) \left( 40 - 1 \right) \left( 40 + 1 \right) \left( 40 + 2 \right) \\ & = \left( 40^2 - 2^2 \right) \left( 40^2 - 1^2 \right) \\ & = \left( 40^2 - 4 \right) \left( 40^2 - 1 \right) \\ & = 40^4 - 40^2 \cdot 5 + 4 \\ & \equiv \boxed{004}\pmod{1000}\ \end{align*}
$*(i-2)(i-1)i(i+1)=\frac{1}{5}[(i-2)(i-1)i(i+1)(i+2)-(i-3)(i-2)(i-1)i(i+1)]$ is how we force the expression to telescope.
|
004
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_10
|
17
|
Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$
|
[asy] defaultpen(fontsize(12)+0.6); size(300); pair A,B,C,D,M,H; real xb=71, xd=121; A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H); draw(B--A--D--C--B); draw(A--M--D^^M--H^^Bp--M--Cp, gray+0.4); draw(rightanglemark(A,H,M,5)); dot("$A$",A,SW); dot("$D$",D,SE); dot("$B$",B,NW); dot("$C$",C,NE); dot("$M$",M,up); dot("$H$",H,down); dot("$B'$",Bp,down); dot("$C'$",Cp,down); [/asy]
According to the problem, we have $AB=AB'=2$, $DC=DC'=3$, $MB=MB'$, $MC=MC'$, and $B'C'=7-2-3=2$
Because $M$ is the midpoint of $BC$, we have $BM=MC$, so: \[MB=MB'=MC'=MC.\]
Then, we can see that $\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$
Therefore, we could start our angle chasing: $\angle{MB'C'}=\angle{MC'B'}=180^\circ-\angle{MC'D}=180^\circ-\angle{MCD}$.
This is when we found that points $M$, $C$, $D$, and $B'$ are on a circle. Thus, $\angle{BMB'}=\angle{CDC'} \Rightarrow \angle{B'MA}=\angle{C'DM}$. This is the time we found that $\bigtriangleup{AB'M} \sim \bigtriangleup{MC'D}$.
Thus, $\frac{AB'}{B'M}=\frac{MC'}{C'D} \Longrightarrow (B'M)^2=AB' \cdot C'D = 6$
Point $H$ is the midpoint of $B'C'$, and $MH \perp AD$. $B'H=HC'=1 \Longrightarrow MH=\sqrt{B'M^2-B'H^2}=\sqrt{6-1}=\sqrt{5}$.
The area of this quadrilateral is the sum of areas of triangles: \[S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{CD'M}}+S_{\bigtriangleup{B'C'M}}\]
\[=S_{\bigtriangleup{AB'M}}\cdot 2 + S_{\bigtriangleup{B'C'M}} + S_{\bigtriangleup{C'DM}}\cdot 2\]
\[=2 \cdot \frac{1}{2} \cdot AB' \cdot MH + \frac{1}{2} \cdot B'C' \cdot MH + 2 \cdot \frac{1}{2} \cdot C'D \cdot MH\]
\[=2\sqrt{5}+\sqrt{5}+3\sqrt{5}=6\sqrt{5}\]
Finally, the square of the area is $(6\sqrt{5})^2=\boxed{180}$
Denote by $M$ the midpoint of segment $BC$.
Let points $P$ and $Q$ be on segment $AD$, such that $AP = AB$ and $DQ = DC$.
Denote $\angle DAM = \alpha$, $\angle BAD = \beta$, $\angle BMA = \theta$, $\angle CMD = \phi$.
Denote $BM = x$. Because $M$ is the midpoint of $BC$, $CM = x$.
Because $AM$ is the angle bisector of $\angle BAD$ and $AB = AP$, $\triangle BAM \cong \triangle PAM$.
Hence, $MP = MB$ and $\angle AMP = \theta$.
Hence, $\angle MPD = \angle MAP + \angle PMA = \alpha + \theta$.
Because $DM$ is the angle bisector of $\angle CDA$ and $DC = DQ$, $\triangle CDM \cong \triangle QDM$.
Hence, $MQ = MC$ and $\angle DMQ = \phi$.
Hence, $\angle MQA = \angle MDQ + \angle QMD = \beta + \phi$.
Because $M$ is the midpoint of segment $BC$, $MB = MC$.
Because $MP = MB$ and $MQ = MC$, $MP = MQ$.
Thus, $\angle MPD = \angle MQA$.
Thus,
\[ \alpha + \theta = \beta + \phi . \hspace{1cm} (1) \]
In $\triangle AMD$, $\angle AMD = 180^\circ - \angle MAD - \angle MDA = 180^\circ - \alpha - \beta$.
In addition, $\angle AMD = 180^\circ - \angle BMA - \angle CMD = 180^\circ - \theta - \phi$.
Thus,
\[ \alpha + \beta = \theta + \phi . \hspace{1cm} (2) \]
Taking $(1) + (2)$, we get $\alpha = \phi$.
Taking $(1) - (2)$, we get $\beta = \theta$.
Therefore, $\triangle ADM \sim \triangle AMB \sim \triangle MDC$.
Hence, $\frac{AD}{AM} = \frac{AM}{AB}$ and $\frac{AD}{DM} = \frac{DM}{CD}$.
Thus, $AM = \sqrt{AD \cdot AD} = \sqrt{14}$ and $DM = \sqrt{AD \cdot CD} = \sqrt{21}$.
In $\triangle ADM$, by applying the law of cosines, $\cos \angle AMD = \frac{AM^2 + DM^2 - AD^2}{2 AM \cdot DM} = - \frac{1}{\sqrt{6}}$.
Hence, $\sin \angle AMD = \sqrt{1 - \cos^2 \angle AMD} = \frac{\sqrt{5}}{\sqrt{6}}$.
Hence, ${\rm Area} \ \triangle ADM = \frac{1}{2} AM \cdot DM \dot \sin \angle AMD = \frac{7 \sqrt{5}}{2}$.
Therefore,
\begin{align*} {\rm Area} \ ABCD & = {\rm Area} \ \triangle AMD + {\rm Area} \ \triangle ABM + {\rm Area} \ \triangle MCD \\ & = {\rm Area} \ \triangle AMD \left( 1 + \left( \frac{AM}{AD} \right)^2 + \left( \frac{MD}{AD} \right)^2 \right) \\ & = 6 \sqrt{5} . \end{align*}
Therefore, the square of ${\rm Area} \ ABCD$ is $\left( 6 \sqrt{5} \right)^2 = \boxed{\textbf{(180) }}$.
|
180
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11
|
18
|
Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that\[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\]Find the least possible value of $a+b.$
|
Denote $P = \left( x , y \right)$.
Because $\frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1$, $P$ is on an ellipse whose center is $\left( 0 , 0 \right)$ and foci are $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$.
Hence, the sum of distance from $P$ to $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ is equal to twice the major axis of this ellipse, $2a$.
Because $\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2} = 1$, $P$ is on an ellipse whose center is $\left( 20 , 11 \right)$ and foci are $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$.
Hence, the sum of distance from $P$ to $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$ is equal to twice the major axis of this ellipse, $2b$.
Therefore, $2a + 2b$ is the sum of the distance from $P$ to four foci of these two ellipses.
To make this minimized, $P$ is the intersection point of the line that passes through $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$, and the line that passes through $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$.
The distance between $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$ is $\sqrt{\left( 20 + 4 \right)^2 + \left( 10 - 0 \right)^2} = 26$.
The distance between $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$ is $\sqrt{\left( 20 - 4 \right)^2 + \left( 12 - 0 \right)^2} = 20$.
Hence, $2 a + 2 b = 26 + 20 = 46$.
Therefore, $a + b = \boxed{\textbf{(023) }}.$
|
023
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_12
|
19
|
There is a polynomial $P(x)$ with integer coefficients such that\[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\]holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$.
|
Because $0 < x < 1$, we have
\begin{align*} P \left( x \right) & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\ & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} x^{2310 a + 105 b + 70 c + 42 d + 30 e} . \end{align*}
Denote by $c_{2022}$ the coefficient of $P \left( x \right)$.
Thus,
\begin{align*} c_{2022} & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} \Bbb I \left\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-0} \Bbb I \left\{ 2310 \cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \Bbb I \left\{ 105 b + 70 c + 42 d + 30 e = 2022 \right\} . \end{align*}
Now, we need to find the number of nonnegative integer tuples $\left( b , c , d , e \right)$ that satisfy
\[ 105 b + 70 c + 42 d + 30 e = 2022 . \hspace{1cm} (1) \]
Modulo 2 on Equation (1), we have $b \equiv 0 \pmod{2}$.
Hence, we can write $b = 2 b'$. Plugging this into (1), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c , d , e \right)$ that satisfy
\[ 105 b' + 35 c + 21 d + 15 e = 1011 . \hspace{1cm} (2) \]
Modulo 3 on Equation (2), we have $2 c \equiv 0 \pmod{3}$.
Hence, we can write $c = 3 c'$. Plugging this into (2), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d , e \right)$ that satisfy
\[ 35 b' + 35 c' + 7 d + 5 e = 337 . \hspace{1cm} (3) \]
Modulo 5 on Equation (3), we have $2 d \equiv 2 \pmod{5}$.
Hence, we can write $d = 5 d' + 1$. Plugging this into (3), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e \right)$ that satisfy
\[ 7 b' + 7 c' + 7 d' + e = 66 . \hspace{1cm} (4) \]
Modulo 7 on Equation (4), we have $e \equiv 3 \pmod{7}$.
Hence, we can write $e = 7 e' + 3$. Plugging this into (4), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e' \right)$ that satisfy
\[ b' + c' + d' + e' = 9 . \hspace{1cm} (5) \]
The number of nonnegative integer solutions to Equation (5) is $\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} = \boxed{\textbf{(220) }}$.
|
220
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_13
|
20
|
For positive integers $a$, $b$, and $c$ with $a < b < c$, consider collections of postage stamps in denominations $a$, $b$, and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$.
|
Notice that we must have $a = 1$; otherwise $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$. Using at most $c-1$ stamps of value $1$ and $b$, it can have all the values from $1$ to $c-1$ cents. Plus $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$, every value up to $1000$ can be represented.
Correction:
This should be $\lfloor \frac{1000}{c} \rfloor$. The current function breaks when $c \mid 1000$ and $b \mid c$. Take $c = 200$ and $b = 20$. Then, we have $\lfloor \frac{999}{200} \rfloor = 4$ stamps of value 200, $\lfloor \frac{199}{20} \rfloor = 9$ stamps of value b, and 19 stamps of value 1. The maximum such a collection can give is $200 \cdot 4 + 20 \cdot 9 +19 \cdot 1 = 999$, just shy of the needed 1000. As for the rest of solution, proceed similarly, except use $1000$ instead of $999$.
Also, some explanation: $b-1$ one cent stamps cover all residues module $b$. Having $\lfloor \frac{c-1}{b} \rfloor$ stamps of value b covers all residue classes modulo $c$. Finally, we just need $\lfloor \frac{1000}{c} \rfloor$ to cover everything up to 1000.
In addition, note that this function sometimes may not always minimize the number of stamps required. This is due to the fact that the stamps of value $b$ and of value $1$ have the capacity to cover values greater than or equal to $c$ (which occurs when $c-1$ has a remainder less than $b-1$ when divided by $b$). Thus, in certain cases, not all $\lfloor \frac{1000}{c} \rfloor$ stamps of value c may be necessary, because the stamps of value $b$ and 1 can replace one $c$.
|
188
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_14
|
21
|
Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = 2$, $O_1O_2 = 15$, $CD = 16$, and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon.
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First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$. Let points $A'$ and $B'$ be the reflections of $A$ and $B$, respectively, about the perpendicular bisector of $\overline{O_1O_2}$. Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2A'$ and $B'O_1C$ are congruent, so $A'D = B'C$ and quadrilateral $A'B'CD$ is an isosceles trapezoid.
[asy] import olympiad; size(180); defaultpen(linewidth(0.7)); pair Ap = dir(105), Bp = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175); draw(unitcircle^^Ap--Bp--O1--C--D--O2--cycle); label("$A'$",Ap,dir(origin--Ap)); label("$B'$",Bp,dir(origin--Bp)); label("$O_1$",O1,dir(origin--O1)); label("$C$",C,dir(origin--C)); label("$D$",D,dir(origin--D)); label("$O_2$",O2,dir(origin--O2)); draw(O2--O1,linetype("4 4")); draw(Ap--D^^Bp--C,linetype("2 2")); [/asy]
Next, remark that $B'O_1 = DO_2$, so quadrilateral $B'O_1DO_2$ is also an isosceles trapezoid; in turn, $B'D = O_1O_2 = 15$, and similarly $A'C = 15$. Thus, Ptolmey's theorem on $A'B'CD$ yields $A'D\cdot B'C + 2\cdot 16 = 15^2$, whence $A'D = B'C = \sqrt{193}$. Let $\alpha = \angle A'B'D$. The Law of Cosines on triangle $A'B'D$ yields
\[\cos\alpha = \frac{15^2 + 2^2 - (\sqrt{193})^2}{2\cdot 2\cdot 15} = \frac{36}{60} = \frac 35,\] and hence $\sin\alpha = \tfrac 45$. Thus the distance between bases $A’B’$ and $CD$ is $12$ (in fact, $\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\tfrac12\cdot 12\cdot(2+16) = 108$.
