Solution stringlengths 200 37.6k | Tag sequencelengths 0 8 | Problem stringlengths 10 19.5k |
|---|---|---|
The $ \cap$ symbol represents an intersection, or the items in common between two lists. The only numbers in common between the three sets given are the positive even numbers $ 20$ and below, which are $ 2$, $ 4$, $ 6$, $ 8$, $ 10$, $ 12$, $ 14$, $ 16$, $ 18$, and $ 20$. Adding them up, $ 2\plus{}4\plus{}6\plus{}8\plus... | [] | Given that $ A$ is the set of positive integers, $ B$ is the set of whole numbers less than or equal to $ 20$, and $ C$ is the set of even integers, what is the sum of the numbers in $ A \cap B \cap C$? |
[hide="Solution"]
A quick and easy way is trying with a sample quadratic equation like: $ x^2\minus{}3x\plus{}2\equal{}0$ with one root double the other.
Substituting $ a,b$ and $ c$ into the possible choices options $ \mathbf{(C),(D)}$ and $ \mathbf{(E)}$ don't satisfy the requirements.
Then using another sample equ... | [
"quadratics",
"Vieta",
"algebra",
"quadratic formula"
] | For one root of $ ax^2 \plus{} bx \plus{} c \equal{} 0$ to be double the other, the coefficients $ a,\,b,\,c$ must be related as follows:
$ \textbf{(A)}\ 4b^2 \equal{} 9c\qquad
\textbf{(B)}\ 2b^2 \equal{} 9ac\qquad
\textbf{(C)}\ 2b^2 \equal{} 9a\qquad \\
\textbf{(D)}\ b^2 \minus{} 8ac \equal{} 0\qquad
\textbf... |
[hide="Alternatively"]
Let the roots be $ r$ and $ s$, with $ r=2s$. By Vieta's, we have $ r+s=3s=-\frac{b}{a}$ and $ rs=2s^2=\frac{c}{a}$. So, $ s=-\frac{b}{3a}$. Substituting, $ 2(\frac{b^2}{9a^2})=\frac{c}{a}$, so $ 2b^2=9ac$. $ \boxed{\textbf{B}}$
[/hide] | [
"quadratics",
"Vieta",
"algebra",
"quadratic formula"
] | For one root of $ ax^2 \plus{} bx \plus{} c \equal{} 0$ to be double the other, the coefficients $ a,\,b,\,c$ must be related as follows:
$ \textbf{(A)}\ 4b^2 \equal{} 9c\qquad
\textbf{(B)}\ 2b^2 \equal{} 9ac\qquad
\textbf{(C)}\ 2b^2 \equal{} 9a\qquad \\
\textbf{(D)}\ b^2 \minus{} 8ac \equal{} 0\qquad
\textbf... |
Find the max value of
$ \sqrt [3]{4x \minus{} 3 \plus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}} \plus{} \sqrt [3]{4x \minus{} 3 \minus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}}$
when x € (-1,1).
$ \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}\equal{}\sqrt{(1\minus{}x)(4\min... | [
"inequalities proposed",
"inequalities"
] | Find the max value of
$ \sqrt[3]{4x\minus{}3\plus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}} \plus{} \sqrt[3]{4x\minus{}3\minus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}}$
when x € (-1,1). |
If I understood it correctly, you proved it is [b]always[/b] 2, right?
$ \sqrt [3]{4x \minus{} 3 \plus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}} \equal{} \sqrt [3]{4x \minus{} 3 \plus{} (4 \minus{} x)\sqrt {1 \minus{} x}} \equal{} \sqrt {1 \minus{} x} \plus{} 1$
This step was a nice observation :lo... | [
"inequalities proposed",
"inequalities"
] | Find the max value of
$ \sqrt[3]{4x\minus{}3\plus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}} \plus{} \sqrt[3]{4x\minus{}3\minus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}}$
when x € (-1,1). |
We set $ A \equal{} 4x \minus{} 3$ and $ B \equal{} 16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}$
Observing that $ \sqrt [3]{A \pm \sqrt {B}} \equal{} \sqrt [3]{(A \pm \sqrt {B})\cdot1\cdot1}$ then we obtain from AM-GM :
$ \sqrt [3]{A \plus{} \sqrt {B}} \plus{} \sqrt [3]{A \minus{} \sqrt {B}} \leq \frac {(A \plu... | [
"inequalities proposed",
"inequalities"
] | Find the max value of
$ \sqrt[3]{4x\minus{}3\plus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}} \plus{} \sqrt[3]{4x\minus{}3\minus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}}$
when x € (-1,1). |
Let $x,y,z$ be nennegative real numbers
The following is also nice \[(x+y+z)^{5}\geq 8(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x+xy^{2}+yz^{2}+zx^{2})?\]
I think it is very nice since,
$(x+y+z)^{5}-8(x^{2}+y^{2}+z^{2})(x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y)=$
$=\sum_{cyc}x(y+z-x)^{4}+32\sum_{cyc}x^{2}y^{2}z.$ :) | [
"inequalities",
"calculus",
"inequalities unsolved"
] | Let $x,y,z$ be nennegative real numbers, find the greatest possible positive value of $k$ such that \[(x+y+z)^{5}\geq k(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x).\] The following is also nice \[(x+y+z)^{5}\geq 8(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x+xy^{2}+yz^{2}+zx^{2})?\] |
Nice solutions, Arqady.
For the first one, if we put $y=1$ and $z=0$, it follows by calculus that $\frac{(x+y+z)^{5}}{ (x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x)}$ attains its minimum $k\approx 11.353$ for $x\approx 3.26953$, the only root of $x^{3}-4x^{2}+3x-2=0.$
It only remains to show (but that should not be to... | [
"inequalities",
"calculus",
"inequalities unsolved"
] | Let $x,y,z$ be nennegative real numbers, find the greatest possible positive value of $k$ such that \[(x+y+z)^{5}\geq k(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x).\] The following is also nice \[(x+y+z)^{5}\geq 8(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x+xy^{2}+yz^{2}+zx^{2})?\] |
Let $ a,b,c\geq 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \frac {1}{6 \minus{} ab} \plus{} \frac {1}{6 \minus{} bc} \plus{} \frac {1}{6 \minus{} ca}\leq \frac {3}{5}$.
$ <\equal{}>16(ab\plus{}bc\plus{}ca) \plus{}a^2b^2c^2 \le 36\plus{}13abc<\equal{}>$
$ a^2(b\plus{}c)\plus{}b^2(c\plus{}a) \plus{}c^2(... | [
"inequalities",
"inequalities unsolved"
] | Let $ a,b,c\geq 0$, s.t. $ a\plus{}b\plus{}c\equal{}3$.Prove that:
$ \frac{1}{6\minus{}ab}\plus{}\frac{1}{6\minus{}bc}\plus{}\frac{1}{6\minus{}ca}\leq \frac{3}{5}$. |
Does there exist a function f such that for all x the following equality holds $ f(f(x)) \equal{} \left\{\begin{array}{l} \sqrt {2009} ,x\in Q \\
2009,x\in\mathbb{R} \end{array}\right.$ :?:
I suppose you mean $ f(x)\equal{}2009$ $ \forall x\in\mathbb R\backslash \mathbb Q$ (and not $ \mathbb R$). If so, no such fu... | [
"function",
"algebra proposed",
"algebra"
] | Does there exist a function $f$ such that for all $x$ the following equality holds?
\[ f(f(x)) = \begin{cases}
\sqrt{2009}, & \mbox{if } x\in \mathbb Q,\\
2009, &\mbox{if } x\in\mathbb{R} \setminus \mathbb Q.
