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Problem: ## Task B-1.3.
A ship traveling along a river has covered $24 \mathrm{~km}$ upstream and $28 \mathrm{~km}$ downstream. For this journey, it took half an hour less than for traveling $30 \mathrm{~km}$ upstream and $21 \mathrm{~km}$ downstream, or half an hour more than for traveling $15 \mathrm{~km}$ upstream and $42 \mathrm{~km}$ downstream, assuming that both the ship and the river move uniformly.
Determine the speed of the ship in still water and the speed of the river.
Answer: ## Solution.
Let $t$ be the time required for the boat to travel $24 \mathrm{~km}$ upstream and $28 \mathrm{~km}$ downstream, $v_{R}$ the speed of the river, and $v_{B}$ the speed of the boat. When the boat is traveling upstream, its speed is $v_{B}-v_{R}$, and when it is traveling downstream, its speed is $v_{B}+v_{R}$.
Since $t=\frac{s}{v}$, from the given data, we obtain the following system of equations:
$\left\{\begin{array}{l}t=\frac{24}{v_{B}-v_{R}}+\frac{28}{v_{B}+v_{R}} \\ t+0.5=\frac{30}{v_{B}-v_{R}}+\frac{21}{v_{B}+v_{R}} \\ t-0.5=\frac{15}{v_{B}-v_{R}}+\frac{42}{v_{B}+v_{R}}\end{array}\right.$
By introducing new variables $x=\frac{3}{v_{B}-v_{R}}, y=\frac{7}{v_{B}+v_{R}}$, the system transforms into:
$\left\{\begin{array}{l}t=8 x+4 y \\ t+0.5=10 x+3 y \\ t-0.5=5 x+6 y\end{array}\right.$
Substituting $t$ from the first equation into the remaining two, we get:
$\left\{\begin{array}{l}8 x+4 y+0.5=10 x+3 y \\ 8 x+4 y-0.5=5 x+6 y\end{array}\right.$
$\left\{\begin{array}{l}2 x-y=0.5 \\ 3 x-2 y=0.5\end{array}\right.$
The solution to the last system is (0.5, 0.5). Then we have:
$\frac{3}{v_{B}-v_{R}}=0.5$, hence, $v_{B}-v_{R}=6 \mathrm{~and}$
$\frac{7}{v_{B}+v_{R}}=0.5$, hence, $v_{B}+v_{R}=14$.
The speed of the river is $v_{R}=4 \mathrm{~km} / \mathrm{h}$, and the speed of the boat is $v_{B}=10 \mathrm{~km} / \mathrm{h}$.
## Note:
By substituting $x=\frac{1}{v_{B}-v_{R}}, y=\frac{1}{v_{B}+v_{R}} \mathrm{~and}$ following the same procedure, the initial system transforms into the system $\left\{\begin{array}{l}6 x-7 y=0.5 \\ 9 x-14 y=0.5\end{array}\right.$
The solution to this system is $\left(\frac{1}{6}, \frac{1}{14}\right)$.
|
2752.
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Problem: 3. (6 points) A construction company was building a tunnel. When $\frac{1}{3}$ of the tunnel was completed at the original speed, they started using new equipment, which increased the construction speed by $20 \%$ and reduced the working hours to $80 \%$ of the original. As a result, it took a total of 185 days to complete the tunnel. If they had not used the new equipment and continued at the original speed, it would have taken $\qquad$ days to complete the tunnel.
Answer: 3. (6 points) A construction company builds a tunnel. When $\frac{1}{3}$ of the tunnel is completed at the original speed, new equipment is used, increasing the construction speed by $20 \%$ and reducing the daily working hours to $80 \%$ of the original. As a result, the tunnel is completed in 185 days. If the new equipment was not used and the original speed was maintained, it would take $\qquad$ 180 days.
【Solution】Solution: $\left(1-\frac{1}{3}\right) \div[(1+20 \%) \times 80 \%]$
$$
\begin{array}{l}
=\frac{2}{3} \div[120 \% \times 80 \%], \\
=\frac{2}{3} \div \frac{24}{25} \\
=\frac{25}{36} ; \\
185 \div\left(\frac{1}{3}+\frac{25}{36}\right) \\
=185 \div \frac{37}{36}, \\
=180 \text { (days). }
\end{array}
$$
Answer: If the original speed was maintained, it would take 180 days.
Therefore, the answer is: 180.
|
2316.
|
Problem: Prove that number $1$ can be represented as a sum of a finite number $n$ of real numbers, less than $1,$ not necessarily distinct, which contain in their decimal representation only the digits $0$ and/or $7.$ Which is the least possible number $n$?
Answer: 1. **Restate the problem in a more manageable form:**
We need to prove that the number \(1\) can be represented as a sum of a finite number \(n\) of real numbers, each less than \(1\), and each containing only the digits \(0\) and \(7\) in their decimal representation. We also need to find the least possible value of \(n\).
2. **Reformulate the problem using a known rational number:**
Consider the rational number \(\frac{1}{7} = 0.\overline{142857}\). We need to express \(\frac{1}{7}\) as a sum of \(n\) real numbers, each less than \(\frac{1}{7}\) and containing only the digits \(0\) and \(1\) in their decimal representation.
3. **Assume \(n < 10\) and find the least value of \(n\):**
We assume that all the real numbers are positive and that \(n < 10\). The least value of \(n\) is found to be \(8\), for which we have the identity:
\[
\frac{1}{7} = 0.\overline{1} + 0.\overline{011111} + 0.\overline{010111} + 0.\overline{010111} + 0.\overline{000111} + 0.\overline{000101} + 0.\overline{000101} + 0.\overline{000100}.
\]
4. **Consider the case with both positive and negative real numbers:**
Suppose now that \(\frac{1}{7}\) is the sum of \(i > 0\) positive real numbers and \(j > 0\) negative real numbers, with \(i + j \leq 7\). Denote by \(u = 0.u_1u_2\ldots\) (respectively by \(-v = -0.v_1v_2\ldots\)) the sum of the positive (respectively negative) real numbers, so that \(\frac{1}{7} + v = u\). Note that \(u\) and \(v\) are necessarily strictly less than \(1\).
5. **Analyze the constraints on \(u\) and \(v\):**
Since \(i, j < 7\), it follows that \(v_6 \geq 3\), so that \(j \geq 3\) and \(i \leq 4\). Then, we must have \(v_5 \geq 4\), so that \(j \geq 4\) and \(i \leq 3\). In this case, we find \(u_3 \geq 3\) and thus \(i = 3\). This finally implies that \(v_2 \geq 6\), which is impossible since \(j = 4\).
6. **Conclude the least value of \(n\):**
The least value of \(n\) is therefore \(8\).
The final answer is \( \boxed{ 8 } \).
|
15848.
|
Problem: 4. Given the three sides of an obtuse triangle are 3, $4, x$, then the range of values for $x$ is ( ).
(A) $1<x<7$.
(B) $5 \ll x<7$.
(C) $1<x<\sqrt{7}$.
(D) $5<x<7$ or $1<x<\sqrt{7}$.
Answer: $D$
|
1937.
|
Problem: 1. Solve the equation: $\frac{8 x+13}{3}=\frac{19-12 x}{2}-\left(16 x-\frac{7-4 x}{6}\right)$.
Answer: 1. $\frac{8 x+13}{3}=\frac{19-12 x}{2}-\left(16 x-\frac{7-4 x}{6}\right)$
$\frac{8 x+13}{3}=\frac{19-12 x}{2}-16 x+\frac{7-4 x}{6} / \cdot 6$
1 BOD
$2 \cdot(8 x+13)=3 \cdot(19-12 x)-96 x+7-4 x \quad 1$ BOD
$16 x+26=57-36 x-96 x+7-4 x \quad 1$ BOD
$16 x+36 x+96 x+4 x=57+7-26$
$152 x=38 \quad /: 152$
1 BOD
$x=\frac{38}{152}$
1 BOD
$x=\frac{1}{4}$
1 BOD
TOTAL 6 POINTS
Note: If the student has not fully simplified the fraction, the solution should be evaluated with 5, not 6 points.
|
1490.
|
Problem: A right-angled triangle has side lengths that are integers. What could be the last digit of the area's measure, if the length of the hypotenuse is not divisible by 5?
Answer: Let the lengths of the legs be $a$ and $b$, and the length of the hypotenuse be $c$. According to the Pythagorean theorem, we then have
$$
a^{2}+b^{2}=c^{2}
$$
A square number can give a remainder of 0, 1, or 4 when divided by 5. If neither $a$ nor $b$ is divisible by 5, then the left side of (1) can give a remainder of 2 (in the form $1+1$), 0 (in the form $1+4$), or 3 (in the form $4+4$) when divided by 5. Since 2 and 3 cannot be the remainder of a square number $-c^{2}$ when divided by 5, and 0 is excluded by the condition, if the length of the hypotenuse is not divisible by 5, then the length of one of the legs will be divisible by 5.
If both legs are even, then their product is divisible by 4. We claim that this is also true if there is an odd leg.
First, note that both legs cannot be odd. In this case, in (1), $a^{2}+b^{2}$ would be even, and thus $c^{2}$, as an even square number, would be divisible by 4, while $a^{2}$ and $b^{2}$, when divided by 4, would each give a remainder of 1, making their sum 2. In this case, however, equality could not hold in (1).
If now one of the legs, say $a$, is odd, then $b$ is even, and $c$ is odd. From (1),
$$
b^{2}=(c+a)(c-a)
$$
Here, both factors on the right side are even. If neither is divisible by 4, then they each give a remainder of 2 when divided by 4, so their sum, $2c$, is divisible by 4. This would, however, make $c$ even, which is not true.
Thus, in (2), the right side is also divisible by $2^{3}=8$. We know that a square number, in this case $b^{2}$, has every prime factor in its prime factorization with an even exponent. This means that $b^{2}$ is divisible by $2^{4}$, and thus $b$ is divisible by 4.
This proves that if (1) holds, then $ab$ is divisible by 4. The area of the triangle, $ab/2$, is thus an even number. Since there is a leg divisible by 5, the area is also divisible by $2 \cdot 5=10$, so the last digit of its measure is 0.
Remark. The statement - that on the one hand, in every Pythagorean triple there is a number divisible by 5, and on the other hand, the product of the legs is divisible by 4 - can also be easily shown based on the known characterization of Pythagorean triples. According to this characterization, Pythagorean triples can be written in the form $a=k\left(u^{2}-v^{2}\right)$, $b=2 k \cdot u v$, and $c=k\left(u^{2}+v^{2}\right)$, where $k$ is any natural number, and $u$ and $v$ are relatively prime numbers of different parities, with $u>v$. See, for example, Rademacher and Toeplitz's book *Numbers and Figures*.
|
10111.
|
Problem: Let $a_n\ (n\geq 1)$ be the value for which $\int_x^{2x} e^{-t^n}dt\ (x\geq 0)$ is maximal. Find $\lim_{n\to\infty} \ln a_n.$
Answer: 1. Define the integral \( I_n(x) = \int_x^{2x} e^{-t^n} \, dt \). We need to find the value of \( x \) that maximizes \( I_n(x) \).
2. To find the maximum, we first compute the derivative of \( I_n(x) \) with respect to \( x \):
\[
I_n'(x) = \frac{d}{dx} \left( \int_x^{2x} e^{-t^n} \, dt \right)
\]
Using the Leibniz rule for differentiation under the integral sign, we get:
\[
I_n'(x) = e^{-(2x)^n} \cdot 2 - e^{-x^n}
\]
Simplifying, we have:
\[
I_n'(x) = 2e^{-(2x)^n} - e^{-x^n}
\]
3. Set the derivative equal to zero to find the critical points:
\[
2e^{-(2x)^n} - e^{-x^n} = 0
\]
\[
2e^{-(2x)^n} = e^{-x^n}
\]
Taking the natural logarithm on both sides:
\[
\ln(2e^{-(2x)^n}) = \ln(e^{-x^n})
\]
\[
\ln 2 - (2x)^n = -x^n
\]
\[
\ln 2 = (2^n - 1)x^n
\]
Solving for \( x^n \):
\[
x^n = \frac{\ln 2}{2^n - 1}
\]
Therefore:
\[
x = \left( \frac{\ln 2}{2^n - 1} \right)^{\frac{1}{n}} = a_n
\]
4. We need to find \( \lim_{n \to \infty} \ln a_n \):
\[
\ln a_n = \ln \left( \left( \frac{\ln 2}{2^n - 1} \right)^{\frac{1}{n}} \right)
\]
\[
\ln a_n = \frac{1}{n} \ln \left( \frac{\ln 2}{2^n - 1} \right)
\]
\[
\ln a_n = \frac{1}{n} \left( \ln (\ln 2) - \ln (2^n - 1) \right)
\]
5. Evaluate the limit:
\[
\lim_{n \to \infty} \ln a_n = \lim_{n \to \infty} \frac{\ln (\ln 2) - \ln (2^n - 1)}{n}
\]
Since \( \ln (\ln 2) \) is a constant, we focus on the term involving \( 2^n \):
\[
\lim_{n \to \infty} \frac{\ln (\ln 2) - \ln (2^n - 1)}{n} = \lim_{n \to \infty} \frac{-\ln (2^n - 1)}{n}
\]
Using the approximation \( \ln (2^n - 1) \approx \ln (2^n) = n \ln 2 \) for large \( n \):
\[
\lim_{n \to \infty} \frac{-n \ln 2}{n} = -\ln 2
\]
The final answer is \(\boxed{-\ln 2}\).
|
3596.
|
Problem: 11. Given that the internal angles $A, B, C$ of $\triangle ABC$ have opposite sides $a, b, c$ respectively, and $\sqrt{3} b \cos \frac{A+B}{2}=c \sin B$.
(1) Find the size of $\angle C$;
(2) If $a+b=\sqrt{3} c$, find $\sin A$.
Answer: (1) $\sqrt{3} \sin B \cos \frac{A+B}{2}=\sin C \sin B \Rightarrow \sqrt{3} \sin \frac{C}{2}=\sin C \Rightarrow \cos \frac{C}{2}=\frac{\sqrt{3}}{2}$,
so $\frac{C}{2}=30^{\circ} \Rightarrow C=60^{\circ}$.
$$
\begin{array}{l}
\text { (2) } \sin A+\sin B=\sqrt{3} \sin C=\frac{3}{2} \Rightarrow \sin A+\sin \left(120^{\circ}-A\right) \\
=\sin A+\frac{\sqrt{3}}{2} \cos A+\frac{1}{2} \sin A=\frac{3}{2} \sin A+\frac{\sqrt{3}}{2} \cos A=\frac{3}{2} \\
\Rightarrow \frac{\sqrt{3}}{2} \sin A+\frac{1}{2} \cos A=\sin \left(A+30^{\circ}\right)=\frac{\sqrt{3}}{2} \Rightarrow A=30^{\circ} \text { or } A=90^{\circ} .
\end{array}
$$
So $\sin A=\frac{1}{2}$ or $\sin A=1$.
|
12314.
|
Problem: Task B-4.2. Let $n$ be the number obtained by writing 2013 zeros between every two digits of the number 14641. Determine all solutions of the equation $x^{4}=n$ in the set $\mathbb{C}$.
Answer: ## Solution.
If between every two digits of the number 14641 we write 2013 zeros, we get a number of the form $1 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 6 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 1$, and we can express it using powers with base 10. Then,
$$
\begin{aligned}
n & =1 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 6 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 1 \\
& =10^{4 \cdot 2013+4}+4 \cdot 10^{3 \cdot 2013+3}+6 \cdot 10^{2 \cdot 2013+2}+4 \cdot 10^{2013+1}+1= \\
& =\left(10^{2013+1}\right)^{4}+4 \cdot\left(10^{2013+1}\right)^{3}+6 \cdot\left(10^{2013+1}\right)^{2}+4 \cdot 10^{2013+1}+1= \\
& =\binom{4}{0}\left(10^{2014}\right)^{4}+\binom{4}{1}\left(10^{2014}\right)^{3} \cdot 1+\binom{4}{2}\left(10^{2014}\right)^{2} \cdot 1^{2}+\binom{4}{3} 10^{2014} \cdot 1^{3}+\binom{4}{4} \cdot 1^{4}
\end{aligned}
$$
Then the solutions to the equation $x^{4}=n$, or $x^{4}=\left(10^{2014}+1\right)^{4}$, are the numbers
$$
x_{1,2}= \pm\left(10^{2014}+1\right), \quad x_{3,4}= \pm\left(10^{2014}+1\right) \cdot i
$$
|
3152.
|
Problem: 6. As shown in Figure 2, let $P$ be a point inside the equilateral $\triangle ABC$ with side length 12. Draw perpendiculars from $P$ to the sides $BC$, $CA$, and $AB$, with the feet of the perpendiculars being $D$, $E$, and $F$ respectively. Given that $PD: PE: PF = 1: 2: 3$. Then, the area of quadrilateral $BDPF$ is
Answer: $$
6.11 \sqrt{3} \text {. }
$$
For the equilateral $\triangle ABC$ with a height of $12 \times \frac{\sqrt{3}}{2}=6 \sqrt{3}$, then
$$
\begin{array}{l}
P D+P E+P F=6 \sqrt{3} . \\
\text { Also, } P D: P E: P F=1: 2: 3, \text { so, } \\
P D=\sqrt{3}, P E=2 \sqrt{3}, P F=3 \sqrt{3} .
\end{array}
$$
As shown in Figure 7, draw
$P S / / B C$, intersecting $A B$ at $S$.
Since $\angle F S P=\angle B$ $=60^{\circ}$, therefore,
$$
\begin{array}{l}
F S=P F \cot 60^{\circ}=3, \\
P S=6, \\
B D=P S+P D \cot 60^{\circ}=7 .
\end{array}
$$
Thus, $S_{\text {quadrilateral BDPF }}=S_{\triangle P S}+S_{\text {trapezoid BDPS }}$
$$
=\frac{1}{2} \times 3 \times 3 \sqrt{3}+\frac{6+7}{2} \times \sqrt{3}=11 \sqrt{3} .
$$
|
6263.
|
Problem: 8.59 For each pair of real numbers $x, y$, the function $f$ satisfies the functional equation
$$
f(x)+f(y)=f(x+y)-x y-1 .
$$
If $f(1)=1$, then the number of integers $n$ (where $n \neq 1$) that satisfy $f(n)=n$ is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) infinitely many.
