Divide the sides of the unit square $ABCD$ into 5 equal parts. The 2nd division point from $A$ on the side $AB$ is denoted by $D'$, and similarly, the 2nd division point from $B$ on the side $BC$, from $C$ on the side $CD$, and from $D$ on the side $DA$ are denoted by $A'$, $B'$, and $C'$, respectively. The lines $AA'$, $BB'$, $CC'$, and $DD'$ form a quadrilateral.

What is the area of the quadrilateral?

Solution:
We denote the vertices of the resulting quadrilateral as $E, F, G, H$, and the center of the original square as $O$. A $90^{\circ}$ rotation around $O$ maps the points $A, B, C, D, A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ to $B, C, D, A, B^{\prime}, C^{\prime}, D^{\prime}, A^{\prime}$, respectively. For example, the intersection point $F$ of segments $A A^{\prime}$ and $B B^{\prime}$ is mapped to the intersection point $G$ of segments $B B^{\prime}$ and $C C^{\prime}$. Similarly, it can be shown that under this rotation, the points $G, H, E$ are mapped to $H, E$, and $F$, respectively, meaning that the quadrilateral $E F G H$ is mapped onto itself. This implies that $E F G H$ is a square, and its center is $O$.

![](https://cdn.mathpix.com/cropped/2024_05_02_95ab49d400f6c2b435b9g-1.jpg?height=485&width=499&top_left_y=392&top_left_x=802)

To calculate the area of the square, it is sufficient to determine the length of one of its sides. In the right triangle $A B A^{\prime}$, $B F$ is the altitude to the hypotenuse $A A^{\prime}$ (since $\angle E F G = 90^{\circ}$), so the triangles $A B A^{\prime}$, $A F B$, and $B F A^{\prime}$ are similar. Therefore, the ratios of corresponding sides of these triangles are equal:

$$
\frac{F A}{A B} = \frac{A B}{A A^{\prime}}
$$

that is, using the fact that by the Pythagorean theorem $A A^{\prime} = \sqrt{1^2 + \left(\frac{2}{5}\right)^2}$,

$$
F A = \frac{1 \cdot 1}{\sqrt{\frac{29}{25}}} = \frac{5 \sqrt{29}}{29}
$$

On the other hand:

$$
\frac{F B}{B A^{\prime}} = \frac{A B}{A A^{\prime}}
$$

so

$$
F B = \frac{\frac{2}{5} \cdot 1}{\sqrt{\frac{29}{25}}} = \frac{2 \sqrt{29}}{29}
$$

But $F B = A E$, so the side length of the small square is:

$$
E F = F A - F B = \frac{3 \sqrt{29}}{29}
$$

Thus, the area of the small square is $\frac{9}{29}$.

Remark. The more general problem, where each side of the original square is divided into $n$ equal parts and the $k$-th division point is chosen on each side, can be solved in a similar manner. In this case, the area of the small square is:

$$
T = \frac{(n-k)^2}{n^2 + k^2}
$$

Answer: \frac{9}{29}


Calculate the definite integral:

$$
\int_{0}^{2 \operatorname{arctan} \frac{1}{2}} \frac{1+\sin x}{(1-\sin x)^{2}} d x
$$

Solution:
## Solution

We will use the universal substitution:

$$
t=\operatorname{tg} \frac{x}{2}
$$

From which:

$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=2 \operatorname{arctg} \frac{1}{2} \Rightarrow t=\operatorname{tg} \frac{2 \operatorname{arctg} \frac{1}{2}}{2}=\operatorname{tg}\left(2 \operatorname{arctg} \frac{1}{2}\right)=\frac{1}{2}
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{0}^{2 \operatorname{arctg} \frac{1}{2}} \frac{1+\sin x}{(1-\sin x)^{2}} d x=\int_{0}^{\frac{1}{2}} \frac{\left(1+\frac{2 t}{1+t^{2}}\right) \cdot \frac{2 d t}{1+t^{2}}}{\left(1-\frac{2 t}{1+t^{2}}\right)^{2}}=\int_{0}^{\frac{1}{2}} \frac{\frac{1+t^{2}+2 t}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{\left(\frac{1+t^{2}-2 t}{1+t^{2}}\right)^{2}}= \\
& =2 \cdot \int_{0}^{\frac{1}{2}} \frac{(1+t)^{2} d t}{(1-t)^{4}}=
\end{aligned}
$$

Substitution:

$$
\begin{aligned}
& u=1-t, d t=-d u \\
& t=0 \Rightarrow u=1 \\
& t=\frac{1}{2} \Rightarrow u=\frac{1}{2}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =-2 \cdot \int_{1}^{\frac{1}{2}} \frac{(2-u)^{2}}{u^{4}} d u=-2 \cdot \int_{1}^{\frac{1}{2}} \frac{4-4 u+u^{2}}{u^{4}} d u= \\
& =2 \cdot \int_{1}^{\frac{1}{2}}\left(\frac{4}{u^{4}}-\frac{4}{u^{3}}+\frac{1}{u^{2}}\right) d u=-\left.2 \cdot\left(-\frac{4}{3 u^{3}}+\frac{2}{u^{2}}-\frac{1}{u}\right)\right|_{1} ^{\frac{1}{2}}= \\
& -2\left(-\frac{4}{3\left(\frac{1}{2}\right)^{3}}+\frac{2}{\left(\frac{1}{2}\right)^{2}}-\frac{1}{\left(\frac{1}{2}\right)}\right)+2\left(-\frac{4}{3 \cdot 1^{3}}+\frac{2}{1^{2}}-\frac{1}{1}\right)= \\
& =-2\left(-\frac{32}{3}+8-2\right)+2\left(-\frac{4}{3}+2-1\right)=\frac{64}{3}-16+4-\frac{8}{3}+4-2= \\
& =\frac{56}{3}-10=\frac{26}{3}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+8-14$ »

Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

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- Last edited: 10:34, May 1, 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals $8-15$

## Material from Plusi

Answer: \frac{26}{3}


For each real number $x$, let $f(x)$ be the minimum of the numbers $4x+1$, $x+2$, and $-2x+4$. Determine the maximum value of $6f(x)+2012$.

Solution:
21. Answer: 2028
Let $L_{1}, L_{2}, L_{3}$ represent the three lines $y=4 x+1, y=x+2$ and $y=-2 x+4$ respectively.
Observe that $L_{1}$ and $L_{2}$ intersects at $\left(\frac{1}{3}, \frac{7}{3}\right), L_{1}$ and $L_{3}$ intersects at $\left(\frac{1}{2}, 3\right)$, and $L_{2}$ and $L_{3}$ intersects at $\left(\frac{2}{3}, \frac{8}{3}\right)$. Thus
$$
f(x)=\left\{\begin{array}{ll}
4 x+1, & x\frac{2}{3} .
\end{array}\right.
$$

Thus the maximum value of $f(x)$ is $\frac{8}{3}$ and the maximum value of $6 f(x)+2012$ is 2028.

Answer: 2028


Given the parabola $y=ax^2+bx+c$ has a line of symmetry at $x=-2$, it is tangent to a certain line at one point, this line has a slope of 2, and a y-intercept of 1, and the parabola intersects the $y=0$ at two points, the distance between which is $2\sqrt{2}$. Try to find the equation of this parabola.

Solution:
[Solution] (1) $y=a x^{2}+b x+c=a\left(x+\frac{b}{2 a}\right)^{2}+c-\frac{b^{2}}{4 a}$, given that $x=-2$ is the axis of symmetry, we get $\frac{b}{2 a}=2$, hence $b=4 a$.
(2) $y=a x^{2}+b x+c$ is tangent to the line $l$,

and $l: y=2 x+1$, so the parabola intersects $l$ at one point,
thus we have $a x^{2}+4 a x+c=2 x+1$,
which simplifies to $a x^{2}+2(2 a-1) x+(c-1)=0$.
and $(2 a-1)^{2}-a(c-1)=0$,
hence $c=\frac{(2 a-1)^{2}+a}{a}$.
(3) $y=a x^{2}+4 a x+\frac{(2 a-1)^{2}+a}{a}$

The intersection points with $y=0$ are given by $a x^{2}+4 a x+\frac{(2 a-1)^{2}+a}{a}=0$,
solving this gives $x_{1,2}=-2 \pm \sqrt{4-\frac{(2 a-1)^{2}+a}{a^{2}}}$.
Given $x_{1}-x_{2}=2 \sqrt{2}$, we get
$$\sqrt{4-\frac{(2 a-1)^{2}+a}{a^{2}}}=\sqrt{2}$$

which simplifies to $4-\frac{(2 a-1)^{2}+a}{a^{2}}=2$,
$$(2 a-1)^{2}+a=2 a^{2}$$

which further simplifies to $2 a^{2}-3 a+1=0,(2 a-1)(a-1)=0$.
Thus, $a=1$ or $a=\frac{1}{2}$.
When $a=1$, the required parabola is

i.e.,
$$y=x^{2}+4 x+2$$
$$y=(x+2)^{2}-2 .$$

When $a=\frac{1}{2}$, the required parabola is
$$y=\frac{1}{2} x^{2}+2 x+1$$

i.e., $y=\frac{1}{2}(x+2)^{2}-1$.

Answer: y=x^{2}+4 x+2 \text{ or } y=\frac{1}{2} x^{2}+2 x+1


Let $x_{i} \geqslant 0(1 \leqslant i \leqslant n, n \geqslant 4), \sum_{i=1}^{n} x_{i}=1$, find the maximum value of $F=\sum_{i=1}^{n} x_{i} x_{i+1}$.

Solution: 2. By the method similar to the above, the maximum value of $F$ can be obtained as $F\left(\frac{1}{2}, \frac{1}{2}, 0,0, \cdots, 0\right)=\frac{1}{4}$.

Answer: \frac{1}{4}


ÃÂ± > 0 is real and n is a positive integer. What is the maximum possible value of ÃÂ± a + ÃÂ± b + ÃÂ± c + ... , where a, b, c, ... is any sequence of positive integers with sum n?

Solution: If α ≤ 1, then we want as many terms as possible and we want each exponent as small as possible. So we must take a = b = c = .... = 1 giving sum nα. If α > 1, then α(α a-1 - 1)(α b-1 ) > 1 for a, b > 1 and hence α a + α b  1. So a maximal expression has at most one exponent greater than 1, in other words it has the form mα + α n-m . Note that α n-1 (α - 1) is an increasing function of n. So for some N(α) we have α k-1 (α - 1) ≥ 1, and hence α + α k ≤ α k+1 iff k ≥ N(α). Hence the maximum value of mα + α n-m as m varies occurs at m = 0 or m = n. Thus the maximum value is the greater of nα and α n . If α ≥ 2, then 2α ≤ α 2 and nα  2, so the maximum is always α n . For 1 < α < 2, the maximum is nα if α < n 1/(n-1) and α n for α ≥ n 1/(n-1) . 8th Chinese 1993 © John Scholes jscholes@kalva.demon.co.uk 25 June 2002

Answer: \max(n\alpha,\alpha^n)


Let $k$ be an integer. If the equation $(x-1)|x+1|=x+\frac{k}{2020}$ has three distinct real roots, how many different possible values of $k$ are there?

Solution:
1. **Substitute \( \frac{k}{2020} = m \):**
   \[
   (x-1)|x+1| = x + m
   \]

2. **Case 1: \( x < -1 \):**
   \[
   |x+1| = -x-1
   \]
   The equation becomes:
   \[
   (x-1)(-x-1) = x + m \implies -x^2 - x + 1 = x + m \implies x^2 + 2x + (m-1) = 0
   \]
   The discriminant of this quadratic equation is:
   \[
   \Delta_1 = 4 - 4(m-1) = 5 - 4m
   \]
   For real roots, we need:
   \[
   \Delta_1 \geq 0 \implies 5 - 4m \geq 0 \implies m \leq \frac{5}{4} \quad \text{(Condition 1)}
   \]
   For the roots \( x_1 \) and \( x_2 \) of this quadratic equation:
   \[
   x_1 + x_2 = -2 \quad \text{and} \quad x_1 x_2 = m-1
   \]
   Since \( x_1 < -1 \) and \( x_2 < -1 \), we need:
   \[
   \frac{-2 - \sqrt{5-4m}}{2} < -1 \implies \sqrt{5-4m} > 1 \implies 5 - 4m > 1 \implies m < 1 \quad \text{(Condition 2)}
   \]

3. **Case 2: \( x \geq -1 \):**
   \[
   |x+1| = x+1
   \]
   The equation becomes:
   \[
   (x-1)(x+1) = x + m \implies x^2 - 1 = x + m \implies x^2 - x - (m+1) = 0
   \]
   The discriminant of this quadratic equation is:
   \[
   \Delta_2 = 1 + 4(m+1) = 4m + 5
   \]
   For distinct real roots, we need:
   \[
   \Delta_2 > 0 \implies 4m + 5 > 0 \implies m > -\frac{5}{4} \quad \text{(Condition 3)}
   \]
   For the roots \( x_3 \) and \( x_4 \) of this quadratic equation:
   \[
   x_3 + x_4 = 1 \quad \text{and} \quad x_3 x_4 = -(m+1)
   \]
   Since \( x_3 \geq -1 \) and \( x_4 \geq -1 \), we need:
   \[
   \frac{1 - \sqrt{4m+5}}{2} \geq -1 \implies \sqrt{4m+5} \leq 3 \implies 4m + 5 \leq 9 \implies m \leq 1 \quad \text{(Condition 4)}
   \]

4. **Combine Conditions:**
   \[
   -\frac{5}{4} < m < 1
   \]
   Converting back to \( k \):
   \[
   -\frac{5}{4} < \frac{k}{2020} < 1 \implies -2525 < k < 2020
   \]
   Since \( k \) is an integer:
   \[
   k \in \{-2524, -2523, \ldots, 2019\}
   \]
   The number of integers in this range is:
   \[
   2019 - (-2524) + 1 = 4544
   \]

The final answer is \(\boxed{4544}\)

Answer: 4544


Let $[x]$ denote the greatest integer not exceeding $x$, then $S=[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\cdots+[\sqrt{99}]$, find the value of $[\sqrt{S}]$.

Solution: For $k^{2} \leqslant i \leqslant k^{2}+2 k$, $[\sqrt{i}]=k, k=1,2, \cdots, 9$, thus $S=\sum_{i=1}^{99}[\sqrt{i}]=\sum_{k=1}^{9} k(2 k+1)=2 \sum_{k=1}^{9} k^{2}+\sum_{k=1}^{9} k=\frac{9 \cdot 10 \cdot 19}{3}+\frac{9 \cdot 10}{2}=615$. Therefore, $[\sqrt{S}]=[\sqrt{615}]=24$.

Answer: 24


If the odd function $y=f(x)$ defined on $\mathbf{R}$ is symmetric about the line $x=1$, and when $0<x \leqslant 1$, $f(x)=\log _{3} x$, then the sum of all real roots of the equation $f(x)=-\frac{1}{3}+f(0)$ in the interval $(0,10)$ is $\qquad$

Solution:
(5) 30 Hint: Given that the graph of the function $y=f(x)$ is symmetric about the line $x=1$, and $f(x)$ is an odd function, we have
$$
f(x+2)=f(-x)=-f(x),
$$

Therefore,
$$
f(x+4)=-f(x+2)=f(x),
$$
$f(x)$ is a periodic function, and 4 is one of its periods.
Since $f(x)$ is an odd function defined on $\mathbf{R}$, we know $f(0)=0$, and the equation $f(x)=-\frac{1}{3}+f(0)$ simplifies to $f(x)=-\frac{1}{3}$.

