Given $\angle AOM=60^{\circ}$, there is a point $B$ on ray $OM$ such that the lengths of $AB$ and $OB$ are both integers, thus $B$ is called an "olympic point". If $OA=8$, then the number of olympic points $B$ is $\qquad$

Solution:
4. 4 .

As shown in Figure 7, draw $A H \perp O M$ at point $H$.
Since $\angle A O M$
$=60^{\circ}$, and $O A$
$=8$, therefore,
$O H=4$,
$A H=4 \sqrt{3}$.
Let $A B=$
$m, H B=n$
($m, n$ are positive integers). Clearly, in the right triangle $\triangle A H B$, we have
$$
m^{2}-n^{2}=(4 \sqrt{3})^{2} \text {, }
$$

which simplifies to $(m+n)(m-n)=48$.
Since $m+n$ and $m-n$ have the same parity, the factorization of 48 can only be the following three:
$$
48=24 \times 2=12 \times 4=8 \times 6 .
$$

Thus, when $(m, n)=(13,11),(8,4),(7,1)$, there are Olympic points $B_{1}, B_{2}, B_{3}$.

Additionally, it is easy to see that the point $B_{3}$ symmetric to the line $A H$ is also an Olympic point, denoted as $B_{3}^{\prime}$.
Therefore, there are four Olympic points: $B_{1}, B_{2}, B_{3}, B_{3}^{\prime}$.

Answer: 4


Let $n$ points be given arbitrarily in the plane, no three of them collinear. Let us draw segments between pairs of these points. What is the minimum number of segments that can be colored red in such a way that among any four points, three of them are connected by segments that form a red triangle?

Solution:
1. **Understanding the Problem:**
   We are given \( n \) points in the plane with no three points collinear. We need to determine the minimum number of segments that must be colored red such that among any four points, three of them form a red triangle.

2. **Initial Assumptions:**
   Assume the edge \( ab \) is not red. This implies that for any four points \(\{a, b, x, y\}\), there must be a red triangle among them.

3. **Implications of the Assumption:**
   Since \( ab \) is not red, the edges \( xy \) must be red. Additionally, either \( ax \) and \( ay \) must be red, or \( bx \) and \( by \) must be red. This ensures that among any four points, three of them form a red triangle.

4. **Red Subgraph:**
   This implies that the complete graph \( K_n \) minus the edge \( ab \) (denoted as \( K_n - \{a, b\} \)) must be red. Let \( A \) be the set of vertices \( x \) such that \( ax \) is red, and \( B \) be the set of vertices \( y \) such that \( by \) is red. It follows that \( A \cup B = K_n \setminus \{a, b\} \).

5. **Contradiction and Conclusion:**
   If we could find \( x \in A \setminus B \) and \( y \in B \setminus A \), then \(\{a, b, x, y\}\) would not satisfy the condition, leading to a contradiction. Therefore, \( B \setminus A = \emptyset \), implying \( A = K_n \setminus \{a, b\} \). Thus, \( K_n - \{b\} = K_{n-1} \) is red.

6. **Final Calculation:**
   The minimum number of red edges required is the number of edges in \( K_{n-1} \), which is given by:
   \[
   m(n) = \binom{n-1}{2} = \frac{(n-1)(n-2)}{2}
   \]

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

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


In a bag, there is $1 \mathrm{~kg}$ of sugar. Using a balance scale with two weighing pans (each large enough for $1 \mathrm{~kg}$ of loose sugar) and exactly one $50 \mathrm{~g}$ weight, $300 \mathrm{~g}$ of sugar should be weighed out.

Show that this is possible with only three weighings!

Solution:
First, one halves the kilogram of sugar without weights and obtains two 500 g portions. In the same way, one halves the 500 g of sugar and obtains two 250 g portions. With the help of the 50 g weight, one determines 50 g of sugar and adds it to the 250 g. Thus, one has 300 g of sugar.

Another solution:

With the 50 g weight, one determines 50 g of sugar. With these 50 g and the 50 g weight, one determines (from the remaining sugar) 100 g of sugar in a second weighing, which together with the 50 g of sugar results in 150 g of sugar.

With these 150 g of sugar, one determines another 150 g of sugar, so that the contents of the two scales together amount to 300 g of sugar.

Answer: 300\mathrm{~}


A triangular pyramid is cut by a plane into two polyhedra. We will find the ratio of the volumes of these polyhedra, given that the cutting plane divides the edges converging at one vertex of the pyramid in the ratio $1: 2, 1: 2, 2: 1$, counting from this vertex.

Construction of the image. Let the quadrilateral $S A B C$ with its diagonals be the image of the given pyramid (Fig. 87), $\triangle P Q R$ - the image of the given section. This image is complete, and what is quite curious, no parameter has been spent on it.

Given: $\quad S A B C$ - pyramid, $\quad S$ - vertex, $\quad P \in[S A]$, $Q \in[A B]$, $\quad R \in[A C]$, $\quad|A Q|:|Q B|=1: 2,|A R|:|R C|=1: 2$, $|A P|:|P S|=2: 1,(P Q R)$ - cutting plane.

To find: $V_{1}: V_{2}$ - the ratio of the volumes of the resulting polyhedra.

Solution:
Solution. Let $V$ be the volume of the pyramid $SABC$, and $V_{1}$ the volume of the pyramid $PAQR$. Then $V_{1}=V-V_{2}$.

Construct $[SO]$ and assume that $[SO]$ is the image of the height of the pyramid $SABC$ (thus, two parameters are used). Construct $(AO)$ and $[PM] \|[SO]$. Then $M \in (AO)$. To simplify the calculations, let $|AB|=a, |AC|=b$, $|AS|=c, |SO|=H$.

Then $|AQ|=\frac{1}{3} a, |AR|=\frac{1}{3} b, |AP|=\frac{2}{3} c$, and from the similarity of triangles $APM$ and $ASO$ we have: $|PM|=\frac{2}{3} H$. Now let's calculate $V$ and $V_{2}$.

We have:

$$
\begin{aligned}
V & =\frac{1}{3} S_{\triangle ABC} \cdot H=\frac{1}{6} ab \sin \widehat{BAC} \cdot H \\
V_{2} & =\frac{1}{3} S_{\triangle AQR} \cdot \frac{2}{3} H=\frac{1}{81} ab \sin \widehat{BAC} \cdot H
\end{aligned}
$$

Then

$$
V_{1}=V-V_{2}=\frac{25}{162} abH \sin \widehat{BAC}
$$

Thus,

$$
\frac{V_{1}}{V_{2}}=\frac{\frac{25}{162} abH \sin \widehat{BAC}}{\frac{1}{81} abH \sin \widehat{BAC}}=\frac{25}{2}
$$

Investigation. By the problem's meaning, $V > V_{2}$. Therefore, $V_{1}=V-V_{2}>0$ and, consequently, $\frac{V_{1}}{V_{2}}>1$. The found value of $\frac{V_{1}}{V_{2}}$ satisfies this inequality.

Thus, $V_{1}: V_{2}=25: 2$.

## Problems for Independent Solving

244. The base of a parallelepiped is a rhombus $ABCD$, with side length $a$ and an acute angle of $60^{\circ}$. Find the volume of the parallelepiped if the lateral edge is equal to $a$, and $\widehat{A_{1}AB}=\widehat{A_{1}AD}=45^{\circ}$.
245. Each edge of the parallelepiped is equal to $a$. Each of the three planar angles at one of the vertices of the parallelepiped is equal to $2\alpha$. Find the volume of the parallelepiped.
246.

Answer: 25:2


In the expression $x_{1}: x_{2}: \cdots: x_{n}$, use parentheses to indicate the order of operations, and the result can be written in fractional form:
$$
\frac{x_{i_{1}} x_{i_{2}} \cdots x_{i_{k}}}{x_{j_{1}} x_{j_{2}} \cdots x_{j_{n-k}}}
$$
(At the same time, each letter in $x_{1}, x_{2}, \cdots, x_{n}$ may appear in the numerator or in the denominator.) How many different fractions can be obtained by adding parentheses in all possible ways?

Solution:
[Solution] Clearly, $x_{1}$ will be in the numerator of the resulting fraction.
When parentheses are added arbitrarily, the division sign before $x_{2}$ is either related to $x_{2}$ or to an expression in the numerator that includes $x_{2}$, so $x_{2}$ will always be in the denominator.

We will use induction to prove that each of the remaining $n-2$ letters $x_{3}, x_{4}, \cdots, x_{n}$ can independently be in the numerator or the denominator, thus yielding $2^{n-2}$ fractions in total.
Indeed, when $n=3$, we can get 2 fractions:
$$
\left(x_{1}: x_{2}\right): x_{3}=\frac{x_{1}}{x_{2} x_{3}} \text { and } x_{1}:\left(x_{2}: x_{3}\right)=\frac{x_{1} x_{3}}{x_{2}} .
$$

So the conclusion holds for $n=3$.
Assume the conclusion holds for $n=k$, we will prove it for $n=k+1$.
Suppose that after adding parentheses in some way, the expression $x_{1}: x_{2}: \cdots: x_{k}$ can be written in the form of some fraction $A$. If in this expression we replace $x_{k}$ with $x_{k}: x_{k+1}$, then $x_{k}$ will still be in its original position in $A$, and $x_{k+1}$ will not be in the original position of $x_{k}$, that is, if $x_{k}$ was originally in the denominator, then $x_{k+1}$ will be in the numerator, and vice versa.

On the other hand, after adding parentheses in the fraction $A$, there will certainly be an expression of the form $\left(p: x_{k}\right)$, where $P$ is either the letter $x_{k-1}$ or some parentheses. Replacing $\left(p: x_{k}\right)$ with $\left[\left(P: x_{k}\right): x_{k+1}\right]=P: \left(x_{k} \cdot x_{k+1}\right)$, it is clear that in the original fraction $A$, $x_{k} \cdot x_{k+1}$ replaces $x_{k}$. That is, $x_{k+1}$ can be in the same position as $x_{k}$, either both in the numerator or both in the denominator.

By the principle of mathematical induction, each of the remaining $n-2$ letters can independently be in the numerator or the denominator.
Therefore, by adding parentheses in all possible ways, we can obtain $2^{n-2}$ different fractions.

Answer: 2^{n-2}


If $y=\sqrt{1-x}+\sqrt{x-\frac{1}{2}}$ has a maximum value of $a$ and a minimum value of $b$, then the value of $a^{2}+b^{2}$ is $\qquad$.

Solution:
9. A. $\frac{3}{2}$.

From $1-x \geqslant 0$ and $x-\frac{1}{2} \geqslant 0$, we get
$$
\frac{1}{2} \leqslant x \leqslant 1 \text {. }
$$

Then $y^{2}=\frac{1}{2}+2 \sqrt{-x^{2}+\frac{3}{2} x-\frac{1}{2}}$ $=\frac{1}{2}+2 \sqrt{-\left(x-\frac{3}{4}\right)^{2}+\frac{1}{16}}$.
Since $\frac{1}{2}<\frac{3}{4}<1$, therefore,
when $x=\frac{3}{4}$, $y^{2}$ reaches its maximum value 1, so $a=1$;
when $x=\frac{1}{2}$ or 1, $y^{2}$ reaches its minimum value $\frac{1}{2}$, so $b=\frac{\sqrt{2}}{2}$.
Thus, $a^{2}+b^{2}=\frac{3}{2}$.

Answer: \frac{3}{2}


Find two natural numbers such that when divided, the quotient is 49, and the remainder is strictly greater than 4 and 294 less than the dividend.

Let $A=n^{2016}+(n+1)^{2016}+(n+2)^{2016}+\ldots+(n+9)^{2016}$, where $n \in \mathbb{N}$. Determine the remainder of the division of $A$ by 10.

Solution:
2.a) Let $a$ and $b$ be the two numbers. Assume $a > b$.

The conditions from the problem can be written as:

$a = 49b + r \quad (1);$

$r > 4 \quad (2)$

$r = a - 294 \quad (3)$.

From equations (1) and (3), we get $a = 49b + a - 294$, which simplifies to $49b = 294$

$\Rightarrow b = 6$. Then $r \in \{0, 1, 2, 3, 4, 5\}$.

Since $r > 4 \Rightarrow r = 5$.

Thus, $a = 299$.

Therefore, the numbers sought are 299 and 6.

b) The remainder of the division of $A$ by 10 is the same as the last digit of $A$.

The numbers $n, n+1, n+2, \ldots, n+9$ have the last digits $0, 1, 2, \ldots, 9$, in some random order.

Thus, $u c(A) = u c\left(0^{2016} + 1^{2016} + 2^{2016} + \ldots + 9^{2016}\right) = $

$= u c(0 + 1 + 6 + 1 + 6 + 5 + 6 + 1 + 6 + 1) = 3$.

Therefore, the remainder of the division of $A$ by 10 is 3.

Answer: 3


In the positive term sequence $\left\{a_{n}\right\}$,
$$
a_{1}=10, a_{k+1}=10 \sqrt{a_{k}}(k=1,2, \ldots) \text {. }
$$

Find the general term formula $a_{n}$.

Solution:
Solution: Let $b_{k}=\lg a_{k}$. Then $b_{k+1}=\frac{1}{2} b_{k}+1$.
Let $x=\frac{1}{2} x+1$, we get $x=2$. Then $b_{k+1}-2=\frac{1}{2}\left(b_{k}-2\right)$.
Therefore, the sequence $\left\{b_{n}-2\right\}$ is a geometric sequence with the first term $b_{1}-2=-1$ and the common ratio $\frac{1}{2}$.

Hence $b_{n}-2=-1 \times\left(\frac{1}{2}\right)^{n-1}$. That is, $a_{n}=10^{2-\left(\frac{1}{2}\right)^{n-1}}\left(n \in \mathbf{N}_{+}\right)$.

Answer: a_{n}=10^{2-(\frac{1}{2})^{n-1}}


For a natural number $N$, if at least six of the nine natural numbers from $1$ to $9$ can divide $N$, then $N$ is called a "Six-Union Number". The smallest "Six-Union Number" greater than 2000 is $\qquad$ .

Solution:
【Solution】Solution: According to the problem, we have:
To satisfy being a hexagonal number. It is divided into being a multiple of 3 and not being a multiple of 3.
If it is not a multiple of 3, then it must be a multiple of $1,2,4,8,5,7$, so their least common multiple is: 8 $\times 5 \times 7=280$. Then the smallest number greater than 2000 that is a multiple of 280 is 2240.

