A certain number, which is written using fewer than 30 digits, starts (if we move from left to right) with the digits 15, that is, it has the form $15 \ldots$. If we multiply it by 5, the result can be obtained simply by moving these two digits to the right end; in the end, we get a number of the form ... 15. Find the original number.

Solution:
139. Let $f$ be a proper fraction, in the decimal periodic expansion of which one period coincides exactly with the initial number $15 \ldots$ By the conditions of this problem, $5 f=$ $0, \ldots 15 \ldots 15 \ldots$, and $100 f=15, \ldots 15 \ldots 15$, from which $95 f=15$ and $f=\frac{3}{19}$. Expanding $\frac{3}{19}$ into a decimal fraction, we find that its period consists of 18 digits: 157894736842105263. This is the number we are looking for. It is unique if we consider numbers not exceeding 30 digits. If, however, we change this restriction and consider, for example, numbers that are written using no more than 50 digits, then there will be another solution to the problem - a 36-digit number, coinciding with two periods of the fraction $f$.

[H. T. R. Aude, A. M. M., 41, 268 (April 1934).]

Answer: 157894736842105263


Three regular polygons are inscribed in a circle, the number of sides of each subsequent one being twice that of the previous one. The areas of the first two are $S_{1}$ and $S_{2}$. Find the area of the third.

Solution:
Let $S$ be the area of the third polygon, $R$ the radius of the circle, $n$ the number of sides of the first polygon, $\alpha=\pi / 2 n$. Then $S_{1}=1 / 2 n R^{2} \sin 4 \alpha$, $S_{2}=n R^{2} \sin 2 \alpha, S=n R^{2} \sin \alpha=S_{2 /} \cos \alpha$. From the first two equations, $S_{1 / S_{2}}=\cos 2 \alpha$ so, $\cos \alpha=\sqrt{\frac{1+\cos 2 \alpha}{2}}=\sqrt{\frac{S_{2}+S_{1}}{2 S_{2}}}, S=\sqrt{\frac{2 S_{2}^{3}}{S_{1}+S_{2}}}$.

## Answer

$S=\sqrt{\frac{2 S_{2}^{3}}{S_{1}+S_{2}}}$

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


Answer: 1


Find the value of $\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$

Solution:
Let $\cos \frac{\pi}{7}=x$, then the original expression $=2 x^{2}-1+3 x$
$$
-4 x^{3}-x=4 x^{3}+2 x^{2}-1 \text {. }
$$

Notice: $\cos \frac{4 \pi}{7}=-\cos \frac{3 \pi}{7}$,
i.e., 2
$$
\begin{array}{l}
\left(2 x^{2}-1\right)^{2}-1=3 x-4 x^{3}, \\
\therefore(x+1)\left(8 x^{3}-4 x+1\right)=0 .
\end{array}
$$

But $x \neq-1,8 x^{3}-4 x^{2}-4 x+1=0$,
$$
\begin{aligned}
\therefore \text { the original expression } & =-\frac{1}{2}\left(8 x^{3}-4 x^{2}-4 x\right)-1 \\
& =-\frac{1}{2} .
\end{aligned}
$$
(Turn to page 41)

Answer: -\frac{1}{2}


A triangle has sides of 13, 17, and 24 units. The 24-unit side is divided into three equal parts. What is the distance from the opposite vertex to the division points?

Solution:
We denote the height of the triangle in question by $m$, the projection of the side of length 13 on the 24-unit side by $x$, so the projection of the 17-unit side is $24-x$, because the foot of the perpendicular from the longest side lies on the side segment. The distance from the vertices opposite the 13 and 17-unit sides to the points that trisect these sides are denoted by $p$ and $q$. The distances from the trisection points to the foot of the perpendicular are $|x-8|$ and $|24-x-8|$. The 13 and 17-unit sides, along with the segments $p$ and $q$, form right triangles with the height.

![](https://cdn.mathpix.com/cropped/2024_05_02_700ff852980fd3bcd20ag-1.jpg?height=320&width=637&top_left_y=385&top_left_x=733)

We apply the Pythagorean theorem to these triangles:

$$
\begin{gathered}
x^{2}+m^{2}=13^{2} \\
(x-8)^{2}+m^{2}=p^{2} \\
(16-x)^{2}+m^{2}=q^{2} \\
(24-x)^{2}+m^{2}=17^{2}
\end{gathered}
$$

From these, we can eliminate $m$ by subtracting (1) from the other equations:

$$
\begin{gathered}
64-16 x=p^{2}-13^{2} \\
256-32 x=q^{2}-13^{2} \\
576-48 x=17^{2}-13^{2}=4 \cdot 30=120
\end{gathered}
$$

From the last equation, we get $x=19 / 2$, and substituting this into (5) and (6) gives:

$$
p^{2}=81 ; \quad q^{2}=121
$$

Thus, the sought distances are $p=9, q=11$.

Katalin Balla (Budapest, Radnóti M. Practical Grammar School II. grade)

Answer: p=9,q=11


Solution:
When $a=b=c$, $S=3$. Conjecture: $S \geqslant 3$.
In fact,
$$\begin{aligned}
S-3 & =\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}-3-\frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c} \\
& =\frac{a^{2}+b^{2}+c^{2}}{a^{2}}+\frac{a^{2}+b^{2}+c^{2}}{b^{2}}+\frac{a^{2}+b^{2}+c^{2}}{c^{2}}-3-2\left(\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}\right) \\
& =a^{2}\left(\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)+b^{2}\left(\frac{1}{a^{2}}+\frac{1}{c^{2}}\right)+c^{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)-2\left(\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}\right) \\
& =a^{2}\left(\frac{1}{b}-\frac{1}{c}\right)^{2}+b^{2}\left(\frac{1}{c}-\frac{1}{a}\right)^{2}+c^{2}\left(\frac{1}{a}-\frac{1}{b}\right)^{2} \\
& \geqslant 0 .
\end{aligned}$$

In conclusion, the minimum value of $S$ is 3.

Answer: 3


The number n is the product of three different prime numbers. If we increase the two smaller ones by 1 and leave the largest one unchanged, the product of the three will increase by 915. Determine the number n.

Solution:
SOLUTION. Let $n=p q r, p<q<r$. The equation $(p+1)(q+1) r=p q r+915$ can be equivalently rearranged to the form $(p+q+1) \cdot r=915=3 \cdot 5 \cdot 61$, from which it follows that the prime number $r$ can only take some of the values 3, 5, and 61. For $r=3$, however, from the last equation we get $(p+q+1) \cdot 3=3 \cdot 5 \cdot 61$, or $p+q=304$. This is in contradiction with the fact that $r$ is the largest. Similarly, we find that $r$ cannot be 5 either. Therefore, $r=61$ and $p+q=14$. By trying all possibilities for $p$ and $q$, we get $p=3, q=11, r=61$ and $n=3 \cdot 11 \cdot 61=2013$.

GUIDING AND SUPPLEMENTARY PROBLEMS:

Answer: 2013


Five identical balls are rolling towards each other on a smooth horizontal surface. The speeds of the first and second are \( v_{1} = v_{2} = 0.5 \) m/s, while the others are \( v_{3} = v_{4} = v_{5} = 0.1 \) m/s. The initial distances between the balls are the same, \( l = 2 \) m. All collisions are perfectly elastic. How much time will pass between the first and last collisions in this system?

1

2

3

4

5

Solution:
# Solution and Evaluation Criteria:

In the case of a perfectly elastic collision, identical balls "exchange" velocities.

Therefore, the situation can be considered as if the balls pass through each other with unchanged speeds. The first collision occurs between the second and third balls. The last collision will occur at the moment when the first ball "passes" by the fifth.

(4 points)

The corresponding time: \( t = \frac{3 l}{v_{1} + v_{5}} = \frac{3 \cdot 2}{0.6} = 10 \text{s} \)

#

Answer: 10


Solution:
【Analysis】First, transform the expression to $481 \frac{1}{6}+265 \frac{1}{12}+904 \frac{1}{20}-185+\frac{1}{30}-161+\frac{1}{42}-704+\frac{1}{56}$, then simplify using the commutative and associative laws of addition and the fraction decomposition formula $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$.
【Solution】Solution: $481 \frac{1}{6}+265 \frac{1}{12}+904 \frac{1}{20}-184 \frac{29}{30}-160 \frac{41}{42}-703 \frac{55}{56}$
$$
\begin{array}{l}
=481 \frac{1}{6}+265 \frac{1}{12}+904 \frac{1}{20}-185+\frac{1}{30}-161+\frac{1}{42}-704+\frac{1}{56} \\
=(481-161)+(265-185)+(904-704)+\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right) \\
=320+80+200+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\cdots+\frac{1}{7}-\frac{1}{8}\right) \\
=600+\frac{1}{2}-\frac{1}{8} \\
=600 \frac{3}{8} .
\end{array}
$$

Therefore, the answer is: $600 \frac{3}{8}$.

Answer: 600\frac{3}{8}


Calculate

$$
\left|\begin{array}{rrr}
2 & -1 & 1 \\
3 & 2 & 2 \\
1 & -2 & 1
\end{array}\right|
$$

Solution:
Solution:

$$
\begin{gathered}
\left|\begin{array}{rrr}
2 & -1 & 1 \\
3 & 2 & 2 \\
1 & -2 & 1
\end{array}\right|=2 \cdot\left|\begin{array}{rr}
2 & 2 \\
-2 & 1
\end{array}\right|-(-1) \cdot\left|\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right|+1 \cdot\left|\begin{array}{rr}
3 & 2 \\
1 & -2
\end{array}\right|= \\
=2 \cdot(2+4)+1 \cdot(3-2)+1 \cdot(-6-2)=12+1-8=5
\end{gathered}
$$

Let's consider the concept of the inverse matrix.

Definition. Let a matrix $A$ of order $n$ be given. If its product with some matrix $B$ of order $n$ equals the identity matrix $E$, i.e., $A \cdot B=E$ or $B \cdot A=E$, then the matrix $B$ is called the inverse of matrix $A$ and is denoted by $A^{-1}$.

It can be proven that if $A \cdot A^{-1}=E$, then $A^{-1} \cdot A=E$, i.e., mutually inverse matrices are commutative.

Theorem 1. Every square matrix $A$, whose determinant $\Delta \neq 0$, has a unique inverse matrix $A^{-1}=B$, the elements of which $b_{i j}$ are found by the formula

$$
b_{i j}=\frac{A_{j i}}{\Delta}
$$

From the theorem, the following rule follows: to find the inverse matrix of matrix (1.2), where $m=n$, the following transformations must be performed:

1. Calculate the determinant $\Delta$ of matrix $A$ ($\Delta \neq 0$).
2. Replace each element of matrix $A$ with its algebraic complement, i.e., form the matrix ($A_{i j}$).
3. Transpose the matrix ($A_{i j}$), i.e., write the matrix ($A_{j i}$).
4. Multiply the matrix ($A_{j i}$) by $\frac{1}{\Delta}$.

As a result, we obtain the matrix $A^{-1}$.

Answer: 5


Let $m>1$ be an integer. The sequence $a_{1}, a_{2}, \cdots$ is defined as follows:
$$
\begin{array}{l}
a_{1}=a_{2}=1, a_{3}=4, \\
a_{n}=m\left(a_{n-1}+a_{n-2}\right)-a_{n-3}(n \geqslant 4) .
\end{array}
$$

Determine all integers $m$ such that every term of the sequence is a perfect square.

Solution:
6. $m=2,10$.

Consider the integer $m>1$, such that the sequence defined as described in the problem contains only perfect squares.
First, prove that $m-1$ is a power of 3.
Assume $m-1$ is even, then $a_{4}=5 m-1$ should be divisible by 4. Hence $m \equiv 1(\bmod 4)$.

But $a_{5}=5 m^{2}+3 m-1 \equiv 3(\bmod 4)$ cannot be a perfect square, a contradiction.
Thus, $m-1$ is odd.
Assume an odd prime $p \neq 3$ divides $m-1$.
Notice that, $a_{n}-a_{n-1} \equiv a_{n-2}-a_{n-3}(\bmod p)$.
Therefore, the sequence modulo $p$ takes the form 1, $1,4,4,7,7,10,10, \cdots$
In fact, by induction, it can be shown:
$a_{2 k} \equiv a_{2 k-1} \equiv 3 k-2(\bmod p)(k \geqslant 1)$.
Since $\operatorname{gcd}(p, 3)=1$, the sequence $a_{n}(\bmod p)$ contains all residues modulo $p$, which contradicts the fact that there are only $\frac{p+1}{2}$ quadratic residues modulo $p$.
This indicates that $m-1$ is a power of 3.
Let $h$ and $k$ be integers, satisfying:
$m=3^{k}+1, a_{4}=h^{2}$.
Then $5 \times 3^{k}=(h-2)(h+2)$.
Since $\operatorname{gcd}(h-2, h+2)=1$, we have
$$
\left\{\begin{array}{l}
h-2=1,3^{k}, 5, \\
h+2=5 \times 3^{k}, 5,3^{k} .
\end{array}\right.
$$

In the first two cases, we get $k=0$, and in the last case, we get $k=2$.
Thus, $m=2$ or 10.
Now, verify the converse.
Assume $m=2$ or 10, let $t=1$ or 3 such that
$$
m=t^{2}+1 \text {. }
$$

The integer sequence $b_{1}, b_{2}, \cdots$ is defined as follows:
$b_{1}=1, b_{2}=1, b_{3}=2$,
$b_{n}=t b_{n-1}+b_{n-2}(n \geqslant 4)$.
Clearly, for $n=1,2,3$, we have $a_{n}=b_{n}^{2}$.
Notice that, if $m=2$, then $a_{4}=9$ and $b_{4}=3$;
if $m=10$, then $a_{4}=49$ and $b_{4}=7$.
In both cases, we have $a_{4}=b_{4}^{2}$.
If $n \geqslant 5$, then
$b_{n}^{2}+b_{n-3}^{2}$
$=\left(t b_{n-1}+b_{n-2}\right)^{2}+\left(b_{n-1}-t b_{n-2}\right)^{2}$
$=\left(t^{2}+1\right)\left(b_{n-1}^{2}+b_{n-2}^{2}\right)$
$=m\left(b_{n-1}^{2}+b_{n-2}^{2}\right)$.
Therefore, by induction, it can be proven that for all $n \geqslant 1$, we have $a_{n}=b_{n}^{2}$.
Hence, the proposition is proven.

Answer: 2,10


A triangle has sides of integer lengths, the radius of its inscribed circle is one unit. What is the radius of the circumscribed circle?