From the graph, we can see that $f(x)=-\frac{1}{3}$ has one real root in each of the intervals $(0,1)$ and $(1,2)$, and the sum of these two roots is $2$; $f(x)=-\frac{1}{3}$ has one real root in each of the intervals $(4,5)$ and $(5,6)$, and the sum of these two roots is $10$; $f(x)=-\frac{1}{3}$ has one real root in each of the intervals $(8,9)$ and $(9,10)$, and the sum of these two roots is $18$.

Therefore, the equation $f(x)=-\frac{1}{3}+f(0)$ has 6 distinct real roots in the interval $(0,10)$, and the sum of these 6 roots is 30.

Answer: 30


Given that $O$ is the circumcenter of acute $\triangle A B C$, $A B=6, A C=10$, if $\overrightarrow{A O}=x \overrightarrow{A B}+y \overrightarrow{A C}$, and $2 x+10 y=5$, then $\cos \angle B A C=$ $\qquad$ .

Solution:
(4) $\frac{1}{3}$ Hint: Since $O$ is the circumcenter of $\triangle A B C$, we have
$$
\begin{array}{l}
\overrightarrow{A O} \cdot \overrightarrow{A C}=|\overrightarrow{A O}| \cdot|\overrightarrow{A C}| \cdot \cos \angle O A C=\frac{1}{2}|\overrightarrow{A C}|^{2}=50 . \\
\text { Also, } \overrightarrow{A O}=x \overrightarrow{A B}+y \overrightarrow{A C} \text {, so } \\
\quad(x \overrightarrow{A B}+y \overrightarrow{A C}) \cdot \overrightarrow{A C}=50,
\end{array}
$$

which means
$$
x \cdot|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cdot \cos \angle B A C+y \cdot|\overrightarrow{A C}|^{2}=50 \text {, }
$$

or
$$
60 x \cdot \cos \angle B A C+100 y=50 .
$$

Since $\triangle A B C$ is an acute triangle, it is easy to see that $x \neq 0$, so
$$
\cos \angle B A C=\frac{5-10 y}{6 x}=\frac{2 x}{6 x}=\frac{1}{3} .
$$

Answer: \frac{1}{3}


Let $K$ be the incenter of $\triangle ABC$, and let points $C_{1}$ and $B_{1}$ be the midpoints of sides $AB$ and $AC$, respectively. The line $AC$ intersects $C_{1}K$ at point $B_{2}$, and the line $AB$ intersects $B_{1}K$ at point $C_{2}$. If the area of $\triangle AB_{2}C_{2}$ is equal to the area of $\triangle ABC$, find the size of $\angle CAB$.

Solution:
18.7 Let the inradius of $\triangle ABC$ be $r$, and the semiperimeter be $p$. The area of $\triangle AC_1B_2 = \frac{1}{2} AC_1 \cdot AB_2 \sin A$; the area of $\triangle ABC = \frac{1}{2} AC \cdot AB \sin A$. Using the area relationship $\triangle AC_1B_2 - \triangle AKB_2 = \triangle AKC_1$, we can derive: $\frac{AB_2 \cdot r p}{2 b} - \frac{AB_2 \cdot r}{2} = \frac{c r}{4}$, from which we can get $AB_2(a-b+c) = bc$ (1); similarly, $AC_2 \cdot (a+b-c) = bc$ (2).
Using the known area relationship, we get $AB_2 \cdot AC_2 = bc$.
Multiplying (1) and (2) yields $(a-b+c)(a+b-c) = bc$, from which we can calculate $\angle CAB = 60^{\circ}$.

Answer: 60


Person A and Person B start from the same point A on a circular track, running in opposite directions. Their speeds are 4 meters per second and 6 meters per second, respectively. If they start at the same time and end when they meet again at point A, then they meet a total of $\qquad$ times from the start to the end.

Solution: answer: 5

Answer: 5


Let trapezium have bases $A B$ and $C D$, measuring $5 \mathrm{~cm}$ and $1 \mathrm{~cm}$, respectively. Let $M$ and $N$ be points on $A D$ and $B C$ such that the segment $M N$ is parallel to the base $A B$, and the area of quadrilateral $A B N M$ is twice the area of quadrilateral $C D M N$. How many centimeters does the segment $M N$ measure?

Solution:
II/3. Let the length of $M N$ be $x$, the height of trapezoid $A B C D$ be $v$, and the height of trapezoid $A B N M$ be $h$. Then, $S_{A B N M}=\frac{(x+5)}{2} \cdot h$ and $S_{C D M N}=\frac{x+1}{2} \cdot(v-h)$. Since $S_{A B N M}=2 S_{C D M N}$ and $S_{A B N M}+S_{C D M N}=S_{A B C D}$, we get the equations

$$
\begin{aligned}
\frac{(x+5)}{2} \cdot h & =2 \frac{x+1}{2} \cdot(v-h) \\
\frac{(x+5)}{2} \cdot h+\frac{x+1}{2} \cdot(v-h) & =\frac{1+5}{2} v
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_e3e00bf556075403c4d7g-08.jpg?height=403&width=588&top_left_y=444&top_left_x=1271)

From the first equation, we express $h=\frac{x+1}{3 x+7} \cdot 2 v$ and substitute it into the second. The resulting equation can be divided by $v$, and a brief calculation shows that $x^{2}=9$ or $x=3$. Therefore, the segment $M N$ measures $3 \mathrm{~cm}$.

If the contestant expresses $S_{A B N M}$ and $S_{C D M N}$ in terms of $x, v$, and $h$ (or other dependent quantities): $\mathbf{1 + 1}$ point. If the contestant expresses the conditions $S_{A B N M}=2 S_{C D M N}$ and $S_{A B N M}+S_{C D M N}=S_{A B C D}$ in terms of $x, v$, and $h: 1+1$ point. Solution $x=3: 3$ points.

Answer: 3


The numbers $2^{n}$ and $5^{n}$ start with the digit $a$. What is $a$?

Solution: 21.3. Answer: 3. By the condition $a \cdot 10^{p}<2^{n}<(a+1) 10^{p}$ and $a \cdot 10^{q}<5^{n}<$ $<(a+1) 10^{q}$. Therefore, $a^{2} 10^{p+q}<10^{n}<(a+1)^{2} 10^{p+q}$, i.e., $a^{2}<10^{n-p-q}<$ $<(a+1)^{2}$. At the same time, $(a+1)^{2} \leqslant 100$. Thus, $a^{2}<10<(a+1)^{2}$, i.e., $a=3$. The number 3 is the leading digit of the numbers $2^{5}$ and $5^{5}$.

Answer: 3


Sven is saving for a bicycle. It costs 330 marks. If he has saved six times as much money as he already has, he can buy the bike.

How much money has Sven already saved?

Solution: Sven has already saved 55 marks, because $6 \cdot 55=330$.

Answer: 55


From the system of equations
$$\left\{\begin{array}{l}
a=\cos u+\cos v+\cos w \\
b=\sin u+\sin v+\sin w \\
c=\cos 2 u+\cos 2 v+\cos w \\
d=\sin 2 u+\sin 2 v+\sin 2 w
\end{array}\right.$$

eliminate \( u, v, w \).

Solution:
[Solution] From the system of equations, we have
$$\begin{array}{l}
a^{2}+b^{2}=3+2[\cos (u-v)+\cos (v-w)+\cos (w-u)], \\
a^{2}-b^{2}=C+2[\cos (u+v)+\cos (v+w)+\cos (w+u)], \\
2 a b=d+2[\sin (u+v)+\sin (v+w)+\sin (w+u)],
\end{array}$$

Therefore, $\left(\frac{a^{2}-b^{2}-c}{2}\right)^{2}+\left(\frac{2 a b-d}{2}\right)^{2}$
$$\begin{array}{l}
=3+2[\cos (u-v)+\cos (v-w)+\cos (w-u)] \\
=a^{2}+b^{2}
\end{array}$$

That is, $\left(a^{2}-b^{2}-c\right)^{2}+(2 a b-d)^{2}=4\left(a^{2}+b^{2}\right)$.

Answer: \left(a^{2}-b^{2}-c\right)^{2}+(2 a b-d)^{2}=4\left(a^{2}+b^{2}\right)


Given the triangle connecting $A(-3,0), B(0,-3)$, $C\left(\frac{15}{7}, \frac{24}{7}\right)$ and the circle $x^{2}+y^{2}=R^{2}(R> 0 )$ always have common points. Then the range of the circle radius $R$ is
(A) $\left(0, \frac{3 \sqrt{10}}{10}\right) \cup \left(\frac{3 \sqrt{89}}{7},+\infty\right)$
(B) $\left(\frac{3 \sqrt{10}}{10}, \frac{3 \sqrt{89}}{7}\right)$
(C) $\left(0, \frac{3 \sqrt{2}}{2}\right) \cup(3,+\infty)$
(D) $\left(\frac{3 \sqrt{2}}{2}, 3\right)$

Solution:
-1 (B).
As shown in Figure 11, from the given information, the equations of the lines on which the three sides lie are
$$
\begin{array}{l}
A B: y=-x-3, \\
A C: y=\frac{2}{3}(x+3), \\
B C: y=3(x-1) .
\end{array}
$$

Further, the distances from the center of the circle to the three vertices and the three sides can be calculated as
$$
\begin{array}{c}
|O A|=3,|O B|= \\
3,|O C|=\frac{3 \sqrt{89}}{7} ; \\
d_{A B}=\frac{3}{\sqrt{2}}, d_{A C}=\frac{6}{\sqrt{13}}, d_{B C}=\frac{3}{\sqrt{10}} .
\end{array}
$$

When $d_{B C} \leqslant R \leqslant|O C|$, the triangle and the circle must have common points, so the range of $R$ is $\left(\frac{3 \sqrt{10}}{10}, \frac{3 \sqrt{89}}{7}\right)$. $)$

Answer: B


Let $n$ be a positive integer and consider an arrangement of $2n$ blocks in a straight line, where $n$ of them are red and the rest blue. A swap refers to choosing two consecutive blocks and then swapping their positions. Let $A$ be the minimum number of swaps needed to make the first $n$ blocks all red and $B$ be the minimum number of swaps needed to make the first $n$ blocks all blue. Show that $A+B$ is independent of the starting arrangement and determine its value.

Solution:
1. **Define the Problem and Variables:**
   - Let \( n \) be a positive integer.
   - Consider an arrangement of \( 2n \) blocks in a straight line, where \( n \) of them are red and the rest are blue.
   - Define \( A \) as the minimum number of swaps needed to make the first \( n \) blocks all red.
   - Define \( B \) as the minimum number of swaps needed to make the first \( n \) blocks all blue.

2. **Define Reddy and Bluey Pairs:**
   - A pair of distinct blocks is called *reddy* if the block nearer to the left is red and the block nearer to the right is blue.
   - A pair of distinct blocks is called *bluey* if the block nearer to the left is blue and the block nearer to the right is red.
   - Note that there are \( n^2 \) *reddy* or *bluey* pairs of blocks. This is because there are \( n \) red blocks and \( n \) blue blocks, and each red block can pair with each blue block exactly once.

3. **Swapping Mechanism:**
   - When we swap two consecutive blocks, we can remove at most one *reddy* pair (which occurs when the two blocks we swap are a *reddy* pair).
   - Similarly, we can remove at most one *bluey* pair when the two blocks we swap are a *bluey* pair.

4. **Counting Reddy and Bluey Pairs:**
   - To make the first \( n \) blocks all red, we need to remove all *bluey* pairs. Hence, \( A \) is the number of *bluey* pairs at the beginning.
   - To make the first \( n \) blocks all blue, we need to remove all *reddy* pairs. Hence, \( B \) is the number of *reddy* pairs at the beginning.

5. **Summing the Minimum Swaps:**
   - Since there are \( n^2 \) pairs in total and each pair is either *reddy* or *bluey*, the sum of the number of *reddy* pairs and *bluey* pairs is always \( n^2 \).
   - Therefore, \( A + B = n^2 \).

\[
A + B = n^2
\]

The final answer is \(\boxed{n^2}\).

Answer: n^2


Find the length of the curve expressed by the polar equation: $ r=1+\cos \theta \ (0\leq \theta \leq \pi)$.

Solution:
To find the length of the curve given by the polar equation \( r = 1 + \cos \theta \) for \( 0 \leq \theta \leq \pi \), we use the formula for the length of a curve in polar coordinates:

\[
L = \int_{\alpha}^{\beta} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta
\]

1. **Calculate \(\frac{dr}{d\theta}\):**

\[
r = 1 + \cos \theta
\]

\[
\frac{dr}{d\theta} = -\sin \theta
\]

2. **Substitute \( r \) and \(\frac{dr}{d\theta}\) into the length formula:**

\[
L = \int_{0}^{\pi} \sqrt{ \left( -\sin \theta \right)^2 + (1 + \cos \theta)^2 } \, d\theta
\]

3. **Simplify the integrand:**

\[
L = \int_{0}^{\pi} \sqrt{ \sin^2 \theta + (1 + \cos \theta)^2 } \, d\theta
\]

\[
= \int_{0}^{\pi} \sqrt{ \sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta } \, d\theta
\]

\[
= \int_{0}^{\pi} \sqrt{ \sin^2 \theta + \cos^2 \theta + 1 + 2\cos \theta } \, d\theta
\]

\[
= \int_{0}^{\pi} \sqrt{ 1 + 1 + 2\cos \theta } \, d\theta
\]

\[
= \int_{0}^{\pi} \sqrt{ 2 + 2\cos \theta } \, d\theta
\]

\[
= \int_{0}^{\pi} \sqrt{ 2(1 + \cos \theta) } \, d\theta
\]

\[
= \int_{0}^{\pi} \sqrt{ 2 \cdot 2\cos^2 \left( \frac{\theta}{2} \right) } \, d\theta \quad \text{(using the double-angle identity: } 1 + \cos \theta = 2\cos^2 \left( \frac{\theta}{2} \right) \text{)}
\]

\[
= \int_{0}^{\pi} \sqrt{ 4\cos^2 \left( \frac{\theta}{2} \right) } \, d\theta
\]

\[
= \int_{0}^{\pi} 2\left| \cos \left( \frac{\theta}{2} \right) \right| \, d\theta
\]

4. **Evaluate the integral:**

Since \( \cos \left( \frac{\theta}{2} \right) \) is non-negative for \( 0 \leq \theta \leq \pi \):

\[
L = \int_{0}^{\pi} 2 \cos \left( \frac{\theta}{2} \right) \, d\theta
\]

Let \( u = \frac{\theta}{2} \), then \( du = \frac{1}{2} d\theta \) or \( d\theta = 2 du \):

\[
L = \int_{0}^{\frac{\pi}{2}} 2 \cos(u) \cdot 2 \, du
\]

\[
= 4 \int_{0}^{\frac{\pi}{2}} \cos(u) \, du
\]

\[
= 4 \left[ \sin(u) \right]_{0}^{\frac{\pi}{2}}
\]

\[
= 4 \left( \sin \left( \frac{\pi}{2} \right) - \sin(0) \right)
\]

\[
= 4 (1 - 0)
\]

\[
= 4
\]

Therefore, the length of the curve for \( 0 \leq \theta \leq \pi \) is \( 4 \).

The final answer is \(\boxed{4}\)

Answer: 4


What shape is cut out of a regular tetrahedron by a plane that is parallel to two opposite edges of the tetrahedron?

