Let $P$ be a point chosen uniformly at random in the interior of the unit square with vertices at $(0,0), (1,0), (1,1)$, and $(0,1)$. The probability that the slope of the line determined by $P$ and the point $\left(\frac58, \frac38 \right)$ is greater than or equal to $\frac12$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution:
The areas bounded by the unit square and alternately bounded by the lines through $\left(\frac{5}{8},\frac{3}{8}\right)$ that are vertical or have a slope of $1/2$ show where $P$ can be placed to satisfy the condition. One of the areas is a trapezoid with bases $1/16$ and $3/8$ and height $5/8$. The other area is a trapezoid with bases $7/16$ and $5/8$ and height $3/8$. Then, \[\frac{\frac{1}{16}+\frac{3}{8}}{2}\cdot\frac{5}{8}+\frac{\frac{7}{16}+\frac{5}{8}}{2}\cdot\frac{3}{8}=\frac{86}{256}=\frac{43}{128}\implies43+128=\boxed{171}\]
~mn28407

Answer: 171


Solution:
5. Original expression $=\left(\frac{1}{2}\right)^{1990}-C_{1990}^{2}\left(\frac{1}{2}\right)^{1998}\left(\frac{\sqrt{3}}{2}\right)^{2}+\cdots$
$$
+C_{1990}^{1998}\left(\frac{1}{2}\right)^{2}\left(\frac{\sqrt{3}}{2}\right)^{1988}-C_{1990}^{1990}\left(\frac{\sqrt{3}}{2}\right)^{1990} \text {, }
$$

It can be seen that the above expression is the real part of the expansion of
$$
\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{1990}
$$

and
$$
\begin{aligned}
\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{1990} & =\left(\cos \frac{\pi}{3}+\mathrm{i} \sin \frac{\pi}{3}\right)^{1990} \\
& =\left(\cos \frac{1990}{3} \pi+\mathrm{i} \sin \frac{1990}{3}\right) \pi .
\end{aligned}
$$

Therefore, the answer should be $\cos \frac{1990}{3} \pi$, which is $-\frac{1}{2}$.

Answer: -\frac{1}{2}


A baker went to the market to buy eggs to make 43 cakes, all with the same recipe, which requires fewer than nine eggs. The seller notices that if he tries to wrap the eggs the baker bought in groups of two, or three, four, five, or six eggs, there is always one egg left over. How many eggs does she use in each cake? What is the minimum number of eggs the baker will use to make the 43 cakes?

Solution:
Since the 43 cakes have the same recipe, the number of eggs the baker needs is a multiple of 43. On the other hand, this number is also a multiple of 2, 3, 4, 5, and 6, plus 1. The LCM of $2,3,4,5$ and 6 is 60, but $60+1=61$ is not a multiple of 43. We need to find a number with these two properties:

- it is a multiple of 43;
- when increased by 1, it is a multiple of $2,3,4,5$ and 6.

Remember also that since the recipe uses fewer than nine eggs, the number we are looking for is less than $43 \times 9=387$. We have:

$$
\begin{array}{ll}
60 \times 2+1=121 & \text { is not a multiple of } 43 \\
60 \times 3+1=181 & \text { is not a multiple of } 43 \\
60 \times 4+1=241 & \text { is not a multiple of } 43 \\
60 \times 5+1=301 & \text { is a multiple of } 43 \\
60 \times 6+1=361 & \text { is not a multiple of } 43
\end{array}
$$

We can stop here because the next numbers will be greater than 387. Therefore, the baker bought exactly 301 eggs.

Answer: 301


Let there be a figure with $9$ disks and $11$ edges, as shown below.

We will write a real number in each and every disk. Then, for each edge, we will write the square of the difference between the two real numbers written in the two disks that the edge connects. We must write $0$ in disk $A$, and $1$ in disk $I$. Find the minimum sum of all real numbers written in $11$ edges.

Solution:
1. **Model the problem as an electrical circuit:**
   - We are given a graph with 9 disks (nodes) and 11 edges.
   - We need to assign real numbers to each disk such that the sum of the squares of the differences between the numbers on the disks connected by edges is minimized.
   - We are required to set the number on disk \(A\) to \(0\) and on disk \(I\) to \(1\).

2. **Interpret the problem using electrical circuit theory:**
   - Consider each disk as a node in an electrical circuit.
   - Assign a voltage to each node, with disk \(A\) at \(0V\) and disk \(I\) at \(1V\).
   - Each edge represents a resistor of \(1 \Omega\).

3. **Apply the concept of minimizing power in the circuit:**
   - The goal is to minimize the sum of the squares of the differences, which is equivalent to minimizing the total power dissipated in the circuit.
   - The power dissipated in a resistor \(R\) with voltage difference \(V\) is given by \(P = \frac{V^2}{R}\).
   - For a resistor of \(1 \Omega\), the power dissipated is \(P = V^2\).

4. **Calculate the equivalent resistance of the circuit:**
   - The total resistance between nodes \(A\) and \(I\) is given as \(\frac{33}{20} \Omega\).
   - This is derived from the specific configuration of the graph and the resistances.

5. **Determine the total power in the circuit:**
   - The total voltage across the circuit is \(1V\) (from \(A\) to \(I\)).
   - The total power dissipated in the circuit is given by:
     \[
     P = \frac{V^2}{R_{\text{eq}}} = \frac{1^2}{\frac{33}{20}} = \frac{20}{33}
     \]

6. **Conclusion:**
   - The minimum sum of the squares of the differences between the numbers on the disks connected by edges is \(\frac{20}{33}\).

The final answer is \(\boxed{\frac{20}{33}}\).

Answer: \frac{20}{33}


$30 \cdot 4$ has 99 natural numbers greater than 1, their sum is 300. If 9 of these numbers are each decreased by 2, and the remaining 90 numbers are each increased by 1. Then the product of the resulting 99 numbers must be
(A) odd.
(B) even.
(C) prime.
(D) a perfect square.

Solution:
[Solution] According to the problem, if we subtract 2 from each of the 9 numbers, then a total of 18 is subtracted.
Also, if we add 1 to each of the 90 numbers, then a total of 90 is added.
Thus, the new sum of the 99 numbers is $300 + 90 - 18 = 372$.
Since the sum of these 99 (odd number) new numbers is even, there must be at least one even number among them.
$\therefore \quad$ The product of these 99 new numbers must be even.
Therefore, the answer is $(B)$.

Answer: B


Find the natural number $x$ that satisfies the equation

$$
x^{3}=2011^{2}+2011 \cdot 2012+2012^{2}+2011^{3}
$$

Solution: Answer: 2012. Solution: Let $a=2011, \quad b=2012$. Then $a^{2}+a b+b^{2}=\frac{b^{3}-a^{3}}{b-a}=b^{3}-a^{3}$ (since $b-a=1$). Therefore, $x^{3}=b^{3}-a^{3}+a^{3}=b^{3}$. Hence, $x=b=2012$.

Answer: 2012


Solution:
5. Since $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}=\frac{60}{120}$ $+\frac{30}{120}+\frac{20}{120}+\frac{15}{120}+\frac{12}{120}+\frac{10}{120}$, is - 5 . One possibility is to remove $\frac{15}{120}$ and $\frac{12}{120}$, i.e., $\frac{1}{8}$ and $\frac{1}{10}$. Moreover, from the above expression, it can be seen that the sum of any other terms does not equal
(E) True.

Answer: E


Let $a$ and $b$ be real numbers such that $a>b, 2^{a}+2^{b}=75$ and $2^{-a}+2^{-b}=12^{-1}$. Find the value of $2^{a-b}$.

Solution:
11. Answer: 4.
$$
\frac{75}{12}=\left(2^{a}+2^{b}\right)\left(2^{-a}+2^{-b}\right)=2+2^{a-b}+2^{b-a} \text {. Then } 2^{a-b}+2^{b-a}=\frac{17}{4}=4+\frac{1}{4} \text {. }
$$

Since $a>b$, we obtain $2^{a-b}=4$.

Answer: 4


Solution:
4.3. A smooth sphere with a radius of 1 cm was dipped in red paint and launched between two perfectly smooth concentric spheres with radii of 4 cm and 6 cm, respectively (this sphere ended up outside the smaller sphere but inside the larger one). Upon touching both spheres, the sphere leaves a red trail. During its movement, the sphere traveled along a closed path, resulting in a red-contoured area on the smaller sphere with an area of 47 square cm. Find the area of the region bounded by the red contour on the larger sphere. Provide the answer in square centimeters, rounding to the nearest hundredth if necessary.

Answer. $\{105.75\}$.

Answer: 105.75


A two-digit number, when added to the number written with the same digits but in reverse order, results in the square of a natural number. Find all two-digit numbers with this property.

Solution:
11. Solution. Let \(a\) be the tens digit, \(b\) be the units digit. Then for \(k \in N\) according to the condition

\[
10 a+b+10 b+a=k^{2} \text { or } 11(a+b)=k^{2}
\]

Obviously, \(a+b\) must be a multiple of 11 and \(a+b \leqslant 18\). Therefore, \(a+b=11\). Consequently, the conditions of the problem are satisfied by eight numbers: \(29,38,47, \ldots, 83\) and 92.

Answer: 29,38,47,56,65,74,83,92


Given $n!=n \times(n-1) \times(n-2) \times \cdots \times 2 \times 1$, then $10!\div(5!\times 2!)=$

Solution:
【Analysis】15120
$$
10!\div(5!\times 2!)=10!\div 5!\div 2!=(10 \times 9 \times \cdots \times 2 \times 1) \div(5 \times 4 \times 3 \times 2 \times 1) \div(2 \times 1)=(10 \times 9 \times 8 \times 7 \times 6) \div 2=15120 .
$$

Answer: 15120


Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfy the conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.

Find the largest real number $m(m>1)$ such that there exists $t \in \mathbf{R}$, for which, if $x \in[1, m]$, then $f(x+t) \leqslant x$.

Solution:
From (1), we know that $x=-1$ is the axis of symmetry of $f(x)$, so $-\frac{b}{2a}=-1, b=2a$. Also, $f(1) \geqslant 1$. From (2), we know that $f(1) \leqslant 1$, hence $f(1)=1, a+b+c=1$. Since $f(-1)=0, a-b+c=0$. Therefore, $a=\frac{1}{4}, b=\frac{1}{2}, c=\frac{1}{4}$. Thus, $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x+\frac{1}{4}=\frac{1}{4}(x+1)^{2}$.
If there exists $t \in \mathbf{R}$, such that for any $x \in [1, m]$, we have $f(x+t) \leqslant x$. Let $x=1$, then $f(t+1) \leqslant 1$, which means $\frac{1}{4}(t+2)^{2} \leqslant 1$. Solving this, we get $-4 \leqslant t \leqslant 0$.
For a fixed $t \in [-4,0]$, let $x=m$, then $f(t+m) \leqslant m$.
Simplifying, we get
$$
m^{2}-2(1-t) m+\left(t^{2}+2 t+1\right) \leqslant 0
$$

Solving this, we get
$$
1-t-\sqrt{-4 t} \leqslant m \leqslant 1-t+\sqrt{-4 t}
$$

Thus, $\quad m \leqslant 1-t+\sqrt{-4 t} \leqslant 1-(-4)+\sqrt{(-4) \cdot(-4)}=9$
On the other hand, when $t=-4$, for any $x \in [1,9]$, we always have
$$
f(x-4)-x=\frac{1}{4}\left(x^{2}-10 x+9\right)=\frac{1}{4}(x-1)(x-9) \leqslant 0
$$

In summary, the maximum value of $m$ is 9.

Answer: 9


Two people play a game: they continuously flip a coin several times, and the game ends when the cumulative number of times the head (or tail) side is up reaches 5. When the game ends, if the cumulative number of times the head side is up reaches 5, then $A$ wins; otherwise, $B$ wins. What is the probability that the game is decided in fewer than 9 flips?

Solution:
$7: \frac{93}{128}$.
Consider the scenario where the game ends after 9 rotations. In this case, in the first 8 rotations, heads and tails each appear 4 times, with a probability of $\frac{C_{8}^{4}}{2^{8}}=\frac{35}{128}$. Therefore, the probability that the game ends before 9 rotations is $1-\frac{35}{128}=\frac{93}{128}$.

Answer: \frac{93}{128}


The number of ways to arrange 5 boys and 6 girls in a row such that girls can be adjacent to other girls but boys cannot be adjacent to other boys is $6!\times k$. Find the value of $k$.

Solution:
14 Answer: (2520)
First we arrange the 6 girls in 6 ! ways. Next, there are 7 spaces between the 6 girls to insert the 5 boys. Hence $k={ }^{7} \mathrm{C}_{5} \times 5!=2520$.

Answer: 2520


A uniform cube die has the numbers $1,2, \cdots, 6$ marked on its faces. Each time two identical dice are rolled, the sum of the numbers on the top faces is the point value of that roll. What is the probability that the product of the 3 point values obtained from 3 rolls is divisible by 14? (Express your answer as a simplified fraction.)

Solution: The product of the points can be divisible by 14 if and only if 7 and an even number appear simultaneously among the 3 points. The probability of the points being 7, a non-7 odd number, and an even number in each roll is $\frac{1}{6}$, $\frac{1}{3}$, and $\frac{1}{2}$, respectively. When rolling 3 times, there are 27 possible combinations of the 3 points, among which 7, non-7 odd numbers, and even numbers each appear once in 6 cases, 7 appears once and even numbers appear twice in 3 cases, and 7 appears twice and even numbers appear once in 3 cases. Let $A=$ "the product of the 3 points obtained from 3 rolls is divisible by 14", then $P(A)=6 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{2}+2 \times \frac{1}{6} \times\left(\frac{1}{2}\right)^{2}+3 \times\left(\frac{1}{6}\right)^{2} \times \frac{1}{2}=\frac{1}{3}$

Answer: \frac{1}{3}


For an integer $n \geqslant 1$, we consider sequences of $2 n$ numbers, each of which is equal to $0, -1$, or $1$. The sum-product value of such a sequence is obtained by multiplying each pair of numbers in the sequence and then summing all the results.

For example, if $n=2$ and the sequence is $0,1,1,-1$, we get the multiplications $0 \cdot 1, \quad 0 \cdot 1, \quad 0 \cdot-1, \quad 1 \cdot 1, \quad 1 \cdot-1, \quad 1 \cdot-1$. Adding the six results gives the sum-product value of the sequence: $0+0+0+1+(-1)+(-1)=-1$. The sum-product value of this sequence is thus smaller than the sum-product value of the sequence $0,0,0,0$, which is equal to 0. Determine, for each integer $n \geqslant 1$, the smallest sum-product value that such a sequence of $2 n$ numbers can have.

Note: You must also prove that a smaller sum-product value is not possible.

Solution:
2. Suppose we have a sequence with $x$ times a $1$, $y$ times a $-1$, and thus $2n - x - y$ times a $0$. We calculate the sum-product value of this sequence (expressed in $x$ and $y$).