(30th American High School Mathematics Examination, 1979)
Answer: [Solution] Substituting $x=1$ into the functional equation, we get
$$
f(y+1)=f(y)+y+2 \text {. }
$$
Since $f(1)=1$, substituting $y=2,3,4, \cdots$ consecutively, we can see that for $y$ being a positive integer, $f(y)>0$. Therefore, for $y$ being a positive integer,
$$
f(y+1)>y+2>y+1 ;
$$
Thus, for integers $n>1$, $f(n)=n$ has no solution. Solving the equation for $f(y)$:
From $\circledast$ we have
$$
f(y)=f(y+1)-(y+2),
$$
Substituting $y=0,-1,-2, \cdots$ consecutively into this equation, we get
$$
\begin{array}{l}
f(0)=-1, \quad f(-1)=-2, \quad f(-2)=-2, \quad f(-3)=-1, \\
f(-4)=1 .
\end{array}
$$
Since $f(-4)>0$ and for $y0$.
Thus, for $y0$.
Therefore, for $n<-4, f(n) \neq n$.
Hence, $f(n)=n$ has only the solutions $n=1,-2$.
Therefore, the answer is $(B)$.
|
4439.
|
Problem: 9.6. Find the minimum value of the expression $(\sqrt{2(1+\cos 2 x)}-\sqrt{36-4 \sqrt{5}} \sin x+2) \cdot(3+2 \sqrt{10-\sqrt{5}} \cos y-\cos 2 y) \cdot$ If the answer is not an integer, round it to the nearest integer.
Answer: Answer: -27 . Instructions. Exact answer: $4 \sqrt{5}-36$.
|
8192.
|
Problem: Given $0 \leqslant x, y, z \leqslant 1$, solve the equation:
$$\frac{x}{1+y+z x}+\frac{y}{1+z+x y}+\frac{z}{1+x+y z}=\frac{3}{x+y+z} .$$
Answer: 3. It is not difficult to prove: $\frac{x}{1+y+z x} \leqslant \frac{1}{x+y+z}$, $\frac{y}{1+z+x y} \leqslant \frac{1}{x+y+z}$, $\frac{z}{1+x+y z} \leqslant \frac{1}{x+y+z}$. Therefore, if the equality holds, it is easy to get $x=y=z=1$.
Note: This problem can also be solved as follows: First prove $\frac{x}{1+y+z x} \leqslant \frac{x}{x+y+z}$, etc., so we can get $x+y+z \geqslant 3$, hence $x=y=z=1$.
|
7436.
|
Problem: ## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{3 / 2} \frac{x^{2} \cdot d x}{\sqrt{9-x^{2}}}
$$
Answer: ## Solution
$$
\int_{0}^{3 / 2} \frac{x^{2} \cdot d x}{\sqrt{9-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& x=3 \sin t \Rightarrow d x=3 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{3}=0 \\
& x=\frac{3}{2} \Rightarrow t=\arcsin \frac{\left(\frac{3}{2}\right)}{3}=\arcsin \frac{1}{2}=\frac{\pi}{6}
\end{aligned}
$$
We get:
$$
\begin{aligned}
& =\int_{0}^{\pi / 6} \frac{9 \sin ^{2} t \cdot 3 \cos t d t}{\sqrt{9-9 \sin ^{2} t}}=\int_{0}^{\pi / 6} \frac{9 \sin ^{2} t \cdot 3 \cos t}{3 \sqrt{1-\sin ^{2} t}} d t=9 \cdot \int_{0}^{\pi / 6} \frac{\sin ^{2} t \cdot \cos t}{\cos t} d t= \\
& =9 \cdot \int_{0}^{\pi / 6} \sin ^{2} t d t=9 \cdot \int_{0}^{\pi / 6} \frac{1-\cos 2 t}{2} d t=\frac{9}{2} \cdot \int_{0}^{\pi / 6}(1-\cos 2 t) d t= \\
& =\left.\frac{9}{2}\left(t-\frac{1}{2} \sin 2 t\right)\right|_{0} ^{\pi / 6}=\frac{9}{2}\left(\frac{\pi}{6}-\frac{1}{2} \sin \frac{\pi}{3}\right)-\frac{9}{2}\left(0-\frac{1}{2} \sin 0\right)= \\
& =\frac{9}{2}\left(\frac{\pi}{6}-\frac{1}{2} \cdot \frac{\sqrt{3}}{2}\right)-0=\frac{3 \pi}{4}-\frac{9 \sqrt{3}}{8}
\end{aligned}
$$
Source — "http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_12-31" Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals
- Content is available under CC-BY-SA 3.0.
|
3209.
|
Problem: Example 6 The rules of a "level-up game" stipulate: On the $n$-th level, a die must be rolled $n$ times. If the sum of the points obtained from these $n$ rolls is greater than $2^{n}$, the level is considered passed. Questions:
(1) What is the maximum number of levels a person can pass in this game?
(2) What is the probability that he can pass the first three levels consecutively?
(Note: A die is a uniform cube with points numbered $1,2,3,4,5,6$ on its faces. The number of points on the face that lands up after rolling the die is the result of the roll.)
Answer: Solution: Since the dice is a uniform cube, the possibility of each number appearing after a roll is equal.
(1) Because the maximum number that can appear on a single dice is 6, and $6 \times 1 > 2^4, 6 \times 5 < 2$.
Therefore, when $n \geqslant 5$, the sum of the numbers that appear after rolling the dice $n$ times cannot be greater than 2, which means this is an impossible event, and the probability of passing is 0.
So, the maximum number of consecutive levels that can be passed is 4.
(2) Let event $A_{n}$ be "failing the $n$-th level", then the complementary event $\overline{A_{n}}$ is "passing the $n$-th level". In the $n$-th level game, the total number of basic events is $6^n$.
Level 1: The number of basic events contained in event $A_{1}$ is 2 (i.e., the cases where the number is 1 or 2),
So the probability of passing this level is: $P\left(\overline{A_{1}}\right)=1-P\left(A_{1}\right)=1-\frac{2}{6}=\frac{2}{3}$
Level 2: The number of basic events contained in event $A_{2}$ is the sum of the number of positive integer solutions to the equation $x+y=a$ when $a$ takes the values $2,3,4$. That is, $C_{1}^{3}+C_{3}^{1}+C_{3}^{1}=1+2+3=6$ (cases).
So the probability of passing this level is $P\left(\bar{A}_{2}\right)=1-P\left(A_{2}\right)=1-\frac{6}{6^{2}}=\frac{5}{6}$
Level 3: The number of basic events contained in event $A_{3}$ is the sum of the number of positive integer solutions to the equation $x+y+z=a$ when $a$ takes the values $3,4,5,6,7,8$. That is, $C_{2}^{3}+C_{3}^{3}+C_{4}^{2}+C_{5}^{3}+C_{6}^{3}+C_{5}^{2}=1+3+6+10+15+21=56$ (cases).
So the probability of passing this level is: $P\left(\bar{A}_{3}\right)=1-P\left(A_{3}\right)=1-\frac{56}{6^{3}}=\frac{20}{27}$
Therefore, the probability of passing the first 3 levels consecutively is: $P\left(\bar{A}_{1}\right) \cdot P\left(\bar{A}_{2}\right) \cdot P\left(\bar{A}_{3}\right)=\frac{2}{3} \times \frac{5}{6} \times \frac{20}{27}=\frac{100}{243}$
|
5569.
|
Problem: 2. (9th Canadian Mathematical Competition) $N$ is an integer, its representation in base $b$ is 777. Find the smallest positive integer $b$ such that $N$ is a fourth power of an integer in decimal notation.
Answer: 2. This problem is equivalent to finding the smallest positive integer $b$, such that the equation $7 b^{2}+7 b+7=x^{4}$, (1) has an integer solution for $x$. Since 7 is a prime number, it follows from equation (1) that 7 is a divisor of $x$. Therefore, let $x=7 k$, then equation (1) becomes $b^{2}+b+1=7^{3} k^{4}$. The smallest $b$ occurs when $k$ is at its minimum. Taking $k=1$, we then have $b^{2}+b+1=343, b^{2}+b-342=0$, which is $(b-18)(b+19)=0$, yielding the positive integer solution $b=18$. Thus, we have $(777)_{18}=\left(7^{4}\right)_{10}$.
|
3821.
|
Problem: Problem 6. (8 points) In the plane, there is a non-closed, non-self-intersecting broken line consisting of 31 segments (adjacent segments do not lie on the same straight line). For each segment, the line defined by it is constructed. It is possible for some of the 31 constructed lines to coincide. What is the minimum number of different lines that can be obtained?
Answer. 9.
Answer: Solution. Evaluation. Besides the ends, the broken line has 30 vertices, each of which is the intersection point of two lines. If the lines are no more than 8, then the intersection points are no more
than 7.8/2 = $28<30$, a contradiction. An example is shown in the drawing.
|
7075.
|
Problem: 7.1. Solve the equation $\frac{n!}{2}=k!+l!$ in natural numbers, where $n!=1 \cdot 2 \cdot \ldots n$. If there are no solutions, write 0; if there is one solution, write $n$; if there are multiple solutions, write the sum of the values of $n$ for all solutions. Recall that a solution is a triplet $(n, k, l)$; if solutions differ in at least one component, they are considered different.
Answer: Answer: 10 (all triples of solutions $(n, k, l):(3,1,2),(3,2,1),(4,3,3)$).
Solution. Note that if $k4$, then $n!>4 \cdot(n-1)!\geqslant 2 \cdot(k!+l!)$, so there are no such solutions. If $n=2$, we get $1=k!+l!$ - no solutions; if $n=3$, the equation $3=k!+l!$ has two solutions: $k=1, l=2$ and $k=2, l=1$; if $n=4$, the equation $12=k!+l!$ gives one more solution $k=l=3$.
|
5080.
|
Problem: Example 1 (Question from the 13th "Hope Cup" Invitational Competition) The real roots of the equations $x^{5}+x+1=0$ and $x+\sqrt[5]{x}+1=0$ are $\alpha, \beta$ respectively, then $\alpha+\beta$ equals ( ).
A. -1
B. $-\frac{1}{2}$
C. $\frac{1}{2}$
D. 1
Answer: Solution: Choose A. Reason: Consider the function $f(x)=x^{5}+x+1$, which is an increasing function on $\mathbf{R}$,
when $x^{5}+x+1=0$,
we get $x=\sqrt[5]{-x-1}$, substituting $-x-1$ for $x$ yields $-x-1=\sqrt[5]{x}$.
From (2), we have $(-x-1)^{5}=x$, which is also $(-x-1)^{5}+(-x-1)+1=0$.
Let $\alpha$ be a root of equation (1), i.e., $f(\alpha)=0$. Let $\beta$ be a root of equation (2), i.e., $-1-\beta=\sqrt[5]{\beta}$, thus from (3) we know that $f(-1-\beta)=0$.
Therefore, we have $f(\alpha)=f(-1-\beta)$, which means $\alpha+\beta=-1$.
|
3146.
|
Problem: Example 5 Given that $x_{1}, x_{2}, \cdots, x_{10}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{10}=2005$, find the maximum and minimum values of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$.
Answer: Solve: The number of ways to write 2005 as the sum of 10 positive integers is finite, so there must be a way that maximizes or minimizes the sum of their squares.
If the positive integers $x_{1}, x_{2}, \cdots, x_{10}$ satisfy $x_{1}+x_{2}+\cdots+x_{10}=2005$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reaches its maximum.
Assume without loss of generality that $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{10}$. If $x_{1}>1$, then $x_{1}+x_{2}=\left(x_{1}-1\right)+\left(x_{2}+1\right)$. $\left[\left(x_{1}-1\right)^{2}+\left(x_{2}+1\right)^{2}\right]-\left(x_{1}^{2}+x_{2}^{2}\right)=-2 x_{1}+2 x_{2}+2=2\left(x_{2}-x_{1}\right)+2 \geqslant 2$. Therefore, replacing $x_{1}, x_{2}$ with $x_{1}-1, x_{2}+1$ keeps their sum unchanged but increases the sum of their squares. This contradicts the assumption that $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is maximized, so $x_{1}=1$.
Similarly, $x_{2}=x_{3}=\cdots=x_{9}=1, x_{10}=1996$.
Thus, the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is: $9+1996^{2}=3984025$.
Now assume $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{10}$, satisfying $x_{1}+x_{2}+\cdots+x_{10}=2005$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reaches its minimum value. If there exist $x_{i}, x_{j}$ and $x_{j}-x_{i} \geqslant 2(j>i)$, then $\left[\left(x_{j}-1\right)^{2}+\left(x_{i}+1\right)^{2}\right]-\left(x_{j}^{2}+x_{i}^{2}\right)=-2 x_{j}+2 x_{i}+2=2\left(x_{i}-x_{j}\right)+2 \leqslant-2$. Therefore, replacing $x_{j}, x_{i}$ with $x_{j}-1, x_{i}+1$ keeps their sum unchanged but decreases the sum of their squares, contradicting the assumption that $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is minimized. Thus, the absolute difference between any two of $x_{1}, x_{2}, \cdots, x_{10}$ is at most 1. Therefore, these 10 numbers can only be 5 instances of 200 and 5 instances of 201. Thus, the minimum value is $5 \times 200^{2}+5 \times 201^{2}=402005$.
|
2177.
|
Problem: Four, (50 points) In an $n \times n$ grid, fill each cell with one of the numbers 1 to $n^{2}$. If no matter how you fill it, there must be two adjacent cells where the difference between the two numbers is at least 1011, find the minimum value of $n$.
---
The translation preserves the original text's formatting and structure.
Answer: Four, take the row $l$ where the number 1 is located and the column $t$ where the number $n^{2}$ is located, and approach $n^{2}$ sequentially along row $l$ and column $t$ from 1. Assume the numbers filled in the small grids passed through are $a_{1}, a_{2}, \cdots, a_{k}$.
From Figure 5, we know $k \leqslant 2 n-3$.
Then make the sum
$$
\begin{array}{l}
\left|1-a_{1}\right|+\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+ \\
\left|a_{k-1}-a_{k}\right|+\left|a_{k}-n^{2}\right| \\
\geqslant \mid\left(1-a_{1}\right)+\left(a_{1}-a_{2}\right)+\left(a_{2}-a_{3}\right)+\cdots+ \\
\quad\left(a_{k-1}-a_{k}\right)+\left(a_{k}-n^{2}\right) \mid \\
=n^{2}-1 .
\end{array}
$$
By the pigeonhole principle, we know that among
$$
\left|a_{i}-a_{i+1}\right|\left(a_{0}=1, a_{k+1}=n^{2}, i=0,1, \cdots, k\right)
$$
the $k+1$ positive integers, at least one number is not less than
$$
\left[\frac{n^{2}-1}{k+1}\right]+1 \text {, and } k+1 \leqslant 2 n-2 \text {. }
$$
According to the problem, we have
$$
\begin{array}{l}
\frac{n^{2}-1}{k+1} \geqslant \frac{n^{2}-1}{2 n-2}>1011-1 \\
\Rightarrow n>2019 .
\end{array}
$$
Therefore, the minimum value of $n$ is 2020.
(Xie Wenxiao, Huanggang High School, Hubei Province, 438000)
|
8300.
|
Problem: 1. If the set
$$
A=\{1,2, \cdots, 10\}, B=\{1,2,3,4\},
$$
$C$ is a subset of $A$, and $C \cap B \neq \varnothing$, then the number of such subsets $C$ is $(\quad)$.
(A) 256
(B) 959
(C) 960
(D) 961
Answer: - 1. C.
From the fact that there are $2^{6}$ subsets $C$ satisfying $C \cap B=\varnothing$, we know that there are $2^{10}-2^{6}=960$ subsets $C$ satisfying $C \cap B \neq \varnothing$.
|
2080.
|
Problem: Augusto has a wire that is $10 \mathrm{~m}$ long. He makes a cut at a point on the wire, obtaining two pieces. One piece has a length of $x$ and the other has a length of $10-x$ as shown in the figure below:
Augusto uses the two pieces of wire to make two squares.
a) What is the length of the side of each of the squares? What is the area of each?
b) What is the length of each of the two pieces of wire so that the sum of the areas of the squares obtained is minimized?
c) Suppose Augusto cuts the wire into ten pieces and uses each one to make a square. What should be the size of each piece so that the sum of the areas of the squares obtained is minimized?
Answer: Solution
In this problem, all lengths are given in meters and areas in square meters.
a) A piece of rope has length $x$ and another piece of rope has length $10-x$. Since a square has four sides of equal length, one square will have a side length of $\frac{x}{4}$ and the other square will have a side length of $\frac{10-x}{4}$.
The area of a square with side length $\ell$ is $\ell^{2}$. Therefore, one square will have an area of $\left(\frac{x}{4}\right)^{2}=\frac{x^{2}}{16}$, while the other square will have an area of $\left(\frac{10-x}{4}\right)^{2}=\frac{100-20 x+x^{2}}{16}$.
b) Let $S(x)$ be the sum of the areas of the two squares. From the previous part, we have
$$
S(x)=\frac{x^{2}}{16}+\frac{100-20 x+x^{2}}{16}=\frac{100-20 x+2 x^{2}}{16}=\frac{1}{8} x^{2}-\frac{5}{4} x+\frac{25}{4}
$$
which is a quadratic function. The minimum of a function of the form
$$
f(x)=a x^{2}+b x+c
$$
with $a>0$ is achieved at $x=\frac{-b}{2 a}$. Thus, the minimum area will be achieved if
$$
x=-\frac{\left(-\frac{5}{4}\right)}{2 \frac{1}{8}}=5
$$
In other words, if the rope is cut exactly in the middle!
c) From the previous part, we know that to minimize the sum of the areas, it is necessary to cut the rope exactly in the middle. Well, we claim that to minimize the area with nine cuts (i.e., creating ten squares), it is necessary that all pieces of rope be equal. To show this, consider the following argument: if two of the ten pieces of rope were different, it would be possible to reduce the area by cutting the pieces of rope so that these two were equal (we are using the previous part). Therefore, any two pieces of rope must be equal. Hence, all must be equal!
|
2753.
|
Problem: 29. Choose any three numbers from $a, b, c, d, e$ and find their sum, exactly obtaining the ten different numbers $7,11,13,14,19,21,22,25,26,28$. Then $a+b+c+d+e=$ ( i). MATHEMATICE YoUTH CLUE A. 25
B. 31
C. 37
D. 43
Answer: Reference answer: B
|
2843.
|
Problem: Problem 3. In the school, there are 50 teachers, of whom 29 drink coffee, 28 drink tea, and 16 do not drink either coffee or tea. How many teachers drink only coffee, and how many drink only tea?