Solution: Some misunderstood the concept of plane section, while others were satisfied with recognizing that the section is a parallelogram, but the answer is: rectangle.

Answer: rectangle


Given that $a, b, c$ are all positive integers, and the parabola $y=ax^2+bx+c$ intersects the $x$-axis at two distinct points $A, B$. If the distances from $A, B$ to the origin are both less than 1. Then the minimum value of $a+b+c$ is $\qquad$ .

Solution:
6. 11. Let $f(x)=a x^{2}+b x+c$.

Suppose the coordinates of $A, B$ are $\left(x_{1}, 0\right),\left(x_{2}, 0\right)$. Then $-10$ has $b>2 \sqrt{a c}, x_{1} x_{2}=\frac{c}{a}0, b2 \sqrt{a c}+1, \sqrt{a}-\sqrt{c}>1, a>$ $(\sqrt{c}+1)^{2} \geqslant(\sqrt{1}+1)^{2}=4$, so $a \geqslant 5 . b>2 \sqrt{a c} \geqslant 2 \sqrt{5}>4, b \geqslant 5$. Therefore, $a+b+c \geqslant 11$, and the quadratic function $f(x)=5 x^{2}+5 x+1$ satisfies the conditions.

Answer: 11


Choose 3 different digits from $0,1,2,3,4,5,6,7,8,9$ to form an ordered array $(x, y, z)$. If $x+y+z$ is a multiple of 4, then the number of arrays that satisfy the condition is $\qquad$.

Answer: 180


Connect the numbers 230, 740, 400, 170, 60 using addition and subtraction so that the result is equal to zero!

Answer: 740+60-230-400-170=0


Solution:
【Analysis】First, decompose 60 into prime factors, $60=2 \times 2 \times 3 \times 5$, then write it in standard form as $2^{2} \times 3 \times 5$, and use the formula for the number of divisors, which is to add 1 to each of the exponents of the different prime factors and then multiply them, and finally subtract 1 to get the answer.

【Solution】 Decompose 60 into prime factors $60=2 \times 2 \times 3 \times 5$, then write it in standard form as $2^{2} \times 3 \times 5$, and use the formula for the number of divisors, which is to add 1 to each of the exponents of the different prime factors and then multiply them.
The number of different divisors of 60 (excluding 1) is $(2+1) \times(1+1) \times(1+1)-1=11$.
Answer: The answer is 11.

Answer: 11


Anita and Boris are shooting at a target with a ball, each 50 times. One part of the target is painted yellow, and the other part is blue. For each hit on the target, a certain number of points is awarded, and if the target is missed, no points are awarded. Anita hit the yellow part 36 times and missed the target 2 times. Boris hit the blue part 6 times and missed the target twice as many times as Anita. At the end of the shooting, the teacher told them that they had scored a total of 716 points, and that Boris scored 4 points less than Anita. How many points does hitting the yellow part, and how many points does hitting the blue part of the target bring?

Solution:
First method:

By multiplying the second equation by 2, we get the equation

$80 x + 12 y = 712$

$36 x + 12 y = 360$

From this, it is clear that the left sides differ by $44 x$, and the right sides by 352,

so $44 x = 352$

$x = 352 : 44$

$x = 8$. A hit in the yellow part of the target scores 8 points.

$6 y = 356 - 320$

$6 y = 36$

$y = 36 : 6$

$y = 6$. A hit in the blue part of the target scores 6 points.

##

Answer: x=8,y=6


In front of the elevator stand people weighing 150, 60, 70, 71, 72, 100, 101, 102, and 103 kg. The elevator's load capacity is 200 kg. What is the minimum number of trips needed to get everyone up?

Solution:
Answer: 5 (or 9)

## Solution

In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $60+70+71=201>200$. Note that no one will be able to go up with the person weighing 150 kg, so a separate trip will be required for his ascent. For the remaining nine people, at least $\frac{8}{2}=4$ trips will be needed, as no more than 2 people can fit in the elevator. Thus, the trips will be: $(60,103),(70,102),(71,101),(72,100)$ and (150).

In this problem, 9 is also considered a correct answer if the solution takes into account not only the elevator's ascents but also its descents to pick up people.

Answer: 5


Given the function $f(x)=\frac{x}{\sqrt{1-x^{2}}}$. Define the function
$$
f_{n}(x)=\underbrace{f(f(f \cdots f(x)) \cdots)}_{n \uparrow},
$$
where the inverse function of $f_{n}(x)$ is $f_{n}^{-1}(x)$. Then
$$
f_{40}\left(\frac{1}{\sqrt{41}}\right) f_{* 0}^{-1}\left(\frac{1}{\sqrt{41}}\right)=
$$

Note: It appears there may have been a typo where $f_{* 0}^{-1}$ should be $f_{40}^{-1}$. I've left the original text as requested.

Solution:
6. $\frac{1}{9}$.
$$
\begin{array}{l}
f_{1}(x)=f(x)=\frac{x}{\sqrt{1-x^{2}}}, \\
f_{2}(x)=f(f(x)) \\
=\frac{\frac{x}{\sqrt{1-x^{2}}}}{\sqrt{1-\left(\frac{x}{\sqrt{1-x^{2}}}\right)^{2}}}=\frac{x}{\sqrt{1-2 x^{2}}}, \\
f_{3}(x)=f(f(f(x))) \\
=\frac{\frac{x}{\sqrt{1-2 x^{2}}}}{\sqrt{1-\left(\frac{x}{\sqrt{1-2 x^{2}}}\right)^{2}}}=\frac{x}{\sqrt{1-3 x^{2}}},
\end{array}
$$

By induction, it is easy to see that $f_{n}(x)=\frac{x}{\sqrt{1-n x^{2}}}$.
From $y=\frac{x}{\sqrt{1-n x^{2}}}$ and the fact that $x, y$ have the same sign, we get
$$
x=\frac{y}{\sqrt{1+n y^{2}}}.
$$

Therefore, $f_{n}^{-1}(x)=\frac{x}{\sqrt{1+n x^{2}}}$.
Thus, $f_{n}(x) f_{n}^{-1}(x)=\frac{x^{2}}{\sqrt{1-n^{2} x^{4}}}$.
Hence, $f_{\infty}\left(\frac{1}{\sqrt{41}}\right) f_{40}^{-1}\left(\frac{1}{\sqrt{41}}\right)=\frac{\frac{1}{41}}{\sqrt{1-\left(\frac{40}{41}\right)^{2}}}=\frac{1}{9}$.

Answer: \frac{1}{9}


Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ be two points on the curve $C: x^{2}-y^{2}=2(x>0)$, then the minimum value of $f=\overrightarrow{O A} \cdot \overrightarrow{O B}$ is $\qquad$ .

Solution:
7.2 Detailed Explanation: From the problem, we know that $x_{1}^{2}-y_{1}^{2}=2, x_{2}^{2}-y_{2}^{2}=2$, where $x_{1}, x_{2}>0$.
$$
\begin{array}{l}
\because f=x_{1} x_{2}+y_{1} y_{2}=\frac{\left(x_{1}+x_{2}\right)^{2}-\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}+y_{2}\right)^{2}-\left(y_{1}-y_{2}\right)^{2}}{4} \\
\quad=\frac{\left(x_{1}+y_{1}+x_{2}-y_{2}\right)\left(x_{1}-y_{1}+x_{2}+y_{2}\right)+\left(x_{1}+y_{1}+y_{2}-x_{2}\right)\left(x_{2}+y_{2}+y_{1}-x_{1}\right)}{4} \\
\quad=\frac{\left(x_{1}+y_{1}\right)\left(x_{2}+y_{2}\right)+\left(x_{1}-y_{1}\right)\left(x_{2}-y_{2}\right)}{2} \geqslant \sqrt{\left(x_{1}^{2}-y_{1}^{2}\right)\left(x_{2}^{2}-y_{2}^{2}\right)}=2, \text{ where the equality holds if } \\
\left(x_{1}, y_{1}\right)=\left(x_{2},-y_{2}\right), \therefore \quad f_{\min }=2 .
\end{array}
$$

Answer: 2


Let $f(x)$ be a function defined on the set of real numbers $\mathbf{R}$, and satisfies the following relations:
$$
f(10+x)=f(10-x), f(20-x)=-f(20+x) \text {. }
$$

Then $f(x)$ is
(A) an even function, and also a periodic function
(B) an even function, but not a periodic function
(C) an odd function, and also a periodic function
(D) an odd function, but not a periodic function

Solution:
(C) $-1-2-1+1$
6. [Analysis and Solution] From the first given equation, we have
$$
\begin{array}{l}
f[10+(10-x)]=f[10-(10-x)] \\
\therefore f(x)=f(20-x)
\end{array}
$$

From the second given equation, we have
$$
\begin{array}{l}
f(x)=-f(20+x) \\
\therefore f(40+x)=f[20+(20+x)] \\
=-f(20+x)=f(x),
\end{array}
$$

which shows that $f(x)$ is a periodic function.
From (1) and (2), we get
$$
f(-x)=f(20+x)=-f(x) \text {. }
$$
$\therefore f(x)$ is an odd function, so the answer is $(\mathbf{C})$.

Answer: C


In the tetrahedron $P-ABC$, edges $PA$, $AB$, and $AC$ are pairwise perpendicular, and $PA=AB=AC$. Points $E$ and $F$ are the midpoints of $AB$ and $PC$, respectively. Then the sine value of the angle $\theta$ formed by $EF$ and the plane $PBC$ is $\qquad$

Solution:
5. $\frac{1}{3}$.

As shown in Figure 4, let $P A=$ $A B=A C=2$.
$$
\begin{array}{l}
\text { Then } E F=\sqrt{2+1}=\sqrt{3} . \\
\text { By } \frac{1}{2}(2 \sqrt{2})^{2} \frac{\sqrt{3}}{2} \times h \\
=\frac{1}{2} \times 2 \times 2 \times 2 \\
\Rightarrow h=\frac{4}{2 \sqrt{3}}=\frac{2}{\sqrt{3}} .
\end{array}
$$

Thus, $h_{E}=\frac{1}{\sqrt{3}}$.
Therefore, $\sin \theta=\frac{1}{\sqrt{3} \times \sqrt{3}}=\frac{1}{3}$.

Answer: \frac{1}{3}


$A B$ is the common perpendicular segment of skew lines $a, b$, $A$ is on line $a$, $B$ is on line $b$, $A B=2$, the skew lines $a, b$ form a $30^{\circ}$ angle, and on line $a$ take $A P=4$, then the distance from point $P$ to line $b$ is

Solution:
8. $2 \sqrt{2}$ Draw $a^{\prime} / / a$ through point $B$, then draw $P Q \perp a^{\prime}$ at $Q, Q H \perp b$ at $H$, and connect $P H$. By the theorem of three perpendiculars, we know that $P H \perp b$, so the length of $P H$ is what we are looking for, and $\angle Q B H=30^{\circ}$.
$$
P H=\sqrt{P Q^{2}+Q H^{2}}=\sqrt{A B^{2}+\left(\frac{B Q}{2}\right)^{2}}=\sqrt{A B^{2}+\left(\frac{A P}{2}\right)^{2}}=2 \sqrt{2} .
$$

Answer: 2\sqrt{2}


A moped rider drives seven kilometers at a speed of $v_{1}=30 \mathrm{~km} / \mathrm{h}$ and then four kilometers at $v_{2}=40 \mathrm{~km} / \mathrm{h}$.

What is his average speed?

Solution:
The total distance $s_{G}$ of the moped rider consists of $s_{1}=7 \mathrm{~km}$ and $s_{2}=4 \mathrm{~km}$, thus $s_{G}=11 \mathrm{~km}$.

To calculate the average speed $\bar{v}$, the total time $t_{G}$ must first be determined: $v=\frac{s}{t}$, from which it follows that $t=\frac{s}{v}$.

$$
t_{1}=\frac{7 \mathrm{~km} \cdot \mathrm{h}}{30 \mathrm{~km}}=\frac{7}{30} \mathrm{~h} \quad, \quad t_{2}=\frac{4 \mathrm{~km} \cdot \mathrm{h}}{40 \mathrm{~km}}=\frac{3}{30} \mathrm{~h}
$$

$t_{G}=t_{1}+t_{2}=\frac{10}{30} \mathrm{~h}=\frac{1}{3} \mathrm{~h}$

For the average speed, we now get:

$$
\bar{v}=\frac{s_{G}}{t_{G}}=33 \frac{\mathrm{km}}{\mathrm{h}}
$$

The moped rider thus travels at an average speed $\bar{v}$ of $33 \mathrm{~km} / \mathrm{h}$.

Answer: 33\mathrm{~}/\mathrm{}


Triangle $A B C$ has a right angle at $B$. Point $D$ lies on side $B C$ such that $3 \angle B A D= \angle B A C$. Given $A C=2$ and $C D=1$, compute $B D$.

Solution:
Answer: $\frac{3}{8}$
Solution: Let $B D=x$. We reflect $D$ over $A B$ to $D^{\prime}$. Then $D D^{\prime}=2 x$, but $A D$ bisects $C A D^{\prime}$, so $4 x=A D^{\prime}=A D$. Also, $A D=\sqrt{x^{2}+A B^{2}}=\sqrt{x^{2}+A C^{2}-B C^{2}}=$ $\sqrt{x^{2}+4-(x+1)^{2}}=\sqrt{3-2 x}$. We have the quadratic $16 x^{2}=3-2 x$ which gives $x=3 / 8$.

Answer: \frac{3}{8}


Find cos x + cos(x + 2Ï/3) + cos(x + 4Ï/3) and sin x + sin(x + 2Ï/3) + sin(x + 4Ï/3).

Solution: Using cos(A+B) = cos A cos B - sin A sin B, we have cos(x + 2π/3) = -(1/2) cos x + (√3)/2 sin x, cos(x + 4π/3) = -(1/2) cos x - (√3)/2 sin x. Hence cos x + cos(x + 2π/3) + cos(x + 4π/3) = 0. Similarly, sin(x + 2π/3) = -1/2 sin x + (√3)/2 cos x, sin(x + 4π/3) = -1/2 sin x - (√3)/2 cos x, so sin x + sin(x + 2π/3) + sin(x + 4π/3) = 0. Thanks to Suat Namli 3rd VMO 1964 © John Scholes jscholes@kalva.demon.co.uk 7 March 2004 Last corrected/updated 7 Mar 04

Answer: 0


Given that $a, b, c, d$ are positive numbers, and
$$
a+20 b=c+20 d=2 \text {. }
$$

Then the minimum value of $\frac{1}{a}+\frac{1}{b c d}$ is

Solution:
8. $\frac{441}{2}$.

From the condition, we have
$$
\begin{aligned}
0 < c d & =\frac{1}{20} \cdot c \cdot 20 d \leqslant \frac{1}{20}\left(\frac{c+20 d}{2}\right)^{2}=\frac{1}{20}. \\
& \text{Then } \frac{1}{a}+\frac{1}{b c d} \geqslant \frac{1}{a}+\frac{20}{b} \\
& =\frac{1}{2}\left(\frac{1}{a}+\frac{20}{b}\right)(a+20 b) \\
& =\frac{1}{2}\left(401+\frac{20 b}{a}+\frac{20 a}{b}\right) \\
& \geqslant \frac{1}{2}\left(401+2 \sqrt{\frac{20 b}{a} \cdot \frac{20 a}{b}}\right) \\
& =\frac{441}{2},
\end{aligned}
$$

Equality holds if and only if $c=20 d$ and $\frac{20 b}{a}=\frac{20 a}{b}$, i.e.,
$$
a=b=\frac{2}{21}, c=1, d=\frac{1}{20}
$$

Thus, the minimum value of $\frac{1}{a}+\frac{1}{b c d}$ is $\frac{441}{2}$.