\end{cases}
\] |
if $ m,n\neq 1$
$ n + m|mn + 1$
$ \iff n + m|n^2 - 1$
$ \iff \exists d\in E = \{k\in\mathbb{N}: \ k|n^2 - 1,k > n\}: \ m = d - n$
$ \iff \exists d\in E = \{k\in\mathbb{N}: \ k|n^2 - 1\}: \ m = max\{d,\frac {n^2 - 1}{d}\} - n$
then :
$ {\{(n,m)\in\mathbb{N}^*^2: \ \frac {mn + 1}{m + n}\in\mathbb{Z}^ + \} = \{(n,ma... | [
"modular arithmetic",
"number theory unsolved",
"number theory"
] | It' s an exercise |
It' s an exercise
Let $ S\equal{}m\plus{}n$. We want $ mn\plus{}1\equal{}n(S\minus{}n)\plus{}1\equal{}0\pmod{S}$ and so $ n^2\equal{}1\pmod{S}$
So, the set of solutions is $ \{(n,d\minus{}n),(d\minus{}n,n)$ where $ d$ is any divisor $ >n$ of $ n^2\minus{}1\}$ | [
"modular arithmetic",
"number theory unsolved",
"number theory"
] | It' s an exercise |
Let's go...
Here is some explanations and a reference:
The question amounts wondering which is the distance between the polynomial function $x \mapsto x^4$ and vector space of the polynomial function of degree $3$ for the "norme" infinite on the interval $[-1,3]$ i.e $\|g\|= \sup_{[-1,3]} |g(x)|$ where $g$ is a ... | [
"algebra",
"polynomial",
"function",
"vector"
] | Let $ f(x)= x^4 + ax^3 + bx^2 + cx +d $
Find $ a,b,c,d $ such that : $ \max_{x\in [-1,3]}/f(x)/ $ get the minimum (/f(x)/ is the absoluted value of f(x)) |
[hide="Solution"]Observe that solutions to $ x^2 \plus{} 7x \plus{} 6 \equal{} x \iff x^2 \plus{} 6x \plus{} 6 \equal{} 0$ are also solutions to the given equation*.
We now wish to rewrite the equation in the form $ (x^2 \plus{} 6x \plus{} 6)(x^2 \plus{} ax \plus{} b) \equal{} 0$.
Equating the constant terms of the o... | [
"quadratics",
"algebra",
"quadratic formula"
] | Find all solutions of: $ (x^2 \plus{} 7x \plus{} 6)^2 \plus{} 7(x^2 \plus{} 7x \plus{} 6)\plus{} 6 \equal{} x$ |
Since $y=2\sin \theta\cos\theta$, it follows that $(x/2)^{2}+(y/x)^{2}=1$.
I think the curve looks like this: $\infty$
Very keen guess:
[url=http://allyoucanupload.webshots.com/v/2005945213650995389][img]http://aycu34.webshots.com/image/6753/2005945213650995389_rs.jpg[/img][/url] | [
"trigonometry",
"parameterization",
"calculus",
"derivative",
"calculus computations"
] | If $x = 2 \cos \theta$, $y = \sin 2\theta$, how would you eliminate the parameter $\theta$? I am trying to find the points on the curve where the tangent is horizontal and vertical. I figure that the cartesian equation is an ellipse. But I tried this:
$x = 2\cos \theta$, $y = 2\sin \theta \cos \theta$, but I cant us... |
Note that since $y = 2\sin \theta \cos \theta$ and $\frac{x}{2}= \cos \theta$, $y = (2\sin \theta )\frac{x}{2}= x \sin \theta$.
Therefore, $\sin \theta = \frac{y}{x}$.
As a result, $\sin^{2}\theta+\cos^{2}\theta = (\frac{y}{x})^{2}+(\frac{x}{2})^{2}= 1$.
That's where mlok's equation came from. | [
"trigonometry",
"parameterization",
"calculus",
"derivative",
"calculus computations"
] | If $x = 2 \cos \theta$, $y = \sin 2\theta$, how would you eliminate the parameter $\theta$? I am trying to find the points on the curve where the tangent is horizontal and vertical. I figure that the cartesian equation is an ellipse. But I tried this:
$x = 2\cos \theta$, $y = 2\sin \theta \cos \theta$, but I cant us... |
[hide="Solution"]
Let us assume the initial length and breadth to be $100l$ and $100b$ respectively.
Initially, the perimeter will be
$\\ P_{1} = 2(100l + 100b) \newline
\Rightarrow P_{1} = 200(l + b) ...........(1)$
Now since length and breadth is increased by 10%, we get the new perimeter to be
$\\ P_{2} = 2(11... | [
"geometry",
"rectangle",
"perimeter"
] | If the length and width of a rectangle are each increased by $ 10 \%$, then the perimeter of the rectangle is increased by
\[ \textbf{(A)}\ 1 \% \qquad
\textbf{(B)}\ 10 \% \qquad
\textbf{(C)}\ 20 \% \qquad
\textbf{(D)}\ 21 \% \qquad
\textbf{(E)}\ 40 \%
\] |
suppose $n$ is the number of students then there are $\frac n 2$ staplers ,$\frac n 3$rulers ,$\frac n 4$glue bottles so $\frac n 2 +\frac n 3 +\frac n 4 =65$ thus we have $\frac{13n}{12}=65$ so $n=60$.
I think you might find the middle school section on this site more suited to your needs. :) | [
"algebra unsolved",
"algebra"
] | In an art class, there were just enough staplers , rulers and glue bottles so that every 2 students had to share a stapler, every 3 students had to share a ruler, and every 4 students had to share a glue bottle. If the sum of the number of staplers, rulers, and glue bottles used by the class was 65, how many students ... |
Hello,
$ 12=2^{2}3$. Therefore we get as possible solutions:
$2^{2}3=12$
$2^{2}*3*5=60$
$2^{2}3^{2}=36$ and
$2^{2}3^{2}*5=180$.
Thus we have $4$ positive integer factors, which are multiplies of $12$.
Best wishes from Leipzig/Germany. | [] | How many positive integer factors of $ 2^2 \times 3^2 \times 5$ are multiples of 12? |
[hide="15"]Is that all you got wrong?
Let the roots be $ r$ and $ s$, then by Vietas we have $ n \equal{} 4rs$ and $ p \equal{} ( \minus{} r \minus{} s)$, so $ \frac {n}{p} \equal{} \frac {4rs}{ \minus{} r \minus{} s}$, but we also know again by Viets and the fact that m is in both equations that $ 2r \plus{} 2s \equal... | [
"algebra",
"polynomial",
"Vieta",
"AMC"
] | I need help with the question and please show ur solution
the AMC 10B questions 21(http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_21) and question 15 (http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_15) |
ok for 15 there are 5 pairs that make 20 dollars : $ (20,(1,5,10,20))$and $ (10,10)$. There are also 4 types of bills and you are choosing 2 of them, which can be done in (4C2) + 4 = 10 ways, so the answer is 1/2.
For 22, we simplify to $ \frac{2(n\minus{}1)!}{n\plus{}1}$ which is not reducible iff n + 1 is an odd pri... | [
"algebra",
"polynomial",
"Vieta",
"AMC"
] | I need help with the question and please show ur solution
the AMC 10B questions 21(http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_21) and question 15 (http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_15) |
For 22, we simplify to $ \frac {2(n \minus{} 1)!}{n \plus{} 1}$ which is not reducible iff n + 1 is an odd prime. There are 8 odd primes less than 24 so the answer is $ 24 \minus{} 8 \equal{} 16$
how did u know that odd primes would not work | [
"algebra",
"polynomial",
"Vieta",
"AMC"
] | I need help with the question and please show ur solution
the AMC 10B questions 21(http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_21) and question 15 (http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_15) |
For 22, we simplify to $ \frac {2(n \minus{} 1)!}{n \plus{} 1}$ which is not reducible iff n + 1 is an odd prime. There are 8 odd primes less than 24 so the answer is $ 24 \minus{} 8 \equal{} 16$
how did u know that odd primes would not work
If $ n \plus{} 1$ is prime, it will not be a divisor of $ (n \minus{} 1)!$... | [
"algebra",
"polynomial",
"Vieta",
"AMC"
] | I need help with the question and please show ur solution
the AMC 10B questions 21(http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_21) and question 15 (http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_15) |
the greater measure is 20 inches, so to be reduced to 5 inches, you divide by 4
the other measure is 16 inches, so we divide that by 4 also, to get 4
now we have our side lengths (4 and 5), so we find the perimeter
$ 4\plus{}4\plus{}5\plus{}5\equal{}18$
our answer is 18 | [
"geometry",
"perimeter",
"rectangle"
] | A photograph measuring 16 inches by 20 inches is reduced uniformly so that the greater measure becomes 5 inches. What is the number of inches in the perimeter of the reduce photo? |
A less random sounding solution:
By "uniformly" they probably mean proportionally, so you're dealing with two rectangles which are the same shape but scaled down.