If it is a multiple of 3. It must also satisfy being a multiple of $1,2,3,6$. It needs to satisfy 2 more numbers.
The smallest number greater than 2000 is 2004 (a multiple of 1, 2, 3, 4, 6) which does not meet the requirement;
2010 is (a multiple of 1, 2, 3, 5, 6) which does not meet the requirement;
2016 is (a multiple of $1,2,3,4,6,7,8,9$) which meets the requirement.
$2016<2240$;
Therefore, the answer is: 2016

Answer: 2016


Find positive integers $x, y$, such that $x+y^{2}+(x, y)^{3}=x y(x, y)$.

Solution:
【Analysis】Let $z=(x, y), x=a z, y=b z$. Then $(a, b)=1$.
The equation transforms into $a+b^{2} z+z^{2}=a b z^{2}$. Hence $z \mid a$. Let $a=c z$, i.e., $x=c z^{2}$. Then the equation becomes $c+b^{2}+z=b c z^{2} \Rightarrow c=\frac{b^{2}+z}{b z^{2}-1}$.
We now discuss the variable $z$.
(1) If $z=1$, then
$$
c=\frac{b^{2}+1}{b-1}=(b+1)+\frac{2}{b-1} \text {, }
$$

the positive integer $b$ can only be 2 or 3, easily yielding two solutions $(x, y)=(5,2)$ or $(5,3)$.
(2) If $z=2$, then $c=\frac{b^{2}+2}{4 b-1}$, i.e.,
$$
16 c=\frac{16 b^{2}+32}{4 b-1}=(4 b+1)+\frac{33}{4 b-1},
$$
$4 b-1$ can only be 3 or 11 (obviously not 33), easily yielding two solutions
$$
(x, y)=(4,2) \text { or }(4,6) \text {. }
$$
(3) If $z \geqslant 3$, then
$$
c z^{2}=\frac{b^{2} z^{2}+z^{3}}{b z^{2}-1}=b+\frac{z^{3}+b}{b z^{2}-1} \text {. }
$$

From $\frac{z^{3}+b}{b z^{2}-1} \geqslant 1$, we get
$$
b \leqslant \frac{z^{3}+1}{z^{2}-1}=\frac{z^{2}-z+1}{z-1}=z+\frac{1}{z-1},
$$

i.e., the positive integer $b \leqslant z$.
Therefore, $c=\frac{b^{2}+z}{b z^{2}-1} \leqslant \frac{z^{2}+z}{z^{2}-1}<2$, the positive integer $c$ can only be 1.
At this point, $b^{2}-b z^{2}+(z+1)=0$.
Noting that $\Delta=z^{4}-4(z+1) \in\left(\left(z^{2}-1\right)^{2}, z^{4}\right)$.
Thus, the equation has no positive integer solutions.
In summary, $(x, y)=(5,2),(5,3),(4,2),(4,6)$.

Answer: (x, y)=(5,2),(5,3),(4,2),(4,6)


A two-digit odd number, if the positions of its unit digit and tens digit are swapped, the resulting number is still an odd number,

there are $\qquad$ such odd numbers.

Solution: Reference answer: 25

Answer: 25


January 1, 2015 was a Thursday, based on this information, we can calculate that February 1, 2015 was a Saturday $\qquad$

Solution:
【Key Point】Cycle problem
【Difficulty】Approximate

【Answer】Sunday.
【Analysis】$31 \div 7=4 \cdots 3$. Starting from Thursday and moving forward three days, it is Sunday.

Answer: Sunday


From the set $\{1,2,3, \cdots, 1000\}$, randomly and without replacement, take 3 numbers $a_{1}, a_{2}, a_{3}$, and from the remaining 997 numbers in the set, randomly and without replacement, take another 3 numbers $b_{1}, b_{2}, b_{3}$. Let the probability of the following event be $p$: a rectangular brick of dimensions $a_{1} \times a_{2} \times a_{3}$ can fit into a rectangular box of dimensions $b_{1} \times b_{2} \times b_{3}$ after appropriate rotation, with the edges of the brick parallel to the edges of the box. If $p$ is written as a reduced fraction, what is the sum of the numerator and the denominator?

Solution:
9. Solution: A rectangular brick of size $a_{1} \times a_{2} \times a_{3}$ can fit into a rectangular box of size $b_{1} \times b_{2} \times b_{3}$ if $a_{1}<b_{1}, a_{2}<b_{2}, a_{3}<b_{3}$. Given $c_{0}<c_{6}$, $c_{1}$ must be one dimension of the brick, and $c_{6}$ must be one dimension of the box. Here, we refer to the length, width, and height of a rectangular prism as its three dimensions.

From the four numbers $c_{2}, c_{3}, c_{4}, c_{5}$, any two (excluding the pair $\left(c_{4}, c_{5}\right)$) can be chosen as the other two dimensions of the brick, and the remaining two numbers can be the other two dimensions of the box. Therefore, there are $C_{4}^{2}-1=5$ (ways). Thus, $P=\frac{C_{1000}^{6} \times 5}{C_{1000}^{3} \cdot C_{997}^{3}}=\frac{1}{4}$.
Therefore, the required probability is $\frac{1}{4}$, and the sum of the numerator and denominator of $P$ is 5.

Answer: 5


Liesl has a bucket. Henri drills a hole in the bottom of the bucket. Before the hole was drilled, Tap A could fill the bucket in 16 minutes, tap B could fill the bucket in 12 minutes, and tap C could fill the bucket in 8 minutes. A full bucket will completely drain out through the hole in 6 minutes. Liesl starts with the empty bucket with the hole in the bottom and turns on all three taps at the same time. How many minutes will it take until the instant when the bucket is completely full?

Solution:
Since taps A, B and C can fill the bucket in 16 minutes, 12 minutes and 8 minutes, respectively, then in 1 minute taps A, B and C fill $\frac{1}{16}, \frac{1}{12}$ and $\frac{1}{8}$ of the bucket, respectively.

Similarly, in 1 minute, $\frac{1}{6}$ of the bucket drains out through the hole.

Therefore, in 1 minute with the three taps on and the hole open, the fraction of the bucket that fills is $\frac{1}{16}+\frac{1}{12}+\frac{1}{8}-\frac{1}{6}=\frac{3+4+6-8}{48}=\frac{5}{48}$.

Thus, to fill the bucket under these conditions will take $\frac{48}{5}$ minutes.

ANSWER: $\frac{48}{5}$ minutes

Answer: \frac{48}{5}


Given 50 numbers. It is known that among their pairwise products, exactly 500 are negative. Determine the number of zeros among these numbers.

Solution: Solution. Let $m, n$ and $p-$ be the number of negative, zero, and positive numbers among the given 50 numbers. Then from the condition of the problem we have: $m+n+p=50$ and $m \cdot p=500$. It follows that $m$ and $p$ are divisors of 500, the sum of which does not exceed 50. Among all divisors of the number 500, only the pair 20 and 25 has this property, i.e. $m+p=45$, hence $n=5$.

Answer: 5


Solution:
Since $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=0 \Rightarrow a b c=0$, then $a, b, c$ at least one is 0. Without loss of generality, let $a=0 \Rightarrow c=-b$,
so $a^{2 n+1}+b^{2 n+1}+c^{2 n+1}=0+b^{2 n+1}+(-b)^{2 n+1}=0$.

Answer: 0


In quadrilateral ABCD, it is known that $\angle A B C=60^{\circ}, B E$ bisects $\angle A B C$, intersecting $A D$ and $C D$ at points $E$ and $F$ respectively; $A E=3, D E=2, O$ is the circumcenter of $\triangle D E F$. The following conclusions are given:

(1) Quadrilateral $O E D F$ is a rhombus;
(2) $\triangle O A C$ is an equilateral triangle;
(3) $\triangle O A E \cong \triangle O C D$.
Among these, the number of correct conclusions is ( ).
(A) 0
(B) 1
(C) 2
(D) 3

Solution:
5. D.

As shown in Figure 3, it is easy to see that
$$
\begin{array}{l}
D E=D F, \\
A E=A B=C D, \\
\angle A D C=60^{\circ} . \\
\text { Since } O E=O F,
\end{array}
$$

we know that $O D$ perpendicularly bisects $E F$, and $\triangle O D E$ and $\triangle O D F$ are both equilateral triangles.
Thus, quadrilateral $O E D F$ is a rhombus.
And $\angle O E A=180^{\circ}-\angle O E D$
$$
=180^{\circ}-\angle O D F=\angle O D C \text {, }
$$

Therefore, $\triangle O A E \cong \triangle O C D$.
By rotating $\triangle O C D$ 60° clockwise around point $O$, it coincides with $\triangle O A C$. Thus, $\angle A O C=60^{\circ}$.
Hence, $\triangle O A C$ is an equilateral triangle.

Answer: D


If real numbers $x, y$ satisfy
$$
x^{3}+y^{3}=26 \text {, and } x y(x+y)=-6 \text {, }
$$

then $|x-y|=$ $\qquad$

Solution:
B. 4 .

From the given information, we have
$$
\begin{array}{l}
(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)=8 \\
\Rightarrow x+y=2 \Rightarrow x y=-3 . \\
\text { Therefore, }|x-y|=\sqrt{(x+y)^{2}-4 x y}=4 .
\end{array}
$$

Answer: 4


From point $P$ outside a circle, with a circumference of $10$ units, a tangent is drawn.
Also from $P$ a secant is drawn dividing the circle into unequal arcs with lengths $m$ and $n$.
It is found that $t_1$, the length of the tangent, is the mean proportional between $m$ and $n$.
If $m$ and $t$ are integers, then $t$ may have the following number of values:
$\textbf{(A)}\ \text{zero}\qquad \textbf{(B)}\ \text{one}\qquad \textbf{(C)}\ \text{two}\qquad \textbf{(D)}\ \text{three}\qquad \textbf{(E)}\ \text{infinitely many}$

Solution:
By definition of [mean proportional](https://artofproblemsolving.com/wiki/index.php/Geometric_mean), $t = \sqrt{mn}$.  Since $m+n=10$, $t = \sqrt{m(10-m)}$.
With trial and error, note that when $m=9$, $t=3$ and when $m=2$, $t=4$.  These values work since another [tangent line](https://artofproblemsolving.com/wiki/index.php/Tangent_line) can be drawn from $P$, and the angle between the [tangent](https://artofproblemsolving.com/wiki/index.php/Tangent_line) and [secant](https://artofproblemsolving.com/wiki/index.php/Secant_line) can decrease to match the values of $m$ and $n$.  Thus, the answer is $\boxed{\textbf{(C)}}$.

Answer: \textbf{(C)}


Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y, z$, we have
$$
\frac{1}{2} f(x y)+\frac{1}{2} f(x z)-f(x) f(y z) \geqslant \frac{1}{4}.
$$

Solution:
21. The given is an inequality, not an equation, and there are three variables, namely $x, y, z$. We attempt to find the result by taking some special values.
Let $x=y=z=1$, substitute into (1), we get
$$
\begin{array}{c}
f(1)-[f(1)]^{2} \geqslant \frac{1}{4} . \\
{\left[f(1)-\frac{1}{2}\right]^{2} \leqslant 0 .} \\
f(1)=\frac{1}{2} .
\end{array}
$$

So
Let $y=z=1$, substitute into (1) and use (2), we get
$$
f(x)-\frac{1}{2} f(x) \geqslant \frac{1}{4} .
$$

So
$$
f(x) \geqslant \frac{1}{2} .
$$

Let $x=y=z=0$, substitute into (1), we get
$$
f(0)-f^{2}(0) \geqslant \frac{1}{4} \text {. }
$$

So
$$
f(0)=\frac{1}{2} \text {. }
$$

Let $x=0$, substitute into (1) and use (4), we get
$$
\frac{1}{2}-\frac{1}{2} f(y z) \geqslant \frac{1}{4} \text {. }
$$

Hence
$$
\begin{array}{l}
f(y z) \leqslant \frac{1}{2}, \\
f(x) \leqslant \frac{1}{2} .
\end{array}
$$

Combining (3) and (5), we get $f(x) \equiv \frac{1}{2}$.

Answer: f(x)\equiv\frac{1}{2}


Place real numbers at the vertices of a pentagon so that the sum of the numbers at the ends of one side is equal to 1, at the ends of another side is equal to 2, ..., and at the ends of the last side is equal to 5.

Solution:
Place the number $x$ at the 1st vertex, $1-x$ at the 2nd, $1+x$ at the 3rd, $-2-x$ at the 4th, and $-2+x$ at the 5th. Then, for any value of $x$, the sums of the numbers on the four sides will be 1, 2, 3, and 4, respectively. To make the sum of the numbers on the 5th side equal to 5, we need to find $x$ from the condition $x+(2+x)=5$. Hence, $x=1.5$.

## Answer

$3 / 2, -1 / 2, 5 / 2, 1 / 2, 7 / 2$.

Answer: \frac{3}{2},-\frac{1}{2},\frac{5}{2},\frac{1}{2},\frac{7}{2}


Some students from a 5th grade class held a chess tournament. Each participant played exactly 2 games against every other participant. In total, 3 games were played each day for 24 days.

Determine the number of participants in this tournament!

Solution:
There were a total of $24 \cdot 3=72$ games.

If everyone plays 2 games against everyone else, there are $n \cdot(n-1)$ games with $n$ participants. This leads to the equation $n \cdot(n-1)=72$, which is only satisfied by $n=9$ in the domain of natural numbers (since there can only be a natural number of participants). This solution is found through systematic trial and error.

To do this, one checks the products $n(n-1)$, i.e., $2 \cdot 1, 3 \cdot 2, \ldots$ until reaching $9 \cdot 8=72$.

A solution using the quadratic formula for $n^{2}-n-72=0$ with $n_{1}=9, n_{2}=-8$ is generally not possible for 5th-grade students.

Answer: 9


At least how many days did the festival last if $11$ theatrical groups participated in a festival, each day some of the groups were scheduled to perform while the remaining groups joined the general audience, and at the conclusion of the festival, each group had seen, during its days off, at least $1$ performance of every other group?

Solution:
To determine the minimum number of days the festival must last, we can use a combinatorial approach based on Sperner's lemma. Here is the detailed solution:

1. **Define the Problem in Terms of Sets:**
   - Let \( n = 11 \) be the number of theatrical groups.
   - Each day, a subset of these groups performs, and the rest are in the audience.
   - By the end of the festival, each group must have seen every other group perform at least once.

2. **Translate the Problem into Set Theory:**
   - Denote the groups by \( g_1, g_2, \ldots, g_{11} \).
   - Let \( G_j \) be the set of groups performing on day \( j \) for \( 1 \leq j \leq k \).
   - Define \( C_i = \{ j \mid g_i \in G_j \} \), which represents the days on which group \( g_i \) performs.