Answer: \frac{441}{2}


A segment $AB$ has a length $AB=12 \mathrm{~cm}$. On $AB$ lie the points $C$ and $D$. The length $AD=9$ $\mathrm{cm}$ and the length $CB=7 \mathrm{~cm}$. Draw. Determine the length of the segment $CD$.

Solution:
$C D=4 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2479.jpg?height=91&width=920&top_left_y=800&top_left_x=471)

### 13.6.2 2. Round 1977, Class 4

Answer: 4\mathrm{~}


Solution:
A
29. Let $\frac{y}{x}=k$, substitute into the given equation, and use the discriminant to find the maximum value of $k$, which is (A).

Answer: A


Let $n$ be a positive integer.  In $n$-dimensional space, consider the $2^n$ points whose coordinates are all $\pm 1$.  Imagine placing an $n$-dimensional ball of radius 1 centered at each of these $2^n$ points.  Let $B_n$ be the largest $n$-dimensional ball centered at the origin that does not intersect the interior of any of the original $2^n$ balls.  What is the smallest value of $n$ such that $B_n$ contains a point with a coordinate greater than 2?

Solution:
1. **Identify the coordinates of the points and the distance from the origin:**
   The $2^n$ points in $n$-dimensional space have coordinates $(\pm 1, \pm 1, \ldots, \pm 1)$. Each of these points is at a distance of $\sqrt{n}$ from the origin, as calculated by the Euclidean distance formula:
   \[
   \text{Distance} = \sqrt{1^2 + 1^2 + \cdots + 1^2} = \sqrt{n}
   \]

2. **Determine the radius of the ball centered at the origin:**
   We need to find the largest $n$-dimensional ball $B_n$ centered at the origin that does not intersect the interior of any of the $2^n$ balls of radius 1 centered at the points $(\pm 1, \pm 1, \ldots, \pm 1)$. For $B_n$ to be tangent to these balls, the radius of $B_n$ plus the radius of any of the $2^n$ balls must equal the distance from the origin to any of these points:
   \[
   r + 1 = \sqrt{n}
   \]
   Solving for $r$, we get:
   \[
   r = \sqrt{n} - 1
   \]

3. **Determine the condition for $B_n$ to contain a point with a coordinate greater than 2:**
   We need $B_n$ to contain a point with at least one coordinate greater than 2. The farthest point from the origin in $B_n$ is at a distance of $r$ from the origin. For a point to have a coordinate greater than 2, the radius $r$ must be at least 2:
   \[
   \sqrt{n} - 1 \geq 2
   \]
   Solving for $n$, we get:
   \[
   \sqrt{n} \geq 3
   \]
   Squaring both sides, we obtain:
   \[
   n \geq 9
   \]

4. **Verify the smallest value of $n$:**
   For $n = 9$, we have:
   \[
   r = \sqrt{9} - 1 = 3 - 1 = 2
   \]
   This means that $B_9$ is tangent to the balls of radius 1 but does not intersect their interiors. However, the problem asks for the smallest $n$ such that $B_n$ contains a point with a coordinate greater than 2. Since $r = 2$ for $n = 9$, it does not strictly contain a point with a coordinate greater than 2.

   For $n = 10$, we have:
   \[
   r = \sqrt{10} - 1 \approx 3.16 - 1 = 2.16
   \]
   This radius is greater than 2, ensuring that $B_{10}$ contains a point with a coordinate greater than 2.

Conclusion:
\[
n \geq 10
\]

The final answer is $\boxed{10}$

Answer: 10


Given that the graph of the function $f(x)=a x^{2}+b x+c$ passes through the point $(-1,0)$. Does there exist constants $a, b, c$ such that the inequality $x \leqslant f(x) \leqslant \frac{1}{2}\left(1+x^{2}\right)$ holds for all real numbers $x$? If they exist, find the values of $a, b, c$; if not, explain the reason.

Solution:
Assume there exist $a, b, c$ that meet the conditions.
Since the graph of $f(x)$ passes through $(-1,0)$,
then $f(-1)=0$, i.e., $a-b+c=0$.
Also, since $x \leqslant f(x) \leqslant \frac{1}{2}\left(1+x^{2}\right)$ holds for all real numbers $x$, let $x=1$, then $1 \leqslant a+b+c \leqslant \frac{1}{2}\left(1+1^{2}\right)=1$.
Thus, $a+b+c=1$.
Therefore, $b=\frac{1}{2}, a+c=\frac{1}{2}$.
Hence, $f(x)=a x^{2}+\frac{1}{2} x+\left(\frac{1}{2}-a\right)$.
From $\left\{\begin{array}{l}f(x) \geqslant x, \\ f(x) \leqslant \frac{1}{2}\left(1+x^{2}\right)\end{array}\right.$,
we get $\left\{\begin{array}{l}a x^{2}-\frac{1}{2} x+\left(\frac{1}{2}-a\right) \geqslant 0, \\ \left(a-\frac{1}{2}\right) x^{2}+\frac{1}{2} x-a \leqslant 0 .\end{array}\right.$
According to the problem, for any real number $x$, (1) and (2) must both hold.
For (1), if $a=0$, then $x \leqslant 1$, which does not meet the requirement; if $a>0$, to make the solution set of (1) $\mathbf{R}$, we need $\left\{\begin{array}{l}a>0, \\ \Delta \leqslant 0,\end{array}\right.$ i.e., $\left\{\begin{array}{l}a>0, \\ \frac{1}{4}-4 a\left(\frac{1}{2}-a\right) \leqslant 0 .\end{array}\right.$
Solving it, we get $a=\frac{1}{4}$.
For $a=\frac{1}{4}$, consider (2) again. Substituting $a=\frac{1}{4}$ into (2) yields
$$
x^{2}-2 x+1 \geqslant 0 \text {. }
$$

Its solution set is $\mathbf{R}$.
Therefore, there exist $a, b, c$ that meet the conditions, where $a=b=c=\frac{1}{4}, b=\frac{1}{2}$.

Answer: =b==\frac{1}{4}


Let the line $l: y=k x+m$ (where $k, m$ are integers), intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ at two distinct points $A, B$, and intersect the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ at two distinct points $C, D$, such that the vector $\overrightarrow{A C}+\overrightarrow{B D}=\overrightarrow{0}$. The number of lines that satisfy the above conditions is $\qquad$

Solution:
8. 9

Analysis: Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$, and suppose $A, B, C, D$ are collinear, then $\overrightarrow{A C}+\overrightarrow{B D}=\overrightarrow{0}$ is equivalent to $x_{1}+x_{2}=x_{3}+x_{4}$. By combining the equation of the line $l$ with the equation of the ellipse, we get: $\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0$. By Vieta's formulas, we know: $x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}$. Similarly, we get: $x_{3}+x_{4}=\frac{2 k m}{3-k^{2}}$. Therefore, from $x_{1}+x_{2}=x_{3}+x_{4}$, we know that $k=0$ or $m=0$ or
$$
\frac{4}{3+4 k^{2}}=\frac{1}{k^{2}-3} \text { . }
$$
(1) When $k=0$, it is easy to see that $m=0, \pm 1, \pm 2, \pm 3$ satisfy the conditions.
(2) When $m=0$, it is easy to see that $k=0, \pm 1$ satisfy the conditions.
(3) When $\frac{8}{3+4 k^{2}}=\frac{2}{k^{2}-3}$, there are no positive integer solutions.

In summary, there are 9 lines $l$ that satisfy the conditions.

Answer: 9


A ship's captain and his ship together are 63 years old. When the captain is twice as old as the ship is now, then the ship will be exactly three times as old as the captain was when the ship's age in years was the square root of the captain's current age. How old is the captain?

Solution:
Let $x$ denote the current age of the ship's captain, and $y$ the age of the ship (naturally, in years). Let $3z$ be the age of the ship when the captain is $2y$ years old. According to the problem, the captain was $z$ years old when the ship was $\sqrt{x}$. Reviewing the three time points and their corresponding ages:

|  | "now" | "future" | "past" |
| :---: | :---: | :---: | :---: |
| captain | $x$ | $2y$ | $z$ |
| ship | $y$ | $3z$ | $\sqrt{x}$ |

We know that $x + y = 63$. Furthermore, it is evident that the difference in age between the captain and the ship is constant, i.e.,

$$
2y - 3z = z - \sqrt{x} = x - y
$$

From these relationships, we can derive an equation for $\sqrt{x}$. Adding the equations $x + y = 63$ and $x - y = z - \sqrt{x}$ and solving for $z$,

$$
z = 2x + \sqrt{x} - 63
$$

Comparing this with $2y - 3z = x - y$,

$$
y = \frac{x + 3z}{3} = \frac{7x + 3\sqrt{x} - 189}{3}
$$

Finally, using the equation $x + y = 63$,

$$
x + \frac{7x + 3\sqrt{x} - 189}{3} = 63
$$

Rearranging,

$$
10x + 3\sqrt{x} - 378 = 0
$$

The left side can be easily factored: $(\sqrt{x} - 6)(10\sqrt{x} + 63)$, from which, given $\sqrt{x} \geq 0$, only $\sqrt{x} = 6$, or $x = 36$, is a valid solution. Therefore, the captain is 36 years old, and the ship is 27 years old. This is indeed a correct solution, and as we have seen, the only possible one.

Based on the solution by Bajszi István (Bonyhád, Petőfi S. Gymnasium, 2nd grade).

Answer: 36


In isosceles $\triangle ABC$, $AB = AC$, the vertex angle $A = 20^{\circ}$, and a point $D$ is taken on side $AB$ such that $AD = BC$. Find the measure of $\angle BDC$.

Solution: Five, Hint: Construct a regular $\triangle A C E$ outside $\triangle A B C$ with $A C$ as a side, and connect $D E$. Then $\triangle A B C \cong \triangle E A D$, so $\triangle E D C$ is an isosceles triangle. It is easy to find that $\angle B D C=30^{\circ}$.

Answer: 30^{\circ}


Find the number which, when multiplied by 3, then divided by 5, increased by 6, after which the square root is extracted, one is subtracted, and the result is squared, will give 4.

Solution:
122. The method of inverse actions involves approaching the unknown by performing actions on numbers in the reverse order of those specified in the problem, and, obviously, all actions are replaced by their inverses. Sometimes this technique is called the "method of inversion."

$$
\sqrt{4}=2 ; 2+1=3 ; 3^{2}=9 ; 9-6=3 ; 3 \cdot 5=15 ; \frac{15}{3}=5
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-105.jpg?height=57&width=1285&top_left_y=1562&top_left_x=428)
calculations. Bhaskara-Acharya refers to it and even borrows problems from it. The time of his life is not precisely known, but it was certainly in the period between the 6th and 10th centuries AD.

Answer: 5


Try to solve the system of equations
$$\left\{\begin{array}{l}
\lg x+\lg y=1 \\
x^{2}+y^{2}-3 x-3 y=8
\end{array}\right.$$

Solution:
[Solution] From (1) we get $x y=10$.
(3)
(2) $+2 \times(3)$, and rearranging terms, we get
$$(x+y)^{2}-3(x+y)-28=0$$

So $x+y=7$ and $x+y=-4$.
Combining (4) with (3), we get the following system of equations:
$$\begin{array}{l}
\left\{\begin{array}{l}
x+y=7 \\
x y=10
\end{array}\right. \\
\left\{\begin{array}{l}
x+y=-4 \\
x y=10
\end{array}\right.
\end{array}$$

From the system of equations (5), we solve to get $\left\{\begin{array}{l}x_{1}=2, \\ y_{1}=5 ;\end{array}\left\{\begin{array}{l}x_{2}=5, \\ y_{2}=2\end{array}\right.\right.$, substituting the values of $x$ and $y$ into the original system of equations, they all satisfy, so they are all solutions to the original system of equations.

From equation (1) in the original system, we know that $x, y$ must be positive numbers, so the solutions of system (6) cannot be solutions to the original system of equations.

Answer: x_1=2, y_1=5; x_2=5, y_2=2


As shown in Figure 6, it is known that $\triangle A B C$ is inscribed in $\odot O$. A line $l$ is drawn through the midpoint $D$ of $B C$, parallel to $A C$. Line $l$ intersects $A B$ at point $E$, and intersects $\odot O$ at points $G$ and $F$, and intersects the tangent line of $\odot O$ at point $A$ at point $P$. If $P E=3, E D=2, E F=3$, then the length of $P A$ is ( ).

(A) $\sqrt{5}$

(B) $\sqrt{6}$

(C) $\sqrt{7}$

(D) $2 \sqrt{2}$

Solution:
Since $A C / / P F$, we have,
$$
\begin{array}{l}
\angle B D E=\angle B C A=\angle P A E \\
\Rightarrow P, A, D, E \text{ are concyclic} \\
\Rightarrow B E \cdot A E=P E \cdot D E=6 \\
\Rightarrow B E=A E=\sqrt{6} .
\end{array}
$$

By the intersecting chords theorem,
$$
G E \cdot E F=B E^{2} \text{. }
$$

Thus, $G E=2, P G=1$.
By the secant-tangent theorem, $P A^{2}=P G \cdot P F$.
Therefore, $P A=\sqrt{6}$.

Answer: B


In triangle $ABC$, the bisector $BE$ and the median $AD$ are equal and perpendicular. Find the area of triangle $ABC$, if $AB=\sqrt{13}$.

Solution:
Answer: 12

Solution. Let $A B=c, B E=A D=2 a$. Since triangle $A B D$ is isosceles (the bisector is perpendicular to the base, $A B=B D=c, B C=2 c$), then by the formula for the length of the bisector (where $\angle A B C=\beta$) $2 a=\frac{4 c^{2}}{3 c} \cos \frac{\beta}{2} \Leftrightarrow \frac{3 a}{2}=c \cos \frac{\beta}{2}$. Consider triangle $A B F$ (where $F$ is the midpoint of segment AD and the intersection point of the bisector and the median). We have $a=c \sin \frac{\beta}{2}$, $c=\frac{a}{\sin \frac{\beta}{2}}$. Therefore, $\operatorname{tg} \frac{\beta}{2}=\frac{2}{3}$ and $\cos \frac{\beta}{2}=\frac{3}{\sqrt{13}}, \sin \frac{\beta}{2}=\frac{2}{\sqrt{13}}, \sin \beta=\frac{12}{13}$. The area of triangle $A B C$ is $S_{A B C}=c^{2} \sin \beta=\frac{12}{13} c^{2}$.

## B-2

In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$ if $A B=\sqrt{26}$.

Answer: 24

## B-3

In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$ if $B E=A D=4$.

Answer: 12

## B-4

In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$ if $B E=A D=6$.

Answer: 27

## Lomonosov High School Mathematics Olympiad

Preliminary stage 2020/21 academic year for 9th grade

## B-1

#

Answer: 12


Solution:
3. Ana has been subscribed to a magazine for eight years. The subscription was $120 \mathrm{kn}$ in the first year. For the next six years, it increased by 4 kn annually. If Ana paid a total of 1084 kn over eight years, by how much did the subscription increase in the eighth year?