Solution:
The area of the triangle $t=r \cdot s$, where $r$ is the radius of the inscribed circle, and $s$ is the semiperimeter. Since $r=1$, we have $t=s$. Using Heron's formula:

$$
s=\sqrt{s \cdot(s-a) \cdot(s-b) \cdot(s-c)}
$$

which simplifies to

$$
s=(s-a)(s-b)(s-c) \cdot(1) \text{ Hence } 8 s=(2 s-2 a)(2 s-2 b)(2 s-2 c) \cdot(2)
$$

Since $a, b, c$ are integers, $2 s$ is also an integer, and for example, $2 s-2 a$ has the same parity as $2 s$. The same can be said about every factor on the right side of (2). Considering that the left side of (2) is even, the factors on the right side must also be even, so the integers $x=s-a, y=s-b, z=s-c$ are all integers. Given that $s-a+s-b+s-c=s$, equation (1) becomes:

$$
x+y+z=x \cdot y \cdot z
$$

We can assume that among $x, y, z$, $z$ is the largest, and then

$$
x+y+z \leq 3 z \cdot(4) H a
$$

If $x, y, z$ were all at least 2, then $x \cdot y \cdot z \geq 4 z .(5) From (3), (4), and (5), we get 3z \geq 4z$, which is a contradiction for $z>0$. Therefore, at least one of $x, y, z$ must be 1. Assuming $x=1$, from (3) we get $1+y+z=y \cdot z$, or $(y-1)(z-1)=2$. Since $y$ and $z$ are positive integers and $z \geq y, y=2$ and $z=3$. Thus, $s=6$ and $a=5, b=4, c=3$.

By the converse of the Pythagorean theorem, the triangle is a right triangle, and by Thales' theorem, the radius of the circumscribed circle is half the length of the hypotenuse, which is $\frac{5}{2}$.

It can be easily verified that the radius of the inscribed circle of a triangle with sides $3, 4, 5$ is indeed 1 unit.

Answer: \frac{5}{2}


A bag contains 5 red balls, 6 black balls, and 7 white balls. Now, 15 balls are drawn from the bag. The probability that exactly 3 of the drawn balls are red is ( ).
(A) $\frac{6}{65}$
(B) $\frac{65}{408}$
(C) $\frac{13}{816}$
(D) $\frac{13}{4896}$

Solution:
6.B.

There are a total of $5+6+7=18$ balls in the bag. Drawing 15 balls is equivalent to leaving 3 balls in the bag. Considering that the balls are indistinguishable in order, the number of different drawing results is $\frac{18 \times 17 \times 16}{3 \times 2 \times 1}=816$.

If exactly 3 red balls are drawn, we first need to draw 3 red balls from the 5 red balls, which has $\frac{5 \times 4 \times 3}{3 \times 2 \times 1}=10$ possible outcomes; then, we need to draw $15-3=12$ balls from the remaining 13 balls (6 black balls and 7 white balls), which means leaving 1 black or white ball in the bag, with 13 different outcomes. Therefore, the number of drawing results where exactly 3 red balls are drawn is $10 \times 13=130$.
Thus, the probability of drawing exactly 3 red balls is
$$
\frac{130}{816}=\frac{65}{408} .
$$

Answer: B


Captain Billy the pirate plundered 1010 gold doubloons and set sail on his ship to a deserted island to bury them as treasure. Every evening of the voyage, he paid each of his pirates one doubloon. On the eighth day of the voyage, the pirates plundered a Spanish caravel, and Billy's treasure doubled, while the number of pirates was halved. On the 48th day of the voyage, the pirates arrived at the deserted island, and Billy buried all his treasure in a spot marked with a cross - exactly 1000 doubloons. How many pirates set out with Billy to the deserted island?

Solution:
Solution. Answer: 30

Before the pirates looted the caravel, Billy managed to pay the pirates their daily wages 7 times. After this, his fortune doubled. This is equivalent to Billy having 2020 doubloons before the voyage, and he paid the wages 14 times. After this, Billy paid the wages 40 times to half of the remaining pirates. This is equivalent to him paying the wages 20 times to all the pirates. In total, Billy paid the full wages 34 times during the voyage. Therefore, initially, there were $(2020-1000) / 34 = 30$ pirates.

If briefly, $1000 = (1010 - 7n) \cdot 2 - 40 \cdot \frac{n}{2}$.

Answer: 30


Let $n \geqslant 2$, find the maximum and minimum value of the product $x_{1} x_{2} \cdots x_{n}$ under the conditions $x_{i} \geqslant \frac{1}{n}, i=1,2, \cdots, n$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$.

Solution:
［Solution］First, find the maximum value. By the AM-GM inequality, we have
$$
\sqrt[n]{x_{1}^{2} \cdot x_{2}^{2} \cdot \cdots \cdot x_{n}^{2}} \leqslant \frac{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}{n}=\frac{1}{n},
$$

and the equality holds when $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{\sqrt{n}}>\frac{1}{n}(n \geqslant 2)$. Therefore, the maximum value sought is $n^{-\frac{n}{2}}$.
Next, find the minimum value. Take any $x_{1}, x_{2}, \cdots, x_{n}$ that satisfy the conditions of the problem. Let
$$
y_{1}=x_{1}, \cdots, y_{n-2}=x_{n-2}, y_{n-1}=\sqrt{x_{n-1}^{2}+x_{n}^{2}-\frac{1}{n^{2}}}, y_{n}=\frac{1}{n} \text {, }
$$

then $y_{i} \geqslant \frac{1}{n}, i=1,2, \cdots, n$, and $y_{1}^{2}+\cdots+y_{n}^{2}=x_{1}^{2}+\cdots+x_{n}^{2}=1$, i.e., $y_{1}$, $y_{2}, \cdots, y_{n}$ satisfy the conditions of the problem. Furthermore, since
$$
\begin{aligned}
y_{n-1}^{2} y_{n}^{2}-x_{n-1}^{2} x_{n}^{2} & =\frac{1}{n^{2}}\left(x_{n-1}^{2}+x_{n}^{2}-\frac{1}{n^{2}}\right)-x_{n-1}^{2} x_{n}^{2} \\
& =-\left(x_{n-1}^{2}-\frac{1}{n^{2}}\right)\left(x_{n}^{2}-\frac{1}{n^{2}}\right) \leqslant 0,
\end{aligned}
$$

we have $y_{1} y_{2} \cdots y_{n-2} y_{n-1} y_{n} \leqslant x_{1} x_{2} \cdots x_{n-2} x_{n-1} x_{n}$.
Repeating this process $n-1$ times, we get
$$
\begin{aligned}
x_{1} x_{2} \cdots x_{n} & \geqslant\left(\frac{1}{n}\right)^{n-1} \sqrt{1-\frac{n-1}{n^{2}}} \\
& =\frac{\sqrt{n^{2}-n+1}}{n^{n}} .
\end{aligned}
$$

Clearly, when $x_{1}=x_{2}=\cdots=x_{n-1}=\frac{1}{n}, x_{n}=\frac{\sqrt{n^{2}-n+1}}{n}$, the equality in the above inequality holds, so under the conditions of the problem, the minimum value of the product $x_{1} x_{2} \cdots x_{n}$ is $\frac{\sqrt{n^{2}-n+1}}{n^{n}}$.

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


Evaluate the sum

$$
\frac{1}{1+\tan 1^{\circ}}+\frac{1}{1+\tan 2^{\circ}}+\frac{1}{1+\tan 3^{\circ}}+\cdots+\frac{1}{1+\tan 89^{\circ}}
$$

(The tangent $(\tan )$ of an angle $\alpha$ is the ratio $BC / AC$ in a right triangle $ABC$ with $\angle C=90^{\circ}$ and $\angle A=\alpha$, and its value does not depend on the triangle used.)

Solution:
By examining a triangle with angles $x, 90-x$, and 90 , it is not hard to see the trigonometric identity

$$
\tan \left(90^{\circ}-x\right)=\frac{1}{\tan x} .
$$

Let us pair up the terms

$$
\frac{1}{1+\tan x} \text { and } \frac{1}{1+\tan \left(90^{\circ}-x\right)}
$$

for $x=1,2,3, \ldots, 44$. The sum of such a pair of terms is

$$
\begin{aligned}
\frac{1}{1+\tan x}+\frac{1}{1+\tan \left(90^{\circ}-x\right)} & =\frac{1}{1+\tan x}+\frac{1}{1+\frac{1}{\tan x}} \\
& =\frac{1}{1+\tan x}+\frac{\tan x}{\tan x+1} \\
& =\frac{1+\tan x}{1+\tan x}=1
\end{aligned}
$$

There is one more term in the sum, the middle term:

$$
\frac{1}{1+\tan 45^{\circ}}=\frac{1}{1+1}=\frac{1}{2} .
$$

Therefore the sum is $44(1)+1 / 2=89 / 2$.

Answer: \frac{89}{2}


There are four cards, each with a number written on both sides. The first card has 0 and 1, the other three cards have 2 and 3, 4 and 5, 7 and 8 respectively. Now, any three of these cards are taken out and placed in a row, forming a total of $\qquad$ different three-digit numbers.

Solution: Answer: 168

Answer: 168


Given the sets
$$
\begin{array}{l}
M=\{(x, y) \mid x(x-1) \leqslant y(1-y)\}, \\
N=\left\{(x, y) \mid x^{2}+y^{2} \leqslant k\right\} .
\end{array}
$$

If $M \subset N$, then the minimum value of $k$ is $\qquad$ .

Solution:
Notice that,
$$
M=\left\{(x, y) \left\lvert\,\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2}\right.\right\}
$$

represents a disk with center $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{\sqrt{2}}{2}$.
By $M \subset N \Rightarrow \sqrt{k} \geqslant \sqrt{2} \times \frac{1}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$
$$
\Rightarrow k_{\min }=2 \text {. }
$$

Answer: 2


Solve the following system of equations:

$$
\sqrt[3]{7^{x}} \cdot \sqrt[5]{5^{y}}=6125
$$

$$
\sqrt[5]{3^{y}}: \sqrt[3]{3^{x}}=3
$$

Solution:
The given equations can also be written as follows:

From (2)

which gives

Substituting (3) into (1):

$$
7^{\frac{x}{3}} \cdot 5^{\frac{y}{5}}=6125
$$

or

$$
3^{\frac{y}{5}}: 3^{\frac{x}{3}}=3
$$

$$
\frac{y}{5}-\frac{x}{3}=1
$$

$$
\frac{y}{5}=1+\frac{x}{3}
$$

$$
7^{\frac{x}{3}} \cdot 5^{1+\frac{x}{3}}=6125
$$

$$
7^{\frac{x}{3}} \cdot 5^{\frac{x}{3}}=1225
$$

which gives

$$
35^{\frac{x}{3}}=35^{2}
$$

From (3) we get

$$
\begin{gathered}
x=6 \\
y=15
\end{gathered}
$$

(Manheim Emil, Beszterczebánya.)

The problem was also solved by: Barna D., Beck F., Brandt D., Dénes A., Erdős A., Freibauer E., Friedmann B., Kisgyörgy S., Krátky Gy., Misángyi V., Petrogalli G., Porkoláb J., Schiffer H., Spitzer Ö., Szabó I., Szabó K., Weisz Á., Weisz J.

Answer: 6,15


If the decimal parts of $7+\sqrt{7}$ and $7-\sqrt{7}$ are $a$ and $b$ respectively, then $a b-3 a+2 b+1=$ $\qquad$ .

Solution:
2.0.

Since $2<\sqrt{7}<3$, therefore, the decimal part of $7+\sqrt{7}$ is $a=\sqrt{7}-2$. Also, because $-3<-\sqrt{7}<-2$, then $0<3-\sqrt{7}<1$. Therefore, the decimal part of $7-\sqrt{7}$ is $b=3-\sqrt{7}$. Hence
$$
\begin{array}{l}
a b-3 a+2 b+1 \\
=(a+2)(b-3)+7=\sqrt{7} \times(-\sqrt{7})+7=0 .
\end{array}
$$

Answer: 0


Given positive integers $a, b, c$ satisfying
$$
1<a<b<c, a+b+c=111, b^{2}=a c \text {. }
$$

then $b=$ . $\qquad$

Solution:
4.36.

Let $(a, c)=d, a=a_{1} d, c=c_{1} d, a_{1}, c_{1}$ be positive integers, and $\left(a_{1}, c_{1}\right)=1, a_{1}<c_{1}$.
Then $b^{2}=a c=d^{2} a_{1} c_{1} \Rightarrow d^{2}\left|b^{2} \Rightarrow d\right| b$.
Let $b=b_{1} d\left(b_{1} \in \mathbf{Z}_{+}\right)$. Then $b_{1}^{2}=a_{1} c_{1}$.
Since $\left(a_{1}, c_{1}\right)=1$, $a_{1}, c_{1}$ are both perfect squares.
Let $a_{1}=m^{2}, c_{1}=n^{2}$. Then
$b_{1}=m n\left(m, n \in \mathbf{Z}_{+}\right.$, and $\left.(m, n)=1, m<n\right)$.
Also, $a+b+c=111$, so
$d\left(a_{1}+b_{1}+c_{1}\right)=111$
$\Rightarrow d\left(m^{2}+n^{2}+m n\right)=111$.
Notice that,
$$
m^{2}+n^{2}+m n \geqslant 1^{2}+2^{2}+1 \times 2=7 \text {. }
$$

Therefore, $d=1$ or 3.
If $d=1$, then $m^{2}+n^{2}+m n=111$.
By trial, only $m=1, n=10$ satisfy the equation, in which case, $a=1$ (discard).
If $d=3$, then $m^{2}+n^{2}+m n=37$.
By trial, only $m=3, n=4$ satisfy the equation, in which case, $a=27, b=36, c=48$, which meets the problem's requirements.
Therefore, $b=36$.

Answer: 36


In the forest, there is a pair of rabbit brothers racing. The younger brother starts running 10 steps ahead, and then the older brother starts chasing. If the younger brother runs 4 steps in the same time the older brother runs 3 steps, and the older brother runs 5 steps for the same distance the younger brother runs 7 steps, then the older brother needs to run $\qquad$ steps to catch up with the younger brother.

Solution:
【Analysis】Assuming the older brother runs 3 steps in 1 second, then the younger brother runs 4 steps in 1 second;
Since the older brother runs 5 steps equal to the younger brother's 7 steps, the distance the older brother runs in 5 steps must be a multiple of 5 and also a multiple of 7;

Assuming the younger brother runs 5 meters per step, and the older brother runs 7 meters per step, we can then derive the speeds of the older and younger brother, and use the speed difference and distance difference to find the time it takes for the older brother to catch up to the younger brother, and finally determine how many steps the older brother needs to run to catch up to the younger brother.

【Solution】Solution: Assuming the older brother runs 3 steps in 1 second, then the younger brother runs 4 steps in 1 second;
(1) Speed of the younger brother and the older brother:

Speed of the younger brother: $4 \times 5=20$ (meters/second);
Speed of the older brother: $3 \times 7=21$ (meters/second).
(2) Time for the older brother to catch up to the younger brother:
$$
\begin{array}{l}
5 \times 10 \div(21-20), \\
=50 \div 1, \\
=50 \text { (seconds). }
\end{array}
$$
(3) Steps the older brother needs to run to catch up to the younger brother:
$$
50 \times 21 \div 7=150 \text { (steps). }
$$

Answer: The older brother needs to run 150 steps to catch up to the younger brother.
The answer is: 150.