3. **Condition Translation:**
   - The condition that each group sees every other group perform translates to \( C_i \not\subseteq C_{i'} \) for any \( 1 \leq i \neq i' \leq 11 \).

4. **Use Sperner's Lemma:**
   - Sperner's lemma states that the largest antichain in the poset \( (\mathcal{P}(\{1, 2, \ldots, k\}), \subseteq) \) has size \( \binom{k}{\lfloor k/2 \rfloor} \).
   - We need \( \binom{k}{\lfloor k/2 \rfloor} \geq 11 \).

5. **Calculate the Minimum \( k \):**
   - Calculate \( \binom{k}{\lfloor k/2 \rfloor} \) for increasing values of \( k \) until the inequality is satisfied.
   - For \( k = 6 \):
     \[
     \binom{6}{3} = \frac{6!}{3!3!} = \frac{720}{6 \cdot 6} = 20
     \]
     Since \( 20 \geq 11 \), \( k = 6 \) works.

6. **Construct the Model:**
   - Choose any 11 elements from the maximal antichain of subsets of \( \{1, 2, \ldots, 6\} \) with 3 elements each.
   - Define \( G_j = \{ g_i \mid j \in C_i \} \) for \( 1 \leq j \leq 6 \).

Thus, the minimum number of days the festival must last is 6.

The final answer is \( \boxed{6} \)

Answer: 6


How many numbers at most can be chosen from the set $M=\{1,2, \ldots, 2018\}$ such that the difference of any two chosen numbers is not equal to a prime number?

Solution:
4. Selecting 505 numbers $1,5,9,13,17, \ldots, 2017 \in \mathrm{M}$, which give a remainder of 1 when divided by four, clearly has the required property, because the difference between any two selected numbers is a multiple of four, thus a composite number. (Since 4 is the smallest composite number, it is not possible to select a larger number of numbers in a similar regular manner.)

We will show that it is not possible to select more numbers with the required property from the given set, regardless of the method used. To do this, we first verify the key statement that from any sequence of eight consecutive numbers, at most two numbers can be selected.

First, we prove that if we select some number $n$, from the next seven numbers $n+1, n+2, \ldots, n+7$ at most one can be selected. Indeed, we cannot then select any of the numbers $n+2, n+3, n+5, n+7$. And from the remaining trio $n+1, n+4, n+6$, only one number can be selected, because their differences are the prime numbers 2, 3, and 5. This proves the statement about the seven numbers that follow any selected number $n$. From this, it is clear that no sequence of eight consecutive numbers can contain more than two selected numbers, as we promised to verify.

The set $\mathrm{M}$ can be divided into the set $\{1,2, \ldots, 10\}$ and 251 subsequent eights. From the set $\{1,2, \ldots, 10\}$, at most three suitable numbers can be selected. If we were to select four suitable numbers from the first ten

$$
\{1,2, \ldots, 10\}=\{1,2\} \cup\{3,4, \ldots, 10\}=\{1,2, \ldots, 8\} \cup\{9,10\}
$$

according to the proven property of each sequence of eight consecutive numbers, we would have to select both numbers 1 and 2, as well as both numbers 9 and 10, but $9-2$ is a prime number. From the set $\mathrm{M}$, then, at most $2 \cdot 251+3=505$ numbers can be selected.

For a complete solution, award 6 points. 1 point for the idea that from a sequence of eight consecutive numbers, at most two can be selected, and 1 point for a correct proof of this idea. 1 point for correctly dividing the numbers into disjoint eights and "remainder". 1 point for correctly deriving that from such a division, at most 505 numbers can be selected. 1 point for constructing the required set itself, and 1 point for justifying that the constructed set meets the requirements.

In the case of a partial solution, award 1 point for constructing an example of 505 numbers that meet the requirements, and an additional point for proving that no additional number can be added to it.

Answer: 505


1 class has 6 boys, 4 girls, and now 3 class leaders need to be selected from them, requiring that there is at least 1 girl among the class leaders, and each person has 1 role, then there are $\qquad$ ways to select.

Solution:
4. First, calculate all possible selections

There are $10 \times 9 \times 8=720$ ways.
Then calculate the number of ways with no girls selected
There are $6 \times 5 \times 4=120$ ways.
Therefore, there are $720-120=600$ ways in total.

Answer: 600


There are 2014 boxes on the table, some of which contain candies, while the others are empty.

On the first box, it is written: âAll boxes are empty.â

On the second - âAt least 2013 boxes are empty.â

On the third - âAt least 2012 boxes are empty.â

...

On the 2014th - âAt least one box is empty.â

It is known that the inscriptions on the empty boxes are false, while those on the boxes with candies are true. Determine how many boxes contain candies.

Solution:
Answer: 1007.

Solution: Suppose that $N$ boxes are empty, then $2014-N$ boxes contain candies. Note that on the box with number $k$, it is written that there are at least $2015-k$ empty boxes. Therefore, the inscriptions on the boxes with numbers $1,2, \ldots, 2014-N$ are false. Consequently, $N=2014-N$, from which $N=1007$.

Answer: 1007


9. (10 points) In $\overline{\mathrm{ABCD}}+\overline{\mathrm{EFG}}=2010$, different letters represent different digits, then $A+B+C+D+E+F+G$
$=$

Solution:
9. (10 points) In $\overline{\mathrm{ABCD}}+\overline{\mathrm{EFG}}=2010$, different letters represent different digits, then $A+B+C+D+E+F+G=$ $\underline{30}$.
【Solution】Solution: Since $A, B, C, D, E, F, G$ are different digits, from the equation $\overline{\mathrm{ABCD}}+\overline{\mathrm{EFG}}=2010$, we can derive:
$$
D+G=10, C+F=10, B+E=9, A=1,
$$

Therefore: $A+B+C+D+E+F+G$
$$
\begin{array}{l}
=A+(B+E)+(C+F)+(D+G) \\
=1+9+10+10 \\
=30
\end{array}
$$

So the answer is: 30.

Answer: 30


Find the smallest natural number that is a multiple of 36 and in whose representation all 10 digits appear exactly once.

Solution: 21. Since the desired number is divisible by 36, it must be divisible by 4 and 9. In this number, the last two digits should represent the largest two-digit number divisible by 4. This number is 96. The remaining digits should be written from the digit 9 to the left in descending order, but since the leftmost digit cannot be 0, the leftmost digit is 1. Since the sum of all single-digit numbers is 45, the desired number is divisible by 9. The desired number is 1023457896.

Answer: 1023457896


Solution:
Solution.

$$
\begin{aligned}
& \sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}= \\
& =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{(2+\sqrt{2+\sqrt{2+\sqrt{3}}})(2-\sqrt{2+\sqrt{2+\sqrt{3}}})}= \\
& =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2^{2}-(\sqrt{2+\sqrt{2+\sqrt{3}}})^{2}}= \\
& =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{4-2-\sqrt{2+\sqrt{3}}}= \\
& =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2-\sqrt{2+\sqrt{3}}}= \\
& =\sqrt{2+\sqrt{3}} \cdot \sqrt{(2+\sqrt{2+\sqrt{3}})(2-\sqrt{2+\sqrt{3}})}=\sqrt{2+\sqrt{3}} \cdot \sqrt{4-2-\sqrt{3}}= \\
& =\sqrt{2+\sqrt{3}} \cdot \sqrt{2-\sqrt{3}}=\sqrt{(2+\sqrt{3})(2-\sqrt{3})}=\sqrt{2^{2}-(\sqrt{3})^{2}}=\sqrt{4-3}= \\
& =\sqrt{1}=1 .
\end{aligned}
$$

Answer: 1.

Answer: 1


Rewrite the equation as

$$
5^{(x-1)^{2}} \log _{7}\left((x-1)^{2}+2\right)=5^{2|x-a|} \log _{7}(2|x-a|+2)
$$

Let $f(t)=5^{t} \log _{7}(t+2)$. This function is increasing. Therefore,

$$
f\left((x-1)^{2}\right)=f(2|x-a|) \Longleftrightarrow(x-1)^{2}=2|x-a|
$$

There will be three solutions in the case of tangency for $a=1 / 2, a=3 / 2$ and when $a=1$, since the vertices of the parabola $y=(x-1)^{2}$ and the curve $y=|x-1|$ coincide.

Solution:
Answer: $a=1 / 2 ; 1 ; 3 / 2$.

Answer to the option: $6-2: a=-3 / 2 ;-1 ;-1 / 2$.

## Lomonosov Moscow State University
## Olympiad "Conquer Sparrow Hills" Variant 7-1

Answer: 1/2;1;3/2


The Microcalculator MK-97 can perform only three operations on numbers stored in memory:

1) check if two selected numbers are equal,
2) add two selected numbers,
3) find the roots of the equation $x^{2}+a x+b=0$ for selected numbers $a$ and $b$, and if there are no roots, display a message about it.

The results of all actions are stored in memory. Initially, one number $x$ is recorded in memory. How can you use MK-97 to determine if this number is equal to one?

Solution:
By adding $x$ to itself, we get $2x$. We compare $x$ and $2x$. If they are equal, then $x=0$. Otherwise, we find the roots of the equation $y^{2}+2xy+x=0$. The discriminant of this equation is $4(x^{2}-x)$, so the roots are equal if and only if $x=1$.

Send a comment

Answer: 1


There are $\displaystyle{2n}$ students in a school $\displaystyle{\left( {n \in {\Bbb N},n \geqslant 2} \right)}$. Each week $\displaystyle{n}$ students go on a trip. After several trips the following condition was fulfilled: every two students were together on at least one trip. What is the minimum number of trips needed for this to happen?

[i]Proposed by Oleksandr Rybak, Kiev, Ukraine.[/i]

Solution:
To solve the problem, we need to determine the minimum number of trips required such that every pair of students is together on at least one trip. We are given that there are \(2n\) students and each trip consists of \(n\) students.

1. **Establishing the minimum number of trips:**
   - Each student must be paired with every other student at least once.
   - There are \(\binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1)\) pairs of students.
   - Each trip can accommodate \(\binom{n}{2} = \frac{n(n-1)}{2}\) pairs of students.

2. **Calculating the lower bound:**
   - Let \(k\) be the number of trips needed.
   - Each student appears in \(k\) trips, and each trip contains \(n\) students.
   - Each student pairs with \(n-1\) other students in each trip.
   - Therefore, each student pairs with \((n-1)k\) other students across all trips.
   - Since each student must pair with \(2n-1\) other students, we have:
     \[
     (n-1)k \geq 2n-1
     \]
   - Solving for \(k\):
     \[
     k \geq \frac{2n-1}{n-1}
     \]
   - For \(n \geq 2\), this simplifies to:
     \[
     k \geq 2 + \frac{1}{n-1}
     \]
   - Since \(k\) must be an integer, the minimum \(k\) is 3 when \(n=2\), and for larger \(n\), \(k\) must be at least 6.

3. **Constructing the solution for specific \(n\):**
   - For \(n=2\), the set of students is \(\{x, y, z, t\}\). The trips can be:
     \[
     \{x, y\}, \{x, z\}, \{x, t\}, \{y, z\}, \{y, t\}, \{z, t\}
     \]
   - For \(n=3\), the set of students is \(\{a, b, c, d, e, f\}\). The trips can be:
     \[
     \{a, b, c\}, \{a, b, d\}, \{c, d, e\}, \{c, d, f\}, \{e, f, a\}, \{e, f, b\}
     \]

4. **Generalizing the solution:**
   - For larger \(n\), we can use the above models and merge them appropriately.
   - For example, for \(n=5\), we can combine the models for \(n=2\) and \(n=3\).

5. **Conclusion:**
   - The minimum number of trips needed is 6 for any \(n \geq 2\).

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

Answer: 6


What is the fraction with a single-digit denominator that is greater than $\frac{7}{9}$ and less than $\frac{8}{9}$?

Solution:
A possible fraction is the arithmetic mean of $\frac{7}{9}$ and $\frac{8}{9}$:

$$
\frac{\frac{7}{9}+\frac{8}{9}}{2}=\frac{\frac{15}{9}}{2}=\frac{15}{30}=\frac{5}{6}
$$

The problem is not uniquely solvable, as for example, $\frac{6}{7}$ is also such a fraction.

Answer: \frac{5}{6}


Solution:
22. Answer: 1
$$
\begin{array}{l}
\sum_{k=0}^{\infty} \frac{2}{\pi} \tan ^{-1}\left(\frac{2}{(2 k+1)^{2}}\right) \\
=\lim _{n \rightarrow \infty} \sum_{k=0}^{n} \frac{2}{\pi} \tan ^{-1}\left(\frac{2}{(2 k+1)^{2}}\right) \\
=\lim _{n \rightarrow \infty} \sum_{k=0}^{n} \frac{2}{\pi} \tan ^{-1}\left(\frac{2 k+2-2 k}{1+(2 k)(2 k+2)}\right) \\
=\lim _{n \rightarrow \infty} \sum_{k=0}^{n} \frac{2}{\pi}\left(\tan ^{-1}(2 k+2)-\tan ^{-1}(2 k)\right) \\
=\lim _{n \rightarrow \infty} \frac{2}{\pi} \tan ^{-1}(2 n+2)=1 .
\end{array}
$$

Answer: 1


Point $A_{1}$ lies on side $B C$ of triangle $A B C$, point $B_{1}$ lies on side $A C$. Let $K$ be the intersection point of segments $A A_{1}$ and $B B_{1}$. Find the area of quadrilateral $A_{1} C B_{1} K$, given that the area of triangle $A_{1} B K$ is 5, the area of triangle $A B_{1} K$ is 8, and the area of triangle $A B K$ is 10.

Solution:
1.II.5. We will use the following consideration. If a line passes through the vertex of a triangle, then the ratio of the areas of the parts into which it divides the triangle is equal to the ratio of the lengths of the segments into which it divides the opposite side, $\frac{S_{1}}{S_{2}}=\frac{m}{n}$. Let's connect vertex $C$ with point $K$ and set $x=S_{B_{1} C K}$ and $y=S_{A_{1} C K}$ (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-037.jpg?height=417&width=491&top_left_y=1930&top_left_x=771)

Then

$$
\frac{10}{8}=\frac{B K}{K B_{1}}=\frac{y+5}{x} \text { and } \frac{10}{5}=\frac{A K}{K A_{1}}=\frac{x+8}{y} .
$$

Thus, $5 x=4 y+20$ and $2 y=x+8$. Solving the obtained system, we get that $x=12$ and $y=10$, so $x+y=22$.