Now let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$; the tangency of circles $\omega_1$ and $\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$, which implies the measure of angle $A'O_2D$ is $180^\circ - \alpha$. Therefore, the Law of Cosines applied to triangle $\triangle A'O_2D$ yields
\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \tfrac45r_1r_2\\ &= (r_1+r_2)^2 - \tfrac45 r_1r_2 = 225 - \tfrac45r_1r_2. \end{align*}
Thus $r_1r_2 = 40$, and so the area of triangle $A'O_2D$ is $\tfrac12r_1r_2\sin\alpha = 16$.
Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\cdot 16 = \boxed{140}$.
|
140
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15
|
22
|
Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\frac23$. When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\frac34$. Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
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Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label.
[2022AIMEIIP2.png](https://artofproblemsolving.com/wiki/index.php/File:2022AIMEIIP2.png)
This problem can be solved by using $2$ cases.
$\textbf{Case 1:}$ $C$'s opponent for the semifinal is $A$
The probability $C$'s opponent is $A$ is $\frac13$. Therefore the probability $C$ wins the semifinal in this case is $\frac13 \cdot \frac13$. The other semifinal game is played between $J$ and $S$, it doesn't matter who wins because $C$ has the same probability of winning either one. The probability of $C$ winning in the final is $\frac34$, so the probability of $C$ winning the tournament in case 1 is $\frac13 \cdot \frac13 \cdot \frac34$
$\textbf{Case 2:}$ $C$'s opponent for the semifinal is $J$ or $S$
It doesn't matter if $C$'s opponent is $J$ or $S$ because $C$ has the same probability of winning either one. The probability $C$'s opponent is $J$ or $S$ is $\frac23$. Therefore the probability $C$ wins the semifinal in this case is $\frac23 \cdot \frac34$. The other semifinal game is played between $A$ and $J$ or $S$. In this case it matters who wins in the other semifinal game because the probability of $C$ winning $A$ and $J$ or $S$ is different.
$\textbf{Case 2.1:}$ $C$'s opponent for the final is $A$
For this to happen, $A$ must have won $J$ or $S$ in the semifinal, the probability is $\frac34$. Therefore, the probability that $C$ won $A$ in the final is $\frac34 \cdot \frac13$.
$\textbf{Case 2.2:}$ $C$'s opponent for the final is $J$ or $S$
For this to happen, $J$ or $S$ must have won $A$ in the semifinal, the probability is $\frac14$. Therefore, the probability that $C$ won $J$ or $S$ in the final is $\frac14 \cdot \frac34$.
In Case 2 the probability of $C$ winning the tournament is $\frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34)$
Adding case 1 and case 2 together we get $\frac13 \cdot \frac13 \cdot \frac34 + \frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34) = \frac{29}{96},$ so the answer is $29 + 96 = \boxed{\textbf{125}}$.
|
125
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_2
|
23
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A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
|
[2022 AIME II 3.png](https://artofproblemsolving.com/wiki/index.php/File:2022_AIME_II_3.png)
Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\frac{6}{\sqrt{2}} = 3\sqrt{2}$). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$. Because of the symmetrical property of the pyramid,
we can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center.
Since the volume is $54 = \frac{1}{3} \cdot S \cdot h = \frac{1}{3} \cdot 6^2 \cdot h$, where $h=\frac{9}{2}$ is the height of this pyramid, we have: $l^2=\left(\frac{9}{2}-l\right)^2+\left(3\sqrt{2}\right)^2$ according to the Pythagorean theorem.
Solve this equation will give us $l = \frac{17}{4},$ therefore $m+n=\boxed{021}.$
To start, we find the height of the pyramid. By the volume of a pyramid formula, we have \[\frac13 \cdot 6^2 \cdot h=54 \implies h=\frac92.\] Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is $\frac12 \cdot 6\sqrt2=3\sqrt2$, we have \[x=\sqrt{\left(\frac92\right)^2+(3\sqrt2)^2}=\sqrt{\frac{153}4}=\frac{3\sqrt{17}}2.\] Taking the cross section of the pyramid and transforming the problem into $2$-d, it suffices to find the radius of the circumcircle of a triangle of side lengths $\frac{3\sqrt{17}}2$, $\frac{3\sqrt{17}}2$, $6\sqrt2$. This turns out to be easy by the formula $R=\frac{abc}{4A}$, and through computing this value (the work has been left out) we find that $R=\frac{17}4$, so our answer is $\boxed{\textbf{021}}$.
|
021
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_3
|
24
|
There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that\[\log_{20x} (22x)=\log_{2x} (202x).\]The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
|
Define $a$ to be $\log_{20x} (22x) = \log_{2x} (202x)$, what we are looking for. Then, by the definition of the logarithm,
\[\begin{cases} (20x)^{a} &= 22x \\ (2x)^{a} &= 202x. \end{cases}\]
Dividing the first equation by the second equation gives us $10^a = \frac{11}{101}$, so by the definition of logs, $a = \log_{10} \frac{11}{101}$. This is what the problem asked for, so the fraction $\frac{11}{101}$ gives us $m+n = \boxed{112}$.
|
112
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4
|
25
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Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points.
|
[isabelchen](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Isabelchen)
Note: This solution seems incorrect.
Although the answer is correct, solution 2 below is a more accurate way to approach this problem. I agree, I don't get how $a + 2 \leq 20$.
As above, we must deduce that the sum of two primes must be equal to the third prime. Then, we can finish the solution using casework.
If the primes are $2,3,5$, then the smallest number can range between $1$ and $15$.
If the primes are $2,5,7$, then the smallest number can range between $1$ and $13$.
If the primes are $2,11,13$, then the smallest number can range between $1$ and $7$.
If the primes are $2,17,19$, then the smallest number can only be $1$.
Adding all cases gets $15+13+7+1=36$. However, due to the commutative property, we must multiply this by 2. For example, in the $2,17,19$ case the numbers can be $1,3,20$ or $1,18,20$. Therefore the answer is $36\cdot2=\boxed{072}$.
Note about solution 1: I don't think that works, because if for example there are 21 points on the circle, your solution would yield $19\cdot4=76$, but there would be $8$ more solutions than if there are $20$ points. This is because the upper bound for each case increases by $1$, but commutative property doubles it to be $4$.
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072
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_5
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26
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Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$. Among all such $100$-tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
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[isabelchen](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Isabelchen)
Define $s_N$ to be the sum of all the negatives, and $s_P$ to be the sum of all the positives.
Since the sum of the absolute values of all the numbers is $1$, $|s_N|+|s_P|=1$.
Since the sum of all the numbers is $0$, $s_N=-s_P\implies |s_N|=|s_P|$.
Therefore, $|s_N|=|s_P|=\frac 12$, so $s_N=-\frac 12$ and $s_P=\frac 12$ since $s_N$ is negative and $s_P$ is positive.
To maximize $x_{76}-x_{16}$, we need to make $x_{16}$ as small of a negative as possible, and $x_{76}$ as large of a positive as possible.
Note that $x_{76}+x_{77}+\cdots+x_{100}=\frac 12$ is greater than or equal to $25x_{76}$ because the numbers are in increasing order.
Similarly, $x_{1}+x_{2}+\cdots+x_{16}=-\frac 12$ is less than or equal to $16x_{16}$.
So we now know that $\frac 1{50}$ is the best we can do for $x_{76}$, and $-\frac 1{32}$ is the least we can do for $x_{16}$.
Finally, the maximum value of $x_{76}-x_{16}=\frac 1{50}+\frac 1{32}=\frac{41}{800}$, so the answer is $\boxed{841}$.
(Indeed, we can easily show that $x_1=x_2=\cdots=x_{16}=-\frac 1{32}$, $x_{17}=x_{18}=\cdots=x_{75}=0$, and $x_{76}=x_{77}=\cdots=x_{100}=\frac 1{50}$ works.)
|
841
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_6
|
27
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A circle with radius $6$ is externally tangent to a circle with radius $24$. Find the area of the triangular region bounded by the three common tangent lines of these two circles.
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Professor Rat's solution, added by @heheman and edited by @megahertz13 and @Yrock for $\LaTeX$.
First, we want to find $O_2D$. We know that $\angle O_1AD = \angle O_2BD = 90^{\circ}$, so by AA similarity, $\triangle O_1AD \sim \triangle O_2BD$. We want to find the length of $x$, and using the similar triangles, we write an equation: $\frac{30 + x}{4} = x$. Solving, we get $x=10$. Therefore, $CD = 10 + 6 = 16$. Next, we find that using AA similarity, $\triangle O_2BD \sim \triangle HO_2D \sim \triangle ECD$ and they are 3-4-5 triangles. We can quickly compute $EF = 2EC = 2 \cdot \left( \frac{3}{4} \cdot 16 \right) = 2 \cdot 12 = 24$. Therefore, the area is $\frac{1}{2} \cdot 16 \cdot 24 = \boxed{192}$.
|
192
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_7
|
28
|
Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$, $\left\lfloor\frac n5\right\rfloor$, and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$.
|
We need to find all numbers between $1$ and $600$ inclusive that are multiples of $4$, $5$, and/or $6$ which are also multiples of $4$, $5$, and/or $6$ when $1$ is added to them.
We begin by noting that the LCM of $4$, $5$, and $6$ is $60$. We can therefore simplify the problem by finding all such numbers described above between $1$ and $60$ and multiplying the quantity of such numbers by $10$ ($600$/$60$ = $10$).
After making a simple list of the numbers between $1$ and $60$ and going through it, we see that the numbers meeting this condition are $4$, $5$, $15$, $24$, $35$, $44$, $54$, and $55$. This gives us $8$ numbers. $8$ * $10$ = $\boxed{080}$.
Solution 1.5
This is Solution 1 with a slick element included. Solution 1 uses the concept that $60k+l$ is a solution for $n$ if $60k+l$ is a multiple of $3$, $4$, and/or $5$ and $60k+l+1$ is a multiple of $3$, $4$, and/or $5$ for positive integer values of $l$ and essentially any integer value of $k$. But keeping the same conditions in mind for $k$ and $l$, we can also say that if $60k+l$ is a solution, then $60k-l-1$ is a solution! Therefore, one doesn't have to go as far as determining the number of values between $1$ and $60$ and then multiplying by $10$. One only has to determine the number of values between $1$ and $30$ and then multiply by $20$. The values of $n$ that work between $1$ and $30$ are $4$, $5$, $15$, and $24$. This gives us $4$ numbers. $4$ * $20$ = $\boxed{080}$.
Note
Soon after the test was administered, a formal request was made to also accept $\boxed{081}$ as an answer and MAA decided to honor this request. The gist of this request stated that the phrasing of the first part of the question could reasonably be interpreted to mean that one is given the condition to begin with that the integer is less than or equal to $600$. In this case, if one was told that the values of $\left\lfloor \frac n4\right\rfloor$, $\left\lfloor\frac n5\right\rfloor$, and $\left\lfloor\frac n6\right\rfloor$ were $150$, $120$, and $100$ respectively, then the only possible choice for $n$ would be $600$ as $601$, $602$, and $603$ do not meet the condition as stated in the first part of the problem. If instead the problem asked for the numbers less than $600$ that met the second condition in the problem, there would have been one unique answer, $\boxed{080}$. ~burkinafaso ~sethl
1. For $n$ to be uniquely determined, $n$ AND $n + 1$ both need to be a multiple of $4, 5,$ or $6.$ Since either $n$ or $n + 1$ is odd, we know that either $n$ or $n + 1$ has to be a multiple of $5.$ We can state the following cases:
1. $n$ is a multiple of $4$ and $n+1$ is a multiple of $5$
2. $n$ is a multiple of $6$ and $n+1$ is a multiple of $5$
3. $n$ is a multiple of $5$ and $n+1$ is a multiple of $4$
4. $n$ is a multiple of $5$ and $n+1$ is a multiple of $6$
Solving for each case, we see that there are $30$ possibilities for cases 1 and 3 each, and $20$ possibilities for cases 2 and 4 each. However, we over-counted the cases where
1. $n$ is a multiple of $24$ and $n+1$ is a multiple of $5$
2. $n$ is a multiple of $5$ and $n+1$ is a multiple of $24$
Each case has $10$ possibilities.
Adding all the cases and correcting for over-counting, we get $30 + 20 + 30 + 20 - 10 - 10 = \boxed {080}.$
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080
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8
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29
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Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$, distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$, and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$. Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$, no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$. The figure shows that there are 8 regions when $m=3$ and $n=2$.
|
We can use recursion to solve this problem:
1. Fix 7 points on $\ell_A$, then put one point $B_1$ on $\ell_B$. Now, introduce a function $f(x)$ that indicates the number of regions created, where x is the number of points on $\ell_B$. For example, $f(1) = 6$ because there are 6 regions.