Call the big rectangle R and the little rectangle r.
The rule of proportionality would tell you: $ \frac{long side of r}{long side of R}\equal{}\fra... | [
"geometry",
"perimeter",
"rectangle"
] | A photograph measuring 16 inches by 20 inches is reduced uniformly so that the greater measure becomes 5 inches. What is the number of inches in the perimeter of the reduce photo? |
if 3/8 must have a cat and we know there are 9 people who have a cat, 9+x+w+z=3/8(32), so 9+x+w+z=12 ->
x+w+z=3, but the 3 numbers must be integer positive, so the solution must be x=w=z=1.... very simply!!!!!
Not so fast, damianoT! ;)
There are 9 people who have [i]only[/i] a cat. Since 12 people have cats, this me... | [] | Jeremy made a Venn diagram showing the number of students in his class who own types of pets. There are 32 students in his class. In addition to the information in the Venn diagram, Jeremy
knows half of the students have a dog, $ \frac{3}{8}$ have a cat, six have some other pet and five have no pet at all. How many st... |
Let $\ f$ be a function such that $\ f(x)=\arctan\left(\frac{1}{x}\right)$.Compute $\int_{0}^{1}f(x)dx$
Let $J: =\int_{0}^{1}\arctan\left(\frac{1}{x}\right) dx$
integrate by parts: let $u=\arctan\left(\frac{1}{x}\right) , du=\frac{-\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}}dx =-\frac{dx}{1+x^{2}}, dv=dx, v=x$
$J=\le... | [
"function",
"integration",
"calculus",
"calculus computations"
] | Let $\ f$ be a function such that $\ f(x)=\arctan\left(\frac{1}{x}\right)$.Compute $\int_{0}^{1}f(x)dx$ |
This is a very useless problem, IMHO: no use of matrices at all, just take $ B\equal{}af(A)\plus{}f(a)I_n$ for suitable $ a$ (to make $ B$ invertible) of modulus $ 1$. Here, $ f$ means complex conjugate. The converse is obvious. Bad problem. | [
"linear algebra",
"matrix",
"linear algebra unsolved"
] | Let $ A$ be a $ n\times n$ matrix with complex elements. Prove that $ A^{\minus{}1} \equal{} \overline{A}$ if and only if there exists an invertible matrix $ B$ with complex elements such that $ A\equal{} B^{\minus{}1} \cdot \overline{B}$. |
Still you should see the results (very bad). And actually after thinking about it $ f(A)\equal{}\overline A$ I think and $ f(a)\equal{}\overline a$, and of course just add the condition for $ B$ invertible. Haven't worked out the last condition (B invertible) but I think this might work, because there are finite values... | [
"linear algebra",
"matrix",
"linear algebra unsolved"
] | Let $ A$ be a $ n\times n$ matrix with complex elements. Prove that $ A^{\minus{}1} \equal{} \overline{A}$ if and only if there exists an invertible matrix $ B$ with complex elements such that $ A\equal{} B^{\minus{}1} \cdot \overline{B}$. |
[hide="Solution"]Use $a$, $b$, and $c$. Clearly, $c=\sqrt{a^{2}+b^{2}}$.
Set up the equation $ab=3a+3b+3\sqrt{a^{2}+b^{2}}$. Solve for the radical and we get $3\sqrt{a^{2}+b^{2}}=ab-3a-3b$. Square both sides to get $a^{2}b^{2}-6a^{2}b-6ab^{2}+9a^{2}+18ab+9b^{2}=9a^{2}+9b^{2}$, or $a^{2}b^{2}-6a^{2}b-6ab^{2}+18ab=0$.... | [] | Source: Romania 1999
7.1 Determine the side lengths of a right trianlge if they are intgers and the product of the leg lengths is equal to three times the perimeter. |
[hide="my way"]
in other words we have:
$2\triangle = 6s$
$\triangle = 3s$
$rs = 3s$
$r = 3$.
Now draw the diagram, since $r = 3$, and tangent lines from the same point are congruent, we deduce that $a+b-6 = c = \sqrt{a^{2}+b^{2}}$. Squaring and using Simon's favorite trick we get $(a-6)(b-6) = 18$ and get the sides ... | [] | Source: Romania 1999
7.1 Determine the side lengths of a right trianlge if they are intgers and the product of the leg lengths is equal to three times the perimeter. |
hello, after squaring both sides of the equation you will get:
$ \sqrt {2(5 \minus{} 9\cos(x) \plus{} 4\cos^2(x))} \equal{} 6\cos(x) \minus{} 6 \plus{} 8\cos^2(x)$
with $ \cos(2x) \equal{} 1 \minus{} 2cos^2(x)$.
Squaring one more time and setting $ t \equal{} \cos(x)$ gives:
$ 32t^4 \plus{} 48t^3 \minus{} 34t^2 \mi... | [
"trigonometry",
"inequalities",
"triangle inequality",
"geometry",
"angle bisector",
"trig identities",
"Law of Cosines"
] | Solve the equation $ \sqrt{2\minus{}2\cos x}\plus{}\sqrt{5\minus{}4\cos x}\equal{}\sqrt{5\minus{}4\cos 2x}.$ |
Exactly
$ \sqrt {2(5 \minus{} 9\cos(x) \plus{} 4\cos^2(x))} \equal{} 3\cos(x) \plus{}1\minus{} 4\cos^2(x)$
with $ \cos(2x) \equal{} 1 \minus{} 2cos^2(x)$.
Squaring one more time and setting $ t \equal{} \cos(x)$ gives:
$ (t\minus{}1)(t\plus{}1)(4t\minus{}3)^2 \equal{} 0$. | [
"trigonometry",
"inequalities",
"triangle inequality",
"geometry",
"angle bisector",
"trig identities",
"Law of Cosines"
] | Solve the equation $ \sqrt{2\minus{}2\cos x}\plus{}\sqrt{5\minus{}4\cos x}\equal{}\sqrt{5\minus{}4\cos 2x}.$ |
$ t \equal{} \cos{x} \equal{} \minus{} 1$ yields $ 2 \plus{} 3 \equal{} 1$ in the original equation, and is therefore a degenerate solution.
[hide="Geometric Solution"]Let $ A,B,C,D$ be points in a plane such that $ AB \equal{} AC \equal{} 1$, $ AD \equal{} 2$, and $ AC$ bisects $ \angle BAD$.