Answer: 22.

Answer: 22


Solution:
Solve the equations which can be transformed into $x_{1}+\frac{1}{2} x_{2}=1-\frac{1}{3} x_{3}, x_{1}^{2}+\frac{1}{2} x_{2}^{2}=3-\frac{1}{3} x_{3}^{2}$, and ingeniously set the vectors $\vec{a}=\left(x_{1}, \frac{1}{\sqrt{2}} x_{2}\right), \vec{b}=\left(1, \frac{1}{\sqrt{2}}\right)$, then
$$
3-\frac{1}{3} x_{3}^{2}=x_{1}^{2}+\frac{1}{2} x_{2}^{2}=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{\left(x_{1}+\frac{1}{2} x_{2}\right)^{2}}{1+\frac{1}{2}}=\frac{2}{3}\left(1-\frac{1}{3} x_{3}\right)^{2} \text {, }
$$

Solve $3-\frac{1}{3} x_{3}^{2} \geqslant \frac{2}{3}\left(1-\frac{1}{3} x_{3}\right)^{2}$, we get $-\frac{21}{11} \leqslant x_{3} \leqslant 3$, then the minimum value of $x_{3}$ is $-\frac{21}{11}$.

Answer: -\frac{21}{11}


Find the maximum value of
$$\int^1_0|f'(x)|^2|f(x)|\frac{1}{\sqrt{x}}dx$$
over all continuously differentiable functions $f:[0,1]\to\mathbb{R}$ with $f(0)=0$ and
$$\int^1_0|f'(x)|^2dx\le1.$$

Solution:
1. **Applying the Cauchy-Schwarz Inequality:**

   Given the function \( f:[0,1] \to \mathbb{R} \) with \( f(0) = 0 \) and \( \int_0^1 |f'(x)|^2 \, dx \leq 1 \), we start by using the Cauchy-Schwarz inequality to bound \( |f(x)| \).

   \[
   |f(x)| = \left| \int_0^x f'(t) \, dt \right| \leq \int_0^x |f'(t)| \, dt
   \]

   By the Cauchy-Schwarz inequality:

   \[
   \int_0^x |f'(t)| \, dt \leq \left( \int_0^x 1 \, dt \right)^{1/2} \left( \int_0^x |f'(t)|^2 \, dt \right)^{1/2}
   \]

   Simplifying the integrals:

   \[
   \int_0^x 1 \, dt = x \quad \text{and} \quad \int_0^x |f'(t)|^2 \, dt \leq \int_0^1 |f'(t)|^2 \, dt \leq 1
   \]

   Therefore:

   \[
   |f(x)| \leq \sqrt{x} \left( \int_0^1 |f'(t)|^2 \, dt \right)^{1/2} \leq \sqrt{x}
   \]

2. **Bounding the Integral:**

   We need to find the maximum value of the integral:

   \[
   \int_0^1 |f'(x)|^2 |f(x)| \frac{1}{\sqrt{x}} \, dx
   \]

   Using the bound \( |f(x)| \leq \sqrt{x} \):

   \[
   |f(x)| \frac{1}{\sqrt{x}} \leq 1
   \]

   Thus:

   \[
   \int_0^1 |f'(x)|^2 |f(x)| \frac{1}{\sqrt{x}} \, dx \leq \int_0^1 |f'(x)|^2 \, dx \leq 1
   \]

3. **Constructing a Function to Achieve the Bound:**

   To see if the bound can be achieved, consider the function \( f(x) = \sqrt{x} \). Then:

   \[
   f'(x) = \frac{1}{2\sqrt{x}}
   \]

   The integral of \( |f'(x)|^2 \) is:

   \[
   \int_0^1 \left( \frac{1}{2\sqrt{x}} \right)^2 \, dx = \int_0^1 \frac{1}{4x} \, dx = \frac{1}{4} \int_0^1 \frac{1}{x} \, dx
   \]

   This integral diverges, so \( f(x) = \sqrt{x} \) is not a valid choice. Instead, consider \( f(x) = x \). Then:

   \[
   f'(x) = 1
   \]

   The integral of \( |f'(x)|^2 \) is:

   \[
   \int_0^1 1^2 \, dx = 1
   \]

   And the integral we are interested in becomes:

   \[
   \int_0^1 1 \cdot x \cdot \frac{1}{\sqrt{x}} \, dx = \int_0^1 \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^1 = \frac{2}{3}
   \]

   Therefore, the maximum value of the integral is \( \frac{2}{3} \).

The final answer is \( \boxed{\frac{2}{3}} \)

Answer: \frac{2}{3}


The perimeter of a right triangle is 18, and the area is 9. What is the length of the hypotenuse of this triangle?

Solution:
## First Solution.

Let $a$ and $b$ be the lengths of the legs, and $c$ be the length of the hypotenuse of the given triangle.

Since $a+b+c=18$, we have

$$
(a+b)^{2}=(18-c)^{2}=18^{2}-36 c+c^{2}
$$

The given triangle is a right triangle, so by the Pythagorean theorem, $a^{2}+b^{2}=c^{2}$. 1 point It follows that

$$
(a+b)^{2}=a^{2}+2 a b+b^{2}=c^{2}+2 a b
$$

Since we have expressed $(a+b)^{2}$ in two ways, equating them gives

$$
2 a b=18^{2}-36 c
$$

Since the area of the triangle is 9, we have $a b=18$. 1 point

We conclude that $36=2 a b=18^{2}-36 c$. From this, $c=8$. 2 points

## Second Solution.

Let $a$ and $b$ be the lengths of the legs, and $c$ be the length of the hypotenuse of the given triangle.

The given triangle is a right triangle, so by the Pythagorean theorem, $a^{2}+b^{2}=c^{2}$. 1 point

Since $a+b+c=18$, we have

$$
a^{2}+b^{2}=c^{2}=(18-a-b)^{2}
$$

After squaring, we get

$$
a^{2}+b^{2}=18^{2}+a^{2}+b^{2}-36 a-36 b+2 a b
$$

or $18^{2}+2 a b=36(a+b)=36 \cdot(18-c)$. 2 points

Since the area of the triangle is 9, we have $a b=18$. 1 point

We conclude that $18^{2}+36=36 \cdot(18-c)$. From this, $c=8$. 2 points

Answer: 8


If three lines $a$, $b$, and $c$ in space are pairwise skew lines, then the number of lines that intersect with $a$, $b$, and $c$ is ( ).
(A) 0
(B) 2
(C) More than 1 but finite
(D) Infinite

Solution:
Regardless of the positional relationship between lines $a$, $b$, and $c$, a parallelepiped $A B C D-A_{1} B_{1} C_{1} D_{1}$ can always be constructed such that $A B$ lies on line $a$, $B_{1} C_{1}$ lies on line $b$, and $D_{1} D$ lies on line $c$, as shown in Figure 4.

Take a point $M$ on the extension of $D D_{1}$. Determine a plane $\alpha$ with point $M$ and line $a$. Plane $\alpha$ intersects lines $B_{1} C_{1}$ and $A_{1} D_{1}$ at points $P$ and $Q$, respectively. Then $P Q \parallel$ plane $\alpha$.
Thus, within plane $\alpha$, line $P M$ does not parallel line $a$,
so $P M$ must intersect line $a$ at a point $N$.
Therefore, line $M N$ intersects lines $a$, $b$, and $c$ simultaneously.
Since there are infinitely many ways to choose point $M$, there are infinitely many lines that intersect lines $a$, $b$, and $c$ simultaneously.

Answer: D


Let $x \in R$, then the minimum value of the function $f(x)=|2 x-1|+|3 x-2|+|4 x-3|+|5 x-4|$ is

Solution:
Solution: 1.
$$
\begin{aligned}
f(x) & =2\left|x-\frac{1}{2}\right|+3\left|x-\frac{2}{3}\right|+4\left|x-\frac{3}{4}\right|+5\left|x-\frac{4}{5}\right| \\
& =2\left(\left|x-\frac{1}{2}\right|+\left|x-\frac{4}{5}\right|\right)+3\left(\left|x-\frac{2}{3}\right|+\left|x-\frac{4}{5}\right|\right)+4\left|x-\frac{3}{4}\right| \\
& \geqslant 2\left|\left(x-\frac{1}{2}\right)-\left(x-\frac{4}{5}\right)\right|+3\left|\left(x-\frac{2}{3}\right)-\left(x-\frac{4}{5}\right)\right| \\
& =2\left|\frac{4}{5}-\frac{1}{2}\right|+3\left|\frac{4}{5}-\frac{2}{3}\right|=1 .
\end{aligned}
$$

Equality holds if and only if $\left(x-\frac{1}{2}\right)\left(x-\frac{4}{5}\right) \leqslant 0,\left(x-\frac{2}{3}\right)\left(x-\frac{4}{5}\right) \leqslant 0, x-\frac{3}{4}=0$, i.e., $x=\frac{3}{4}$. Therefore, $f(x)_{\text {min }}=f\left(\frac{3}{4}\right)=1$.

Answer: 1


If the geometric sequence $\left\{a_{n}\right\}$ satisfies $a_{1}-a_{2}=3, a_{1}-a_{3}=2$, then the common ratio of $\left\{a_{n}\right\}$ is $\qquad$

Solution: Let $\left\{a_{n}\right\}$ have a common ratio of $q$, then $\left\{\begin{array}{l}a_{1}(1-q)=3, \\ a_{1}\left(1-q^{2}\right)=2\end{array} \Rightarrow 1+q=\frac{2}{3} \Rightarrow q=-\frac{1}{3}\right.$.

Answer: -\frac{1}{3}


Suppose $N$ is any positive integer. Add the digits of $N$ to obtain a smaller integer. Repeat this process of digit-addition till you get a single digit number. Find the number of positive integers $N \le 1000$, such that the final single-digit number $n$ is equal to $5$.
Example: $N = 563\to (5 + 6 + 3) = 14 \to(1 + 4) = 5$ will be counted as one such integer.

Solution:
1. **Understanding the Problem:**
   We need to find the number of positive integers \( N \leq 1000 \) such that the final single-digit number obtained by repeatedly adding the digits of \( N \) is equal to 5.

2. **Digit Sum and Modulo 9:**
   The process of repeatedly adding the digits of a number until a single digit is obtained is equivalent to finding the number modulo 9. This is because the sum of the digits of a number \( N \) is congruent to \( N \) modulo 9. This is a well-known property in number theory.

3. **Finding Numbers Congruent to 5 Modulo 9:**
   We need to find the number of integers \( N \) such that \( N \equiv 5 \pmod{9} \). This means we are looking for numbers of the form:
   \[
   N = 9k + 5
   \]
   where \( k \) is a non-negative integer.

4. **Range of \( N \):**
   Since \( N \leq 1000 \), we need to find the values of \( k \) such that:
   \[
   9k + 5 \leq 1000
   \]
   Solving for \( k \):
   \[
   9k + 5 \leq 1000 \implies 9k \leq 995 \implies k \leq \frac{995}{9} \implies k \leq 110.555\ldots
   \]
   Since \( k \) must be an integer, the largest possible value of \( k \) is 110.

5. **Counting the Values:**
   The values of \( k \) range from 0 to 110, inclusive. Therefore, there are:
   \[
   110 - 0 + 1 = 111
   \]
   possible values of \( k \).

6. **Conclusion:**
   Thus, there are 111 positive integers \( N \leq 1000 \) such that the final single-digit number obtained by repeatedly adding the digits of \( N \) is equal to 5.

The final answer is \(\boxed{111}\).

Answer: 111


The diagram shows a semicircle with diameter $P Q$ inscribed in a rhombus $A B C D$. The rhombus is tangent to the arc of the semicircle in two places. Points $P$ and $Q$ lie on sides $B C$ and $C D$ of the rhombus respectively. The line of symmetry of the semicircle is coincident with the diagonal $A C$ of the rhombus. It is given that $\angle C B A=60^{\circ}$. The semicircle has radius 10 . The area of the rhombus can be written in the form $a \sqrt{b}$ where $a$ and $b$ are integers and $b$ is prime. What is the value of $a b+a+b ?$

Solution:
18. 603 Let $O$ be the centre of the semicircle and let $M$ and $N$ be the feet of the perpendiculars drawn from $O$ to $A B$ and $A D$ respectively. Let $G$ be the intersection of the diagonals of the rhombus.
$$
\begin{array}{l}
P O=10 \text { and } \angle O P C=30^{\circ} . \text { So } O C=10 \tan 30^{\circ}=\frac{10}{\sqrt{3}} . \\
M O=10 \text { and } \angle O A M=60^{\circ} . \text { So } A O=\frac{10}{\sin 60^{\circ}}=\frac{20}{\sqrt{3}} .
\end{array}
$$

Therefore $A C=\frac{10}{\sqrt{3}}+\frac{20}{\sqrt{3}}=\frac{30}{\sqrt{3}}=10 \sqrt{3}$.
Hence $A G=5 \sqrt{3}$ and $\angle G B A=30^{\circ}$. So $B G=\frac{5 \sqrt{3}}{\tan 30^{\circ}}=15$.
Therefore the area of triangle $B G A$ is $\frac{1}{2} \times 15 \times 5 \sqrt{3}=\frac{75}{2} \sqrt{3}$.
So the area of the rhombus is $4 \times \frac{75}{2} \sqrt{3}=150 \sqrt{3}$.
Therefore $a=150$ and $b=3$, so $a b+a+b=450+150+3=603$.

Answer: 603


Using three 2s, the largest number that can be written is ( ).
(A) $2^{2^{2}}$
(B) 222
(C) 242
(D) greater than 1000

Solution:
6. D.

The largest number is $2^{22}>1000$.