2. Now, put the second point $B_2$ on $\ell_B$. Join $A_1~A_7$ and $B_2$ will create $7$ new regions (and we are not going to count them again), and split the existing regions. Let's focus on the spliting process: line segment formed between $B_2$ and $A_1$ intersect lines $\overline{B_1A_2}$, $\overline{B_1A_3}$, ..., $\overline{B_1A_7}$ at $6$ points $\Longrightarrow$ creating $6$ regions (we already count one region at first), then $5$ points $\Longrightarrow$ creating $5$ regions (we already count one region at first), 4 points, etc. So, we have: \[f(2) = f(1) + 7 + (6+5+...+1) = 34.\]
3. If you still need one step to understand this: $A_1~A_7$ and $B_3$ will still create $7$ new regions. Intersecting \[\overline{A_2B_1}, \overline{A_2B_2};\] \[\overline{A_3B_1}, \overline{A_3B_2};\] \[...\] \[\overline{A_7B_1}, \overline{A_7B_2}\] at $12$ points, creating $12$ regions, etc. Thus, we have: \[f(3) = f(2)+7+(12+10+8+...+2)=34+7+6\cdot 7=83.\]
Yes, you might already notice that: \[f(n+1) = f(n)+7+(6+5+...+1)\cdot n = f(n) + 7 + 21n.\]
5. Finally, we have $f(4) = 153$, and $f(5)=244$. Therefore, the answer is $\boxed{244}$.
Note: we could deduce a general formula of this recursion: $f(n+1)=f(n)+N_a+\frac{n\cdot (N_a) \cdot (N_a-1)}{2}$, where $N_a$ is the number of points on $\ell_A$
We want to derive a general function $f(m,n)$ that indicates the number of bounded regions. Observing symmetry, we know this is a symmetric function about $m$ and $n$. Now let's focus on $f(m+1, n)-f(m, n)$, which is the difference caused by adding one point to the existing $m$ points of line $\ell_A$. This new point, call it #m, when connected to point #1 on $\ell_B$, crosses $m*(n-1)$ lines, thus making additional $m*(n-1)+1$ bounded regions; when connected to point #2 on $\ell_B$, it crosses $m*(n-2)$ lines, thus making additional $m*(n-2)+1$ bounded regions; etc. By simple algebra/recursion methods, we see
$f(m+1, n)-f(m, n)=m*\frac{n(n-1)}{2} +n$
Notice $f(1,n)=n-1$. Not very difficult to figure out:
$f(m, n)=\frac{m(m-1)n(n-1)}{4} +mn-1$
The fact that $f(3,2)=8$ makes us more confident about the formula. Now plug in $m=5, n=7$, we get the final answer of $\boxed{244}$.
[AIME 2022 II 9-min.png](https://artofproblemsolving.com/wiki/index.php/File:AIME_2022_II_9-min.png)
Let some number of segments be constructed. We construct a new segment. We start from the straight line $l_B.$ WLOG from point $B_3.$ Segment will cross several existing segments (points $A,B,C,...$) and enter one of the points of the line $l_A (A_1).$
Each of these points adds exactly 1 new bounded region (yellow bounded regions).
The exception is the only first segment $(A_1 B_1),$ which does not create any bounded region.
Thus, the number of bounded regions is $1$ less than the number of points of intersection of the segments plus the number of points of arrival of the segments to $l_A.$
Each point of intersection of two segments is determined uniquely by the choice of pairs of points on each line.
The number of such pairs is $\dbinom{n}{2} \cdot \dbinom{m}{2}.$
Exactly one segment comes to each of the $n$ points of the line $l_A$ from each of the $m$ points of the line $l_B.$ The total number of arrivals is equal to $mn.$
Hence, the total number of bounded regions is $N = \dbinom{n}{2} \cdot \dbinom{m}{2} + mn – 1.$
We plug in $m=5, n=7$, we get the final answer of $\boxed{244}$.
[email protected], vvsss
When a new point is added to a line, the number of newly bounded regions it creates with each line segment will be one more than the number of intersection points the line makes with other lines.
Case 1: If a new point $P$ is added to the right on a line when both lines have an equal amount of points.
WLOG, let the point be on line $\ell_A$. We consider the complement, where new lines don't intersect other line segments. Simply observing, we see that the only line segments that don't intersect with the new lines are lines attached to some point that a new line does not pass through. If we look at a series of points on line $\ell_B$ from left to right and a line connects $P$ to an arbitrary point, then the lines formed with that point and with remaining points on the left of that point never intersect with the line with $P$. Let there be $s$ points on lines $\ell_A$ and $\ell_B$ before $P$ was added. For each of the $s$ points on $\ell_B$, we subtract the total number of lines formed, which is $s^2$, not counting $P$. Considering all possible points on $\ell_B$, we get $(s^2-s)+(s^2-2s)\cdots(s^2-s^2)$ total intersections. However, for each of the lines, there is one more bounded region than number of intersections, so we add $s$. Simplifying, we get $s^3-s\sum_{i=1}^{s}{i}+s\Longrightarrow s(s^2-\sum_{i=1}^{s}{i}+1)$. Note that this is only a recursion formula to find the number of new regions added for a new point $P$ added to $\ell_A$.
Case 2: If a new point $P$ is added to the right of a line that has one less point than the other line.
Continuing on case one, let this point $P$ be on line $\ell_B$. With similar reasoning, we see that the idea remains the same, except $s+1$ lines are formed with $P$ instead of just $s$ lines. Once again, each line from $P$ to a point on line $\ell_A$ creates $s$ non-intersecting lines for that point and each point to its left. Subtracting from $s(s+1)$ lines and considering all possible lines created by $P$, we get $(s(s+1)-s)+(s(s+1)-2s)\cdots(s(s+1)-s(s+1)$ intersections. However, the number of newly bounded regions is the number of intersections plus the number of points on line $\ell_A$. Simplying, we get $s(s+1)^2-s\sum_{i=1}^{s+1}{i}+(s+1)$ newly bounded regions.
For the base case $s=2$ for both lines, there are $4$ bounded regions. Next, we plug in $s=2,3,4$ for both formulas and plug $s=5$ for the first formula to find the number of regions when $m=6$ and $n=5$. Notice that adding a final point on $\ell_A$ is a variation of our Case 1. The only difference is for each of the $s$ lines formed by $P$, there are $s+1$ points that can form a non-intersecting line. Therefore, we are subtracting a factor of $s+1$ lines instead of $s$ lines from a total of $s(s+1)$ lines. However, the number of lines formed by $P$ remains the same so we still add $s$ at the end when considering intersection points. Thus, the recursive equation becomes $(s(s+1)-(s+1))+(s(s+1)-2(s+1))\cdots(s(s+1)-s(s+1))+s\Longrightarrow s^2(s+1)-(s+1)\sum_{i=1}^{s}{i}+s$. Plugging $s=5$ into this formula and adding the values we obtained from the other formulas, the final answer is $4+4+9+12+22+28+45+55+65=\boxed{244}$.
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244
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9
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30
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are $14!$ arrangements without restrictions.
First, there are $\binom{7}{5}$ ways to choose the man-woman diameters. Then, there are $10\cdot8\cdot6\cdot4\cdot2$ ways to place the five men each in a man-woman diameter. Finally, there are $9!$ ways to place the nine women without restrictions.
Together, the requested probability is \[\frac{\tbinom{7}{5}\cdot(10\cdot8\cdot6\cdot4\cdot2)\cdot9!}{14!} = \frac{21\cdot(10\cdot8\cdot6\cdot4\cdot2)}{14\cdot13\cdot12\cdot11\cdot10} = \frac{48}{143},\] from which the answer is $48+143 = \boxed{191}.$
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191
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1
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31
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There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$. For that unique $a$, find $a+U$.
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$.)
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minor edit by KevinChen_Yay
We define $U' = \sum^{2023}_{n=1} {\frac{n^2-na}{5}}$. Since for any real number $x$, $\lfloor x \rfloor \le x \le \lfloor x \rfloor + 1$, we have $U \le U' \le U + 2023$. Now, since $-1000 \le U \le 1000$, we have $-1000 \le U' \le 3023$.
Now, we can solve for $U'$ in terms of $a$. We have:
\begin{align*} U' &= \sum^{2023}_{n=1} {\frac{n^2-na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5} - \frac{na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5}} - \sum^{2023}_{n=1} {\frac{na}{5}} \\ &= \frac{\sum^{2023}_{n=1} {{n^2}} - \sum^{2023}_{n=1} {na}}{5} \\ &= \frac{\frac{2023(2023+1)(2023 \cdot 2 + 1)}{6} - \frac{a \cdot 2023(2023+1)}{2} }{5} \\ &= \frac{2023(2024)(4047-3a)}{30} \\ \end{align*}
So, we have $U' = \frac{2023(2024)(4047-3a)}{30}$, and $-1000 \le U' \le 3023$, so we have $-1000 \le \frac{2023(2024)(4047-3a)}{30} \le 3023$, or $-30000 \le 2023(2024)(4047-3a) \le 90690$. Now, $2023 \cdot 2024$ is much bigger than $90690$ or $30000$, and since $4047-3a$ is an integer, to satsify the inequalities, we must have $4047 - 3a = 0$, or $a = 1349$, and $U' = 0$.
Now, we can find $U - U'$. We have:
\begin{align*} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor} - \sum^{2023}_{n=1} {\frac{n^2-1349n}{5}} \\ &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \end{align*}.
Now, if $n^2-1349n \equiv 0 \text{ (mod 5)}$, then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = 0$, and if $n^2-1349n \equiv 1 \text{ (mod 5)}$, then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = -\frac{1}{5}$, and so on. Testing with $n \equiv 0,1,2,3,4, \text{ (mod 5)}$, we get $n^2-1349n \equiv 0,2,1,2,0 \text{ (mod 5)}$ respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for $U - U'$, we get:
\begin{align*} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \\ &= 404 \cdot 0 - 405 \cdot \frac{2}{5} - 405 \cdot \frac{1}{5} - 405 \cdot \frac{2}{5} - 404 \cdot t0 \\ &= -405(\frac{2}{5}+\frac{1}{5}+\frac{2}{5}) \\ &= -405 \end{align*}
Since $U' = 0$, this gives $U = -405$, and we have $a + U = 1349-405 = \boxed{944}$.
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944
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10
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32
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Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$
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Define $f(x)$ to be the number of subsets of $\{1, 2, 3, 4, \ldots x\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair.
Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the conditions. It is easy to see that \[f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + f(3)^2.\]
We see that $f(1) = 2$, $f(2) = 3$, and $f(n) = f(n-1) + f(n-2)$. This is because if the element $n$ is included in our subset, then there are $f(n-2)$ possibilities for the rest of the elements (because $n-1$ cannot be used), and otherwise there are $f(n-1)$ possibilities. Thus, by induction, $f(n)$ is the $n+1$th Fibonacci number.
This means that $f'(10) = 2(34) + 2(21) + 2(2)(13) + 2(3)(8) + 5^2 = \boxed{235}$.
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235
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11
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33
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Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\angle AEP).$
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By Miquel's theorem, $P=(AEF)\cap(BFD)\cap(CDE)$ (intersection of circles). The law of cosines can be used to compute $DE=42$, $EF=35$, and $FD=13$. Toss the points on the coordinate plane; let $B=(-7, 0)$, $D=(0, 0)$, and $C=(48, 0)$, where we will find $\tan^{2}\left(\measuredangle CDP\right)$ with $P=(BFD)\cap(CDE)$.
By the extended law of sines, the radius of circle $(BFD)$ is $\frac{13}{2\sin 60^{\circ}}=\frac{13}{3}\sqrt{3}$. Its center lies on the line $x=-\frac{7}{2}$, and the origin is a point on it, so $y=\frac{23}{6}\sqrt{3}$.
The radius of circle $(CDE)$ is $\frac{42}{2\sin 60^{\circ}}=14\sqrt{3}$. The origin is also a point on it, and its center is on the line $x=24$, so $y=2\sqrt{3}$.
The equations of the two circles are \begin{align*}(BFD)&:\left(x+\tfrac{7}{2}\right)^{2}+\left(y-\tfrac{23}{6}\sqrt{3}\right)^{2}=\tfrac{169}{3} \\ (CDE)&:\left(x-24\right)^{2}+\left(y-2\sqrt{3}\right)^{2}=588\end{align*} These equations simplify to \begin{align*}(BFD)&:x^{2}+7x+y^{2}-\tfrac{23}{3}\sqrt{3}y=0 \\ (CDE)&: x^{2}-48x+y^{2}-4\sqrt{3}y=0\end{align*} Subtracting these two equations gives that both their points of intersection, $D$ and $P$, lie on the line $55x-\tfrac{11}{3}\sqrt{3}y=0$. Hence, $\tan^{2}\left(\measuredangle AEP\right)=\tan^{2}\left(\measuredangle CDP\right)=\left(\frac{55}{\tfrac{11}{3}\sqrt{3}}\right)^{2}=3\left(\tfrac{55}{11}\right)^{2}=\boxed{075}$. To scale, the configuration looks like the figure below:
Denote $\theta = \angle AEP$.
In $AFPE$, we have $\overrightarrow{AF} + \overrightarrow{FP} + \overrightarrow{PE} + \overrightarrow{EA} = 0$.