Answer: 150


Find all integers $a$ for which the modulus of the number $\left|a^{2}-3 a-6\right|$ is a prime number. Answer. $-1 ; 4$.

Solution:
Solution. The number $a^{2}-3 a$ is always even (as the difference of two even or two odd numbers), so the prime number $\left|a^{2}-3 a-6\right|$ can only be 2. The equation $a^{2}-3 a-6=2$, or $a^{2}-3 a-8=0$, has no integer solutions. Solving the equation $a^{2}-3 a-6=-2$, we find the roots $a_{1}=-1, a_{2}=4$.

Comment. Correct solution - 20 points. There is a gap in the justification - 15 points. Only established that the prime number is 2 - 2 points. The solution is started, there is some progress - 5 points. The solution is started, but the progress is insignificant - 1 point. Only the correct answer without a solution - 1 point.

Answer: -1,4


Given two circles $C_{1}: x^{2}+y^{2}=1$ and $C_{2}$ : $(x-2)^{2}+y^{2}=16$. Then the locus of the center of the circle that is externally tangent to $C_{1}$ and internally tangent to $C_{2}$ is $\qquad$
$\qquad$

Solution: 4. $\frac{(x-1)^{2}}{\frac{25}{4}}+\frac{y^{2}}{\frac{21}{4}}=1$

Answer: \frac{(x-1)^{2}}{\frac{25}{4}}+\frac{y^{2}}{\frac{21}{4}}=1


Calculate the maximum volume of a tetrahedron contained within a hemisphere of radius 1.

Solution:
Let us denote the given hemisphere by $H$, the whole sphere by $K$, and the circle forming the flat part of the boundary of $H$ by $E$. Let $O$ be any tetrahedron contained in $H$. Its interior is the intersection of four open half-spaces determined by the planes of the faces. The center of the sphere $K$ does not belong to the interior of the tetrahedron; hence it does not belong to at least one of these half-spaces. Therefore, there is a face $S$ of the tetrahedron $Q$ such that the opposite vertex of the tetrahedron and the center of the sphere $K$ lie on opposite sides of the plane of this face (or, in the limiting case, the center of the sphere lies on this plane).

Thus, $Q$ is contained in the smaller of the two parts (or in one of the two equal parts) into which the plane of the face $S$ divides the sphere $K$. Therefore, the height of the tetrahedron $Q$ dropped onto the face $S$ has a length not greater than $1$, so the volume of $Q$ does not exceed one third of the area of this face. The area of the face $S$, being a triangle, is not greater than the area of an equilateral triangle inscribed in the circle which is the section of the sphere $K$ by the plane of this face (see Note); and even more so, it is not greater than the area of an equilateral triangle inscribed in the great circle $E$. The latter area equals $3\sqrt{3}/4$. Therefore, the volume of $Q$ is less than or equal to $\sqrt{3}/4$.

In all the estimates mentioned, we can achieve equality by taking $O$ to be a tetrahedron whose base is an equilateral triangle inscribed in $E$, and the fourth vertex is placed at the farthest point from $E$ on the hemisphere $H$ (at the "north pole").

The number $\sqrt{3}/4$ is the sought maximum.

Note: For completeness, we will provide an elementary proof of the fact that among all triangles inscribed in a fixed circle of radius $r$, the one with the largest area is the equilateral triangle.

Let $ABC$ be any such triangle. By placing point $C$ at the midpoint of the larger of the two arcs into which points $A$ and $B$ divide the circle circumscribed around triangle $ABC$, we obtain an isosceles triangle $ABC$, acute or right, with an area not less than that of triangle $ABC$. Therefore, in searching for a triangle with the maximum area (inscribed in a given circle), we can limit our attention to the class of isosceles triangles without obtuse angles (inscribed in this circle).

If now $ABC$ is a triangle from this class ($|AC| = |BC|$), then denoting by $x$ the distance from side $AB$ to the center of the circle, we get the relationships: $|AB| = 2\sqrt{r^2 - x^2}$, $h_C = r + x$ (the height dropped from vertex $C$), and thus

We have used here the inequality between the geometric mean and the arithmetic mean of four numbers: $a_1 = a_2 = a_3 = r + x$, $a_4 = 3r - 3x$. This inequality becomes an equality if and only if these numbers are equal, i.e., for $x = r/2$. It remains to note that the last equality means that triangle $ABC$ is equilateral.

Answer: \frac{\sqrt{3}}{4}


In the cells of an $8 \times 8$ chessboard, there are 8 white and 8 black chips such that no two chips are in the same cell. Additionally, no column or row contains chips of the same color. For each white chip, the distance to the black chip in the same column is calculated. What is the maximum value that the sum of these distances can take? The distance between chips is the distance between the centers of the cells they occupy.

Solution:
Answer: 32

Solution. An example of chip placement where the sum of distances is 32 is shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_06_856de67b2aabfad709d8g-4.jpg?height=361&width=359&top_left_y=970&top_left_x=547)

We will prove that the sum of distances cannot be greater than 32. For this, consider an arbitrary placement of chips that satisfies the condition. Let \( S \) be the sum of the distances specified in the condition.

Let the white and black chips in the \( i \)-th column be in rows \( a_i \) and \( b_i \), respectively. We estimate the distance between them:

\[
|a_i - b_i| = |(a_i - 4.5) - (b_i - 4.5)| \leq |a_i - 4.5| + |b_i - 4.5|
\]

(essentially, this is the triangle inequality for points \( a_i, b_i \), and 4.5 on a line). Summing this estimate for all differences \( |a_i - b_i| \), we get:

\[
S \leq |a_1 - 4.5| + |a_2 - 4.5| + \ldots + |a_8 - 4.5| + |b_1 - 4.5| + |b_2 - 4.5| + \ldots + |b_8 - 4.5|
\]

Notice that the right-hand side does not depend on the placement of the chips, as the numbers \( a_i, 0 \leq i \leq 8 \), are some permutation of the numbers from 1 to 8. Similarly for \( b_i \). We have:

\[
\begin{aligned}
S & \leq |1 - 4.5| + |2 - 4.5| + \ldots + |8 - 4.5| + |1 - 4.5| + |2 - 4.5| + \ldots + |8 - 4.5| = \\
& = 2 \cdot (3.5 + 2.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5 + 3.5) = 32
\end{aligned}
\]

Remark. From the proof of the estimate, it is not difficult to understand how any example with the required sum of distances is constructed: in each column, a chip of one color is in the upper half, and a chip of the other color is in the lower half. In any other case, equality in the triangle inequality will not be achieved.

## Criteria

A complete solution to the problem is worth 7 points. In the absence of such a solution, the following criteria are summed:

6 points. It is proven that the sum of distances cannot exceed 32.

If the solution is unjustifiably reduced to some "optimal" case or class of cases (for example, it is claimed without proof that in each column, the black and white chips must be in different halves of the column), then points for this part are not given.

1 point. An example of chip placement is provided where the sum of distances is 32.

Answer: 32


Find all $(k, z) \in \mathbb{N}^{2}$ such that $2^{k}+1=z^{2}$.

Solution:
Let $(k, z)$ be a solution. We factorize: $2^{k}=(z-1)(z+1)$, so $z-1$ and $z+1$ are powers of 2. Let $a<b$ such that $z-1=2^{a}$ and $z+1=2^{b}$. The difference between these two equations gives: $2^{b-a}\left(2^{a}-1\right)=2$. Therefore, $2^{b-a}=1$ and $2^{a}=1$ (by the uniqueness of the prime factorization). Therefore, $b=a+1$ and $a=0$. Thus, $z=3$, and we verify conversely that this is indeed a solution.

## 5 Tuesday 23 morning: Colin Davalo

This course has not yet been integrated.

## 6 Tuesday 23 afternoon: Florent Noisette

This course has not yet been integrated.

## 3 Group C: Geometry and Functional Equations

## 1 Sunday 21 morning: Cécile Gachet

Study Principle In this course, we focus on different geometric transformations that map points to points, lines to lines, and circles to circles. The study plan is as follows:

1. We consider a quantity that we would like to see preserved by our transformation (lengths, ratios of lengths, geometric angles, oriented angles);
2. We define the transformations that preserve this quantity and study their structure:

- Center of a transformation: does such a transformation have a fixed point? If so, how to construct it?
- "Degrees of freedom": can we determine a transformation from the given data of a certain number of points and their images? Conversely, for which $n$ can we fix $A_{1}, \ldots, A_{n}$ and $B_{1}, \ldots, B_{n}$ such that there exists an appropriate transformation that maps $A_{1}$ to $B_{1}, \ldots, A_{n}$ to $B_{n}$? (and what additional constraints must be imposed on these points to increase the degree of freedom $n$?)
- Composition of two transformations of the same type
- Decomposition of a transformation as a product of "elementary" factors.

3. If we have two types of transformations (those that preserve the schtroumpfs and those that preserve the bagadas, for example), we study the relative position of the set of transformations that preserve the schtroumpfs and the set of transformations that preserve the bagadas, and how our composition law behaves with respect to them (see the diagram below).

Additional points covered/not covered We emphasized the difference between a direct transformation and an indirect transformation (starting observation: axial symmetry is an isometry, it preserves angles but multiplies all oriented angle values by -1), this question playing an important role in the study of the number of degrees of freedom of a transformation. We studied the notion of interior similarity. We provided the construction of the Miquel point, from which we could deduce the theorem of the concurrence of the four circles at the Miquel point.

The general study of axial symmetries (especially their composition) was not carried out. The existence of a center for every similarity was assumed.

## Summary Diagram

Exercises covered In addition to the study of the various transformations that we tried to conduct as several small exercises, the following exercises were posed:

Answer: (k,z)=(3,3)


Solution:
## Solution.

Notice that $x=14$ and $x=-14$ are not solutions to the equation.

Let $a=\sqrt[3]{14-x}, b=\sqrt[3]{14+x}$. Then $a b=\sqrt[3]{196-x^{2}}$, and the original equation can be written as $3 a^{2}-2 b^{2}=-5 a b$.

If $b$ were zero, then $x$ would be -14, which we have verified is not a solution to the equation. Therefore, $b \neq 0$.

Dividing by $b^{2}$, we get

$$
3 \cdot \frac{a^{2}}{b^{2}}-2=-5 \cdot \frac{a}{b}
$$

Substituting $t=\frac{a}{b}$ and rearranging, we get $3 t^{2}+5 t-2=0$.

The solutions to this quadratic equation are $t_{1}=\frac{1}{3}, t_{2}=-2$.

For $\frac{\sqrt[3]{14-x}}{\sqrt[3]{14+x}}=\frac{1}{3}$, we get $\frac{14-x}{14+x}=\frac{1}{27}$, which gives $x_{1}=13$.

Similarly, for $\frac{\sqrt[3]{14-x}}{\sqrt[3]{14+x}}=-2$, we get the second solution $x_{2}=-18$.

Note: Instead of dividing by $b^{2}$ and introducing the substitution $t$, the equation $3 a^{2}-2 b^{2}+5 a b=0$ can be written as $(3 a-b)(a+2 b)=0$, which is worth 3 points. After that, the case $3 a=b$ and obtaining the solution $x=13$ is worth 2 points, and the case $a=-2 b$ and $x=-18$ is worth another 2 points.

Answer: x_1=13,x_2=-18


Let real numbers $x, y$ satisfy the equation $2 x^{2}+3 y^{2}=6 y$. Then the maximum value of $x+y$ is $\qquad$ .

Solution:
2. $1+\frac{\sqrt{10}}{2}$.

Let $x+y=t$. Then $x=t-y$.
From $2(t-y)^{2}+3 y^{2}=6 y$, we get
$5 y^{2}-2(2 t+3) y+2 t^{2}=0$.
Since $y$ is a real number, therefore,
$$
\begin{array}{l}
\Delta=4(2 t+3)^{2}-4 \times 5 \times 2 t^{2} \geqslant 0 \\
\Rightarrow \frac{2-\sqrt{10}}{2} \leqslant t \leqslant \frac{2+\sqrt{10}}{2}
\end{array}
$$

Thus, the maximum value of $x+y$ is $1+\frac{\sqrt{10}}{2}$.

Answer: 1+\frac{\sqrt{10}}{2}


Solution:
Solution.

From the condition we have $\left\{\begin{array}{l}x^{2} y^{2}(y+x)=12, \\ x^{2} y^{2}(y-x)=4 .\end{array}\right.$

Dividing the first equation by the second, we get

$\frac{x^{2} y^{2}(y+x)}{x^{2} y^{2}(y-x)}=\frac{12}{4} \Leftrightarrow \frac{y+x}{y-x}=3 \Leftrightarrow y+x=3 y-3 x \Leftrightarrow y=2 x$.

From the first equation of the system, we find

$x^{2}(2 x)^{3}+x^{3}(2 x)^{2}=12 \Leftrightarrow 8 x^{5}+4 x^{5}=12 \Leftrightarrow x^{5}=1, x=1$.

Then $y=2$.

Answer: $(1 ; 2)$

Answer: (1;2)


In a mathematics competition, the sum of the scores of Bill and Dick equalled the sum of the scores of Ann and Carol.
If the scores of Bill and Carol had been interchanged, then the sum of the scores of Ann and Carol would have exceeded
the sum of the scores of the other two. Also, Dick's score exceeded the sum of the scores of Bill and Carol.
Determine the order in which the four contestants finished, from highest to lowest. Assume all scores were nonnegative.
$\textbf{(A)}\ \text{Dick, Ann, Carol, Bill} \qquad \textbf{(B)}\ \text{Dick, Ann, Bill, Carol} \qquad \textbf{(C)}\ \text{Dick, Carol, Bill, Ann}\\ \qquad \textbf{(D)}\ \text{Ann, Dick, Carol, Bill}\qquad \textbf{(E)}\ \text{Ann, Dick, Bill, Carol}$

Solution: Let the scores be $A$, $B$, $C$, and $D$. Then we have $A + C = B + D$, $A + B > C + D$, and $D > B + C$. Call these equations $(1)$, $(2)$, and $(3)$ respectively. Then adding $(1)$ and $(2)$ gives $2A + B + C > B + C + 2D \implies 2A > 2D \implies A > D$. Now subtracting $(1)$ from $(2)$ gives $A + B - A - C > C + D - B - D \implies B - C > C - B \implies 2B > 2C \implies B > C$. Finally from $(3)$, since  $D > B + C$ and $C$ is non-negative, we must have $D > B$, so putting our results together we get $A > D > B > C$, which is answer $\boxed{\text{E}}$.