Answer: D


Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$. The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. A parallelepiped is a solid with six parallelogram faces such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy]

Solution:
Denote $\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}$.
Denote by $d$ the length of each side of a rhombus.
Now, we put the solid to the 3-d coordinate space.
We put the bottom face on the $x-O-y$ plane.
For this bottom face, we put a vertex with an acute angle $2 \alpha$ at the origin, denoted as $O$.
For two edges that are on the bottom face and meet at $O$, we put one edge on the positive side of the $x$-axis. The endpoint is denoted as $A$. Hence, $A = \left( d , 0 , 0 \right)$.
We put the other edge in the first quadrant of the $x-O-y$ plane. The endpoint is denoted as $B$. Hence, $B = \left( d \cos 2 \alpha , d \sin 2 \alpha , 0 \right)$.
For the third edge that has one endpoint $O$, we denote by $C$ its second endpoint.
We denote $C = \left( u , v , w \right)$.
Without loss of generality, we set $w > 0$.
Hence,
\[ u^2 + v^2 + w^2 = d^2 . \hspace{1cm} (1) \]
We have
\begin{align*} \cos \angle AOC & = \frac{\overrightarrow{OA} \cdot \overrightarrow{OC}}{|OA| \cdot |OC|} \\ & = \frac{u}{d} , \hspace{1cm} (2) \end{align*}
and
\begin{align*} \cos \angle BOC & = \frac{\overrightarrow{OB} \cdot \overrightarrow{OC}}{|OB| \cdot |OC|} \\ & = \frac{u \cos 2 \alpha + v \sin 2 \alpha}{d} . \hspace{1cm} (3) \end{align*}
Case 1: $\angle AOC = \angle BOC = 2 \alpha$ or $2 \left( 90^\circ - \alpha \right)$.
By solving (2) and (3), we get
\begin{align*} u & = \pm d \cos 2 \alpha , \\ v & = \pm d \cos 2 \alpha \frac{1 - \cos 2 \alpha}{\sin 2 \alpha} \\ & = \pm d \cos 2 \alpha \tan \alpha . \end{align*}
Plugging these into (1), we get
\begin{align*} w & = d \sqrt{1 - \cos^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha} \\ & = d \sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha} . \hspace{1cm} (4) \end{align*}
Case 2: $\angle AOC = 2 \alpha$ and $\angle BOC = 2 \left( 90^\circ - \alpha \right)$, or $\angle BOC = 2 \alpha$ and $\angle AOC = 2 \left( 90^\circ - \alpha \right)$.
By solving (2) and (3), we get
\begin{align*} u & = \pm d \cos 2 \alpha , \\ v & = \mp d \cos 2 \alpha \frac{1 + \cos 2 \alpha}{\sin 2 \alpha} \\ & = \mp d \cos 2 \alpha \cot \alpha . \end{align*}
Plugging these into (1), we get
\begin{align*} w & = d \sqrt{1 - \cos^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha} \\ & = d \sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha} . \hspace{1cm} (5) \end{align*}
We notice that $(4) > (5)$. Thus, (4) (resp. (5)) is the parallelepiped with a larger (resp. smaller) height.
Therefore, the ratio of the volume of the larger parallelepiped to the smaller one is
\begin{align*} \frac{(4)}{(5)} & = \frac{\sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha}} {\sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha}} \\ & = \sqrt{\frac{\tan^2 2 \alpha - \tan^2 \alpha}{\tan^2 2 \alpha - \cot^2 \alpha}} . \end{align*}
Recall that $\tan \alpha = \frac{\sqrt{21}}{\sqrt{31}}$.
Thus, $\tan 2 \alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{\sqrt{21 \cdot 31}}{5}$.
Plugging this into the equation above, we get
\begin{align*} \frac{(4)}{(5)} & = \frac{63}{62}. \end{align*}
Therefore, the answer is $63 + 62 = \boxed{\textbf{(125) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Answer: 125


Ma Peng and Li Hu calculated the product of two two-digit numbers, Jia and Yi. Ma Peng misread the unit digit of Jia, resulting in a product of 473; Li Hu misread the tens digit of Jia, resulting in a product of 407. What should the product of Jia and Yi be? $\qquad$

Solution: Answer: 517

Answer: 517


Replace the 26 English letters with numbers 0 to 25, according to the following correspondence:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline $\mathrm{A}$ & $\mathrm{B}$ & $\mathrm{C}$ & $\mathrm{D}$ & $\mathrm{E}$ & $\mathrm{F}$ & $\mathrm{G}$ & $\mathrm{H}$ & $\mathrm{I}$ & $\mathrm{J}$ & $\mathrm{K}$ & $\mathrm{L}$ & $\mathrm{M}$ \\
\hline 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline $\mathrm{N}$ & 0 & $\mathrm{P}$ & $\mathrm{Q}$ & $\mathrm{R}$ & $\mathrm{S}$ & $\mathrm{T}$ & $\mathrm{U}$ & $\mathrm{V}$ & $\mathrm{W}$ & $\mathrm{X}$ & $\mathrm{Y}$ & $\mathrm{Z}$ \\
\hline 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 \\
\hline
\end{tabular}

Replace the letters in the pinyin Ã¢ÂÂ $x_{1} x_{2} x_{3} x_{4} x_{5}$ Ã¢ÂÂ with the corresponding numbers from the table above, then $x_{1}+x_{2}, x_{1}+x_{4}, x_{3}$, $x_{2}+x_{5}, x_{3}+x_{5}$ modulo 26 are $25,15,20,11,24$ respectively. Find the Chinese pinyin $x_{1} x_{2} x_{3} x_{4} x_{3} x_{5}$.

Solution: Reference answer: SHUXUE

Answer: SHUXUE


Determine all ordered pairs of natural numbers $(n ; m)$ that satisfy the equation $2^{n}+65=m^{2}$!

Solution:
For odd $n$, $2^{n} \equiv \pm 2(\bmod 10)$, so $2^{m}+65 \equiv (5 \pm 2)(\bmod 10)$ and thus is not a square of a natural number [for the squares of natural numbers, it is well-known that one of the congruences $\left.m^{2} \equiv 0 ; 1 ; 4 ; 5 ; 6 ; 9(\bmod 10)\right]$ always holds]. Therefore, there is no solution for odd $n$.

Now let $n=2 k$ with $k \in \mathbb{N}$. Then

$$
2^{n}5$ follows

$$
\left(2^{k}\right)^{2}+2 \cdot 2^{5}+15$ the following chain of inequalities:

$$
2^{n}=\left(2^{k}\right)^{2}<2^{n}+65<\left(2^{k}+1\right)^{2}
$$

Thus, $m^{2}=2^{n}+65$ lies between the squares of two consecutive natural numbers and therefore cannot itself be the square of a natural number.

The examination of the cases $k=0 ; 1 ; 2 ; 3 ; 4 ; 5$ yields the two solutions $\left(n_{1}=2 k_{1} ; m_{1}\right)=(4 ; 9)$, $\left(n_{2}=2 k_{2} ; m_{2}\right)=(10 ; 33)$. The verification confirms the correctness.

Answer: (4,9),(10,33)


There are $n$ parking spaces along a road, and $n$ drivers each driving a car. Each driver parks their car in front of their favorite parking space. If that space is already occupied, they park in the nearest available space down the road. If that space and all the spaces below it are occupied, they drive away without parking. Question: How many arrays $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ can ensure that no parking space is empty? Here, $a_{i}$ is the favorite parking space of the $i$-th driver, and $a_{1}, a_{2}, \cdots, a_{n}$ do not have to be distinct.

Solution:
There are $(n+1)^{n-1}$ arrays that satisfy the conditions.
For any array $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ that satisfies the conditions, define an associated array $\left(b_{2}, \cdots, b_{n}\right)$. Here, $b_{i}$ is the difference between the $i$-th driver's intended parking spot and the position of the car parked by the previous person (i.e., the $(i-1)$-th person), under the modulo $(n+1)$ operation. Clearly, two different arrays $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ correspond to different associated arrays.

Next, we prove: For any array $\left(b_{2}, \cdots, b_{n}\right), b_{i} \in\{0,1, \cdots, n\}$, there is an array that satisfies the conditions with it as the associated array.

Imagine $n$ parking spots arranged in a circle, with the last position (the $(n+1)$-th) being a virtual spot. Place the first driver in any position. For $i=2, \cdots, n$, the $i$-th driver parks in the next available spot that is $b_{i}$ positions away from the spot where the $(i-1)$-th driver parked. The array $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ satisfies the conditions if and only if the last empty spot is the virtual spot. Note that by rotating the position of the first driver, we can ensure that the empty spot is the virtual spot. Therefore, the number of arrays $\left(b_{2}, \cdots, b_{n}\right)$ is the same as the number of arrays $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ that satisfy the conditions, which is $(n+1)^{n-1}$.

Answer: (n+1)^{n-1}


For Eeyore's Birthday, Winnie-the-Pooh, Owl, and Piglet decided to give balloons. Winnie-the-Pooh prepared twice as many balloons as Piglet, and Owl prepared three times as many balloons as Piglet. When Piglet was carrying his balloons, he was in a hurry, stumbled, and some of the balloons burst. Eeyore received a total of 31 balloons for his party. How many balloons did Piglet end up giving?

Solution:
# Answer: 1

Solution. If Piglet hadn't burst his balloons, then Eeyore would have received exactly 6 times more balloons than Piglet prepared. The next number after 31 that is a multiple of 6 is $36, 36: 6=6$, which means Piglet prepared no less than 6 balloons.

If all of Piglet's balloons had burst, then Eeyore would have received exactly 5 times more than Piglet prepared. If Piglet initially had 7 or more balloons, then even now Eeyore would have received at least 35 balloons, which is not the case. Therefore, Piglet prepared exactly 6 balloons. If everything had been fine, Eeyore would have received 36 balloons, but since 5 burst, Piglet gave 1 balloon.

Answer: 1


Given the function $f(x)=a x^{2}-c$, satisfying $-4 \leqslant f(1) \leqslant -1, -1 \leqslant f(2) \leqslant 5$, $f(3)$'s range of values is $\qquad$.

Solution:
2. $-1 \leqslant f(3) \leqslant 20$.

Let $f(3)=m f(1)+n f(2)$, i.e., $9a-c=m(a-c)+n(4a-c), a, c$ have corresponding coefficients respectively $\frac{8}{3} f(2) \leqslant 20$.

Answer: -1\leqslantf(3)\leqslant20


Solution:
To solve the given problem, we need to find the limit of the integral as \( n \) approaches infinity:

\[
\lim_{n \to \infty} \int_0^1 f(x) g_n(x) \, dx
\]

where \( g_n(x) = n - 2n^2 \left| x - \frac{1}{2} \right| + \left| n - 2n^2 \left| x - \frac{1}{2} \right| \right| \).

First, let's simplify \( g_n(x) \):

1. **Simplify \( g_n(x) \):**
   \[
   g_n(x) = n - 2n^2 \left| x - \frac{1}{2} \right| + \left| n - 2n^2 \left| x - \frac{1}{2} \right| \right|
   \]

   Consider the term \( n - 2n^2 \left| x - \frac{1}{2} \right| \):
   - When \( x = \frac{1}{2} \), \( \left| x - \frac{1}{2} \right| = 0 \), so \( n - 2n^2 \left| x - \frac{1}{2} \right| = n \).
   - As \( x \) moves away from \( \frac{1}{2} \), \( \left| x - \frac{1}{2} \right| \) increases, making \( 2n^2 \left| x - \frac{1}{2} \right| \) large.

   For large \( n \), the term \( 2n^2 \left| x - \frac{1}{2} \right| \) dominates, and \( n - 2n^2 \left| x - \frac{1}{2} \right| \) becomes negative for \( x \neq \frac{1}{2} \).

2. **Behavior of \( g_n(x) \) as \( n \to \infty \):**
   \[
   g_n(x) = n - 2n^2 \left| x - \frac{1}{2} \right| + \left| n - 2n^2 \left| x - \frac{1}{2} \right| \right|
   \]

   For \( x \neq \frac{1}{2} \), \( n - 2n^2 \left| x - \frac{1}{2} \right| \) is negative, so:
   \[
   \left| n - 2n^2 \left| x - \frac{1}{2} \right| \right| = 2n^2 \left| x - \frac{1}{2} \right| - n
   \]

   Thus,
   \[
   g_n(x) = n - 2n^2 \left| x - \frac{1}{2} \right| + 2n^2 \left| x - \frac{1}{2} \right| - n = 0 \quad \text{for} \quad x \neq \frac{1}{2}
   \]

   For \( x = \frac{1}{2} \):
   \[
   g_n\left( \frac{1}{2} \right) = n
   \]

3. **Integral of \( f(x) g_n(x) \):**
   \[
   \int_0^1 f(x) g_n(x) \, dx
   \]

   As \( n \to \infty \), \( g_n(x) \) behaves like a Dirac delta function centered at \( x = \frac{1}{2} \) with height \( n \). Therefore, we can use the properties of the Dirac delta function to evaluate the integral.

   By the theorem stated in the problem, if \( f_n(x) \) is a sequence of functions that converges to the Dirac delta function, then:
   \[
   \lim_{n \to \infty} \int_{-M}^M f_n(x) g(x) \, dx = g(a)
   \]

   Here, \( f_n(x) = g_n(x) \) and \( g(x) = f(x) \), and the Dirac delta is centered at \( x = \frac{1}{2} \). Thus:
   \[
   \lim_{n \to \infty} \int_0^1 f(x) g_n(x) \, dx = f\left( \frac{1}{2} \right)
   \]

\(\blacksquare\)

The final answer is \( \boxed{ f\left( \frac{1}{2} \right) } \)

Answer: f\left( \frac{1}{2} \right)


Let $a, b, c$ be three distinct real numbers, and $p(x)$ be a polynomial with real coefficients. It is known that: (1) $p(x)$ divided by $x-a$ yields a remainder of $a$; (2) $P(x)$ divided by $x-b$ yields a remainder of $b$; (3) $p(x)$ divided by $x-c$ yields a remainder of $c$. Find the remainder when the polynomial $p(x)$ is divided by $(x-a)(x-b)(x-c)$.

Solution:
Given that the remainder of $p(x)$ divided by $x-a$ is $a$, we can set
$$
p(x)=(x-a) \cdot Q_{1}(x)+a \text {. }
$$

Thus, $p(b)=(b-a) \cdot Q_{1}(b)+a=b$. Noting that $a \neq b$,
we have $Q_{1}(b)=1$, which means $Q_{1}(x)=(x-b) \cdot Q_{2}(x)+1$.
Substituting the above into (1), we get $p(x)=(x-a)(x-b) \cdot Q_{2}(x)+x$.
From condition (3), we have $p(c)=(c-a)(c-b) \cdot Q_{2}(c)+c=c$,
which gives $Q_{2}(c)=0$, so $Q_{2}(x)=(x-c) \cdot Q_{3}(x)$.
Substituting the above into (2), we get $p(x)=(x-a)(x-b)(x-c) \cdot Q_{3}(x)+x$.
Therefore, the remainder of $p(x)$ divided by $(x-a)(x-b)(x-c)$ is $x$.

Answer: x


10 $50 \mathrm{~g}$ weights and 5 $100 \mathrm{~g}$ weights are placed on both sides of the balance at the same time to keep the balance, then if 2 $1 \mathrm{~kg}$ weights are placed on the left side of the balance, and 6 $300 \mathrm{~g}$ weights are placed on the right side, how many $50 \mathrm{~g}$ weights need to be added to the right side to keep the balance?