Answer: Solution. Since 16 teachers do not drink either coffee or tea, we get that $50-16=34$ teachers drink either coffee or tea. However, 28 teachers drink tea, and 29 drink coffee, so $(28+29)-34=57-34=23$ teachers drink
both coffee and tea. Only coffee is drunk by $29-23=6$ teachers, and only tea is drunk by $28-23=5$ teachers.
|
1156.
|
Problem: 12.180. A side of the triangle is equal to 15, the sum of the other two sides is 27. Find the cosine of the angle opposite the given side, if the radius of the inscribed circle in the triangle is 4.
Answer: ## Solution.
Let $a$ be the given side of the triangle, $a=15$, $b$ and $c$ be the other two sides, $b+c=27$, $c=27-b$. For definiteness, we will assume that $b>c$. The semiperimeter of the triangle $p=\frac{a+b+c}{2}=21$, its area $S=p r=84$.
By Heron's formula $S=\sqrt{p(p-a)(p-b)(p-c)}$, thus
$$
\begin{aligned}
& \sqrt{21 \cdot(21-15)(21-b)(21-27+b)}=84 \Leftrightarrow \sqrt{21 \cdot 6(21-b)(b-6)}=84 \Leftrightarrow \\
& \Leftrightarrow(21-b)(b-6)=56 \Leftrightarrow b^{2}-27 b+182=0, \\
& b_{1}=14, b_{2}=13 . \\
& \text { We have } b=14, c=13 \text {. By the cosine rule }
\end{aligned}
$$
$$
\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{5}{13}
$$
Answer: $\frac{5}{13}$.
|
2203.
|
Problem: 3. Let $AB$ be a chord of the unit circle $\odot O$. If the area of $\odot O$ is exactly equal to the area of the square with side $AB$, then $\angle AOB=$ $\qquad$ (to 0.001 degree).
Answer: 3. $124.806^{\circ}$
|
9811.
|
Problem: 83. Fill in the following squares with $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ respectively, so that the sum of the two five-digit numbers is 99999. Then the number of different addition equations is $\qquad$. $(a+b$ and $b+a$ are considered the same equation)
Answer: Answer: 1536
|
12876.
|
Problem: On a board, the numbers from 1 to 2009 are written. A couple of them are erased and instead of them, on the board is written the remainder of the sum of the erased numbers divided by 13. After a couple of repetition of this erasing, only 3 numbers are left, of which two are 9 and 999. Find the third number.
Answer: 1. **Define the invariant:**
Let \( I \) be the remainder when the sum of all the integers on the board is divided by 13. Initially, the numbers from 1 to 2009 are written on the board. The sum of these numbers is:
\[
S = \sum_{k=1}^{2009} k = \frac{2009 \times 2010}{2} = 2009 \times 1005
\]
We need to find \( S \mod 13 \).
2. **Calculate \( 2009 \mod 13 \) and \( 1005 \mod 13 \):**
\[
2009 \div 13 = 154 \quad \text{remainder} \quad 7 \quad \Rightarrow \quad 2009 \equiv 7 \pmod{13}
\]
\[
1005 \div 13 = 77 \quad \text{remainder} \quad 4 \quad \Rightarrow \quad 1005 \equiv 4 \pmod{13}
\]
3. **Calculate \( S \mod 13 \):**
\[
S = 2009 \times 1005 \equiv 7 \times 4 = 28 \equiv 2 \pmod{13}
\]
Therefore, \( I = 2 \).
4. **Determine the third number \( a \):**
After several operations, only three numbers are left on the board: 9, 999, and \( a \). The sum of these three numbers modulo 13 must be equal to the invariant \( I \):
\[
a + 9 + 999 \equiv 2 \pmod{13}
\]
5. **Calculate \( 999 \mod 13 \):**
\[
999 \div 13 = 76 \quad \text{remainder} \quad 11 \quad \Rightarrow \quad 999 \equiv 11 \pmod{13}
\]
6. **Set up the equation:**
\[
a + 9 + 11 \equiv 2 \pmod{13}
\]
Simplify the equation:
\[
a + 20 \equiv 2 \pmod{13}
\]
Subtract 20 from both sides:
\[
a \equiv 2 - 20 \pmod{13}
\]
\[
a \equiv -18 \pmod{13}
\]
Since \(-18 \equiv -18 + 26 \equiv 8 \pmod{13}\), we have:
\[
a \equiv 8 \pmod{13}
\]
7. **Find the possible values of \( a \) between 1 and 2009:**
The possible values of \( a \) are of the form:
\[
a = 8 + 13k \quad \text{for integer} \quad k
\]
We need \( 1 \leq a \leq 2009 \):
\[
1 \leq 8 + 13k \leq 2009
\]
\[
-7 \leq 13k \leq 2001
\]
\[
0 \leq k \leq 154
\]
Therefore, the possible values of \( a \) are:
\[
a = 8, 21, 34, \ldots, 2002
\]
The final answer is \( \boxed{8} \).
|
4301.
|
Problem: 4. The number of real solutions to the equation $\left|x^{2}-3 x+2\right|+\left|x^{2}+2 x-3\right|=11$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 4
Answer: 4. C.
The original equation is $|x-1|(|x-2|+|x+3|)=11$.
By discussing four cases: $x \leqslant -3$, $-3 < x \leqslant 2$, $2 < x \leqslant 1$, and $x > 1$, we know that there are 2 real solutions that satisfy the equation.
|
7448.
|
Problem: ## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}$
Answer: ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\left((n+1)^{2}-(n-1)^{2}\right) \cdot\left((n+1)^{2}+(n-1)^{2}\right)}{(n+1)^{3}+(n-1)^{3}}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(n^{2}+2 n+1-n^{2}+2 n-1\right) \cdot\left(n^{2}+2 n+1+n^{2}-2 n+1\right)}{(n+1)^{3}+(n-1)^{3}}= \\
& =\lim _{n \rightarrow \infty} \frac{4 n\left(2 n^{2}+2\right)}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{3}} 8 n\left(n^{2}+1\right)}{n^{3}\left((n+1)^{3}+(n-1)^{3}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{8\left(1+\frac{1}{n^{2}}\right)}{\left(1+\frac{1}{n}\right)^{3}+\left(1-\frac{1}{n}\right)^{3}}=\frac{8 \cdot 1}{1^{3}+1^{3}}=4
\end{aligned}
$$
## Problem Kuznetsov Limits 3-24
|
3516.
|
Problem: Condition of the problem
Find the derivative.
$$
y=\frac{1}{24}\left(x^{2}+8\right) \sqrt{x^{2}-4}+\frac{x^{2}}{16} \arcsin \frac{2}{x}, x>0
$$
Answer: ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{1}{24}\left(x^{2}+8\right) \sqrt{x^{2}-4}+\frac{x^{2}}{16} \arcsin \frac{2}{x}\right)^{\prime}= \\
& =\frac{1}{24} \cdot 2 x \cdot \sqrt{x^{2}-4}+\frac{1}{24}\left(x^{2}+8\right) \cdot \frac{1}{2 \sqrt{x^{2}-4}} \cdot 2 x+\frac{2 x}{16} \cdot \arcsin \frac{2}{x}+\frac{x^{2}}{16} \cdot \frac{1}{\sqrt{1-\left(\frac{2}{x}\right)^{2}}} \cdot\left(-\frac{2}{x^{2}}\right)= \\
& =\frac{2 x\left(x^{2}-4\right)}{24 \sqrt{x^{2}-4}}+\frac{x\left(x^{2}+8\right)}{24 \sqrt{x^{2}-4}}+\frac{x}{8} \cdot \arcsin \frac{2}{x}-\frac{1}{8} \cdot \frac{x}{\sqrt{x^{2}-4}}= \\
& =\frac{3 x^{3}-3 x}{24 \sqrt{x^{2}-4}}+\frac{x}{8} \cdot \arcsin \frac{2}{x}=\frac{x^{3}-x}{8 \sqrt{x^{2}-4}}+\frac{x}{8} \cdot \arcsin \frac{2}{x}
\end{aligned}
$$
## Problem Kuznetsov Differentiation 13-1
|
6223.
|
Problem: \section*{Problem 5 - 071225}
All ordered pairs of real numbers \((x, y)\) are to be determined for which the system of equations
\[
\begin{aligned}
x \cdot\left(a x^{2}+b y^{2}-a\right) & =0 \\
y \cdot\left(a x^{2}+b y^{2}-b\right) & =0
\end{aligned}
\]
is satisfied. Here, \(a\) and \(b\) are real numbers with \(a \neq 0, b \neq 0\) and \(a \neq b\).
Answer: }
We perform a case distinction:
1. Case: \(x=0\).
Then the second equation becomes \(y \cdot b \cdot\left(y^{2}-1\right)=0\), which, due to \(b \neq 0\), leads to \(y=0\) or \(y= \pm 1\). For all three elements \((x, y) \in\{(0,-1),(0,0),(0,1)\}\), verification confirms that they are indeed solutions to the system of equations.
2. Case: \(x \neq 0\).
Then it follows from the first equation that \(a x^{2}+b y^{2}-a=0\), and thus, due to \(a \neq b\), \(a x^{2}+b y^{2}-b \neq 0\). Therefore, from the second equation, we directly get \(y=0\). Substituting this into the equation just obtained, we get \(a x^{2}-a=0\) or \(x= \pm 1\). Here, too, all elements of the set \(\{(-1,0),(1,0)\}\) are solutions to the system of equations, as the verification confirms.
Thus, the given system of equations has a total of five solutions, which are noted in the two cases.
|
4072.
|
Problem: 6. Let $[x]$ denote the greatest integer not exceeding the real number $x$,
$$
\begin{array}{c}
S=\left[\frac{1}{1}\right]+\left[\frac{2}{1}\right]+\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\left[\frac{3}{2}\right]+ \\
{\left[\frac{4}{2}\right]+\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{3}{3}\right]+\left[\frac{4}{3}\right]+} \\
{\left[\frac{5}{3}\right]+\left[\frac{6}{3}\right]+\cdots}
\end{array}
$$
up to 2016 terms, where, for a segment with denominator $k$, there are $2 k$ terms $\left[\frac{1}{k}\right],\left[\frac{2}{k}\right], \cdots,\left[\frac{2 k}{k}\right]$, and only the last segment may have fewer than $2 k$ terms. Then the value of $S$ is
Answer: 6.1078.
$$
2+4+\cdots+2 \times 44=1980 \text {. }
$$
For any integer \( k \) satisfying \( 1 \leqslant k \leqslant 44 \), the sum includes \( 2k \) terms with the denominator \( k \): \(\left[\frac{1}{k}\right],\left[\frac{2}{k}\right], \cdots,\left[\frac{2 k}{k}\right]\), whose sum is \( k+2 \).
Also, \( 2016-1980=36 \), so the sum includes 36 terms with the denominator 45: \(\left[\frac{1}{45}\right],\left[\frac{2}{45}\right], \cdots,\left[\frac{36}{45}\right]\), whose sum is zero.
$$
\text { Therefore, } S=\sum_{k=1}^{44}(k+2)=1078 \text {. }
$$
|
4126.
|
Problem: 40. The sum of three consecutive odd numbers is equal to the fourth power of a single-digit number. Find all such triples of numbers.
Answer: 40. The sum of three consecutive odd numbers is an odd number, and if the fourth power is an odd number, then the base is also odd. Therefore, single-digit numbers should be sought among the numbers $3,5,7,9$.
1) $(2 x-1)+(2 x+1)+(2 x+3)=3^{4}, 6 x+3=3^{4}$, from which $2 x+1=3^{3}$ and $x=$ $=13 ; 2 x-1=25 ; 2 x+1=27 ; 2 x+3=29 ; 25+27+29=81=3^{4}$;
2) $6 x+3=5^{4}$ (no integer solutions);
3) $6 x+3=7^{4}=2401$ (no integer solutions);
4) $6 x+3=9^{4} ; 2 x+1=2187 ; 2 x-1=2185 ; 2 x+3=2189$.
Thus, the problem has two solutions:
a) $25,27,29$; b) $2185,2187,2189$.
|
2434.
|
Problem: Two is $10 \%$ of $x$ and $20 \%$ of $y$. What is $x - y$?
$(\mathrm {A}) \ 1 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 5 \qquad (\mathrm {D}) \ 10 \qquad (\mathrm {E})\ 20$
Answer: $2 = \frac {1}{10}x \Longrightarrow x = 20,\quad 2 = \frac{1}{5}y \Longrightarrow y = 10,\quad x-y = 20 - 10=10 \mathrm{(D)}$.
|
1036.
|
Problem: 19. Given $m \in\{11,13,15,17,19\}$, $n \in\{1999,2000, \cdots, 2018\}$.
Then the probability that the unit digit of $m^{n}$ is 1 is ( ).
(A) $\frac{1}{5}$
(B) $\frac{1}{4}$
(C) $\frac{3}{10}$
(D) $\frac{7}{20}$
(E) $\frac{2}{5}$
Answer: 19. E.
Consider different cases.
When $m=11$, $n \in\{1999,2000, \cdots, 2018\}$, the unit digit of $m^{n}$ is 1, there are 20 ways to choose;
When $m=13$, $n \in\{2000,2004, \cdots, 2016\}$, the unit digit of $m^{n}$ is 1, there are 5 ways to choose;
When $m=15$, $n \in\{1999,2000, \cdots, 2018\}$, the unit digit of $m^{n}$ is 5;
When $m=17$, $n \in\{2000,2004, \cdots, 2016\}$, the unit digit of $m^{n}$ is 1, there are 5 ways to choose;
When $m=19$, $n \in\{2000,2002, \cdots, 2018\}$, the unit digit of $m^{n}$ is 1, there are 10 ways to choose.
In summary, the required result is $\frac{20+5+5+10}{5 \times 20}=\frac{2}{5}$.
|
3280.
|
Problem: 1. (6 points) Today is January 31, 2015, welcome to the 2015 "Spring Cup" Final. The calculation result of the expression $\frac{\frac{2015}{1}+\frac{2015}{0.31}}{1+0.31}$ is
Answer: 【Solution】Solve: $\frac{\frac{2015}{1}+\frac{2015}{0.31}}{1+0.31}$
$$
\begin{array}{l}
=\frac{\left(\frac{2015}{1}+\frac{2015}{0.31}\right) \times 0.31}{(1+0.31) \times 0.31} \\
=\frac{(1+0.31) \times 2015}{(1+0.31) \times 0.31} \\
=6500 ;
\end{array}
$$
Therefore, the answer is: 6500.
|
1761.
|
Problem: Consider a regular hexagon with an incircle. What is the ratio of the area inside the incircle to the area of the hexagon?
Answer: 1. **Assume the side length of the regular hexagon is 1.**
- Let the side length of the hexagon be \( s = 1 \).
2. **Calculate the area of the regular hexagon.**
- The formula for the area of a regular hexagon with side length \( s \) is:
\[
A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} s^2
\]
- Substituting \( s = 1 \):
\[
A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} \cdot 1^2 = \frac{3\sqrt{3}}{2}
\]
3. **Determine the radius of the incircle.**
- The radius \( r \) of the incircle of a regular hexagon with side length \( s \) is given by:
\[
r = \frac{s \sqrt{3}}{2}
\]
- Substituting \( s = 1 \):
\[
r = \frac{1 \cdot \sqrt{3}}{2} = \frac{\sqrt{3}}{2}
\]
4. **Calculate the area of the incircle.**
- The area of a circle with radius \( r \) is:
\[
A_{\text{incircle}} = \pi r^2
\]
- Substituting \( r = \frac{\sqrt{3}}{2} \):
\[
A_{\text{incircle}} = \pi \left( \frac{\sqrt{3}}{2} \right)^2 = \pi \cdot \frac{3}{4} = \frac{3\pi}{4}
\]
5. **Find the ratio of the area of the incircle to the area of the hexagon.**
- The ratio is given by:
\[
\text{Ratio} = \frac{A_{\text{incircle}}}{A_{\text{hexagon}}} = \frac{\frac{3\pi}{4}}{\frac{3\sqrt{3}}{2}} = \frac{3\pi}{4} \cdot \frac{2}{3\sqrt{3}} = \frac{3\pi \cdot 2}{4 \cdot 3\sqrt{3}} = \frac{6\pi}{12\sqrt{3}} = \frac{\pi}{2\sqrt{3}} = \frac{\pi \sqrt{3}}{6}
\]
The final answer is \(\boxed{\frac{\pi \sqrt{3}}{6}}\).
|
2413.
|
Problem: Let $ABC$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $ABC$ such that $AP{\cdot}AG + BP{\cdot}BG + CP{\cdot}CG$ is a minimum, and express this minimum value in terms of the side lengths of $ABC$.
Answer: 1. **Define the function \( f(P) \):**
For any point \( P \) in the plane, define the function:
\[
f(P) = AP \cdot AG + BP \cdot BG + CP \cdot CG
\]
We aim to find the point \( P \) that minimizes \( f(P) \).
2. **Prove that \( P \) must be the centroid \( G \):**
Let \( G \) be the centroid of triangle \( ABC \). We will show that \( f(P) \) is minimized when \( P \equiv G \).
3. **Claim: \( P \) must be contained in \( \triangle ABC \):**
- Let \( \mathcal{R}_A \) be the region of all points that lie on the same side of line \( BC \) as point \( A \) (including all points on line \( BC \)).
- Define \( \mathcal{R}_B \) and \( \mathcal{R}_C \) similarly.