Let $ x \equal{} \an... | [
"trigonometry",
"inequalities",
"triangle inequality",
"geometry",
"angle bisector",
"trig identities",
"Law of Cosines"
] | Solve the equation $ \sqrt{2\minus{}2\cos x}\plus{}\sqrt{5\minus{}4\cos x}\equal{}\sqrt{5\minus{}4\cos 2x}.$ |
Squaring both sides and let $ t\equal{}\cos x$ and squaring again gives $ (t\minus{}1)(t\plus{}1)(4t\minus{}3)^2\equal{}0 \left(\minus{}\frac{1}{4}\leq t\leq 1\right)\Longleftrightarrow t\equal{}1,\ \frac{3}{4}$. The answer is $ x\equal{}2m\pi\ ,\ \cos ^ {\minus{}1} \frac{3}{4}\plus{}2n\pi$ for $ m,\ n\in\mathbb{Z}$. | [
"trigonometry",
"inequalities",
"triangle inequality",
"geometry",
"angle bisector",
"trig identities",
"Law of Cosines"
] | Solve the equation $ \sqrt{2\minus{}2\cos x}\plus{}\sqrt{5\minus{}4\cos x}\equal{}\sqrt{5\minus{}4\cos 2x}.$ |
[hide="Answer"]Label the intersection of the diagonals $E$. $AE^2+BE^2=11$, $BE^2+CE^2=BC^2$, $AE^2+DE^2=1001$, so we have $11-BE^2=1001-DE^2\Rightarrow BE^2=DE^2-990$. Substituting, $CE^2+DE^2-990=BC^2=CD^2-990$. Drawing the perpendicular $AF$ to $DC$, we have $CF=\sqrt{11}$. $AF^2=BC^2=1001-(CD-\sqrt{11})^2=990-CD^2+... | [
"geometry",
"trapezoid",
"ratio",
"quadratics",
"trigonometry",
"algebra",
"quadratic formula"
] | In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$ |
[hide]Let the perpendicular from $A$ to $CD$ intersect $BD$ at $F$. Let the two diagonals intersect at $E$, and let $\angle{EAF}=\theta$. Then \[ \angle{EAF} \cong \angle{EBA} \cong \angle{ECB} \cong \angle{EDC} = \theta \] Thus, with all four of these triangles having an angle of theta and a right angle, we have fo... | [
"geometry",
"trapezoid",
"ratio",
"quadratics",
"trigonometry",
"algebra",
"quadratic formula"
] | In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$ |
[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=699&sid=0410aaa1bd30b9a182e2c716eee02a44[/img]
Let $BC=x$. This means that $AC=\sqrt{11+x^2}$. Now notice that $\triangle ABC \sim \triangle BCD$. Therefore, $\frac{BD}{BC}=\frac{AC}{AB} \implies BD=\frac{x\sqrt{x^2+11}}{\sqrt{11}}$. Another valu... | [
"geometry",
"trapezoid",
"ratio",
"quadratics",
"trigonometry",
"algebra",
"quadratic formula"
] | In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$ |
[hide="Answer"]Label the intersection of the diagonals $ E$. $ AE^2 \plus{} BE^2 \equal{} 11$, $ BE^2 \plus{} CE^2 \equal{} BC^2$, $ AE^2 \plus{} DE^2 \equal{} 1001$, so we have $ 11 \minus{} BE^2 \equal{} 1001 \minus{} DE^2\Rightarrow BE^2 \equal{} DE^2 \minus{} 990$. Substituting, $ CE^2 \plus{} DE^2 \minus{} 990 \eq... | [
"geometry",
"trapezoid",
"ratio",
"quadratics",
"trigonometry",
"algebra",
"quadratic formula"
] | In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$ |
[hide]Let $ E$ be the foot of the perpendicular from $ A$ to $ CD$.
Note that since $ AC \perp BD$, $ \triangle ABC \sim \triangle BCD \implies \frac {AB}{BC} \equal{} \frac {BC}{CD} \implies BC^2 \equal{} CD \cdot \sqrt {11}$. Also, by the Pythagorean theorem, $ AE^2 \plus{} ED^2 \equal{} AD^2$, but $ AE \equal{} BC$... | [
"geometry",
"trapezoid",
"ratio",
"quadratics",
"trigonometry",
"algebra",
"quadratic formula"
] | In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$ |
[hide=quick]
Set the trapezoid on the coordinate plane with $C$ as the origin and $CD$ lying on the $x$ axis. Furthemore denote $CD=e$ and $BC=a$
Then looking at slopes we have $(\frac{a}{e})(\frac{a}{\sqrt{11}})=1$. Thus $a^2=e\sqrt{11}$. Now doing distance formula on
$AD$ gives $a^2+(e-\sqrt{11})^2=1001$. Now just su... | [
"geometry",
"trapezoid",
"ratio",
"quadratics",
"trigonometry",
"algebra",
"quadratic formula"
] | In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$ |
[hide=quick from this lemma] lemma: if $AB \perp BC$ and $CD\perp BC$, such that $AB=h_{1} , DC=h_{2}$ and $BC=d$ then $d^2=h_{1}h_{2}$
proof of lemma is easy from similarity, then we have from this lemma $x^2=y\sqrt{11}$ and from orthogonal diagonal theorem we have $x^2+1001=y^2+11$, solving for $x^2$ we get $x^2=\b... | [
"geometry",
"trapezoid",
"ratio",
"quadratics",
"trigonometry",
"algebra",
"quadratic formula"
] | In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$ |
[hide=quick from this lemma] lemma: if $AB \perp BC$ and $CD\perp BC$, such that $AB=h_{1} , DC=h_{2}$ and $BC=d$ then $d^2=h_{1}h_{2}$
proof of lemma is easy from similarity, then we have from this lemma $x^2=y\sqrt{11}$ and from orthogonal diagonal theorem we have $x^2+1001=y^2+11$, solving for $x^2$ we get $x^2=\b... | [
"geometry",
"trapezoid",
"ratio",
"quadratics",
"trigonometry",
"algebra",
"quadratic formula"
] | In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$ |
This is probably stupid, but whenever I see a problem like this I'm strongly tempted to approximate it by hand.
Intuition tells us that $e^\pi$ is probably bigger, so we need to do a low estimate. Call $e = 27/10$ and $\pi = 22/7$. That should still be a small estimate. So we have $3^{66/7} / 10^{22/7}$. That's abo... | [
"inequalities",
"function",
"logarithms",
"real analysis",
"real analysis theorems"
] | Which is bigger, $e^\pi$ or $\pi^e$?
[hide="solution"]
Let's consider the maximum of $x^{1/x}$. First we take a natural logarithm. This gives
$\frac{\ln(x)}{x}$. The maximum of this function is the maximum of the first function since $\ln(x)$ is increasing. The maximum is easily found to be at $x=e$. So now we have:
... |
Define : $x\ s.s\ y\Longleftrightarrow sgn(x)=sgn(y)$, i.e. $x=y=0$ or $xy>0\ .$
Here are some examples :
$1.\blacktriangleright\frac ab\ s.s\ ab$ for $b\ne 0\ .$
$2.\blacktriangleright 0<a\ne 1\ ,\ \{x,y\}\subset R\Longrightarrow (a^x-a^y)\ s.s\ (a-1)(x-y)\ .$
$3.\blacktriangleright 0<a,b\ne 1\ ,\ x\in R\Longright... | [
"inequalities",
"function",
"logarithms",
"real analysis",
"real analysis theorems"
] | Which is bigger, $e^\pi$ or $\pi^e$?
[hide="solution"]
Let's consider the maximum of $x^{1/x}$. First we take a natural logarithm. This gives
$\frac{\ln(x)}{x}$. The maximum of this function is the maximum of the first function since $\ln(x)$ is increasing. The maximum is easily found to be at $x=e$. So now we have:
... |
Your integral can be calculated exactly.
Let consider the case when $n=2p$. Then I obtained, assuming that I made no computational mistakes, that:
$\lim\int_{0}^{\infty}\frac{(\sin(x))^{2p}}{(x)^{2p}}dx=\lim_{p\rightarrow \infty}\int_{0}^{1}\frac{2p}{(1+4y^{2})(1+42^{2}y^{2})\cdots (1+4p^{2}y^{2})}dy$.
Why th... | [
"calculus",
"integration",
"trigonometry",
"limit",
"function",
"real analysis",
"real analysis unsolved"
] | Find $\int_{0}^{\infty}\frac{(\sin{x})^{n}}{x^{n}}dx,n\in \mathbb{N}$ (this integral is motivated by another problem mentioned here ).
Also evaluate; $a_{n}: =\int_{0}^{\infty}\frac{(\sin(x) )^{n}}{x^{n}}dx,$
$\lim_{n\rightarrow \infty}a_{n}=?$ |
Why this limit is 0?. Any ideas?
For $n\ge 2,\ \left|\frac{\sin^{n}x}{x^{n}}\right|\le \frac{\sin^{2}x}{x^{2}},$ and the latter is an absolutely integrable function on $[0,\infty).$
For $x\ne0,\lim_{n\to\infty}\frac{\sin^{n}x}{x^{n}}=0,$ pointwise.