Thus,
\[ AF + FP e^{i \theta} + PE e^{i \left( \theta + 60^\circ \right)} + EA e^{- i 120^\circ} = 0. \]
Taking the real and imaginary parts, we get
\begin{align*} AF + FP \cos \theta + PE \cos \left( \theta + 60^\circ \right) + EA \cos \left( - 120^\circ \right) & = 0 \hspace{1cm} (1) \\ FP \sin \theta + PE \sin \left( \theta + 60^\circ \right) + EA \sin \left( - 120^\circ \right) & = 0 \hspace{1cm} (2) \end{align*}
In $BDPF$, analogous to the analysis of $AFPE$ above, we get
\begin{align*} BD + DP \cos \theta + PF \cos \left( \theta + 60^\circ \right) + FB \cos \left( - 120^\circ \right) & = 0 \hspace{1cm} (3) \\ DP \sin \theta + PF \sin \left( \theta + 60^\circ \right) + FB \sin \left( - 120^\circ \right) & = 0 \hspace{1cm} (4) \end{align*}
Taking $(1) \cdot \sin \left( \theta + 60^\circ \right) - (2) \cdot \cos \left( \theta + 60^\circ \right)$, we get
\[ AF \sin \left( \theta + 60^\circ \right) + \frac{\sqrt{3}}{2} FP - EA \sin \theta = 0 . \hspace{1cm} (5) \]
Taking $(3) \cdot \sin \theta - (4) \cdot \cos \theta$, we get
\[ BD \sin \theta - \frac{\sqrt{3}}{2} FP + FB \sin \left( \theta + 120^\circ \right) . \hspace{1cm} (6) \]
Taking $(5) + (6)$, we get
\[ AF \sin \left( \theta + 60^\circ \right) - EA \sin \theta + BD \sin \theta + FB \sin \left( \theta + 120^\circ \right) . \]
Therefore,
\begin{align*} \tan \theta & = \frac{\frac{\sqrt{3}}{2} \left( AF + FB \right)} {\frac{FB}{2} + EA - \frac{AF}{2} - BD} \\ & = 5 \sqrt{3} . \end{align*}
Therefore, $\tan^2 \theta = \boxed{075}$.
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075
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_12
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34
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Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$.
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$, where $m$ and $n$
are relatively prime positive integers. Find $m + n$. A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy]
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Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Let one of the vertices be at the origin and the three adjacent vertices be $u$, $v$, and $w$. For one of the parallelepipeds, the three diagonals involving the origin have length $\sqrt {21}$. Hence, $(u+v)\cdot (u+v)=u\cdot u+v\cdot v+2u\cdot v=21$ and $(u-v)\cdot (u-v)=u\cdot u+v\cdot v-2u\cdot v=31$. Since all of $u$, $v$, and $w$ have equal length, $u\cdot u=13$, $v\cdot v=13$, and $u\cdot v=-2.5$. Symmetrically, $w\cdot w=13$, $u\cdot w=-2.5$, and $v\cdot w=-2.5$. Hence the volume of the parallelepiped is given by $\sqrt{\operatorname{det}\begin{pmatrix}13&-2.5&-2.5\\-2.5&13&-2.5\\-2.5&-2.5&13\end{pmatrix}}=\sqrt{\operatorname{det}\begin{pmatrix}15.5&-15.5&0\\-2.5&13&-2.5\\0&-15.5&15.5\end{pmatrix}}=\sqrt{15.5^2\operatorname\det\begin{pmatrix}1&-1&0\\-2.5&13&-2.5\\0&-1&1\end{pmatrix}}=\sqrt{15.5^2\cdot 8}$.
For the other parallelepiped, the three diagonals involving the origin are of length $\sqrt{31}$ and the volume is $\sqrt{\operatorname{det}\begin{pmatrix}13&2.5&2.5\\2.5&13&2.5\\2.5&2.5&13\end{pmatrix}}=\sqrt{\operatorname{det}\begin{pmatrix}10.5&-10.5&0\\2.5&13&2.5\\0&-10.5&10.5\end{pmatrix}}=\sqrt{10.5^2\operatorname\det\begin{pmatrix}1&-1&0\\2.5&13&2.5\\0&-1&1\end{pmatrix}}=\sqrt{10.5^2\cdot 18}$.
Consequently, the answer is $\sqrt\frac{10.5^2\cdot 18}{15.5^2\cdot 8}=\frac{63}{62}$, giving $\boxed{125}$.
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125
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13
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35
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The following analog clock has two hands that can move independently of each other.
Initially, both hands point to the number $12$. The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let $N$ be the number of sequences of $144$ hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the $144$ movements, the hands have returned to their initial position. Find the remainder when $N$ is divided by $1000$.
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This problem is, in essence, the following: A $12\times12$ coordinate grid is placed on a (flat) torus; how many loops are there that pass through each point while only moving up or right? In other words, Felix the frog starts his journey at $(0,0)$ in the coordinate plane. Every second, he jumps either to the right or up, until he reaches an $x$- or $y$-coordinate of $12$. At this point, if he tries to jump to a coordinate outside the square from $(0,0)$ to $(11,11)$, he "wraps around" and ends up at an $x$- or $y$- coordinate of $0$. How many ways are there for Felix to jump on every grid point in this square, so that he ends at $(0,0)$? This is consistent with the construction of the flat torus as $\mathbb Z^2/12\mathbb Z^2$ (2-dimensional modular arithmetic. $(\mathbb{Z}_{12})^2$)
Moving on, define a $\textit{path}$ from point $A$ to point $B$ to be a sequence of "up"s and "right"s that takes Felix from $A$ to $B$. The $\textit{distance}$ from $A$ to $B$ is the length of the shortest path from $A$ to $B$. At the crux of this problem is the following consideration: The points $A_i=(i,12-i), i\in{0,...,11}$ are pairwise equidistant, each pair having distance of $12$ in both directions.
[asy] size(7cm); for (int x=0; x<12; ++x){ for (int y=0; y<12; ++y){ fill(circle((x,y),0.05));}} for (int i=0; i<12; ++i){ fill(circle((i,11-i),0.1),red);} pen p=green+dashed; path u=(3,8)--(4,8)--(4,9)--(4,10)--(4,11)--(5,11)--(5,11.5); path v=(5,-0.5)--(5,0)--(5,1)--(6,1)--(6,2)--(6,3)--(6,4)--(7,4); draw(u,p); draw(v,p); pen p=blue+dashed; path u=(4,7)--(5,7)--(5,8)--(5,9)--(5,10)--(6,10)--(6,11)--(6,11.5); path v=(6,-0.5)--(6,0)--(7,0)--(7,1)--(7,2)--(7,3)--(8,3); draw(u,p); draw(v,p); [/asy]
A valid complete path then joins two $A_i$'s, say $A_i$ and $A_j$. In fact, a link between some $A_i$ and $A_j$ fully determines the rest of the cycle, as the path from $A_{i+1}$ must "hug" the path from $A_i$, to ensure that there are no gaps. We therefore see that if $A_0$ leads to $A_k$, then $A_i$ leads to $A_{i+k}$. Only the values of $k$ relatively prime to $12$ result in solutions, though, because otherwise $A_0$ would only lead to $\{A_i:\exists n\in \mathbb Z:i\equiv kn\quad\text{mod 12}\}$. The number of paths from $A_0$ to $A_k$ is ${12\choose k}$, and so the answer is
\[{12\choose1}+{12\choose5}+{12\choose7}+{12\choose11}=1\boxed{608}.\]
Notes:
- One can prove that the path from $A_{i+1}$ must "hug" the path from $A_i$ by using techniques similar to those in Solution 2.
- One can count the paths as follows: To get from $A_0$ to $A_i$, Felix takes $k$ rights and $12-k$ ups, which can be done in ${12\choose k}$ ways.
This is more of a solution sketch and lacks rigorous proof for interim steps, but illustrates some key observations that lead to a simple solution.
Note that one can visualize this problem as walking on a $N \times N$ grid where the edges warp. Your goal is to have a single path across all nodes on the grid leading back to $(0,\ 0)$. For convenience, any grid position are presumed to be in $\mod N$.
Note that there are exactly two ways to reach node $(i,\ j)$, namely $(i - 1,\ j)$ and $(i,\ j - 1)$.
As a result, if a path includes a step from $(i,\ j)$ to $(i + 1,\ j)$, there cannot be a step from $(i,\ j)$ to $(i,\ j + 1)$. However, a valid solution must reach $(i,\ j + 1)$, and the only valid step is from $(i - 1,\ j + 1)$.
So a solution that includes a step from $(i,\ j)$ to $(i + 1,\ j)$ dictates a step from $(i - 1,\ j + 1)$ to $(i,\ j + 1)$ and by extension steps from $(i - a,\ j + a)$ to $(i - a + 1,\ j + a)$. We observe the equivalent result for steps in the orthogonal direction.
This means that in constructing a valid solution, taking one step in fact dictates N steps, thus it's sufficient to count valid solutions with $N = a + b$ moves of going right $a$ times and $b$ times up the grid. The number of distinct solutions can be computed by permuting 2 kinds of indistinguishable objects $\binom{N}{a}$.
Here we observe, without proof, that if $\gcd(a, b) \neq 1$, then we will return to the origin prematurely. For $N = 12$, we only want to count the number of solutions associated with $12 = 1 + 11 = 5 + 7 = 7 + 5 = 11 + 1$.
(For those attempting a rigorous proof, note that $\gcd(a, b) = \gcd(a + b, b) = \gcd(N, b) = \gcd(N, a)$).
The total number of solutions, noting symmetry, is thus
\[2\cdot\left(\binom{12}{1} + \binom{12}{5}\right) = 1608\]
This yields $\boxed{\textbf{608}}$ as our desired answer.
|
608
|
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_14
|
36
|
Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying
the real and imaginary part of $z$ are both integers;
$|z|=\sqrt{p},$ and
there exists a triangle whose three side lengths are $p,$ the real part of $z^{3},$ and the imaginary part of $z^{3}.$
|
Assume that $z=a+bi$. Then,
\[z^3=(a^3-3ab^2)+(3a^2b-b^3)i\]Note that by the Triangle Inequality,
\[|(a^3-3ab^2)-(3a^2b-b^3)|<p\implies |a^3+b^3-3ab^2-3a^2b|<a^2+b^2\]Thus, we know
\[|a+b||a^2+b^2-4ab|<a^2+b^2\]Without loss of generality, assume $a>b$ (as otherwise, consider $i^3\overline z=b+ai$). If $|a/b|\geq 4$, then
\[17b^2\geq a^2+b^2>|a+b||a^2+b^2-4ab|\geq |b-4b||16b^2-16b^2+b^2|=3b^3\]`Thus, this means $b\leq\frac{17}3$ or $b\leq 5$. Also note that the roots of $x^2-4x+1$ are $2\pm\sqrt 3$, so thus if $b\geq 6$,
\[2\sqrt 3b=(2(2-\sqrt 3)-4)b<a<4b\]Note that
\[1000>p=a^2+b^2\geq 12b^2+b^2=13b^2\]so $b^2<81$, and $b<9$. If $b=8$, then $16\sqrt 3\leq a\leq 32$. Note that $\gcd(a,b)=1$, and $a\not\equiv b\pmod 2$, so $a=29$ or $31$. However, then $5\mid a^2+b^2$, absurd.
If $b=7$, by similar logic, we have that $14\sqrt 3 <a< 28$, so $b=26$. However, once again, $5\mid a^2+b^2$. If $b=6$, by the same logic, $12\sqrt3<a<24$, so $a=23$, where we run into the same problem. Thus $b\leq 5$ indeed.
If $b=5$, note that
\[(a+5)(a^2+25-20a)<a^2+25\implies a<20\]We note that $p=5^2+18^2=349$ works. Thus, we just need to make sure that if $b\leq 4$, $a\leq 18$. But this is easy, as
\[p>(a+b)(a^2+b^2-4ab)\geq (4+18)(4^2+18^2-4\cdot 4\cdot 18)>1000\]absurd. Thus, the answer is $\boxed{349}$.
Denote $z = a + i b$. Thus, $a^2 + b^2 = p$.
Thus,
\[z^3 = a \left( a^2 - 3 b^2 \right) + i b \left( - b^2 + 3 a^2 \right) .\]
Because $p$, ${\rm Re} \left( z^3 \right)$, ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have ${\rm Re} \left( z^3 \right) > 0$ and ${\rm Im} \left( z^3 \right) > 0$.
Thus,
\begin{align*} a \left( a^2 - 3 b^2 \right) & > 0 , \hspace{1cm} (1) \\ b \left( - b^2 + 3 a^2 \right) & > 0. \hspace{1cm} (2) \end{align*}
Because $p$, ${\rm Re} \left( z^3 \right)$, ${\rm Im} \left( z^3 \right)$ are three sides of a triangle, we have the following triangle inequalities:
\begin{align*} {\rm Re} \left( z^3 \right) + {\rm Im} \left( z^3 \right) & > p \hspace{1cm} (3) \\ p + {\rm Re} \left( z^3 \right) & > {\rm Im} \left( z^3 \right) \hspace{1cm} (4) \\ p + {\rm Im} \left( z^3 \right) & > {\rm Re} \left( z^3 \right) \hspace{1cm} (5) \end{align*}
We notice that $| z^3 | = p^{3/2}$, and ${\rm Re} \left( z^3 \right)$, ${\rm Im} \left( z^3 \right)$, and $| z^3 |$ form a right triangle. Thus, ${\rm Re} z^3 + {\rm Im} z^3 > p^{3/2}$.
Because $p > 1$, $p^{3/2} > p$.
Therefore, (3) holds.
Conditions (4) and (5) can be written in the joint form as
\[\left| {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) \right| < p . \hspace{1cm} (4)\]
We have
\begin{align*} {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) & = \left( a^3 - 3 a b^2 \right) - \left( - b^3 + 3 a^2 b \right) \\ & = \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \end{align*}
and $p = a^2 + b^2$.
Thus, (5) can be written as
\[\left| \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \right| < a^2 + b^2 . \hspace{1cm} (6)\]
Therefore, we need to jointly solve (1), (2), (6).
From (1) and (2), we have either $a, b >0$, or $a, b < 0$.
In (6), by symmetry, without loss of generality, we assume $a, b > 0$.
