The sum of positive numbers $a, b, c$ and $d$ is 4. Find the minimum value of the expression

$$
\frac{a^{8}}{\left(a^{2}+b\right)\left(a^{2}+c\right)\left(a^{2}+d\right)}+\frac{b^{8}}{\left(b^{2}+c\right)\left(b^{2}+d\right)\left(b^{2}+a\right)}+\frac{c^{8}}{\left(c^{2}+d\right)\left(c^{2}+a\right)\left(c^{2}+b\right)}+\frac{d^{8}}{\left(d^{2}+a\right)\left(d^{2}+b\right)\left(d^{2}+c\right)}
$$

Solution:
Answer: $\frac{1}{2}$

First solution. By the inequality of means for four numbers, we have

$$
\begin{aligned}
& \frac{a^{8}}{\left(a^{2}+b\right)\left(a^{2}+c\right)\left(a^{2}+d\right)}+\frac{a^{2}+b}{16}+\frac{a^{2}+c}{16}+\frac{a^{2}+d}{16} \geqslant \\
& \quad \geqslant 4 \sqrt[4]{\frac{a^{8}}{\left(a^{2}+b\right)\left(a^{2}+c\right)\left(a^{2}+d\right)} \cdot \frac{a^{2}+b}{16} \cdot \frac{a^{2}+c}{16} \cdot \frac{a^{2}+d}{16}}=\frac{a^{2}}{2}
\end{aligned}
$$

Therefore,

$$
\frac{a^{8}}{\left(a^{2}+b\right)\left(a^{2}+c\right)\left(a^{2}+d\right)} \geqslant \frac{a^{2}}{2}-\left(\frac{a^{2}+b}{16}+\frac{a^{2}+c}{16}+\frac{a^{2}+d}{16}\right)=\frac{5 a^{2}}{16}-\frac{b+c+d}{16}
$$

Summing this inequality with three similar ones, we get

$$
\begin{gathered}
\frac{a^{8}}{\left(a^{2}+b\right)\left(a^{2}+c\right)\left(a^{2}+d\right)}+\frac{b^{8}}{\left(b^{2}+c\right)\left(b^{2}+d\right)\left(b^{2}+a\right)}+\frac{c^{8}}{\left(c^{2}+d\right)\left(c^{2}+a\right)\left(c^{2}+b\right)}+\frac{d^{8}}{\left(d^{2}+a\right)\left(d^{2}+b\right)\left(d^{2}+c\right)} \geqslant \\
\geqslant\left(\frac{5 a^{2}}{16}-\frac{b+c+d}{16}\right)+\left(\frac{5 b^{2}}{16}-\frac{c+d+a}{16}\right)+\left(\frac{5 c^{2}}{16}-\frac{d+a+b}{16}\right)+\left(\frac{5 d^{2}}{16}-\frac{a+b+c}{16}\right)= \\
=\frac{5\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}{16}-\frac{3(a+b+c+d)}{16}=\frac{5\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}{16}-\frac{3}{4} \geqslant \frac{5}{4}-\frac{3}{4}=\frac{1}{2} .
\end{gathered}
$$

The last inequality is true since $4\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \geqslant(a+b+c+d)^{2}=16$, which can be verified, for example, by expanding the brackets.

If $a=b=c=d=1$, then the sum of the fractions in the problem condition is $\frac{1}{2}$, so the minimum value of the expression is $\frac{1}{2}$.

Second solution. For brevity, let

$$
K=\frac{a^{8}}{\left(a^{2}+b\right)\left(a^{2}+c\right)\left(a^{2}+d\right)}+\frac{b^{8}}{\left(b^{2}+c\right)\left(b^{2}+d\right)\left(b^{2}+a\right)}+\frac{c^{8}}{\left(c^{2}+d\right)\left(c^{2}+a\right)\left(c^{2}+b\right)}+\frac{d^{8}}{\left(d^{2}+a\right)\left(d^{2}+b\right)\left(d^{2}+c\right)}
$$