Hence by the Lebesgue Dominated Convergence Theorem, we can inte... | [
"calculus",
"integration",
"trigonometry",
"limit",
"function",
"real analysis",
"real analysis unsolved"
] | Find $\int_{0}^{\infty}\frac{(\sin{x})^{n}}{x^{n}}dx,n\in \mathbb{N}$ (this integral is motivated by another problem mentioned here ).
Also evaluate; $a_{n}: =\int_{0}^{\infty}\frac{(\sin(x) )^{n}}{x^{n}}dx,$
$\lim_{n\rightarrow \infty}a_{n}=?$ |
wlog assume $ x\geq y$ than we have $ x^2 < x^2 \plus{} 3y < x^2 \plus{} 4x \plus{} 4 \equal{} (x \plus{} 2)^2$ so $ x^2 \plus{} 3y \equal{} (x \plus{} 1)^2$ and further $ 3y \equal{} 2x \plus{} 1$
by solving this linear diophan equation we get $ x \equal{} 1 \plus{} 2k$ and $ y \equal{} 1 \plus{} 3k$ for a non-negati... | [
"number theory unsolved",
"number theory"
] | Find all pairs of positive integers $ (x,y)$ such that both $ x^2 \plus{} 3y$ and $ y^2 \plus{} 3x$ are perfect squares. |
[hide]
To have a right triangle in a circle, you have to have one of the sides a diameter. If $X$ is odd, then this can't happen... If $X$ is even ($X\ge4$), then there are $\frac{X(X-2)}{2}$ ways to choose a right triangle (there are $\frac{X}{2}$ ways to choose a diameter, and $X-2$ ways to choose another point of ... | [
"geometry",
"probability"
] | if X points are equally spaced on a circle, find the probability that 3 points chosen at random will form a right triangle.[/code] |
If $ R$ is a finite Boolean ring with identity $ 1 \neq 0$ then $ R \cong \mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}$. To see this, you can use the Chinese Remainder Theorem and notice that for $ e \in R$ with $ e^2 \equal{} e$ one has $ R \cong Re \times R(1 \minus{} e)$ where $ e$ and $ 1 \minus{... | [
"Ring Theory",
"superior algebra",
"superior algebra unsolved"
] | A ring is called boolean if $ x^2\equal{}x$ for all $ x$ in $ A$.
Prove that one can define a structure of boolean ring of order $ n$ if and only if $ n\equal{}2^k$ for some $ k \in N$. |
what you want to prove is $\displaystyle\alpha(G)\le n-\frac{\sum_i d_i}{n}$
where $\alpha(G)$ is the independence number of $G$ and $d_i$'s are the degrees of the vertices of $G$.(the rest follows from the fact that $\chi(G)\ge \frac{n}{\alpha(G)}$)
or more simply what you need is that $\frac{\sum_i d_i}{n}\le n... | [
"combinatorics unsolved",
"combinatorics"
] | prove that in graph with $v$ virtex and $\epsilon$ edge :
$\frac {v^2}{v^2-2\epsilon} \leq \chi$ |
$ F(x) \equal{} 0$ has local maximum $ \Longrightarrow F'(x) \equal{} 0$ has 3 distinct real roots $ \alpha ,\ \beta ,\ \gamma\ (\alpha <\beta <\gamma)$.
Under this condition, We can rewrite the given inequality as follows.
$ \frac{3}{10}\cdot\frac{16}{81}(\alpha^{2}\plus{}\beta^{2}\plus{}\gamma^{2}\plus{}\alpha\be... | [
"algebra",
"polynomial",
"inequalities",
"inequalities proposed"
] | $ F(x)$ is polynomial with real coefficients. $ F(x) \equal{} x^{4}\plus{}a_{1}x^{3}\plus{}a_{2}x^{2}\plus{}a_{1}x^{1}\plus{}a_{0}$. $ M$ is local maximum and $ m$ is minimum. Prove that $ \frac{3}{10}(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^{2}< M\minus{}m < 3(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^... |
$ F(x)$ is polynomial with real coefficients. $ F(x) \equal{} x^{4} \plus{} a_{1}x^{3} \plus{} a_{2}x^{2} \plus{} a_{1}x^{1} \plus{} a_{0}$. $ M$ is local maximum and $ m$ is minimum. Prove that $ \frac {3}{10}(\frac {a_{1}^{2}}{4} \minus{} \frac {2a_{2}}{3^{2}})^{2} < M \minus{} m < 3(\frac {a_{1}^{2}}{4} \minus{} \fr... | [
"algebra",
"polynomial",
"inequalities",
"inequalities proposed"
] | $ F(x)$ is polynomial with real coefficients. $ F(x) \equal{} x^{4}\plus{}a_{1}x^{3}\plus{}a_{2}x^{2}\plus{}a_{1}x^{1}\plus{}a_{0}$. $ M$ is local maximum and $ m$ is minimum. Prove that $ \frac{3}{10}(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^{2}< M\minus{}m < 3(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^... |
Decomposing into columns isn't a very informative way of looking at row operations.
Row operations correspond to multiplication on the left by elementary matrices. If you show those latter three statements, you will have proved this in the case that $ A$ is an elementary matrix of one of those three types. The next ... | [
"linear algebra",
"matrix",
"algorithm",
"linear algebra unsolved"
] | det(AB)=det(A)det(B)
Maybe there is intuitive proving of this using the following
- $ AB \equal{} \left[Ab_1\cdots Ab_n\right]$
following is actually just def of determinant
-linear property
$ \begin{vmatrix}
{r_1}\\ {.}\\ {.}\\ {.}
\end{vmatrix}
\equal{}
\begin{vmatrix}
{k \cdot r'_1 \plus{} r''_1}\\ {.}... |
Umm... I think you can't construct a triangle with sides $ 2$, $ 1$, $ 1$. ($ b \plus{} c > a$ isn't true).
Oh, no. In my definition a triangle is definited by the condition $ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b) \ge 0$. ;) | [
"inequalities",
"function",
"calculus",
"derivative",
"inequalities proposed"
] | Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :) |
I guess it depends on what you call a triangle - it the case of triangle sides being $ 2$, $ 1$, $ 1$, you would only have a line, but I understand this is a degenerative case and as such, may be considered a triangle.
Anyway, even if you consider $ (2,1,1)$ a triangle, the statement is still not true - take any tripl... | [
"inequalities",
"function",
"calculus",
"derivative",
"inequalities proposed"
] | Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :) |
Using the substitutions $ a\equal{}y\plus{}z$, $ b\equal{}z\plus{}x$, $ c\equal{}x\plus{}y$ and $ x\plus{}y\plus{}z\equal{}p$, $ x^2\plus{}y^2\plus{}z^2\equal{}1$, the inequality becomes
$ \frac 1{1\plus{}2px\minus{}x^2}\plus{}\frac 1{1\plus{}2py\minus{}y^2}\plus{}\frac 1{1\plus{}2pz\minus{}z^2} \le \frac 9{3p^2\minus... | [
"inequalities",
"function",
"calculus",
"derivative",
"inequalities proposed"
] | Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :) |
Solution is here:
http://www.toanthpt.net/forums/showthread.php?p=66534#post66534
I think it' s not WRONG :D
I have studied in detail this solution and I think it is wrong. Now I will explain my reasoning.
WLOG we can assume $ a\plus{}b\plus{}c\equal{}3$ We put $ r\equal{}abc$ and $ q\equal{}ab\plus{}bc\plus{}ca$... | [
"inequalities",
"function",
"calculus",
"derivative",
"inequalities proposed"
] | Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :) |
Solution is here:
http://www.toanthpt.net/forums/showthread.php?p=66534#post66534
I think it' s not WRONG :D
I have studied in detail this solution and I think it is wrong. Now I will explain my reasoning.