Solution: Reference answer: 4

Answer: 4


If the equation $x^{2}-x y-2 y^{2}+x+a=0$ represents two straight lines, then $a=$

Solution:
7. $\frac{2}{9}$.

Notice that, $x^{2}-x y-2 y^{2}=(x+y)(x-2 y)$.
Let $x^{2}-x y-2 y^{2}+x+a$
$$
=(x+y+m)(x-2 y+2 m) \text {. }
$$

Expanding the right side of the above equation and comparing the coefficients on both sides, we get
$$
\left\{\begin{array} { l }
{ a = 2 m ^ { 2 } , } \\
{ 3 m = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=\frac{1}{3}, \\
a=\frac{2}{9} .
\end{array}\right.\right.
$$

Answer: \frac{2}{9}


For positive integers $n(n \geqslant 2)$, let $a_{n}=\sum_{k=1}^{n-1} \frac{n}{(n-k) 2^{k-1}}$.
Find the maximum value in the sequence $\left\{a_{n}\right\}$.

Solution:
Sure, here is the translated text:

```
9. Calculate the following:
$$
a_{2}=2, a_{3}=3, a_{4}=a_{5}=\frac{10}{3} \text {. }
$$

We will prove by mathematical induction that:
When $n \geqslant 5$, $a_{n} \leqslant \frac{10}{3}$.
Assume $a_{n} \leqslant \frac{10}{3}(n \geqslant 5)$.
Then $a_{n+1}=\sum_{k=1}^{n} \frac{n+1}{(n+1-k) 2^{k-1}}$
$$
\begin{array}{l}
=\frac{n+1}{n}+\frac{n+1}{2 n} a_{n} \leqslant \frac{n+1}{n} \times \frac{8}{3} \\
\leqslant \frac{6}{5} \times \frac{8}{3}<\frac{10}{3} .
\end{array}
$$

Therefore, the maximum value in the sequence $\left\{a_{n}\right\}$ is
$$
a_{4}=a_{5}=\frac{10}{3} \text {. }
$$
```

Answer: \frac{10}{3}


For the prevention and control of infectious diseases, the "four early" measures refer to $\qquad$
A. early detection, early reporting, early isolation, early treatment
B. early detection, early disinfection, early diagnosis, early treatment
C. early detection, early diagnosis, early reporting, early isolation
D. early detection, early diagnosis, early disinfection, early isolation

Solution:
【Answer】A
【Analysis】Early detection, early reporting, early isolation, and early treatment are the key measures to control the spread of infectious diseases, abbreviated as the “four early actions”. The spread of infectious diseases in the population must have three basic links: the source of infection, the transmission route, and the susceptible population. The absence of any one of these links will not lead to new infections and outbreaks. Preventive and control measures for infectious diseases can be divided into three categories: controlling the source of infection, cutting off the transmission route, and protecting the susceptible population. Adhering to the “four early actions”, ensuring early detection, early reporting, early isolation, and early treatment of patients and asymptomatic carriers, prevents a larger-scale outbreak and protects the health of a broader population; at the same time, through the “four early actions”, timely discovery and treatment of patients can be achieved, minimizing the damage of the epidemic to people's lives and health.

【Comment】“The best doctor treats the disease before it occurs,” preventing problems before they arise. The country implements a prevention-oriented policy for the prevention and control of infectious diseases, strengthening the learning of knowledge about the prevention and treatment of infectious diseases, enhancing self-protection awareness, and fostering a civilized and healthy lifestyle.

Answer: A


In a rectangular box $ABCD EFGH$ with edge lengths $AB=AD=6$ and $AE=49$, a plane slices through point $A$ and intersects edges $BF, FG, GH, HD$ at points $P, Q, R, S$ respectively. Given that $AP=AS$ and $PQ=QR=RS$, find the area of pentagon $APQRS$.

Solution:
Answer:
$$
\frac{141 \sqrt{11}}{2}
$$

Let $A D$ be the positive $x$-axis, $A B$ be the positive $y$-axis, and $A E$ be the positive $z$-axis, with $A$ the origin. The plane, which passes through the origin, has equation $k_{1} x+k_{2} y=z$ for some undetermined parameters $k_{1}, k_{2}$. Because $A P=A S$ and $A B=A D$, we get $P B=S D$, so $P$ and $S$ have the same $z$-coordinate. But $P\left(0,6,6 k_{2}\right)$ and $S\left(6,0,6 k_{1}\right)$, so $k_{1}=k_{2}=k$ for some $k$. Then $Q$ and $R$ both have $z$-coordinate 49 , so $Q\left(\frac{49}{k}-6,6,49\right)$ and $R\left(6, \frac{49}{k}-6,49\right)$. The equation $Q R^{2}=R S^{2}$ then gives
$$
\left(\frac{49}{k}-6\right)^{2}+(49-6 k)^{2}=2\left(12-\frac{49}{k}-12\right)^{2} \text {. }
$$

This is equivalent to
$$
(49-6 k)^{2}\left(k^{2}+1\right)=2(49-12 k)^{2},
$$
which factors as
$$
(k-7)\left(36 k^{3}-336 k^{2}-203 k+343\right)=0 .
$$

This gives $k=7$ as a root. Note that for $Q$ and $R$ to actually lie on $F G$ and $G H$ respectively, we must have $\frac{49}{6} \geq k \geq \frac{49}{12}$. Via some estimation, one can show that the cubic factor has no roots in this range (for example, it's easy to see that when $k=1$ and $k=\frac{336}{36}=\frac{28}{3}$, the cubic is negative, and it also remains negative between the two values), so we must have $k=7$.
Now consider projecting $A P Q R S$ onto plane $A B C D$. The projection is $A B C D$ save for a triangle $Q^{\prime} C R^{\prime}$ with side length $12-\frac{49}{k}=5$. Thus the projection has area $36-\frac{25}{2}=\frac{47}{2}$. Since the area of the projection equals $[A P Q R S] \cdot \cos \theta$, where $\theta$ is the (smaller) angle between planes $A P Q R S$ and $A B C D$, and since the planes have normal vectors $(k, k,-1)$ and $(0,0,1)$ respectively, we get $\cos \theta=\frac{(k, k,-1) \cdot(0,0,1)}{\sqrt{k^{2}+k^{2}+1}}=\frac{1}{\sqrt{2 k^{2}+1}}=\frac{1}{\sqrt{99}}$ and so
$$
[A P Q R S]=\frac{47 \sqrt{99}}{2}=\frac{141 \sqrt{11}}{2} .
$$

Answer: \frac{141\sqrt{11}}{2}


Determine all triples of real numbers $(x ; y ; z)$ that are solutions to the following system of equations (1), (2), (3):

$$
\begin{aligned}
x \cdot y & =2 \\
x \cdot z & =3 \\
x^{2}+y^{2} & =5
\end{aligned}
$$

Solution:
I. If a triple $(x ; y ; z)$ of real numbers is a solution to the system of equations (1), (2), (3), then: From (1), $y \neq 0$, so from (1) we have $x=\frac{2}{y}$. Substituting this into (3) yields $x^{2}+\frac{4}{x^{2}}=5$, thus

$$
x^{4}-5 x^{2}+4=0
$$

Therefore, the number $u=x^{2} \quad$ (5) satisfies the equation $u^{2}-5 u+4=0 \quad$ (6).

From (6), it follows that $u$ is one of the numbers

$$
u_{1,2}=\frac{1}{2}(5 \pm \sqrt{25-16}) \quad \Rightarrow u_{1}=4, \quad u_{2}=1
$$

Thus, $x$ is, according to (5), one of the numbers

$$
x_{1}=2 ; \quad x_{2}=-2 ; \quad x_{3}=1 ; \quad x_{4}=-1
$$

From (1), the corresponding value for $y$ is

$$
y_{1}=1 ; \quad y_{2}=-1 ; \quad y_{3}=2 ; \quad y_{4}=-2
$$

and from (2), the corresponding value for $z$ is

$$
z_{1}=\frac{3}{2} ; \quad z_{2}=-\frac{3}{2} ; \quad z_{3}=3 ; \quad z_{4}=-3
$$

Therefore, only the triples

$$
\left(2 ; 1 ; \frac{3}{2}\right) ; \quad\left(-2 ;-1 ;-\frac{3}{2}\right) ; \quad(1 ; 2 ; 3) ; \quad(-1 ;-2 ;-3)
$$

can satisfy the system of equations (1), (2), (3). They do satisfy it, as can be verified by substitution.

Adapted from $[5]$

2. Solution:

If we consider the equation "(3)+2 (1)", we obtain $9=x^{2}+2 x y+y^{2}=(x+y)^{2}$, thus $x+y= \pm 3$. Similarly, from "(3)+2 (1)" we get the equation $1=x^{2}-2 x y+y^{2}=(x-y)^{2}$, thus $x-y= \pm 1$.

We perform a complete case analysis based on the signs of $x+y$ and $x-y$:

Case 1: $x+y=3$.

Case 1.1: $x-y=1$. It follows that $(x, y, z)=\left(2,1, \frac{3}{2}\right)$.

Case 1.2: $x-y=-1$. It follows that $(x, y, z)=(1,2,3)$.

Case 2: $x+y=-3$.

Case 2.1: $x-y=1$. It follows that $(x, y, z)=(-1,-2,-3)$.

Case 2.2: $x-y=-1$. It follows that $(x, y, z)=\left(-2,-1,-\frac{3}{2}\right)$.

Thus, exactly the four mentioned triples solve the system of equations, as confirmed by substitution.

Answer: (2;1;\frac{3}{2});(-2;-1;-\frac{3}{2});(1;2;3);(-1;-2;-3)


Agnese, Beatrice, Claudio, and Dario are playing with 53 stacks of coins. No matter which two stacks are chosen, they have a different number of coins. On each turn, a player chooses a stack and removes one coin from it. The player loses if, by removing a coin from a stack, they make that stack the same height as another stack on the table.

A stack can have 0 coins, and two stacks with 0 coins are considered equal. Agnese starts, followed by Beatrice, Claudio, and Dario, after which it is Agnese's turn again, and the order continues.

If at the beginning of the game there are 2020 coins in total and if all players play optimally, who loses?
(A) Agnese
(B) Beatrice
(C) Claudio
(D) Dario
(E) It is not possible to determine with the given information

Solution:
10. The answer is (C). Let $a_{1}, a_{2}, \ldots, a_{53}$ be the piles ordered in increasing height.

We observe that the game ends. With each move, the total number of tokens decreases by 1, and if there are 51 tokens in total, the game is already over because there are at least two piles with 0 tokens.

Certainly, if the game has not already been lost, $a_{i}$ will have fewer tokens than $a_{j}$ if $i<j$: if this were not the case, since the number of tokens decreases by 1 with each move, there is a moment when the two piles have the same number of tokens, but this implies that the game is over, which is absurd.

Moreover, the only configuration in which each move leads the player to lose (let's call it the decisive configuration) is the one where the heights of the piles are $0,1,2, \ldots, 52$, in some order. Indeed, if the height difference between $a_{i+1}$ and $a_{i}$ were at least 2 for some $i$, then we could remove a token from $a_{i+1}$; if $a_{1}$ had more than 0 tokens, then we could remove a token from this pile.

Thus, the pile $a_{i}$ at the end of the game will have $i-1$ tokens for every $i \in\{1,2, \ldots, 53\}$. The number $N$ of tokens that will be removed to reach the decisive configuration is equal to the difference between the number of tokens initially present and the number of tokens in the final configuration, that is,

$$
N=2020-(0+1+\cdots+52)=2020-\frac{52 \cdot 53}{2}=642
$$

Since we can write $N=4 \cdot 160+2$ and there are 4 players, when the decisive configuration is reached, it will be Claudio's turn, so Claudio loses.

Answer: C


Inside a square of side length 1, four quarter-circle arcs are traced with the edges of the square serving as the radii. It is known that these arcs intersect pairwise at four distinct points, which in fact are the vertices of a smaller square. Suppose this process is repeated for the smaller square, and so on and so forth. What is the sum of the areas of all squares formed in this manner?

Solution:
Answer: $\frac{1+\sqrt{3}}{2}$.
Solution: By similarity, we note that the areas of the squares are in geometric progression. Hence, we need only to find out the area of the first smaller square. Note that the diagonal of the smaller square is the overlap of two line segments of length $\frac{\sqrt{3}}{2}$, with a total length of 1 . This is going to be $\sqrt{3}-1$, and so the area of the smaller square is $\frac{1}{2}(\sqrt{3}-1)^{2}=2-\sqrt{3}$.
Thus, the sum of the areas of all squares is given by $\frac{1}{1-(2-\sqrt{3})}=\frac{1+\sqrt{3}}{2}$.

Answer: \frac{1+\sqrt{3}}{2}


The numbers $a, b, c, d$ belong to the interval $[-10.5,10.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$.

Solution:
Answer: 462

Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-22 \leqslant y-1 \leqslant 20$ and $-19 \leqslant 2-x \leqslant 23$. Therefore, $(y-1)(2-x)+2 \leqslant 20 \cdot 23+2=462$. The maximum value of the expression is achieved when $a=c=-10.5, b=d=10.5$.

Answer: 462


6. $a, b, c$ are all positive integers, and satisfy $a b+b c=3984$, $a c+b c=1993$. Then the maximum value of $a b c$ is $\qquad$

Solution: 6. Since 1993 is a prime number, from $c(a+b)=1993$, we get $c=1, a+b=1993$. Substituting $b=1993-a$ into the other equation, we get $a^{2}-1992 a+1991=0$. Solving this, we get $a_{1}=1, a_{2}=1991$, then $b_{1}=1992, b_{2}=2$. Therefore, $(a, b, c)$ has two sets $(1,1992,1),(1991,2,1)$. Thus, the maximum value of $abc$ is 3982.

Answer: 3982


Let $f(x)$ represent a quartic polynomial in $x$. If $f(1)=f(2)=f(3)=0, f(4)=6$, $f(5)=72$, then the last digit of $f(2010)$ is $\qquad$.