Thus, (1) and (2) are reduced to
\[a > \sqrt{3} b . \hspace{1cm} (7)\]
Let $a = \lambda b$. Plugging this into (6), we get
\begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| < \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} . \hspace{1cm} (8) \end{align*}
Because $p= a^2 + b^2$ is a prime, $a$ and $b$ are relatively prime.
Therefore, we can use (7), (8), $a^2 + b^2 <1000$, and $a$ and $b$ are relatively prime to solve the problem.
To facilitate efficient search, we apply the following criteria:
\begin{enumerate*}
\item To satisfy (7) and $a^2 + b^2 < 1000$, we have $1 \leq b \leq 15$.
In the outer layer, we search for $b$ in a decreasing order.
In the inner layer, for each given $b$, we search for $a$.
\item Given $b$, we search for $a$ in the range $\sqrt{3} b < a < \sqrt{1000 - b^2}$.
\item We can prove that for $b \geq 9$, there is no feasible $a$.
The proof is as follows.
For $b \geq 9$, to satisfy $a^2 + b^2 < 1000$, we have $a \leq 30$.
Thus, $\sqrt{3} < \lambda \leq \frac{30}{9}$.
Thus, the R.H.S. of (8) has the following upper bound
\begin{align*} \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} & < \frac{1}{b} \frac{\lambda^2 + \lambda}{\lambda + 1} \\ & = \frac{\lambda}{b} \\ & \leq \frac{\frac{30}{9}}{9} \\ & < \frac{10}{27} . \end{align*}
Hence, to satisfy (8), a necessary condition is
\begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| < \frac{10}{27} . \end{align*}
However, this cannot be satisfied for $\sqrt{3} < \lambda \leq \frac{30}{9}$.
Therefore, there is no feasible solution for $b \geq 9$.
Therefore, we only need to consider $b \leq 8$.
\item We eliminate $a$ that are not relatively prime to $b$.
\item We use the following criteria to quickly eliminate $a$ that make $a^2 + b^2$ a composite number.
For $b \equiv 1 \pmod{2}$, we eliminate $a$ satisfying $a \equiv 1 \pmod{2}$.
For $b \equiv \pm 1 \pmod{5}$ (resp. $b \equiv \pm 2 \pmod{5}$), we eliminate $a$ satisfying $a \equiv \pm 2 \pmod{5}$ (resp. $a \equiv \pm 1 \pmod{5}$).
\item For the remaining $\left( b, a \right)$, check whether (8) and the condition that $a^2 + b^2$ is prime are both satisfied.
The first feasible solution is $b = 5$ and $a = 18$.
Thus, $a^2 + b^2 = 349$.
\item For the remaining search, given $b$, we only search for $a \geq \sqrt{349 - b^2}$.
Following the above search criteria, we find the final answer as $b = 5$ and $a = 18$.
Thus, the largest prime $p$ is $p = a^2 + b^2 = \boxed{\textbf{(349) }}$.
|
349
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_15
|
37
|
Positive real numbers $b \not= 1$ and $n$ satisfy the equations \[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\] The value of $n$ is $\frac{j}{k},$ where $j$ and $k$ are relatively prime positive integers. Find $j+k.$
|
Denote $x = \log_b n$.
Hence, the system of equations given in the problem can be rewritten as
\begin{align*} \sqrt{x} & = \frac{1}{2} x , \\ bx & = 1 + x . \end{align*}
Solving the system gives $x = 4$ and $b = \frac{5}{4}$.
Therefore,
\[n = b^x = \frac{625}{256}.\]
Therefore, the answer is $625 + 256 = \boxed{881}$.
|
881
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_2
|
38
|
A plane contains $40$ lines, no $2$ of which are parallel. Suppose that there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Find the number of points where exactly $2$ lines intersect.
|
In this solution, let $\boldsymbol{n}$-line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$-line points.
There are $\binom{40}{2}=780$ pairs of lines. Among them:
The $3$-line points account for $3\cdot\binom32=9$ pairs of lines.
The $4$-line points account for $4\cdot\binom42=24$ pairs of lines.
The $5$-line points account for $5\cdot\binom52=50$ pairs of lines.
The $6$-line points account for $6\cdot\binom62=90$ pairs of lines.
It follows that the $2$-line points account for $780-9-24-50-90=\boxed{607}$ pairs of lines, where each pair intersect at a single point.
|
607
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_3
|
39
|
The sum of all positive integers $m$ such that $\frac{13!}{m}$ is a perfect square can be written as $2^a3^b5^c7^d11^e13^f,$ where $a,b,c,d,e,$ and $f$ are positive integers. Find $a+b+c+d+e+f.$
|
We first rewrite $13!$ as a prime factorization, which is $2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.$
For the fraction to be a square, it needs each prime to be an even power. This means $m$ must contain $7\cdot11\cdot13$. Also, $m$ can contain any even power of $2$ up to $2^{10}$, any odd power of $3$ up to $3^{5}$, and any even power of $5$ up to $5^{2}$. The sum of $m$ is \[(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)(7^1)(11^1)(13^1) =\] \[1365\cdot273\cdot26\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4.\] Therefore, the answer is $1+2+1+3+1+4=\boxed{012}$.
|
012
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_4
|
40
|
Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$
|
[Ptolemy's theorem](https://artofproblemsolving.com/wiki/index.php/Ptolemy%27s_theorem) states that for cyclic quadrilateral $WXYZ$, $WX\cdot YZ + XY\cdot WZ = WY\cdot XZ$.
We may assume that $P$ is between $B$ and $C$. Let $PA = a$, $PB = b$, $PC = c$, $PD = d$, and $AB = s$. We have $a^2 + c^2 = AC^2 = 2s^2$, because $AC$ is a diameter of the circle. Similarly, $b^2 + d^2 = 2s^2$. Therefore, $(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112$. Similarly, $(b+d)^2 = 2s^2 + 180$.
By Ptolemy's Theorem on $PCDA$, $as + cs = ds\sqrt{2}$, and therefore $a + c = d\sqrt{2}$. By Ptolemy's on $PBAD$, $bs + ds = as\sqrt{2}$, and therefore $b + d = a\sqrt{2}$. By squaring both equations, we obtain
\begin{alignat*}{8} 2d^2 &= (a+c)^2 &&= 2s^2 + 112, \\ 2a^2 &= (b+d)^2 &&= 2s^2 + 180. \end{alignat*}
Thus, $a^2 = s^2 + 90$, and $d^2 = s^2 + 56$. Plugging these values into $a^2 + c^2 = b^2 + d^2 = 2s^2$, we obtain $c^2 = s^2 - 90$, and $b^2 = s^2 - 56$. Now, we can solve using $a$ and $c$ (though using $b$ and $d$ yields the same solution for $s$).
\begin{align*} ac = (\sqrt{s^2 - 90})(\sqrt{s^2 + 90}) &= 56 \\ (s^2 + 90)(s^2 - 90) &= 56^2 \\ s^4 &= 90^2 + 56^2 = 106^2 \\ s^2 &= \boxed{106}. \end{align*}
|
106
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5
|
41
|
Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
We break the problem into stages, one for each card revealed, then further into cases based on the number of remaining unrevealed cards of each color. Since [expected value](https://artofproblemsolving.com/wiki/index.php/Expected_value) is linear, the expected value of the total number of correct card color guesses across all stages is the sum of the expected values of the number of correct card color guesses at each stage; that is, we add the probabilities of correctly guessing the color at each stage to get the final answer (See [https://brilliant.org/wiki/linearity-of-expectation/](https://artofproblemsolving.comhttps://brilliant.org/wiki/linearity-of-expectation/))
At any stage, if there are $a$ unrevealed cards of one color and $b$ of the other color, and $a \geq b$, then the optimal strategy is to guess the color with $a$ unrevealed cards, which succeeds with probability $\frac{a}{a+b}.$
Stage 1:
There are always $3$ unrevealed cards of each color, so the probability of guessing correctly is $\frac{1}{2}$.
Stage 2:
There is always a $3$-$2$ split ($3$ unrevealed cards of one color and $2$ of the other color), so the probability of guessing correctly is $\frac{3}{5}$.
Stage 3:
There are now $2$ cases:
The guess from Stage 2 was correct, so there is now a $2$-$2$ split of cards and a $\frac{1}{2}$ probability of guessing the color of the third card correctly.
The guess from Stage 2 was incorrect, so the split is $3$-$1$ and the probability of guessing correctly is $\frac{3}{4}$.
Thus, the overall probability of guessing correctly is $\frac{3}{5} \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{3}{4} = \frac{3}{5}$.
Stage 4:
This stage has $2$ cases as well:
The guesses from both Stage 2 and Stage 3 were incorrect. This occurs with probability $\frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10}$ and results in a $3$-$0$ split and a certain correct guess at this stage.
Otherwise, there must be a $2$-$1$ split and a $\frac{2}{3}$ probability of guessing correctly.
The probability of guessing the fourth card correctly is therefore $\frac{1}{10} \cdot 1 + \frac{9}{10} \cdot \frac{2}{3} = \frac{7}{10}$.
Stage 5:
Yet again, there are $2$ cases:
In Stage 4, there was a $2$-$1$ split and the guess was correct. This occurs with probability $\frac{9}{10} \cdot \frac{2}{3} = \frac{3}{5}$ and results in a $1$-$1$ split with a $\frac{1}{2}$ chance of a correct guess here.
Otherwise, there must be a $2$-$0$ split, making a correct guess certain.
In total, the fifth card can be guessed correctly with probability $\frac{3}{5} \cdot \frac{1}{2} + \frac{2}{5} \cdot 1 = \frac{7}{10}$.
Stage 6:
At this point, only $1$ card remains, so the probability of guessing its color correctly is $1$.
In conclusion, the expected value of the number of cards guessed correctly is \[\frac{1}{2} + \frac{3}{5} + \frac{3}{5} + \frac{7}{10} + \frac{7}{10} + 1 = \frac{5+6+6+7+7+10}{10} = \frac{41}{10},\] so the answer is $41 + 10 = \boxed{051}.$
|
051
|
https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_6
|
42
|
Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$.
|
$n$ can either be $0$ or $1$ (mod $2$).
Case 1: $n \equiv 0 \pmod{2}$
Then, $n \equiv 2 \pmod{4}$, which implies $n \equiv 1 \pmod{3}$ and $n \equiv 4 \pmod{6}$, and therefore $n \equiv 3 \pmod{5}$. Using [CRT](https://artofproblemsolving.com/wiki/index.php/Chinese_Remainder_Theorem), we obtain $n \equiv 58 \pmod{60}$, which gives $16$ values for $n$.
Case 2: $n \equiv 1 \pmod{2}$
$n$ is then $3 \pmod{4}$. If $n \equiv 0 \pmod{3}$, $n \equiv 3 \pmod{6}$, a contradiction. Thus, $n \equiv 2 \pmod{3}$, which implies $n \equiv 5 \pmod{6}$. $n$ can either be $0 \pmod{5}$, which implies that $n \equiv 35 \pmod{60}$ by CRT, giving $17$ cases; or $4 \pmod{5}$, which implies that $n \equiv 59 \pmod{60}$ by CRT, giving $16$ cases.
The total number of extra-distinct numbers is thus $16 + 16 + 17 = \boxed{049}$.
|
049
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_7
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43
|
Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$
|
This solution refers to the Diagram section.
Let $O$ be the incenter of $ABCD$ for which $\odot O$ is tangent to $\overline{DA},\overline{AB},$ and $\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\overleftrightarrow{DA},\overleftrightarrow{AB},$ and $\overleftrightarrow{BC},$ respectively, such that $\overline{RT}$ intersects $\odot O$ at $P$ and $Q.$
We obtain the following diagram:
Note that $\angle RXZ = \angle TZX = 90^\circ$ by the properties of tangents, so $RTZX$ is a rectangle. It follows that the diameter of $\odot O$ is $XZ = RT = 25.$
Let $x=PQ$ and $y=RX=TZ.$
We apply the Power of a Point Theorem to $R$ and $T:$
\begin{align*} y^2 &= 9(9+x), \\ y^2 &= 16(16-x). \end{align*}
We solve this system of equations to get $x=7$ and $y=12.$ Alternatively, we can find these results by the symmetry on rectangle $RTZX$ and semicircle $\widehat{XPZ}.$
We extend $\overline{SP}$ beyond $P$ to intersect $\odot O$ and $\overleftrightarrow{CD}$ at $E$ and $F,$ respectively, where $E\neq P.$ So, we have $EF=SP=5$ and $PE=25-SP-EF=15.$ On the other hand, we have $PX=15$ by the Pythagorean Theorem on right $\triangle PRX.$ Together, we conclude that $E=X.$ Therefore, points $S,P,$ and $X$ must be collinear.