- If \( P \notin \mathcal{R}_A \), then let \( P' \) be the reflection of \( P \) in line \( BC \). We have \( f(P') < f(P) \), contradicting the minimality of \( f(P) \). Thus, \( P \in \mathcal{R}_A \).
- Similarly, \( P \in \mathcal{R}_B \) and \( P \in \mathcal{R}_C \).
- Therefore, \( P \) must be contained in \( \triangle ABC \).
4. **Claim: \( AK \sin A : BK \sin B : CK \sin C = AG : BG : CG \):**
- Let \( K \) be the isogonal conjugate of \( G \) with respect to \( \triangle ABC \).
- Let \( D, E, F \) be the projections of \( K \) onto sides \( BC, CA, AB \), respectively.
- We need to show:
\[
\frac{AK}{BK} = \frac{AG}{BG} \cdot \frac{\sin B}{\sin A}
\]
- Let \( Z = \overline{CG} \cap \overline{AB} \). Then,
\[
\frac{AK}{BK} = \frac{\sin \angle ABK}{\sin \angle BAK} = \frac{\sin \angle CBG}{\sin \angle CAG} = \frac{\sin \angle GCB}{\sin \angle GCA} \cdot \frac{\sin \angle CBG / \sin \angle GCB}{\sin \angle CAG / \sin \angle GCA}
\]
\[
= \frac{\sin \angle BCZ / BZ}{\sin \angle ACZ / AZ} \cdot \frac{CG / CB}{CG / CA} = \frac{\sin B / CZ}{\sin A / CZ} \cdot \frac{AG}{BG} = \frac{AG}{BG} \cdot \frac{\sin B}{\sin A}
\]
5. **Using the Sine Rule:**
- By the Sine Rule, we have:
\[
EF = AK \cdot \sin A, \quad FD = BK \cdot \sin B, \quad DE = CK \cdot \sin C
\]
- From the previous claim, there exists a positive real number \( t \) such that:
\[
EF = t \cdot AG, \quad FD = t \cdot BG, \quad DE = t \cdot CG
\]
6. **Area considerations:**
- Denote by \( [\lambda] \) the area of any figure \( \lambda \). Then,
\[
AP \cdot AG + BP \cdot BG + CP \cdot CG = AP \cdot \frac{EF}{t} + BP \cdot \frac{FD}{t} + CP \cdot \frac{DE}{t}
\]
\[
\ge \frac{[AEPF]}{t/2} + \frac{[BFPD]}{t/2} + \frac{[CDPE]}{t/2} = \frac{[ABC]}{t/2}
\]
- Equality holds if and only if \( \overline{AP} \perp \overline{EF} \), \( \overline{BP} \perp \overline{FD} \), \( \overline{CP} \perp \overline{DE} \), which is true if and only if \( P \equiv G \).
7. **Finding the value of \( f(G) \):**
- We need to find the value of:
\[
f(G) = AG^2 + BG^2 + CG^2
\]
- Using Apollonius's theorem and the fact that \( G \) divides each median in the ratio \( 2:1 \), we obtain:
\[
f(G) = \frac{a^2 + b^2 + c^2}{3}
\]
The final answer is \( \boxed{\frac{a^2 + b^2 + c^2}{3}} \).
|
5004.
|
Problem: 1. Four points $A, B, C, D$ in space are pairwise 1 unit apart, and points $P, Q$ move on line segments $AB, CD$ respectively. The minimum distance between point $P$ and $Q$ is
Answer: $-1 . \frac{\sqrt{2}}{2}$.
From the problem, we know that the tetrahedron $ABCD$ is a regular tetrahedron. Therefore, finding the minimum distance between points $P$ and $Q$ is equivalent to finding the distance between $AB$ and $CD$, as shown in Figure 1.
Take the midpoints $E$ and $F$ of $AB$ and $CD$, respectively. Then $AF \perp CD, BF \perp CD$ $\Rightarrow CD \perp$ plane $ABF \Rightarrow EF \perp CD$. Also, $EF \perp AB$, so $EF$ is the distance between $AB$ and $CD$. Hence, $AE = BE = \frac{1}{2}, AF = \frac{\sqrt{3}}{2}$ $\Rightarrow EF = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2} - \left(\frac{1}{2}\right)^{2}} = \frac{\sqrt{2}}{2}$.
|
5097.
|
Problem: The function $f$ maps the set of positive integers into itself, and satisfies the equation
$$
f(f(n))+f(n)=2 n+6
$$
What could this function be?
Answer: I. solution. The function $f(n)=n+2$ is suitable because its values are positive integers, and for any positive integer $n$,
$$
f(f(n))+f(n)=((n+2)+2)+(n+2)=2 n+6
$$
We will show that there is no other solution.
The function $f$ is certainly injective, that is, if $n \neq m$, then $f(n) \neq f(m)$. Indeed, if $f(n)=f(m)$, then $f(f(n))=f(f(m))$ and $2 n+6=f(f(n))+f(n)=f(f(m))=f(m)=2 m+6$, from which it follows that $n=m$. The injectivity implies that if $n \leq h$, then $f(n)=1, f(n)=2$ or $f(n) \geq h$.
Now we determine the value of $f(1)$. If we write (1) for $n=1$, we can establish an upper bound for $f(1)$:
$$
f(1)=8-f(f(1)) \leq 8-1=7
$$
To prove that $f(1)=3$, we need to exclude six cases.
Case I: $f(1)=1$. Then $f(f(1))=1$ and $f(f(1))+f(1)=2 \neq 2 \cdot 1+6$.
Case II: $f(1)=2$. Then
$f(2)=f(f(1))=2 \cdot 1+6-f(1)=6, f(6)=f(f(2))=2 \cdot 2+6-f(2)=4, f(4)=f(f(6))=2 \cdot 6+6-f(6)=14, f(14)=f(f$
This contradicts the fact that $f(14)$ is positive.
Similarly, starting with $f(1)=4,5,6$ and 7 leads to a contradiction, either a negative or a multi-valued function within at most three steps. (See the 1st Remark at the end of the problem.)
From $f(1)=3$, it follows that for every odd $n$, $f(n)=n+2$. This is true for $n=1$, as we have seen. If for some $n$, $f(n)=n+2$, then $f(n+2)=f(f(n))=2 n+6-f(n)=2 n+6-(n+2)=n+4$.
Now we determine the value of $f(2)$ as well. If we write (1) for $n=2$, we get that $f(2)=10-f(f(2)) \leq 10-1=9$. It is also certain that $f(2)$ cannot be an odd number greater than 1, because the function takes these values at odd places, which would contradict the injectivity. The possible values of $f(2)$ are therefore 4, and 1, 2, 6, and 8. Now we exclude these:
Case I: $f(2)=1$. Then $f(1)=f(f(2))=2 \cdot 2+6-f(2)=9$, which contradicts $f(1)=3$.
Similarly, we reach a contradiction for $f(2)=2,6$ and 8.
From $f(2)=4$, it again follows that for every even $n$, $f(n)=n+2$.
II. solution. That $f(n)=n+2$ is suitable can be verified according to the previous solution.
Let $n$ be any fixed positive integer. We will show that $f(n)=n+2$.
Define the following sequence: $a_{0}=n, a_{k+1}=f\left(a_{k}\right)$. The values of the function $f$ are positive integers, so this sequence consists of positive integers. Let $b_{k}=a_{k}-(n+2 k)$. By definition, $b_{0}=0$, and we want to show that $b_{1}$ is also 0.
According to equation (1),
$$
a_{k+2}+a_{k+1}=2 a_{k}+6
$$
from which
$b_{k+2}+b_{k+1}=a_{k+2}-(n+2 k+4)+a_{k+1}-(n+2 k+2)=a_{k+2}+a_{k+1}-(2 n+4 k+6)=2 a_{k}+6-(2 n+4 k+6)=2\left(a_{k}-(n\right.$
We prove by induction that for any non-negative integer $k$,
$$
b_{k}=\frac{1-(-2)^{k}}{3} b_{1}
$$
This is true for $k=0$ and $k=1$, because the equations $b_{0}=0 \cdot b_{1}=0$ and $b_{1}=b_{1}$ hold. If (3) holds for $k=m$ and $k=m+1$, then it also holds for $k=m+2$:
$$
\begin{aligned}
& b_{m+2}=2 b_{m}-b_{m+1}=2 \frac{1-(-2)^{m}}{3} b_{1}-\frac{1-(-2)^{m+1}}{3} b_{1}= \\
& \quad=\frac{2-2 \cdot(-2)^{m}-1+(-2)^{m+1}}{3} b_{1}=\frac{1-(-2)^{m+2}}{3} b_{1}
\end{aligned}
$$
This proves (3).
From the definition of the sequence $b_{k}$, $b_{k} \geq 1-n-2 k$. From this, using (3), we can obtain lower and upper estimates for $b_{1}$ by substituting some odd and even number for $k$. Let $m$ be a positive integer; substituting $k=2 m+1$ and $k=2 m$,
$$
-1-n-4 m \leq b_{2 m+1}=\frac{1+2^{2 m+1}}{3} b_{1}, 1-n-4 m \leq b_{2 m}=\frac{1+2^{2 m}}{3} b_{1}
$$
The coefficient of $b_{1}$ in the first inequality is positive, in the second it is negative. If we divide by it, the direction of the first inequality changes, but not the second. This gives us the following estimate:
$$
-\frac{3(n+4 m+1)}{1+2^{2 m+1}} \leq b_{1} \leq \frac{3(n+4 m-1)}{2^{2 m}-1}
$$
If we now let $m \rightarrow \infty$, then both the lower and upper estimates tend to 0, so $b_{1}=0$.
Based on the work of Gábor Zsolt Tóth (Budapest, Árpád Gimn., III. o.t.) and Benedek Valkó (Fazekas M. Fóv. Gyak. Gimn., IV. o.t.)
Remarks. 1. The second solution is actually a generalized version of the first. The case distinctions of the first solution were calculated simultaneously. For example, if $f(1)=x$, then based on the functional equation, we can determine that $f(x)=8-x, f(8-x)=3 x-2, f(3 x-2)=24-5 x, f(24-5 x)=11 x-22, f(11 x-22)=76-21 x$ and so on. For the last two numbers to be positive integers, it is necessary that $11 x-22 \geq 1$ and $76-21 x \geq 1$, that is, $2 \frac{1}{11} \leq x \leq 3 \frac{4}{7}$. The only integer in this interval is 3.
2. The second solution also works if the function $f$ maps the set of positive real numbers to itself. In this case, $f(x)=x+2$ is also the only solution.
|
3680.
|
Problem: 30. Find the remainder when the 2018-digit number $\underbrace{\overline{55 \cdots}}_{2018 \text { 555 }}$ is divided by 13.
Answer: Reference answer: 3
|
2367.
|
Problem: 1. A line $l$ intersects a hyperbola $c$, then the maximum number of intersection points is ( ).
A. 1
B. 2
C. 3
D. 4
Answer: 一、1. B. The intersection point refers to the solution of the system of equations
$$
\left\{\begin{array}{l}
a x+b y+c=0 \\
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
\end{array}\right.
$$
The solution to such a system of equations has at most two solutions. Therefore, the answer is B.
|
2453.
|
Problem: 1. Let the universal set be the set of real numbers. If $A=\{x \mid \sqrt{x-2} \leqslant 0\}, B=\left\{x \mid 10^{x^{2}-2}=10^{x}\right\}$, then $A \cap \bar{B}$ is
A. $\{2\}$
B. $\{-1\}$
C. $\{x \mid x \leqslant 2\}$
D. $\varnothing$
Answer: $$
A=\{2\}, x^{2}-2=x \Rightarrow x^{2}-x-2=(x+1)(x-2)=0 \Rightarrow B=\{-1,2\} \text {. }
$$
Thus, $A \cap \complement_{\mathbf{R}} B=\varnothing$. Therefore, the answer is $D$.
|
1597.
|
Problem: 4. As shown in Figure 1, in the right triangular prism $A B C-A_{1} B_{1} C_{1}$, $A A_{1}=A B=A C$, and $M$ and $Q$ are the midpoints of $C C_{1}$ and $B C$ respectively. If for any point $P$ on the line segment $A_{1} B_{1}$, $P Q \perp A M$, then $\angle B A C$ equals ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $90^{\circ}$
Answer: 4.D.
Let the midpoint of $A C$ be $R$. It is easy to know that $A_{1} R \perp A M$. Also, $P Q \perp A M$, so $A M \perp$ plane $A_{1} B_{1} Q R$. Therefore, $A M \perp Q R$, which means $A M \perp A B$. Also, $A B \perp A A_{1}$, so $A B \perp$ plane $A C C_{1} A_{1}$. Therefore, $A B \perp A C$.
|
3545.
|
Problem: 7.242. $\left(16 \cdot 5^{2 x-1}-2 \cdot 5^{x-1}-0.048\right) \lg \left(x^{3}+2 x+1\right)=0$.
Answer: Solution.
Domain: $x^{3}+2 x+1>0$.
From the condition $16 \cdot 5^{2 x-1}-2^{x-1}-0.048=0$ or $\lg \left(x^{3}+2 x+1\right)=0$. Rewrite the first equation as
$\frac{16}{5} \cdot 5^{2 x}-\frac{2}{5} \cdot 5^{x}-0.048=0 \Leftrightarrow 16 \cdot 5^{2 x}-2 \cdot 5^{x}-0.24=0$.
Solving this equation as a quadratic in terms of $5^{x}$, we get $5^{x}=-\frac{3}{40}$ (no solutions), or $5^{x}=5^{-1} \Leftrightarrow x_{1}=-1$ (does not satisfy the domain). From the second equation, we have
$x^{3}+2 x+1=1 \Leftrightarrow x^{3}+2 x=0 \Leftrightarrow x\left(x^{2}+2\right)=0, x_{3}=0, x^{2}+2 \neq 0$.
Answer: 0.
|
3122.
|
Problem: Example 1 In $\triangle ABC$, it is known that $x \sin A + y \sin B + z \sin C = 0$. Find the value of $(y + z \cos A)(z + x \cos B)(x + y \cos C) + (y \cos A + z)(z \cos B + x)(x \cos C + y)$.
Answer: In $\triangle A B C$, $\sin C=\sin (A+B)=\sin A \cos B+\cos A \sin B$. Substituting the known conditions, we get $x \sin A+y \sin B+z(\sin A \cos B+\cos A \sin B)=0$, which simplifies to $\sin A(x+z \cos B)=-\sin B(y+z \cos A)$.
Similarly, $\sin B(y+x \cos C)=-\sin C(z+x \cos B)$, $\sin C(z+y \cos A)=-\sin A(x+y \cos C)$,
Multiplying the three equations, we have:
$(x+z \cos B)(y+x \cos C)(z+y \cos A)=-(y+z \cos A)(z+x \cos B)(x+y \cos C)$, which simplifies to $(y+z \cos A)(z+x \cos B)(x+y \cos C)+(y \cos A+z)(z \cos B+x)(x \cos C+y)=0$.
|
10098.
|
Problem: In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
$\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000$
Answer: The length $L$ of the rectangle is $\frac{1000}{25}=40$ meters. The perimeter $P$ is $\frac{1000}{10}=100$ meters. Since $P_{rect} = 2L + 2W$, we plug values in to get:
$100 = 2\cdot 40 + 2W$
$100 = 80 + 2W$
$2W = 20$
$W = 10$ meters
Since $A_{rect} = LW$, the area is $40\cdot 10=400$ square meters or $\boxed{C}$.
|
1494.
|
Problem: 11.005. The plane angle at the vertex of a regular triangular pyramid is $90^{\circ}$. Find the ratio of the lateral surface area of the pyramid to the area of its base.
Answer: Solution.
For a regular pyramid $AB=BC=AC=a \quad$ (Fig. 11.7). $\angle ASC=90^{\circ}, SA=SC=SB$. This means that $\angle SAC=45^{\circ}, AD=DC$.
Fig. 11.7 Then from $\triangle SAC: SA=\frac{AD}{\cos 45^{\circ}}=\frac{a \sqrt{2}}{2}$ (since $\triangle ASC$ is isosceles). $SD$ is the apothem. Then $SD=AS \sin 45^{\circ}=\frac{a}{2}$. The lateral surface area $S_{\text {lat }}=p \cdot SD$, where $p$ is the semiperimeter of the base $-p=\frac{3}{2} a$; $S_{\text {lat }}=\frac{3}{2} a \cdot \frac{a}{2}=\frac{3 a^{2}}{4}$. The base area $S_{\text {base }}=\frac{a^{2} \sqrt{3}}{4}$. The ratio $\frac{S_{\text {lat }}}{S_{\text {base}}}=\frac{3 a^{2} \cdot 4}{4 a^{2} \cdot \sqrt{3}}=\sqrt{3}$.
Answer: $\sqrt{3}$.
|
5905.
|
Problem: The knights in a certain kingdom come in two colors. $\frac{2}{7}$ of them are red, and the rest are blue. Furthermore, $\frac{1}{6}$ of the knights are magical, and the fraction of red knights who are magical is $2$ times the fraction of blue knights who are magical. What fraction of red knights are magical?
$\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{3}{13}\qquad\textbf{(C) }\frac{7}{27}\qquad\textbf{(D) }\frac{2}{7}\qquad\textbf{(E) }\frac{1}{3}$
Answer: Let $k$ be the number of knights: then the number of red knights is $\frac{2}{7}k$ and the number of blue knights is $\frac{5}{7}k$.
Let $b$ be the fraction of blue knights that are magical - then $2b$ is the fraction of red knights that are magical. Thus we can write the equation $b \cdot \frac{5}{7}k + 2b \cdot \frac{2}{7}k = \frac{k}{6}\implies \frac{5}{7}b + \frac{4}{7}b = \frac{1}{6}$
$\implies \frac{9}{7}b = \frac{1}{6} \implies b=\frac{7}{54}$
We want to find the fraction of red knights that are magical, which is $2b = \frac{7}{27} = \boxed{\textbf{(C) }\frac{7}{27}}$
~KingRavi
|
1443.
|
Problem: A father wants to divide his property among his children: first, he gives 1000 yuan and one-tenth of the remaining property to the eldest child, then 2000 yuan and one-tenth of the remaining property to the second child, then 3000 yuan and one-tenth of the remaining property to the third child, and so on. It turns out that each child receives the same amount of property. The father has $\qquad$ children.