WLOG we can assume $ a \plus{} b \plus{} c \equal{} 3$ We put $ r \equal{} abc$ and $ q \equal{} ab \plus{} bc \... | [
"inequalities",
"function",
"calculus",
"derivative",
"inequalities proposed"
] | Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :) |
Very nice inequality and very nive proof, can_hang2007! :lol:
Even, we can prove that $ S_b \plus{} S_c\geq0.$ :wink:
$ S_b \plus{} S_c \equal{} \frac {1}{a^2 \plus{} c^2} \plus{} \frac {2(b^2 \minus{} ac)}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \plus{} \frac {1}{a^2 \plus{} b^2} \plus{} \frac {2(c^2 \minus{} ab)}{(a^... | [
"inequalities",
"function",
"calculus",
"derivative",
"inequalities proposed"
] | Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :) |
$ \frac {3c^2 \minus{} 2bc \minus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2bc \minus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$
$ \equal{} (b \minus{} c)\left(\frac {3b \minus{} c}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \minus{} \frac {3c \minus{} b}{(a^2 \plus{} c^2)(b^2 \plus{... | [
"inequalities",
"function",
"calculus",
"derivative",
"inequalities proposed"
] | Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :) |
$ \frac {3c^2 \minus{} 2bc \minus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2bc \minus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$
$ \equal{} (b \minus{} c)\left(\frac {3b \minus{} c}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \minus{} \frac {3c \minus{} b}{(a^2 \plus{} c^2)(b^2 \plus{... | [
"inequalities",
"function",
"calculus",
"derivative",
"inequalities proposed"
] | Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :) |
Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)
onl... | [
"inequalities",
"function",
"calculus",
"derivative",
"inequalities proposed"
] | Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :) |
In a different topic, someone asked why the following is true:
A_1 is disjoint from at least 50 subsets, so it has at most $\frac{n}{51}$ elements.
In fact, the reason is simple: Since the vertex $A_{1}$ of the graph $G$ has degree at most 50, there are at most 50 subsets $A_{i}$ (with $i\neq 1$) which are not disjo... | [
"pigeonhole principle",
"linear algebra",
"matrix",
"inequalities",
"combinatorics solved",
"combinatorics"
] | Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements.
Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element. |
My idea is slightly different. Define the graph $ G$ like Harazi's solution. Suppose that $ G$ is triangle-free.
I will prove that all vertices in $ G$ are of degree at most $ 50$. If this is false, then there is a vertex $ v$ adjacent to at least $ 51$ other vertices: $ v_{1},v_{2},\dots,v_{51}$. Consider the s... | [
"pigeonhole principle",
"linear algebra",
"matrix",
"inequalities",
"combinatorics solved",
"combinatorics"
] | Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements.
Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element. |
The bounds are incredibly weak. This question was used at a UK selection camp, and I produced the following solution, which disregards $ A_{101}$ entirely.
Let $ S_1 \equal{} A_1\cup A_2 \cup ... \cup A_{25}$ and $ S_2,S_3,S_4$ three more unions of 25 $ A_i$ such that all 100 are covered.
Let $ T_{i,j} \equal{} X \... | [
"pigeonhole principle",
"linear algebra",
"matrix",
"inequalities",
"combinatorics solved",
"combinatorics"
] | Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements.
Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element. |
Hi all. Sorry for digging this very old post back up, but I think there's one method that hasn't been covered by the above discussion, which is pretty simple (I think) but powerful enough to give the sharp bound mentioned by Ilthigore 2 posts up (3 years ago?!). I'm guessing that this is probably the method that Freddi... | [
"pigeonhole principle",
"linear algebra",
"matrix",
"inequalities",
"combinatorics solved",
"combinatorics"
] | Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements.
Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element. |
@hello123: Please be more precise. I think you are confusing sets and elements, but I am not sure because your posting is too brief.
@angyansheng: A little correction:
On the other hand, for each choice of 50 $A_i$, their union contains at least
more than
$\frac{50}{51}n$ elements of $X$. Hence if we consider the ... | [
"pigeonhole principle",
"linear algebra",
"matrix",
"inequalities",
"combinatorics solved",
"combinatorics"
] | Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements.
Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element. |
Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements.
Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element.
[b][co... | [
"pigeonhole principle",
"linear algebra",
"matrix",
"inequalities",
"combinatorics solved",
"combinatorics"
] | Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements.
Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element. |
It is known that $X$ is a set with $102$ elements. Suppose $A_1, A_2, ..., A_{101}$ is a set of subset of $X$ such that the sum of every $50$ of them has more than $100$ elements. Prove that there are $1 \le i <j <k \le 101$ such that $A_i \cap A_j$, $A_i \cap A_k $and $A_k \cap A_j$ are not empty.
[url=https://artofpr... | [
"pigeonhole principle",
"linear algebra",
"matrix",
"inequalities",
"combinatorics solved",
"combinatorics"
] | Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements.
Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element. |
Let $ Z$ be the total number of points and set
$ Z \equal{} X_1 \plus{} ... \plus{} X_5$
where $ X_i$ is the number of points shown on the ith die.
The probability generating function of each $ X_i$ is
$ G_{X_i}(s) \equal{} \frac {s \plus{} ... \plus{} s^6}{6}$.
Since the $ X's$ are independent:
$ G_Z(t) \equal{}... | [
"function",
"probability",
"algebra",
"binomial theorem",
"probability and stats"
] | I'm not really sure how to go about this, so any help would be appreciated!
Suppose you throw a six-sided die five times. Find the probability that the sum of the outcomes of the throws is 15 using generating functions. |
$ \frac {s^5(1 \minus{} s^6)^5}{6^5(1 \minus{} s)^5}\equal{}\frac{1}{6^5}s^5\left({5 \choose 0}\minus{}{5 \choose 1}s^6\plus{}...\right)\left({\minus{}5 \choose 0}\minus{}{\minus{}5 \choose 1}s\plus{}...\right)$.
The coefficient of $ s^{15}$ is
$ \frac{1}{6^5}\left({5 \choose 0}{\minus{}5 \choose 10}\minus{}{5 \choos... | [
"function",
"probability",
"algebra",
"binomial theorem",
"probability and stats"
] | I'm not really sure how to go about this, so any help would be appreciated!
Suppose you throw a six-sided die five times. Find the probability that the sum of the outcomes of the throws is 15 using generating functions. |
[hide="See if you know the answer."]$ x^3 \minus{} x \equal{} x(x^2 \minus{} 1)$
$ \Rightarrow x(x \minus{} 1)(x \plus{} 1)$
As it is three consecutive number so it is divisible by 3! and hence divisible by 3.[/hide] | [
"modular arithmetic",
"factorial"
] | If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before.
If $ x$ is an integer, prove that $ x^3\minus{}x$ is divisible by 3. |
brute force it by observing it in mod 3.
This is essentially what you do when you claim that three consecutive numbers must contain a multiple of $ 3$; I would hardly consider it brute forcing. :wink:
Another problem very similar to this that is solved along the same lines is:
(Zeitz) Show that $ n(n \plus{}... | [
"modular arithmetic",
"factorial"
] | If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before.
If $ x$ is an integer, prove that $ x^3\minus{}x$ is divisible by 3. |
If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before.
If $ x$ is an integer, prove that $ x^3 \minus{} x$ is divisible by 3.
By fermat, we have $ x^3\equiv x\pmod{3}\Longleftrightarrow 3|(x^3\minus{}x)$ | [
"modular arithmetic",
"factorial"
] | If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before.
If $ x$ is an integer, prove that $ x^3\minus{}x$ is divisible by 3. |
Another problem very similar to this that is solved along the same lines is:
(Zeitz) Show that $ n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)$ is never a perfect square for all natural $ n$.
Could someone help me out on this one? :oops: The only way I know is to expand $ n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)\plus{... | [
"modular arithmetic",
"factorial"
] | If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before.
If $ x$ is an integer, prove that $ x^3\minus{}x$ is divisible by 3. |
I don't think that's the right solution
This is what the book says but I don't understand.