Answer: E


Given the three sides of an obtuse triangle are $3, 4, x$, then the range of $x$ is
(A) $1<x<7$.
(B) $5<x<7$.
(C) $1<x<\sqrt{7}$.
(D) $5<x<7$ or $1<x<\sqrt{7}$.

Solution:
[Solution] Let $x$ be the longest side, and its opposite angle be $\alpha$, then $\cos \alpha25$, so $x>5$.
Also, $x4-3=1, \quad \therefore \quad 1<x<\sqrt{7}$.
Therefore, the answer is $(D)$.

Answer: D


Petya climbed up a moving upward escalator, counting 75 steps, and then descended the same escalator (i.e., moving against the direction of the escalator), counting 150 steps. During the descent, Petya walked three times faster than during the ascent. How many steps are there on the stopped escalator?

Solution:
Answer: 120.

For convenience, let's introduce a unit of time during which Petya took one step while ascending the escalator. We will measure all speeds in steps per unit of time. Petya's speed while ascending is 1 step per unit of time, and his speed while descending is 3 steps per unit of time. Let the speed of the escalator be $x$. The absolute speed of Petya was $1+x$ while ascending and $3-x$ while descending. Petya spent 75 units of time ascending and 50 units of time descending (he took steps three times faster while descending). The distance covered while ascending and descending was the same - the length of the escalator. Therefore, $75 \cdot(1+x)=50 \cdot(3-x)$, from which $125 x=75, x=0.6$.

The length of the escalator is $75 \cdot(1+0.6)=120$ steps. This is the number of steps on the stopped escalator.

Answer: 120


The smallest positive period of the function $f(x)=|\sin x|-|\cos x|$ is

Solution:
8. $\pi$. It is easy to know that $\pi$ is a period of $f(x)$. Now, we will use proof by contradiction to show that $\pi$ is its smallest positive period.

Assume $T(0<T<\pi)$ is a period of $f(x)$, then
$$
|\sin (x+T)|-|\cos (x+T)|=|\sin x|-|\cos x|
$$

holds for any $x \in \mathbf{R}$. Taking $x=0$, we get
$$
|\sin T|-|\cos T|=|\sin 0|-|\cos 0|=1
$$

However, when $0<T<\pi$, $0<|\sin T| \leqslant 1, 0<|\cos T|<1$, thus $-1<|\sin T|-|\cos T| \leqslant 1$, which is a contradiction. This shows that there does not exist a positive number $T$ that satisfies (1) and is less than $\pi$.

Answer: \pi


Let $N$ be the number of functions $f:\{1,2,3,4,5,6,7,8,9,10\} \rightarrow \{1,2,3,4,5\}$ that have the property that for $1\leq x\leq 5$ it is true that $f(f(x))=x$. Given that $N$ can be written in the form $5^a\cdot b$ for positive integers $a$ and $b$ with $b$ not divisible by $5$, find $a+b$.

Solution:
To solve the problem, we need to count the number of functions \( f: \{1,2,3,4,5,6,7,8,9,10\} \rightarrow \{1,2,3,4,5\} \) that satisfy the condition \( f(f(x)) = x \) for \( 1 \leq x \leq 5 \).

1. **Mapping for \( \{6,7,8,9,10\} \):**
   - The values \( 6, 7, 8, 9, 10 \) can map to any of \( 1, 2, 3, 4, 5 \) without any restrictions.
   - There are \( 5 \) choices for each of these \( 5 \) elements.
   - Therefore, the number of ways to assign values to \( 6, 7, 8, 9, 10 \) is \( 5^5 \).

2. **Mapping for \( \{1,2,3,4,5\} \) with the condition \( f(f(x)) = x \):**
   - The condition \( f(f(x)) = x \) implies that \( f \) must be an involution on \( \{1,2,3,4,5\} \). An involution is a function that is its own inverse.
   - Involutions on a set can be decomposed into fixed points and 2-cycles (pairs \((a, b)\) such that \( f(a) = b \) and \( f(b) = a \)).

3. **Counting the involutions:**
   - We need to count the number of ways to partition \( \{1,2,3,4,5\} \) into fixed points and 2-cycles.
   - We can have:
     - No pairs (all fixed points): \( 1 \) way.
     - One pair and three fixed points: \( \binom{5}{2} = 10 \) ways.
     - Two pairs and one fixed point: \( \frac{\binom{5}{2} \cdot \binom{3}{2}}{2!} = \frac{10 \cdot 3}{2} = 15 \) ways.
     - Three pairs (not possible since we only have 5 elements).

4. **Total number of involutions:**
   - Summing these, we get \( 1 + 10 + 15 = 26 \) involutions.

5. **Combining the counts:**
   - The total number of functions \( f \) is the product of the number of ways to assign values to \( 6, 7, 8, 9, 10 \) and the number of involutions on \( \{1,2,3,4,5\} \):
   \[
   N = 5^5 \cdot 26
   \]

6. **Expressing \( N \) in the form \( 5^a \cdot b \):**
   - We have \( N = 5^5 \cdot 26 \).
   - Here, \( a = 5 \) and \( b = 26 \).

7. **Finding \( a + b \):**
   - \( a + b = 5 + 26 = 31 \).

The final answer is \( \boxed{31} \).

Answer: 31


The maximal correlation of two random variables $X$ and $Y$ is defined to be the supremum of the correlations of $f(X)$ and $g(Y)$ where $f,g:\mathbb{R} \to \mathbb{R}$ are measurable functions such that $f(X)$ and $g(Y)$ is (almost surely?) finite.
Let $U$ be the uniformly distributed random variable on $[0,2\pi]$ and let $m,n$ be positive integers. Compute the maximal correlation of $\sin(nU)$ and $\sin(mU)$.
(Remark: It seems that to make sense we should require that $E[f(X)]$ and $E[g(Y)]$ as well as $E[f(X)^2]$ and $E[g(Y)^2]$ are finite.
In fact, we may then w.l.o.g. assume that $E[f(X)]=E[g(Y)]=0$ and $E[f(Y)^2]=E[g(Y)^2]=1$.)

Solution:
1. **Understanding the Problem:**
   The problem asks us to compute the maximal correlation of $\sin(nU)$ and $\sin(mU)$, where $U$ is a uniformly distributed random variable on $[0, 2\pi]$, and $m$ and $n$ are positive integers. The maximal correlation of two random variables $X$ and $Y$ is defined as the supremum of the correlations of $f(X)$ and $g(Y)$ over all measurable functions $f$ and $g$ such that $f(X)$ and $g(Y)$ are finite.

2. **Correlation Definition:**
   The correlation $\rho$ between two random variables $A$ and $B$ is given by:
   \[
   \rho(A, B) = \frac{\text{Cov}(A, B)}{\sqrt{\text{Var}(A) \text{Var}(B)}}
   \]
   where $\text{Cov}(A, B)$ is the covariance of $A$ and $B$, and $\text{Var}(A)$ and $\text{Var}(B)$ are the variances of $A$ and $B$, respectively.

3. **Uniform Distribution Properties:**
   Since $U$ is uniformly distributed on $[0, 2\pi]$, its probability density function is:
   \[
   f_U(u) = \frac{1}{2\pi}, \quad 0 \leq u \leq 2\pi
   \]

4. **Expectation and Variance Calculations:**
   We need to calculate the expectations and variances of $\sin(nU)$ and $\sin(mU)$:
   \[
   E[\sin(nU)] = \int_0^{2\pi} \sin(nu) \cdot \frac{1}{2\pi} \, du = 0
   \]
   because the integral of $\sin(nu)$ over a full period is zero.

   Similarly,
   \[
   E[\sin(mU)] = 0
   \]

   The variances are:
   \[
   \text{Var}(\sin(nU)) = E[\sin^2(nU)] - (E[\sin(nU)])^2 = E[\sin^2(nU)]
   \]
   \[
   E[\sin^2(nU)] = \int_0^{2\pi} \sin^2(nu) \cdot \frac{1}{2\pi} \, du = \frac{1}{2}
   \]
   because $\sin^2(x)$ has an average value of $\frac{1}{2}$ over a full period.

   Thus,
   \[
   \text{Var}(\sin(nU)) = \frac{1}{2}
   \]
   and similarly,
   \[
   \text{Var}(\sin(mU)) = \frac{1}{2}
   \]

5. **Covariance Calculation:**
   The covariance $\text{Cov}(\sin(nU), \sin(mU))$ is:
   \[
   \text{Cov}(\sin(nU), \sin(mU)) = E[\sin(nU) \sin(mU)] - E[\sin(nU)]E[\sin(mU)]
   \]
   Since $E[\sin(nU)] = 0$ and $E[\sin(mU)] = 0$, we have:
   \[
   \text{Cov}(\sin(nU), \sin(mU)) = E[\sin(nU) \sin(mU)]
   \]

   Using the product-to-sum identities:
   \[
   \sin(nU) \sin(mU) = \frac{1}{2} [\cos((n-m)U) - \cos((n+m)U)]
   \]

   Therefore,
   \[
   E[\sin(nU) \sin(mU)] = \frac{1}{2} \left( E[\cos((n-m)U)] - E[\cos((n+m)U)] \right)
   \]

   Since $U$ is uniformly distributed over $[0, 2\pi]$, the expectations of $\cos((n-m)U)$ and $\cos((n+m)U)$ are zero:
   \[
   E[\cos((n-m)U)] = 0 \quad \text{and} \quad E[\cos((n+m)U)] = 0
   \]

   Thus,
   \[
   \text{Cov}(\sin(nU), \sin(mU)) = 0
   \]

6. **Correlation Calculation:**
   Finally, the correlation $\rho(\sin(nU), \sin(mU))$ is:
   \[
   \rho(\sin(nU), \sin(mU)) = \frac{\text{Cov}(\sin(nU), \sin(mU))}{\sqrt{\text{Var}(\sin(nU)) \text{Var}(\sin(mU))}} = \frac{0}{\sqrt{\frac{1}{2} \cdot \frac{1}{2}}} = 0
   \]

7. **Maximal Correlation:**
   Since the correlation is zero, the maximal correlation is also zero.

\[
\boxed{0}
\]

Answer: 0


In decimal, find the smallest natural number: its square number starts with 19 and ends with 89.

Solution:
[Solution] The last digit of the required number can only be 3 or 7. We write out the squares of all two-digit numbers ending in 3 and 7, among which only $17^{2}, 33^{2}, 67^{2}, 83^{2}$ end with 89. To make the first two digits of the natural number $x$ be 19, the inequality must hold:
$19 \leqslant x^{2} \cdot 10^{-N}<20$, where $N$ is a natural number. When $N=2 k$ and $N=2 k+1$, we get the inequalities
$19 \leqslant x^{2} \cdot 10^{-2 k}<20$ and $190 \leqslant x^{2} \cdot 10^{-2 k}<200$, taking the square root, in the first case we get the inequality
$$
4.3588989 \cdots \leqslant x \cdot 10^{-k}<4.4721359 \cdots ;
$$

and in the second case, we have
$13.784048 \cdots \leqslant x \cdot 10^{-k}<14.142135 \cdots$.
It is not difficult to see that, due to the change in $k$, there will be infinitely many possibilities. We examine $k=1$ and $k=2$.
If $4.358 \leqslant x \cdot 10^{-1}<4.472$, then $x=44$;
If $4.358 \leqslant x \cdot 10^{-2}<4.472$, then $436 \leqslant x \leqslant 447$;
If $13.784 \cdots \leqslant x \cdot 10^{-1}<14.142 \cdots$, then $138 \leqslant x \leqslant 141$;
If $13.784 \cdots \leqslant x \cdot 10^{-2}<14.142 \cdots$, then $1379 \leqslant x \leqslant 1414$.
When $k=3$, we similarly examine the two possibilities, and so on. Among the $x$ values we obtained, the smallest number whose last two digits are $17,33,67,83$ is $1383\left(1383^{2}=1912689\right)$. Therefore, the required number is 1383.

Answer: 1383


Xiaoming is playing a new type of flying chess on the computer. If the dice roll shows a black number, it means he needs to move backward; if it shows a red number, it means he needs to move forward (for example, black 2 means move back 2 steps, red 3 means move forward 3 steps). He rolled the dice 5 times, getting black 5, red 4, red 2, black 3, and red 1. After all the moves, he found himself on the 6th square. Before the first roll, he was on the $\qquad$ square

Solution:
【Answer】 7

【Solution】 Restoration problem, $6-1+3-2-4+5=7$

Answer: 7


Find all functions $f: Z_{+} \rightarrow Z_{+}$, such that for any positive integer $n$, we have
$$
f(f(f(n)))+f(f(n))+f(n)=3 n \text {. }
$$
(2008, Dutch National Team Selection Exam)

Solution:
If $f(m)=f(n)$, then
$$
3 m=3 n \Rightarrow m=n \text{. }
$$

Thus, $f$ is injective.
Next, we prove by induction on $n$ that:
$$
f(n)=n \text{. }
$$
(1) When $n=1$, since $f$ is a function from $\mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, we have $f(1) \geqslant 1$.
Therefore, $f(f(1)) \geqslant 1, f(f(f(1))) \geqslant 1$.
Also, $f(f(f(1)))+f(f(1))+f(1)=3$, so $f(1)=f(f(1))=f(f(f(1)))=1$.
(2) Suppose $k \geqslant 2$, and for $n<k$, $f(n)=n$.
By the injective property, for all $m(m \geqslant k)$, we have $f(m) \geqslant k$.
In particular,
$$
f(k) \geqslant k, f(f(k)) \geqslant k, f(f(f(k))) \geqslant k \text{. }
$$

From (1), we know
$$
f(f(f(k)))+f(f(k))+f(k)=3 k \text{. }
$$

Thus, $f(k)=f(f(k))=f(f(f(k)))=k$.
In conclusion, $f(n)=n$ for all $n \in \mathbf{Z}_{+}$.
Upon verification, $f(n)=n$ is a solution to the original equation.

Answer: f(n)=n


Quadrilateral $ABCD$ has vertices $A(0,3), B(0, k), C(t, 10)$, and $D(t, 0)$, where $k>3$ and $t>0$. The area of quadrilateral $ABCD$ is 50 square units. What is the value of $k$?

Solution:
Using the information given, the quadrilateral looks like

![](https://cdn.mathpix.com/cropped/2024_04_17_a164435ebfb99ea5f1c4g-23.jpg?height=688&width=707&top_left_y=1938&top_left_x=812)

We note that the diagram has been drawn with $k3$, so $k-3$ is positive and makes sense as a length.

The area of the trapezoid is

$$
\frac{t}{2}(A B+C D)=\frac{t}{2}(k-3+10)=\frac{t}{2}(k+7) .
$$

Since the area is 50 , this means $50=\frac{t}{2}(k+7)$ or $k+7=\frac{100}{t}$ so $k=\frac{100}{t}-7$.

The answer to part (b) is 5 , so $t=5$. This means $k=\frac{100}{5}-7=20-7=13$.