Answer: (Method 1)
Let the father's property be $x$ yuan;
Since the eldest and the second son receive the same amount of property, we have
$$
1000+(x-1000) \times \frac{1}{10}=2000+\left[(x-1000) \times\left(1-\frac{1}{10}\right)-2000\right] \times \frac{1}{10} ;
$$
Solving this, we get $x=81000$;
The eldest son receives $1000+(81000-1000) \times \frac{1}{10}=9000$ yuan;
That is, each child receives 9000 yuan;
This father has a total of $81000 \div 9000=9$ children.
(Method 2)
Since each child receives the same amount of property;
The eldest and the second son receive the same amount of property;
The eldest son receives 1000 yuan plus $\frac{1}{10}$ of the remaining property (let's denote this remaining as $A$); the second son receives 2000 yuan plus $\frac{1}{10}$ of the remaining property (let's denote this remaining as $B$); thus, $\frac{1}{10}$ of $A$ is 1000 yuan more than $\frac{1}{10}$ of $B$;
$A$ is $1000 \div \frac{1}{10}=10000$ yuan more than $B$;
From $A$ to $B$, it decreases by $\frac{1}{10}$ of $A$ and 2000 yuan;
Therefore, $\frac{1}{10}$ of $A$ is $10000-2000=8000$ yuan;
$A$ is $8000 \div \frac{1}{10}=80000$ yuan;
This father's property is $1000+80000=81000$ yuan;
The eldest son receives $1000+(81000-1000) \times \frac{1}{10}=9000$ yuan;
That is, each child receives 9000 yuan;
This father has a total of $81000 \div 9000=9$ children.
(Method 3)
Let this father have $n$ children;
Then the second-to-last child receives $1000(n-1)$ yuan plus $\frac{1}{10}$ of the remainder;
The last child receives $1000n$ yuan;
These $1000n$ yuan are $\frac{9}{10}$ of what remains after the second-to-last child receives $1000(n-1)$; each child receives the same amount of property;
From the equality of the property received by the second-to-last and the last child, we have
$$
1000(n-1)+1000n \times \frac{1}{9}=1000n \text {; }
$$
Solving this, we get $n=9$;
This father has a total of 9 children.
|
2918.
|
Problem: Example 11 Let $x>0, y>0, \sqrt{x}(\sqrt{x}+2 \sqrt{y})$ $=\sqrt{y}(6 \sqrt{x}+5 \sqrt{y})$. Find the value of $\frac{x+\sqrt{x y}-y}{2 x+\sqrt{x y}+3 y}$.
Answer: Solution: From the given, we have
$$
(\sqrt{x})^{2}-4 \sqrt{x} \sqrt{y}-5(\sqrt{y})^{2}=0,
$$
which is $(\sqrt{x}-5 \sqrt{y})(\sqrt{x}+\sqrt{y})=0$.
$$
\begin{array}{l}
\because \sqrt{x}+\sqrt{y}>0, \\
\therefore \sqrt{x}-5 \sqrt{y}=0,
\end{array}
$$
which means $x=25 y$.
Substituting $x=25 y$ into the original fraction, we get
the original expression $=\frac{1}{2}$.
|
1878.
|
Problem: 9. Given is a regular tetrahedron of volume 1 . We obtain a second regular tetrahedron by reflecting the given one through its center. What is the volume of their intersection?
Answer: Solution: $1 / 2$
Imagine placing the tetrahedron $A B C D$ flat on a table with vertex $A$ at the top. By vectors or otherwise, we see that the center is $3 / 4$ of the way from $A$ to the bottom face, so the reflection of this face lies in a horizontal plane halfway between $A$ and $B C D$. In particular, it cuts off the smaller tetrahedron obtained by scaling the original tetrahedron by a factor of $1 / 2$ about $A$. Similarly, the reflections of the other three faces cut off tetrahedra obtained by scaling $A B C D$ by $1 / 2$ about $B, C$, and $D$. On the other hand, the octahedral piece remaining remaining after we remove these four smaller tetrahedra is in the intersection of $A B C D$ with its reflection, since the reflection sends this piece to itself. So the answer we seek is just the volume of this piece, which is
$$
\begin{array}{l}
\text { (volume of } A B C D)-4 \cdot(\text { volume of } A B C D \text { scaled by a factor of } 1 / 2) \\
=1-4(1 / 2)^{3}=1 / 2
\end{array}
$$
|
12311.
|
Problem: Let's determine all the triples of numbers $(x, y, m)$ for which
$$
-2 x + 3 y = 2 m, \quad x - 5 y = -11
$$
and $x$ is a negative integer, $y$ is a positive integer, and $m$ is a real number.
Answer: From the second equation, $x=5 y-11$. Since $x$ is negative according to our condition, the equality can only hold if $y<\frac{11}{5}$. From the constraint on $y$, we get $y=2$ or $y=1$, and therefore $x=-1$ or $x=-6$.
Now we can determine the values of $m$ from the first equation:
\[
\begin{aligned}
\text { if } y=2 \text { and } x=-1, \quad \text { then } m=4 \\
\text { and if } y=1 \text { and } x=-6, \quad \text { then } m=7.5
\end{aligned}
\]
Remark. During the solution, we only assumed that $x$ is negative, no constraint is needed for $m$, and for any real number pair $(x ; y)$, $-2 x+3 y$ is always real.
|
1668.
|
Problem: ## Zadatak B-1.2.
Na slici su prikazani pravilni peterokut. $A B C D E$ i kvadrat $A B F G$. Odredite mjeru kuta $F A D$.
Answer: ## Rješenje.
Zbroj mjera unutrašnjih kutova pravilnog peterokuta iznosi $(5-2) 180^{\circ}=540^{\circ}$ pa je mjera unutrašnjeg kuta pravilnog peterokuta jednaka $\frac{540^{\circ}}{5}=108^{\circ}$.
Budući da je trokut $A E D$ jednakokračan, $\varangle D A E=\frac{180^{\circ}-108^{\circ}}{2}=36^{\circ}$
Kut $\varangle B A F=45^{\circ}$ jer je to kut između dijagonale i stranice kvadrata.
Tada je $\varangle F A D=\varangle B A E-\varangle B A F-\varangle D A E=108^{\circ}-45^{\circ}-36^{\circ}=27^{\circ}$.
|
8705.
|
Problem: The equation of line $\ell_1$ is $24x-7y = 319$ and the equation of line $\ell_2$ is $12x-5y = 125$. Let $a$ be the number of positive integer values $n$ less than $2023$ such that for both $\ell_1$ and $\ell_2$ there exists a lattice point on that line that is a distance of $n$ from the point $(20,23)$. Determine $a$.
[i]Proposed by Christopher Cheng[/i]
[hide=Solution][i]Solution. [/i] $\boxed{6}$
Note that $(20,23)$ is the intersection of the lines $\ell_1$ and $\ell_2$. Thus, we only care about lattice points on the the two lines that are an integer distance away from $(20,23)$. Notice that $7$ and $24$ are part of the Pythagorean triple $(7,24,25)$ and $5$ and $12$ are part of the Pythagorean triple $(5,12,13)$. Thus, points on $\ell_1$ only satisfy the conditions when $n$ is divisible by $25$ and points on $\ell_2$ only satisfy the conditions when $n$ is divisible by $13$. Therefore, $a$ is just the number of positive integers less than $2023$ that are divisible by both $25$ and $13$. The LCM of $25$ and $13$ is $325$, so the answer is $\boxed{6}$.[/hide]
Answer: 1. **Intersection of Lines**:
We start by finding the intersection of the lines $\ell_1$ and $\ell_2$. The equations are:
\[
24x - 7y = 319
\]
\[
12x - 5y = 125
\]
To find the intersection, we solve this system of linear equations. We can use the method of elimination or substitution. Here, we will use elimination.
2. **Elimination Method**:
Multiply the second equation by 2 to align the coefficients of $x$:
\[
24x - 10y = 250
\]
Now subtract the second equation from the first:
\[
(24x - 7y) - (24x - 10y) = 319 - 250
\]
Simplifying, we get:
\[
3y = 69 \implies y = 23
\]
Substitute $y = 23$ back into the first equation:
\[
24x - 7(23) = 319 \implies 24x - 161 = 319 \implies 24x = 480 \implies x = 20
\]
Therefore, the intersection point is $(20, 23)$.
3. **Distance from $(20, 23)$**:
We need to find lattice points on $\ell_1$ and $\ell_2$ that are an integer distance $n$ from $(20, 23)$. The distance formula is:
\[
d = \sqrt{(x - 20)^2 + (y - 23)^2}
\]
For $d$ to be an integer, $(x - 20)^2 + (y - 23)^2$ must be a perfect square, say $k^2$. Thus:
\[
(x - 20)^2 + (y - 23)^2 = n^2
\]
4. **Lattice Points on $\ell_1$**:
The line $\ell_1$ can be rewritten in slope-intercept form:
\[
y = \frac{24}{7}x - \frac{319}{7}
\]
Notice that $24$ and $7$ are part of the Pythagorean triple $(7, 24, 25)$. Therefore, the distance $n$ must be a multiple of $25$ for there to be lattice points on $\ell_1$.
5. **Lattice Points on $\ell_2$**:
The line $\ell_2$ can be rewritten in slope-intercept form:
\[
y = \frac{12}{5}x - \frac{125}{5}
\]
Notice that $12$ and $5$ are part of the Pythagorean triple $(5, 12, 13)$. Therefore, the distance $n$ must be a multiple of $13$ for there to be lattice points on $\ell_2$.
6. **Common Multiples**:
We need $n$ to be a multiple of both $25$ and $13$. The least common multiple (LCM) of $25$ and $13$ is:
\[
\text{LCM}(25, 13) = 325
\]
We need to count the multiples of $325$ that are less than $2023$:
\[
\left\lfloor \frac{2023}{325} \right\rfloor = 6
\]
Conclusion:
The number of positive integer values $n$ less than $2023$ such that there exists a lattice point on both $\ell_1$ and $\ell_2$ at a distance $n$ from $(20, 23)$ is $\boxed{6}$.
|
4287.
|
Problem: Let's determine $m$ such that the expression
$$
m x^{2}+(m-1) x+m-1
$$
is negative for all values of $x$.
---
Determine $m$ so that the expression
$$
m x^{2}+(m-1) x+m-1
$$
is negative for all values of $x$.
Answer: For the given expression to be negative for all values of $x$, it is necessary that the coefficient of $x^{2}$ be negative and that the curve representing the expression remain below the abscissa for all values of $x$. The equation $y=0$ must not have real or equal roots; for the roots to be complex, it is necessary that the discriminant
$$
y_{1}=(m-1)^{2}-4 m(m-1)
$$
be negative. If we geometrically represent this $y_{1}$ function, we see that the curve corresponding to the function intersects the abscissa at points whose abscissas are +1 and $-\frac{1}{3}$. The discriminant is therefore negative if $m>1$ or if $m<-\frac{1}{3}$. But $m$ cannot be positive according to the first condition, so the given expression is negative for all values of $x$ if $m<-\frac{1}{3}$.
The problem was solved by: Détshy K., Erdős A., Freibauer E., Goldziher K., Schwartz E., Szabó I., Szabó K., Weisz
J.
|
3490.
|
Problem: 2. How many integers $b$ exist such that the equation $x^{2}+b x-9600=0$ has an integer solution that is a multiple of both 10 and 12? Specify the largest possible $b$.
Answer: 2. Solution. Since the desired integer solution $x$ is divisible by 10 and 12, it is divisible by 60, hence it can be written in the form $x=60 t, t \in Z$. Substitute $x=60 t$ into the original equation:
$3600 t^{2}+60 b t-9600=0$. Express $b: b=\frac{160}{t}-60 t$. For $b$ to be an integer, $t$ must be a divisor of the number 160. $160=2^{5} \cdot 5$. The number $160=2^{5} \cdot 5$ has 12 positive divisors: $1,2,2^{2}, 2^{3}, 2^{4}, 2^{5}, 5,2 \cdot 5,2^{2} \cdot 5,2^{3} \cdot 5,2^{4} \cdot 5,2^{5} \cdot 5$. Taking the sign into account, we get 24 divisors. Each of them corresponds to a solution $x=60 t$ and an integer $b=\frac{160}{t}-60 t$. Therefore, the number of integers $b$ that satisfy the condition of the problem is 24. Since $b$ decreases as $t$ increases, the maximum $b$ corresponds to the smallest $t=-160$, so $b_{\max }=-1+60 \cdot 160=9599$.
|
8527.
|
Problem: Example 1 The range of the function $y=-x^{2}-2 x+3(-5 \leqslant x \leqslant 0)$ is $(\quad)$.
(A) $(-\infty, 4]$
(B) $[3,12]$
(C) $[-12,4]$
(D) $[4,12]$
Answer: Solve $y=-(x+1)^{2}+4$, and $-5 \leqslant x \leqslant 0$, so, the range of the function $y=-x^{2}-2 x+3(-5 \leqslant$ $x \leqslant 0)$ is $[-12,4]$, the answer is $\mathrm{C}$.
|
1244.
|
Problem: 4.206 There are two forces $f_{1}$ and $f_{2}$ acting on the origin $O$ of the coordinate axis,
$$\begin{array}{l}
\vec{f}_{1}=\overrightarrow{O A}=\sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right) \\
\vec{f}_{2}=\overrightarrow{O B}=2\left[\cos \left(-30^{\circ}\right)+i \sin \left(-30^{\circ}\right)\right]
\end{array}$$
(1) Find the magnitude and direction of their resultant force;
(2) Find the distance between points $A$ and $B$ (accurate to 0.1).
Answer: [Solution]
$$\text { (1) Resultant force } \begin{aligned}
\vec{f}= & \overrightarrow{f_{1}}+\overrightarrow{f_{2}} \\
= & \sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right) \\
& +2\left[\cos \left(-30^{\circ}\right)+i \sin \left(-30^{\circ}\right)\right] \\
= & 1+i+\sqrt{3}-i \\
= & \sqrt{3}+1 \\
= & (\sqrt{3}+1)\left(\cos 0^{\circ}+i \sin 0^{\circ}\right) .
\end{aligned}$$
So the magnitude of the resultant force is $\sqrt{3}+1$, and its direction is the same as the $x$-axis.
(2) Since $\overrightarrow{O A}=1+i$, the Cartesian coordinates of point $A$ are $(1,1)$, and $\overrightarrow{O B}=\sqrt{3}-i$, so the Cartesian coordinates of point $B$ are $(\sqrt{3},-1)$.
$$\text { Therefore } \begin{aligned}
|A B| & =\sqrt{(\sqrt{3}-1)^{2}+(-1-1)^{2}} \\
& =\sqrt{8-2 \sqrt{3}} \approx 2.1
\end{aligned}$$
|
6611.
|
Problem: 6. Given that $\overrightarrow{O A} \perp \overrightarrow{O B}$, and $|\overrightarrow{O A}|=|\overrightarrow{O B}|=24$. If $t \in[0,1]$, then
$$
|t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right|
$$
the minimum value is ( ).
(A) $2 \sqrt{193}$
(B) 26
(C) $24 \sqrt{2}$
(D) 24
Answer: 6. B.
Construct a square $O A C B$, connect the diagonal $A B$, and let $D$ and $E$ be points on the diagonal $A B$ and the side $O B$ respectively, such that
$$
\begin{array}{l}
t \overrightarrow{A B}-\overrightarrow{A O}=\overrightarrow{O D}, \\
\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}=\overrightarrow{E D}, \\
E B=10, O D=D C . \\
\text { Therefore }|t \overrightarrow{A B}-\overrightarrow{A O}|+\left|\frac{5}{12} \overrightarrow{B O}-(1-t) \overrightarrow{B A}\right| \\
=|\overrightarrow{E D}|+|\overrightarrow{D C}| \leqslant|\overrightarrow{E C}|=26 .
\end{array}
$$
|
10436.
|
Problem: All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$, with $A$ at the origin and $\overline{BC}$ parallel to the $x$-axis. The area of the triangle is $64$. What is the length of $BC$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$
Answer: By: Albert471
Plotting points $B$ and $C$ on the graph shows that they are at $\left( -x,x^2\right)$ and $\left( x,x^2\right)$, which is isosceles. By setting up the triangle area formula you get: $64=\frac{1}{2}*2x*x^2 = 64=x^3$ Making x=4, and the length of $BC$ is $2x$, so the answer is $\boxed{\textbf{(C)}\ 8}$.
|
1976.
|
Problem: 1B. If for the non-zero real numbers $a, b$ and $c$ the equalities $a^{2}+a=b^{2}, b^{2}+b=c^{2}$ and $c^{2}+c=a^{2}$ hold, determine the value of the expression $(a-b)(b-c)(c-a)$.
Answer: Solution. By adding the three equations, we obtain
$$
a^{2}+b^{2}+c^{2}+a+b+c=a^{2}+b^{2}+c^{2}, \text { i.e., } a+b+c=0
$$
Accordingly, $a+b=-c, b+c=-a, c+a=-b$. From the equation $a^{2}+a=b^{2}$, it follows that $(a-b)(a+b)=-a$, and since $a+b=-c \neq 0$, we get $a-b=-\frac{a}{a+b}$. Similarly, $b-c=-\frac{b}{b+c}$ and $c-a=-\frac{c}{c+a}$. Therefore,
$$
(a-b)(b-c)(c-a)=\frac{-a}{a+b} \cdot \frac{-b}{b+c} \cdot \frac{-c}{c+a}=\frac{-a}{-c} \cdot \frac{-b}{-a} \cdot \frac{-c}{-b}=1
$$
2AB. Prove that for any real numbers $a, b$, and $c$, the equation
$$
3(a+b+c) x^{2}+4(a b+b c+c a) x+4 a b c=0
$$
has real solutions. Under what conditions are the solutions of the equation equal to each other?
|
8464.
|
Problem: 2. As shown in Figure 1, the side length of rhombus $A B C D$ is $a$, and $O$ is a point on the diagonal $A C$, with $O A=a, O B=$ $O C=O D=1$. Then $a$ equals ( ).