## Solution.

The subscription was $120 \mathrm{kn}$ in the first year,

so it was $120+4=124 \mathrm{kn}$ in the second year, $124+4=128 \mathrm{kn}$ in the third year. \quad 1 POINT

It was $128+4=132 \mathrm{kn}$ in the fourth year, $132+4=136 \mathrm{kn}$ in the fifth year, \quad 1 POINT

$136+4=140 \mathrm{kn}$ in the sixth year, and $140+4=144 \mathrm{kn}$ in the seventh year. \quad 1 POINT

Over these seven years, she paid a total of $120+124+128+132+136+140+144=924 \mathrm{kn}$. \quad 1 POINT

This means the subscription in the eighth year was $1084 \mathrm{kn}-924 \mathrm{kn}=160 \mathrm{kn}$. \quad 1 POINT

In the eighth year, the subscription increased by $160-144=16$ kuna.

1 POINT

TOTAL 6 POINTS

Answer: 16


Solution:
## Solution

The desired decomposition of vector $x$ is:

$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$

Or in the form of a system:

$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \cdot q_{3}+\gamma \cdot r_{3}=x_{3}
\end{array}\right.
$$

We obtain:

$$
\left\{\begin{array}{l}
3 \beta+4 \gamma=0 \\
-2 \alpha+\beta=-8 \\
\alpha-\beta+\gamma=9
\end{array}\right.
$$

Add the third row multiplied by 2 to the second row:

$$
\left\{\begin{array}{l}
3 \beta+4 \gamma=0 \\
-\beta+2 \gamma=10 \\
\alpha-\beta+\gamma=9
\end{array}\right.
$$

Add the second row multiplied by 3 to the first row:

$$
\left\{\begin{array}{l}
10 \gamma=30 \\
-\beta+2 \gamma=10 \\
\alpha-\beta+\gamma=9
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
\gamma=3 \\
-\beta+2 \gamma=10 \\
\alpha-\beta+\gamma=9
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=3 \\
-\beta+2 \cdot 3=10 \\
\alpha-\beta+3=9
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=3 \\
\beta=-4 \\
\alpha-\beta=6
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=3 \\
\beta=-4 \\
\alpha-(-4)=6
\end{array}\right. \\
& \left\{\begin{array}{l}
\gamma=3 \\
\beta=-4 \\
\alpha=2
\end{array}\right.
\end{aligned}
$$

The desired decomposition:

$x=2 p-4 q+3 r$

## Problem Kuznetsov Analytic Geometry $2-28$

Answer: 2p-4q+3r


A sphere is inscribed in a cone. The ratio of the radius of the circle of contact between the spherical and conical surfaces to the radius of the base of the cone is $k$. Find the cosine of the angle between the slant height of the cone and the plane of the base.

Solution:
Solution.

From the problem statement, the radius of the circle of tangency between the spherical and conical surfaces will be $K O_{1}$, and the radius of the base of the cone will be $L O_{2}$ (Fig. 12.76). $\angle S L O_{2}=\alpha$. From the diagram, it is clear that $\triangle S K O \sim \Delta S O_{2} L$, and $\triangle S K O \sim \Delta K O_{1} O$, so $\angle K O O_{1}=\angle S L O_{2}=\alpha$. Then from $\Delta K O_{1} O$, $\mathrm{KO}=\frac{\mathrm{KO}_{1}}{\sin \alpha}$. From $\Delta \mathrm{LOO}_{2}$, $\mathrm{OO}_{2}=\mathrm{LO}_{2} \cdot \operatorname{tg} \frac{\alpha}{2}$. Since $\mathrm{KO}$ and $O \mathrm{O}_{2}$ are equal, then $\frac{K O_{1}}{\sin \alpha}=L O_{2} \operatorname{tg} \frac{\alpha}{2} \cdot \frac{K O_{1}}{L O_{2}}=\sin \alpha \operatorname{tg} \frac{\alpha}{2}=2 \sin ^{2} \frac{\alpha}{2}=k$. Therefore, $\sin \frac{\alpha}{2}=\sqrt{\frac{k}{2}}$. Hence, $\cos \alpha=1-2 \sin ^{2} \frac{\alpha}{2}=1-2 \cdot \frac{k}{2}=1-k$.

Answer: $1-k$.

Answer: 1-k


Let $ n $ be a natural number. How many numbers of the form $ \pm 1\pm 2\pm 3\pm\cdots\pm n $ are there?

Solution:
1. First, we need to understand the sum of the first \( n \) natural numbers. This sum is given by:
   \[
   S_n = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}
   \]

2. We are interested in the number of distinct sums that can be formed by taking each number from 1 to \( n \) and either adding or subtracting it. Let's denote this sum as \( T_n \):
   \[
   T_n = \pm 1 \pm 2 \pm 3 \pm \cdots \pm n
   \]

3. The range of possible values for \( T_n \) is from \( -S_n \) to \( S_n \). This is because the smallest possible sum is when all terms are negative, and the largest possible sum is when all terms are positive:
   \[
   -S_n \leq T_n \leq S_n
   \]

4. Additionally, \( T_n \) must have the same parity (even or odd) as \( S_n \). This is because the sum of an even number of terms will always be even, and the sum of an odd number of terms will always be odd. Since \( S_n = \frac{n(n+1)}{2} \), we need to consider its parity:
   - If \( n \) is even, \( n(n+1) \) is even, so \( S_n \) is an integer.
   - If \( n \) is odd, \( n(n+1) \) is even, so \( S_n \) is an integer.

5. Therefore, \( T_n \) can take on all integer values from \( -S_n \) to \( S_n \) that have the same parity as \( S_n \). The number of such values is:
   \[
   \text{Number of values} = \frac{S_n - (-S_n)}{2} + 1 = \frac{2S_n}{2} + 1 = S_n + 1
   \]

6. Substituting \( S_n = \frac{n(n+1)}{2} \) into the equation, we get:
   \[
   \text{Number of values} = \frac{n(n+1)}{2} + 1
   \]

The final answer is \(\boxed{\frac{n(n+1)}{2} + 1}\).

Answer: \frac{n(n+1)}{2} + 1


The graph on the right shows the fuel consumption per 100 kilometers for four different types of cars: $U, V, W, X$. If each car has 50 liters of fuel, what is the maximum total distance that these four cars can travel in kilometers?

Solution: $\begin{array}{l}\text { Analysis: }(50 \div 20) \times 100+(50 \div 25) \times 100+(50 \div 5) \times 100+(50 \div 10) \times 100 \\ =250+200+1000+500 \\ =1950\end{array}$

Answer: 1950


We are looking for all pairs $\left(p_{1} ; p_{2}\right)$ of (in the decimal system) two-digit prime numbers $p_{1}$ and $p_{2}$ with $p_{1}<p_{2}$, that satisfy the following conditions:

1. Neither the units nor the tens digits of $p_{1}$ and $p_{2}$ are prime numbers.
2. The tens digits of $p_{1}$ and $p_{2}$ are not divisible by 3.
3. The sum $p_{1}+p_{2}$ is neither divisible by 10 nor by 13.

Solution:
Since $p_{1}$ and $p_{2}$ are prime numbers, the only possible digits for their units place are $1, 3, 7, 9$. According to condition 1, 3 and 7 are excluded. For the tens place, only the numbers $1, 4, 8$ remain after conditions 1 and 2. Therefore, $p_{1}$ and $p_{2}$ can take the values $11, 19, 41, 89$ (49 and 81 are not prime numbers!).

Since $p_{1}<p_{2}$, only the pairs $(11, 19), (11, 41), (11, 89), (41, 89)$ satisfy conditions 1 and 2. Of these, only the pair $(19, 89)$ also satisfies condition 3, making it the only solution to the problem.

Answer: (19,89)


In the Cartesian coordinate system, if the equation $m\left(x^{2}+y^{2}+2 y+1\right)=(x-2 y+3)^{2}$ represents an ellipse, then the range of values for $m$ is ( ).
(A) $(0,1)$
(B) $(1,+\infty)$
(C) $(0,5)$
(D) $(5,+\infty)$

Solution: Solution: From the given, we have $\frac{\sqrt{x^{2}+(y+1)^{2}}}{\left|\frac{x-2 y+3}{\sqrt{1^{2}+(-2)^{2}}}\right|}=\sqrt{\frac{5}{m}}$, which indicates that the ratio of the distance from $(x, y)$ to the fixed point $(0,-1)$ to the distance from $(x, y)$ to the line $x-2 y+3=0$ is a constant $\sqrt{\frac{5}{m}}$. By the definition of an ellipse, we get $\sqrt{\frac{5}{m}}5$. Therefore, the correct choice is (D).

Answer: D


There are $\mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D}, \mathbf{E}$ five people, each of whom always tells lies or always tells the truth, and they all know each other's behavior. $\mathrm{A}$ says $\mathrm{B}$ is a liar, $\mathrm{B}$ says $\mathrm{C}$ is a liar, $\mathrm{C}$ says $\mathrm{D}$ is a liar, $\mathrm{D}$ says $\mathbf{E}$ is a liar. Then, among these five people, what is the maximum number of liars?

Solution:
【Analysis】If A tells the truth, from what A says, we know B is lying, from what B says, we know C is telling the truth, and so on, we find D is lying and E is telling the truth, which means 2 people are lying.
If A is lying, then B is telling the truth, C is lying, D is telling the truth, and E is lying, which means 3 people are lying. Therefore, there are at most 3 liars.

Answer: 3


How many integers $n$ with $10 \le n \le 500$ have the property that the hundreds digit of $17n$ and $17n+17$ are different?

Solution:
1. We need to find the number of integers \( n \) such that \( 10 \leq n \leq 500 \) and the hundreds digit of \( 17n \) and \( 17n + 17 \) are different.
2. Let \( 17n = k \). Then \( 17n + 17 = k + 17 \).
3. We need the hundreds digit of \( k \) and \( k + 17 \) to be different. This means that \( k \) must be close to a multiple of 100, specifically within 17 units of a multiple of 100.
4. We can write \( k \) as \( 100x + y \) where \( 0 \leq y < 100 \). For the hundreds digit to change, \( y \) must be such that \( 100x + y \) and \( 100x + y + 17 \) have different hundreds digits.
5. This happens when \( y \geq 83 \) because \( 100x + 83 \) and \( 100x + 100 \) have different hundreds digits.
6. Therefore, \( 17n \) must be in the form \( 100x + y \) where \( y \geq 83 \).
7. We need to find the range of \( n \) such that \( 10 \leq n \leq 500 \). This translates to \( 170 \leq 17n \leq 8500 \).
8. Dividing by 17, we get \( 10 \leq n \leq 500 \).
9. For each \( x \) from 2 to 85 (since \( 170 \leq 17n \leq 8500 \)), there is exactly one \( n \) such that \( 17n \) is in the form \( 100x + y \) with \( y \geq 83 \).
10. Therefore, there are \( 85 - 2 + 1 = 84 \) such integers \( n \).

The final answer is \( \boxed{84} \)

Answer: 84


To obtain phosphorus, calcium phosphate was previously treated with sulfuric acid, the resulting orthophosphoric acid was mixed with carbon and calcined. In this process, orthophosphoric acid turned into metaphosphoric acid, which, upon interaction with carbon, yielded phosphorus, hydrogen, and carbon monoxide. Illustrate all stages of phosphorus production by this method in the form of reaction equations. Calculate the mass of the water-soluble salt that forms when 31 g of calcium phosphate reacts with 17.75 ml of a $69\%$ sulfuric acid solution (density of the solution 1.60 g/ml). (10 points)

Solution: 13. $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+3 \mathrm{H}_{2} \mathrm{SO}_{4}=2 \mathrm{H}_{3} \mathrm{PO}_{4}+3 \mathrm{CaSO}_{4} \cdot \mathrm{H}_{3} \mathrm{PO}_{4}=\mathrm{HPO}_{3}+\mathrm{H}_{2} \mathrm{O}$ (heating). $2 \mathrm{HPO}_{3}+6 \mathrm{C}=2 \mathrm{P}+\mathrm{H}_{2}+6 \mathrm{CO}$. Let's find the mass and number of moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ $17.75 * 1.60 * 0.69=19.6 \mathrm{r} ; \mathrm{n}\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)=19.6 / 98=0.2$ moles. Let's find the number of moles of $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \mathrm{n}\left(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right)=31 / 310=0.1$ moles. Write the equation according to the reagent ratio, taking into account that the water-soluble salt is $\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}-$ calcium dihydrogen phosphate $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+2 \mathrm{H}_{2} \mathrm{SO}_{4}=\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}+$ $2 \mathrm{CaSO}_{4}$. Let's find the mass of $\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2} \mathrm{~m}\left(\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}\right)=0.1 * 234=23.4$ g.

Answer: 23.4


Find all values of the parameter $a$ for which the equation $|x+a|=\frac{1}{x}$ has exactly two roots.

Solution:
Answer: $a=-2$. Solution. The intersection of the graphs of the right and left parts can only be in the first

![](https://cdn.mathpix.com/cropped/2024_05_06_b17fb121c45cb400e742g-1.jpg?height=97&width=1602&top_left_y=2236&top_left_x=250)
the graph $y=|x+a|$ in the first quadrant represents the line $y=x+a$, which intersects with the hyperbola $y=\frac{1}{x}$ at exactly one point (since $y=\frac{1}{x}$ is a decreasing function). Now consider the values $a<0$. For $x \geq -a$, we have (reasoning similarly) one point of intersection of the graphs. Therefore, we need to find $a$ such that there is exactly one more point of intersection for $x \in (0, -a)$. We have $-x-a=\frac{1}{x} \Leftrightarrow x^{2}+a x+1=0 \Leftrightarrow x=\left(-a \pm \sqrt{a^{2}-4}\right) / 2$, hence $a=-2$ (for $-2<a<0$ the quadratic equation has no roots, and for $a<-2$ it has two positive roots, both less than $|a|$).

Answer: -2


Let $f(x, y, z)=\sin ^{2}(x-y)+\sin ^{2}(y-z)+\sin ^{2}(z-x), x, y, z \in \mathbf{R}$, find the maximum value of $f(x, y, z)$. (2007 Zhejiang Province Mathematics Competition Problem)

Solution:
30.
$$\begin{array}{l}
f(x, y, z)=\sin ^{2}(x-y)+\sin ^{2}(y-z)+\sin ^{2}(z-x)= \\
\frac{1}{2}[1-\cos 2(x-y)+1-\cos 2(y-z)+1-\cos 2(z-x)]= \\
\frac{3}{2}-\frac{1}{2}[(\cos 2 x \cos 2 y+\sin 2 x \sin 2 y)+ \\
(\cos 2 y \cos 2 z+\sin 2 y \sin 2 z)+ \\
(\cos 2 z \cos 2 x+\sin 2 z \sin 2 x)] = \\
\frac{3}{2}-\frac{1}{4}\left[(\cos 2 x+\cos 2 y+\cos 2 z)^{2}+(\sin 2 x+\sin 2 y+\sin 2 z)^{2}-3\right]= \\
\frac{9}{4}-\frac{1}{4}\left[(\cos 2 x+\cos 2 y+\cos 2 z)^{2}+(\sin 2 x+\sin 2 y+\sin 2 z)^{2}\right] \leqslant \frac{9}{4}
\end{array}$$

When $\cos 2 x+\cos 2 y+\cos 2 z=\sin 2 x+\sin 2 y+\sin 2 z=0$, $f(x, y, z)$ reaches its maximum value $\frac{9}{4}$. For example, when $x=0, y=\frac{\pi}{3}, z=\frac{2 \pi}{3}$, $f(x, y, z)$ reaches its maximum value $\frac{9}{4}$.