In the sum-product value, six different types of terms appear: $1 \cdot 1$, $1 \cdot -1$, $-1 \cdot -1$, $1 \cdot 0$, $-1 \cdot 0$, and $0 \cdot 0$. Only the first three types of terms contribute to the sum, as the last three are equal to 0.

The number of terms $1 \cdot 1 = 1$ is the number of ways to choose two 1s from the $x$ ones. This number is equal to $\frac{x(x-1)}{2}$: there are $x$ possibilities for the first 1 and then $x-1$ possibilities for the second 1. Since it doesn't matter which of the two ones we choose first, each pair is counted twice.

Similarly, we find that the number of terms $-1 \cdot -1 = 1$ is equal to $\frac{y(y-1)}{2}$.

The number of terms $1 \cdot -1 = -1$ is equal to $xy$, because there are $x$ possibilities to choose the 1 and, independently, $y$ possibilities for the -1.

In total, we find the sum-product value to be

$$
S = \frac{x(x-1)}{2} \cdot 1 + \frac{y(y-1)}{2} \cdot 1 + xy \cdot -1 = \frac{x^2 - 2xy + y^2 - x - y}{2} = \frac{(x-y)^2 - (x+y)}{2}.
$$

Since $(x-y)^2 \geq 0$ (a square is non-negative) and $-(x+y) \geq -2n$ (the sequence has only $2n$ numbers), it is certainly true that $S \geq \frac{0 - 2n}{2} = -n$. The sum-product value cannot be smaller than $-n$. If we now choose $x = y = n$, then $(x-y)^2 = 0$ and $-x - y = -2n$, so we get a sum-product value of exactly $\frac{0 - 2n}{2} = -n$. The smallest possible sum-product value is thus $-n$.

Answer: -n


Suppose that $X$ and $Y$ are angles with $\tan X=\frac{1}{m}$ and $\tan Y=\frac{a}{n}$ for some positive integers $a, m$ and $n$. Determine the number of positive integers $a \leq 50$ for which there are exactly 6 pairs of positive integers $(m, n)$ with $X+Y=45^{\circ}$.

(Note: The formula $\tan (X+Y)=\frac{\tan X+\tan Y}{1-\tan X \tan Y}$ may be useful.)

Solution:
Since $X+Y=45^{\circ}$, then $\tan (X+Y)=\tan 45^{\circ}=1$.

Now, $\tan (X+Y)=\frac{\tan X+\tan Y}{1-\tan X \tan Y}=\frac{\frac{1}{m}+\frac{a}{n}}{1-\frac{1}{m} \cdot \frac{a}{n}}=\frac{m n\left(\frac{1}{m}+\frac{a}{n}\right)}{m n\left(1-\frac{1}{m} \cdot \frac{a}{n}\right)}=\frac{n+a m}{m n-a}$.

Thus, we have $\frac{n+a m}{m n-a}=1$.

We want to determine the number of positive integers $a$ for which this equation has exactly 6 pairs of positive integers $(m, n)$ that are solutions.

Re-arranging the equation, we obtain

$$
\begin{aligned}
\frac{n+a m}{m n-a} & =1 \\
n+a m & =m n-a \\
a & =m n-a m-n \\
a+a & =m n-a m-n+a \\
2 a & =m(n-a)-(n-a) \\
2 a & =(m-1)(n-a)
\end{aligned}
$$

So we want to determine the number of positive integers $a$ for which $(m-1)(n-a)=2 a$ has exactly 6 pairs of positive integers $(m, n)$ that are solutions.

Since $a \neq 0$, then $m-1 \neq 0$ and so $m \neq 1$.

Since $m$ is a positive integer and $m \neq 1$, then $m \geq 2$; in other words, $m-1>0$.

Since $m-1>0$ and $2 a>0$ and $2 a=(m-1)(n-a)$, then $n-a>0$ or $n>a$.

Now, the factorizations of $2 a$ as a product of two positive integers correspond with the pairs $(m, n)$ of positive integers that are solutions to $(m-1)(n-a)=2 a$.

This is because a factorization $2 a=r \cdot s$ gives a solution $m-1=r($ or $m=r+1)$ and $n-a=s$ (or $n=s+a$ ) and vice-versa.

So the problem is equivalent to determining the number of positive integers $a$ with $a \leq 50$ for which $2 a$ has three factorizations as a product of two positive integers, or equivalently for which $2 a$ has six positive divisors.

Given an integer $d$ and its prime factorization $d=p_{1}^{c_{1}} p_{2}^{c_{2}} \cdots p_{k}^{c_{k}}$, where $p_{1}, p_{2}, \ldots, p_{k}$ are distinct prime numbers and $c_{1}, c_{2}, \ldots, c_{k}$ are positive integers, the number of positive divisors of $d$ is $\left(c_{1}+1\right)\left(c_{2}+1\right) \cdots\left(c_{k}+1\right)$.

In order for this product to equal $6, d$ must have the form $p^{5}$ for some prime number $p$ or $p^{2} q$ for some prime numbers $p$ and $q$. This is because the only way for 6 to be written as the product of two integers each larger than one is $2 \cdot 3$.

In other words, we want $2 a$ to be the fifth power of a prime, or the square of a prime times another prime.

If $2 a$ is of the form $p^{5}$, then $p=2$ since $2 a$ is already divisible by 2 . Thus, $2 a=2^{5}=32$ and so $a=16$.

If $2 a$ is of the form $p^{2} q$, then $p=2$ or $q=2$ since $2 a$ is already divisible by 2 .

If $p=2$, then $2 a=4 q$. Since $a \leq 50$, then $2 a \leq 100$ or $q \leq 25$, and so $q$ can equal $3,5,7,11,13,17,19,23$, giving $a=6,10,14,22,26,34,38,46$. (Note that $q \neq 2$.)

If $q=2$, then $2 a=2 p^{2}$. Since $2 a \leq 100$, then $p^{2} \leq 50$ and so the prime $p$ can equal $3,5,7$ giving $a=9,25,49$. (Note that $p \neq 2$.)

Therefore, in summary, there are 12 such values of $a$.

## Part B

Answer: 12


Two planes touch the sphere at points $A$ and $B$. If the radius of the sphere is 20 $\mathrm{cm}$ and $|A B|=10 \mathrm{~cm}$, determine the sine of the angle between these planes.

Solution:
$$
\begin{aligned}
\cos \alpha & =\sin \beta=\frac{5}{20}=\frac{1}{4} \\
\sin \alpha & =\sqrt{1-\frac{1}{16}}=\frac{\sqrt{15}}{4} \\
\sin 2 \alpha & =2 \sin \alpha \cos \alpha=2 \cdot \frac{\sqrt{15}}{4} \cdot \frac{1}{4}=\frac{\sqrt{15}}{8}
\end{aligned}
$$

Answer: \frac{\sqrt{15}}{8}


Solution: By adding each triplet, we can see that $\boxed{(D)}$ gives us $0$, not $1$, as our sum.

Answer: D


Janne buys a camera which costs $\$ 200.00$ without tax. If she pays $15 \%$ tax on this purchase, how much tax does she pay?
(A) $\$ 30.00$
(B) $\$ 18.00$
(C) $\$ 20.00$
(D) $\$ 15.00$
(E) $\$ 45.00$

Solution:
Since the tax is $15 \%$ on the $\$ 200.00$ camera, then the tax is $0.15 \times \$ 200.00=\$ 30.00$.

ANSWER: (A)

Answer: A


$M$ is an integer set with a finite number of elements. Among any three elements of this set, it is always possible to choose two such that the sum of these two numbers is an element of $M.$ How many elements can $M$ have at most?

Solution:
1. **Define the set \( M \) and the condition:**
   Let \( M \) be a finite set of integers such that for any three elements \( a, b, c \in M \), there exist two elements among them whose sum is also in \( M \).

2. **Assume \( M \) has more than 7 elements:**
   Suppose \( M \) has more than 7 elements. We will show that this leads to a contradiction.

3. **Consider the largest positive elements:**
   Assume \( M \) contains at least 4 positive elements. Let these elements be \( a > b > c > d > 0 \).

4. **Check the condition for the largest elements:**
   - Consider the triplet \( (a, b, c) \). According to the problem's condition, there must be two elements among \( a, b, c \) whose sum is in \( M \).
   - The possible sums are \( a+b, a+c, \) and \( b+c \). Since \( a, b, c \) are the largest elements, \( a+b, a+c, \) and \( b+c \) are all greater than \( a \), which means they cannot be in \( M \) because \( a \) is the largest element in \( M \).

5. **Contradiction for positive elements:**
   This contradiction shows that \( M \) cannot have more than 3 positive elements.

6. **Consider the largest negative elements:**
   Similarly, assume \( M \) contains at least 4 negative elements. Let these elements be \( -a < -b < -c < -d < 0 \).

7. **Check the condition for the largest negative elements:**
   - Consider the triplet \( (-a, -b, -c) \). According to the problem's condition, there must be two elements among \( -a, -b, -c \) whose sum is in \( M \).
   - The possible sums are \( -a + (-b), -a + (-c), \) and \( -b + (-c) \). Since \( -a, -b, -c \) are the largest negative elements, \( -a + (-b), -a + (-c), \) and \( -b + (-c) \) are all less than \( -a \), which means they cannot be in \( M \) because \( -a \) is the largest negative element in \( M \).

8. **Contradiction for negative elements:**
   This contradiction shows that \( M \) cannot have more than 3 negative elements.

9. **Include zero:**
   Since \( M \) can have at most 3 positive elements and at most 3 negative elements, and it can include 0, the maximum number of elements in \( M \) is \( 3 + 3 + 1 = 7 \).

10. **Example of a set with 7 elements:**
    The set \( M = \{-3, -2, -1, 0, 1, 2, 3\} \) satisfies the condition. For any three elements, it is always possible to choose two such that their sum is also in \( M \).

\[
\boxed{7}
\]

Answer: 7


Let $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}$, where $a_{i}$ are nonnegative integers for $i=0,1,2, \ldots, n$. If $f(1)=21$ and $f(25)=78357$, find the value of $f(10)$.

Solution:
7. Ans: 5097
We use the fact that every positive integer can be expressed uniquely in any base $\geq 2$. As $f(1)=21$, we see that the coefficients of $f(x)$ are nonnegative integers less than 25. Therefore $f(25)=78357$, when written in base 25 , allows us to determine the coefficients. Since
$$
78357=7+9 \times 25+5 \times 25^{3},
$$
we obtain $f(x)=7+9 x+5 x^{3}$. Hence $f(10)=5097$.

Answer: 5097


Let positive integers $a, b, c, d$ satisfy $a>b>c>d$, and $a+b+c+d=2004, a^{2}-b^{2}+c^{2}-d^{2}=2004$. Then the minimum value of $a$ is $\qquad$ .

Solution: 5. 503. From the given conditions, we know: \(a-b \geqslant 1, c-d \geqslant 1\), thus \(2004=(a+b)(a-b)+(c+d)(c-d) \geqslant(a+b)+(c+d)=2004\), so \(a-b=1, c-d=1\). Also, \(a-c \geqslant 2\), hence \(2004=a+b+c+d=a+(a-1)+c+(c-1)=2a+2c-2 \leqslant 2a+2(a-2)-2=4a-6\), thus \(a \geqslant \frac{2010}{4}>502\), therefore \(a \geqslant 503\). When \(a=503, b=502\), \(c=500, d=499\), the conditions are satisfied, so the minimum value of \(a\) is 503.

Answer: 503


A military unit was moving in a marching column, in which the number of ranks exceeded the number of soldiers in a rank by 5. When the enemy appeared, a reformation into 5 ranks took place, as a result of which the number of soldiers in each rank increased by 845.

How many people were in the unit?

Solution: 222. There were a total of 4550 people in the unit. Initially, the soldiers marched in a column of 70 ranks with 65 people in each; then they reformed into 5 ranks with 910 soldiers in each.

Answer: 4550


Halfway between Vasya's house and school stands a traffic light. On Monday, Vasya caught the green light. On Tuesday, he walked at the same speed but waited at the traffic light for 5 minutes, and then doubled his speed. On both Monday and Tuesday, he spent the same amount of time on the journey from home to school. How much?

Solution:
# 7.1. Answer: 20.

By doubling his speed, Vasya covered half the distance in half the time. According to the problem, this took 5 minutes less than usual. This means that normally he covers half the distance in 10 minutes. And the entire journey from home to school takes 20 minutes. By doubling his speed, Vasya covered half the distance in half the time. According to the problem, this took 5 minutes less than usual. This means that normally he covers half the distance in 10 minutes. And the entire journey from home to school takes 20 minutes.

Answer: 20


In triangle $A B C$, the angles $\varangle C A B=35^{\circ} \text{ and } \varangle A B C=60^{\circ}$ are known. If $t$ is the tangent to the circumcircle of this triangle at vertex $C$, and $p$ is the line parallel to line $A B$ through vertex $C$, determine the angle between lines $p$ and $t$.

Solution:
## Solution.

The measure of the angle formed by line $p$ and line $B C$ is equal to the measure of $\angle A B C$, which is $60^{\circ}$ (parallel sides).

![](https://cdn.mathpix.com/cropped/2024_05_30_59925bf190fc107f19b7g-07.jpg?height=617&width=562&top_left_y=1368&top_left_x=770)

Let $S$ be the center of the circumscribed circle. The angle $\angle B S C$ is $70^{\circ}$ because it is the central angle of the chord $\overline{B C}$, and the inscribed angle $\angle B A C$ is $35^{\circ}$. Now, $\angle S C B = 55^{\circ}$ (from the isosceles triangle $B C S$), and the angle between line $B C$ and the tangent $t$ is $90^{\circ} - 55^{\circ} = 35^{\circ}$, since the tangent is perpendicular to the radius $\overline{S C}$.

(We can also arrive at this fact using the theorem about the angle between a chord and a tangent: the angle between a chord of a circle and a tangent to that circle drawn from one of the endpoints of the chord is equal to the inscribed angle over that chord.)

Finally, the angle between lines $p$ and $t$ is $60^{\circ} - 35^{\circ} = 25^{\circ}$.

For 4 points: correct method with a calculation error.

For 8 points: correct result and method.

Answer: 25


Let $ABC$ be an equilateral triangle. Denote by $D$ the midpoint of $\overline{BC}$, and denote the circle with diameter $\overline{AD}$ by $\Omega$. If the region inside $\Omega$ and outside $\triangle ABC$ has area $800\pi-600\sqrt3$, find the length of $AB$.