Let $G$ be the foot of the perpendicular from $D$ to $\overline{AB}.$ Note that $\overline{DG}\parallel\overline{XP},$ as shown below:
As $\angle PRX = \angle AGD = 90^\circ$ and $\angle PXR = \angle ADG$ by the AA Similarity, we conclude that $\triangle PRX \sim \triangle AGD.$ The ratio of similitude is \[\frac{PX}{AD} = \frac{RX}{GD}.\] We get $\frac{15}{AD} = \frac{12}{25},$ from which $AD = \frac{125}{4}.$
Finally, the perimeter of $ABCD$ is $4AD = \boxed{125}.$
|
125
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8
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44
|
Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$
|
Plugging $2$ and $m$ into $P(x)$ and equating them, we get $8+4a+2b+c = m^3+am^2+bm+c$. Rearranging, we have \[(m^3-8) + (m^2 - 4)a + (m-2)b = 0.\] Note that the value of $c$ won't matter as it can be anything in the provided range, giving a total of $41$ possible choices for $c.$ So what we just need to do is to just find the number of ordered pairs $(a, b)$ that work, and multiply it by $41.$
We can start by first dividing both sides by $m-2.$ (Note that this is valid since $m\neq2:$ \[m^2 + 2m + 4 + (m+2)a + b = 0.\] We can rearrange this so it is a quadratic in $m$: \[m^2 + (a+2)m + (4 + 2a + b) = 0.\] Remember that $m$ has to be unique and not equal to $2.$ We can split this into two cases: case $1$ being that $m$ has exactly one solution, and it isn't equal to $2$; case $2$ being that $m$ has two solutions, one being equal to $2,$ but the other is a unique solution not equal to $2.$
$\textbf{Case 1:}$
There is exactly one solution for $m,$ and that solution is not $2.$ This means that the discriminant of the quadratic equation is $0,$ using that, we have $(a+2)^2 = 4(4 + 2a + b),$ rearranging in a neat way, we have \[(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2\sqrt{4+b}.\] Using the fact that $4+b$ must be a perfect square, we can easily see that the values for $b$ can be $-4, -3, 0, 5,$ and $12.$ Also since it's a "$\pm$" there will usually be $2$ solutions for $a$ for each value of $b.$ The two exceptions for this would be if $b = -4$ and $b = 12.$ For $b=-4$ because it would be a $\pm0,$ which only gives one solution, instead of two. And for $b=12$ because then $a = -6$ and the solution for $m$ would equal to $2,$ and we don't want this. (We can know this by putting the solutions back into the quadratic formula).
So we have $5$ solutions for $b,$ each of which give $2$ values for $a,$ except for $2,$ which only give one. So in total, there are $5*2 - 2 = 8$ ordered pairs of $(a,b)$ in this case.
$\textbf{Case 2:}$
$m$ has two solutions, but exactly one of them isn't equal to $2.$ This ensures that $1$ of the solutions is equal to $2.$
Let $r$ be the other value of $m$ that isn't $2.$ By Vieta:
\begin{align*} r+2 &= -a-2\\ 2r &= 4+2a+b. \end{align*} From the first equation, we subtract both sides by $2$ and double both sides to get $2r = -2a - 8$ which also equals to $4+2a+b$ from the second equation. Equating both, we have $4a + b + 12 = 0.$ We can easily count that there would be $11$ ordered pairs $(a,b)$ that satisfy that.
However, there's an outlier case in which $r$ happens to also equal to $2,$ and we don't want that. We can reverse engineer and find out that $r=2$ when $(a,b) = (-6, 12),$ which we overcounted. So we subtract by one and we conclude that there are $10$ ordered pairs of $(a,b)$ that satisfy this case.
This all shows that there are a total of $8+10 = 18$ amount of ordered pairs $(a,b).$ Multiplying this by $41$ (the amount of values for $c$) we get $18\cdot41=\boxed{738}$ as our final answer.
|
738
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_9
|
45
|
The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples growing on any of the six trees.
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In the arithmetic sequence, let $a$ be the first term and $d$ be the common difference, where $d>0.$ The sum of the first six terms is \[a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.\]
We are given that
\begin{align*} 6a+15d &= 990, \\ 2a &= a+5d. \end{align*}
The second equation implies that $a=5d.$ Substituting this into the first equation, we get
\begin{align*} 6(5d)+15d &=990, \\ 45d &= 990 \\ d &= 22. \end{align*}
It follows that $a=110.$ Therefore, the greatest number of apples growing on any of the six trees is $a+5d=\boxed{220}.$
|
220
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_1
|
46
|
Let $N$ be the number of ways to place the integers $1$ through $12$ in the $12$ cells of a $2 \times 6$ grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by $3.$ One way to do this is shown below. Find the number of positive integer divisors of $N.$
\[\begin{array}{|c|c|c|c|c|c|} \hline \,1\, & \,3\, & \,5\, & \,7\, & \,9\, & 11 \\ \hline \,2\, & \,4\, & \,6\, & \,8\, & 10 & 12 \\ \hline \end{array}\]
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We replace the numbers which are 0 mod(3) to 0, 1 mod(3) to 1, and 2 mod(3) to 2.
Then, the problem is equivalent to arranging 4 0's,4 1's, and 4 2's into the grid (and then multiplying by $4!^3$ to account for replacing the remainder numbers with actual numbers) such that no 2 of the same numbers are adjacent.
Then, the numbers which are vertically connected must be different. 2 1s must be connected with 2 2s, and 2 1s must be connected with 2 2s (vertically), as if there are less 1s connected with either, then there will be 2s or 3s which must be connected within the same number.
For instance if we did did:
- 12 x 3
- 13 x 1
Then we would be left with 23 x1, and 2 remaining 3s which aren't supposed to be connected, so it is impossible. Similar logic works for why all 1s can't be connected with all 2s.
Thus, we are left with the problem of re-arranging 2x 12 pairs, 2x 13 pairs, 2x 23 pairs. Notice that the pairs can be re-arranged horizontally in any configuration, but when 2 pairs are placed adjacent there is only 1 configuration for the rightmost pair to be set after the leftmost one has been placed.
We have $\frac{6!}{2!2!2!}=90$ ways to horizontally re-arrange the pairs, with 2 ways to set the initial leftmost column. Thus, there are 180 ways to arrange the pairs. Accounting for the initial simplification of the problem with 1-12 -> 0-3, we obtain the answer is:
\[2488320=2^{11}\cdot3^5\cdot5^1\]
The number of divisors is: $12\cdot6\cdot2=\boxed{144}.$
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144
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_10
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47
|
Find the number of collections of $16$ distinct subsets of $\{1,2,3,4,5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X \cap Y \not= \emptyset.$
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Denote by $\mathcal C$ a collection of 16 distinct subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$.
Denote $N = \min \left\{ |S|: S \in \mathcal C \right\}$.
Case 1: $N = 0$.
This entails $\emptyset \in \mathcal C$.
Hence, for any other set $A \in \mathcal C$, we have $\emptyset \cap A = \emptyset$. This is infeasible.
Case 2: $N = 1$.
Let $\{a_1\} \in \mathcal C$.
To get $\{a_1\} \cap A \neq \emptyset$ for all $A \in \mathcal C$.
We must have $a_1 \in \mathcal A$.
The total number of subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain $a_1$ is $2^4 = 16$.
Because $\mathcal C$ contains 16 subsets.
We must have $\mathcal C = \left\{ \{a_1\} \cup A : \forall \ A \subseteq \left\{ 1, 2, 3, 4, 5 \right\} \backslash \left\{a_1 \right\} \right\}$.
Therefore, for any $X, Y \in \mathcal C$, we must have $X \cap Y \supseteq \{a_1\}$.
So this is feasible.
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $a_1$.
Therefore, the number of solutions is 5.
Case 3: $N = 2$.
Case 3.1: There is exactly one subset in $\mathcal C$ that contains 2 elements.
Denote this subset as $\left\{ a_1, a_2 \right\}$.
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$, except $\left\{ a_3, a_4, a_5 \right\}$.
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$.
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$.
Therefore, the number of solutions is $\binom{5}{2} = 10$.
Case 3.2: There are exactly two subsets in $\mathcal C$ that contain 2 elements.
They must take the form $\left\{ a_1, a_2 \right\}$ and $\left\{ a_1, a_3 \right\}$.
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$, except $\left\{ a_3, a_4, a_5 \right\}$ and $\left\{ a_2, a_4, a_5 \right\}$.
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$.
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$ and $\left\{ a_1, a_3 \right\}$.
Therefore, the number of solutions is $5 \cdot \binom{4}{2} = 30$.
Case 3.3: There are exactly three subsets in $\mathcal C$ that contain 2 elements.
They take the form $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_1, a_4 \right\}$.
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$, except $\left\{ a_3, a_4, a_5 \right\}$, $\left\{ a_2, a_4, a_5 \right\}$, $\left\{ a_2, a_3, a_5 \right\}$.
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$.
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_1, a_4 \right\}$.
Therefore, the number of solutions is $5 \cdot \binom{4}{3} = 20$.
Case 3.4: There are exactly three subsets in $\mathcal C$ that contain 2 elements.
They take the form $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_2, a_3 \right\}$.
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$, except $\left\{ a_3, a_4, a_5 \right\}$, $\left\{ a_2, a_4, a_5 \right\}$, $\left\{ a_1, a_4, a_5 \right\}$.
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$.
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_2, a_3 \right\}$.
Therefore, the number of solutions is $\binom{5}{3} = 10$.
Case 3.5: There are exactly four subsets in $\mathcal C$ that contain 2 elements.
They take the form $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_1, a_4 \right\}$, $\left\{ a_1, a_5 \right\}$.
We then put all subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements into $\mathcal C$, except $\left\{ a_3, a_4, a_5 \right\}$, $\left\{ a_2, a_4, a_5 \right\}$, $\left\{ a_1, a_4, a_5 \right\}$, $\left\{ a_2, a_3, a_4 \right\}$.
This satisfies $X \cap Y \neq \emptyset$ for any $X, Y \in \mathcal C$.
Now, we count the number of $\mathcal C$ in this case.
We only need to determine $\left\{ a_1, a_2 \right\}$, $\left\{ a_1, a_3 \right\}$, $\left\{ a_1, a_4 \right\}$, $\left\{ a_1, a_5 \right\}$.
Therefore, the number of solutions is 5.
Putting all subcases together, the number of solutions is this case is $10 + 30 + 20 + 10 + 5 = 75$.
Case 4: $N \geq 3$.
The number of subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$ that contain at least three elements is $\sum_{i=3}^5 \binom{5}{i} = 16$.
Because $\mathcal C$ has 16 elements, we must select all such subsets into $\mathcal C$.
Therefore, the number of solutions in this case is 1.
Putting all cases together, the total number of $\mathcal C$ is $5 + 75 + 1 = \boxed{\textbf{(081) }}$.
Denote the $A$ as $\{ 1,2,3,4,5 \}$ and the collection of subsets as $S$.
Case 1: There are only sets of size $3$ or higher in $S$:
Any two sets in $S$ must have at least one element common to both of them (since $3+3>5$). Since there are $16$ subsets of $A$ that have size $3$ or higher, there is only one possibility for this case.
Case 2: There are only sets of size $2$ or higher in $S$:
Firstly, there cannot be a no size $2$ set $S$, or it will be overcounting the first case.
If there is only one such size $2$ set, there are $10$ ways to choose it. That size $2$ set, say $X$, cannot be in $S$ with $Y = A/X$ (a three element set). Thus, there are only $15$ possible size $3$ subsets that can be in $S$, giving us $10$ for this case.
If there are two sets with size $2$ in $S$, we can choose the common elements of these two subsets in $5$ ways, giving us a total of $5\cdot 6 = 30$.
If there are three sets with size $2$, they can either share one common element, which can be done in $5 \cdot 4 = 20$ ways, or they can share pairwise common elements (sort of like a cycle), which can be done $\binom{5}{2} = 10$ ways. In total, we have $30$ possibilities.
If there are four sets with size $2$, they all have to share one common element, which can be done in $5\cdot 1$ ways.
Thus, summing everything up, this will give us $75$ possible sets $S$
Case 3: There is a set with size $1$ in $S$:
Notice that be at most one size $1$ subset. There are $5$ ways to choose that single element set. Say it's $\{ 1\}$. All other subsets in $S$ must have a $1$ in them, but there are only $2^4-1 = 15$ of them. Thus this case yields $5\cdot 1 = 5$ possibilities.
Thus, the total number of sets $S$ would be $1+75+5 = \boxed{\textbf{(081)}}$.
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081
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_11
|
48
|
In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Define $L_1$ to be the foot of the altitude from $A$ to $BC$. Furthermore, define $L_2$ to be the foot of the altitude from $Q$ to $BC$. From here, one can find $AL_1=12$, either using the 13-14-15 triangle or by calculating the area of $ABC$ two ways. Then, we find $BL_1=5$ and $L_1C = 9$ using Pythagorean theorem. Let $QL_2=x$. By AA similarity, $\triangle{AL_1M}$ and $\triangle{QL_2M}$ are similar. By similarity ratios, \[\frac{AL_1}{L_1M}=\frac{QL_2}{L_2M}\]
\[\frac{12}{2}=\frac{x}{L_2M}\]
\[L_2M = \frac{x}{6}\]
Thus, $BL_2=BM-L_2M=7-\frac{x}{6}$. Similarly, $CL_2=7+\frac{x}{6}$. Now, we angle chase from our requirement to obtain new information.
\[\angle{PBQ}=\angle{PCQ}\]
\[\angle{QCM}+\angle{PCM}=\angle{QBM}+\angle{PBM}\]
\[\angle{QCL_2}+\angle{PCM}=\angle{QBL_2}+\angle{PBM}\]
\[\angle{PCM}-\angle{PBM}=\angle{QBL_2}-\angle{QCL_2}\]
\[\angle{MAB}-\angle{MAC}=\angle{QBL_2}-\angle{QCL_2}\text{(By inscribed angle theorem)}\]
\[(\angle{MAL_1}+\angle{L_1AB})-(\angle{CAL_1}-\angle{MAL_1})=\angle{QBL_2}-\angle{QCL_2}\]
\[2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1}=\angle{QBL_2}-\angle{QCL_2}\]
Take the tangent of both sides to obtain
\[\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\tan(\angle{QBL_2}-\angle{QCL_2})\]
By the definition of the tangent function on right triangles, we have $\tan{MAL_1}=\frac{7-5}{12}=\frac{1}{6}$, $\tan{CAL_1}=\frac{9}{12}=\frac{3}{4}$, and $\tan{L_1AB}=\frac{5}{12}$. By abusing the tangent angle addition formula, we can find that
\[\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\frac{196}{2397}\]
By substituting $\tan{\angle{QBL_2}}=\frac{6x}{42-x}$, $\tan{\angle{QCL_2}}=\frac{6x}{42+x}$ and using tangent angle subtraction formula we find that
\[x=\frac{147}{37}\]
Finally, using similarity formulas, we can find
\[\frac{AQ}{AM}=\frac{12-x}{x}\]. Plugging in $x=\frac{147}{37}$ and $AM=\sqrt{148}$, we find that \[AQ=\frac{99}{\sqrt{148}}\] Thus, our final answer is $99+148=\boxed{247}$.