Answer: \frac{9}{4}


In a right triangular prism $A B C-A_{1} B_{1} C_{1}$, $\angle A C B=90^{\circ}, A C=2 B C, A_{1} B \perp B_{1} C$, find the sine of the angle formed by $B_{1} C$ and the lateral face $A_{1} A B B_{1}$.

Solution:
16. Taking $C$ as the origin, $C A, C B, C C_{1}$ as the $x, y, z$ axes to establish a rectangular coordinate system, with $|B C|=2, \left|C C_{1}\right|=a$, then $A(4,0,0), A_{1}(4,0, a), B(0,2,0), B_{1}(0,2, a)$. Since $A_{1} B \perp B_{1} C$, we have $\overrightarrow{\mathrm{BA}_{1}} \cdot \overrightarrow{\mathrm{CB}_{1}}=(4,-2, a) \cdot(0,2, a)=a^{2}-4=0, a=2$.

Let $\vec{n}=(x, y, z)$ be a normal vector to the plane $A_{1} A B B_{1}$, then $\vec{n} \cdot \overrightarrow{A_{1} B_{1}}=-4 x+2 y=0$,
$\vec{n} \cdot \overrightarrow{B B_{1}}=2 z=0$, yielding $\vec{n}=(1,2,0)$.
Since $B_{1} C=(0,-2,-2)$,
the sine of the angle $\alpha$ between $B_{1} C$ and the plane $A_{1} A B B_{1}$ is
$$
\sin \alpha=\left|\vec{n} \cdot \overrightarrow{B_{1} C}\right| /\left(|\vec{n}|\left|\overrightarrow{B C_{1}}\right|\right)=\frac{\sqrt{10}}{5} .
$$

Answer: \frac{\sqrt{10}}{5}


Given that the perimeter of $\triangle ABC$ is 20, the radius of the inscribed circle is $\sqrt{3}$, and $BC=7$, then the value of $\tan A$ is $\qquad$

Solution: As shown in the figure, $x+y+z=10 \Rightarrow x=10-7=3$, thus $\tan \frac{A}{2}=\frac{r}{x}=\frac{\sqrt{3}}{3}$, hence $\frac{A}{2}=30^{\circ} \Rightarrow A=60^{\circ}$. Therefore, $\tan A=\sqrt{3}$.

Answer: \sqrt{3}


Find the real-coefficient polynomial $f(x)$ that satisfies the following conditions:
(1) For any real number $a$, $f(a+1)=f(a)+f(1)$.
(2) There exists a real number $k_{1} \neq 0$, such that $f\left(k_{1}\right)=k_{2}, f\left(k_{2}\right)=k_{3}, \cdots, f\left(k_{n-1}\right)=k_{n}, f\left(k_{n}\right)=k_{1}$, where $n$ is the degree of $f(x)$.

Solution:
Let $a=0$, from (1) we get $f(1)=f(0)+f(1)$, thus $f(0)=0$, and $f(x)$ does not contain a constant term. Let
$$
f(x)=a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n} .
$$

From condition (1), we know
$$
\begin{aligned}
& a_{1}(1+x)+a_{2}(1+x)^{2}+\cdots+a_{n}(1+x)^{n} \\
= & a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}+a_{1}+a_{2}+\cdots+a_{n} .
\end{aligned}
$$

Expanding and comparing the coefficients on both sides, we get
$$
a_{2}=a_{3}=\cdots=a_{n}=0 .
$$

Therefore, $f(x)=a_{1} x$.
From (2), we get $f\left(k_{1}\right)=a_{1} k_{1}=k_{1}$, but $k_{1} \neq 0$ so $a_{1}=1$. The required polynomial is $f(x)=x$.

Answer: f(x)=x


Find the number of all integer solutions of the inequality $\sqrt{1+\sin \frac{\pi x}{4}-3 \cos \frac{\pi x}{2}}+\sqrt{6} \cdot \sin \frac{\pi x}{4} \geq 0$, belonging to the interval [1991; 2013].

Solution: Answer: 9. Comments. Solution of the inequality: $\frac{2}{3}+8 k \leq x \leq \frac{10}{3}+8 k$. From the given interval, the following numbers will be included in the answer: 1993, 1994, 1995, 2001, 2002, 2003, 2009, 2010, 2011 - 9 numbers.

Answer: 9


Two plane mirrors serve as the faces of a dihedral angle. A light ray, perpendicular to the edge of the angle and parallel to the first mirror, is reflected from the second mirror, then from the first, then again from the second, again from the first, and finally, reflecting for the fifth time from the second mirror, returns along the same straight line. What is the magnitude of the dihedral angle between the mirrors?

(Remember, according to the laws of reflection, the incident and reflected rays lie in a plane perpendicular to the mirror plane and form equal angles with the mirror plane.)

Solution: 2. Since the light ray falls perpendicular to the edge of the dihedral angle, all reflections occur in one plane perpendicular to the edge (Fig. 81). Since after the third reflection the ray returns along the same path, at this reflection it must be directed perpendicular to the mirror plane. The exterior angle of the isosceles triangle $A_{1} A_{2} O$ is $2 \alpha$ ( $\alpha$ is the given angle). Therefore, in the right triangle $A_{2} A_{3} O$, one angle is $\alpha$, the other is $2 \alpha$. Thus, $3 \alpha=90^{\circ} ; \alpha=30^{\circ}$.

Answer: 30


The maximum volume of a regular tetrahedron inscribed in a sphere with radius $R$ is

Solution: $\frac{8 \sqrt{3}}{27} R^{3} ;$

Answer: \frac{8\sqrt{3}}{27}R^{3}


Insert Ã¢ÂÂ+Ã¢ÂÂ or Ã¢ÂÂ-Ã¢ÂÂ signs between the numbers $1, 2, 3, \cdots, 1989$. What is the smallest non-negative number that can be obtained from the resulting sum?

Solution:
12. Except for 995, all numbers $1,2,3, \cdots, 1989$ can be divided into 994 pairs: $(1,1989),(2,1988), \cdots,(994,996)$. Since the parity of the two numbers in each pair is the same, the result of the operation is always even regardless of how the “+” or “-” signs are placed before each pair. Since 995 is odd, the sum of the numbers $1,2,3, \cdots, 1989$ with any placement of “+” or “-” signs is always odd. Therefore, the smallest non-negative number that can be obtained is not less than 1.
On the other hand, the number 1 can be obtained in the following way:
$$
1=1+(2-3-4+5)+(6-7-8+9)+\cdots+(1986-1987-1988+1989) \text {. }
$$

Therefore, the smallest non-negative number that can be obtained from the sum is 1.

Answer: 1


Factorize $n^{5}-5 n^{3}+4 n$. What can we conclude in terms of divisibility?

Solution:
The idea is to try to make squares and differences of squares appear in order to factorize the proposed expression.

$$
\begin{aligned}
n^{5}-5 n^{3}+4 n & =n\left(n^{4}-5 n^{2}+4\right) \\
& =n\left(\left(n^{4}-4 n^{2}+4\right)-n^{2}\right) \\
& =n\left(\left(n^{2}-2\right)^{2}-n^{2}\right) \\
& =n\left(n^{2}-n-2\right)\left(n^{2}+n-2\right) \\
& =(n-2)(n-1) n(n+1)(n+2)
\end{aligned}
$$

Thus, for any integer $n, n^{5}-5 n^{3}+4 n$ is the product of 5 consecutive integers. Among these five integers, at least one of them will be divisible by 3, at least one by 5, at least one by 2, and another by 4. In this way, $n^{5}-5 n^{3}+4 n$ will always be divisible by $3 \times 5 \times 2 \times 4=120$.

Remark: More generally, it can be shown (for example using binomial coefficients or Legendre's formula) that the product of $k$ consecutive integers is divisible by $k!$, that is, by $1 \times 2 \times 3 \times \cdots \times k$.

Answer: 120


$$
y=\left|\sqrt{x^{2}+4 x+5}-\sqrt{x^{2}+2 x+5}\right|
$$

The maximum value is $\qquad$.

Solution:
2. $\sqrt{2}$.

Notice that
$$
\begin{array}{l}
y= \mid \sqrt{(x+2)^{2}+(0-1)^{2}}- \\
\sqrt{(x+1)^{2}+(0-2)^{2}} \mid .
\end{array}
$$

The above expression represents the absolute value of the difference in distances from point $P(x, 0)$ to points $A(-2,1)$ and $B(-1,2)$ (as shown in Figure 3). Therefore,
$$
\begin{array}{l}
|P A-P B| \\
\leqslant|A B|=\sqrt{2} .
\end{array}
$$

Thus, the maximum value of $y$ is $\sqrt{2}$, which occurs when $x=-3$.

Answer: \sqrt{2}


In an international meeting of $n \geq 3$ participants, 14 languages are spoken. We know that:

- Any 3 participants speak a common language.

- No language is spoken more that by the half of the participants.

What is the least value of $n$?

Solution:
To find the least value of \( n \) such that any 3 participants speak a common language and no language is spoken by more than half of the participants, we can proceed as follows:

1. **Define the problem constraints:**
   - There are \( n \) participants.
   - There are 14 languages.
   - Any 3 participants speak a common language.
   - No language is spoken by more than \( \left\lfloor \frac{n}{2} \right\rfloor \) participants.

2. **Case 1: \( n \) is even, say \( n = 2k \):**
   - Each language can be spoken by at most \( k \) participants.
   - The number of ways to choose 3 participants from \( 2k \) participants is given by:
     \[
     \binom{2k}{3} = \frac{(2k)(2k-1)(2k-2)}{6}
     \]
   - Each language can cover at most \( \binom{k}{3} \) trios of participants:
     \[
     \binom{k}{3} = \frac{k(k-1)(k-2)}{6}
     \]
   - Since there are 14 languages, the total number of trios covered by all languages is at most:
     \[
     14 \binom{k}{3} = 14 \cdot \frac{k(k-1)(k-2)}{6}
     \]
   - For the condition to hold, the total number of trios covered by the languages must be at least the total number of trios of participants:
     \[
     14 \cdot \frac{k(k-1)(k-2)}{6} \geq \frac{(2k)(2k-1)(2k-2)}{6}
     \]
   - Simplifying this inequality:
     \[
     14k(k-1)(k-2) \geq (2k)(2k-1)(2k-2)
     \]
     \[
     14k(k-1)(k-2) \geq 4k(2k-1)(k-1)
     \]
     \[
     14(k-2) \geq 4(2k-1)
     \]
     \[
     14k - 28 \geq 8k - 4
     \]
     \[
     6k \geq 24
     \]
     \[
     k \geq 4
     \]
   - Therefore, the smallest possible value of \( n \) when \( n \) is even is \( n = 2k = 2 \times 4 = 8 \).

3. **Case 2: \( n \) is odd:**
   - If \( n \) is odd, say \( n = 2k + 1 \), the analysis is similar but does not provide a better bound than the even case. The constraints on the number of participants speaking each language and the number of trios remain the same, leading to the same conclusion.

Thus, the least value of \( n \) that satisfies the given conditions is \( n = 8 \).

The final answer is \( \boxed{8} \).

Answer: 8


Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.

Solution: Let $r$ be the common ratio, where $r>1$. We then have $ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16$. We now have, letting, subtracting the 2 equations, $ar^{2}+-2ar+a=12$, so we have $3ar=432,$ or $ar=144$, which is how much Betty had. Now we have $144+\dfrac{144}{r}+144r=444$, or $144(r+\dfrac{1}{r})=300$, or $r+\dfrac{1}{r}=\dfrac{25}{12}$, which solving for $r$ gives $r=\dfrac{4}{3}$, since $r>1$, so Alex had $\dfrac{3}{4} \cdot 144=\boxed{108}$ peanuts.

Answer: 108


Given that $A$ is a subset of $S=\{1,2,3,4,5,6\}$ with at least 2 elements, and $a, b$ are two distinct elements in $A$. When $A$ ranges over $S$ and $a, b$ range over $A$, the total sum of the product $ab$ is $M=$ $\qquad$

Solution:
6.2800 .

For any $\{a, b\} \subset S$ (assuming $a<b$), the number of $A$ that satisfies $\{a, b\} \subseteq A \subseteq S$ is $2^{6-2}=2^{4}$. Therefore, the contribution of $\{a, b\}$ to the total sum $M$ is $a b \times 2^{4}$.
When $a$ takes all values of $1,2,3,4,5$ and $b$ takes all values of $2,3,4,5,6$, we have
$$
\begin{aligned}
M= & 2^{4}(1 \times 2+1 \times 3+1 \times 4+1 \times 5+1 \times 6+2 \times 3+ \\
& 2 \times 4+2 \times 5+2 \times 6+3 \times 4+3 \times 5+3 \times 6+ \\
& 4 \times 5+4 \times 6+5 \times 6) \\
= & 2^{4}(2+9+24+50+90)=2800 .
\end{aligned}
$$

Answer: 2800


Find the sum of all primes $p$ for which there exists a prime $q$ such that $p^{2}+p q+q^{2}$ is a square.

Solution:
Answer: 83 and 5 both work, because $3^{2}+3 \cdot 5+5^{2}=49$. Now, say $p^{2}+p q+q^{2}=k^{2}$, for a positive integer $k$. Then $(p+q)^{2}-k^{2}=p q$, or:
$$
(p+q+k)(p+q-k)=p q
$$

Since $p+q+k$ is a divisor of $p q$, and it is greater than $p$ and $q, p+q+k=p q$. Then $p+q-k=1$. So:
$$
2 p+2 q=p q+1 \Leftrightarrow p q-2 p-2 q+4=3 \Leftrightarrow(p-2)(q-2)=3
$$

This shows that one of $p$ and $q$ is 3 and the other is 5 .

Answer: 8


Find the largest real number $\lambda$, such that for a real-coefficient polynomial $f(x)=x^{3}+a x^{2}+c$ with all roots being non-negative real numbers, if $x \geqslant 0$, then $f(x) \geqslant \lambda(x-a)^{3}$, and find the condition for equality.

Solution: 33. Let the three roots of $f(x)$ be $\alpha, \beta, \gamma$, and assume $0 \leqslant \alpha \leqslant \beta \leqslant \gamma$, then $x-a=x+\alpha+$ $\beta+\gamma, f(x)=(x-\alpha)(x-\beta)(x-\gamma)$. (1) When $0 \leqslant x \leqslant \alpha$, we have $-f(x)=$ $(\alpha-x)(\beta-x)(\gamma-x) \leqslant\left(\frac{\alpha+\beta+\gamma-3 x}{3}\right)^{3} \leqslant \frac{1}{27}(x+\alpha+\beta+\gamma)^{3}=\frac{1}{27}(x-$ $a)^{3}$, so $f(x) \geqslant-\frac{1}{27}(x-a)^{3}$. The equality holds when $x=0, \alpha=\beta=\gamma$; (2) When $\alpha \leqslant x \leqslant \beta$ or $x>\gamma$, $f(x)=(x-\alpha)(x-\beta)(x-\gamma)>0>-\frac{1}{27}(x-a)^{3}$. (3) When $\beta \leqslant x \leqslant \gamma$, $-f(x)=(x-\alpha)(x-\beta)(\gamma-x) \leqslant\left(\frac{x+\gamma-\alpha-\beta}{3}\right)^{3} \leqslant$ $\frac{1}{27}(x+\alpha+\beta+\gamma)^{3}=\frac{1}{27}(x-a)^{3}$. So $f(x) \geqslant-\frac{1}{27}(x-a)^{3}$. The equality holds when $\alpha=\beta=0$, $\gamma=2 x$. In summary, the maximum value of $\lambda$ is $-\frac{1}{27}$.