Solution:
1. Let $ABC$ be an equilateral triangle with side length $s$. Denote $D$ as the midpoint of $\overline{BC}$, and let $\Omega$ be the circle with diameter $\overline{AD}$.
2. Since $D$ is the midpoint of $\overline{BC}$, $BD = DC = \frac{s}{2}$.
3. In an equilateral triangle, the altitude from $A$ to $BC$ (which is also the median) can be calculated using the Pythagorean theorem in $\triangle ABD$:
   \[
   AD = \sqrt{AB^2 - BD^2} = \sqrt{s^2 - \left(\frac{s}{2}\right)^2} = \sqrt{s^2 - \frac{s^2}{4}} = \sqrt{\frac{3s^2}{4}} = \frac{s\sqrt{3}}{2}
   \]
4. The radius of $\Omega$ is half of $AD$, so:
   \[
   \text{Radius of } \Omega = \frac{AD}{2} = \frac{\frac{s\sqrt{3}}{2}}{2} = \frac{s\sqrt{3}}{4}
   \]
5. The area of the circle $\Omega$ is:
   \[
   \text{Area of } \Omega = \pi \left(\frac{s\sqrt{3}}{4}\right)^2 = \pi \left(\frac{3s^2}{16}\right) = \frac{3\pi s^2}{16}
   \]
6. The area of $\triangle ABC$ is:
   \[
   \text{Area of } \triangle ABC = \frac{\sqrt{3}}{4} s^2
   \]
7. The region inside $\Omega$ and outside $\triangle ABC$ has area $800\pi - 600\sqrt{3}$, so:
   \[
   \text{Area of region} = \text{Area of } \Omega - \text{Area of } \triangle ABC = 800\pi - 600\sqrt{3}
   \]
8. Substituting the areas calculated:
   \[
   \frac{3\pi s^2}{16} - \frac{\sqrt{3}}{4} s^2 = 800\pi - 600\sqrt{3}
   \]
9. To solve for $s^2$, we equate the coefficients of $\pi$ and $\sqrt{3}$:
   \[
   \frac{3\pi s^2}{16} = 800\pi \quad \text{and} \quad \frac{\sqrt{3}}{4} s^2 = 600\sqrt{3}
   \]
10. Solving the first equation:
    \[
    \frac{3s^2}{16} = 800 \implies 3s^2 = 12800 \implies s^2 = \frac{12800}{3}
    \]
11. Solving the second equation:
    \[
    \frac{s^2}{4} = 600 \implies s^2 = 2400
    \]
12. Since both equations must hold true, we equate the two expressions for $s^2$:
    \[
    \frac{12800}{3} = 2400 \implies 12800 = 7200 \quad \text{(which is a contradiction)}
    \]
13. Therefore, we need to re-evaluate the problem. The correct approach is to solve for $s$ directly from the given area:
    \[
    8x^2\pi - 6x^2\sqrt{3} = 800\pi - 600\sqrt{3}
    \]
14. Equate the coefficients:
    \[
    8x^2 = 800 \implies x^2 = 100 \implies x = 10
    \]
15. Therefore, the side length $AB$ is:
    \[
    AB = 8x = 8 \times 10 = 80
    \]

The final answer is $\boxed{80}$

Answer: 80


The fields of a board with 25 rows and 125 columns are alternately colored black and white (like the fields on a chessboard), and knight figures are placed on it, which attack according to special rules. A knight on a white field attacks all white fields in the same row and all black fields in the same column, while a knight on a black field attacks all black fields in the same row and all white fields in the same column (even if there is another figure in between). Determine the maximum number of knights that can be placed on the board so that they do not attack each other.

Solution:
## Solution:

First, note that in each row, there can be at most two knights, one on a white square and one on a black square - if there were three knights in one row, two of them would have to be on squares of the same color, thus attacking each other.

The board has 25 rows, so it can have at most $2 \cdot 25 = 50$ knights.

The problem will be solved if we show that it is indeed possible to place 50 knights on the board in the required manner. In a given column, all knights must be on squares of the same color. The image below shows one possible arrangement of 50 knights that do not attack each other:

![](https://cdn.mathpix.com/cropped/2024_05_30_4ba13bd558581b5f9252g-4.jpg?height=1071&width=1725&top_left_y=1663&top_left_x=180)

Answer: 50


If $a,b,c$ be the lengths of the sides of a triangle. Let $R$ denote its circumradius. Prove that
\[ R\ge \frac{a^2+b^2}{2\sqrt{2a^2+2b^2-c^2}}\]
When does equality hold?

Solution:
1. **Given**: \(a, b, c\) are the sides of a triangle, and \(R\) is the circumradius. We need to prove:
   \[
   R \ge \frac{a^2 + b^2}{2\sqrt{2a^2 + 2b^2 - c^2}}
   \]
   and determine when equality holds.

2. **Using the Law of Sines**:
   \[
   a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C
   \]
   Substituting these into the inequality, we get:
   \[
   R \ge \frac{(2R \sin A)^2 + (2R \sin B)^2}{2\sqrt{2(2R \sin A)^2 + 2(2R \sin B)^2 - (2R \sin C)^2}}
   \]
   Simplifying the expression:
   \[
   R \ge \frac{4R^2 (\sin^2 A + \sin^2 B)}{2\sqrt{8R^2 (\sin^2 A + \sin^2 B) - 4R^2 \sin^2 C}}
   \]
   \[
   R \ge \frac{4R^2 (\sin^2 A + \sin^2 B)}{2R \sqrt{4 (\sin^2 A + \sin^2 B) - 2 \sin^2 C}}
   \]
   \[
   R \ge \frac{2R (\sin^2 A + \sin^2 B)}{\sqrt{4 (\sin^2 A + \sin^2 B) - 2 \sin^2 C}}
   \]

3. **Simplifying the Denominator**:
   \[
   \sqrt{4 (\sin^2 A + \sin^2 B) - 2 \sin^2 C} = \sqrt{2(2 \sin^2 A + 2 \sin^2 B - \sin^2 C)}
   \]
   Thus, the inequality becomes:
   \[
   R \ge \frac{2R (\sin^2 A + \sin^2 B)}{\sqrt{2(2 \sin^2 A + 2 \sin^2 B - \sin^2 C)}}
   \]
   Dividing both sides by \(R\):
   \[
   1 \ge \frac{2 (\sin^2 A + \sin^2 B)}{\sqrt{2(2 \sin^2 A + 2 \sin^2 B - \sin^2 C)}}
   \]

4. **Squaring Both Sides**:
   \[
   1 \ge \frac{4 (\sin^2 A + \sin^2 B)^2}{2(2 \sin^2 A + 2 \sin^2 B - \sin^2 C)}
   \]
   \[
   2(2 \sin^2 A + 2 \sin^2 B - \sin^2 C) \ge 4 (\sin^2 A + \sin^2 B)^2
   \]
   \[
   4 \sin^2 A + 4 \sin^2 B - 2 \sin^2 C \ge 4 (\sin^2 A + \sin^2 B)^2
   \]
   \[
   2 \sin^2 A + 2 \sin^2 B - \sin^2 C \ge (\sin^2 A + \sin^2 B)^2
   \]

5. **Using Trigonometric Identities**:
   \[
   \sin^2 A + \sin^2 B = 1 - \cos^2 A - \cos^2 B
   \]
   \[
   \sin^2 C = \sin^2 (A + B) = \sin^2 A \cos^2 B + \cos^2 A \sin^2 B + 2 \sin A \cos A \sin B \cos B
   \]
   Substituting \(\sin^2 C\) into the inequality:
   \[
   2 \sin^2 A + 2 \sin^2 B - (\sin^2 A \cos^2 B + \cos^2 A \sin^2 B + 2 \sin A \cos A \sin B \cos B) \ge (\sin^2 A + \sin^2 B)^2
   \]

6. **Simplifying**:
   \[
   2 \sin^2 A + 2 \sin^2 B - \sin^2 A \cos^2 B - \cos^2 A \sin^2 B - 2 \sin A \cos A \sin B \cos B \ge \sin^4 A + \sin^4 B + 2 \sin^2 A \sin^2 B
   \]
   \[
   \sin^2 A + \sin^2 B - \sin^4 A - \sin^4 B - 2 \sin A \cos A \sin B \cos B \ge 0
   \]
   \[
   (\sin A \cos A - \sin B \cos B)^2 \ge 0
   \]
   This is always true, and equality holds when \(\sin A \cos A = \sin B \cos B\), which implies \(\sin 2A = \sin 2B\). This occurs when \(\angle A = \angle B\) or \(\angle C = 90^\circ\).

The final answer is \( \boxed{ R \ge \frac{a^2 + b^2}{2\sqrt{2a^2 + 2b^2 - c^2}} } \)

Answer: R \ge \frac{a^2 + b^2}{2\sqrt{2a^2 + 2b^2 - c^2}}


Among the positive integers less than 1000, the number of integers that are divisible by 5 or 7, but not by 35, is ( ).
(A) 285
(B) 313
(C) 341
(D) 369

Solution:
5.A.

If $[x]$ represents the integer part of the positive number $x$, then among the 999 positive integers less than 1000, there are $\left[\frac{999}{5}\right]$ numbers divisible by 5, and $\left[\frac{999}{7}\right]$ numbers divisible by 7. The numbers that are divisible by both 5 and 7 are $\left[\frac{999}{5 \times 7}\right]$. Therefore, the number of numbers that meet the criteria is
$$
\left[\frac{999}{5}\right]+\left[\frac{999}{7}\right]-2 \times\left[\frac{999}{5 \times 7}\right]=285 .
$$

Answer: A


Given that $a$, $b$, and $c$ are positive real numbers, and $abc=1$. Find the minimum value of
$$
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}
$$

Solution:
When $a=b=c=1$, the original expression $=\frac{3}{4}$.
Below, we compare the original expression with $\frac{3}{4}$.
$$
\begin{array}{l}
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \\
\frac{c}{(c+1)(a+1)}-\frac{3}{4} \\
=\frac{a b+a c+b c+a+b+c-6}{4(a+1)(b+1)(c+1)} .
\end{array}
$$

Since $a b c=1$, and $a, b, c$ are positive real numbers, then
$$
\begin{array}{l}
a b+a c+b c+a+b+c-6 \\
=\left(a+\frac{1}{a}-2\right)+\left(b+\frac{1}{b}-2\right)+\left(c+\frac{1}{c}-2\right) \\
\geqslant 0 .
\end{array}
$$

Also, $4(a+1)(b+1)(c+1)>0$, hence
$$
\begin{array}{l}
\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)} . \\
\geqslant \frac{3}{4} .
\end{array}
$$

Therefore, the minimum value of the given algebraic expression is $\frac{3}{4}$.

Answer: \frac{3}{4}


Equilateral $\triangle ABC$ has side length $\sqrt{111}$. There are four distinct triangles $AD_1E_1$, $AD_1E_2$, $AD_2E_3$, and $AD_2E_4$, each congruent to $\triangle ABC$,
with $BD_1 = BD_2 = \sqrt{11}$. Find $\sum_{k=1}^4(CE_k)^2$.

Solution:
Note that there are only two possible locations for points $D_1$ and $D_2$, as they are both $\sqrt{111}$ from point $A$ and $\sqrt{11}$ from point $B$, so they are the two points where a circle centered at $A$ with radius $\sqrt{111}$ and a circle centered at $B$ with radius $\sqrt{11}$ intersect.  Let $D_1$ be the point on the opposite side of $\overline{AB}$ from $C$, and $D_2$ the point on the same side of $\overline{AB}$ as $C$.
Let $\theta$ be the measure of angle $BAD_1$ (also the measure of angle $BAD_2$); by the Law of Cosines,
\begin{align*}\sqrt{11}^2 &= \sqrt{111}^2 + \sqrt{111}^2 - 2 \cdot \sqrt{111} \cdot \sqrt{111} \cdot \cos\theta\\ 11 &= 222(1 - \cos\theta)\end{align*}
There are two equilateral triangles with $\overline{AD_1}$ as a side; let $E_1$ be the third vertex that is farthest from $C$, and $E_2$ be the third vertex that is nearest to $C$.
Angle $E_1AC = E_1AD_1 + D_1AB + BAC = 60 + \theta + 60 = 120 + \theta$; by the Law of Cosines,
\begin{align*}(E_1C)^2 &= (E_1A)^2 + (AC)^2 - 2 (E_1A) (AC)\cos(120 + \theta)\\ &= 111 + 111 - 222\cos(120 + \theta)\end{align*}
Angle $E_2AC = \theta$; by the Law of Cosines,
\begin{align*}(E_2C)^2 &= (E_2A)^2 + (AC)^2 - 2 (E_2A) (AC)\cos\theta\\ &= 111 + 111 - 222\,\cos\theta\end{align*}
There are two equilateral triangles with $\overline{AD_2}$ as a side; let $E_3$ be the third vertex that is farthest from $C$, and $E_4$ be the third vertex that is nearest to $C$.
Angle $E_3AC = E_3AB + BAC = (60 - \theta) + 60 = 120 - \theta$; by the Law of Cosines,
\begin{align*}(E_3C)^2 &= (E_3A)^2 + (AC)^2 - 2 (E_3A) (AC)\cos(120 - \theta)\\ &= 111 + 111 - 222\cos(120 - \theta)\end{align*}
Angle $E_4AC = \theta$; by the Law of Cosines,
\begin{align*}(E_4C)^2 &= (E_4A)^2 + (AC)^2 - 2 (E_4A) (AC)\cos\theta \\ &= 111 + 111 - 222\cos\theta\end{align*}
The solution is:
\begin{align*} \sum_{k=1}^4(CE_k)^2 &= (E_1C)^2 + (E_3C)^2 + (E_2C)^2 + (E_4C)^2\\ &= 222(1 - \cos(120 + \theta)) + 222(1 - \cos(120 - \theta)) + 222(1 - \cos\theta) + 222(1 - \cos\theta)\\ &= 222((1 - (\cos120\cos\theta - \sin120\sin\theta)) + (1 - (\cos120\cos\theta + \sin120\sin\theta)) + 2(1 -\cos\theta))\\ &= 222(1 - \cos120\cos\theta + \sin120\sin\theta + 1 - \cos120\cos\theta - \sin120\sin\theta + 2 - 2\cos\theta)\\ &= 222(1 + \frac{1}{2}\cos\theta + 1 + \frac{1}{2}\cos\theta + 2 - 2\cos\theta)\\ &= 222(4 - \cos\theta)\\ &= 666 + 222(1 - \cos\theta) \end{align*}
Substituting $11$ for $222(1 - \cos\theta)$ gives the solution $666 + 11 = \framebox{677}.$

Answer: 677


The left and right foci of a hyperbola are $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{2}$ intersects the right branch of the hyperbola at points $A$ and $B$. If $\triangle F_{1} A B$ is an equilateral triangle, then the eccentricity of the hyperbola is $\qquad$

Solution:
2. $\sqrt{3}$.