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247
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12
|
49
|
Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$
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Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
\[\tan A = 2 \cos A \implies \sin A = 2 \cos^2 A \implies \sin^2 A + \cos^2 A = 4 \cos^4 A + \cos^2 A = 1\]
\[\implies \cos^2 A = \frac {\sqrt {17} - 1}{8}.\]
\[c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\frac {n}{2}}+(4 \cos^2 A)^{\frac {n}{2}} =\]
\[= \left(\frac {\sqrt {17} + 1}{2}\right)^{\frac {n}{2}}+ \left(\frac {\sqrt {17} - 1}{2}\right)^{\frac {n}{2}}.\]
It is clear, that $c_n$ is not integer if $n \ne 4k, k > 0.$
Denote $x = \frac {\sqrt {17} + 1}{2}, y = \frac {\sqrt {17} - 1}{2} \implies$
\[x \cdot y = 4, x + y = \sqrt{17}, x - y = 1 \implies x^2 + y^2 = (x - y)^2 + 2xy = 9 = c_4.\]
\[c_8 = x^4 + y^4 = (x^2 + y^2)^2 - 2x^2 y^2 = 9^2 - 2 \cdot 16 = 49.\]
\[c_{4k+4} = x^{4k+4} + y^{4k+4} = (x^{4k} + y^{4k})(x^2+y^2)- (x^2 y^2)(x^{4k-2}+y^{4k-2}) = 9 c_{4k}- 16 c_{4k – 4} \implies\]
\[c_{12} = 9 c_8 - 16 c_4 = 9 \cdot 49 - 16 \cdot 9 = 9 \cdot 33 = 297.\]
\[c_{16} = 9 c_{12} - 16 c_8 = 9 \cdot 297 - 16 \cdot 49 = 1889.\]
\[c_{12m + 4} \pmod{10} = 9 \cdot c_{12m} \pmod{10} - 16 \pmod{10} \cdot c_{12m - 4} \pmod{10} =\]
\[= (9 \cdot 7 - 6 \cdot 9) \pmod{10} = (3 - 4) \pmod{10} = 9.\]
\[c_{12m + 8}\pmod{10} = 9 \cdot c_{12m+4} \pmod{10} - 16 \pmod{10} \cdot c_{12m } \pmod{10} =\]
\[= (9 \cdot 9 - 6 \cdot 7) \pmod{10} = (1 - 2)\pmod{10} = 9.\]
\[c_{12m + 12} \pmod{10} = 9 \cdot c_{12m + 8} \pmod{10} - 16 \pmod{10} \cdot c_{12m + 4} \pmod{10} =\]
\[= (9 \cdot 9 - 6 \cdot 9) \pmod{10} = (1 - 4) \pmod{10} = 7 \implies\]
The condition is satisfied iff $n = 12 k + 4$ or $n = 12k + 8.$
If $n \le N$ then the number of possible n is $\left\lfloor \frac{N}{4} \right\rfloor - \left\lfloor \frac{N}{12} \right\rfloor.$
For $N = 1000$ we get $\left\lfloor \frac{1000}{4} \right\rfloor - \left\lfloor \frac{1000}{12} \right\rfloor = 250 - 83 = \boxed{167}.$
[email protected], vvsss
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167
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_13
|
50
|
A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangle $ABDC$ is perpendicular to $\mathcal{P},$ vertex $B$ is $2$ meters above $\mathcal{P},$ vertex $C$ is $8$ meters above $\mathcal{P},$ and vertex $D$ is $10$ meters above $\mathcal{P}.$ The cube contains water whose surface is parallel to $\mathcal{P}$ at a height of $7$ meters above $\mathcal{P}.$ The volume of water is $\frac{m}{n}$ cubic meters, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
Let's first view the cube from a direction perpendicular to $ABDC$, as illustrated above. Let $x$ be the cube's side length. Since $\triangle CHA \sim \triangle AGB$, we have
\[\frac{CA}{CH} = \frac{AB}{AG}.\]
We know $AB = x$, $AG = \sqrt{x^2-2^2}$, $AC = \sqrt{2}x$, $CH = 8$. Plug them into the above equation, we get
\[\frac{\sqrt{2}x}{8} = \frac{x}{\sqrt{x^2-2^2}}.\]
Solving this we get the cube's side length $x = 6$, and $AC = 6\sqrt{2}.$
Let $PQ$ be the water's surface, both $P$ and $Q$ are $7$ meters from $\mathcal P$. Notice that $C$ is $8$ meters from $\mathcal P$, this means
\[CP = \frac{1}{8}CA = \frac{3\sqrt{2}}{4}.\]
Similarly,
\[DQ = \frac{3}{8}CA = \frac{9\sqrt{2}}{4}.\]
Now, we realize that the 3D space inside the cube without water is a frustum, with $P$ on its smaller base and $Q$ on its larger base. To find its volume, all we need is to find the areas of both bases and the height, which is $x = 6$. To find the smaller base, let's move our viewpoint onto the plane $ABDC$ and view the cube from a direction parallel to $ABDC$, as shown above. The area of the smaller base is simply
\[S_1 = CP^2 = \Bigl(\frac{3\sqrt{2}}{4}\Bigr)^2 = \frac{9}{8}.\]
Similarly, the area of the larger base is
\[S_2 = DQ^2 = \Bigl(\frac{9\sqrt{2}}{4}\Bigr)^2 = \frac{81}{8}.\]
Finally, applying the formula for a frustum's volume,
\[V = \frac{1}{3} \cdot x \cdot (S_1 + \sqrt{S_1S_2} + S_2) = \frac{1}{3} \cdot 6 \cdot \Bigl(\frac{9}{8} + \sqrt{\frac{9}{8}\cdot\frac{81}{8}} + \frac{81}{8}\Bigl) = \frac{117}{4}.\]
The water's volume is thus
\[6^3 - \frac{117}{4} = \frac{747}{4},\]
giving $\boxed{751}$.
[2023 AIME II 14.png](https://artofproblemsolving.com/wiki/index.php/File:2023_AIME_II_14.png)
Denote $h(X)$ the distance from point $X$ to $\mathcal{P}, h(A) = 0, h(B) = 2,$
$h(C) = 8, h(D) = 10, h(G) = h(I) = h(H) = 7, AB = a, AC = a \sqrt{2}.$
Let slope $AB$ to $\mathcal{P}$ be $\alpha.$ Notation is shown in the diagram.
\[\tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {h(B)}{AB}\cdot \frac {AC}{h(C)} = \frac {\sqrt{2}}{4} \implies a = 6.\]
Let $S = GI \cap CD \implies h(S) = h(G) = 7.$
\[h(C) – h(G) = 8 - 7 = 1, h(D)- h(I) = 10 - 7 = 3.\]
\[h(E) = h(F) = \frac {h(D) +h(B)}{2} = 6 \implies\]
\[\frac {DI}{DE} = \frac {h(D) - h(I)}{h(D)-h(E)} = \frac {3}{4} \implies DI = DH = \frac {9}{2}.\]
Similarly $CG = \frac {3}{2} \implies SD = 9.$
Let the volume without water be $V,$ volume of the pyramid $SCGJ$ be $U.$
It is clear that $U + V = 27U = \frac {SD}{6} \cdot DI^2 = \frac {243}{8} \implies$
$V = \frac {243 \cdot 26}{8 \cdot 27 } = \frac {117}{4} = 6^3 - \frac {747}{4}$ from which $\boxed{751}.$
[email protected], vvsss
We introduce a Cartesian coordinate system to the diagram.
We put the origin at $A$. We let the $z$-components of $B$, $C$, $D$ be positive.
We set the $x$-axis in a direction such that $B$ is on the $x-O-z$ plane.
The coordinates of $A$, $B$, $C$ are $A = \left( 0, 0, 0 \right)$, $B = \left( x_B, 0 , 2 \right)$, $C = \left( x_C, y_C, 8 \right)$.
Because $AB \perp AC$, $\overrightarrow{AB} \cdot \overrightarrow{AC} = 0$.
Thus,
\[ x_B x_C + 16 = 0 . \hspace{1cm} (1) \]
Because $AC$ is a diagonal of a face, $AC^2 = 2 AB^2$.
Thus,
\[ x_C^2 + y_C^2 + 8^2 = 2 \left( x_B^2 + 2^2 \right) . \hspace{1cm} (2) \]
Because plane $ABCD$ is perpendicular to plan $P$, $\hat z \cdot \left( \overrightarrow{AB} \times \overrightarrow{AC} \right) = 0$.
Thus,
\[ \begin{vmatrix} 0 & 0 & 1 \\ x_B & 0 & 2 \\ x_C & y_C & 8 \end{vmatrix} = 0 . \hspace{1cm} (3) \]
Jointly solving (1), (2), (3), we get one solution $x_B = 4 \sqrt{2}$, $x_C = - 2 \sqrt{2}$, $y_C = 0$.
Thus, the side length of the cube is $|AB| = \sqrt{x_B^2 + 2^2} = 6$.
Denote by $P$ and $Q$ two vertices such that $AP$ and $AQ$ are two edges, and satisfy the right-hand rule that $\widehat{AB} \times \widehat{AP} = \widehat{AQ}$.
Now, we compute the coordinates of $P$ and $Q$.
Because $|AB| = 6$, we have $\overrightarrow{AP} \times \overrightarrow{AQ} = 6 \overrightarrow{AB}$, $\overrightarrow{AQ} \times \overrightarrow{AB} = 6 \overrightarrow{AP}$, $\overrightarrow{AB} \times \overrightarrow{AP} = 6 \overrightarrow{AQ}$.
Hence,
\begin{align*} \begin{bmatrix} \hat i & \hat j & \hat k \\ x_P & y_P & z_P \\ x_Q & y_Q & z_Q \end{bmatrix} & = 6 \left( 4 \sqrt{2} \hat i + 2 \hat k \right) , \\ \begin{vmatrix} \hat i & \hat j & \hat k \\ x_Q & y_Q & z_Q \\ 4 \sqrt{2} & 0 & 2 \end{vmatrix} & = 6 \left( x_P \hat i + y_P \hat j + z_P \hat k \right) , \\ \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 \sqrt{2} & 0 & 2 \\ x_P & y_P & z_P \end{vmatrix} & = 6 \left( x_Q \hat i + y_Q \hat j + z_Q \hat k \right) . \end{align*}
By solving these equations, we get
$y_P^2 + y_Q^2 = 36 .$
In addition, we have $\overrightarrow{AC} = \overrightarrow{AP} + \overrightarrow{AQ}$.
Thus, $P = \left( - \sqrt{2} , 3 \sqrt{2} , 4 \right)$, $Q = \left( - \sqrt{2} , - 3 \sqrt{2} , 4 \right)$.
Therefore, the volume of the water is
\begin{align*} V = & 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ z_B u + z_P v + z_Q w \leq 7 \right\} dw dv du \\ & = 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u + 4 v + 4 w \leq 7 \right\} dw dv du \\ & = 6^3 - 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u + 4 v + 4 w > 7 \right\} dw dv du . \end{align*}
Define $u' = 1 - u$, $v' = 1 - v$, $w' = 1 - w$.
Thus,
\begin{align*} V & = 6^3 - 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u' + 4 v' + 4 w' < 3 \right\} dw dv du \\ & = 6^3 - 6^3 \int_{u'=0}^1 \left( \int_{v'=0}^1 \int_{w'=0}^1 \mathbf 1 \left\{ v' + w' < \frac{3}{4} - \frac{u'}{2} \right\} dw' dv' \right) du' \\ & = 6^3 - 6^3 \int_{u'=0}^1 \frac{1}{2} \left( \frac{3}{4} - \frac{u'}{2} \right)^2 du' . \end{align*}
Define $u'' = \frac{3}{4} - \frac{u'}{2}$.
Thus,
\begin{align*} V & = 6^3 - 6^3 \int_{u'' = 1/4}^{3/4} \left( u'' \right)^2 du'' \\ & = 6^3 - 6^3 \frac{1}{3} \left( \left(\frac{3}{4}\right)^3 - \left(\frac{1}{4}\right)^3 \right) \\ & = 216 - \frac{117}{4} \\ & = \frac{747}{4} . \end{align*}
Therefore, the answer is $747 + 4 = \boxed{\textbf{(751) }}$.