Solution:
【Analysis】It is given in the problem that $1,2,3$ are three zeros of the function $f(x)$, so we consider starting from the zero point form of the function.
Solution From the problem, we know
$$
f(1)=f(2)=f(3)=0 \text {. }
$$

Let the one-variable quartic polynomial function be
$$
f(x)=(x-1)(x-2)(x-3)(a x+b) \text {, }
$$

where $a, b \in \mathbf{R}$.
Therefore, $\left\{\begin{array}{l}f(4)=6(4 a+b)=6, \\ f(5)=24(5 a+b)=72 .\end{array}\right.$
Solving these, we get $a=2, b=-7$.
Thus, $f(x)=(x-1)(x-2)(x-3)(2 x-7)$.
Hence, $f(2010)$
$$
\begin{array}{l}
=(2010-1)(2010-2)(2010-3)(4020-7) \\
\equiv 2(\bmod 10) .
\end{array}
$$

Therefore, the last digit of $f(2010)$ is 2.

Answer: 2


A right circular cone is inscribed in a cylinder such that its base coincides with one base of the cylinder. The vertex of the cone is located at the center of the other base of the cylinder. If the measure of the central angle of the cone's lateral surface is $120^{\circ}$, determine the ratio of the surface areas of the cone and the cylinder.

Solution:
## Solution.

The surface area of a cylinder with radius $R$ and height $H$ is $O_{V}=2 R \pi(R+H)$.

The surface area of a cone with the same base, height $H$, and slant height $s$ is $O_{S}=R \pi(R+s)$.

Then, $\frac{O_{S}}{O_{V}}=\frac{R+s}{2(R+H)}$.

Since the central angle of the lateral surface $\alpha=120^{\circ}$, we have:

$\frac{360^{\circ}}{120^{\circ}}=\frac{s}{R}$, so $R=\frac{1}{3} s$.

$H=\sqrt{s^{2}-R^{2}}$, so $H=\frac{2 \sqrt{2}}{3} s$.

1 point

Therefore:

$\frac{O_{S}}{O_{V}}=\frac{R+s}{2(R+H)}=\frac{2}{1+2 \sqrt{2}}=\frac{-2+4 \sqrt{2}}{7}$.

Note: The student does not need to rationalize the denominator in the final solution to receive full marks.

Answer: \frac{-2+4\sqrt{2}}{7}


In a small town, there is only one tram line. It is circular, and trams run along it in both directions. There are stops on the loop: Circus, Park, and Zoo. The journey from Park to Zoo via Circus is three times longer than not via Circus. The journey from Circus to Zoo via Park is twice as short as not via Park. Which route from Park to Circus is shorter - via Zoo or not via Zoo - and by how many times?

(A.V. Shapovalov)

Solution:
Answer: The path not through the Zoo is 11 times shorter.

Solution. Let's board the tram at the Zoo stop and travel through the Circus to the Park, and then, without leaving the tram, return to the Zoo. The second part of the journey is three times shorter than the first, meaning the first part takes up three quarters of the full circle, and the second part takes up one quarter. Let's mark the Zoo and the Park on the map, and somewhere on the longer arc between them, mark the Circus (see the diagram). Now, on the same tram, let's travel from the Circus to the Zoo (passing the Park, as shown on the map).

![](https://cdn.mathpix.com/cropped/2024_05_06_79cfe44052259f2639fcg-04.jpg?height=357&width=392&top_left_y=302&top_left_x=581)

Having arrived at the Zoo, we will return to the Circus on the same tram, completing a full circle. The first part of the journey is twice as short as the second, meaning it takes up one third of the circle. This means that the path (still on the same tram) from the Circus to the Park will not pass through the Zoo and will constitute $\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$ of the full circle. The path through the Zoo is $1-\frac{1}{12}=\frac{11}{12}$ of the circle, which is 11 times longer.

Answer: 11


Let $n$ be a positive integer. A regular hexagon with side $n$ is divided into equilateral triangles with side 1 by lines parallel to its sides. Find the number of regular hexagons all of whose vertices are among the vertices of equilateral triangles.

Solution:
4. Denote the given hexagon by $P$. We call the vertices of the equilateral triangles simply nodes. For every regular hexagon $Q$ with the vertices in nodes, consider the hexagon $\bar{Q}$ circumscribed about $Q$ with sides parallel to the sides of $P$. The vertices of $\bar{Q}$ are nodes as well.

For $0 \leqslant m<n$, hexagon $\bar{Q}$ of side $n-m$

![](https://cdn.mathpix.com/cropped/2024_06_04_6f458ef4962c28de5a12g-13.jpg?height=332&width=393&top_left_y=1597&top_left_x=1129)
can be chosen in $3 m^{2}+3 m+1=(m+1)^{3}-m^{3}$ ways. Also, given $\bar{Q}$, hexagon $Q$ can be chosen in $n-m$ ways. Thus the total number of hexagons $Q$ equals

$$
\begin{aligned}
& \sum_{m=0}^{n-1}(n-m)\left((m+1)^{3}-m^{3}\right)=\sum_{m=0}^{n-1}(n-m)(m+1)^{3}-\sum_{m=0}^{n-1}(n-m) m^{3} \\
& \quad=\sum_{m=1}^{n}(n-m+1) m^{3}-\sum_{m=0}^{n-1}(n-m) m^{3}=\sum_{m=1}^{n} m^{3}=\frac{n^{2}(n+1)^{2}}{4}
\end{aligned}
$$

The SMO was sponsored by

![](https://cdn.mathpix.com/cropped/2024_06_04_6f458ef4962c28de5a12g-15.jpg?height=108&width=222&top_left_y=344&top_left_x=895)

![](https://cdn.mathpix.com/cropped/2024_06_04_6f458ef4962c28de5a12g-16.jpg?height=1040&width=1426&top_left_y=222&top_left_x=215)

Mathematical Competitions in Serbia

$h t t p: / /$ srb.imomath.com/

$\sim \sim \sim \sim \sim \sim \sim$

The IMO Compendium Olympiad Archive http://www.imocompendium.com/
Mathematical Society of Serbia $h t t p: / / w w w . d m s . r s /$

![](https://cdn.mathpix.com/cropped/2024_06_04_6f458ef4962c28de5a12g-16.jpg?height=451&width=296&top_left_y=1757&top_left_x=194)

## The IMO Compendium - 2nd Edition: 1959-2009

Until the first edition of this book appearing in 2006, it has been almost impossible to obtain a complete collection of the problems proposed at the IMO in book form. "The IMO Compendium" is the result of a collaboration between four former IMO participants from Yugoslavia, now Serbia, to rescue these problems from old and scattered manuscripts, and produce the ultimate source of IMO practice problems. This book attempts to gather all the problems and solutions appearing on the IMO through 2009. This second edition contains 143 new problems, picking up where the 1959-2004 edition has left off, accounting for 2043 problems in total.

Publisher: Springer (2011); Hardcover, 823 pages; Language: English; ISBN: 1441998535

For information on how to order, visit http://www.imocompendium.com/

Answer: \frac{n^{2}(n+1)^{2}}{4}


A highway running from west to east intersects with $n$ equal roads, numbered from 1 to $n$ in order. Cars travel on these roads from south to north and from north to south. The probability that a car will approach the highway from each of these roads is $\frac{1}{n}$. Similarly, the probability that a car will turn off the highway onto each of these roads is $\frac{1}{n}$. The road by which a car leaves the highway is independent of the road by which it entered the highway.

Find the probability that a random car that has entered the highway will pass the $k$-th intersection (either drive through it or turn at it).

Solution:
Answer: $\frac{2 k n-2 k^{2}+2 k-1}{n^{2}}$.

Solution. The specified event will occur in one of three cases.

Answer: \frac{2kn-2k^{2}+2k-1}{n^{2}}


Find the maximal value of
$$ S=\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{c+7}}+\sqrt[3]{\frac{c}{d+7}}+\sqrt[3]{\frac{d}{a+7}} $$
where \(a, b, c, d\) are nonnegative real numbers which satisfy \(a+b+c+d=100\).

Solution:
Since the value $8 / \sqrt[3]{7}$ is reached, it suffices to prove that $S \leqslant 8 / \sqrt[3]{7}$.  Assume that $x, y, z, t$ is a permutation of the variables, with $x \leqslant y \leqslant z \leqslant t$. Then, by the rearrangement inequality,
\[ S \leqslant\left(\sqrt[3]{\frac{x}{t+7}}+\sqrt[3]{\frac{t}{x+7}}\right)+\left(\sqrt[3]{\frac{y}{z+7}}+\sqrt[3]{\frac{z}{y+7}}\right) \]
Claim. The first bracket above does not exceed $\sqrt[3]{\frac{x+t+14}{7}}$.
Proof. Since
\[ X^{3}+Y^{3}+3 X Y Z-Z^{3}=\frac{1}{2}(X+Y-Z)\left((X-Y)^{2}+(X+Z)^{2}+(Y+Z)^{2}\right) \]
the inequality $X+Y \leqslant Z$ is equivalent (when $X, Y, Z \geqslant 0$) to $X^{3}+Y^{3}+3 X Y Z \leqslant Z^{3}$. Therefore, the claim is equivalent to
\[ \frac{x}{t+7}+\frac{t}{x+7}+3 \sqrt[3]{\frac{x t(x+t+14)}{7(x+7)(t+7)}} \leqslant \frac{x+t+14}{7} \]
Notice that
\[ \begin{array}{r} 3 \sqrt[3]{\frac{x t(x+t+14)}{7(x+7)(t+7)}}=3 \sqrt[3]{\frac{t(x+7)}{7(t+7)} \cdot \frac{x(t+7)}{7(x+7)} \cdot \frac{7(x+t+14)}{(t+7)(x+7)}} \\ \leqslant \frac{t(x+7)}{7(t+7)}+\frac{x(t+7)}{7(x+7)}+\frac{7(x+t+14)}{(t+7)(x+7)} \end{array} \]
by the AM-GM inequality, so it suffices to prove
\[ \frac{x}{t+7}+\frac{t}{x+7}+\frac{t(x+7)}{7(t+7)}+\frac{x(t+7)}{7(x+7)}+\frac{7(x+t+14)}{(t+7)(x+7)} \leqslant \frac{x+t+14}{7} . \]
A straightforward check verifies that the last inequality is in fact an equality.
The claim leads now to
\[ S \leqslant \sqrt[3]{\frac{x+t+14}{7}}+\sqrt[3]{\frac{y+z+14}{7}} \leqslant 2 \sqrt[3]{\frac{x+y+z+t+28}{14}}=\frac{8}{\sqrt[3]{7}} \]
the last inequality being due to the AM-CM inequality (or to the fact that $\sqrt[3]{ }$ is concave on $[0, \infty))$.

Answer: \frac{8}{\sqrt[3]{7}}


Let $D$ be the set of points consisting of the equilateral $\triangle P_{1} P_{2} P_{3}$ and its interior, and let point $P_{0}$ be the center of $\triangle P_{1} P_{2} P_{3}$. If the set $S=$ $\left\{P|P \in D|, P P_{0}|\leqslant| P P_{i} \mid, i=1,2,3\right\}$, then the planar region represented by the set $S$ is ( )
A. Triangular region
B. Quadrilateral region
C. Pentagonal region
D. Hexagonal region

Solution: As shown in the figure, $A, B, C, D, E, F$ are the trisection points of each side. From $\left|P P_{0}\right| \leqslant\left|P P_{1}\right|$, we know that point $P$ is on the perpendicular bisector of segment $P_{0} P_{i}$ or on the side closer to point $P_{0}$. Since point $P \in D$, then $S$ represents the regular hexagon $A B C D E F$ as shown in Figure 2. Choose D.

Answer: D


Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}$, define
$$
S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2} \text {. }
$$
(1) Find the minimum value of $S_{n}$;
(2) Under the condition $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, find the minimum value of $S_{n}$;
(3) Under the condition $x_{1}+x_{2}+\cdots+x_{n}=1$, find the minimum value of $S_{n}$, and provide a proof.

Solution:
14. (1) $S_{n} \geqslant \sum_{i=1}^{n}\left(2 \sqrt{\frac{n-1}{n^{2}}}\right)^{2}$
$$
=4 \sum_{i=1}^{n} \frac{n-1}{n^{2}}=4 \times \frac{n-1}{n} \text {. }
$$

When $x_{1}=x_{2}=\cdots=x_{n}=\frac{\sqrt{n-1}}{n}$, the minimum value $4 \times \frac{n-1}{n}$ is achieved.
$$
\begin{array}{l}
\text { (2) } S_{n}=\sum_{i=1}^{n}\left(x_{i}^{2}+2 \times \frac{n-1}{n^{2}}+\frac{(n-1)^{2}}{n^{4}} \cdot \frac{1}{x_{i}^{2}}\right) \\
=1+2 \times \frac{n-1}{n}+\frac{(n-1)^{2}}{n^{4}} \sum_{i=1}^{n} \frac{1}{x_{i}^{2}} \\
\geqslant 1+2 \times \frac{n-1}{n}+\frac{(n-1)^{2}}{n^{2}}=\left(1+\frac{n-1}{n}\right)^{2} .
\end{array}
$$

When $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{\sqrt{n}}$, the minimum value
$$
\left(1+\frac{n-1}{n}\right)^{2}
$$
is achieved.
(3) Since $\left[\sum_{i=1}^{n} 1 \times\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)\right]^{2}$
$$
\leqslant\left(\sum_{i=1}^{n} 1^{2}\right) \cdot \sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2} \text {, }
$$

Therefore, $S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)^{2}$
$$
\begin{array}{l}
\geqslant \frac{1}{n}\left[\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \cdot \frac{1}{x_{i}}\right)\right]^{2} \\
\geqslant \frac{1}{n}\left[1+\frac{n-1}{n^{2}} \cdot n^{2}\right]^{2}=n .
\end{array}
$$

When $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{n}$, the minimum value $n$ is achieved.

Answer: n


Fold a square piece of paper to form a triangle and a quadrilateral. It is known that the ratio of their areas is $3: 5$. Then the ratio of their perimeters is

Solution:
2. $\frac{6}{7}$.

As shown in Figure 5, $ABCD$ is a square, and point $M$ is on side $AB$. Let the side length of the square be $1$, $BM=x$, then $AM=1-x$. Therefore,
$$
\frac{3}{5}=\frac{S_{\triangle ADM}}{S_{\text{quadrilateral } BCDM}}=\frac{\frac{1}{2} AM \cdot AD}{\frac{1}{2}(BM+DC) \cdot BC}=\frac{1-x}{1+x} \text{.}
$$

Solving for $x$ gives $x=\frac{1}{4}$. At this point, $DM=\frac{5}{4}$, thus,
$$
\frac{\text{Perimeter of } \triangle ADM}{\text{Perimeter of quadrilateral } BCDM}=\frac{1+\frac{3}{4}+\frac{5}{4}}{\frac{1}{4}+1+1+\frac{5}{4}}=\frac{6}{7} \text{.}
$$

Answer: \frac{6}{7}


Given $\sin \alpha=\frac{5}{7}$, $\cos (\alpha+\beta)=\frac{11}{14}$, and $\alpha, \beta$ are acute angles, find $\cos \beta$.