(A) $\frac{\sqrt{5}+1}{2}$
(B) $\frac{\sqrt{5}-1}{2}$
(C) 1
(D) 2
Answer: 2. A. Since $\triangle B O C \sim \triangle A B C$, we have $\frac{B O}{A B}=\frac{B C}{A C}$, which means
$$
\frac{1}{a}=\frac{a}{a+1} \Rightarrow a^{2}-a-1=0 .
$$
Given $a>0$, solving yields $a=\frac{1+\sqrt{5}}{2}$.
|
3581.
|
Problem: V-2 If one side of the rectangle is reduced by $3 \mathrm{~cm}$, and the other side is reduced by $2 \mathrm{~cm}$, we get a square whose area is $21 \mathrm{~cm}^{2}$ less than the area of the rectangle. Calculate the dimensions of the rectangle.
Answer: Solution. If we denote the length of the side of the square by $a$, then the dimensions of the rectangle are $a+3$ and $a+2$. From the diagram, it is seen that the rectangle is divided into a square with area $a^{2}$ and three rectangles with areas $2a$, $3a$, and 6. Since the area of the square is $21 \mathrm{~cm}^{2}$ less than the area of the rectangle, it follows that $2a + 3a + 6 = 21$, or $5a = 15$. From this, we get $a = 3 \mathrm{~cm}$. Therefore, the dimensions of the rectangle are $6 \mathrm{~cm}$ and $5 \mathrm{~cm}$.
Note: For a diagram from which the equation $2a + 3a + 6 = 21$ is derived, 5 points are awarded.
|
1714.
|
Problem: 1. Given $a, b>0, a \neq 1$, and $a^{b}=\log _{a} b$, then the value of $a^{a^{b}}-\log _{a} \log _{a} b^{a}$ is
Answer: 1. -1 Explanation: From $a^{b}=\log _{a} b$ we know: $b=a^{a^{b}}, b=\log _{a} \log _{a} b$,
$$
a^{a^{b}}-\log _{a} \log _{a} b^{a}=b-\log _{a}\left(a \log _{a} b\right)=b-\log _{a} a-\log _{a} \log _{a} b=b-1-b=-1
$$
|
3679.
|
Problem: 4. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=0, a_{2}=1$, and for all $n \geqslant 3, a_{n}$ is the smallest positive integer greater than $a_{n-1}$ such that there is no subsequence of $a_{1}, a_{2}, \cdots, a_{n}$ that forms an arithmetic sequence. Find $a_{2014}$.
Answer: 4. First, prove a lemma using mathematical induction.
Lemma A non-negative integer appears in the sequence if and only if its ternary expansion contains only 0 and 1.
Proof It is obvious that the proposition holds for 0.
Assume the proposition holds for all non-negative integers less than \( N \), and consider \( N \).
If the ternary expansion of \( N \) contains the digit 2, replace all occurrences of 2 with 0 to get the number \( N_{0} \); replace all occurrences of 2 with 1 to get the number \( N_{1} \). Thus, the ternary expansions of \( N_{0} \) and \( N_{1} \) do not contain the digit 2.
By the induction hypothesis, the numbers \( N_{0} \) and \( N_{1} \) are in the sequence.
Since \( N_{0} \), \( N_{1} \), and \( N \) form an arithmetic sequence, \( N \) does not appear in the sequence. If the ternary expansion of \( N \) does not contain the digit 2, then we need to prove that \( N \) must appear in the sequence. If not, then there exist terms \( N_{0} \) and \( N_{1} \) in the sequence such that \( N_{0} \), \( N_{1} \), and \( N \) form an arithmetic sequence. Let the common difference be \( d \), and \( 3^{k} \| d \). Then the ternary expansions of these three numbers have the same lowest \( k-1 \) digits, and the \( k \)-th digit is different for each pair, so one of the numbers must have a 2 in the \( k \)-th digit, which contradicts the assumption.
Returning to the original problem.
We only need to find the 2014th non-negative integer whose ternary expansion does not contain the digit 2.
Notice that under this restriction, the ternary carry-over method is equivalent to binary, so we just need to write 2013 (note \( a_{1}=0 \)) in binary \((11111011101)_{2}\) and then convert it to ternary \( a_{2014}=88327 \).
|
5496.
|
Problem: In a stairwell, there are 10 mailboxes. One distributor drops a flyer into 5 mailboxes. Later, another distributor also drops a flyer into 5 mailboxes. What is the probability that this way, at least 8 mailboxes will contain a flyer?
Answer: Solution. The first distributor can choose 5 out of 10 mailboxes in $\binom{10}{5}$ ways, and the same applies to the second distributor. Therefore, the total number of cases is $\binom{10}{5}^{2}$.
If we want at least 8 mailboxes to have flyers, then the second distributor must place flyers in at least 3 of the empty mailboxes. However, they can also place flyers in 4 or 5 empty mailboxes, as this would still meet the requirement of the problem.
Let's calculate how many good solutions there are in total. The case with 5 additional empty mailboxes can occur in $\binom{10}{5}\binom{5}{5}$ ways. (The first 5 mailboxes are chosen from 10, and the other 5 are chosen from the empty mailboxes.) The case with 4 additional empty mailboxes has $\binom{10}{5}\binom{5}{4}\binom{5}{1}$ possibilities. Finally, the case with 3 additional empty mailboxes can be chosen in $\binom{10}{5}\binom{5}{3}\binom{5}{2}$ ways.
The number of favorable cases:
$$
\binom{10}{5}\left[\binom{5}{5}+\binom{5}{4}\binom{5}{1}+\binom{5}{3}\binom{5}{2}\right]=\binom{10}{5}(1+25+100)=\binom{10}{5} \cdot 126
$$
The desired probability:
$$
P=\frac{\binom{10}{5} \cdot 126}{\binom{10}{5}^{2}}=\frac{126}{\binom{10}{5}}=\frac{126}{252}=\frac{1}{2}
$$
|
4360.
|
Problem: 18. (3 points) Li Shuang rides a bike at a speed of 320 meters per minute from location $A$ to location $B$. On the way, due to a bicycle malfunction, he pushes the bike and walks for 5 minutes to a place 1800 meters from $B$ to repair the bike. After 15 minutes, he continues towards $B$ at 1.5 times his original riding speed, and arrives at $B$ 17 minutes later than the expected time. What is Li Shuang's walking speed in meters per minute?
Answer: 【Solution】Solve: $1800 \div 320-1800 \div(320 \times 1.5)$
$$
\begin{array}{l}
=5.625-3.75 \\
=1.875 \text { (minutes) } \\
320 \times[5-(17-15+1.875)] \div 5 \\
=320 \times[5-3.875] \div 5 \\
=320 \times 1.125 \div 5 \\
=360 \div 5 \\
=72 \text { (meters/minute) }
\end{array}
$$
Answer: Li Shuang's speed of pushing the cart while walking is 72 meters/minute.
Therefore, the answer is: 72.
|
2987.
|
Problem: 53. How many four-digit numbers contain at least one even digit?
Answer: 53. From four-digit numbers, we need to discard all those numbers that do not have a single even digit. We will get: $9 \cdot 10 \cdot 10 \cdot 10-5 \cdot 5 \cdot 5 \cdot 5 \cdot 5=8375$ numbers.
|
3067.
|
Problem: 1. The range of the function $f(x)=\sin x+\cos x+\tan x+$ $\arcsin x+\arccos x+\arctan x$ is $\qquad$ .
Answer: $=、 1 \cdot\left[-\sin 1+\cos 1-\tan 1+\frac{\pi}{4}, \sin 1+\cos 1+\tan 1+\frac{3 \pi}{4}\right]$.
Obviously, $f(x)=\sin x+\cos x+\tan x+\arctan x+\frac{\pi}{2}$, $x \in[-1,1]$.
Below we prove: $g(x)=\cos x+\tan x$ is an increasing function on $[-1,1]$.
Let $x_{1} 、 x_{2} \in[-1,1]$, and $x_{1}0$, thus,
$$
\frac{\cos \frac{x_{1}-x_{2}}{2}}{\cos x_{1} \cdot \cos x_{2}} \geqslant \frac{1}{\cos \frac{x_{1}-x_{2}}{2}} \geqslant 1 .
$$
Note that the equality cannot hold simultaneously, therefore,
$$
\frac{\cos \frac{x_{1}-x_{2}}{2}}{\cos x_{1} \cdot \cos x_{2}}>1 \text {. }
$$
Thus, $g\left(x_{1}\right)<g\left(x_{2}\right)$.
Hence $g(x)$ is an increasing function on $[-1,1]$.
Since $y=\sin x, y=\arctan x$ are increasing functions on $[-1,1]$, therefore, $f(x)$ is an increasing function on $[-1,1]$, and its range is $[f(-1), f(1)]$.
|
8193.
|
Problem: Let $a_1,a_2,\ldots,a_n$ be a permutation of the numbers $1,2,\ldots,n$, with $n\geq 2$. Determine the largest possible value of the sum \[ S(n)=|a_2-a_1|+ |a_3-a_2| + \cdots + |a_n-a_{n-1}| . \]
[i]Romania[/i]
Answer: To determine the largest possible value of the sum
\[ S(n) = |a_2 - a_1| + |a_3 - a_2| + \cdots + |a_n - a_{n-1}|, \]
we need to consider the arrangement of the permutation \(a_1, a_2, \ldots, a_n\) that maximizes the sum of the absolute differences between consecutive terms.
1. **Understanding the Problem:**
- We are given a permutation of the numbers \(1, 2, \ldots, n\).
- We need to maximize the sum of the absolute differences between consecutive terms in this permutation.
2. **Strategy for Maximization:**
- To maximize the sum of absolute differences, we should place the numbers in such a way that the difference between consecutive terms is as large as possible.
- This can be achieved by alternating between the largest and smallest remaining numbers.
3. **Constructing the Permutation:**
- For an odd \(n\), we can arrange the numbers as follows:
\[
\left(\frac{n+1}{2}, n, 1, n-1, 2, n-2, \ldots, \frac{n+1}{2} + 1\right)
\]
- For an even \(n\), we can arrange the numbers as follows:
\[
\left(\frac{n}{2}, n, 1, n-1, 2, n-2, \ldots, \frac{n}{2} + 1\right)
\]
4. **Calculating the Sum \(S(n)\):**
- For both odd and even \(n\), the sum \(S(n)\) can be calculated by summing the absolute differences between consecutive terms in the constructed permutation.
5. **Example Calculation:**
- For \(n = 5\) (odd):
\[
\left(3, 5, 1, 4, 2\right)
\]
\[
S(5) = |5 - 3| + |1 - 5| + |4 - 1| + |2 - 4| = 2 + 4 + 3 + 2 = 11
\]
- For \(n = 4\) (even):
\[
\left(2, 4, 1, 3\right)
\]
\[
S(4) = |4 - 2| + |1 - 4| + |3 - 1| = 2 + 3 + 2 = 7
\]
6. **General Formula:**
- The general formula for the maximum sum \(S(n)\) is:
\[
S(n) = \left\lfloor \frac{n^2}{2} \right\rfloor - 1
\]
The final answer is \( \boxed{ \left\lfloor \frac{n^2}{2} \right\rfloor - 1 } \).
|
15154.
|
Problem: There is a regular $17$-gon $\mathcal{P}$ and its circumcircle $\mathcal{Y}$ on the plane.
The vertices of $\mathcal{P}$ are coloured in such a way that $A,B \in \mathcal{P}$ are of diff erent colour, if the shorter arc connecting $A$ and $B$ on $\mathcal{Y}$ has $2^k+1$ vertices, for some $k \in \mathbb{N},$ including $A$ and $B.$
What is the least number of colours which suffices?
Answer: 1. **Identify the relevant values of \(2^k + 1\):**
Since the minor arc \(AB\) has at most 9 vertices on it, the relevant values of \(2^k + 1\) are \(3, 5,\) and \(9\). This is because:
\[
2^1 + 1 = 3, \quad 2^2 + 1 = 5, \quad 2^3 + 1 = 9
\]
2. **Enumerate the vertices:**
Enumerate the vertices of the 17-gon as \(V_0, V_1, \ldots, V_{16}\).
3. **Determine the conditions for different colors:**
We have that \(V_i\) and \(V_j\) are of different colors if \(i - j \equiv \pm 2, \pm 4, \pm 8 \pmod{17}\). This is derived from the relevant values of \(2^k + 1\).
4. **Establish a lower bound for the number of colors:**
Since \(V_0, V_2, V_4\) must have pairwise distinct colors, we have \(n \geq 3\).
5. **Check if 4 colors suffice:**
If there are 4 colors \(A, B, C,\) and \(D\), we can try to color the vertices such that the condition is satisfied. One possible coloring is:
\[
\{ V_0, V_1, \cdots, V_{16} \} = \{ A, D, B, B, C, C, D, D, B, B, C, A, D, D, B, B, C \}
\]
This coloring ensures that no two vertices \(V_i\) and \(V_j\) with \(i - j \equiv \pm 2, \pm 4, \pm 8 \pmod{17}\) have the same color. Therefore, \(n \leq 4\).
6. **Prove that 3 colors are insufficient:**
Assume there exists a solution with \(n = 3\). Without loss of generality, let \(V_0 = A\) and \(V_2 = B\). Then:
\[
V_4 = C, \quad V_6 = A, \quad V_8 = B, \quad V_{10} = C, \quad V_{12} = A, \quad V_{14} = B, \quad V_{16} = C
\]
Continuing this pattern, we find that \(V_{13} = A\), which contradicts the assumption that \(n = 3\) is a solution because \(V_0\) and \(V_{13}\) would be the same color, violating the condition for different colors.
Therefore, the least number of colors required is \(4\).
The final answer is \(\boxed{4}\).
|
16385.
|
Problem: 10.319. The diagonals of an isosceles trapezoid are perpendicular to each other, and its area is $a^{2}$. Determine the height of the trapezoid.
Answer: Solution.
Let in trapezoid $A B C D$ (Fig. 10.108) $A B=C D, A C \perp B D, O$ - the intersection point of $A C$ and $B D, C K$ - the height of the trapezoid. Since the trapezoid is isosceles, then $A K=\frac{A D+B C}{2}, A O=D O$, and since $\angle A O D=90^{\circ}$, then $\angle O A D=45^{\circ}$. Therefore, in the right $\triangle A K C \quad A K=C K$. The area of the trapezoid $S=\frac{A D+B C}{2} \cdot C K=A K \cdot C K=C K^{2}=a^{2}$. Thus, $C K=a$.
Answer: $a$.
|
4659.
|
Problem: 8. Find the last four digits of $7^{7^{-7}}$ (100 sevens).
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Answer: 8. Solution: $\because 7 \equiv-1(\bmod 4), \therefore 7^{7^{7}} \equiv$ $-1(\bmod 4)(98$ sevens$)$, let $7^{\gamma^{\prime 7}}=4 k+3(98$ sevens$), k \in \mathbf{Z}^{+}$, then $7^{7^{\prime 7}}=7^{4 k+3}$ (99 sevens), $\because 7^{4} \equiv$ $1(\bmod 100), \therefore 7^{4 k} \equiv 1(\bmod 100), 7^{7^{\prime 2}} \equiv 7^{4 k+3}$ $\equiv 7^{3} \equiv 43(\bmod 100)$. Let $7^{\gamma^{7}}=7^{100 m+43}$ (100 sevens), $m \in \mathbf{Z}^{+}$.
$$
\because 7^{4} \equiv 2401(\bmod 10000), \therefore 7^{8} \equiv 4801
$$
$(\bmod 10000), 7^{16} \equiv 9601(\bmod 10000)$,
$$
7^{32} \equiv 9201(\bmod 10000), \quad 7^{64} \equiv 8401
$$
$(\bmod 10000), 7^{100} \equiv 7^{4} \times 7^{32} \times 7^{64} \equiv 2401 \times 9201$ $\times 8401 \equiv 1(\bmod 10000), 7^{100 m} \equiv 1(\bmod 10000)$, $7^{100 m+43} \equiv 7^{43} \equiv 7^{3} \times 7^{8} \times 7^{32} \equiv 2343$ $(\bmod 10000), \therefore 7^{\gamma^{7}}(100$ sevens$) \equiv 7^{100 m+43} \equiv$ $2343(\bmod 10000), 7^{7^{77}}(100$ sevens$)$ has the last four digits as 2343.
|
10251.
|
Problem: 1. Let $S=\{1,2, \cdots, n\}, A$ be an arithmetic sequence with at least two terms, a positive common difference, all of whose terms are in $S$, and such that adding any other element of $S$ does not form an arithmetic sequence with the same common difference as $A$. Find the number of such $A$. (Here, a sequence with only two terms can also be considered an arithmetic sequence.)
Answer: Let the common difference of $A$ be $d$, then $1 \leqslant d \leqslant n-1$. We discuss in two cases:
(1) If $n$ is even, then when $1 \leqslant d \leqslant \frac{n}{2}$, there are $d$ $A$s with a common difference of $d$; when $\frac{n}{2}+1 \leqslant d \leqslant n-1$, there are $n-d$ $A$s with a common difference of $d$. Therefore, when $n$ is even, there are $\left(1+2+\cdots+\frac{n}{2}\right)+\left[1+2+\cdots+n-\left(\frac{n}{2}+1\right)\right]=\frac{n^{2}}{4}$ such $A$s.
(2) If $n$ is odd, then when $1 \leqslant d \leqslant \frac{n-1}{2}$, there are $d$ $A$s with a common difference of $d$; when $\frac{n+1}{2} \leqslant d \leqslant n-1$, there are $n-d$ $A$s with a common difference of $d$. Therefore, when $n$ is odd, there are $\left(1+2+\cdots+\frac{n-1}{2}\right)+\left(1+2+\cdots+\frac{n-1}{2}\right)=\frac{n^{2}-1}{4}$ such $A$s.