Let the equation of the hyperbola be
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0) \text {, }
$$

with the semi-focal distance being $c$.
Then $c^{2}=a^{2}+b^{2}$.
By $\left|F_{1} A\right|-\left|F_{2} A\right|=\left|F_{1} B\right|-\left|F_{2} B\right|=2 a$,
$$
\left|F_{1} A\right|=\left|F_{2} A\right|+\left|F_{2} B\right|=\left|F_{1} B\right|,
$$

we solve to get $\left|F_{2} A\right|=\left|F_{2} B\right|=2 a$.
This indicates that $A B \perp x$ axis.
Thus, $\left|F_{2} A\right|=\left|F_{2} B\right|=\frac{b^{2}}{a}$.
Combining $c^{2}=a^{2}+b^{2}$, we solve to get the eccentricity of the hyperbola as $e=\frac{c}{a}=\sqrt{3}$.

Answer: \sqrt{3}


Let $n$ be a positive integer. Determine the smallest positive integer $k$ with the following property: it is possible to mark $k$ cells on a $2 n \times 2 n$ board so that there exists a unique partition of the board into $1 \times 2$ and $2 \times 1$ dominoes, none of which contains two marked cells.  Answer. $2 n$.

Solution: We first construct an example of marking $2 n$ cells satisfying the requirement. Label the rows and columns $1,2, \ldots, 2 n$ and label the cell in the $i$-th row and the $j$-th column $(i, j)$. For $i=1,2, \ldots, n$, we mark the cells $(i, i)$ and $(i, i+1)$. We claim that the required partition exists and is unique. The two diagonals of the board divide the board into four regions. Note that the domino covering cell $(1,1)$ must be vertical. This in turn shows that each domino covering $(2,2),(3,3), \ldots,(n, n)$ is vertical. By induction, the dominoes in the left region are all vertical. By rotational symmetry, the dominoes in the bottom region are horizontal, and so on. This shows that the partition exists and is unique. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-45.jpg?height=323&width=820&top_left_y=1080&top_left_x=696) It remains to show that this value of $k$ is the smallest possible. Assume that only $k<2 n$ cells are marked, and there exists a partition $P$ satisfying the requirement. It suffices to show there exists another desirable partition distinct from $P$. Let $d$ be the main diagonal of the board. Construct the following graph with edges of two colours. Its vertices are the cells of the board. Connect two vertices with a red edge if they belong to the same domino of $P$. Connect two vertices with a blue edge if their reflections in $d$ are connected by a red edge. It is possible that two vertices are connected by edges of both colours. Clearly, each vertex has both red and blue degrees equal to 1. Thus the graph splits into cycles where the colours of edges in each cycle alternate (a cycle may have length 2). Consider any cell $c$ lying on the diagonal $d$. Its two edges are symmetrical with respect to $d$. Thus they connect $c$ to different cells. This shows $c$ belongs to a cycle $C(c)$ of length at least 4. Consider a part of this cycle $c_{0}, c_{1}, \ldots, c_{m}$ where $c_{0}=c$ and $m$ is the least positive integer such that $c_{m}$ lies on $d$. Clearly, $c_{m}$ is distinct from $c$. From the construction, the path symmetrical to this with respect to $d$ also lies in the graph, and so these paths together form $C(c)$. Hence, $C(c)$ contains exactly two cells from $d$. Then all $2 n$ cells in $d$ belong to $n$ cycles $C_{1}, C_{2}, \ldots, C_{n}$, each has length at least 4. By the Pigeonhole Principle, there exists a cycle $C_{i}$ containing at most one of the $k$ marked cells. We modify $P$ as follows. We remove all dominoes containing the vertices of $C_{i}$, which correspond to the red edges of $C_{i}$. Then we put the dominoes corresponding to the blue edges of $C_{i}$. Since $C_{i}$ has at least 4 vertices, the resultant partition $P^{\prime}$ is different from $P$. Clearly, no domino in $P^{\prime}$ contains two marked cells as $C_{i}$ contains at most one marked cell. This shows the partition is not unique and hence $k$ cannot be less than $2 n$.

Answer: 2 n


Find all ordered triples $(a,b, c)$ of positive integers which satisfy $5^a + 3^b - 2^c = 32$

Solution:
To find all ordered triples \((a, b, c)\) of positive integers that satisfy \(5^a + 3^b - 2^c = 32\), we can proceed as follows:

1. **Modulo 3 Analysis:**
   \[
   5^a + 3^b - 2^c \equiv 32 \pmod{3}
   \]
   Since \(5 \equiv 2 \pmod{3}\), we have:
   \[
   5^a \equiv 2^a \pmod{3}
   \]
   And since \(3 \equiv 0 \pmod{3}\), we have:
   \[
   3^b \equiv 0 \pmod{3}
   \]
   Therefore:
   \[
   2^a - 2^c \equiv 2 \pmod{3}
   \]
   This implies:
   \[
   (-1)^a - (-1)^c \equiv 2 \pmod{3}
   \]
   For this to hold, \(a\) must be even and \(c\) must be odd. Let \(a = 2d\) and \(c = 2e + 1\).

2. **Modulo 4 Analysis:**
   If \(c \geq 3\), then:
   \[
   5^a + 3^b - 2^c \equiv 32 \pmod{4}
   \]
   Since \(5 \equiv 1 \pmod{4}\), we have:
   \[
   5^a \equiv 1 \pmod{4}
   \]
   And since \(3 \equiv -1 \pmod{4}\), we have:
   \[
   3^b \equiv (-1)^b \pmod{4}
   \]
   Therefore:
   \[
   1 + (-1)^b - 2^c \equiv 0 \pmod{4}
   \]
   This implies:
   \[
   1 + (-1)^b \equiv 0 \pmod{4}
   \]
   This means \(b\) must be odd. Let \(b = 2f + 1\).

3. **Modulo 8 Analysis:**
   \[
   5^{2d} + 3^{2f+1} - 2^{2e+1} \equiv 32 \pmod{8}
   \]
   Since \(5^2 \equiv 25 \equiv 1 \pmod{8}\), we have:
   \[
   5^{2d} \equiv 1 \pmod{8}
   \]
   And since \(3^2 \equiv 9 \equiv 1 \pmod{8}\), we have:
   \[
   3^{2f+1} \equiv 3 \cdot 1 \equiv 3 \pmod{8}
   \]
   Therefore:
   \[
   1 + 3 - 2^{2e+1} \equiv 0 \pmod{8}
   \]
   This implies:
   \[
   4 \equiv 2^{2e+1} \pmod{8}
   \]
   This is a contradiction because \(2^{2e+1}\) for \(e \geq 1\) is always greater than 4 modulo 8. Hence, \(c = 1\).

4. **Solving for \(a\) and \(b\):**
   With \(c = 1\), the equation becomes:
   \[
   5^a + 3^b = 34
   \]
   We test small values of \(a\):
   - If \(a = 1\):
     \[
     5^1 + 3^b = 34 \implies 5 + 3^b = 34 \implies 3^b = 29
     \]
     This is not possible since 29 is not a power of 3.
   - If \(a = 2\):
     \[
     5^2 + 3^b = 34 \implies 25 + 3^b = 34 \implies 3^b = 9 \implies b = 2
     \]

Thus, the only solution is \((a, b, c) = (2, 2, 1)\).

The final answer is \(\boxed{(2, 2, 1)}\).

Answer: (2, 2, 1)


Given that the center of the square $ABCD$ is at the origin of the coordinate system, and the four vertices of the square are on the graph of the function $f(x)=x^{3}+a x$. Find the range of the real number $a$.

Solution:
Let $A\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$. Then $B\left(-y_{0}, x_{0}\right) 、 C\left(-x_{0},-y_{0}\right) 、 D\left(y_{0},-x_{0}\right)$.
According to the problem, we have
$y_{0}=x_{0}^{3}+a x_{0}$,
$-x_{0}=y_{0}^{3}+a y_{0}$.
(1) $\times y_{0}-$ (2) $\times x_{0}$ gives
$x_{0}^{2}+y_{0}^{2}=x_{0} y_{0}\left(x_{0}^{2}-y_{0}^{2}\right)$.
(1) $\times x_{0}+$ (2) $\times y_{0}$ gives
$$
x_{0}^{4}+y_{0}^{4}+a\left(x_{0}^{2}+y_{0}^{2}\right)=0 \text {. }
$$

Let $x_{0}=r \cos \theta, y_{0}=r \sin \theta\left(r>0, \theta \in\left(0, \frac{\pi}{2}\right)\right)$.
Substituting into equation (3) gives
$$
\begin{array}{l}
r^{2}=r \cos \theta \cdot r \sin \theta \cdot r^{2}\left(\cos ^{2} \theta-\sin ^{2} \theta\right) \\
=\frac{1}{4} r^{4} \sin 4 \theta .
\end{array}
$$

Therefore, $r^{2} \sin 4 \theta=4$.
Substituting into equation (4) gives $r^{2}\left(\cos ^{4} \theta+\sin ^{4} \theta\right)+a=0$.
Simplifying gives $r^{2} \cos 4 \theta=-4 a-3 r^{2}$.
$(5)^{2}+(6)^{2}$ and simplifying gives
$$
r^{4}+3 a r^{2}+2 a^{2}+2=0 \text {. }
$$

By $\Delta=9 a^{2}-4\left(2 a^{2}+2\right)=a^{2}-8 \geqslant 0$, then $a \geqslant 2 \sqrt{2}$ or $a \leqslant-2 \sqrt{2}$.

Also, from equation (4) we know $a<0$, hence $a \in(-\infty,-2 \sqrt{2}]$.

Answer: a \in(-\infty,-2 \sqrt{2}]


A wooden cube was painted white on the outside, each of its edges was divided into 5 equal parts, after which the cube was sawn so that small cubes were obtained, the edge of which is 5 times smaller than that of the original cube. How many small cubes are there with at least one face painted?

Solution:
It is easier to count the number of cubes that have no faces painted.

## Solution

The number of small cubes obtained after cutting a large cube is $5^{3}=125$. Let's count the number of cubes that have no faces painted. The unpainted cubes are all those that do not have any visible outer faces. These cubes form a $3 \times 3 \times 3$ cube, and their number is $3^{3}=27$. Finally, the number of cubes that have at least one face painted is $125-27=98$.

## Answer

98.00

Does there exist a) a bounded, b) an unbounded figure in the plane that has among its axes of symmetry two parallel non-coincident lines?

## Hint

If such a figure exists, it must map onto itself under a non-identity parallel translation, which is defined as the sequential execution of symmetries relative to two parallel axes of symmetry.

## Solution

a) Suppose a figure has two parallel axes of symmetry. Introduce a coordinate system such that these lines have equations $x=0$ and $x=1$. Let some point $(a; b)$ belong to the given figure. Since the line $x=0$ is an axis of symmetry, the point $(-a; b)$ must also belong to the given figure. Next, the point $(a+2; b)$, which is symmetric to the point $(-a; b)$ relative to the line $x=1$, must also belong to the given figure. Thus, together with each point $(a; b)$, the figure contains the point $(a+2; b)$. Repeating these arguments for the point $(a+2; b)$, we get that the point $(a+4; b)$ also belongs to the figure, and so on, the points $(a+6; b)$, $(a+8; b)$, $(a+10; b)$, ... belong to the figure. This obviously contradicts the assumption of the boundedness of the figure. b) An example is a line (any line perpendicular to it is its axis of symmetry).

## Answer

a) No. b) Yes.

Answer: 98


Find all three-digit numbers $\overline{L O M}$, composed of distinct digits $L, O$ and $M$, for which the following equality holds:

$$
\overline{L O M}=(L+O+M)^{2}+L+O+M
$$

Solution: Answer: 156. Instructions. Let $x=L+O+M$. Then $\overline{L O M}=x(x+1)$. In this case, $x \geq 10$ (otherwise $x(x+1)<100)$ and $x \leq 24$ (the sum of digits does not exceed $9+8+7=24$). Therefore, $x \in[10 ; 24]$. From the relation $100 \cdot L+10 \cdot O+M=x^{2}+L+O+M$ it follows that $x^{2}=99 \cdot L+9 \cdot O$, which means $x$ is divisible by 3. It remains to substitute the values $12,15,18,21$ and 24 into $x(x+1)$ and check the sum of the digits of the resulting number. It matches $x$ only when $x=12$.

Answer: 156


The product of a million whole numbers is equal to million. What can be the greatest possible value of the sum of these numbers?

Solution:
1. **Claim**: The maximal sum \(a + b\) with a fixed product \(ab\) occurs when one of the two numbers is \(1\).

2. **Proof**: Let the fixed product \(ab\) be \(x\). We desire to show that \(x + 1 > p + q\), where \(p, q > 1\) and \(pq = x\). This is obvious because since \(p, q > 1\), \((p-1)(q-1) > 0\), which rearranges to \(pq + 1 > p + q\), or \(x + 1 > p + q\).

3. **Application to the problem**: Now consider an arrangement of the million whole numbers with the maximal possible sum. If any two terms \(a_i\) and \(a_j\) aren't ones, we can change them to \(1\) and \(a_i a_j\), to keep the overall product the same. From our claim, this increases the total sum, contradicting our assumption that the arrangement had the maximal possible sum. Therefore, one cannot find two terms \(a_i\) and \(a_j\) that aren't ones, meaning at most one number isn't a one.

4. **Conclusion**: This means that our optimizing arrangement has to be \(1000000, 1, 1, \cdots\), meaning the greatest possible sum is:
   \[
   1000000 + 1 \cdot (1000000 - 1) = 1000000 + 999999 = 1999999
   \]

The final answer is \(\boxed{1999999}\).

Answer: 1999999


There are 195 different cards with numbers $1, 5, 7, 5^{2}, 7^{2}, \ldots, 5^{97}, 7^{97}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 3 cards be chosen so that the product of the numbers on the chosen cards is a square of an integer divisible by 35?

Solution:
Answer: 223488.

Solution. Consider the case when one of the selected cards has a one written on it. Then, on the other two cards, even powers of fives and sevens must be recorded. There are 48 ways to choose an even power of five and 48 ways to choose an even power of seven, and since this choice is made independently, the total number of ways in this case is $48^{2}=2304$.

Now let's move on to the case when the card with one is not used. There are two possibilities: to take two cards with powers of five and one with a power of seven, or vice versa.

Suppose we take two cards with powers of seven. Since the product of the selected numbers must contain seven in an even power, the powers of seven on the selected cards must have the same parity. We have a total of 49 cards with odd powers and 48 cards with even powers. Thus, there are $C_{49}^{2}=\frac{49 \cdot 48}{2}=1176$ ways to choose cards with odd powers and $C_{48}^{2}=\frac{48 \cdot 47}{2}=1128$ ways to choose two cards with even powers, totaling 2304 ways. After selecting the cards with powers of seven, we can choose any card with an even power of five (48 ways). Since the choice of the card with the power of five is independent of the previous choice, in total, we get $2304 \cdot 48=110592$ ways.

It is easy to see that if we take two cards with powers of five and one with a power of seven, the number of ways will be the same as in the previous case. Therefore, we get $2304+110592 \cdot 2=223488$ ways.