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751
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_14
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51
|
For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n \equiv 1 \pmod{2^n}.$ Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n = a_{n+1}.$
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[Bxiao31415](https://artofproblemsolving.com/wiki/index.php/User:Bxiao31415)
(small changes by bobjoebilly and [IraeVid13](https://artofproblemsolving.com/wiki/index.php/User:Iraevid13))
We first check that $\gcd(23, 2^n) = 1$ hence we are always seeking a unique modular inverse of $23$, $b_n$, such that $a_n \equiv 23b_n \equiv 1 \mod{2^n}$.
Now that we know that $b_n$ is unique, we proceed to recast this problem in binary. This is convenient because $x \mod{2^n}$ is simply the last $n$-bits of $x$ in binary, and if $x \equiv 1 \mod{2^n}$, it means that of the last $n$ bits of $x$, only the rightmost bit (henceforth $0$th bit) is $1$.
Also, multiplication in binary can be thought of as adding shifted copies of the multiplicand. For example:
\begin{align} 10111_2 \times 1011_2 &= 10111_2 \times (1000_2 + 10_2 + 1_2) \\ &= 10111000_2 + 101110_2 + 10111_2 \\ &= 11111101_2 \end{align}
Now note $23 = 10111_2$, and recall that our objective is to progressively zero out the $n$ leftmost bits of $a_n = 10111_2 \times b_n$ except for the $0$th bit.
Write $b_n = \underline{c_{n-1}\cdots c_2c_1c_0}_2$, we note that $c_0$ uniquely defines the $0$th bit of $a_n$, and once we determine $c_0$, $c_1$ uniquely determines the $1$st bit of $a_n$, so on and so forth.
For example, $c_0 = 1$ satisfies $a_1 \equiv10111_2 \times 1_2 \equiv 1 \mod{10_2}$
Next, we note that the second bit of $a_1$ is $1$, so we must also have $c_1 = 1$ in order to zero it out, giving
\[a_2 \equiv 10111_2 \times 11_2 \equiv 101110_2 + a_1 \equiv 1000101_2 \equiv 01_2 \mod{100_2}\]
$a_{n+1} = a_{n}$ happens precisely when $c_n = 0$. In fact we can see this in action by working out $a_3$. Note that $a_2$ has 1 on the $2$nd bit, so we must choose $c_2 = 1$. This gives
\[a_3 \equiv 10111_2 \times 111_2 \equiv 1011100_2 + a_2 \equiv 10100001_2 \equiv 001_2 \mod{1000_2}\]
Note that since the $3$rd and $4$th bit are $0$, $c_3 = c_4 = 0$, and this gives $a_3 = a_4 = a_5$.
It may seem that this process will take forever, but note that $23 = 10111_2$ has $4$ bits behind the leading digit, and in the worst case, the leading digits of $a_n$ will have a cycle length of at most $16$. In fact, we find that the cycle length is $11$, and in the process found that $a_3 = a_4 = a_5$, $a_6 = a_7$, and $a_{11} = a_{12}$.
Since we have $90$ complete cycles of length $11$, and the last partial cycle yields $a_{993} = a_{994} = a_{995}$ and $a_{996} = a_{997}$, we have a total of $90 \times 4 + 3 = \boxed{363}$ values of $n \le 1000$ such that $a_n = a_{n+1}$
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363
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_15
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52
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Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than $1000$ that is a palindrome both when written in base ten and when written in base eight, such as $292 = 444_{\text{eight}}.$
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Assuming that such palindrome is greater than $777_8 = 511,$ we conclude that the palindrome has four digits when written in base $8.$ Let such palindrome be \[(\underline{ABBA})_8 = 512A + 64B + 8B + A = 513A + 72B.\]
It is clear that $A=1,$ so we repeatedly add $72$ to $513$ until we get palindromes less than $1000:$
\begin{align*} 513+72\cdot0 &= 513, \\ 513+72\cdot1 &= \boxed{585}, \\ 513+72\cdot2 &= 657, \\ 513+72\cdot3 &= 729, \\ 513+72\cdot4 &= 801, \\ 513+72\cdot5 &= 873, \\ 513+72\cdot6 &= 945, \\ 513+72\cdot7 &= 1017. \\ \end{align*}
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585
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_2
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53
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Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$
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This solution refers to the Diagram section.
Let $\angle PAB = \angle PBC = \angle PCA = \theta,$ from which $\angle PAC = 90^\circ-\theta,$ and $\angle APC = 90^\circ.$
Moreover, we have $\angle PBA = \angle PCB = 45^\circ-\theta,$ as shown below:
Note that $\triangle PAB \sim \triangle PBC$ by AA Similarity. The ratio of similitude is $\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},$ so $\frac{10}{PB} = \frac{1}{\sqrt2}$ and thus $PB=10\sqrt2.$ Similarly, we can figure out that $PC=20$.
Finally, $AC=\sqrt{10^2+20^2}=10\sqrt{5}$, so the area of $\triangle ABC$ is \[\frac12\cdot AB\cdot AC = \frac12\cdot (10\sqrt{5})^2 = \boxed{250}.\]
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250
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3
|
54
|
Let $x,y,$ and $z$ be real numbers satisfying the system of equations
\begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*}
Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$
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We first subtract the second equation from the first, noting that they both equal $60$.
\begin{align*} xy+4z-yz-4x&=0 \\ 4(z-x)-y(z-x)&=0 \\ (z-x)(4-y)&=0 \end{align*}
Case 1: Let $y=4$.
The first and third equations simplify to:
\begin{align*} x+z&=15 \\ xz&=44 \end{align*}
from which it is apparent that $x=4$ and $x=11$ are solutions.
Case 2: Let $x=z$.
The first and third equations simplify to:
\begin{align*} xy+4x&=60 \\ x^2+4y&=60 \end{align*}
We subtract the following equations, yielding:
\begin{align*} x^2+4y-xy-4x&=0 \\ x(x-4)-y(x-4)&=0 \\ (x-4)(x-y)&=0 \end{align*}
We thus have $x=4$ and $x=y$, substituting in $x=y=z$ and solving yields $x=6$ and $x=-10$.
Then, we just add the squares of the solutions (make sure not to double count the $4$), and get \[4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}.\]
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273
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4
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55
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Let $S$ be the set of all positive rational numbers $r$ such that when the two numbers $r$ and $55r$ are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of $S$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$
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Denote $r = \frac{a}{b}$, where $\left( a, b \right) = 1$.
We have $55 r = \frac{55a}{b}$.
Suppose $\left( 55, b \right) = 1$, then the sum of the numerator and the denominator of $55r$ is $55a + b$.
This cannot be equal to the sum of the numerator and the denominator of $r$, $a + b$.
Therefore, $\left( 55, b \right) \neq 1$.
Case 1: $b$ can be written as $5c$ with $\left( c, 11 \right) = 1$.
Thus, $55r = \frac{11a}{c}$.
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same,
\[ a + 5c = 11a + c . \]
Hence, $2c = 5 a$.
Because $\left( a, b \right) = 1$, $\left( a, c \right) = 1$.
Thus, $a = 2$ and $c = 5$.
Therefore, $r = \frac{a}{5c} = \frac{2}{25}$.
Case 2: $b$ can be written as $11d$ with $\left( d, 5 \right) = 1$.
Thus, $55r = \frac{5a}{c}$.
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same,
\[ a + 11c = 5a + c . \]
Hence, $2a = 5 c$.
Because $\left( a, b \right) = 1$, $\left( a, c \right) = 1$.
Thus, $a = 5$ and $c = 2$.
Therefore, $r = \frac{a}{11c} = \frac{5}{22}$.
Case 3: $b$ can be written as $55 e$.
Thus, $55r = \frac{a}{c}$.
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same,
\[ a + 55c = a + c . \]
Hence, $c = 0$. This is infeasible.
Thus, there is no solution in this case.
Putting all cases together, $S = \left\{ \frac{2}{25}, \frac{5}{22} \right\}$.
Therefore, the sum of all numbers in $S$ is
\[ \frac{2}{25} + \frac{5}{22} = \frac{169}{550} . \]
Therefore, the answer is $169 + 550 = \boxed{\textbf{(719) }}$.
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719
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_5
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56
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Consider the L-shaped region formed by three unit squares joined at their sides, as shown below. Two points $A$ and $B$ are chosen independently and uniformly at random from inside the region. The probability that the midpoint of $\overline{AB}$ also lies inside this L-shaped region can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
[asy] unitsize(2cm); draw((0,0)--(2,0)--(2,1)--(1,1)--(1,2)--(0,2)--cycle); draw((0,1)--(1,1)--(1,0),dashed); [/asy]
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We proceed by calculating the complement.
Note that the only configuration of the 2 points that makes the midpoint outside of the L shape is one point in the top square, and one in the right square. This occurs with $\frac{2}{3} \cdot \frac{1}{3}$ probability.
Let the topmost coordinate have value of: $(x_1,y_1+1)$, and rightmost value of: $(x_2+1,y_2)$.
The midpoint of them is thus: \[\left(\frac{x_1+x_2+1}{2}, \frac{y_1+y_2+1}{2} \right)\]
It is clear that $x_1, x_2, y_1, y_2$ are all between 0 and 1. For the midpoint to be outside the L-shape, both coordinates must be greater than 1, thus:
\[\frac{x_1+x_2+1}{2}>1\]
\[x_1+x_2>1\]
By symmetry this has probability $\frac{1}{2}$. Also by symmetry, the probability the y-coordinate works as well is $\frac{1}{2}$. Thus the probability that the midpoint is outside the L-shape is:
\[\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{18}\]
We want the probability that the point is inside the L-shape however, which is $1-\frac{1}{18}=\frac{17}{18}$, yielding the answer of $17+18=\boxed{35}$
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35
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_6
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57
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Each vertex of a regular dodecagon ($12$-gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle.
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Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.
Case 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.
Case 2: There is one pair. Again start with 2 options for each vertex in 1-6, but now multiply by 6 since there are 6 possibilities for which pair can have the same color assigned instead of the opposite. Thus, the cases are: $2^6*6=384$
case 3: There are two pairs, but oppositely colored. Start with $2^6$ for assigning 1-6, then multiply by 6C2=15 for assigning which have repeated colors. Divide by 2 due to half the cases having the same colored opposites. $2^6*15/2=480$
It is apparent that no other cases exist, as more pairs would force there to be 2 pairs of same colored oppositely spaced vertices with the same color. Thus, the answer is: $64+384+480=\boxed{928}$
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928
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7
|
58
|
Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product\[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\]
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For any $k\in Z$, we have,
\begin{align*} & \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = \omega^{3 \cdot 7} + \omega^{2k + 7} + \omega^{3k} + \omega^{-2k + 3 \cdot 7} + \omega^7 + \omega^k + \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \\ & = 1 + \omega^{2k} + \omega^{3k} + \omega^{-2k} + 1 + \omega^k + \omega^{-3k} + \omega^{-k} + 1 \\ & = 2 + \omega^{-3k} \sum_{j=0}^6 \omega^{j k} \\ & = 2 + \omega^{-3k} \frac{1 - \omega^{7 k}}{1 - \omega^k} \\ & = 2 . \end{align*}
The second and the fifth equalities follow from the property that $\omega^7 = 1$.
Therefore,
\begin{align*} \Pi_{k=0}^6 \left( \omega^{3k} + \omega^k + 1 \right) & = \left( \omega^{3 \cdot 0} + \omega^0 + 1 \right) \Pi_{k=1}^3 \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = 3 \cdot 2^3 \\ & = \boxed{\textbf{024}}. \end{align*}
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024
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_8
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59
|
Circles $\omega_1$ and $\omega_2$ intersect at two points $P$ and $Q,$ and their common tangent line closer to $P$ intersects $\omega_1$ and $\omega_2$ at points $A$ and $B,$ respectively. The line parallel to $AB$ that passes through $P$ intersects $\omega_1$ and $\omega_2$ for the second time at points $X$ and $Y,$ respectively. Suppose $PX=10,$ $PY=14,$ and $PQ=5.$ Then the area of trapezoid $XABY$ is $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n.$
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Mathkiddie
Notice that line $\overline{PQ}$ is the [radical axis](https://artofproblemsolving.com/wiki/index.php/Radical_axis) of circles $\omega_1$ and $\omega_2$. By the radical axis theorem, we know that the tangents of any point on line $\overline{PQ}$ to circles $\omega_1$ and $\omega_2$ are equal. Therefore, line $\overline{PQ}$ must pass through the midpoint of $\overline{AB}$, call this point M. In addition, we know that $AM=MB=6$ by circle properties and midpoint definition.
Then, by Power of Point,
\begin{align*} AM^2&=MP*MQ \\ 36&=MP(MP+5) \\ MP&=4 \\ \end{align*}
Call the intersection point of line $\overline{A\omega_1}$ and $\overline{XY}$ be C, and the intersection point of line $\overline{B\omega_2}$ and $\overline{XY}$ be D. $ABCD$ is a rectangle with segment $MP=4$ drawn through it so that $AM=MB=6$, $CP=5$, and $PD=7$. Dropping the altitude from $M$ to $\overline{XY}$, we get that the height of trapezoid $XABY$ is $\sqrt{15}$. Therefore the area of trapezoid $XABY$ is
\begin{align*} \frac{1}{2}\cdot(12+24)\cdot(\sqrt{15})=18\sqrt{15} \end{align*}
Giving us an answer of $\boxed{033}$.
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033
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https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_9
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