Combining (1) and (2), we get that the number of such $A$s is $\left[\frac{n^{2}}{4}\right]$.
|
4981.
|
Problem: Let $a_1, a_2, a_3, a_4$ be integers with distinct absolute values. In the coordinate plane, let $A_1=(a_1,a_1^2)$, $A_2=(a_2,a_2^2)$, $A_3=(a_3,a_3^2)$ and $A_4=(a_4,a_4^2)$. Assume that lines $A_1A_2$ and $A_3A_4$ intersect on the $y$-axis at an acute angle of $\theta$. The maximum possible value for $\tan \theta$ can be expressed in the form $\dfrac mn$ for relatively prime positive integers $m$ and $n$. Find $100m+n$.
[i]Proposed by James Lin[/i]
Answer: 1. **Given Information and Setup:**
- We have integers \(a_1, a_2, a_3, a_4\) with distinct absolute values.
- Points in the coordinate plane are \(A_1 = (a_1, a_1^2)\), \(A_2 = (a_2, a_2^2)\), \(A_3 = (a_3, a_3^2)\), and \(A_4 = (a_4, a_4^2)\).
- Lines \(A_1A_2\) and \(A_3A_4\) intersect on the \(y\)-axis at an acute angle \(\theta\).
- We need to find the maximum possible value for \(\tan \theta\) expressed as \(\frac{m}{n}\) for relatively prime positive integers \(m\) and \(n\), and then find \(100m + n\).
2. **Slope Calculation:**
- The slope of line \(A_1A_2\) is:
\[
m_1 = \frac{a_2^2 - a_1^2}{a_2 - a_1} = a_2 + a_1
\]
- The slope of line \(A_3A_4\) is:
\[
m_2 = \frac{a_4^2 - a_3^2}{a_4 - a_3} = a_4 + a_3
\]
3. **Intersection on the \(y\)-axis:**
- Since the lines intersect on the \(y\)-axis, their \(x\)-coordinates at the intersection point are zero. Thus, the \(y\)-intercepts of the lines are:
\[
c_1 = a_1^2 \quad \text{and} \quad c_2 = a_3^2
\]
4. **Angle Between Lines:**
- The tangent of the angle \(\theta\) between the two lines is given by:
\[
\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{(a_1 + a_2) - (a_3 + a_4)}{1 + (a_1 + a_2)(a_3 + a_4)} \right|
\]
5. **Maximizing \(\tan \theta\):**
- Let \(x = a_1 + a_2\) and \(y = a_3 + a_4\). We need to maximize:
\[
f(x, y) = \left| \frac{x - y}{1 + xy} \right|
\]
6. **Case Analysis:**
- **Case 1: \(x, y > 0\)**
- Here, \(x, y \geq 1\) since they are integers. In this case, \(|f(x, y)| < 1\), which cannot exceed our current maximum.
- **Case 2: \(x > 0, y < 0\)**
- The numerator is positive, and the denominator is negative, so \(|f(x, y)|\) cannot be our maximum.
- **Case 3: \(x < 0, y > 0\)**
- Let \(x = -u\) and \(y = v\). We want to maximize:
\[
\frac{u + v}{uv - 1}
\]
- We need to check when:
\[
\frac{u + v}{uv - 1} > \frac{3}{2} \implies 3(uv - 1) < 2(u + v) \implies 3uv - 3 < 2u + 2v \implies 3uv - 2u - 2v < 3
\]
- If \(u, v > 1\), then it's impossible because \(3uv - 2u - 2v \geq 3\) is false. Thus, we can let \(v = 1\) and need \(u < 5\).
7. **Checking Specific Values:**
- For \(u = 4\) and \(v = 1\):
- We have \(a_1 = -6, a_2 = 2, a_3 = 4, a_4 = -3\).
- This gives:
\[
\tan \theta = \left| \frac{(-6 + 2) - (4 - 3)}{1 + (-6 + 2)(4 - 3)} \right| = \left| \frac{-4 - 1}{1 - 4} \right| = \left| \frac{-5}{-3} \right| = \frac{5}{3}
\]
The final answer is \(\boxed{503}\).
|
16385.
|
Problem: 9.171. $0.6^{\lg ^{2}(-x)+3} \leq\left(\frac{5}{3}\right)^{2 \lg x^{2}}$.
Answer: ## Solution.
Domain of definition: $x<0$.
Since $\lg x^{2 k}=2 k \lg |x|$, taking into account the domain of definition, we can rewrite the given inequality as
$\left(\frac{3}{5}\right)^{\lg ^{2}(-x)+3} \leq\left(\frac{3}{5}\right)^{-4 \lg (-x)} \Leftrightarrow \lg ^{2}(-x)+3 \geq-4 \lg (-x), \lg ^{2}(-x)+4 \lg (-x)+3 \geq 0$.
Solving it as a quadratic inequality in terms of $\lg (-x)$, we get
$\left[\begin{array}{l}\lg (-x) \geq-1, \\ \lg (-x) \leq-3\end{array} \Leftrightarrow\left[\begin{array}{l}-x \geq 0.1, \\ -x \leq 0.001\end{array} \Leftrightarrow\left[\begin{array}{l}x \leq-0.1, \\ x \geq-0.001\end{array}\right.\right.\right.$
Answer: $x \in(-\infty ;-0.1] \cup[-0.001 ; 0)$.
|
5262.
|
Problem: 2. Find all integer solutions of the inequality
$$
x^{2} y^{2}+y^{2} z^{2}+x^{2}+z^{2}-38(x y+z)-40(y z+x)+4 x y z+761 \leq 0
$$
Answer: # Solution:
$\left(x^{2} y^{2}+2 x y z+z^{2}\right)+\left(y^{2} z^{2}+2 x y z+x^{2}\right)-38(x y+z)-40(y z+x)+761 \leq 0$.
$(x y+z)^{2}+(y z+x)^{2}-38(x y+z)-40(y z+x)+761 \leq 0$,
$(x y+z)^{2}-2 \cdot 19(x y+z)+361-361+(y z+x)^{2}-2 \cdot 20(y z+x)+400-400+761 \leq 0$,
$(x y+z-19)^{2}+(y z+x-20)^{2} \leq 0,\left\{\begin{array}{l}x y+z=19, \\ y z+x=20,\end{array} \Leftrightarrow\left\{\begin{array}{c}x y+z=19, \\ y(z-x)-(z-x)=20,\end{array} \Leftrightarrow\left\{\begin{array}{c}x y+z=19, \\ (z-x)(y-1)=1 .\end{array}\right.\right.\right.$
Since we need to find integer solutions to the inequality, there are only two cases:
1) $\left\{\begin{array}{c}x y+z=19, \\ z-x=1, \\ y=2,\end{array} \Leftrightarrow\left\{\begin{array}{c}2 x+x+1=19, \\ z=x+1, \\ y=2,\end{array} \Leftrightarrow\left\{\begin{array}{l}x=6, \\ y=2, \\ z=7 .\end{array}\right.\right.\right.$
2) $\left\{\begin{array}{c}x y+z=19, \\ z-x=-1, \\ y=0,\end{array} \Leftrightarrow\left\{\begin{array}{c}x-1=19, \\ z=x-1, \\ y=0,\end{array} \Leftrightarrow\left\{\begin{array}{c}x=20, \\ y=0, \\ z=19 .\end{array}\right.\right.\right.$
Answer: $(6 ; 2 ; 7),(20 ; 0 ; 19)$.
|
12172.
|
Problem: Solve the triangle whose area $t=357.18 \mathrm{~cm}^{2}$, where the ratio of the sides is $a: b: c=4: 5: 6$.
Answer: If $2 s=a+b+c$, then the area of the triangle is:
$$
t=\sqrt{s \cdot(s-a) \cdot(s-b) \cdot(s-c)}
$$
Since
$$
a: b: c=4: 5: 6
$$
therefore
$$
b=\frac{5 a}{4}, c=\frac{6 a}{4} \text { and } s=\frac{15 a}{8}
$$
Substituting these values into the area formula, we get:
$$
357.18=\frac{15 a^{2}}{64} \sqrt{7}
$$
from which
$$
a=\sqrt{\frac{357.18 \times 64}{15 \sqrt{7}}}=24 \text{ cm}
$$
furthermore
$$
b=30 \text{ cm}, \quad c=36 \text{ cm}
$$
The angles can be most conveniently obtained from the formulas
$$
\operatorname{tg}=\frac{\alpha}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}
$$
etc. Performing the calculations, we get:
$$
\alpha=41^{\circ} 24^{\prime} 34^{\prime \prime}, \beta=55^{\circ} 46^{\prime} 16^{\prime \prime}, \gamma=82^{\circ} 49^{\prime} 10^{\prime \prime}
$$
(Andor Szenes, Kaposvár.)
Number of solutions: 53.
|
4484.
|
Problem: Find all triples $(m,p,q)$ where $ m $ is a positive integer and $ p , q $ are primes.
\[ 2^m p^2 + 1 = q^5 \]
Answer: We are given the equation:
\[ 2^m p^2 + 1 = q^5 \]
where \( m \) is a positive integer and \( p \) and \( q \) are primes. We need to find all possible triples \((m, p, q)\).
1. **Rewrite the equation:**
\[ 2^m p^2 = q^5 - 1 \]
Notice that \( q^5 - 1 \) can be factored using the difference of powers:
\[ q^5 - 1 = (q-1)(q^4 + q^3 + q^2 + q + 1) \]
Therefore, we have:
\[ 2^m p^2 = (q-1)(q^4 + q^3 + q^2 + q + 1) \]
2. **Analyze the factors:**
Since \( p \) and \( q \) are primes, and \( 2^m p^2 \) is a product of powers of 2 and a prime squared, we need to consider the possible values of \( q \).
3. **Case 1: \( q = 2 \)**
If \( q = 2 \), then:
\[ 2^m p^2 = (2-1)(2^4 + 2^3 + 2^2 + 2 + 1) = 1 \cdot 31 = 31 \]
This implies:
\[ 2^m p^2 = 31 \]
Since 31 is a prime number and not a power of 2 times a square of a prime, this case fails.
4. **Case 2: \( q \neq 2 \)**
Since \( q \) is an odd prime, \( q-1 \) is even. Let \( q-1 = 2^a \cdot b \) where \( b \) is odd. Then:
\[ 2^m p^2 = 2^a \cdot b \cdot (q^4 + q^3 + q^2 + q + 1) \]
Since \( 2^m p^2 \) is a product of powers of 2 and a prime squared, we need to consider the greatest common divisor (gcd) of \( q-1 \) and \( q^4 + q^3 + q^2 + q + 1 \).
5. **GCD Analysis:**
\[ \gcd(q-1, q^4 + q^3 + q^2 + q + 1) \]
Since \( q^4 + q^3 + q^2 + q + 1 \equiv 5 \pmod{q-1} \), the gcd can be either 1 or 5.
6. **Subcase 1: \( \gcd(q-1, q^4 + q^3 + q^2 + q + 1) = 1 \)**
This implies \( q-1 = 2^m \) and \( q^4 + q^3 + q^2 + q + 1 = p^2 \).
- If \( q = 3 \):
\[ q^4 + q^3 + q^2 + q + 1 = 3^4 + 3^3 + 3^2 + 3 + 1 = 81 + 27 + 9 + 3 + 1 = 121 = 11^2 \]
Thus, \( p = 11 \) and \( q = 3 \), and \( q-1 = 2 \), so \( m = 1 \).
Therefore, we have the solution:
\[ (m, p, q) = (1, 11, 3) \]
- For \( q > 3 \):
\[ (2q^2 + q)^2 < 4p^2 < (2q^2 + q + 1)^2 \]
This inequality shows that \( q^4 + q^3 + q^2 + q + 1 \) cannot be a perfect square for \( q > 3 \).
7. **Subcase 2: \( \gcd(q-1, q^4 + q^3 + q^2 + q + 1) = 5 \)**
This implies \( p = 5 \) and \( q-1 = 2^a \cdot 5 \).
- Since \( q^4 + q^3 + q^2 + q + 1 \) is odd, it must be equal to 5, which is a contradiction because \( q^4 + q^3 + q^2 + q + 1 \) is much larger than 5 for any prime \( q \).
Therefore, the only solution is:
\[ (m, p, q) = (1, 11, 3) \]
The final answer is \( \boxed{ (1, 11, 3) } \)
|
7635.
|
Problem: 15. As shown in the figure, the area of square $\mathrm{ABCD}$ is 196 square centimeters, and it contains two partially overlapping smaller squares. The larger of the two smaller squares has an area that is 4 times the area of the smaller one, and the overlapping area of the two squares is 1 square centimeter. Therefore, the area of the shaded part is $\qquad$ square centimeters.
Answer: 【Answer】 72
【Analysis】Key point: Skillful area calculation
The area of the square is 196, so the side length is 14. The overlapping area is 1, so the side length is 1; the area of the larger square is 4 times that of the smaller square, so the side length of the larger square is twice that of the smaller square, and the sum of the side lengths of the larger and smaller squares is $14+1=15$.
Therefore, the side length of the smaller square is $15 \div 3=5$, and the side length of the larger square is $5 \times 2=10$.
The area of the smaller rectangle is $(5-1) \times(10-1)=36$, so the area of the two smaller rectangles is $36 \times 2=72\left(\mathrm{~cm}^{2}\right)$.
|
7875.
|
Problem: 25. Anna, Bridgit and Carol run in a $100 \mathrm{~m}$ race. When Anna finishes, Bridgit is $16 \mathrm{~m}$ behind her and when Bridgit finishes, Carol is $25 \mathrm{~m}$ behind her. The girls run at constant speeds throughout the race. How far behind was Carol when Anna finished?
A $37 \mathrm{~m}$
B $41 \mathrm{~m}$
C $50 \mathrm{~m}$
D $55 \mathrm{~m}$
E $60 \mathrm{~m}$
Answer: 25. A Carol finishes 25 metres behind Bridgit, so she travels 75 metres while Bridgit runs 100 metres. Therefore she runs 3 metres for every 4 metres Bridgit runs. When Anna finishes, Bridgit has run 84 metres, so that at that time Carol has run $\frac{3}{4} \times 84$ metres $=63$ metres. Hence Carol finishes $(100-63)$ metres $=37$ metres behind Anna.
|
3577.
|
Problem: Given the complex number $z$ has a modulus of 1. Find
$$
u=\frac{(z+4)^{2}-(\bar{z}+4)^{2}}{4 i} \text {. }
$$
the maximum value.
Answer: $$
\begin{array}{l}
\text { Given }|z|=1 \text {, we can set } z=\cos x+i \sin x \text {. Then, } \\
u=(4+\cos x) \sin x \text {. } \\
\text { By } u^{2}=(4+\cos x)^{2} \sin ^{2} x \\
=\frac{(4+\cos x)(4+\cos x)(\sqrt{6}+1)(1+\cos x) \cdot(\sqrt{6}+3)(1-\cos x)}{(\sqrt{6}+1)(\sqrt{6}+3)} \\
\leqslant \frac{1}{9+4 \sqrt{6}} \\
\cdot\left[\frac{(1+\cos x)+(1+\cos x)+(\sqrt{6}+1)(1+\cos x)+(\sqrt{6}+3)(1-\cos x)}{4}\right]^{4} \\
=\frac{1}{9+4 \sqrt{6}}\left(\frac{6+\sqrt{6}}{2}\right)^{4} \\
=\frac{9+24 \sqrt{6}}{4} \text {. } \\
\end{array}
$$
When $4+\cos x=$ ( $\sqrt{6}+1)(1+\cos x)$
$$
=(\sqrt{6}+3)(1-\cos x) \text {, }
$$
i.e., $\cos x=\frac{\sqrt{6}-2}{2}$, the maximum value is
$$
u_{\max }=\frac{\sqrt{9+24 \sqrt{6}}}{2}
$$
The coefficients $\sqrt{6}+1, \sqrt{6}+3$ are determined by the method of undetermined coefficients.
|
8192.
|
Problem: ## Problem 1
Perform the calculations:
a) $7 \cdot 147 - 7 \cdot 47$ (1p)
b) $(2+4+6+8+\cdots+50)-(1+3+5+7+\cdots+49)$ (2p)
c) $10 \cdot 9^{2} : 3^{2} - 3^{4} \quad(2 \text{p})$
d) $(\overline{a b} + \overline{b c} + \overline{c a}) : (a + b + c) \quad$ (2p)
Answer: ## Problem 1
a) 7400 ..................................................................................................................... 1p
c) \(10 \cdot(9: 3)^{2}-3^{4}=\)..................................................................................... 1p
= 11 ...................................................................................................................... 1p
|
2126.
|
Problem: 1. The curve $(x+2 y+a)\left(x^{2}-y^{2}\right)=0$ represents three straight lines intersecting at one point on the plane if and only if
A. $a=0$
B. $a=1$
C. $a=-1$
D. $a \in \mathbf{R}$
Answer: Obviously, the three lines are $x+2y+a=0$, $x+y=0$, and $x-y=0$. Since $x+y=0$ and $x-y=0$ intersect at point $(0,0)$, substituting into $x+2y+a=0 \Rightarrow a=0$, so the answer is $A$.
|
1619.
|
Problem: 9.27 In the metro train at the initial stop, 100 passengers entered. How many ways are there to distribute the exit of all these passengers at the next 16 stops of the train?
Answer: 9.27 The first passenger can exit at any of the 16 stops, as can the second, i.e., for two passengers there are $16^{2}$ possibilities. Therefore, for 100 passengers, there are $16^{100}$ ways.
Answer: $16^{100}$ ways.
|
2566.
|
Problem: 2. In the complex plane, there are 7 points corresponding to the 7 roots of the equation $x^{7}=$ $-1+\sqrt{3} i$. Among the four quadrants where these 7 points are located, only 1 point is in ( ).
(A) the I quadrant
(B) the II quadrant
(C) the III quadrant
(D) the IV quadrant
Answer: 2. (C).
$$
\begin{array}{l}
\because x^{7}=-1+\sqrt{3} \mathrm{i}=2\left(\cos 120^{\circ} + i \sin 120^{\circ}\right), \\
\therefore x_{n}=2^{\frac{1}{7}}\left(\cos \frac{\left(\frac{2}{3}+2 n\right) \pi}{7}+\sin \frac{\left(\frac{2}{3}+2 n\right) \pi}{7}\right) .
\end{array}
$$
where $n=0,1,2, \cdots, 6$.