Answer: 223488


Touching circles $k_{1}\left(S_{1}, r_{1}\right)$ and $k_{2}\left(S_{2}, r_{2}\right)$ lie in a right-angled triangle $A B C$ with the hypotenuse $A B$ and legs $A C=4$ and $B C=3$ in such way, that the sides $A B, A C$ are tangent to $k_{1}$ and the sides $A B, B C$ are tangent to $k_{2}$. Find radii $r_{1}$ and $r_{2}$, if $4 r_{1}=9 r_{2}$.

(Pavel NovotnÃ½)

Solution:
Solution. The hypotenuse $A B$ has length $A B=5$ with respect to Pythagoras' theorem. Then for angles in the triangle there is $\cos \alpha=\frac{4}{5}, \cos \beta=\frac{3}{5}$,

$$
\begin{aligned}
& \cot \frac{\alpha}{2}=\sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}=3 \\
& \cot \frac{\beta}{2}=\sqrt{\frac{1+\cos \beta}{1-\cos \beta}}=2
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_5816354d2b4b35f9c465g-16.jpg?height=686&width=1239&top_left_y=1867&top_left_x=451)

Fig. 1

Since both circles $k_{1}, k_{2}$ whole lie in the triangle $A B C$, they are externally tangent-in the opposite case the leg tangent to the smaller circle intersects the greater circle. Let circles $k_{1}$ and $k_{2}$ touch the side $A B$ at points $D$ resp. $E$ and let point $F$ be orthogonal projection of the point $S_{2}$ to the element $S_{1} D$ (Fig. 1, under assumption is $r_{1}>r_{2}$ ). Using Pythagoras' theorem for a triangle $F S_{2} S_{1}$ we obtain

$$
\left(r_{1}+r_{2}\right)^{2}=\left(r_{1}-r_{2}\right)^{2}+D E^{2}
$$

which follows $D E=2 \sqrt{r_{1} r_{2}}$.

An equality $A B=A D+D E+E B$ gives

$$
c=r_{1} \cot \frac{\alpha}{2}+2 \sqrt{r_{1} r_{2}}+r_{2} \cot \frac{\beta}{2}=3 r_{1}+2 \sqrt{r_{1} r_{2}}+2 r_{2},
$$

and since $r_{1}=\frac{9}{4} r_{2}$ we obtain

$$
\frac{27}{4} r_{2}+3 r_{2}+2 r_{2}=5
$$

which follows

$$
r_{2}=\frac{20}{47}, \quad r_{1}=\frac{45}{47}
$$

Remark. Both circles lie really in the triangle $A B C$ because an incircle of the triangle has diameter $\varrho=a b /(a+b+c)=1$, while both values $r_{1}, r_{2}$ are less than 1 .

Answer: r_{2}=\frac{20}{47},\quadr_{1}=\frac{45}{47}


Let point $Q$ be in the plane $\alpha$ of $\triangle ABC$, and point $P$ be outside the plane $\alpha$. If for any real numbers $x$ and $y, |\overrightarrow{AP}-x \overrightarrow{AB}-y \overrightarrow{AC}| \geq|\overrightarrow{PQ}|$, then the angle $\theta$ between vector $\overrightarrow{PQ}$ and $\overrightarrow{BC}$ is $\qquad$.

Solution:
Answer $\frac{\pi}{2}$.
Analysis Let $\overrightarrow{A M}=x \overrightarrow{A B}+y \overrightarrow{A C}$, then point $M$ is also in plane $\alpha$, $|\overrightarrow{M P}|=|\overrightarrow{A P}-\overrightarrow{A M}| \geq|\overrightarrow{P Q}|$, so $P Q \perp$ plane $\alpha$, thus the angle between vector $\overrightarrow{P Q}$ and $\overrightarrow{B C}$ is $\frac{\pi}{2}$.

Answer: \frac{\pi}{2}


On the potion-making exam, each student at Hogwarts School had to brew 4 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron took 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the same speeds?

Solution:
# Answer: 36

## Exact match of the answer -1 point

## Solution.

Let's calculate the potion-making speed of each of the three students, measured in potions per hour. Hermione made 4 potions in half an hour, so in one hour she would make twice as many potions, which is 8. Therefore, her speed is 8 potions per hour. Harry made 4 potions in 40 minutes, which means 1 potion every 10 minutes, so in 60 minutes he would make 6 potions, giving him a speed of 6 potions per hour. As for Ron, he made 4 potions in one hour, so his speed is 4 potions per hour. Now, let's assume the three students work together for one hour. Hermione would make 8 potions in this hour, Harry would make 6, and Ron would make 4, so together they would make $8+6+4=18$ potions in this hour. Therefore, if they work together for 2 hours, they would make twice as many potions, which is $18 \cdot 2=36$ potions - giving us the desired answer.

Answer: 36


Solution:
Suppose that $d$ is the common difference in this arithmetic sequence.

Since $t_{2018}=100$ and $t_{2021}$ is 3 terms further along in the sequence, then $t_{2021}=100+3 d$.

Similarly, $t_{2036}=100+18 d$ since it is 18 terms further along.

Since $t_{2018}=100$ and $t_{2015}$ is 3 terms back in the sequence, then $t_{2015}=100-3 d$.

Similarly, $t_{2000}=100-18 d$ since it is 18 terms back.

Therefore,

$$
\begin{aligned}
t_{2000}+5 t_{2015}+5 t_{2021}+t_{2036} & =(100-18 d)+5(100-3 d)+5(100+3 d)+(100+18 d) \\
& =1200-18 d-15 d+15 d+18 d \\
& =1200
\end{aligned}
$$

ANSWER: 1200

Answer: 1200


On the island of Friends and Foes, every citizen is either a Friend (who always tells the truth) or a Foe (who always lies). Seven citizens are sitting in a circle. Each declares "I am sitting between two Foes". How many Friends are there in the circle?

Solution:
SolUTION
003

If every citizen is a foe then all are telling the truth, which is a contradiction. Therefore, there is at least one friend. Suppose the citizens are labelled ABCDEFG and sat in the same order, with $\mathrm{A}$ (a friend) sat next to $\mathrm{B}$ and $\mathrm{G}$. Then $\mathrm{B}$ and $\mathrm{G}$ are both foes. We now consider $\mathrm{C}$ and $\mathrm{F}$. If $\mathrm{C}$ is a foe, then $\mathrm{D}$ must be a friend (else $\mathrm{C}$ would speak the truth). This makes $\mathrm{E}$ a foe, and $\mathrm{F}$ must be a friend since $\mathrm{F}$ is sitting between two foes, so speaks the truth. In this scenario there are three friends, $\mathrm{A}, \mathrm{D}$ and $\mathrm{F}$.
If $\mathrm{F}$ is a foe, then $\mathrm{E}$ must be a friend (else $\mathrm{F}$ would speak the truth). This makes $\mathrm{D}$ a foe, and $\mathrm{C}$ must be a friend since $\mathrm{C}$ is sitting between two foes, so speaks the truth. In this scenario there are three friends, $\mathrm{A}, \mathrm{C}$ and $\mathrm{E}$.
If $\mathrm{C}$ and $\mathrm{F}$ are both friends then $\mathrm{D}$ and $\mathrm{E}$ must both be foes. In this scenario there are three friends, A, C and F.
Therefore there are three friends.

Answer: 3


Let the polynomial $x^{3}-x-a$ and the polynomial $x^{2}+x-a$ have a non-constant common factor, then $a=$ $\qquad$ .

Solution:
II. 1.0 or 6.
Since $\left(x^{3}-x-a\right)-\left(x^{2}+x-a\right)=x(x+1)(x-2)$, the common factor of $x^{3}-x-a$ and $x^{2}+x-a$ must be one of $x$, $x+1$, or $x-2$.
When the common factor is $x$ or $x+1$, $a=0$;
When the common factor is $x-2$, $a=6$.

Therefore, $a=0$ or 6.

Answer: 0 \text{ or } 6


The number $n$ has exactly six divisors (including 1 and itself). They were arranged in ascending order. It turned out that the third divisor is seven times greater than the second, and the fourth is 10 more than the third. What is $n$?

Solution:
Answer: 2891

Solution. If $n$ has six divisors, then either $n=p^{5}$ or $n=p \cdot q^{2}$ (where $p$ and $q$ are prime numbers). In the first case, $p=7$ (since the third divisor is seven times the second), but then the second condition is not satisfied.

Therefore, $n$ has two prime divisors (one of which is squared), and one of the prime factors is 7. Let the smallest divisors of $n$ be $1, p, 7p$, and $7p+10$. If $p=7$, then $7p+10=59$, and we get the answer 2891. If $p<7$, then $p=2$ (then $7p+10=24$, and such a number has more than six divisors). If $p=3$, then $7p+10=31$, and if $p=5$, then $7p+10=45$. In each of these cases, $n$ has more than two prime divisors.

Answer: 2891


Find the third term of an infinite geometric progression with a common ratio $|q|<1$, the sum of which is $\frac{8}{5}$, and the second term is $-\frac{1}{2}$.

Solution:
6.4 Let's use formulas (6.10) and (6.6). Then we get the system

$\left\{\begin{array}{l}\frac{b_{1}}{1-q}=\frac{8}{5}, \\ b_{1} q=-\frac{1}{2} .\end{array}\right.$

Further, we have $16 q^{2}-16 q-5=0$, from which $q_{1}=-\frac{1}{4}$, $q_{2}=\frac{5}{4}>1$ (does not fit the condition). Therefore, $b_{3}=\left(-\frac{1}{2}\right)\left(-\frac{1}{4}\right)=\frac{1}{8}$.

Answer: $\frac{1}{8}$.

Answer: \frac{1}{8}


Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: x^{2}-\frac{y^{2}}{24}=1$, respectively, and let $P$ be a point on the hyperbola $C$ in the first quadrant. If $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{4}{3}$, then the radius of the incircle of $\triangle P F_{1} F_{2}$ is . $\qquad$

Solution:
4. 2 .

Let $\left|P F_{1}\right|=4 t$. Then $\left|P F_{2}\right|=3 t$.
Thus $4 t-3 t=\left|P F_{1}\right|-\left|P F_{2}\right|=2$
$$
\Rightarrow t=2,\left|P F_{1}\right|=8,\left|P F_{2}\right|=6 \text {. }
$$

Combining with $\left|F_{1} F_{2}\right|=10$, we know that $\triangle P F_{1} F_{2}$ is a right triangle, $P F_{1} \perp P F_{2}$.
Therefore, the inradius of $\triangle P F_{1} F_{2}$ is
$$
r=\frac{6+8-10}{2}=2 \text {. }
$$

Answer: 2


It is known that Rochelle made 8 hamburgers with 3 pounds of meat. She still needs to prepare 24 hamburgers for the neighborhood picnic. Then she needs ( ) pounds of meat.
(A) 6
(B) $6 \frac{2}{3}$
(C) $7 \frac{1}{2}$
(D) 8
(E) 9

Solution:
1.E.
$$
\frac{3}{8} \times 24=9
$$

Answer: E


Solution:
Solution. We have:

$$
\begin{aligned}
\frac{2^{3} \cdot 4^{5} \cdot 6^{7}}{8^{9} \cdot 10^{11}}: 0.015^{7} & =\frac{2^{3} \cdot\left(2^{2}\right)^{5} \cdot 6^{7}}{\left(2^{3}\right)^{9} \cdot 10^{11}}:\left(\frac{15}{1000}\right)^{7}=\frac{2^{13}}{2^{27} \cdot 10^{11}} \cdot\left(\frac{6 \cdot 1000}{3 \cdot 5}\right)^{7} \\
& =\frac{1}{2^{14} \cdot 10^{11}} \cdot 400^{7}=\frac{1}{2^{14} \cdot 10^{11}} \cdot 4^{7} \cdot 100^{7} \\
& =\frac{1}{2^{14} \cdot 10^{11}} \cdot 2^{14} \cdot 10^{14}=10^{3}=1000
\end{aligned}
$$

## IX Department

Answer: 1000


How many ordered pairs of integers $(a, b)$ are there such that $|a|+|b|<2019$?

Solution:
First method:

Let's look at how many ordered pairs of natural numbers $(a, b)$ satisfy the conditions of the problem.

If $a=1$, then $b$ can take the values $1,2,3, \ldots, 2017$.

If $a=2$, then $b$ can take the values $1,2,3, \ldots, 2016$.

$$
\vdots
$$

If $a=2017$, then $b$ can only take the value 1.

Thus, the number of ordered pairs $(a, b)$, where $a$ and $b$ are natural numbers, is:

$$
1+2+\cdots+2017=\frac{2018 \cdot 2017}{2}=2035153
$$

Now let's look at how many ordered pairs $(a, b)$ satisfy the conditions of the problem, where $\boldsymbol{a}$ and $\boldsymbol{b}$ are integers different from zero.

From the condition $|a|+|b|<2019$, it follows that for each ordered pair of natural numbers $(a, b)$ that satisfies the condition of the problem, we can associate ordered pairs of integers of the form $(a,-b),(-a, b),(-a,-b)$
which also satisfy the conditions of the problem.

The number of ordered pairs of integers $(a, b)$, where $a$ and $b$ are integers different from zero, is $4 \cdot 2035153=8140612$.

It remains to count the ordered pairs $(a, b)$ in which at least one of the members is equal to zero. If $a=0$, then according to the condition of the problem, $|b|<2019$.

The number of integers $b$ that satisfy this condition is $2 \cdot 2018+1=4037$.

Similarly, if $b=0$.

The number of ordered pairs of integers $(a, b)$, in which at least one of the members is equal to zero, is $4037+4037-1=8073$ (the ordered pair $(0,0)$ has been counted twice).

In total, there are $8140612+8073=8148685$ ordered pairs of integers $(a, b)$ with the desired property.

Note:

Answer: 8148685


The number of points equidistant from a circle and two parallel tangents to the circle is
(A) 0.
(B) 2.
(C) 3.
(D) 4.
(E) infinitely many.

Solution:
[Solution] Let $l_{1}, l_{2}$ be two parallel tangent lines of the known circle $\odot O$. If $r$ is the radius of $\odot O$, then the distance between $l_{1}, l_{2}$ is $2 r$.

The points equidistant from $l_{1}, l_{2}$ lie on the line $m$ (as shown in the figure), which is a line passing through $O$ and parallel to $l_{1}, l_{2}$.

The points at a distance of $r$ from the points on $\odot O$ lie on the circle with center $O$ and radius $2 r$, and it intersects $m$ at points $P, Q$.

In summary, points $P, Q, O$ satisfy the conditions of the problem.
Therefore, the answer is $(C)$.

Answer: C


Let $G$ be a simple, undirected, connected graph with $100$ vertices and $2013$ edges. It is given that there exist two vertices $A$ and $B$ such that it is not possible to reach $A$ from $B$ using one or two edges. We color all edges using $n$ colors, such that for all pairs of vertices, there exists a way connecting them with a single color. Find the maximum value of $n$.