Answer: -\frac{1}{27}


Given $n+2$ real numbers
$$
a_{1}, a_{2}, \cdots, a_{n}, 16, a_{n+2} \text {, }
$$

where the average of the first $n$ numbers is 8, the average of the first $n+1$ numbers is 9, and the average of these $n+2$ numbers is 10. Then the value of $a_{n+2}$ is $\qquad$

Solution:
2. 18 .

From the condition, $\frac{8 n+16}{n+1}=9 \Rightarrow n=7$.
Also, $\frac{8 \times 7+16+a_{n+2}}{7+1+1}=10 \Rightarrow a_{n+2}=18$.

Answer: 18


The number of perfect inhabitants of a city was a perfect square, in other words, a whole number squared. With $100$ people plus the new number of inhabitants turned out to be a perfect square plus one. Now, with another increase of $100$ people, the number of inhabitants is again a perfect square. What was the number of inhabitants original city?

Solution:
1. Let the original number of inhabitants be \( n^2 \), where \( n \) is an integer.
2. After increasing the population by 100, the new number of inhabitants is \( n^2 + 100 \).
3. According to the problem, \( n^2 + 100 \) is one more than a perfect square. Let this perfect square be \( m^2 \). Therefore, we have:
   \[
   n^2 + 100 = m^2 + 1
   \]
   Simplifying, we get:
   \[
   n^2 + 99 = m^2
   \]
4. After another increase of 100 people, the number of inhabitants becomes \( n^2 + 200 \).
5. According to the problem, \( n^2 + 200 \) is a perfect square. Let this perfect square be \( k^2 \). Therefore, we have:
   \[
   n^2 + 200 = k^2
   \]
6. We now have two equations:
   \[
   m^2 = n^2 + 99
   \]
   \[
   k^2 = n^2 + 200
   \]
7. Subtract the first equation from the second:
   \[
   k^2 - m^2 = (n^2 + 200) - (n^2 + 99)
   \]
   Simplifying, we get:
   \[
   k^2 - m^2 = 101
   \]
8. We can factor the left-hand side using the difference of squares:
   \[
   (k - m)(k + m) = 101
   \]
9. Since 101 is a prime number, the only factor pairs are (1, 101) and (-1, -101). We consider the positive pair:
   \[
   k - m = 1
   \]
   \[
   k + m = 101
   \]
10. Solving these two equations simultaneously:
    \[
    k - m = 1
    \]
    \[
    k + m = 101
    \]
    Adding these equations, we get:
    \[
    2k = 102 \implies k = 51
    \]
    Subtracting the first equation from the second, we get:
    \[
    2m = 100 \implies m = 50
    \]
11. Substitute \( m = 50 \) back into the equation \( m^2 = n^2 + 99 \):
    \[
    50^2 = n^2 + 99
    \]
    \[
    2500 = n^2 + 99
    \]
    \[
    n^2 = 2401
    \]
    \[
    n = 49
    \]
12. Therefore, the original number of inhabitants was \( n^2 = 49^2 \).

The final answer is \( \boxed{49^2} \).

Answer: 49^2


Calculate $\int_{4}^{9}\left(\frac{2 x}{5}+\frac{1}{2 \sqrt{x}}\right) d x$.

Solution:
Solution. Based on properties 5 and 6 of the definite integral and formula (5.3), we obtain

$$
\begin{aligned}
& \int_{4}^{9}\left(\frac{2 x}{5}+\frac{1}{2 \sqrt{x}}\right) d x=\frac{2}{5} \int_{4}^{9} x d x+\int_{4}^{9} \frac{1}{2 \sqrt{x}} d x=\left.\frac{2}{5} \cdot \frac{x^{2}}{2}\right|_{4} ^{9}+ \\
& +\left.\sqrt{x}\right|_{4} ^{9}=\frac{1}{5}\left(9^{2}-4^{2}\right)+(\sqrt{9}-\sqrt{4})=\frac{1}{5} \cdot 65+1=14
\end{aligned}
$$

Answer: 14


Teacher Guo has a cake to share with 4 or 5 children, so Teacher Guo cuts the cake into several pieces, which may not be of the same size; in this way, whether 4 or 5 children come, they can take some pieces so that each person gets the same amount. Therefore, Teacher Guo must cut the cake into at least $\qquad$ pieces.

Solution:
【Answer】Solution: According to the problem, cut the cake three times, one cut each horizontally, vertically, and longitudinally, the four large pieces each occupy $\frac{1}{5}$, the sum of the four small pieces occupies $\frac{1}{5}$, answer: Teacher Guo divides the cake into at least 8 pieces.

Therefore, the answer is 8.

Answer: 8


The function $f(n)$ is defined for all positive integer $n$ and take on non-negative integer values such that $f(2)=0, f(3)>0$ and $f(9999)=3333$. Also, for all $m, n$,
$$
f(m+n)-f(m)-f(n)=0 \quad \text { or } \quad 1 .
$$

Determine $f(2005)$.

Solution:
9. Ans: 668
The given relation implies that
$$
f(m+n) \geq f(m)+f(n) .
$$

Putting $m=n=1$ we obtain $f(2) \geq 2 f(1)$. Since $f(2)=0$ and $f(1)$ is nonnegative, $f(1)=0$. Next, since
$$
f(3)=f(2+1)=f(2)+f(1)+\{0 \text { or } 1\}=\left\{\begin{array}{ll}
0 & \text { or } 1
\end{array}\right\},
$$
where $f(3)>0$, it follows that $f(3)=1$. Now, it can be proved inductively that $f(3 n) \geq n$ for all $n$. Also, if $f(3 n)>n$ for some $n$, then $f(3 m)>m$ for all $m \geq n$. So since $f(3 \times 3333)=f(9999)=3333$, it follows that $f(3 n)=n$ for $n \leq 3333$. In particular $f(3 \times 2005)=2005$. Consequently,
$$
\begin{aligned}
2005 & =f(3 \times 2005) \\
& \geq f(2 \times 2005)+f(2005) \\
& \geq 3 f(2005)
\end{aligned}
$$
and so $f(2005) \leq 2005 / 3<669$. On the other hand,
$$
f(2005) \geq f(2004)+f(1)=f(5 \times 668)=668 .
$$

Therefore $f(2005)=668$.

Answer: 668


Determine the smallest natural number $n$ having the following property: For every integer $p, p \geq n$, it is possible to subdivide (partition) a given square into $p$ squares (not necessarily equal).

Solution:
To determine the smallest natural number \( n \) such that for every integer \( p \geq n \), it is possible to subdivide a given square into \( p \) squares (not necessarily equal), we will follow these steps:

1. **Check for \( p = 1 \) to \( p = 5 \):**
   - For \( p = 1 \), the given square itself is the only square.
   - For \( p = 2 \), it is impossible to partition a square into exactly 2 squares.
   - For \( p = 3 \), it is impossible to partition a square into exactly 3 squares.
   - For \( p = 4 \), it is possible to partition a square into 4 smaller squares of equal size.
   - For \( p = 5 \), it is impossible to partition a square into exactly 5 squares.

2. **Find models for \( p = 6, 7, 8 \):**
   - For \( p = 6 \), we can partition a square into 6 smaller squares by first partitioning it into 4 equal squares and then further subdividing one of these squares into 3 smaller squares.
   - For \( p = 7 \), we can partition a square into 7 smaller squares by first partitioning it into 4 equal squares and then further subdividing one of these squares into 4 smaller squares.
   - For \( p = 8 \), we can partition a square into 8 smaller squares by first partitioning it into 4 equal squares and then further subdividing two of these squares into 2 smaller squares each.

3. **Apply induction \( p \to p + 3 \):**
   - Assume that for some \( k \geq 6 \), it is possible to partition a square into \( k \) squares.
   - To partition a square into \( k + 3 \) squares, we can take the partition of \( k \) squares and subdivide one of the squares into 4 smaller squares. This increases the total number of squares by 3.

By induction, if it is possible to partition a square into \( k \) squares for \( k \geq 6 \), then it is also possible to partition a square into \( k + 3 \) squares. Therefore, starting from \( p = 6 \), we can partition a square into any number of squares \( p \geq 6 \).

Conclusion:
The smallest natural number \( n \) such that for every integer \( p \geq n \), it is possible to subdivide a given square into \( p \) squares is \( n = 6 \).

The final answer is \( \boxed{ n = 6 } \).

Answer: n = 6


If the parabola $y=x^{2}+mx+2$ intersects the line segment $MN$ (including points $M$ and $N$) connecting points $M(0,1)$ and $N(2,3)$ at two distinct points, find the range of values for $m$.

Solution:
Solution: The function expression of line segment $MN$ is $y=x+1$. Therefore, the original problem is equivalent to the equation $x^{2}+m x+2=x+1$ having two distinct real roots in the interval $[0,2]$. Organizing, we get
$$
\begin{array}{l}
x^{2}+(m-1) x+1=0 . \\
\text { Let } f(x)=x^{2}+(m-1) x+1 .
\end{array}
$$

To ensure that $f(x)=0$ has two distinct real roots in $[0,2]$, we need to consider the endpoints, the discriminant, and the axis of symmetry, which gives us
$$
\left\{\begin{array}{l}
\Delta=(m-1)^{2}-4>0, \\
f(0)=1>0, \\
f(2)=4+2(m-1)+1 \geqslant 0, \\
0<-\frac{m-1}{2}<2 .
\end{array}\right.
$$

Solving this, we get $-\frac{3}{2} \leqslant m<-1$.

Answer: -\frac{3}{2} \leqslant m < -1


KÃ¡ja was supposed to multiply two two-digit numbers. Due to inattention, he switched the digits in one of the factors and got a product that was 4248 less than the correct result.

What should KÃ¡ja have correctly obtained?

(L. HozovÃ¡)

Solution:
Let $x$ and $y$ be Kája's two-digit numbers, and suppose he swapped the digits in the first number. If the digits of the number $x$ are denoted by $a$ and $b$, then we have

$$
(10 a+b) y-(10 b+a) y=4248 .
$$

After simplification, we get $9(a-b) y=4248$ or $(a-b) y=472$. This implies that the number $y$ is a two-digit divisor of 472.

The prime factorization of 472 is $472=2^{3} \cdot 59$, so its only two-digit divisor is 59. This means that $a-b=8$. The following two possibilities satisfy this condition:

- $a=8, b=0$, so $(10 a+b) y=80 \cdot 59=4720$,
- $a=9, b=1$, so $(10 a+b) y=91 \cdot 59=5369$.

Kája should have correctly obtained either 4720 or 5369.

Note. Swapping the digits in the first case results in 08. This does give a valid solution $(80 \cdot 59-8 \cdot 59=4248)$, but it is not a two-digit number. It seems unlikely that Kája, even being inattentive, would not notice this. Therefore, excluding this possibility is not considered an error.

Answer: 5369


The number $x$ satisfies the condition $\frac{\sin 3 x}{\sin x}=\frac{5}{3}$. Find the value of the expression $\frac{\cos 5 x}{\cos x}$ for such $x$.

Solution:
Answer: $-\frac{11}{9}$.

Solution.

$$
\begin{aligned}
& \sin x \neq 0 \rightarrow \frac{\sin 3 x}{\sin x}=\frac{3 \sin x-4 \sin ^{3} x}{\sin x}=3-4 \sin ^{2} x \rightarrow 3-4 \sin ^{2} x=\frac{5}{3} \rightarrow \sin ^{2} x=\frac{1}{3} \rightarrow \\
& \rightarrow \cos ^{2} x=\frac{2}{3} \rightarrow \cos 2 x=\frac{1}{3} ; \\
& \frac{\cos 5 x}{\cos x}=\frac{\cos (3 x+2 x)}{\cos x}=\frac{\cos (3 x+2 x)}{\cos x}=\frac{\cos 3 x \cos 2 x-\sin 3 x \sin 2 x}{\cos x}= \\
& \frac{\left(4 \cos ^{3} x-3 \cos x\right) \cos 2 x-2 \sin 3 x \sin x \cos x}{\cos x}=\left(4 \cos ^{2} x-3\right) \cos 2 x-2 \sin 3 x \sin x= \\
& =(2 \cos 2 x-1) \cos 2 x-\cos 2 x+\cos 4 x==(2 \cos 2 x-1) \cos 2 x-\cos 2 x+2 \cos ^{2} 2 x-1= \\
& =\left(\frac{2}{3}-1\right) \cdot \frac{1}{3}-\frac{1}{3}+\frac{2}{9}-1=-\frac{11}{9} .
\end{aligned}
$$

Answer: -\frac{11}{9}


Solution:
Answer: $-18910$

Change the order of summation and simplify the inner sum:
$$
\begin{aligned}
\sum_{k=1}^{60} \sum_{n=1}^{k} \frac{n^{2}}{61-2 n} & =\sum_{n=1}^{60} \sum_{k=n}^{60} \frac{n^{2}}{61-2 n} \\
& =\sum_{n=1}^{60} \frac{n^{2}(61-n)}{61-2 n}
\end{aligned}
$$

Then, we rearrange the sum to add the terms corresponding to $n$ and $61-n$ :
$$
\begin{aligned}
\sum_{n=1}^{60} \frac{n^{2}(61-n)}{61-2 n} & =\sum_{n=1}^{30}\left(\frac{n^{2}(61-n)}{61-2 n}+\frac{(61-n)^{2}(61-(61-n))}{61-2(61-n)}\right) \\
& =\sum_{n=1}^{30} \frac{n^{2}(61-n)-n(61-n)^{2}}{61-2 n} \\
& =\sum_{n=1}^{30} \frac{n(61-n)(n-(61-n))}{61-2 n} \\
& =\sum_{n=1}^{30}-n(61-n) \\
& =\sum_{n=1}^{30} n^{2}-61 n
\end{aligned}
$$

Finally, using the formulas for the sum of the first $k$ squares and sum of the first $k$ positive integers, we conclude that this last sum is
$$
\frac{30(31)(61)}{6}-61 \frac{30(31)}{2}=-18910
$$

So, the original sum evaluates to -18910 .

Answer: -18910


Let two positive numbers $x$ and $y$ be inversely proportional. If $x$ increases by $P \%$, then $y$ decreases by
(A) $p \%$;
(B) $\frac{p}{1+p} \%$;
(C) $\frac{100}{p} \%$
(D) $\frac{p}{100+p} \%$;
(E) $\frac{100 p}{100+p} \%$

Solution:
26. When $x$ and $y$ are inversely proportional, it means that if $x$ is multiplied by $k$, then $y$ should be divided by $k$. Let $x^{\prime}$ and $y^{\prime}$ be the new values of $x$ and $y$ after $x$ increases by $p \%$, then
$$
\begin{aligned}
x^{\prime} & =\left(1+\frac{p}{100}\right) x, \\
y^{\prime} & =\frac{y}{1+\frac{p}{100}}=\frac{100 y}{100+p} .
\end{aligned}
$$

By definition, the percentage decrease in $y$ is
$$
\begin{array}{l}
1\left(10\left(-\frac{y-y^{\prime}}{y}\right)=100\left(1-\frac{100}{100+p}\right)\right. \\
=\frac{100}{100+p} .
\end{array}
$$

Answer: E


Riley has 64 cubes with dimensions $1 \times 1 \times 1$. Each cube has its six faces labelled with a 2 on two opposite faces and a 1 on each of its other four faces. The 64 cubes are arranged to build a $4 \times 4 \times 4$ cube. Riley determines the total of the numbers on the outside of the $4 \times 4 \times 4$ cube. How many different possibilities are there for this total?