[hide="Solution"]
$ x^{100} \minus{} 2x^{51} \plus{}1\equal{} (x^2 \minus{} 1)q(x) \plus{} ax \plus{} b$. Putting $ x \equal{} 1$ into this relation gives $ b \equal{} 0$. Putting $ x \equal{} \minus{} 1$ gives $ a \equal{} ... | [
"modular arithmetic"
] | Find the remainder on dividing $ x^{100} \minus{} 2x^{51} \plus{} 1$ by $ x^2 \minus{} 1$ |
I don't think that's the right solution
This is what the book says but I don't understand.
[hide="Solution"]
$ x^{100} \minus{} 2x^{51} \plus{} 1 \equal{} (x^2 \minus{} 1)q(x) \plus{} ax \plus{} b$. Putting $ x \equal{} 1$ into this relation gives $ b \equal{} 0$. Putting $ x \equal{} \minus{} 1$ gives $ a \equal{} \... | [
"modular arithmetic"
] | Find the remainder on dividing $ x^{100} \minus{} 2x^{51} \plus{} 1$ by $ x^2 \minus{} 1$ |
$ f(x)\equal{}x^{100}\minus{}2x^{51}\plus{}1\equal{}(x^2\minus{}1)Q(x)\plus{}l(x)$, $ l(x)\equal{}ax\plus{}b$
$ \Longleftrightarrow f(x)\minus{}l(x)\equal{}(x\plus{}1)(x\minus{}1)Q(x)$, which means that $ y\equal{}f(x)$ has intersection points with $ y\equal{}l(x)$ at $ x\equal{}\pm 1$, or
$ y\equal{}l(x)$ passes t... | [
"modular arithmetic"
] | Find the remainder on dividing $ x^{100} \minus{} 2x^{51} \plus{} 1$ by $ x^2 \minus{} 1$ |
We can now find the relation between u and v
$\sin\theta \frac{u'(r\cos\theta)}{u(r\cos\theta)} = \cos\theta \frac{v'(r\sin\theta)}{v(r\sin\theta)}$
where $(r, \theta)$ is the radial coord.
it seems to me that we r in need of another differential equation to completely specify the forms of u and v. | [
"function",
"trigonometry",
"real analysis",
"real analysis unsolved"
] | Hi all,
How can we prove that the unique 2D function which is separable and circularly symmetric is a 2D Gaussian, i.e.
$G(x, y) = a e^{\frac{x^2+y^2}{b}}$.
Thanks in advance. |
Suppose $|f'| \leq M$.
Given $\epsilon > 0$. Since $\{a_n\}$ is Cauchy, then there exists $N$ such that for all $n,m>N$, $|a_n-a_m| < \frac\epsilon M$. Then $|f(a_n)-f(a_m)| = |f'(\xi)|\cdot|a_n-a_m| < M \cdot \frac\epsilon M = \epsilon$. So $f\{(a_n)\}$ is Cauchy too.
If $f'$ is unbounded, take $f(x)=\frac 1x$ fo... | [
"function",
"algebra",
"domain",
"calculus",
"derivative",
"real analysis",
"real analysis solved"
] | Let $f(x)$ be the real- valued function which is differentiable on the opened-interval $(0,1).$ If $f'(x)$ is bounded, then prove that for the Cauchy sequence $\{a_n\}_{n=1}^{\infty}$ in the opened-interval $(0,1),\ f(a_n)$ will become Cauchy sequence as well.How about in case that $f$ is simply differentiable? |
The integral:
$\int_{1/2}^{1}\int_{0}^{\sqrt{1-x^2}}dydx$
(LaTeX isn't hard, trust me. Mostly you just put computations between 2 $\$$)
The line $x=\frac{1}{2}$ is $r cos\theta=\frac{1}{2}$
Hope that helps! :D | [
"calculus",
"integration",
"LaTeX",
"function",
"calculus computations"
] | Hi can someone help me with the double integration:
Sorry I dont know how to use the LATEX code.
it is a double integration
The first integration goes [b]from 1/2 to1[/b]. the second integration goes from 0[b] to sqrt(1-x^2)[/b]. there is no function after ,just[b] dy dx[/b].
I am asked to calculate the integral us... |
The integral:
$\int_{1/2}^{1}\int_{0}^{\sqrt{1-x^2}}dydx$
(LaTeX isn't hard, trust me. Mostly you just put computations between 2 $\$$)
The line $x=\frac{1}{2}$ is $r cos\theta=\frac{1}{2}$
Hope that helps! :D
All righty!
It makes sense now
B | [
"calculus",
"integration",
"LaTeX",
"function",
"calculus computations"
] | Hi can someone help me with the double integration:
Sorry I dont know how to use the LATEX code.
it is a double integration
The first integration goes [b]from 1/2 to1[/b]. the second integration goes from 0[b] to sqrt(1-x^2)[/b]. there is no function after ,just[b] dy dx[/b].
I am asked to calculate the integral us... |
$a_{n+1}=\frac{1}{3}a_n+\frac{1}{3}\Longleftrightarrow a_{n+1}-\frac{1}{2}=\frac{1}{3}\left(a_n-\frac{1}{2}\right)$, thus $a_n=\left(\frac{1}{3}\right)^{n-1}\left(x-\frac{1}{2}\right)+\frac{1}{2}$.
From $a_n-\frac{1}{6}=\left(\frac{1}{3}\right)^{n-1}\left(x-\frac{1}{2}\right)+\frac{1}{3},\ a_n+\frac{1}{6}=\left(\fra... | [
"algebra unsolved",
"algebra"
] | The sequence $(a_n)$ is given by $a_1=x\in R$ and $3a_{n+1}=a_n+1$ for $n\geq 1$. Set
$A=\sum_{n=1}^\infty [a_n-\frac 16]$ and $B=\sum_{n=1}^\infty [a_n+\frac 16]$. (where $[x]$ denote the greatest integer less that or equal to $x$).Compute the sum $A+B$ in terms of $x$. :lol: |
I think you can actually determine when the first player wins and when the second does, but for a quick proof of what you want here, assume $B$ (the second player) only wins when $n=n_1,\ldots,n_k$. Now, by the Chinese Remainder Theorem, you can choose some large $n$ (larger than all $n_i$) such that $n-n_i\equiv p_i\p... | [
"modular arithmetic",
"combinatorics proposed",
"combinatorics"
] | two player play a game on the board .first there exists number $n$ in positive integer on board .after that first player subtract $x^2$ such that x^2<n .(x is positive integer)then second player keep on this rule .each person who reach 0 win. prove there exist infinitive many $n$ such that
second player has strate... |
Is the following considered a valid proof?
$\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by:
$x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$
Prove that $\lim x_{n}= \sqrt{\alpha}$.
.
1)First show that this sequence convergences. Once you show that then $\lim x_{n}= \lim x_{n+1}$
... | [
"limit",
"induction",
"inequalities",
"topology",
"real analysis",
"real analysis unsolved"
] | Is the following considered a valid proof?
$\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by:
$x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$
Prove that $\lim x_{n}= \sqrt{\alpha}$.
By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\... |
Have you shown monotonicity? I can see how you might argue that, but it's not obvious - and if it's not obvious, it's incumbent on you to explain how it follows. A proof has an audience - you must reach out to your reader and explain yourself.
Yes, I understand this. But I'm certain that my argument for monotonici... | [
"limit",
"induction",
"inequalities",
"topology",
"real analysis",
"real analysis unsolved"
] | Is the following considered a valid proof?
$\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by:
$x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$
Prove that $\lim x_{n}= \sqrt{\alpha}$.
By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\... |
Essentially, you've said that because $y_{n}>b,$ and $z_{n}<b,$ and $y_{n+1}$ is the mean of $y_{n}$ and $z_{n},$ then $y_{n}$ tends to $b$. That's not convincing. In particular, you haven't given a good reason for the sequence to converge at all.
Actually, I would give a few points for that (say, 3 out of 10). The ... | [
"limit",
"induction",
"inequalities",
"topology",
"real analysis",
"real analysis unsolved"
] | Is the following considered a valid proof?
$\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by:
$x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$
Prove that $\lim x_{n}= \sqrt{\alpha}$.