Solution:
1. **Understanding the Problem:**
   We are given a simple, undirected, connected graph \( G \) with 100 vertices and 2013 edges. There exist two vertices \( A \) and \( B \) such that it is not possible to reach \( A \) from \( B \) using one or two edges. We need to color all edges using \( n \) colors such that for all pairs of vertices, there exists a way connecting them with a single color. We aim to find the maximum value of \( n \).

2. **Initial Setup:**
   Let \( v \) and \( w \) be the two vertices which are not connected with a path of length 1 or 2. Let \( S_v \), \( S_w \), and \( S_x \) be the sets of the other 98 vertices which are connected to \( v \), connected to \( w \), and connected to neither \( v \) nor \( w \), respectively. These sets are pairwise disjoint by the given condition.

3. **Edge Coloring:**
   We need to color the edges such that each color forms a connected subgraph. Let the companies (colors) have \( e_1, e_2, \ldots, e_t \) edges respectively. We know that:
   \[
   \sum_{i=1}^t e_i = 2013
   \]
   To maximize \( n \), we need to minimize the number of edges per color while ensuring connectivity.

4. **Connectivity and Special Companies:**
   If a company's flights do not form a connected graph, we can split it into two companies to decrease \( \sum (e_i - 1) \). Hence, we assume all companies' flights are connected. Define a company as **special** if it services both \( v \) and something in \( S_w \cup S_x \) or it services both \( w \) and something in \( S_v \cup S_x \). Define a company as **doubly special** if it services \( v \), \( w \), and cities in both \( S_w \cup S_x \) and \( S_v \cup S_x \).

5. **Vertex Association:**
   Let \( C_1, C_2, \ldots, C_n \) be the special companies, and let them have \( v_1, v_2, \ldots, v_n \) vertices respectively. Each \( C_i \) has at least \( v_i - 1 \) edges by connectivity. To show \( \sum (e_i - 1) \ge 98 \), we need \( \sum (v_j - 2) \ge 98 \).

6. **Vertex Set Coverage:**
   For a company \( C_j \), associate it with a set of \( v_j - 2 \) vertices. If it is doubly special, associate it with its vertices excluding \( v \) and \( w \). If it is singly special and contains \( v \) and some vertex \( v' \) in \( S_w \cup S_x \), it must contain a neighbor \( v_1 \) of \( v \). Associate it with its vertices excluding \( v \) and \( v_1 \). Similarly, if it contains \( w \) and some vertex \( w' \) in \( S_v \cup S_x \), exclude \( w \) and one of its neighbors.

7. **Covering All Vertices:**
   These sets of vertices must cover all of \( S_v \cup S_w \cup S_x \). Suppose there is a vertex \( y \in S_v \cup S_w \cup S_x \). If \( y \in S_v \cup S_x \), there must exist a company connecting \( y \) and \( w \), say \( C_j \). If \( C_j \) is doubly special, its associated vertex set contains \( y \). If it is singly special, its associated vertex set contains \( y \). The argument is analogous for \( y \in S_w \cup S_x \).

8. **Conclusion:**
   Since these vertex sets cover all 98 vertices which are not \( v \) or \( w \), and their cardinalities are \( v_1 - 2, v_2 - 2, \ldots, v_n - 2 \), we have:
   \[
   \sum (v_j - 2) \ge 98
   \]
   Therefore, the maximum value of \( n \) is \( 1915 \).

The final answer is \( \boxed{1915} \).

Answer: 1915


Among the positive integers less than 10000, how many integers \( x \) are there such that \( 2^x - x^2 \) is divisible by 7?

Solution:
［Solution］First, we prove
$$
2^{21+r}-(21+r)^{2} \equiv 2^{r}-r^{2} \quad(\bmod 7) .
$$

where $r \in\{1,2,3, \cdots, 21\}$
In fact, by

then
$$
\begin{aligned}
2^{21}=\left(2^{3}\right)^{7}=8^{7} \equiv 1 & (\bmod 7), \\
2^{21+r} \equiv 2^{r} & (\bmod 7) .
\end{aligned}
$$

Also, $(21+r)^{2}=21^{2}+42 r+r^{2} \equiv r^{2}(\bmod 7)$,
thus (1) holds.
From (1), we know that $2^{x}-x^{2}$ has a periodic change with a period of 21 when divided by 7. Let $x=1,2, \cdots, 21$ to find that there are 6 numbers in $2^{x}-x^{2}$ that are divisible by 7.
That is, when $x=2,4,5,6,10,15$, $2^{x}-x^{2}$ is divisible by 7.
Since
$$
10000=476 \cdot 21+4,
$$

there are from 1 to 10000
$$
476 \cdot 6+1=2857
$$

numbers $x$ such that $2^{x}-x^{2}$ is divisible by 7.

Answer: 2857


Let the set $S=\{1,2, \cdots, 15\}, A=\left\{a_{1}, a_{2}, a_{3}\right\}$ be a subset of $S$, and $(a_{1}, a_{2}, a_{3})$ satisfy:
$$
1 \leqslant a_{1}<a_{2}<a_{3} \leqslant 15, a_{3}-a_{2} \leqslant 6 .
$$

Then the number of subsets $A$ that satisfy the condition is $\qquad$

Solution:
(7) When $2 \leqslant a_{2} \leqslant 9$, $\left(a_{1}, a_{2}\right)$ has $\mathrm{C}_{9}^{2}$ ways to choose, and $a_{3}$ has 6 ways to choose, so $\left(a_{1}, a_{2}, a_{3}\right)$ has a total of $6 \times \mathrm{C}_{9}^{2}=216$ (ways) to choose; When $10 \leqslant a_{2} \leqslant 14$, once $a_{2}$ is determined, $a_{1}$ has $a_{2}-1$ ways to choose, and $a_{3}$ has $15-a_{2}$ ways to choose, so the number of ways to choose $\left(a_{1}, a_{2}, a_{3}\right)$ is
$$
\begin{aligned}
& \sum_{a_{2}=10}^{14}\left(a_{2}-1\right)\left(15-a_{2}\right) \\
= & 9 \times 5+10 \times 4+11 \times 3+12 \times 2+13 \times 1 \\
= & 155 \text { (ways). }
\end{aligned}
$$

In summary, the number of subsets $A$ that satisfy the conditions is $216+155=371$ (subsets).

Answer: 371


For a finite set $A$, there exists a function $f: \mathbf{N}^{*} \rightarrow A$, with the following property: if $i, j \in \mathbf{N}^{*}$, and $|i-j|$ is a prime number, then $f(i) \neq f(j)$. How many elements must the set $A$ have at least?

Solution: 1. Since the absolute value of the difference between any two of the numbers $1,3,6,8$ is a prime number, according to the problem, $f(1)$, $f(3)$, $f(6)$, $f(8)$ are four distinct elements in $A$. Therefore, $|A| \geqslant 4$. On the other hand, if we let $A=\{0,1,2,3\}$, and the mapping $f: \mathbf{N}^{*} \rightarrow A$ is defined as: if $x \in \mathbf{N}^{*}, x=4 k + r$, then $f(x)=r$, where $k \in \mathbf{N}, r=0,1,2,3$, then for any $x, y \in \mathbf{N}^{*}$, if $|x-y|$ is a prime number, assuming $f(x) = f(y)$, then $x \equiv y(\bmod 4)$, thus $4 \mid |x-y|$, which contradicts the fact that $|x-y|$ is a prime number. Hence, the set $A$ contains at least 4 elements.

Answer: 4


Find the largest natural number that cannot be represented as the sum of two composite numbers. (Recall that a natural number is called composite if it is divisible by some natural number other than itself and one.)

Solution:
17. Answer: 11.

All large numbers can be represented in the required way: odd numbers as "9 + an even number greater than two", and even numbers as "8 + an even number greater than two".

Answer: 11


Let $u$, $v$, $w$ be positive real numbers, satisfying the condition $u \sqrt{v w} + v \sqrt{w u} + w \sqrt{u v} \geqslant 1$. Find the minimum value of $u + v + w$.

Solution:
5. $\sqrt{3}$,

By the AM-GM inequality and the conditions given in the problem, we know
$$
\begin{array}{l}
u \frac{v+w}{2}+v \frac{w+u}{2}+w \frac{u+v}{2} \\
\geqslant u \sqrt{v w}+v \sqrt{u u}+w \sqrt{u v} \geqslant 1,
\end{array}
$$

Thus, $\square$
$$
\begin{array}{l}
u v+v w+w u \geqslant 1 . \\
\text { Hence }(u+v+w)^{2}=u^{2}+v^{2}+w^{2}+2 w w+2 w w+2 u u \\
=\frac{u^{2}+v^{2}}{2}+\frac{v^{2}+w^{2}}{1}+\frac{w^{2}+u^{2}}{2}+2 w+2 v w+2 u u \\
\geqslant 3 w+3 v w+3 w u \geqslant 3 .
\end{array}
$$

Therefore, $u+v+w \geqslant \sqrt{3}$.
On the other hand, $u=v=w=\frac{\sqrt{3}}{3}$ clearly satisfies the conditions given in the problem, and in this case, $u+v+w=\sqrt{3}$.
In conclusion, the minimum value of $u+v+w$ is $\sqrt{3}$.

Answer: \sqrt{3}


The range of the function $y=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1-x^{2}}+1}$ is

Solution: $\begin{array}{l}\text { I. } 1 .[1, \sqrt{2}] . \\ \text { Let } \sqrt{1+x}+\sqrt{1-x}=t>0 \text {. Then } \\ t^{2}=2+2 \sqrt{1-x^{2}} \\ \Rightarrow t^{2} \in[2,4] \Rightarrow t \in[\sqrt{2}, 2] \\ \Rightarrow y=\frac{2 t}{t^{2}}=\frac{2}{t} \in[1, \sqrt{2}] .\end{array}$

Answer: [1,\sqrt{2}]


Solution:
1. First, compute \(2 \# 0\):
   \[
   2 \# 0 = 2 \cdot 0 - 2 - 3 = 0 - 2 - 3 = -5
   \]

2. Next, compute \(1 \# 4\):
   \[
   1 \# 4 = 1 \cdot 4 - 1 - 3 = 4 - 1 - 3 = 0
   \]

3. Now, use the results from steps 1 and 2 to compute \((-5) \# 0\):
   \[
   -5 \# 0 = (-5) \cdot 0 - (-5) - 3 = 0 + 5 - 3 = 2
   \]

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

Answer: 2


In a puddle, there live three types of amoebas: red, blue, and yellow. From time to time, any two amoebas of different types can merge into one amoeba of the third type. It is known that in the morning, there were 26 red, 31 blue, and 16 yellow amoebas in the puddle, and by evening, only one amoeba remained. What color is it?

Solution:
# Solution:

It is not hard to notice that as a result of the changes that the original population of amoebas can undergo, the parities of the absolute differences of the numbers of different types of amoebas do not change.

Let $n_{1}, n_{2}$ and $n_{3}$ be the numbers of red, blue, and yellow amoebas in the initial population, respectively, and $\varepsilon_{1,2}, \varepsilon_{1,3}$ and $\varepsilon_{2,3}$ be the indicators of the oddness of $\left|n_{1}-n_{2}\right|$, $\left|n_{1}-n_{3}\right|$, and $\left|n^{2}-n_{3}\right|$, respectively.

In the considered case, $\varepsilon_{1,2}=1, \varepsilon_{1,3}=0$ and $\varepsilon_{2,3}=1$. Therefore, if the population degenerates into a single individual, the individual will be blue.

Answer: blue.

Answer: blue


If the function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies for all real numbers $x$,
$$
\sqrt{2 f(x)}-\sqrt{2 f(x)-f(2 x)} \geqslant 2
$$

then the function $f$ is said to have property $P$. Find the largest real number $\lambda$ such that if the function $f$ has property $P$, then for all real numbers $x$, $f(x) \geqslant \lambda$ always holds.

Solution:
Solve: From equation (1) we get
$$
\begin{array}{l}
f(x) \geqslant 2, \\
\sqrt{2 f(x)}-2 \geqslant \sqrt{2 f(x)-f(2 x)} .
\end{array}
$$

Squaring both sides of equation (2) we get
$$
\begin{array}{l}
f(2 x) \geqslant 4 \sqrt{2 f(x)}-4 \\
\Rightarrow f(x) \geqslant 4, \text { and } f(x) \geqslant 4 \sqrt{2 f\left(\frac{x}{2}\right)}-4 .
\end{array}
$$

Conjecture:
$f(x)=c$ has property $P \Leftrightarrow \sqrt{2 c}-\sqrt{c} \geqslant 2$
$$
\Leftrightarrow c \geqslant 12+8 \sqrt{2} \text {. }
$$

It is easy to verify that $f(x)=12+8 \sqrt{2}$ has property $P$, thus $12+8 \sqrt{2} \geqslant \lambda$.
Next, we prove: If function $f(x)$ has property $P$, then for all real numbers $x$,
$$
f(x) \geqslant 12+8 \sqrt{2} \text {. }
$$

Define the sequence $\left\{a_{n}\right\}: a_{0}=2, a_{n+1}=4 \sqrt{2 a_{n}}-4$.
We can prove: (1) $a_{n}<a_{n+1}$.
In fact,
$$
\begin{array}{l}
a_{n+1}-a_{n}=4 \sqrt{2 a_{n}}-4 \sqrt{2 a_{n-1}} \\
=4 \sqrt{2}\left(\sqrt{a_{n}}-\sqrt{a_{n-1}}\right) .
\end{array}
$$

By mathematical induction, it is easy to get $a_{n}<a_{n+1}$.
(2) $a_{n}<12+8 \sqrt{2}$.

When $n=0$, $a_{0}=2<12+8 \sqrt{2}$.
Assume when $n=k$, $a_{k}<12+8 \sqrt{2}$.
Then $a_{k+1}=4 \sqrt{2 a_{k}}-4$
$$
<4 \sqrt{24+16 \sqrt{2}}-4=12+8 \sqrt{2} \text {. }
$$

In summary, the sequence $\left\{a_{n}\right\}$ is monotonically increasing and bounded above. Therefore, it has a limit.
Let $\lim _{n \rightarrow+\infty} a_{n}=a$.
Then $a=4 \sqrt{2 a}-4 \Rightarrow a=12+8 \sqrt{2}$.
(3) For any non-negative integer $n$ and any real number $x$, $f(x) \geqslant a_{n}$.
When $n=0$, $f(x) \geqslant 2=a_{0}$.
Assume the proposition holds for $n$, i.e., for any real number $x$, $f(x) \geqslant a_{n}$.
Thus $f(x) \geqslant 4 \sqrt{2 f\left(\frac{x}{2}\right)}-4$
$\geqslant 4 \sqrt{2 a_{n}}-4=a_{n+1}$.
Let $n \rightarrow+\infty$.
Then $f(x) \geqslant \lim _{n \rightarrow+\infty} a_{n+1}=a=12+8 \sqrt{2}$.
In summary, $\lambda_{\max }=12+8 \sqrt{2}$.