ANSWER: $6,5,13$

Answer: 13


A bag of peanuts contains 1988 peanuts. A monkey takes one peanut on the first day, and from the second day onwards, the number of peanuts it takes each day is the total of all the peanuts taken in previous days. If on a certain day the number of peanuts left in the bag is less than the total number of peanuts already taken, on that day it starts over by taking one peanut and follows the original rule for a new cycle. If this continues, on which day will the monkey have taken all the peanuts from the bag?

Solution:
［Solution］Suppose the monkey used $k$ days in the first round, meaning that on the $k+1$-th day, it started taking one peanut again.
At this point, in the first $k$ days, a total of
$$
1+1+2+2^{2}+2^{3}+\cdots+2^{k-2}=2^{k-1}
$$

peanuts were taken, and it satisfies $1988-2^{k-1}<2^{k-1}$.
If in the second round it used $t$ days, then in the second round, it took $2^{t-1}$ peanuts, and
$$
1988-2^{k-1}-2^{t-1}<2^{t-1}, \quad t<k .
$$

Therefore, we only need to decompose 1988 into the sum of powers of 2, i.e.,
$$
1988=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+2^{2} .
$$

So, the total number of days used is
$$
\begin{aligned}
& (10+1)+(9+1)+(8+1)+(7+1)+(6+1)+(2+1) \\
= & 11+10+9+8+7+3 \\
= & 48 \text { (days). }
\end{aligned}
$$

Answer: 48


Given the hyperbola $\frac{x^{2}}{t}-\frac{y^{2}}{t}=1$ with the right focus $F$, any line passing through $F$ intersects the right branch of the hyperbola at $M, N$, and the perpendicular bisector of $M N$ intersects the $x$-axis at $P$. When $t$ takes any positive real number except 0, $\frac{|F P|}{|M N|}=$ $\qquad$ .

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

Obviously, the eccentricity $e=\sqrt{2}$. Let $M\left(x_{1}, y_{1}\right), N\left(x_{3}\right.$, $\left.y_{2}\right)$, the slope angle of $M N$ be $a$, the midpoint of $M N$ be $Q$, and the distance from $M$ to the right directrix be $d$. Then, by the definition of the hyperbola, we have $\frac{|M F|}{d}=e$, i.e., $\frac{|M F|}{x_{1}-\frac{a^{2}}{c}}=e,|M F|=$ $e x_{1}-a$.
Similarly, $|N F|=e x_{2}-$
$a$,
$$
2|F Q|=\mid M F-
$$
$N F \mid$
$$
\begin{array}{l}
=\left|e x_{1}-a-e x_{2}+a\right|=e\left|x_{1}-x_{2}\right| . \\
|F P|=\frac{|F Q|}{\cos \alpha}=\frac{e\left|x_{1}-x_{2}\right|}{\cos \alpha} . \\
\text { Also, }\left|x_{1}-x_{2}\right|=M N \cdot \cos \alpha,
\end{array}
$$

Thus, $\frac{|F P|}{|M N|}=\frac{1}{2} e=\frac{\sqrt{2}}{2}$.

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


What is the biggest shadow that a cube of side length $1$ can have, with the sun at its peak?

Note: "The biggest shadow of a figure with the sun at its peak" is understood to be the biggest possible area of the orthogonal projection of the figure on a plane.

Solution:
1. **Understanding the Problem:**
   We need to find the maximum possible area of the orthogonal projection (shadow) of a cube with side length 1 when the sun is at its peak. This means we are looking for the largest possible area of the shadow of the cube on a plane.

2. **Simplifying the Problem:**
   Consider a cube with side length 1. The cube has 6 faces, each with an area of 1. When the sun is directly overhead, the shadow of the cube on the ground is the orthogonal projection of the cube onto the plane.

3. **Projection of the Cube:**
   The orthogonal projection of a cube onto a plane can vary depending on the orientation of the cube. The maximum shadow area occurs when the projection is maximized.

4. **Choosing a Vertex and Adjacent Faces:**
   Let's pick a vertex \( P \) of the cube and consider the three faces that meet at \( P \). These faces are adjacent and form a corner of the cube. The projection of these three faces will give us an insight into the maximum shadow area.

5. **Analyzing the Projection:**
   The three faces adjacent to vertex \( P \) can be considered as forming a tetrahedron \( PABC \), where \( A, B, \) and \( C \) are the vertices adjacent to \( P \). The shadow of the remaining three faces is double the surface area of the tetrahedron \( PABC \).

6. **Calculating the Area of the Triangle \( \triangle ABC \):**
   The area of \( \triangle ABC \) can be calculated as follows:
   \[
   \text{Area of } \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Since \( A, B, \) and \( C \) are vertices of the cube, the length of each side of \( \triangle ABC \) is \( \sqrt{2} \) (diagonal of a face of the cube).

7. **Using the Formula for the Area of an Equilateral Triangle:**
   The area of an equilateral triangle with side length \( s \) is given by:
   \[
   \text{Area} = \frac{s^2 \sqrt{3}}{4}
   \]
   For \( \triangle ABC \), \( s = \sqrt{2} \):
   \[
   \text{Area of } \triangle ABC = \frac{(\sqrt{2})^2 \sqrt{3}}{4} = \frac{2 \sqrt{3}}{4} = \frac{\sqrt{3}}{2}
   \]

8. **Maximum Shadow Area:**
   The maximum shadow area of the cube is double the area of \( \triangle ABC \):
   \[
   \text{Maximum Shadow Area} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}
   \]

The final answer is \( \boxed{ \sqrt{3} } \).

Answer: \sqrt{3}


Solution:
Since each $\square$ has length $\frac{2}{3}$, then a strip of three $\square$ will have length $3 \times \frac{2}{3}=2$.

We need two strips of length 2 to get a strip of length 4 .

Since $3 \square$ make a strip of length 2 , then $6 \square$ make a strip of length 4 .

ANSWER: (A)

Answer: A


Solution:
Solution. We have
$$
a_{2}=a_{1}-\frac{1}{2}+\frac{2}{3}-\frac{1}{4}
$$
and
$$
a_{3}=a_{2}-\frac{1}{3}+\frac{2}{4}-\frac{1}{5}=a_{1}-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}
$$

Further
$$
a_{4}=a_{1}-\frac{1}{2}+\frac{1}{3}+\frac{1}{5}-\frac{1}{6}
$$

The pattern is now clear and (ideally we can prove it by by induction)
$$
a_{k}=a_{1}-\frac{1}{3}+\frac{1}{4}+\frac{1}{k+1}-\frac{1}{k+2} .
$$

In particular,
$$
a_{100}=\frac{1}{6}-\frac{1}{3}+\frac{1}{4}+\frac{1}{101}-\frac{1}{102}=\frac{1}{10302} .
$$

Answer: \frac{1}{10302}


The price for a ride on the "Voskhod" carousel in February 2020 was 300 rubles per person. In March, the price was reduced, and the number of visitors increased by $50 \%$, while the revenue increased by $25 \%$. By how many rubles was the price reduced?

Solution:
Answer: 50.

Solution. The entry fee for two people in February 2020 was 600 rubles. In March, instead of every two people, the stadium is visited by three people, as the number of visitors increased by $50 \%$. Since the total collection increased by $25 \%$, they will pay $600+0.25 \cdot 600=750$ (rub.), and thus one ticket will cost $750: 3=250$ (rub.), which is 50 rubles less than in February.

Comment. Answer without justification - 1 point.

Answer: 50


Given that point $P$ is a moving point on the line $x+2 y=4$, and through point $P$ two tangents are drawn to the ellipse $x^{2}+4 y^{2}=4$, with the points of tangency being $A$ and $B$, when point $P$ moves, the line $AB$ passes through a fixed point with coordinates

Solution:
Answer $\left(1, \frac{1}{2}\right)$.
Analysis Let $P\left(x_{0}, y_{0}\right)$, then $x_{0}+2 y_{0}=4$, i.e., $y_{0}=\frac{1}{2}\left(4-x_{0}\right)$, and the chord equation of the tangent points $A B$ is $x_{0} x+4 y_{0} y=4$, i.e.,
$$
x_{0} x+4 y \cdot \frac{1}{2}\left(4-x_{0}\right)=4, \quad(x-2 y) x_{0}+(8 y-4)=0
$$

Thus $x-2 y=0,8 y-4=0$, solving gives $x=1, y=\frac{1}{2}$, i.e., the coordinates of the fixed point through which the line $A B$ passes are $\left(1, \frac{1}{2}\right)$, hence fill in $\left(1, \frac{1}{2}\right)$.

Answer: (1,\frac{1}{2})


In the division problem $a: b=c$, $a, b, c$ are to be replaced by natural numbers so that a correctly calculated division problem results, where only the digits 1, 2, 3, 4, 5, and each exactly once, are to be used.

Determine all triples $(a ; b ; c)$ of natural numbers that meet these requirements!

Solution:
Given that there are 5 digits available for three numbers, at least one must be a single-digit number. Since the dividend $a$ in a division problem (within the realm of natural numbers) is the largest value, $a$ must be at least a two-digit number. If both $b$ and $c$ were single-digit, their product $a$ would be at most $5 \cdot 4 = 20$, meaning not all five digits would be used. Therefore, $a$ must be a two-digit number, and exactly one of the two values $b$ or $c$ must also be a two-digit number, while the other is a single-digit number.

Since the digit 5 can only be used once, it cannot be used as a unit digit, as this would make the corresponding number divisible by 5, which would in turn require at least one of the other three numbers to also be divisible by 5. This is not possible due to the lack of available unit digits 0 or another 5.

Therefore, the digit 5 must be used as a tens digit. However, it cannot be the tens digit of the quotient or divisor, as this would make the value at least 51, while the largest possible value for the dividend, which must be larger, would be 43. Thus, the 5 must be the tens digit of $a$.

Neither $b$ nor $c$ can be 1, as this would make the other value equal to $a$, leading to the reuse of digits.

Since 53 is a prime number and $51 = 3 \cdot 17$ as the only non-trivial factorization uses the unavailable digit 7, these two possibilities for $a$ are excluded. This leaves two cases:

1. Case: $a = 52$. Then, because $52 = 2 \cdot 26 = 4 \cdot 13$, there are exactly two possibilities: $b = 4$ and $c = 13$ or $b = 13$ and $c = 4$. No other factorizations exist in this case.
2. Case: $a = 54$. Then, because $54 = 2 \cdot 27 = 3 \cdot 18 = 6 \cdot 9$, there are no non-trivial factorizations that meet the requirements of the problem.

Therefore, there are exactly two such triples, namely $(a, b, c) \in \{(52, 4, 13), (52, 13, 4)\}$.

Answer: (52,4,13),(52,13,4)


Let $n\geq 2$ be a positive integer and let $a_1,a_2,...,a_n\in[0,1]$ be real numbers. Find the maximum value of the smallest of the numbers: \[a_1-a_1a_2, \ a_2-a_2a_3,...,a_n-a_na_1.\]

Solution:
1. **Define the problem and variables:**
   Let \( n \geq 2 \) be a positive integer and let \( a_1, a_2, \ldots, a_n \in [0,1] \) be real numbers. We need to find the maximum value of the smallest of the numbers:
   \[
   a_1 - a_1 a_2, \ a_2 - a_2 a_3, \ \ldots, \ a_n - a_n a_1.
   \]

2. **Express each term in a simplified form:**
   Each term \( a_i - a_i a_{i+1} \) can be rewritten as:
   \[
   a_i (1 - a_{i+1}).
   \]
   Since \( a_i \in [0,1] \), it follows that \( a_i (1 - a_{i+1}) \geq 0 \).

3. **Consider the product of all terms:**
   We look at the product of these \( n \) expressions:
   \[
   \prod_{i=1}^n (a_i - a_i a_{i+1}) = \prod_{i=1}^n a_i (1 - a_{i+1}).
   \]

4. **Apply the AM-GM inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   \sqrt[n]{\prod_{i=1}^n a_i (1 - a_{i+1})} \leq \frac{1}{n} \sum_{i=1}^n a_i (1 - a_{i+1}).
   \]
   Since \( a_i \in [0,1] \), the maximum value of \( a_i (1 - a_{i+1}) \) occurs when \( a_i = \frac{1}{2} \) and \( 1 - a_{i+1} = \frac{1}{2} \), giving:
   \[
   a_i (1 - a_{i+1}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.
   \]

5. **Calculate the maximum value:**
   If all \( a_i = \frac{1}{2} \), then:
   \[
   \prod_{i=1}^n a_i (1 - a_{i+1}) = \left( \frac{1}{4} \right)^n.
   \]
   Taking the \( n \)-th root, we get:
   \[
   \sqrt[n]{\left( \frac{1}{4} \right)^n} = \frac{1}{4}.
   \]

6. **Verify the conditions for equality:**
   The equality in the AM-GM inequality holds when all terms are equal, i.e., \( a_i = \frac{1}{2} \) for all \( i \). This indeed gives:
   \[
   a_i - a_i a_{i+1} = \frac{1}{2} \left( 1 - \frac{1}{2} \right) = \frac{1}{4}.
   \]

7. **Conclusion:**
   Therefore, the maximum value of the smallest of the numbers \( a_1 - a_1 a_2, a_2 - a_2 a_3, \ldots, a_n - a_n a_1 \) is \( \frac{1}{4} \).

The final answer is \(\boxed{\frac{1}{4}}\).

Answer: \frac{1}{4}


Given two sets of real numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$ and $B=\left\{b_{1}, b_{2}, \cdots, b_{50}\right\}$. If the mapping $f$ from $A$ to $B$ makes every element in $B$ have a preimage, and $f\left(a_{1}\right) \leqslant f\left(a_{2}\right) \leqslant \cdots \leqslant$ $f\left(a_{100}\right)$, then the number of such mappings is ( ).
(A) $\mathrm{C}_{100}^{50}$
(B) $\mathrm{C}_{90}^{50}$
(C) $\mathrm{C}_{99}^{49}$
(D) $\mathrm{C}_{99}^{49}$

Solution: (Tip: Let $b_{1}<b_{2}<\cdots<b_{50}$, and divide the elements $a_{1}, a_{2}, \cdots, a_{100}$ of $A$ into 50 non-empty groups in order. Define the mapping $f$: $A \rightarrow B$, such that the image of the elements in the $i$-th group under $f$ is $b_{i}(i=1,2, \cdots, 50)$. It is easy to see that such an $f$ satisfies the problem's requirements, and each such grouping corresponds one-to-one to a mapping that satisfies the conditions. Therefore, the number of mappings $f$ that satisfy the problem's requirements is equal to the number of ways to divide $A$ into 50 groups in order. The number of ways to divide $A$ is $\mathrm{C}_{99}^{49}$, so there are $\mathrm{C}_{99}^{99}$ such mappings. Hence, the answer is (D).)