By the Cauchy-Bunyakovsky inequality for the sets of numbers

$$
\begin{aligned}
& \frac{a^{8}}{\left(a^{2}+b\right)\left(a^{2}+c\right)\left(a^{2}+d\right)}, \frac{b^{8}}{\left(b^{2}+c\right)\left(b^{2}+d\right)\left(b^{2}+a\right)}, \frac{c^{8}}{\left(c^{2}+d\right)\left(c^{2}+a\right)\left(c^{2}+b\right)}, \frac{d^{8}}{\left(d^{2}+a\right)\left(d^{2}+b\right)\left(d^{2}+c\right)} \text { and } \\
& a^{2}+d, \quad b^{2}+a, \quad c^{2}+b, \quad d^{2}+c
\end{aligned}
$$

we have

$$
\begin{aligned}
& \left(a^{2}+b^{2}+c^{2}+d^{2}+4\right) K=\left(\left(a^{2}+d\right)+\left(b^{2}+a\right)+\left(c^{2}+b\right)+\left(d^{2}+c\right)\right) K \geqslant \\
& \geqslant\left(\frac{a^{4}}{\sqrt{\left(a^{2}+b\right)\left(a^{2}+c\right)}}+\frac{b^{4}}{\sqrt{\left(b^{2}+c\right)\left(b^{2}+d\right)}}+\frac{c^{4}}{\sqrt{\left(c^{2}+d\right)\left(c^{2}+a\right)}}+\frac{d^{4}}{\sqrt{\left(d^{2}+a\right)\left(d^{2}+b\right)}}\right)^{2}
\end{aligned}
$$

Let the last expression in the parentheses be denoted by $L$ and estimate it by the Cauchy-Bunyakovsky inequality for the sets

$$
\begin{aligned}
& \frac{a^{4}}{\sqrt{\left(a^{2}+b\right)\left(a^{2}+c\right)}}, \frac{b^{4}}{\sqrt{\left(b^{2}+c\right)\left(b^{2}+d\right)}}, \quad \frac{c^{4}}{\sqrt{\left(c^{2}+d\right)\left(c^{2}+a\right)}}, \quad \frac{d^{4}}{\sqrt{\left(d^{2}+a\right)\left(d^{2}+b\right)}} \text { and } \\
& \sqrt{\left(a^{2}+b\right)\left(a^{2}+c\right)}, \quad \sqrt{\left(b^{2}+c\right)\left(b^{2}+d\right)}, \quad \sqrt{\left(c^{2}+d\right)\left(c^{2}+a\right)}, \quad \sqrt{\left(d^{2}+a\right)\left(d^{2}+b\right)}
\end{aligned}
$$

Then $L M \geqslant\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}$, where

$$
M=\sqrt{\left(a^{2}+b\right)\left(a^{2}+c\right)}+\sqrt{\left(b^{2}+c\right)\left(b^{2}+d\right)}+\sqrt{\left(c^{2}+d\right)\left(c^{2}+a\right)}+\sqrt{\left(d^{2}+a\right)\left(d^{2}+b\right)}
$$

Thus,

$$
K \geqslant \frac{L^{2}}{a^{2}+b^{2}+c^{2}+d^{2}+4} \geqslant \frac{1}{a^{2}+b^{2}+c^{2}+d^{2}+4} \cdot \frac{\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{4}}{M^{2}}
$$

Finally, by the inequality of means for two numbers $\sqrt{\left(x^{2}+y\right)\left(x^{2}+z\right)} \leqslant \frac{\left(x^{2}+y\right)+\left(x^{2}+z\right)}{2}$, therefore

$$
M \leqslant \frac{2 a^{2}+b+c}{2}+\frac{2 b^{2}+c+d}{2}+\frac{2 c^{2}+d+a}{2}+\frac{2 d^{2}+a+b}{2}=a^{2}+b^{2}+c^{2}+d^{2}+4
$$

Thus,

$$
K \geqslant \frac{\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{4}}{\left(a^{2}+b^{2}+c^{2}+d^{2}+4\right)^{3}}=\frac{a^{2}+b^{2}+c^{2}+d^{2}}{\left(1+\frac{4}{a^{2}+b^{2}+c^{2}+d^{2}}\right)^{3}}=
$$