Solution:
We can categorize the 64 small cubes into four groups: 8 cubes that are completely in the interior of the larger cube (and are completely invisible), $4 \times 6=24$ "face cubes" that have exactly one of their faces showing, $2 \times 12=24$ "edge cubes" that have exactly two faces showing, and 8 "corner cubes" that have 3 faces showing.

Notice that $8+24+24+8=64$, so this indeed accounts for all 64 small cubes.

The diagram below has all visible faces of the smaller cubes labelled so that faces of edge cubes are labelled with an $\mathbf{E}$, faces of face cubes are labelled with an $\mathbf{F}$, and faces of corner cubes are labelled with a $\mathbf{C}$ :

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-16.jpg?height=905&width=984&top_left_y=163&top_left_x=625)

Each of the 8 corner cubes has 3 of its faces exposed.

Because of the way the faces are labelled on the small cubes, each of the 8 corner cubes will always contribute a total of $1+1+2=4$ to the sum of the numbers on the outside.

Regardless of how the small cubes are arranged, the corner cubes contribute a total of $8 \times 4=32$ to the total of all numbers on the outside of the larger cube.

The edge cubes can each show either a 1 and a 1 or a 1 and a 2 , for a total of either 2 or 3 .

Thus, each edge cube contributes a total of either 2 or 3 to the total on the outside of the larger cube.

This means the smallest possible total that the edge cubes contribute is $24 \times 2=48$, and the largest possible total that they contribute is $24 \times 3=72$.

Notice that it is possible to arrange the edge cubes to show any total from 48 to 72 inclusive.

This is because if we want the total to be $48+k$ where $k$ is any integer from 0 to 24 inclusive, we simply need to arrange exactly $k$ of the edge cubes to show a total of 3 .

Each of the face cubes shows a total of 1 or a total of 2 . There are 24 of them, so the total showing on the face cubes is at least 24 and at most $24 \times 2=48$.

Similar to the reasoning for the edge cubes, every integer from 24 to 48 inclusive is a possibility for the total showing on the face cubes.

Therefore, the smallest possible total is $32+48+24=104$, and the largest possible total is $32+72+48=152$.

By earlier reasoning, every integer between 104 and 152 inclusive is a possibility, so the number of possibilities is $152-103=49$.

ANSWER: 49

Answer: 49


The necessary and sufficient condition for the line $a x+b y+c=0(a, b, c \neq 0)$ to be symmetric to the line $p x+q y+m=0(p, q, m \neq 0)$ with respect to the $y$-axis is ( ).
A. $\frac{b}{q}=\frac{c}{m}$
B. $-\frac{a}{p}=\frac{b}{q}$
C. $\frac{a}{p}=\frac{b}{q} \neq \frac{c}{m}$
D. $-\frac{a}{p}=\frac{b}{q}=\frac{c}{m}$

Solution: 4. The line symmetric to the line $a x+b y+c=0(a, b, c \neq 0)$ with respect to the $y$-axis is obtained by replacing $x$ with $-x$ to get $-a x+b y+c=0$. Note that the line $-a x+b y+c=0$ coincides with the line $p x+q y+m=0$ if and only if the necessary and sufficient conditions are met, hence the answer is D.

Answer: D


Given the parabola
$$
y=x^{2}-2 m x+4 m-8
$$
with vertex $A$, construct an inscribed equilateral $\triangle A M N$ (points $M, N$ are on the parabola). Find the area of $\triangle A M N$;
If the parabola $y=x^{2}-2 m x+4 m-8$ intersects the $x$-axis at points with integer coordinates, find the integer value of $m$.

Solution:
(1) From the symmetry of the parabola and the equilateral triangle, we know that $M N \perp y$-axis.

As shown in Figure 8, let the axis of symmetry of the parabola intersect $M N$ at point $B$. Then
$$
A B=\sqrt{3} B M .
$$

Let $M(a, b)$ $(m<a)$. Then
$$
\begin{aligned}
& B M=a-m . \\
& \text { Also, } A B=y_{B}-y_{A} \\
= & \left(a^{2}-m^{2}\right)-2 m(a-m) \\
= & (a-m)^{2},
\end{aligned}
$$

Thus, $B M^{2}=A B=\sqrt{3} B M$, which means $B M=\sqrt{3}$.
Therefore, $S_{\triangle A M N}=\sqrt{3} B M^{2}=3 \sqrt{3}$.
(2) According to the problem, $\Delta=4\left[(m-2)^{2}+4\right]$ is a perfect square.
Let $(m-2)^{2}+4=n^{2}$, then
$$
(n+m-2)(n-m+2)=4 \text {. }
$$

Since $n+m-2$ and $n-m+2$ have the same parity, we have
$$
\left\{\begin{array} { l }
{ n + m - 2 = 2 , } \\
{ n - m + 2 = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
n+m-2=-2, \\
n-m+2=-2 .
\end{array}\right.\right.
$$

Solving these, we get $\left\{\begin{array}{l}m=2, \\ n=2\end{array}\right.$ or $\left\{\begin{array}{l}m=2, \\ n=-2 .\end{array}\right.$
In summary, $m=2$.

Answer: 2


Given an integer $n \geqslant 3$. Find the minimum value that $\sum_{i=1}^{n}\left(\frac{1}{x_{i}}-x_{i}\right)$ can achieve, where $x_{1}, x_{2}, \cdots, x_{n}$ are positive real numbers satisfying $\sum_{i=1}^{n} \frac{x_{i}}{x_{i}+n-1}=1$. Also find the values of $x_{i}$ when the minimum value is achieved.

Solution:
5. The minimum value is 0, if and only if all $x_{i}$ are 1.
$$
\text{Let } y_{i}=\frac{x_{i}}{x_{i}+n-1}(i=1,2, \cdots, n) \text{. }
$$

Notice that, $y_{i}$ are all positive real numbers and their sum is 1.
$$
\begin{aligned}
\text{From } x_{i} & =\frac{(n-1) y_{i}}{1-y_{i}} \\
\Rightarrow & \sum_{i=1}^{n} \frac{1}{x_{i}}=\frac{1}{n-1} \sum_{i=1}^{n} \frac{1-y_{i}}{y_{i}} \\
& =\frac{1}{n-1} \sum_{i=1}^{n} \frac{1}{y_{i}} \sum_{j \neq i} y_{j} \\
& =\frac{1}{n-1} \sum_{i=1}^{n} \sum_{i \neq j} \frac{y_{j}}{y_{i}} \\
= & \frac{1}{n-1} \sum_{i=1}^{n} \sum_{i \neq j} \frac{y_{i}}{y_{j}} \\
= & \frac{1}{n-1} \sum_{i=1}^{n} y_{i} \sum_{j \neq i} \frac{1}{y_{j}} \\
\geqslant & \frac{1}{n-1} \sum_{i=1}^{n}\left(y_{i} \frac{(n-1)^{2}}{\sum_{j \neq i} y_{j}}\right) \\
= & \sum_{i=1}^{n} \frac{(n-1) y_{i}}{1-y_{i}} \\
= & \sum_{i=1}^{n} x_{i},
\end{aligned}
$$

The equality holds when all $y_{i}=\frac{1}{n}$, i.e., all $x_{i}=1$.

Answer: 0


In a triangular pyramid, the perimeters of all its faces are equal. Find the area of the total surface of this pyramid if the area of one of its faces is $S$.

Solution:
Suppose the area of the face $A B C$ is $S$ (fig.). Let's write that the perimeters of all faces are equal:

$$
A K+K B+A B=A B+A C+C B,(1)
$$

$$
A K+K C+A C=K C+C B+K B,(2)
$$

$$
A K+K B+A B=K B+B C+K C .
$$

By adding equalities (1) and (2) term by term, we find that $A K=B C$. Substituting $B C$ for $A K$ in equality (3), we find that $A B=K C$. Then from (1), $A C=K B$. From this, it easily follows that all faces of the tetrahedron are equal to each other, and the area of its surface is $4 S$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-22.jpg?height=569&width=652&top_left_y=1528&top_left_x=703)

Answer

$4 S$.

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Answer: 4S


Solution:
Solution. The arithmetic mean will be found using the simplified formula

$$
\bar{X}=k \bar{U}+x_{0} \quad\left(U=\frac{X-x_{0}}{k}\right)
$$

by setting $x_{0}=1500$ and $k=200$. All necessary calculations are presented in the following table:

| Interval $X_{i}^{\prime}-X_{i}^{\prime \prime}$ | Frequency  $m_{i}$ | Midpoint of interval $x_{i}$ | $u_{i}=\frac{x_{i}-x_{0}}{k}$ | $m_{i} u_{i}$ |
| :---: | :---: | :---: | :---: | :---: |
| $400-600$ | 1 | 500 | -5 | -5 |
| $600-800$ | 3 | 700 | -4 | -12 |
| $800-1000$ | 6 | 900 | -3 | -18 |
| $1000-1200$ | 11 | 1100 | -2 | -22 |
| $1200-100$ | 15 | 1300 | 0 | -15 |
| $1400-1600$ | 20 | 1500 | 0 | 14 |
| $1600-1800$ | 14 | 1700 | 2 | 24 |
| $1800-2000$ | 12 | 1900 | 4 | 30 |
| $2000-200$ | 10 | 2100 | 5 | 24 |
| $2200-2400$ | 6 | 2300 | 5 | 10 |
| 2400 and above | 2 | 2500 | 5 | 30 |
| Sum | 100 |  |  |  |

$$
\bar{U}=\frac{30}{100}=0.3, \quad \bar{X}=k \bar{U}+x_{0}=200 \cdot 0.3+1500=1560
$$

Answer: 1560


Solution:
4. $\sqrt{6}-2+\sqrt{3}-\sqrt{2}$.

As shown in Figure 2, construct an isosceles right triangle $\triangle ABC$ such that
$$
\begin{array}{l}
\angle C=90^{\circ}, \\
A C=B C=1 .
\end{array}
$$

Then $A B=\sqrt{2}$.
Construct $\angle CAD=30^{\circ}$.
Then $C D=\frac{\sqrt{3}}{3}$,
$$
A D=\frac{2 \sqrt{3}}{3} \text {. }
$$

Construct the angle bisector $AE$ of $\angle BAD$. Let $CE=x$. Then
$$
\begin{array}{l}
B E=1-x, D E=x-\frac{\sqrt{3}}{3} . \\
\text { Hence } \frac{x-\frac{\sqrt{3}}{3}}{\frac{2 \sqrt{3}}{3}}=\frac{1-x}{\sqrt{2}} \\
\Rightarrow x=\frac{\sqrt{2}+1}{\sqrt{3}+\sqrt{2}}=\sqrt{6}-2+\sqrt{3}-\sqrt{2} .
\end{array}
$$

Therefore, $\tan 37.5^{\circ}=\frac{CE}{AC}=\frac{x}{1}=\sqrt{6}-2+\sqrt{3}-\sqrt{2}$.

Answer: \sqrt{6}-2+\sqrt{3}-\sqrt{2}


Solution:
Solution: By differentiating both sides of the equation, we get
$$
\frac{9-x}{30 x^{2}-x-20}=\frac{9-x}{20 x^{2}+9 x-20}
$$

By the property of fractions, when two fractions are equal, if the numerators are equal, then the fractions are zero or the denominators are equal.
$$
\therefore 9-x=0
$$

or $30 x^{2}-x-20=20 x^{2}+9 x-20$.
Solving, we get $x_{1}=9, x_{2}=0, x_{3}=1$.

Answer: x_{1}=9, x_{2}=0, x_{3}=1


Using only the digit 7, numbers should be formed that, when combined, result in the number 1964. The following types of operations are allowed: addition, subtraction, multiplication, and division. Fractions with the same numerator and denominator are not to be used.

Provide one of the possible solutions!

Solution: For example: $(777+777+777): 7+77+777+777=1964$

Answer: 1964


Find the sum $\sum_{k=1}^{19} k\binom{19}{k}$.

Solution:
(ans. $19 \cdot 2^{18}$.
$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1} \Rightarrow k\binom{n}{k}=n\binom{n-1}{k-1}$. Then, $n=19 \Rightarrow$ $\left.\sum_{k=1}^{19} k\binom{19}{k}=\sum_{k=0}^{18} 19\binom{18}{k-1}=19 \cdot 2^{18}.\right)$

Answer: 19\cdot2^{18}


Solution:
7.B.
$$
a+\frac{1}{a}=14, a^{2}+\frac{1}{a^{2}}=14^{2}-2=194 \text{, }
$$
- $a^{4}+\frac{1}{a^{4}}=194^{2}-2$.

Then $194^{2}-2=(200-6)^{2}-2 \equiv 34(\bmod 100)$.

Answer: B


Given real numbers $x, y$ satisfy: $1+\cos ^{2}(x+y-1)=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}$, find the minimum value of $x y$.

Solution:
Solve $1+\cos ^{2}(x+y-1)=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}=\frac{\left(x^{2}+y^{2}-2 x y\right)+2(x-y)+1+1}{x-y+1}$
$$
=\frac{(x-y+1)^{2}+1}{x-y+1}=x-y+1+\frac{1}{x-y+1} \text {. }
$$

Since $00$, thus $x-y+1+\frac{1}{x-y+1} \geq 2 \ldots \ldots \ldots \ldots$.
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array} { l }
{ 1 + \operatorname { c o s } ^ { 2 } ( x + y - 1 ) = 2 } \\
{ x - y + 1 = 1 }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ \operatorname { c o s } ^ { 2 } ( x + y - 1 ) = 1 } \\
{ x = y }
\end{array} \Rightarrow \left\{\begin{array}{l}
x+y-1=k \pi, k \in \mathbf{Z} \\
x=y
\end{array}\right.\right.\right. \\
\Rightarrow x=y=\frac{1+k \pi}{2}, k \in \mathbf{Z} \Rightarrow x y=\left(\frac{1+k \pi}{2}\right)^{2} \geq \frac{1}{4}, k \in \mathbf{Z} \text {, hence }(x y)_{\min }=\frac{1}{4} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .20 \text { points } \\
\end{array}
$$

Answer: \frac{1}{4}


How many four-digit numbers exist that are divisible by 17 and end in 17?

Solution:
1. Let's denote the four-digit number as $\overline{x y 17}$. Then the number $\overline{x y 17}-17$ is also divisible by 17. But $\overline{x y 17}-17=100 \cdot \overline{x y}+17-17=100 \cdot \overline{x y}$. Since the numbers 100 and 17 are coprime, the two-digit number $\overline{x y}$ is divisible by 17. By enumeration, we find all two-digit numbers divisible by 17: 17, 34, 51, 68, 85. There are five of them. Therefore, there will also be five such numbers: 1717, 3417, 5117, 6817, 8517.

Answer: 5.

Answer: 5


Consider integers $\{1,2, \ldots, 10\}$. A particle is initially at 1. It moves to an adjacent integer in the next step. What is the expected number of steps it will take to reach 10 for the first time?

Solution:
35. Answer: 81 .
Let $E_{k}$ denote the expected number of steps it takes to go from $k-1$ to $k, k=2, \ldots, 100$. Then $E_{k+1}=\frac{1}{2}\left(1+E_{k}+E_{k+1}\right)+\frac{1}{2}$, which implies $E_{k+1}=E_{k}+2$.
It is clear that $E_{2}=1$. Then $E_{3}=3, E_{4}=5, \ldots, E_{10}=17$. So
$$
E=E_{2}+E_{3}+\cdots+E_{10}=1+3+\cdots+17=81 .
$$

Answer: 81