Answer: D


In the cash register of ÃÂurÃÂe, there are banknotes of 100 kuna and 10 kuna, as well as coins of 5 kuna, 2 kuna, and 1 kuna. The total value of the money in the cash register is 1844 kuna. The number of 100 kuna banknotes is one-sixth of the number of 10 kuna banknotes, and the number of 5 kuna coins is half the number of 10 kuna banknotes. The number of 1 kuna coins is one-third of the number of 2 kuna and 5 kuna coins combined, and the number of 2 kuna coins is six more than the number of 5 kuna coins. How many banknotes and coins are there in total in the cash register?

Result: $\quad 158$

Solution:
## Solution.

Let's denote the number of 100 kuna banknotes as follows:

From the conditions of the problem, we sequentially obtain:

![](https://cdn.mathpix.com/cropped/2024_05_30_adbfbded7af8db94c594g-6.jpg?height=517&width=1453&top_left_y=935&top_left_x=153)

The total value of the money in the cash register can be represented as follows:

| label | value | total |
| :--- | :--- | :--- |
| U-_ and 6 more | 100 | 100 |

Since $183 \bigcirc$ and 14 more kuna amount to 1844 kuna,

then 183

$\bigcirc$ is worth $1844-14=1830 \mathrm{kn}$.

This means that

it is worth $1830: 183=10 \mathrm{kn}$.

In the cash register, there were 10 banknotes of $100 \mathrm{kn}$, 60 banknotes of $10 \mathrm{kn}$, 30 coins of 5 kuna, 36 coins of 2 kuna, and 22 coins of 1 kuna.

In total, there were $10+60+30+36+22=158$ banknotes and coins.

Answer: 158


Solution:
Let $x_{1}=\cos \frac{\pi}{9}, x_{2}=\cos \frac{5 \pi}{9}, x_{3}=\cos \frac{7 \pi}{9}$, then
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=\cos \frac{\pi}{9}+\cos \frac{5 \pi}{9}+\cos \frac{7 \pi}{9}=2 \cos \frac{4 \pi}{9} \cos \frac{\pi}{3}+\cos \frac{5 \pi}{9}=\cos \frac{4 \pi}{9}+\cos \frac{5 \pi}{9}=0 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=\cos \frac{\pi}{9} \cos \frac{5 \pi}{9}+\cos \frac{5 \pi}{9} \cos \frac{7 \pi}{9}+\cos \frac{7 \pi}{9} \cos \frac{\pi}{9} \\
=\frac{1}{2}\left(\cos \frac{2 \pi}{3}+\cos \frac{4 \pi}{9}+\cos \frac{4 \pi}{3}+\cos \frac{2 \pi}{9}+\cos \frac{8 \pi}{9}+\cos \frac{2 \pi}{3}\right) \\
=\frac{1}{2}\left[-\frac{3}{2}+\cos \frac{4 \pi}{9}+\left(\cos \frac{2 \pi}{9}+\cos \frac{8 \pi}{9}\right)\right] \\
=-\frac{3}{4}+\frac{1}{2}\left(\cos \frac{4 \pi}{9}+2 \cos \frac{5 \pi}{9} \cos \frac{\pi}{3}\right) \\
=-\frac{3}{4}+\frac{1}{2}\left(\cos \frac{4 \pi}{9}+\cos \frac{5 \pi}{9}\right)=-\frac{3}{4} ; \\
x_{1} \cdot x_{2} \cdot x_{3}=\cos \frac{\pi}{9} \cos \frac{5 \pi}{9} \cos \frac{7 \pi}{9}=\cos \frac{\pi}{9} \cos \frac{4 \pi}{9} \cos \frac{2 \pi}{9} \\
=\frac{8 \sin \frac{\pi}{9} \cos \frac{\pi}{9} \cos \frac{2 \pi}{9} \cos \frac{4 \pi}{9}}{8 \sin \frac{\pi}{9}}=\frac{\sin \frac{8 \pi}{9}}{8 \sin \frac{\pi}{9}}=\frac{1}{8}
\end{array}
$$

Therefore, $x_{1}, x_{2}, x_{3}$ are the roots of the equation $x^{3}-\frac{3}{4} x-\frac{1}{8}=0$.
Thus, $x^{5}=\frac{3}{4} x^{3}+\frac{1}{8} x^{2}=\frac{3}{4}\left(\frac{3}{4} x+\frac{1}{8}\right)+\frac{1}{8} x^{2}=\frac{1}{8} x^{2}+\frac{9}{16} x+\frac{3}{32}$,
$$
\begin{array}{l}
\text { Hence, } \cos ^{5} \frac{\pi}{9}+\cos ^{5} \frac{5 \pi}{9}+\cos ^{5} \frac{7 \pi}{9}=\frac{1}{8}\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)+\frac{9}{16}\left(x_{1}+x_{2}+x_{3}\right)+\frac{3}{32} \times 3 \\
=\frac{1}{8}\left[\left(x_{1}+x_{2}+x_{3}\right)^{2}-2\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)\right]+\frac{9}{16}\left(x_{1}+x_{2}+x_{3}\right)+\frac{9}{32} \\
=\frac{1}{8} \times(-2) \times\left(-\frac{3}{4}\right)+\frac{9}{32} \\
=\frac{15}{32} .
\end{array}
$$

Answer: \frac{15}{32}


For which value of $\lambda$ is the area of the resulting polygon the smallest?

Solution:
Let's determine the value of $\lambda$ at which the area of the triangle cut off by the line connecting two adjacent division points of sides $a$ and $b$ of a polygon is the largest!

![](https://cdn.mathpix.com/cropped/2024_05_02_f60a8259d33d10e0461eg-1.jpg?height=399&width=486&top_left_y=256&top_left_x=814)

The area of such a triangle is:

$$
t=\frac{(a-\lambda a) \cdot \lambda b \cdot \sin \gamma}{2}=(1-\lambda) \cdot \lambda \frac{a b \sin \gamma}{2}
$$

We need to examine the sum of such expressions.

Since the sum of the expressions $\frac{a b \cdot \sin \gamma}{2}$ is constant, it is sufficient to determine the value of $\lambda$ at which the expression $y=(1-\lambda) \lambda=-\lambda^{2}+\lambda$ is maximal. This function can be represented by a parabola with a vertical axis in the $\lambda, y$ coordinate system. This parabola intersects the $\lambda$ axis at $\lambda=1$ and $\lambda=0$, and its maximum value is at $\lambda=0.5$.

$$
\text { ifj. Csonka Pal (Budapest, III. o.) }
$$

Remark. $1^{\circ}$. The maximum value of $y=\frac{1}{4}-\left(1-\frac{1}{2}\right)^{2}$ is $\frac{1}{4}$, which is attained where the square term is 0, i.e., at $\lambda=\frac{1}{2}$.

$2^{\circ}$. The statement is not true for degenerate polygons, because in such cases, some of the triangles involved in the calculation represent a decrease, while others represent an increase in the area of the given polygon. Thus, with the increase of the product $\lambda(\lambda-1)$, the area of the new polygon may either increase or decrease. For example, if we place $45^{\circ}$ angles at the vertices of a square such that the diagonals are the angle bisectors, we get a star-shaped quadrilateral, for which the polygons created in the above manner all have the same area.

Answer: \lambda=\frac{1}{2}


As shown in Figure 4, there are five circles that are externally tangent to each other in sequence, and all are tangent to lines $a$ and $b$. If the diameters of the smallest and largest circles are 18 and 32, respectively, then the diameter of $\odot \mathrm{O}_{3}$ is $\qquad$

Solution:
8.24.

As shown in Figure 7, let the radii of the five circles $\odot O_{1}, \odot O_{2}, \odot O_{3}, \odot O_{4}, \odot O_{5}$ be $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$, respectively. Draw perpendiculars from points $O_{1}, O_{2}, O_{3}$ to line $a$, with the feet of the perpendiculars being $A_{1}, A_{2}, A_{3}$, respectively. Connect $O_{1} O_{3}$. Clearly, $O_{2}$ lies on $O_{1} O_{3}$. Draw $O_{1} B_{1} \perp O_{2} A_{2}$ at point $B_{1}$, and $O_{2} B_{2} \perp O_{3} A_{3}$ at point $B_{2}$.
Then $\triangle O_{1} O_{2} B_{1} \backsim \triangle O_{2} O_{3} B_{2}$
$$
\begin{array}{l}
\Rightarrow \frac{O_{1} O_{2}}{O_{2} O_{3}}=\frac{O_{2} B_{1}}{O_{3} B_{2}} \\
\Rightarrow \frac{r_{1}+r_{2}}{r_{2}+r_{3}}=\frac{r_{2}-r_{1}}{r_{3}-r_{2}} \Rightarrow \frac{r_{2}}{r_{1}}=\frac{r_{3}}{r_{2}} .
\end{array}
$$

Similarly, $\frac{r_{3}}{r_{2}}=\frac{r_{4}}{r_{3}}, \frac{r_{4}}{r_{3}}=\frac{r_{5}}{r_{4}}$.
Let $\frac{r_{2}}{r_{1}}=k$. Then
$$
r_{2}=k r_{1}, r_{3}=k^{2} r_{1}, r_{4}=k^{3} r_{1}, r_{5}=k^{4} r_{1} \text {. }
$$

Since $r_{1}=\frac{18}{2}=9, r_{5}=\frac{32}{2}=16$, we have
$$
k^{2}=\frac{4}{3} \text {. }
$$

Thus, $r_{3}=12$.
Therefore, the diameter of $\odot O_{3}$ is 24.

Answer: 24


If the integers from 1 to 10 are divided into 5 pairs, and the sums of the individual pairs are 9, 10, 11, 12, and 13, then how many different groupings would result in these sums? Someone claims that the ten numbers can be arranged into five pairs in 945 different ways. Did they calculate correctly?

Solution:
I. Solution. I. The representation of all numbers from 9 to 13 as the sum of two different numbers chosen from the integers 1 to 10:

$$
\begin{aligned}
& 9=1+8=2+7=3+6=4+5 \\
& 10=1+9=2+8=3+7=4+6 \\
& 11=1+10=2+9=3+8=4+7=5+6 \\
& 12=2+10=3+9=4+8=5+7 \\
& 13=3+10=4+9=5+8=6+7
\end{aligned}
$$

Representing 9 as $1+8$, 10 can be obtained from the unused numbers as $3+7$ and $4+6$. In the first case, 12 and 13 can only be given by $2+10$ and $4+9$, respectively, and the remaining $5+6$ forms 11, since the sum of the numbers 1 to 10 is equal to the sum of the numbers 9 to 13, so the sum of the two unused numbers in the first four representations always gives the remaining sum. In the second case, only the representations $11=2+9$ and $13=3+10$ are free for both terms.

Representing 9 as $2+7$, there are again two possibilities for 10: $1+9$ and $4+6$. In the first case, 12 can only be represented as $4+8$, and therefore 13 only as $3+10$, in the second case, $12=3+9$ is uniquely determined, and therefore $13=5+8$.

Representing 9 as $3+6$, either $10=1+9$ or $10=2+8$. In the first case, only $11=4+7$ and $13=5+8$ remain possible, in the second case, only $12=5+7$ and $13=4+9$. Finally, starting from the representation $9=4+5$, three representations remain for 10, but only two for 12, making it more efficient to continue with the pairing. Taking $12=2+10$, only $11=3+8$ and $13=6+7$ remain; if $12=3+9$, then $13=6+7,11=1+10$.

Thus, we have developed every representation of the number 9 into complete pairs in every possible way, generating all desired groupings, and we have obtained the following 8 groupings:

|  | 9 | 10 | 11 | 12 | 13 |
| ---: | ---: | ---: | ---: | ---: | ---: |
| I. | $1+8$ | $3+7$ | $5+6$ | $2+10$ | $4+9$ |
| II. | $1+8$ | $4+6$ | $2+9$ | $5+7$ | $3+10$ |
| III. | $2+7$ | $1+9$ | $5+6$ | $4+8$ | $3+10$ |
| IV. | $2+7$ | $4+6$ | $1+10$ | $3+9$ | $5+8$ |
| V. | $3+6$ | $1+9$ | $4+7$ | $2+10$ | $5+8$ |
| VII. | $3+6$ | $2+8$ | $1+10$ | $5+7$ | $4+9$ |
| VII. | $4+5$ | $1+9$ | $3+8$ | $2+10$ | $6+7$ |
| VIII. | $4+5$ | $2+8$ | $1+10$ | $3+9$ | $6+7$ |

Accordingly, the sums of the number pairs do not always uniquely determine the pairing of the numbers.

II. The pair for the number 1 can be chosen in 9 different ways from the other numbers. Each of these pairings can be continued in 7 different ways by choosing the smallest of the remaining numbers as the first element of the next pair, because in this way the second number of this pair can always be chosen in 7 different ways from the remaining 7 numbers, so the number of such further developments is $9 \cdot 7$. Similarly, by choosing the smallest of the unpaired numbers as the first element of the third and then the fourth pair, the second element of the third pair can be chosen in 5 different ways, and the second element of the fourth pair in 3 different ways, so the number of further developments first rises to $9 \cdot 7 \cdot 5$ and then to $9 \cdot 7 \cdot 5 \cdot 3=945$. This is the number of possibilities for the 5-pair arrangement, because we always considered every possibility for further development, these all differ from each other in at least one pair, and finally because the 4 pairs formed always determine the fifth pair as well. - According to this, the person calculated correctly.

Pál László (Pécs, Zipernovszky K. Mechanical Engineering High School I. class) Csur Endre (Budapest, Könyves Kálmán High School II. class)

Second solution to the second question of the problem. Two elements can be selected from $n$ different elements in $n(n-1) / 2$ ways, because after selecting the first element of the pair in $n$ ways, the second element can be chosen in $n-1$ ways, and the $n(n-1)$ pairings thus obtained give every pair twice, e.g., as $a, b$ and $b, a$. Therefore, if we perform the pairing of the 10 numbers written on the cards in such a way that we draw 2 cards at a time from a bag 4 times, each as the two elements of a pair, and also consider the remaining 2 numbers as a pair, this can be done in

$$
\frac{10 \cdot 9}{2} \cdot \frac{8 \cdot 7}{2} \cdot \frac{6 \cdot 5}{2} \cdot \frac{4 \cdot 3}{2}
$$

ways.

However, in this way we will get each of the 5-pair arrangements multiple times, specifically as many times as 5 different objects - here the pairs - can be arranged in a row. There are $2 \cdot 3 \cdot 4 \cdot 5$ possibilities for this, because by setting aside one object, the second can be placed in 2 ways relative to it: before it and after it, the third can be placed in 3 ways: before the first, before the second, and after the second, and so on. [^0]

Now we can get the number of ways to arrange the 10 numbers into 5 pairs - if we do not consider the order of the pairs - by dividing the above number by the latter:

$$
\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{2^{4} \cdot 2 \cdot 3 \cdot 4 \cdot 5}=5 \cdot 9 \cdot 7 \cdot 3=945
$$

Bárd Péter András (Budapest, Fazekas M. Grammar School I. class)

[^0]:    ${ }^{1}$ This problem was a puzzle in the 49th issue of Élet és Tudomány in 1966.