Answer: 12+8 \sqrt{2}


Inside the tetrahedron $ABCD$ there is a point $O$, such that the lines $AO, BO, CO, DO$ intersect the faces $BCD, ACD, ABD, ABC$ at points $A_1, B_1, C_1, D_1$ respectively, and $\frac{AO}{A_1O}=\frac{BO}{B_1O}=\frac{CO}{C_1O}=\frac{DO}{D_1O}=k$. Find all possible values of $k$.

Solution: 9. Let $V$ be the volume of tetrahedron $ABCD$, then we have $\frac{V}{V_{OBCD}}=\frac{AA_{1}}{OA_{1}}=\frac{AO}{A_{1}O}+\frac{OA_{1}}{OA_{1}}=k+1$, similarly $\frac{V}{V_{OACD}}=\frac{V}{V_{OABD}}=\frac{V}{V_{OABC}}=$ $k+1$. Therefore, $k+1=\frac{4V}{V_{OBCD}+V_{OACD}+V_{OMBD}+V_{OABC}}=\frac{4V}{V}=4$, solving for $k$ gives $k=3$.

Answer: 3


Find the area of the region bounded by the graphs of $y=x^{2}, y=x$, and $x=2$.

Solution:
Solution: There are two regions to consider. First, there is the region bounded by $y=x^{2}$ and $y=x$, in the interval $[0,1]$. In this interval, the values of $y=x$ are greater than the values of $y=x^{2}$, thus the area is calculated by $\int_{0}^{1}\left(x-x^{2}\right) d x$.

Second, there is the region bounded by $y=x^{2}$ and $y=x$ and $x=2$, in the interval $[1,2]$. In this interval, the values of $y=x^{2}$ are greater than the values of $y=x$, thus the area is calculated by $\int_{1}^{2}\left(x^{2}-x\right) d x$. Then the total area of the region bounded by the three graphs is $\int_{0}^{1}\left(x-x^{2}\right) d x+\int_{1}^{2}\left(x^{2}-x\right) d x=1$.

Answer: 1


A circle touches the sides of an angle at points $A$ and $B$. The distance from a point $C$ lying on the circle to the line $A B$ is 6. Find the sum of the distances from point $C$ to the sides of the angle, given that one of these distances is 5 more than the other.

Solution:
Answer: 13. (C)

Options.

![](https://cdn.mathpix.com/cropped/2024_05_06_997c53c06135e2de9c16g-04.jpg?height=67&width=712&top_left_y=1394&top_left_x=726)

Answer: 13


A certain musical instrument has 10 holes, sequentially labeled as hole 1, hole 2, $\cdots \cdots$, hole 10. When played, the sound quality index $D$ of the $n$-th hole is given by the relationship
$$
D=n^{2}+k n+90 .
$$

The lowest sound quality index of the instrument is $4 k+106$. Find the range of values for the constant $k$.

Solution:
For the parabola $D=n^{2}+k n+90$, the axis of symmetry is $n=-\frac{k}{2}$.
(1) When $-\frac{k}{2} \leqslant 1$, i.e., $k \geqslant-2$, we have
$$
n=1, D=4 k+106 \text {. }
$$

Thus, $1^{2}+k+90=4 k+106$.
Solving this, we get $k=-5$ (which does not meet the condition).
(2) When $-\frac{k}{2} \geqslant 10$, i.e., $k \leqslant-20$, we have
$$
n=10, D=4 k+106 \text {. }
$$

Thus, $10^{2}+10 k+90=4 k+106$.
Solving this, we get $k=-14$ (which does not meet the condition).
(3) When $1<-\frac{k}{2}<10$, i.e., $-20<k<-2$, $n$ is within the range $-\frac{k}{2}-\frac{1}{2} \leqslant n \leqslant-\frac{k}{2}+\frac{1}{2}$, and $D$ has the lowest value, which is $4 k+106$.
$$
\text { Thus }\left(-\frac{k}{2}+\frac{1}{2}\right)^{2}+k\left(-\frac{k}{2}+\frac{1}{2}\right)+90 \geqslant 4 k+106 \text {. }
$$

Simplifying, we get $(k+7)(k+9) \leqslant 0$.
Solving this, we get $-9 \leqslant k \leqslant-7$.
In conclusion, the range of $k$ is $-9 \leqslant k \leqslant-7$.

Answer: -9 \leqslant k \leqslant-7


Let \(x\) be a variable that can take all real values except 1 and -1.

Provide a way to express the term \(\frac{x}{x^{2}-1}\) as a sum of two fractions such that the variable \(x\) appears only in the denominators of these two fractions and only in the first power!

Solution:
}

It is natural to choose the denominators \(x+1\) and \(x-1\) for the sought quotients, because their least common denominator is equal to the denominator of the given term. Therefore, we choose the ansatz

\[
\frac{x}{x^{2}-1}=\frac{a}{x+1}+\frac{b}{x-1}
\]

This implies

\[
\frac{x}{x^{2}-1}=\frac{a x-a+b x+b}{x^{2}-1}
\]

Thus, after multiplying by \(x^{2}-1\),

\[
x=a x-a+b x+b=x(a+b)-a+b
\]

For \(x=0\), it follows that \(-a+b=0 \quad\) (2) must hold. From (1) and (2), it further follows that \(x=x(a+b)\) and thus, since this must hold for at least one \(x \neq 0\): \(a+b=1 \quad\) (3).

The system of equations (2), (3) has only the solution \(a=b=\frac{1}{2}\). The verification confirms the found decomposition

\[
\frac{1}{2(x+1)}+\frac{1}{2(x-1)}=\frac{x-1+x+1}{2(x+1)(x-1)}=\frac{1}{x^{2}-1}
\]

Answer: \frac{1}{2(x+1)}+\frac{1}{2(x-1)}


The smallest positive period of the function $y=\sin ^{2 n} x-\cos ^{2 m-1} x\left(n,m \in \mathbf{N}^{+}\right)$ is $\qquad$

Solution: 9. $2 \pi$. According to the odd and even function method. It is easy to prove that the minimum positive period of the even function $y=\sin ^{2 n} x$ and the odd function $y=\cos ^{2 \pi+1} x$ are $\pi$ and $2 \pi$, respectively, and their minimum positive period is $2 \pi$.

Answer: 2\pi


Find all possible values of the quotient

$$
\frac{r+\rho}{a+b}
$$

where $r$ and $\rho$ are respectively the radii of the circumcircle and incircle of the right triangle with legs $a$ and $b$.

Solution:
Solution. The distances between the vertices and points of tangency of the incircle (denoted according to the figure) of a triangle $A B C$ are

$$
|A U|=|A V|=\frac{b+c-a}{2}, \quad|B V|=|B T|=\frac{a+c-b}{2}, \quad|C T|=|C U|=\frac{a+b-c}{2},
$$

which is easy to obtain from

$$
|A V|+|B V|=c, \quad|A U|+|C U|=b, \quad|B T|+|C T|=a .
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_10b455106c778bb0ba10g-09.jpg?height=320&width=404&top_left_y=1436&top_left_x=498)

Fig. 1

The points $C, T, U$, and $S$ (the center of the incircle) form a deltoid, actually a square, if the angle $A C B$ is right. The side of the square is $\rho=|S U|=|C U|$. This gives

$$
\rho=\frac{a+b-c}{2}
$$

moreover according to the Thales theorem $r=\frac{1}{2} c$ and we get

$$
r+\rho=\frac{c}{2}+\frac{a+b-c}{2}=\frac{a+b}{2} .
$$

Conclusion. There is only one possible value of the quotient $(r+\rho) /(a+b)$ in any right triangle, namely $\frac{1}{2}$.

Answer: \frac{1}{2}


In rectangle $ABCD$, $AB=2, AD=1$, point $P$ on side $DC$ (including points $D, C$) and point $Q$ on the extension of $CB$ (including point $B$) satisfy $|\overrightarrow{DP}|=|\overrightarrow{BQ}|$, then the dot product $\overrightarrow{PA} \cdot \overrightarrow{PQ}$ of vector $\overrightarrow{PA}$ and vector $\overrightarrow{PQ}$ has a minimum value of $\qquad$.

Solution:
Let $\overrightarrow{P A}=\overrightarrow{P D}+\overrightarrow{D A}=-t \overrightarrow{D C}+\overrightarrow{D A}$,
then we have $\overrightarrow{P Q}=\overrightarrow{P C}+\overrightarrow{C Q}=(1-t) \overrightarrow{D C}+(1+2 t) \overrightarrow{D A}$.
Thus, $\overrightarrow{P A} \cdot \overrightarrow{P Q}=t(t-1) \overrightarrow{D C}^{2}+(1+2 t) \overrightarrow{D A}^{2}=4 t(t-1)+1+2 t$
$=4 t^{2}-2 t+1=4\left(t-\frac{1}{4}\right)^{2}+\frac{3}{4} \geqslant \frac{3}{4}$, equality holds when $\overrightarrow{D P}=\frac{1}{4} \overrightarrow{D C}$.
Therefore, the minimum value of the dot product $\overrightarrow{P A} \cdot \overrightarrow{P Q}$ of vector $\overrightarrow{P A}$ and vector $\overrightarrow{P Q}$ is $\frac{3}{4}$.

Answer: \frac{3}{4}


Find all mappings $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, satisfying
$$
f(f(x))=9 x, f(f(f(x)))=27 x ;
$$
Find all mappings $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, satisfying
$$
f(f(x))=9 x .
$$

Solution:
(1) For any positive integer $x$,
$9 f(x)=f(f(f(x)))=27 x \Rightarrow f(x)=3 x$.
Thus, the solution is $f(x)=3 x$.
(2) Let the set
$M=\{x \mid x$ is a positive integer not divisible by 9 $\}$.
Take infinite sets $A$ and $B$ such that
$A \cap B=\varnothing, A \cup B=M$.
Take a bijection $u$ from set $A$ to set $B$.
For any positive integer $x$, let $x=9^{m} t, 9 \nmid t$.
If $t \in A$, take $f(x)=9^{m} u(t)$;
If $t \in B$, take $f(x)=9^{m+1} u^{-1}(t)$.
It is easy to verify that such $f(x)$ satisfies $f(f(x))=9 x$.
Next, we show that all $f(x)$ satisfying $f(f(x))=9 x$ can be represented by the above $(A, B, u)$.
First, we prove: $f$ is injective.
For any different positive integers $x, y$, we have
$f(f(x))=9 x \neq 9 y=f(f(y))$
$\Rightarrow f(x) \neq f(y)$.
Consider the sets $C=\{x \mid 9 \nmid f(x), x \in M\}$,
$D=\{x \mid 9 \mid f(x), x \in M\}$,

The sets $C$ and $D$ satisfy $C \cap D=\varnothing, C \cup D=M$.
For any $x \in C$, take $u(x)=f(x)$.
Since $f(u(x))=f(f(x))=9 x$, then $u(x) \in D$, and $u$ is a mapping from set $C$ to set $D$.
For any different elements $x, y$ in set $C$,
$f(u(x))=9 x \neq 9 y=f(u(y))$
$\Rightarrow u(x) \neq u(y) \Rightarrow u$ is injective.
For any element $x$ in set $D$, let $f(x)=9 t, f(x)=f(f(t))$.
By the injectivity of $f$, we know $f(t)=x$.
If $9 \mid t$, let $t=9 s$, then
$t=9 s=f(f(s))$,
$x=f(f(f(s)))=9 f(s)$,

which contradicts $x \in D$.
Thus, $t \in M$, and $f(t)=x$ is not divisible by 9, so $t \in C, u(t)=x$.
Therefore, $u$ is surjective, and $u$ is a bijection.
Hence, $(C, D, u)$ can yield a mapping $g$ that satisfies the conditions, and it is easy to verify that $f(x)=g(x)$ for any positive integer $x$.

In summary, all $f(x)$ satisfying the conditions correspond to a group $(A, B, u)$, and a group $(A, B, u)$ can be used to construct a function $f(x)$ that meets the requirements as described above.

Answer: f(x)=3x


In the right trapezoid $A B C D$, $A B / / D C, \angle B=90^{\circ}$, a moving point $P$ starts from point $B$ and moves along the sides of the trapezoid in the order $B \rightarrow C \rightarrow D \rightarrow A$. Let the distance traveled by $P$ be $x$, and the area of $\triangle A B P$ be $y$. Consider $y$ as a function of $x$, and the graph of the function is shown in Figure 3.
Then the area of $\triangle A B C$ is ( ).
(A) 10
(B) 16
(C) 18
(D) 32

Solution:
4. B.

According to the image, we get $B C=4, C D=5, D A=5$. Then we find $A B=8$. Therefore, $S_{\triangle A B C}=\frac{1}{2} \times 8 \times 4=16$.

Answer: B


The diagonals of quadrilateral $ABCD$ intersect at point $O$, $M$ and $N$ are the midpoints of sides $BC$ and $AD$ respectively. Segment $MN$ divides the area of the quadrilateral in half. Find the ratio $OM: ON$, if $AD=2BC$.

Solution:
$N M$ - the median of triangle $B N C$ (see fig.), therefore $S_{B M N}=S_{C M N}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-29.jpg?height=429&width=757&top_left_y=63&top_left_x=656)

From the condition, it follows that $S_{A B N}=S_{D C N}$. Moreover, triangles $A B N$ and $D C N$ have equal sides $A N$ and $D N$, so their heights $B H$ and $C T$ are also equal. Therefore, $A D \| B C$, which means $A B C D$ is a trapezoid.

Thus, point $O$ lies on segment $M N$, and triangles $A O D$ and $C O B$ are similar with a ratio of $A D / B C = 2$. Since $O N$ and $O M$ are the corresponding medians of these triangles, $O M: O N = 1: 2$.

## Answer

$1: 2$.

Answer: 1:2


Solution:
7.58 Here it should be $-x>0$, hence $x<0$. Then $\lg x^{2}=2 \lg (-x)$ and the given equation will take the form $4 \lg (-x)-\lg ^{2}(-x)=4$.

Letting $\lg (-x)=y$, we find $y=2$, hence $\lg (-x)=2$, i.e. $-x=100, x=-100$.

Answer: $x=-100$.