After calculation, we get $\arg x_{0}=\frac{2 \pi}{21}, \arg x_{1}=\frac{8 \pi}{21}, \arg x_{2}=\frac{14 \pi}{21}$, $\arg x_{3}=\frac{20 \pi}{21}, \arg x_{4}=\frac{26 \pi}{21}, \arg x_{5}=\frac{32 \pi}{21}, \arg x_{6}=\frac{38 \pi}{21}$.
Clearly, there is only 1 root corresponding to a point in the third quadrant.
|
3146.
|
Problem: Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?
$ \text{(A)}\ 10\qquad\text{(B)}\ 20\qquad\text{(C)}\ 30\qquad\text{(D)}\ 50\qquad\text{(E)}\ 72 $
Answer: 1. First, we need to determine the number of students who prefer each type of pie. We know:
- 12 students prefer chocolate pie.
- 8 students prefer apple pie.
- 6 students prefer blueberry pie.
2. Calculate the total number of students who prefer chocolate, apple, or blueberry pie:
\[
12 + 8 + 6 = 26
\]
3. Subtract this number from the total number of students in the class to find the number of students who prefer either cherry or lemon pie:
\[
36 - 26 = 10
\]
4. Since half of the remaining students prefer cherry pie and the other half prefer lemon pie, we divide the remaining students by 2:
\[
\frac{10}{2} = 5
\]
Therefore, 5 students prefer cherry pie.
5. To find the number of degrees Richelle should use for cherry pie in her pie graph, we need to calculate the proportion of students who prefer cherry pie and then convert this proportion to degrees. The proportion of students who prefer cherry pie is:
\[
\frac{5}{36}
\]
6. Multiply this proportion by the total number of degrees in a circle (360 degrees):
\[
\frac{5}{36} \times 360 = 50
\]
7. Therefore, Richelle should use 50 degrees for cherry pie in her pie graph.
The final answer is $\boxed{50}$.
|
1038.
|
Problem: Father played chess with uncle. For a won game, the winner received 8 crowns from the opponent, and for a draw, nobody got anything. Uncle won four times, there were five draws, and in the end, father earned 24 crowns.
How many games did father play with uncle?
(M. Volfová)
Answer: Father lost four times, so he had to pay his uncle $4 \cdot 8=32$ crowns.
However, Father won so many times that even after paying these 32 crowns, he still gained 24 crowns. His total winnings were $32+24=56$ crowns, so he won $56: 8=7$ games.
Father won seven times, lost four times, and drew five times, so he played a total of $7+4+5=16$ games with his uncle.
Suggested scoring. 2 points for determining Father's total winnings; 2 points for the number of games Father won; 2 points for the total number of games played.
|
1091.
|
Problem: ## Problem 4
Given the numbers $1,2,3, \ldots, 1000$. Find the largest number $m$ with the property that by removing any $m$ numbers from these 1000 numbers, among the $1000-m$ remaining numbers, there exist two such that one divides the other.
Selected problems by Prof. Cicortas Marius
Note: a) The actual working time is 3 hours.
b) All problems are mandatory.
c) Each problem is graded from 0 to 7.
## NATIONAL MATHEMATICS OLYMPIAD
Local stage - 15.02.2014
## Grade IX
## Grading Rubric
Answer: ## Problem 4
If $m \geq 500$ then, for example, by removing the numbers from 1 to 500, among the remaining numbers, from 501 to 1000, there is no pair of numbers that can be divisible by each other, because their ratio is $<2$. 2 points We will show that $m=499$ has the required property. 1 point We show that among any 501 numbers from 1 to 1000, there is one that divides another. Each of the 501 numbers is of the form $2^{k}(2 t+1), 0 \leq k \leq 9, t$ is a natural number; to each such number, we associate $2 t+1$. There are 500 odd numbers less than 1000, so two of the 501 numbers will correspond to the same odd number. Among these two numbers, one will be obtained from the other by multiplying by a power of 2. 4 points
|
5478.
|
Problem: Problem 5.3. In five of the nine circles in the picture, the numbers 1, 2, 3, 4, 5 are written. Replace the digits $6, 7, 8, 9$ in the remaining circles $A, B, C, D$ so that the sums of the four numbers along each of the three sides of the triangle are the same.
Construct the correspondence
- In circle $A$
- In circle $B$
- In circle $C$
- In circle $D$
- the number 6 is placed.
- the number 7 is placed.
- the number 8 is placed.
- the number 9 is placed.
Answer: Answer: $A=6, B=8, C=7, D=9$.
Solution. From the condition, it follows that $A+C+3+4=5+D+2+4$, from which $D+4=A+C$. Note that $13 \geqslant D+4=A+C \geqslant 6+7$. Therefore, this is only possible when $D=9$, and $A$ and $C$ are 6 and 7 in some order. Hence, $B=8$.
The sum of the numbers along each side is $5+9+3+4=20$. Since $5+1+8+A=20$, then $A=6$. Since $6+C+3+4=20$, then $C=7$.
|
12237.
|
Problem: The function $f: \mathbb{R}\rightarrow \mathbb{R}$ is such that $f(x+1)=2f(x)$ for $\forall$ $x\in \mathbb{R}$ and $f(x)=x(x-1)$ for $\forall$ $x\in (0,1]$. Find the greatest real number $m$, for which the inequality $f(x)\geq -\frac{8}{9}$ is true for
$\forall$ $x\in (-\infty , m]$.
Answer: 1. **Given Conditions and Initial Analysis:**
- The function \( f: \mathbb{R} \rightarrow \mathbb{R} \) satisfies \( f(x+1) = 2f(x) \) for all \( x \in \mathbb{R} \).
- For \( x \in (0,1] \), \( f(x) = x(x-1) \).
2. **Behavior of \( f(x) \) in \( (0,1] \):**
- We need to find the minimum and maximum values of \( f(x) \) in the interval \( (0,1] \).
- \( f(x) = x(x-1) \) is a quadratic function opening downwards.
- The vertex of the parabola \( f(x) = x(x-1) \) is at \( x = \frac{1}{2} \).
- Evaluating at \( x = \frac{1}{2} \):
\[
f\left(\frac{1}{2}\right) = \frac{1}{2} \left(\frac{1}{2} - 1\right) = \frac{1}{2} \left(-\frac{1}{2}\right) = -\frac{1}{4}
\]
- Evaluating at the endpoints \( x = 0 \) and \( x = 1 \):
\[
f(0) = 0 \quad \text{and} \quad f(1) = 0
\]
- Therefore, \( \min_{x \in (0,1]} f(x) = -\frac{1}{4} \) and \( \max_{x \in (0,1]} f(x) = 0 \).
3. **General Form of \( f(x) \) for \( x \in (k, k+1] \):**
- For \( x \in (k, k+1] \), where \( k \in \mathbb{Z} \), we have:
\[
f(x) = 2^k f(x-k)
\]
- Since \( f(x-k) = (x-k)(x-k-1) \) for \( x-k \in (0,1] \), we get:
\[
f(x) = 2^k (x-k)(x-k-1)
\]
4. **Analyzing \( f(x) \) for \( x \in (1,2] \):**
- For \( x \in (1,2] \):
\[
f(x) = 2 \cdot (x-1)(x-2)
\]
- The minimum value in this interval is at \( x = \frac{3}{2} \):
\[
f\left(\frac{3}{2}\right) = 2 \left(\frac{3}{2} - 1\right) \left(\frac{3}{2} - 2\right) = 2 \cdot \frac{1}{2} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{2}
\]
- Therefore, \( f(x) \geq -\frac{1}{2} \) for \( x \in (1,2] \).
5. **Analyzing \( f(x) \) for \( x \in (2,3] \):**
- For \( x \in (2,3] \):
\[
f(x) = 4 \cdot (x-2)(x-3)
\]
- The minimum value in this interval is at \( x = \frac{5}{2} \):
\[
f\left(\frac{5}{2}\right) = 4 \left(\frac{5}{2} - 2\right) \left(\frac{5}{2} - 3\right) = 4 \cdot \frac{1}{2} \cdot \left(-\frac{1}{2}\right) = -1
\]
- Therefore, \( f(x) \in [-1, 0] \) for \( x \in (2,3] \).
6. **Finding \( m \) such that \( f(x) \geq -\frac{8}{9} \):**
- We need to solve \( 4(x-2)(x-3) = -\frac{8}{9} \):
\[
4(x-2)(x-3) = -\frac{8}{9}
\]
\[
36(x-2)(x-3) = -8
\]
\[
9(x-2)(x-3) = -2
\]
\[
9(x^2 - 5x + 6) = -2
\]
\[
9x^2 - 45x + 54 = -2
\]
\[
9x^2 - 45x + 56 = 0
\]
- Solving this quadratic equation:
\[
x = \frac{45 \pm \sqrt{2025 - 2016}}{18} = \frac{45 \pm 3}{18}
\]
\[
x_1 = \frac{48}{18} = \frac{8}{3}, \quad x_2 = \frac{42}{18} = \frac{7}{3}
\]
- Therefore, \( f(x) \geq -\frac{8}{9} \) for \( 2 < x \leq \frac{7}{3} \).
7. **Conclusion:**
- The greatest real number \( m \) for which \( f(x) \geq -\frac{8}{9} \) for all \( x \in (-\infty, m] \) is \( m = \frac{7}{3} \).
The final answer is \( \boxed{\frac{7}{3}} \)
|
5537.
|
Problem: 18.3.19 $\star \star$ Find all positive integer triples $(a, b, c)$ that satisfy $a^{2}+b^{2}+c^{2}=2005$ and $a \leqslant b \leqslant c$.
Answer: Given $44<\sqrt{2005}<45,25<\sqrt{\frac{2005}{3}}<26$, so
$$
26 \leqslant c \leqslant 44 .
$$
$a^{2}+b^{2}=0(\bmod 3)$, if and only if $a=b=0(\bmod 3)$. Therefore, $3 \mid a^{2}+b^{2}$ implies $9 \mid a^{2}+b^{2}$.
If $3 \nmid c$, then $c^{2} \equiv 1(\bmod 3), a^{2}+b^{2}=2005-c^{2} \equiv 0(\bmod 3)$, thus $9 \mid a^{2}+b^{2}, c^{2} \equiv 2005 \equiv 7(\bmod 9), c \equiv 4,5(\bmod 9)$.
Among (1), the only values of $c$ that satisfy $3 \mid c$ or $c \equiv 4,5(\bmod 9)$ are
$$
\begin{aligned}
c & =27,30,33,36,39,42,31,32,40,41 . \\
2005-27^{2} & =1276=4 \times 319, \\
2005-39^{2} & =484=4 \times 121, \\
2005-41^{2} & =324=4 \times 81,
\end{aligned}
$$
$319,121,81$ cannot be expressed as the sum of two positive integers squared $\left(319 \equiv 3 \neq a^{2}+b^{2}(\bmod 4) .121\right.$ and 81 can be easily verified). And $4 \mid a^{2}+b^{2}$ implies $a, b$ are both even. But $\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}$ cannot equal $319,121,81$, so $a^{2}+b^{2} \neq 4 \times 319,4 \times 121,4 \times 51$.
$$
\begin{array}{l}
2005-30^{2}=1105=529+576=23^{2}+24^{2}, \\
2005-31^{2}=1044=144+900=12^{2}+30^{2}, \\
2005-32^{2}=981=81+900=9^{2}+30^{2}, \\
2005-33^{2}=916=16+900=4^{2}+30^{2}, \\
2005-36^{2}=709=225+484=15^{2}+22^{2}, \\
2005-40^{2}=405=81+324=9^{2}+18^{2}, \\
2005-42^{2}=241=16+225=4^{2}+15^{2} .
\end{array}
$$
Thus, there are 7 solutions to this problem, which are
$$
(a, b, c)=(23,24,30),(12,30,31),(9,30,32),(4,30,33),(15,22,
$$
|
9916.
|
Problem: 5. Through the vertex $M$ of some angle, a circle is drawn, intersecting the sides of the angle at points $N$ and $K$, and the bisector of this angle at point $L$. Find the sum of the lengths of segments $M N$ and $M K$, if the area of $M N L K$ is 49, and the angle $L M N$ is $30^{\circ}$.
Answer: Answer: $14 \sqrt[4]{3}$.
## Solution.
Let $L P$ and $L Q$ be the perpendiculars to $M N$ and $M K$ respectively.
Point $\mathrm{L}$ lies on the angle bisector and is therefore equidistant from the sides of the angle, which means $L P = L Q$.
Right triangles $N P L$ and $L Q K$ are congruent (by leg and hypotenuse), as well as right triangles $M P L$ and $M Q L$ (similarly, by leg and hypotenuse).
On one hand,
$$
M N + M K = M P + P N + M K = M P + K Q + M K = M P + M Q = 2 M P
$$
On the other hand,
$$
49 = S_{M N L K} = S_{M P L} + S_{M Q L} = 2 S_{M P L} = M P \cdot P L = M P^2 \cdot \operatorname{tg} 30^{\circ} = \frac{M P^2}{\sqrt{3}} \Rightarrow M P = 7 \sqrt[4]{3}
$$
Thus, we have $\quad M N + M K = 2 M P = 14 \sqrt[4]{3}$.
## Grading criteria are provided in the table:
| Points | Grading Criteria |
| :---: | :--- |
| $\mathbf{7}$ | Complete and justified solution. |
| $\mathbf{6}$ | Justified solution with minor flaws. |
| $\mathbf{5 - 6}$ | The solution contains minor errors, gaps in justification, but is generally correct and can become fully correct with minor corrections or additions. |
| $\mathbf{4}$ | The problem is more solved than not, for example, one of the two (more complex) significant cases is correctly considered. |
| $\mathbf{2 - 3}$ | The problem is not solved, but formulas, diagrams, considerations, or auxiliary statements relevant to the solution of the problem are provided. |
| $\mathbf{1}$ | The problem is not solved, but an attempt at a solution has been made, for example, individual (partial) cases have been considered in the absence of a solution or in the case of an incorrect solution. |
| $\mathbf{0}$ | The solution is absent or does not meet any of the criteria listed above. |
|
11684.
|
Problem: 4. Let $A$ and $B$ be $n$-digit numbers, where $n$ is odd, which give the same remainder $r \neq 0$ when divided by $k$. Find at least one number $k$, which does not depend on $n$, such that the number $C$, obtained by appending the digits of $A$ and $B$, is divisible by $k$.
Answer: Solution. Let $A=k a+r, B=k b+r, 0<r<k$; then we have:
$$
C=10^{n} A+B=10^{n}(k a+r)+(k b+r)=k\left(10^{n} a+b\right)+\left(10^{n}+1\right) r .
$$
The number $C$ is divisible by $k$ if and only if the number $10^{n}+1$ is divisible by $k$. Since $n=2 m+1$, we have
$$
10^{n}+1=10^{2 m+1}+1=(10+1)\left(10^{2 m}-10^{m-1}+\ldots+1\right)=11 D
$$
which means that a number $k$ that satisfies the condition of the problem is the number 11.
## 4th year
|
2238.
|
Problem: 7. Given that $z$ is a complex number, and $|z|=1$. When $\mid 1+z+$ $3 z^{2}+z^{3}+z^{4}$ | takes the minimum value, the complex number $z=$ $\qquad$ or . $\qquad$
Answer: 7. $-\frac{1}{4} \pm \frac{\sqrt{15}}{4}$ i.
Notice,
$$
\begin{array}{l}
\left|1+z+3 z^{2}+z^{3}+z^{4}\right| \\
=\left|\frac{1}{z^{2}}+\frac{1}{z}+3+z+z^{2}\right| \\
=\left|\left(z+\frac{1}{z}\right)^{2}+z+\frac{1}{z}+1\right| \\
=\left|(z+\bar{z})^{2}+z+\bar{z}+1\right| \\
=\left|4 \operatorname{Re}^{2} z+2 \operatorname{Re} z+1\right| \\
\geqslant \frac{16-4}{16}=\frac{3}{4},
\end{array}
$$
Here, $\operatorname{Re} z=-\frac{1}{4}$.
Thus, $z=-\frac{1}{4} \pm \frac{\sqrt{15}}{4}$ i.
|
4426.
|
Problem: Example: Given the radii of the upper and lower bases of a frustum are 3 and 6, respectively, and the height is $3 \sqrt{3}$, the radii $O A$ and $O B$ of the lower base are perpendicular, and $C$ is a point on the generatrix $B B^{\prime}$ such that $B^{\prime} C: C B$ $=1: 2$. Find the shortest distance between points $A$ and $C$ on the lateral surface of the frustum.
Answer: Given $B^{\prime} D \perp O B$, then $B^{\prime} D=3 \sqrt{3}$.
$$
\begin{array}{l}
\because O^{\prime} B^{\prime}=3, O B=6, \\
\therefore B^{\prime} B=\sqrt{(3 \sqrt{3})^{2}+(6-3)^{2}}=6,
\end{array}
$$
and $O^{\prime} B^{\prime}: O B=3: 6=1: 2$,
$\therefore$ In the development diagram, $B^{\prime}$ is the midpoint of $V B$, $V B=12$. Given $B^{\prime} C: C B=1: 2$,
$$
\begin{array}{c}
\therefore B^{\prime} C=\frac{1}{3} B^{\prime} \cdot B=2, \\
V C=6+2=8 .
\end{array}
$$
Let the central angle of the sector ring in the frustum's lateral development be $\theta$.
$$
\begin{array}{l}
\theta=\frac{6-3}{6} \times 360^{\circ}=180^{\circ}, \\
\alpha=\frac{1}{4} \theta=45^{\circ} .
\end{array}
$$
In $\triangle V A C$,
$$
\begin{aligned}
A C^{2} & =V A^{2}+V C^{2}-2 V A \cdot V C \cos \alpha \\
& =12^{2}+8^{2}-2 \times 12 \times 8 \times \cos 45^{\circ} \\
& =208-96 \sqrt{2}, \\
\therefore A C & =4 \sqrt{13-6 \sqrt{2}} .
\end{aligned}
$$
|
5386.
|
Problem: 5. Given positive real numbers $a$ and $b$ satisfy $a+b=1$, then $M=$ $\sqrt{1+a^{2}}+\sqrt{1+2 b}$ the integer part is
Answer: 5. 2
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3225.
|
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