Answer: 945


The number of lattice points on the line segment with endpoints $(3,17),(48,281)$ (including the endpoints) is
(A) 2.
(B) 4.
(C) 6.
(D) 16.
(E) 46.

Solution:
［Solution］The slope of the line passing through the two given points is
$$
k=\frac{281-17}{48-3}=\frac{264}{45}=\frac{3 \cdot 2 \cdot 2 \cdot 2 \cdot 11}{3 \cdot 3 \cdot 5}=\frac{3 \cdot 2^{3} \cdot 11}{3^{2} \cdot 5}
$$

Obviously, the only common factor of the numerator and denominator is 3, i.e., $k=\frac{88}{15}$.
Thus, apart from the endpoints, there are only two integer points on the line segment:
and
$$
\begin{array}{l}
(48-15,281-88)=(33,193), \\
(15+3,88+17)=(18,105) .
\end{array}
$$

Therefore, the answer is $(B)$.

Answer: B


III. Find all pairs of positive numbers $x, y$ that satisfy the equation

$$
\begin{aligned}
& \frac{4 x^{2} y+6 x^{2}+2 x y-4 x}{3 x-y-2}+\sin \left(\frac{3 x^{2}+x y+x-y-2}{3 x-y-2}\right)=2 x y+y^{2}+\frac{x^{2}}{y^{2}}+\frac{2 x}{y}+ \\
& \quad+\frac{2 x y\left(x^{2}+y^{2}\right)}{(3 x-y-2)^{2}}+\frac{1}{(x+y)^{2}}\left(x^{2} \sin \frac{(x+y)^{2}}{x}+y^{2} \sin \frac{(x+y)^{2}}{y^{2}}+2 x y \sin \frac{(x+y)^{2}}{3 x-y-2}\right)
\end{aligned}
$$

In the answer, write the sum of the values $x^{2}+y^{2}$ for all obtained pairs of solutions $(x, y)$, rounding to two decimal places if necessary. If there are no solutions, write -1; if there are infinitely many solutions, write -2.

Solution:
Solution. The proposed equation can be rewritten as

$$
f\left(\alpha_{1} u_{1}+\alpha_{2} u_{2}+\alpha_{3} u_{3}\right)=\alpha_{1} f\left(u_{1}\right)+\alpha_{2} f\left(u_{2}\right)+\alpha_{3} f\left(u_{3}\right),
$$

where

$$
\begin{gathered}
\alpha_{1}=\frac{x^{2}}{(x+y)^{2}}, \alpha_{2}=\frac{y^{2}}{(x+y)^{2}}, \alpha_{3}=\frac{2 x y}{(x+y)^{2}}, \quad \alpha_{1}+\alpha_{2}+\alpha_{3}=1, \quad \alpha_{1}>0, \alpha_{2}>0, \alpha_{3}>0 \forall x, y>0 ; \\
u_{1}=\frac{(x+y)^{2}}{x}, u_{2}=\frac{(x+y)^{2}}{y^{2}}, u_{3}=\frac{(x+y)^{2}}{3 x-y-2}
\end{gathered}
$$

and the function $f(t)=t^{2}+\sin t-$ is strictly convex on the entire number line. By Jensen's inequality, such equality is possible only if $u_{1}=u_{2}=u_{3}$, from which we have, taking into account $x>0, y>0$

$$
x=y^{2}=3 x-y-2 \Rightarrow 2 y^{2}-y-2=0 \Rightarrow y=\frac{1+\sqrt{17}}{4}, x=\frac{9+\sqrt{17}}{8}
$$

Therefore, $x^{2}+y^{2}=\frac{85+13 \sqrt{17}}{32}=4.331 \ldots$

Answer: 4.33

III. 1 Find all pairs of positive numbers $x, y$ that satisfy the equation

$$
\begin{aligned}
& \frac{4 x^{2} y+6 x^{2}+2 x y-4 x}{3 x-y-2}+\sin \left(\frac{3 x^{2}+x y+x-y-2}{3 x-y-2}\right)=2 x y+y^{2}+\frac{x^{2}}{y^{2}}+\frac{2 x}{y}+ \\
& \quad+\frac{2 x y\left(x^{2}+y^{2}\right)}{(3 x-y-2)^{2}}+\frac{1}{(x+y)^{2}}\left(x^{2} \sin \frac{(x+y)^{2}}{x}+y^{2} \sin \frac{(x+y)^{2}}{y^{2}}+2 x y \sin \frac{(x+y)^{2}}{3 x-y-2}\right)
\end{aligned}
$$

In the answer, write the sum of the values $x^{2}+y^{2}$ for all obtained pairs of solutions $(x, y)$, rounding to two decimal places if necessary. If there are no solutions, write -1; if there are infinitely many solutions, write -2.

Answer: 4.33

III. 2 Find all pairs of positive numbers $x, y$ that satisfy the equation

$$
\begin{aligned}
& \frac{2 x^{2}+14 x y-4 y^{2}+2 x y^{2}-12 y}{2 x-y-3}+\cos \left(\frac{x^{2}+4 x y-2 y^{2}-7 y+2 x-3}{2 x-y-3}\right)=\frac{x^{2}}{y^{2}}+\frac{2 y^{3}}{x}+\frac{2 x}{y}+ \\
& \quad+\frac{x y\left(x y+2 x^{2}\right)}{(2 x-y-3)^{2}}+\frac{1}{(x+y)^{2}}\left(x^{2} \cos \frac{(x+y)^{2}}{2 x-y-3}+2 x y \cos \frac{(x+y)^{2}}{x}+y^{2} \cos \frac{(x+y)^{2}}{y^{2}}\right)
\end{aligned}
$$

In the answer, write the sum of the values $x^{2}+y$ for all obtained pairs of solutions $(x, y)$, rounding to two decimal places if necessary. If there are no solutions, write -1; if there are infinitely many solutions, write -2.

Answer: $16+4 \sqrt{13} \approx 30.42$

Solutions $x=(7+\sqrt{13}) / 2, y=(1+\sqrt{13}) / 2$.

III. 3 Find all pairs of positive numbers $x, y$ that satisfy the equation

$$
\begin{aligned}
& \frac{14 x y+4 y^{2}-2 x^{2} y-8 y}{x+2 y-4}+\cos \left(\frac{2 y^{2}+3 x y+x-2 y-4}{x+2 y-4}\right)=\frac{y^{2}}{x^{2}}+\frac{2 y}{x}+x^{2}+ \\
& \quad+\frac{2 x y\left(x^{2}+y^{2}\right)}{(x+2 y-4)^{2}}+\frac{1}{(x+y)^{2}}\left(x^{2} \cos \frac{(x+y)^{2}}{x^{2}}+y^{2} \cos \frac{(x+y)^{2}}{y}+2 x y \cos \frac{(x+y)^{2}}{x+2 y-4}\right)
\end{aligned}
$$

In the answer, write the sum of the values $x+y^{2}$ for all obtained pairs of solutions $(x, y)$, rounding to two decimal places if necessary. If there are no solutions, write -1; if there are infinitely many solutions, write -2.

Answer: $24-4 \sqrt{17} \approx 7.51$

Solutions $x=(\sqrt{17}-1) / 2, y=(9-\sqrt{17}) / 2$.

III. 4 Find all pairs of positive numbers $x, y$ that satisfy the equation

$$
\begin{aligned}
& \frac{6 x y^{2}-6 x^{2} y+12 x^{2}+2 y^{2}-10 x y+12 x}{3 x-y+3}+\sin \left(\frac{6 x^{2}-2 x y+y^{2}+9 x-y+3}{3 x-y+3}\right)=\frac{y^{2}}{x^{2}}+\frac{2 x^{3}}{y}+\frac{2 y}{x}+ \\
& +\frac{x y\left(x y+2 y^{2}\right)}{(3 x-y+3)^{2}}+\frac{1}{(x+y)^{2}}\left(x^{2} \sin \frac{(x+y)^{2}}{x^{2}}+2 x y \sin \frac{(x+y)^{2}}{y}+y^{2} \sin \frac{(x+y)^{2}}{3 x-y+3}\right)
\end{aligned}
$$

In the answer, write the sum of the values $x+2 y$ for all obtained pairs of solutions $(x, y)$, rounding to two decimal places if necessary. If there are no solutions, write -1; if there are infinitely many solutions, write -2.

Answer: $6+\sqrt{33} \approx 11.74$

Solutions $x=(3+\sqrt{33}) / 4, y=(21+3 \sqrt{33}) / 8$.

Answer: 4.33


In the Cartesian coordinate system, the coordinates of two vertices of the square $O A B C$ are $O(0,0), A(4,3)$, and point $C$ is in the fourth quadrant. Then the coordinates of point $B$ are $\qquad$

Solution:
Answer: $(7,-1)$

Solution: By drawing a grid diagram, we get $B(7,-1)$

Answer: (7,-1)


Solution: 16. $\frac{1365}{4}$

Answer: \frac{1365}{4}


Let $X$ be a set containing $n$ elements. Find the number of ordered triples $(A,B, C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C$.

Solution:
1. **Count the number of ordered triples $(A, B, C)$ such that $A \subseteq B \subseteq C$:**

   For each element $x \in X$, there are 4 possibilities:
   - $x$ belongs to all three sets $A$, $B$, and $C$.
   - $x$ belongs to $B$ and $C$ but not to $A$.
   - $x$ belongs to $C$ only.
   - $x$ belongs to none of the sets.

   Since there are $n$ elements in $X$, the total number of such triples is:
   \[
   4^n
   \]

2. **Count the number of ordered triples $(A, B, C)$ such that $A \subseteq B = C$:**

   For each element $x \in X$, there are 3 possibilities:
   - $x$ belongs to both $A$ and $B = C$.
   - $x$ belongs to $B = C$ but not to $A$.
   - $x$ belongs to none of the sets.

   Since there are $n$ elements in $X$, the total number of such triples is:
   \[
   3^n
   \]

3. **Subtract the number of triples where $B = C$ from the total number of triples:**

   The number of ordered triples $(A, B, C)$ such that $A \subseteq B \subseteq C$ and $B \neq C$ is:
   \[
   4^n - 3^n
   \]

The final answer is $\boxed{4^n - 3^n}$

Answer: 4^n - 3^n


In a Cartesian coordinate system, a point whose both vertical and horizontal coordinates are integers is called an integer point. The number of integer points $(x, y)$ that satisfy the inequality $(|x|-1)^{2}+(|y|-1)^{2}<2$ is $\qquad$ .

Solution: 6. 16 Detailed Explanation: From $(|x|-1)^{2}+(|y|-1)^{2}<2$, we can deduce that $(|x|-1,|y|-1)$ are $(0,0),(0,1),(0,-1)$, $(1,0),(-1,0)$, thus yielding 16 pairs of $(x, y)$.

Answer: 16


Determine the largest three-digit number $\overline{a b c}$ such that $\overline{a b c}+\overline{b c a}+\overline{c a b}=1221$. Different letters represent different digits.

Solution:
5. The equality $\overline{a b c}+\overline{b c a}+\overline{c a b}=1221$ corresponds to the equality:

$(100 a+10 b+c)+(100 b+10 c+a)+(100 c+10 a+b)=1221, \quad 2$ points

which is $111 a+111 b+111 c=1221, \quad 2$ points

or $111 \cdot(a+b+c)=1221, \quad 2$ points

from which $a+b+c=1221: 111=11$. 2 points

Since the largest three-digit number is sought such that

$\overline{a b c}+\overline{b c a}+\overline{c a b}=1221$, and the digits $a, b$ and $c$ must be different,

it follows that $a=8, b=2$ and $c=1$,

so the desired number is 821.

2 points

TOTAL 10 POINTS

Answer: 821


We call a pair $(a,b)$ of positive integers, $a<391$, [i]pupusa[/i] if

$$\textup{lcm}(a,b)>\textup{lcm}(a,391)$$

Find the minimum value of $b$ across all [i]pupusa[/i] pairs.

Solution:
1. **Identify the prime factorization of 391:**
   \[
   391 = 17 \times 23
   \]
   Therefore, the possible values for \(\gcd(a, 391)\) are \(1\), \(17\), or \(23\).

2. **Define the least common multiple (LCM) and greatest common divisor (GCD):**
   \[
   \text{lcm}(a, 391) = \frac{391a}{\gcd(a, 391)}
   \]
   \[
   \text{lcm}(a, b) = \frac{ab}{\gcd(a, b)}
   \]

3. **Set up the inequality given in the problem:**
   \[
   \text{lcm}(a, b) > \text{lcm}(a, 391)
   \]
   Substituting the LCM expressions, we get:
   \[
   \frac{ab}{\gcd(a, b)} > \frac{391a}{\gcd(a, 391)}
   \]

4. **Simplify the inequality:**
   \[
   \frac{ab}{e} > \frac{391a}{d}
   \]
   where \(e = \gcd(a, b)\) and \(d = \gcd(a, 391)\).

5. **Cancel \(a\) from both sides (assuming \(a \neq 0\)):**
   \[
   \frac{b}{e} > \frac{391}{d}
   \]

6. **Rearrange to solve for \(b\):**
   \[
   b > \frac{391e}{d}
   \]

7. **Maximize \(d\) and minimize \(e\):**
   - To minimize \(b\), we need to maximize \(d\) and minimize \(e\).
   - The maximum value of \(d\) is \(23\) (since \(d\) can be \(1\), \(17\), or \(23\)).
   - The minimum value of \(e\) is \(1\) (since \(e\) is a divisor of both \(a\) and \(b\)).

8. **Substitute \(d = 23\) and \(e = 1\):**
   \[
   b > \frac{391 \times 1}{23} = 17
   \]

9. **Determine the minimum integer value of \(b\):**
   - Since \(b\) must be greater than \(17\), the smallest integer value for \(b\) is \(18\).

10. **Verify with an example:**
    - Let \(a = 23\) (so \(\gcd(a, 391) = 23\)).
    - Let \(b = 18\) (so \(\gcd(a, b) = 1\)).
    - Check the LCMs:
      \[
      \text{lcm}(23, 391) = \frac{23 \times 391}{23} = 391
      \]
      \[
      \text{lcm}(23, 18) = \frac{23 \times 18}{1} = 414
      \]
    - Since \(414 > 391\), the pair \((23, 18)\) is indeed a *pupusa* pair.

The final answer is \(\boxed{18}\).

Answer: 18


In a sequence of natural numbers $ a_1 $, $ a_2 $, $ \dots $, $ a_ {1999} $, $ a_n-a_ {n-1} -a_ {n-2} $ is divisible by $ 100 (3 \leq n \leq 1999) $. It is known that $ a_1 = 19$ and $ a_2 = 99$. Find the remainder of $ a_1 ^ 2 + a_2 ^ 2 + \dots + a_ {1999} ^ 2 $ by $8$.