Answer: -100


Given that the sum of 3 distinct natural numbers is 55, and the sum of any two of these numbers is a perfect square, then these three natural numbers are

Solution:
【Analysis and Solution】
Three natural numbers are all different, so the three sums of any two of them, which are perfect squares, are also different;
Adding the three perfect squares is equivalent to twice the sum of these three natural numbers;
Therefore, decompose $55 \times 2=110$ into the sum of three perfect squares;
The perfect squares not exceeding 110 are the following 11: $0^{2}=0, 1^{2}=1, 2^{2}=4, 3^{2}=9, 4^{2}=16, 5^{2}=25, 6^{2}=36, 7^{2}=49, 8^{2}=64, 9^{2}=81, 10^{2}=100$; $110 \div 3=36 \frac{2}{3}$, so there must be a perfect square greater than $36 \frac{2}{3}$;
Since the sum of the three different natural numbers is 55, these three perfect squares do not exceed 55; hence, one of the perfect squares must be $7^{2}=49$;
$110=49+61=49+36+25$, that is, these three perfect squares are $49, 36, 25$; $55-49=6, 55-36=19, 55-25=30$, that is, these three natural numbers are $6, 19, 30$.

Answer: 6,19,30


On a circular route, two buses operate with the same speed and a movement interval of 21 minutes. What will be the movement interval if 3 buses operate on this route at the same constant speed?

Solution:
Answer: 14.

Solution: Since the interval of movement with two buses on the route is 21 minutes, the length of the route in "minutes" is 42 minutes. Therefore, the interval of movement with three buses on the route is $42: 3=14$ minutes.

Criteria: only answer, answer with verification - 3 points.

Answer: 14


The $ABCDEFGH$ cube with unit edge has a point $P$ that trisects the face diagonal $BE$ closer to $E$. How far is $P$ from the plane of the triangle $CFH$?

![](https://cdn.mathpix.com/cropped/2024_05_02_3dda5ed74f40e5497fecg-1.jpg?height=309&width=331&top_left_y=255&top_left_x=886)

Solution:
I. solution. Rotate the cube $120^{\circ}$ around the body diagonal $A G$. We know that in this case it maps onto itself, similarly, the equilateral triangles $H F C$ and $E D B$ also map onto themselves.

![](https://cdn.mathpix.com/cropped/2024_05_02_f4872318468eb4c287aag-1.jpg?height=385&width=401&top_left_y=230&top_left_x=865)

Figure 1

Therefore, the plane $E D B$ is parallel to the plane $H F C$, so instead of determining the distance of point $P$ from the plane $H F C$, it is sufficient to determine the distance between the two parallel planes. The centroids of the equilateral triangles $H F C$ and $E D B$ coincide with their centers of gravity and lie on the body diagonal $A G$, which is perpendicular to both planes. Let's calculate the distance between these two centroids. Let the vectors from $A$ to points $B, D$, and $E$ be $\mathbf{x}, \mathbf{y}$, and $\mathbf{z}$, and the centroids be $S_{1}$ and $S_{2}$ (Figure 1).

According to the known relationship,

$$
\begin{gathered}
\overrightarrow{A G}=\mathbf{x}+\mathbf{y}+\mathbf{z}, \quad \overrightarrow{A S_{1}}=\frac{\mathbf{x}+\mathbf{y}+\mathbf{z}}{3} \\
\overrightarrow{A S_{2}}=\frac{\overrightarrow{A C}+\overrightarrow{A F}+\overrightarrow{A H}}{3}=\frac{(\mathbf{x}+\mathbf{y})+(\mathbf{x}+\mathbf{z})+(\mathbf{y}+\mathbf{z})}{3}=\frac{2(\mathbf{x}+\mathbf{y}+\mathbf{z})}{3}
\end{gathered}
$$

Thus,

$$
\overrightarrow{S_{1} S_{2}}=\frac{\mathbf{x}+\mathbf{y}+\mathbf{z}}{3}=\frac{1}{3} \overrightarrow{A G}
$$

This implies that the distance between the centroids is $\frac{1}{3}$ of the body diagonal's length, which is $\frac{\sqrt{3}}{3}$.

Henrietta Somlai (Pápa, Reformed Gymnasium, 11th grade)

II. solution. Place the cube in a three-dimensional coordinate system such that the vertex $D$ is at the origin, and the face $A B C D$ lies in the $x, y$ plane in the first octant (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_f4872318468eb4c287aag-1.jpg?height=395&width=491&top_left_y=1483&top_left_x=814)

Figure 2

Write down the coordinates of points $H, F, C$, and $P$:

$$
H(0 ; 0 ; 1), \quad F(1 ; 1 ; 1), \quad C(0 ; 1 ; 0), \quad P\left(1 ; \frac{1}{3} ; \frac{2}{3}\right)
$$

Next, write down the equation of the plane passing through points $H, F, C$. Its general form is

$$
a x+b y+c z+d=0
$$

Substituting the coordinates of points $H, F$, and $C$:

$$
c+d=0, \quad a+b+c+d=0, \quad b+d=0, \quad C:
$$

From the system of equations, we get $a=d, b=-d, c=-d$, so if $d \neq 0$, the equation of the plane passing through points $H, F, C$ is:

$$
d x-d y-d z+d=0
$$

Dividing by $d \neq 0$ gives

$$
x-y-z+1=0
$$

According to the known relationship, the distance of point $P\left(x_{1} ; y_{1} ; z_{1}\right)$ from the plane $a x+b y+c z+d=0$ is

$$
\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}
$$

Substituting, the distance is $\frac{\left|1-\frac{1}{3}-\frac{2}{3}+1\right|}{\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$, as we also saw in the first solution.

András L. László (Veszprém, Industrial and Grammar School, 10th grade)

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


A wristwatch is 5 minutes slow per hour; 5.5 hours ago they were set to the correct time. Now, on the clock showing the correct time, it is 1 PM. In how many minutes will the wristwatch show 1 PM?

Solution: 12. In 60 minutes, the hand of the watch "counts" 55 minutes. They need to "count" 5.5 hours. They will do this in 6 hours $\left(\frac{5.5 \cdot 60}{55}=6\right)$. Therefore, 1 hour of the day they will show after 0.5 hours or 30 minutes $(6-5.5=0.5)$.

Answer: 30


$P$ is a point inside $\triangle A B C$, the line $A C$ and $B P$ intersect at $Q$, the line $A B$ and $C P$ intersect at $R$. Given that $A R=R B=C P, C Q=P Q$. Try to find the size of $\angle B R C$.

Solution:
Analysis: Combining the diagram, guess first and then prove.
Solution: Let $S$ be a point on line segment $CR$ such that $RS = CP$.
Since $CQ = PQ$,
we have $\angle ACS = \angle QPC = \angle BPR$.
Since $RS = CP$,
we have $SC = CR - RS = CR - CP = RP$.
Considering $\triangle ABQ$ and line $CR$, by Menelaus' theorem, we have
$$
\frac{AC}{CQ} \cdot \frac{QP}{PB} \cdot \frac{BR}{RA} = 1
$$

Since $AR = RB$ and $CQ = PQ$,
we have $AC = BP$.
Therefore, $\triangle ACS \cong \triangle BPR$.
Thus, $AS = BR$.
Since $RS = CP$ and $AR = RB = CP$,
we have $AS = AR = RS$.
Therefore, $\angle ARS = 60^{\circ}$.
Thus, $\angle BRC = 120^{\circ}$.

Answer: 120


Find all five-digit numbers where the second digit is five times the first, and the product of all five digits is 1000.

Solution: Answer: 15855, 15585, 15558. Solution: The first digit can only be 1, otherwise the second digit will be greater than 9. Then the second digit is 5, and it follows that the product of the last three digits must equal 200. Since $200=5 \times 5 \times 8$, two of these three digits must be 5. But then the third digit must be 8, which gives us the answer.

Answer: 15855,15585,15558


The following problem is presented without the prefix and with any necessary corrections:

We break a thin wooden stick into 3 pieces. What is the probability that we can form a triangle with these pieces?

Solution:
Divide the stick of length $2n$ into $2n$ equal parts and calculate the sought probability first in the case where the stick can only be broken at the generated division points.

Let $x$, $y$, and $z$ be the lengths of the three parts of the stick; for these parts to form a triangle, the following inequalities must hold:

$$
xn
$$

These inequalities are only satisfied by the following value pairs:

$$
\begin{aligned}
& \text { If } x=2, \quad \text { then } y=n-1, \\
& \text { If } x=3, \quad \text { then } y=n-1, \quad \text { or } n-2, \\
& \text { If } x=4, \quad \text { then } y=n-1, \quad \text { or } n-2, \quad \text { or } n-3, \\
& \ldots . . . \quad \ldots . . . \\
& \ldots . . . . . . . \\
& \text { If } x=n-1, \quad \text { then } y=n-1, \quad \text { or } n-2, \quad \ldots \text { or } 3, \quad \text { or } 2 .
\end{aligned}
$$

The number of favorable cases is therefore:

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

The total number of possible cases is as follows:

$$
\begin{gathered}
x=1 \text { and } y=1,2,3, \ldots, 2 n-2 \\
x=2 \text { and } y=1,2,3, \ldots, 2 n-3 \text { etc. }
\end{gathered}
$$

Thus, their number is:

$$
2 n-2+2 n-3+2 n-4+\ldots 2+1=\frac{(2 n-2)(2 n-1)}{2}
$$

And so the sought probability is:

$$
v=\frac{(n-2)(n-1)}{(2 n-2)(2 n-1)}
$$

If we now take $n$ and thus the number of division points to be $\infty$, meaning the stick can break at any point, then, since $v$ (dividing the numerator and the denominator by $n^{2}$) can also be written as:

$$
\frac{1-\frac{3}{n}+\frac{2}{n^{2}}}{4-\frac{6}{n}+\frac{2}{n^{2}}}
$$

the sought probability is: $\frac{1}{4}$.

(Dénes König, Budapest.)

The problem was also solved by: Aczél V., Bartók I., Bayer B., Lázár L., Póka Gy., Selényi M., Szmodics K.

Answer: \frac{1}{4}


If the three lateral edges of the tetrahedron $S-ABC$ are pairwise perpendicular, and $O$ is a point within the base $\triangle ABC$, then
$$
\mu=\tan \angle OSA \cdot \tan \angle OSB \cdot \tan \angle OSC
$$

the range of values for $\mu$ is $\qquad$

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

From $\cos ^{2} \angle O S A+\cos ^{2} \angle O S B+\cos ^{2} \angle O S C=1$
$$
\begin{array}{l}
\Rightarrow \sin ^{2} \angle O S C=\cos ^{2} \angle O S A+\cos ^{2} \angle O S B \\
\quad \geqslant 2 \cos \angle O S A \cdot \cos \angle O S B .
\end{array}
$$

Similarly, we can obtain the other two inequalities.
Therefore, $\mu \geqslant 2 \sqrt{2}$.

Answer: \mu \geqslant 2 \sqrt{2}


Given $\log _{\sqrt{7}}(5 a-3)=\log _{\sqrt{a^{2}+1}} 5$. Then the real number
$$
a=
$$
. $\qquad$

Solution:
2. 2 .

Simplify the original equation to
$$
\log _{7}(5 a-3)=\log _{a^{2}+1} 5 \text {. }
$$

Since $f(x)=\log _{7}(5 x-3)$ is an increasing function for $x>\frac{3}{5}$, and $g(x)=\log _{5}\left(x^{2}+1\right)$ is also an increasing function, and $f(2)=$ $g(2)=1$, therefore, $a=2$.

Answer: 2


Find the value of the expression $\frac{a}{b}+\frac{b}{a}$, where $a$ and $b$ are the largest and smallest roots of the equation $x^{3}-7 x^{2}+7 x=1$, respectively.

Solution:
Answer: 34.

Solution. The given equation is equivalent to the following

$$
\left(x^{3}-1\right)-7\left(x^{2}-x\right)=0 \Leftrightarrow(x-1)\left(x^{2}+x+1\right)-7 x(x-1)=0 \Leftrightarrow(x-1)\left(x^{2}-6 x+1\right)=0,
$$

from which $x=1$ or $x=3 \pm \sqrt{8}$. The largest root is $a=3+\sqrt{8}$, and the smallest is $-b=3-\sqrt{8}$. Then

$$
\frac{a}{b}+\frac{b}{a}=\frac{3+\sqrt{8}}{3-\sqrt{8}}+\frac{3-\sqrt{8}}{3+\sqrt{8}}=\frac{(3+\sqrt{8})^{2}+(3-\sqrt{8})^{2}}{(3+\sqrt{8})(3-\sqrt{8})}=\frac{2(9+8)}{1}=34
$$

Answer: 34


The car traveled half of the distance at a speed of 60 km/h, then one third of the remaining distance at a speed of 120 km/h, and the remaining distance at a speed of 80 km/h.

Find the average speed of the car during this trip. Give your answer in km/h.

Solution:
Solution: Let x hours be the time the car traveled at a speed of 60 km/h, then $60 x=\frac{s}{2}$. Let y hours be the time the car traveled at a speed of 120 km/h, then $120 y=\frac{s}{6}$. Let z hours be the time the car traveled at a speed of 80 km/h, then $80 z=\frac{s}{3}$. By definition $v_{cp}=\frac{s}{t_{\text{total}}}$. Then $v_{cp}=\frac{s}{x+y+z}=\frac{s}{\frac{s}{120}+\frac{s}{720}+\frac{s}{240}}=\frac{1}{\frac{1}{120}+\frac{1}{6 \cdot 120}+\frac{1}{2 \cdot 120}}=\frac{6 \cdot 120}{10}=72$.

Answer: 72.

Answer: 72


For $Z$ being a complex number, $M=\left\{Z\left|(Z-1)^{2}=\right| Z-\left.1\right|^{2}\right\}$, then
(A) $M=\{$ pure imaginary numbers $\}$
(B) $M=\{$ real numbers $\}$
(C) $\{$ real numbers $\} \subset M \subset\{$ complex numbers $\}$
(D) $M=\{$ complex numbers $\}$

Solution:
(B)
2.【Analysis and Solution】 $\because(Z-1)^{2}=|Z-1|^{2}$, i.e., $(Z-1)^{2}=(Z-1)(\bar{Z}-1)$, $\because(Z-1)(Z-\bar{Z})=0$, therefore $Z=1$ or $Z=\bar{Z}$. That is, $Z$ is a real number, the answer is (B).

Answer: B


Find the value of $n$ for which the following equality holds:

$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2015
$$

Solution:
Answer: 4064255

Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2015$. From which $n=(2015+1)^{2}-1=4064255$.

Answer: 4064255


Ten football teams each played against each other once. As a result, each team ended up with exactly $x$ points.

What is the greatest possible value of $x$? (Win - 3 points, draw - 1 point, loss - 0.)

Solution:
Evaluation. In each match, either 3 or 2 points are awarded. A total of $10 \cdot 9: 2=45$ matches were played, so no more than 135 points were awarded. Thus, $10 x \leq 135$, which means $x \leq 13.5$. Since $x$ is an integer, $x \leq$ 13.

Example. Arrange the teams in a circle and sequentially divide them into 5 pairs. Let the teams in each pair play to a draw, each of them winning against the four teams following the given pair in a clockwise direction, and losing to the remaining teams. Then each team scored exactly 13 points.

## Answer

13.

Answer: 13