Solution:
Solving, we get $\cos \alpha=\frac{2 \sqrt{6}}{7}, \sin (\alpha+\beta)$ $=\frac{5 \sqrt{8}}{14}$, thus
$$
\begin{aligned}
\cos \beta & =\cos [(\alpha+\beta)-\alpha]=\cos (\alpha \\
& +\beta) \cos \alpha+\sin (\alpha+\beta) \sin \alpha \\
& =\frac{1}{98}(22 \sqrt{6}+25 \sqrt{3}) .
\end{aligned}
$$

Analysis. The result does not show any contradiction, but since $\frac{2 \sqrt{6}}{7}$
$$
\alpha+\beta, \beta<0$ contradicts the problem statement. It is evident that the original problem is incorrect. If we change it to $\cos (\alpha-\beta)=\frac{11}{14}$, the contradiction can be eliminated.

Answer: \frac{1}{98}(22 \sqrt{6}+25 \sqrt{3})


In the first quadrant of the Cartesian coordinate system, number the points (lattice points) with integer coordinates in the following manner: (1) $(0,0)$, (2) $(1,0)$, (3) $(1,1)$, (4) $(0,1)$, (5) $(0,2)$, (6) $(1,2)$, (7) $(2,2)$, (8) $(2,1)$, (9) $(2,0), \cdots \cdots$ (as shown in the order of the arrows in the figure). Find the coordinates of the 2000th point.

Solution: Parse Note that the lattice points $(x, y)$ satisfying the conditions $0 \leqslant x \leqslant k, 0 \leqslant y \leqslant k$ total $(k+1)^{2}$. Considering the inequality $(k+1)^{2}$ 44, we know that the x-coordinate of the point is 44, and its y-coordinate is $44-(64-$ $45)=25$. Therefore, the coordinates of the 2000th point are $(44,25)$.

Answer: (44,25)


There are $n$ integers, whose product is $n$, and whose sum is zero. Then $n$ ( ).
(A) must be even
(B) must be odd
(C) may be even or may be odd
(D) does not exist

Solution:
8. (A).
$1, 2, -1, -2$ satisfy the condition, that is, $n$ can be 4.
If $n$ is odd, then $n$ numbers are all odd, the sum of an odd number of odd numbers cannot be zero.

Answer: A


Find the range of the function $y=5 \sin x-12 \cos x$.

Solution:
11.51 In the coordinate plane $x O y$, let's construct the point $A(5 ; 12)$ and consider the vector $\overline{O A}$ (Fig. 11.18). Suppose this vector forms an angle $\varphi$ with the x-axis. Then, from the figure, it is clear that $\quad \cos \varphi=\frac{5}{\sqrt{25+144}}=\frac{5}{13}, \quad \sin \varphi=\frac{12}{\sqrt{25+144}}=\frac{12}{13}$, $\operatorname{tg} \varphi=\frac{12}{5}, \quad$ i.e., $\varphi=\operatorname{arctg} \frac{12}{5}$. Let's transform the given function as follows:

$$
\begin{aligned}
& y=13\left(\frac{5}{13} \sin x-\frac{12}{13} \cos x\right)= \\
& =13(\cos \varphi \sin x-\sin \varphi \cos x)=13 \sin (x-\varphi)
\end{aligned}
$$

where $\varphi=\operatorname{arctg} \frac{12}{5}$. Clearly, $-1 \leq \sin (x-\varphi) \leq 1$, hence $-13 \leq y \leq 13$.

Answer: $-13 \leq y \leq 13$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-367.jpg?height=377&width=313&top_left_y=861&top_left_x=501)

Fig. 11.18

Answer: -13\leqy\leq13


Let the sum of all positive divisors of the positive integer $\mathrm{n}$ be $\sigma(\mathrm{n})$, find the value: $\left[\sqrt[3]{\sum_{\mathrm{n}=1}^{2020} \frac{\sigma(\mathrm{n})}{\mathrm{n}}}\right]$, here $[\mathrm{x}]$ represents the greatest integer not exceeding $x$.

Solution:
Question 215, Solution: Note that
$$
\begin{array}{l}
\sum_{\mathrm{n}=1}^{2020} \frac{\sigma(\mathrm{n})}{\mathrm{n}}=\sum_{\mathrm{n}=1}^{2020} \frac{\sum_{\mathrm{d} \mid \mathrm{n}} \mathrm{d}}{\mathrm{n}}=\sum_{\mathrm{n}=1}^{2020} \sum_{\mathrm{d} \mid \mathrm{n}} \frac{1}{\mathrm{~d}} \\
=\sum_{\mathrm{d}=1}^{2020} \sum_{1 \leq \mathrm{n} \leq 2020, \mathrm{d} \mid \mathrm{n}} \frac{1}{\mathrm{~d}}=\sum_{\mathrm{d}=1}^{2020} \frac{1}{\mathrm{~d}} \cdot\left[\frac{2020}{\mathrm{~d}}\right] \\
\end{array}
$$

On the one hand, according to (*) we know:
$$
\begin{array}{l}
\sum_{n=1}^{2020} \frac{\sigma(n)}{n} \cdot 2020 + \frac{1}{2} \times 1010 + \frac{1}{3} \times 673 + \frac{1}{4} \times 505 > 14^{3} \\
\text { Therefore, }\left[\sqrt[3]{\sum_{n=1}^{2020} \frac{\sigma(n)}{n}}\right]=14 .
\end{array}
$$

Answer: 14


Calculate: $\frac{\left(1+\frac{1}{2}\right)^{2} \times\left(1+\frac{1}{3}\right)^{2} \times\left(1+\frac{1}{4}\right)^{2} \times\left(1+\frac{1}{5}\right)^{2} \times \cdots \times\left(1+\frac{1}{10}\right)^{2}}{\left(1-\frac{1}{2^{2}}\right) \times\left(1-\frac{1}{3^{2}}\right) \times\left(1-\frac{1}{4^{2}}\right) \times\left(1-\frac{1}{5^{2}}\right) \times \cdots \times\left(1-\frac{1}{10^{2}}\right)}$

Solution:
【Solution】Solve: $\frac{\left(1+\frac{1}{2}\right)^{2} \times\left(1+\frac{1}{3}\right)^{2} \times\left(1+\frac{1}{4}\right)^{2} \times\left(1+\frac{1}{5}\right)^{2} \times \cdots \times\left(1+\frac{1}{10}\right)^{2}}{\left(1-\frac{1}{2^{2}}\right) \times\left(1-\frac{1}{3^{2}}\right) \times\left(1-\frac{1}{4^{2}}\right) \times\left(1-\frac{1}{5^{2}}\right) \times \cdots \times\left(1-\frac{1}{10^{2}}\right)}$
$$
\begin{array}{l}
=\frac{\left(1+\frac{1}{2}\right) \times\left(1+\frac{1}{3}\right) \times\left(1+\frac{1}{4}\right) \times \cdots \times\left(1+\frac{1}{10}\right)}{\left(1-\frac{1}{2}\right) \times\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{4}\right) \times \cdots \times\left(1-\frac{1}{10}\right)} \\
=\frac{\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \cdots \times \frac{11}{10}}{\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{9}{10}} \\
=\frac{\frac{11}{2}}{\frac{1}{10}} \\
=55
\end{array}
$$

Therefore, the answer is: 55.

Answer: 55


Patrick asks SpongeBob: "What is your lucky number?" SpongeBob says: "There is a sequence of numbers: $0,1,3,8$, $21,55,144,377,987, \cdots \cdots$ Starting from the third number, each number multiplied by 3 is exactly equal to the sum of the numbers immediately before and after it. My lucky number is the remainder when the 2020th number in this sequence is divided by 6." So, SpongeBob's lucky number is $\qquad$ . (If it is divisible, then the remainder is 0)

Answer: 2


How many ways can the integers from -7 to 7 be arranged in a sequence such that the absolute values of the numbers in the sequence are nonincreasing?

Solution:
Answer:
128
Solution: Each of the pairs $a,-a$ must occur in increasing order of $a$ for $a=1, \ldots, 7$, but $a$ can either occur before or after $-a$, for a total of $2^{7}=128$ possible sequences.

Answer: 128


Calculate the arithmetic mean of all four-digit numbers written using the digits $1,3,5,7$, if all digits are used and only once.

Solution:
## Solution.

There are $4 \cdot 3 \cdot 2 \cdot 1=24$ such four-digit numbers.

1 point

For example, if the digit 1 is in the first position, we have 6 different numbers with the given property: $1357, 1375, 1537, 1573, 1735, 1753$.

Similarly, the remaining numbers can be listed.

It is important to note that each of the four digits will appear exactly 6 times in the thousands, hundreds, tens, and units place.

The sum of all such numbers is:

$$
\begin{gathered}
6(1000+3000+5000+7000)+6(100+300+500+700)+6(10+30+50+70)+6(1+3+5+7) \\
\quad=(6000+600+60+6)(1+3+5+7)=6666 \cdot(1+3+5+7)=6666 \cdot 16 . \quad 2 \text { points }
\end{gathered}
$$

The arithmetic mean of the desired sequence of numbers is $\frac{6666 \cdot 16}{24}=4444 . \quad 1$ point

Note: The desired sum can be calculated as follows:

If we sum all the digits in the units place, we get $6(1+3+5+7)=96$.

The same sum is obtained by summing the tens, hundreds, and thousands places.

Therefore, the total sum of all 24 numbers is $96+960+9600+96000=106656$.

Answer: 4444


If $P(x, y)$ is a point on the hyperbola $\frac{x^{2}}{8}-\frac{y^{2}}{4}=1$, then the minimum value of $|x-y|$ is $\qquad$ .

Solution:
Let $P(2 \sqrt{2} \sec \theta, 2 \tan \theta)$, then $|x-y|=2\left|\frac{\sqrt{2}-\sin \theta}{\cos \theta}\right| \geqslant 2$,
equality holds when $\sin \theta+\cos \theta=\sqrt{2} \Rightarrow \theta=\frac{\pi}{4} \Rightarrow P(4,2)$. Therefore, the minimum value of $|x-y|$ is 2.

Answer: 2


The three vertices of a triangle are $(0,0)$, $(2,3)$, and $(3,-2)$. When the triangle is translated, the translation path of $(0,0)$ is $y^{2}=2x$. Find the translation paths of the other two points.

Solution:
Solve: Write $\mathrm{y}^{2}=2 \mathrm{x}$ as a parametric equation
$$
\left\{\begin{array}{l}
x=-\frac{t^{2}}{2} \\
y=t .
\end{array}\right.
$$
$\left(-\frac{t^{2}}{2}, t\right)$ is the translation increment of $(0,0)$, and also the translation increment of the triangle. Therefore, the translation trajectory of (2,3) is
$$
\left\{\begin{array}{l}
x=2+\frac{t^{2}}{2} \\
y=3+t
\end{array}\right.
$$

Eliminating $t$, we get $(y-3)^{2}=2(x-2)$.
Similarly, the translation trajectory of $(3,-2)$ is
$$
(y+2)^{2}=2(x-3) \text {. }
$$
$$
\text { Let } f(x, y)=0 \text { . }
$$

The translation point of $(x, y)$ with respect to $(h, k)$ is $\left(x^{\prime}, y^{\prime}\right)$, then we have
$$
f\left(x^{\prime}-\dot{n}, y^{\prime}-k\right)=0 \text {. }
$$

Removing the dot
$$
f(x-h, y-k)=0 .
$$

We call (B) the translation curve of (A) with respect to $(\mathrm{h}, \mathrm{k})$.

This result provides us with a method for plotting the translation transformation of function curves and equation curves.

Answer: (y-3)^{2}=2(x-2) \text{ and } (y+2)^{2}=2(x-3)


What is the sum of the exterior angles at three vertices of a triangle?

Solution: $\Delta$ Together with the three interior angles (which sum up to $180^{\circ}$), this results in three straight angles, that is, $540^{\circ}$, so the sum of the exterior angles is $540-180=360^{\circ} . \triangleleft$

Answer: 360


Let the sequence $\left\{a_{n}\right\}$ be defined as
$$
a_{1}=a, a_{n+1}=1+\frac{1}{a_{1}+a_{2}+\cdots+a_{n}-1}(n \geqslant 1)
$$

Find all real numbers $a$ such that $0<a_{n}<1(n \geqslant 2)$.

Solution:
20. From the given conditions, we have
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n-1}=\frac{a_{n}}{a_{n}-1} . \\
\text { Also, } a_{n+1}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{a_{1}+a_{2}+\cdots+a_{n}-1} \\
=\frac{a_{n}^{2}}{a_{n}^{2}-a_{n}+1}=\frac{a_{n}^{2}}{\left(a_{n}-\frac{1}{2}\right)^{2}+\frac{3}{4}}(n \geqslant 2) .
\end{array}
$$

Since \( a_{1}=a, a_{2}=\frac{a_{1}}{a_{1}-1} \), therefore, when \( a_{1}=a \neq 0 \), \( a_{n}>0(n \geqslant 3) \), and when \( 0<a_{2}=\frac{a}{a-1}<1 \), i.e., \( a<0 \), we have
$$
\begin{array}{l}
a_{n+1}-a_{n} \\
=\frac{1}{a_{1}+a_{2}+\cdots+a_{n-1}+a_{n}-1}-\frac{1}{a_{1}+a_{2}+\cdots+a_{n-1}-1} \\
<0(n \geqslant 2),
\end{array}
$$

i.e., \( a_{n+1}<a_{n}(n \geqslant 2) \).
Thus, \( 0<a_{n}<1(n \geqslant 2) \).
Therefore, \( a<0 \).

Answer: a<0


Given a four-digit number $N$ is a perfect square, and each digit of $N$ is less than 7. Increasing each digit by 3 results in another four-digit number that is also a perfect square. Find $N$.

Solution: Let $N=a \cdot 10^{3}+b \cdot 10^{2}+c \cdot 10+d$, where integers $a, b, c, d$ satisfy $1 \leqslant a \leqslant 6$, $0 \leqslant b, c, d \leqslant 6$. From the problem, we know $N=n^{2} \leqslant 6666$, so $n \leqslant 81$. Also, $n^{2}+3333=m^{2}$, which means $(m+n)(m-n)=3 \times 11 \times 101$. Since $m+n>m-n, n \leqslant 81$, we have $m+n=101$, $m-n=33$, thus $n=34, N=1156$.

Answer: 1156