By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\... |
Same question, different proof. I'm suspicious of my proof because I don't use the fact that the sets are bounded. But it says that the diameter of the metric spaces approach 0, which I do use, and is this in essence the same as bounded?
If $\{E_{n}\}$ is a sequence of closed and bounded sets in a complete metric ... | [
"limit",
"induction",
"inequalities",
"topology",
"real analysis",
"real analysis unsolved"
] | Is the following considered a valid proof?
$\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by:
$x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$
Prove that $\lim x_{n}= \sqrt{\alpha}$.
By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\... |
Is the following considered a valid proof?
$\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by:
$x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$
Prove that $\lim x_{n}= \sqrt{\alpha}$.
By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\alpha}$,... | [
"limit",
"induction",
"inequalities",
"topology",
"real analysis",
"real analysis unsolved"
] | Is the following considered a valid proof?
$\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by:
$x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$
Prove that $\lim x_{n}= \sqrt{\alpha}$.
By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\... |
we suppose that $ h\equal{}a\minus{}a'>0,g\equal{}b\minus{}b'$; $ f(x\plus{}h)\minus{}f(x)\equal{}f(x\plus{}h)\plus{}f(a\minus{}x\minus{}h)\minus{}(f(x)\plus{}f(a'\minus{}x))\equal{}b\minus{}b'\equal{}g$
now for $ r\in[0,h[: \ \forall x\in S_r\equal{}\{nh\plus{}r: \ n\in \mathbb{N}\}: \ f(x)\equal{}kx\plus{}m(r)$ wher... | [
"function",
"inequalities",
"algebra proposed",
"algebra"
] | Let $ a,b,a',b'$ be real numbers such that $ a\neq a'$ and $ f: \mathbb R\to\mathbb R$ be a function satisfying
\[ f(x)\plus{}f(a\minus{}x)\equal{}b\qquad\text{and}\qquad f(x)\plus{}f(a'\minus{}x)\equal{}b'\]
for all $ x\in\mathbb R$. Show that there exist real numbers $ k,c,C$ such that
\[ |kx\plus{}c\minus{}f(x)|\... |
Given a set $ \{1,2,3,....n\}$ of natural numbers, find the number of triangles possible with distinct sides.
Use the fact that in a triangle with sides $ a,b,c$ that
$ a \plus{} b > c$
$ b \plus{} c > a$
$ c \plus{} a > b$ | [] | Given a set $ \{1,2,3,....n\}$ of natural numbers, find the number of triangles possible with distinct sides. |
The probability that you get $ 60 \%$ or higher is $ \frac{1}{2} \left (1\minus{}\binom{10}{5} \times \frac{1}{1024} \right )\equal{} \frac{1}{2} \left ( \frac{193}{256} \right )\equal{}\boxed{\frac{193}{512}}$. | [
"percent",
"probability",
"counting",
"distinguishability",
"ratio",
"FTW",
"symmetry"
] | If a ten question True/False test is given and $ 60\%$ is a passing grade, what is your percent change, to the nearest whole percent, of passing if you guess on each question? |
Let's first assign some variables:
$ R$=getting question correct
$ N$=getting question incorrect
First, let's find the probability fraction, then convert it to a percent. I'll present two ways to solve this problem; the first is for those who know combinatorics, and the second is to clarify for those that don't ... | [
"percent",
"probability",
"counting",
"distinguishability",
"ratio",
"FTW",
"symmetry"
] | If a ten question True/False test is given and $ 60\%$ is a passing grade, what is your percent change, to the nearest whole percent, of passing if you guess on each question? |
AIME15's solution is perfectly rigorous and much slicker. By symmetry, it is clearly equally likely for a person to get $ k$ questions right and $ 10 \minus{} k$ questions wrong as it is for a person to get $ k$ questions wrong and $ 10 \minus{} k$ questions right. However, we must be careful to account for the fact th... | [
"percent",
"probability",
"counting",
"distinguishability",
"ratio",
"FTW",
"symmetry"
] | If a ten question True/False test is given and $ 60\%$ is a passing grade, what is your percent change, to the nearest whole percent, of passing if you guess on each question? |
Excuse me, math154:
1) please notice that not everybody who goes to this website would know the combinatorics of probability, and those who do know it could always look at my first solution
2) AIME15 does indeed have an incorrect solution, as the question asked for a percent, and AIME15 gave a fraction
Sorry, I d... | [
"percent",
"probability",
"counting",
"distinguishability",
"ratio",
"FTW",
"symmetry"
] | If a ten question True/False test is given and $ 60\%$ is a passing grade, what is your percent change, to the nearest whole percent, of passing if you guess on each question? |
[hide]Factor $x^3-2x^2-x+2=(x^2-1)(x-2)=(x+1)(x-1)(x-2)$. Let $R(x)$ be the remainder, so it has degree 2 or less, and let $P(x)=x^{100}-4x^{98}+5x+6$. $R(1)=P(1)=1-4+5+6=8$, $R(-1)=P(-1)=1-4-5+6=-2$, and $R(2)=P(2)=5*2+6=16$.
Let $R(x)=ax^2+bx+c$, so that $a+b+c=8$, $a-b+c=-2$, $4a+2b+c=16$. Subtracting the second e... | [] | Find the remainder when $\displaystyle x^{100}-4x^{98}+5x+6$ is divided by $\displaystyle x^3-2x^2-x+2$. |
[hide]Factor $x^3-2x^2-x+2=(x^2-1)(x-2)=(x+1)(x-1)(x-2)$. Let $R(x)$ be the remainder, so it has degree 2 or less, and let $P(x)=x^{100}-4x^{98}+5x+6$. $R(1)=P(1)=1-4+5+6=8$, $R(-1)=P(-1)=1-4-5+6=-2$, and $R(2)=P(2)=5*2+6=16$.
Let $R(x)=ax^2+bx+c$, so that $a+b+c=8$, $a-b+c=-2$, $4a+2b+c=16$. Subtracting the second e... | [] | Find the remainder when $\displaystyle x^{100}-4x^{98}+5x+6$ is divided by $\displaystyle x^3-2x^2-x+2$. |
Answer
$ 3(xy+y-x-y)-2xy=2$
$ xy-3x=2$
$ x(y-3)=2$
When $ x$ equals $ 2$ or $ 1$ (only positive divisors of 2) then $ y-3$ respectively equals $ 1$ and $ 2$
Hence pairs $ (x,\,y)$ are:
$ (2,\,4)$ and $ (1,\,5)$ | [] | Find all pairs of nonnegative integers x and y satisfying the equation:
3[y(x+1)-x-y]-2xy=2 |
Answer
$ 3(xy+y-x-y)-2xy=2$
$ xy-3x=2$
$ x(y-3)=2$
When $ x$ equals $ 2$ or $ 1$ (only positive divisors of 2) then $ y-3$ respectively equals $ 1$ and $ 2$
Hence pairs $ (x,\,y)$ are:
$ (2,\,4)$ and $ (1,\,5)$
Your solution is correct and is the same as mine :lol: | [] | Find all pairs of nonnegative integers x and y satisfying the equation:
3[y(x+1)-x-y]-2xy=2 |
End of preview. Expand in Data Studio
Dataset Card
imo_lq_filtered
Dataset Details
We scraped conversations and their tags from topics posted on Art of Problem Solving's High School Olympiads section, then normalized the data and removed duplicates. We treated the first post in each topic as the Problem, and posts following it that potentially contained answers as Solutions.
Dataset Description
Using open-web-math/filtering-models, we removed text data with a perplexity greater than 15,000 in accordance with the paper's methodology. Additionally, for posts after the first one in each topic, we considered text data unlikely to be answers and removed it if it was 200 characters or less, or if its LaTeX ratio was 0.1 or lower.
Reference: https://huggingface.co/open-web-math/filtering-models/blob/main/example/perplexity.py
Dataset Sources
https://artofproblemsolving.com/community/c6h
Uses
We believe this dataset will be useful for fine-tuning language models on Mathematical Olympiad problems.
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