Solution:
1. Given the sequence \( a_1, a_2, \dots, a_{1999} \) where \( a_n - a_{n-1} - a_{n-2} \) is divisible by \( 100 \) for \( 3 \leq n \leq 1999 \), we start by noting that this implies:
   \[
   a_n \equiv a_{n-1} + a_{n-2} \pmod{100}
   \]
   We are given \( a_1 = 19 \) and \( a_2 = 99 \).

2. To find the remainder of \( a_1^2 + a_2^2 + \dots + a_{1999}^2 \) modulo \( 8 \), we first reduce the sequence modulo \( 4 \):
   \[
   a_1 \equiv 19 \equiv 3 \pmod{4}
   \]
   \[
   a_2 \equiv 99 \equiv 3 \pmod{4}
   \]
   Using the recurrence relation \( a_n \equiv a_{n-1} + a_{n-2} \pmod{4} \), we compute the next few terms:
   \[
   a_3 \equiv a_2 + a_1 \equiv 3 + 3 \equiv 6 \equiv 2 \pmod{4}
   \]
   \[
   a_4 \equiv a_3 + a_2 \equiv 2 + 3 \equiv 5 \equiv 1 \pmod{4}
   \]
   \[
   a_5 \equiv a_4 + a_3 \equiv 1 + 2 \equiv 3 \pmod{4}
   \]
   \[
   a_6 \equiv a_5 + a_4 \equiv 3 + 1 \equiv 4 \equiv 0 \pmod{4}
   \]
   \[
   a_7 \equiv a_6 + a_5 \equiv 0 + 3 \equiv 3 \pmod{4}
   \]
   We observe that the sequence is periodic with a period of 6: \( 3, 3, 2, 1, 3, 0 \).

3. Next, we calculate the sum of the squares of one period modulo \( 8 \):
   \[
   3^2 + 3^2 + 2^2 + 1^2 + 3^2 + 0^2 = 9 + 9 + 4 + 1 + 9 + 0 = 32 \equiv 0 \pmod{8}
   \]

4. Since the sequence is periodic with a period of 6, we need to determine how many complete periods fit into 1999 terms and the remainder:
   \[
   1999 \div 6 = 333 \text{ complete periods with a remainder of } 1
   \]
   Therefore, the sequence up to \( a_{1999} \) consists of 333 complete periods plus one additional term.

5. The sum of the squares of the complete periods is:
   \[
   333 \times 0 \equiv 0 \pmod{8}
   \]
   The additional term is \( a_1^2 \equiv 3^2 \equiv 9 \equiv 1 \pmod{8} \).

6. Adding the contributions from the complete periods and the additional term:
   \[
   0 + 1 \equiv 1 \pmod{8}
   \]

The final answer is \( \boxed{1} \).

Answer: 1


Misha, Anton, Katya, and Natasha organized a table tennis tournament. When asked who took which place, they answered:

Misha: - I was neither first nor last.

Anton: - I was not last.

Katya: - I was first.

Natasha: - I was last.

It is known that one of the kids lied, while the other three told the truth. Who took the third place, given that it was a boy?

Solution:
Answer: Misha took third place.

Solution. First, let's find out who among the children lied. If Katya and Natasha told the truth, then they took the first and last places, so Misha and Anton cannot have taken these places, which means they also told the truth and no one lied, which is impossible. Therefore, either Katya or Natasha lied. If Natasha lied, then no one was last, which is impossible. Therefore, Katya lied. From this, we can see that Anton took the first place, Natasha the last, and Misha and Katya the second and third places. Remembering that a boy took third place, we conclude that it was Misha.

Answer: Misha


Let $a,b,c,d,e$ be positive real numbers. Find the largest possible value for the expression
$$\frac{ab+bc+cd+de}{2a^2+b^2+2c^2+d^2+2e^2}.$$

Solution:
To find the largest possible value for the expression
\[ \frac{ab+bc+cd+de}{2a^2+b^2+2c^2+d^2+2e^2}, \]
we will use the method of inequalities and optimization.

1. **Inequality Setup:**
   We start by setting up inequalities for each term in the numerator and denominator. We use the AM-GM inequality:
   \[ 2a^2 + \alpha b^2 \geq 2\sqrt{2\alpha}ab, \]
   \[ (1-\alpha)b^2 + \beta c^2 \geq 2\sqrt{(1-\alpha)\beta}bc, \]
   \[ (2-\beta)c^2 + \gamma d^2 \geq 2\sqrt{(2-\beta)\gamma}cd, \]
   \[ (1-\gamma)d^2 + 2e^2 \geq 2\sqrt{2(1-\gamma)}de. \]

2. **Parameter Constraints:**
   We need to find the values of \(\alpha\), \(\beta\), and \(\gamma\) such that:
   \[ 2\alpha = (1-\alpha)\beta = (2-\beta)\gamma = 2(1-\gamma). \]

3. **Solving for \(\alpha\):**
   From the first equality:
   \[ 2\alpha = (1-\alpha)\beta. \]
   Solving for \(\beta\):
   \[ \beta = \frac{2\alpha}{1-\alpha}. \]

4. **Solving for \(\beta\):**
   From the second equality:
   \[ (1-\alpha)\beta = (2-\beta)\gamma. \]
   Substituting \(\beta\):
   \[ (1-\alpha)\left(\frac{2\alpha}{1-\alpha}\right) = (2-\frac{2\alpha}{1-\alpha})\gamma. \]
   Simplifying:
   \[ 2\alpha = \left(2 - \frac{2\alpha}{1-\alpha}\right)\gamma. \]
   \[ 2\alpha = \left(\frac{2(1-\alpha) - 2\alpha}{1-\alpha}\right)\gamma. \]
   \[ 2\alpha = \left(\frac{2 - 4\alpha}{1-\alpha}\right)\gamma. \]
   Solving for \(\gamma\):
   \[ \gamma = \frac{2\alpha(1-\alpha)}{2 - 4\alpha}. \]

5. **Solving for \(\gamma\):**
   From the third equality:
   \[ (2-\beta)\gamma = 2(1-\gamma). \]
   Substituting \(\beta\):
   \[ (2 - \frac{2\alpha}{1-\alpha})\gamma = 2(1-\gamma). \]
   Simplifying:
   \[ \left(\frac{2(1-\alpha) - 2\alpha}{1-\alpha}\right)\gamma = 2(1-\gamma). \]
   \[ \left(\frac{2 - 4\alpha}{1-\alpha}\right)\gamma = 2(1-\gamma). \]
   Solving for \(\gamma\):
   \[ \gamma = \frac{2(1-\gamma)(1-\alpha)}{2 - 4\alpha}. \]

6. **Finding \(\alpha\):**
   Solving the system of equations, we find:
   \[ \alpha = \frac{1}{3}. \]

7. **Verification:**
   Substituting \(\alpha = \frac{1}{3}\) into the inequalities, we verify that equality holds:
   \[ 2a^2 + \frac{1}{3}b^2 \geq 2\sqrt{2 \cdot \frac{1}{3}}ab, \]
   \[ \left(1 - \frac{1}{3}\right)b^2 + \beta c^2 \geq 2\sqrt{\left(1 - \frac{1}{3}\right)\beta}bc, \]
   \[ \left(2 - \beta\right)c^2 + \gamma d^2 \geq 2\sqrt{\left(2 - \beta\right)\gamma}cd, \]
   \[ \left(1 - \gamma\right)d^2 + 2e^2 \geq 2\sqrt{2\left(1 - \gamma\right)}de. \]

8. **Final Calculation:**
   The largest value is:
   \[ \frac{1}{2\sqrt{2 \cdot \frac{1}{3}}} = \sqrt{\frac{3}{8}}. \]

The final answer is \(\boxed{\sqrt{\frac{3}{8}}}\).

Answer: \sqrt{\frac{3}{8}}


Given positive integers $a, b$ satisfy
$$
|b-2|+b-2=0,|a-b|+a-b=0,
$$

and $a \neq b$. Then the value of $a b$ is $\qquad$ .

Solution:
2.1.2.
From the given conditions, we know that $a>0, b-2 \leqslant 0, a-b \leqslant 0$. Therefore, $a<b \leqslant 2$. Hence, $a=1, b=2$.

Answer: 2


Some circles are placed on the plane in such a way that there are exactly 12 points where at least two circles intersect. What is the minimum number of regions into which these circles divide the plane?

Solution:
Let's call a set of circles connected if from any point on any circle, we can reach every other circle by traveling along the arcs. In the following, we will prove the following statement:

If the set of circles $M$ breaks down into exactly $m$ connected components, and exactly $k$ points lie on at least two circles, then these circles divide the plane into at least $k+m+1$ regions.

First, let $m=1$, that is, $M$ is connected. Start drawing the circles one by one so that the resulting set of circles is always connected. This is possible because of the connectedness of $M$; at each step, there will be an un-drawn circle that intersects or touches one of the already drawn circles.

![](https://cdn.mathpix.com/cropped/2024_05_02_ad8ed6bcd1c9967242d6g-1.jpg?height=659&width=694&top_left_y=451&top_left_x=694)

Figure 1

Let the $n$-th circle drawn have $l$ points in common with the previously drawn circles; then $l \geq 1$. This circle is divided into $l$ arcs by these $l$ points, and each arc divides a closed region of the plane into two, or encloses a closed region from the infinite region (Figure 1). Therefore, the number of regions in the plane increases by $l$ in the $n$-th step, while the number of intersection or tangency points of the circles increases by at most $l$, since among the $l$ points of intersection or tangency created by the $n$-th circle with the previously drawn ones, there may be points that already belonged to at least two circles before the $n$-th step.

In the first step, the single circle drawn divides the plane into two regions, and the number of points lying on at least two circles is 0. According to the above, the number of regions increases by at least as much as the number of intersection or tangency points of the circles in each step, so - after a total of $k$ points lie on at least two circles - the number of regions created is at least $k+2$.

Switching to the case $m>1$, if the $i$-th connected component $(i=1,2, \ldots, m)$ has $k_{i}$ intersection or tangency points of the circles, then at least $k_{i}+2$ regions are created in this component (including the infinite region). Considering that $\sum_{i=1}^{m} k_{i}=k$, and that the infinite region was counted in each of the $m$ connected components, the total number of regions created is at least

$$
=\sum_{i=1}^{m} k_{i}+2 m-(m-1)=k+m+1
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_ad8ed6bcd1c9967242d6g-1.jpg?height=466&width=480&top_left_y=2071&top_left_x=801)

In the problem, $k=12$, so the minimum number of regions is 14. This many regions are created, for example, when we have four circles, each pair of which intersects (Figure 2).

Answer: 14


Suppose that the roots of $x^3+3x^2+4x-11=0$ are $a$, $b$, and $c$, and that the roots of $x^3+rx^2+sx+t=0$ are $a+b$, $b+c$, and $c+a$. Find $t$.

Solution:
By [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas) on the polynomial $P(x) = x^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0$, we have $a + b + c = s = -3$, $ab + bc + ca = 4$, and $abc = 11$. Then

$t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)$
This is just the definition for $-P(-3) = \boxed{023}$.
Alternatively, we can expand the expression to get
\begin{align*} t &= -(-3-a)(-3-b)(-3-c)\\  &= (a+3)(b+3)(c+3)\\  &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\ t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}

Answer: 23


In a keyless padlock, the bolt part with four identical and equally spaced teeth on one side is inserted into a sleeve with four identical rings that can rotate independently around the bolt axis (see figure). This is only possible in a specific position of the rings, as is the opening of the lock.

On the rings, six letters are engraved; four of them (one per ring) form the known key word when in the open position.

a) How many different key words are possible with this construction at each lock? A "key word" is any (even nonsensical) combination of four letters.

b) Compare the security of this lock with that of a similar lock that has six rings with four letters each.

c) How many different rings with six different letters from the 26 letters of the alphabet can the manufacturing company produce?

Rings are considered the same if they have the same letters (regardless of order) and the notch is under the same letter.

d) How many locks with different key words can be made from these rings?

Solution:
a) First, one can set the first ring in six different positions. Then, for each of these positions, six positions of the second ring are possible. Thus, there are already 36 different possibilities for the first two rings.

For each of these, the third ring can again be set in six different ways, resulting in 216 settings.

Finally, this number is multiplied by six again when the last ring is set, so that a total of 1296 different settings result.

In general, it can be shown that with $n$ rings each with $m$ numbers, $m^{n}$ different keywords can be formed.

b) From the solution of a) it follows immediately:

1. A lock with four rings, each with six letters, contains 1296 keywords,
2. A lock with six rings, each with four letters, contains 4096 keywords.

The security of the second lock is therefore in the ratio of $4096: 1296 \approx 3: 1$ to that of the first, i.e., the second lock is about three times as secure as the first.

c) It is necessary to determine how many possibilities there are to select $k$ (in our case $k=6$) different elements from $n$ (in our case $n=26$) different elements.

First, one can select a letter from the 26 letters in 26 different ways.

For each of these 26 possibilities, there are now 25 possibilities for selecting a second letter, making a total of $26 \cdot 25$.

However, it is easy to see that each combination of letters appears twice: for example, to $c$ the letter $d$ is chosen once and to $d$ the letter $c$ is chosen once. Therefore, the product 26 $\cdot$ 25 must still be divided by two to obtain the number of different possibilities. For each of these possibilities, there are again 24 new selection possibilities for the third letter, but each combination of letters appears multiple times: for example, to $(a b)$ the letter $c$ is added once, to $(b c)$ the letter $a$ is added once, and to $(a c)$ the letter $b$ is added once. There are no other combinations of the three elements $a, b$, and $c$. Thus, the number of $26 \cdot 25 \cdot 24$ combinations is now reduced to $\frac{26 \cdot 25 \cdot 24}{1 \cdot 2 \cdot 3}=2600$.

One can see how the development continues:

In general, the number of combinations of $k$ elements from $n$ elements is given by

$$
x=\frac{n \cdot(n-1) \cdot(n-2) \cdot \ldots \cdot(n-k+1)}{1 \cdot 2 \cdot 3 \cdot \ldots \cdot k}
$$

in our case, therefore, $x=230230$.

Since each of the rings can have the notch under any letter, this number must still be multiplied by six: $230230 \cdot 6=1381380$. Thus, there are 1381380 different rings.

d) Since each of the 26 letters can appear on each of the four rings of a lock, the task is to determine how many different combinations of 4 out of 26 letters there are, where the order matters but each letter can repeat up to four times.

First, one can select 26 letters; for each of these 26 possibilities, one can again choose a second letter in 26 different ways, so that one already has $26 \cdot 26=26^{2}$ possibilities.

One then considers further that the number is again multiplied by 26 when selecting the third letter: $26^{3}$. When selecting the fourth, there are then $26^{4}=456976$ different possibilities. Clearly, the same formula as in a) applies generally.

Answer: 456976