It remains to note that by the Cauchy-Bunyakovsky inequality for the sets $a, b, c, d$ and 1,1 , 1,1

$$
4\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \geqslant(a+b+c+d)^{2}=16
$$

therefore $a^{2}+b^{2}+c^{2}+d^{2} \geqslant 4$. Consequently, the numerator in the right-hand side of (*) is at least 4, and the denominator is at most $2^{3}$, so $K \geqslant \frac{1}{2}$. If $a=b=c=d=1$, then $K=\frac{1}{2}$, so the minimum value of $K$ is $\

Answer: \frac{1}{2}


Find the smallest positive integer n so that a cube with side n can be divided into 1996 cubes each with side a positive integer. Solution

Solution: : 13. Divide all the cubes into unit cubes. Then the 1996 cubes must each contain at least one unit cube, so the large cube contains at least 1996 unit cubes. But 12 3 = 1728 < 1996 < 2197 = 13 3 , so it is certainly not possible for n < 13. It can be achieved with 13 by 1.5 3 + 11.2 3 + 1984.1 3 = 13 3 (actually packing the cubes together to form a 13 x 13 x 13 cube is trivial since there are so many unit cubes). 11th Ibero 1996 © John Scholes jscholes@kalva.demon.co.uk 22 Oct 2000

Answer: 13


In a non-decreasing sequence of positive integers $a_{1}, a_{2}, \cdots, a_{m}, \cdots$, for any positive integer $m$, define $b_{m}=\min \left\{n \mid a_{n} \geqslant m\right\}$. It is known that $a_{19}=85$. Find the maximum value of $S=a_{1}+a_{2}+\cdots+a_{19}+b_{1}+b_{2}+\cdots+b_{85}$.

Solution: 2. First, the largest number must exist. From the condition, we have $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{19}=85$. We conjecture that the extremum point is when each $a_{i}$ is as large as possible and all $a_{i}$ are equal, and all $b_{j}$ are equal. In fact, if there is $a_{i}a_{i}$, so $a_{i+1} \geqslant a_{i}+1$. But $a_{i}<a_{i}+1$, so $b_{a_{i}+1}=i+1$, $b_{a_{i}+1}^{\prime}=i=b_{a_{i}+1}-1, b_{j}^{\prime}=b_{j}$ (when $j \neq a_{i}+1$). From this, we can see that the adjustment makes $b_{a_{i}+1}$ decrease by 1, and the rest of the $b_{i}$ remain unchanged. Thus, the value of $S$ does not decrease. In summary, we have $S \leqslant 19 \times 85 + 1 \times 85 = 1700$, with equality when $a_{i}=85, b_{j}=1(1 \leqslant i \leqslant 19,1 \leqslant j \leqslant 85)$.

Answer: 1700


There are 9 representatives, each from different countries, with 3 people from each country. They randomly sit at a round table with 9 chairs. The probability that each representative is seated next to at least one representative from another country is $\qquad$ .

Solution:
8. $\frac{41}{56}$.
9 The number of ways to arrange the chairs by country is the multinomial coefficient $\frac{9!}{3!3!3!}=1680$. The required probability is 1 minus the probability that at least three representatives from one country sit together.

Let the sets of seating arrangements where the representatives from the first, second, and third countries sit together be $A, B, C$ respectively. By the principle of inclusion-exclusion and the symmetry of sets $A, B, C$, we have
$$
\begin{aligned}
|A \cup B \cup C| & =(|A|+|B|+|C|)-(|A \cap B|+|A \cap C|+|B \cap C|)+|A \cap B \cap C| \\
& =3|A|-3|A \cap B|+|A \cap B \cap C| .
\end{aligned}
$$

To determine the number of elements in $A$, we can choose a chair for the rightmost representative from the first country. The two chairs to the left of this chair are also occupied by representatives from the first country. The remaining 6 chairs are then arranged for the representatives from the other two countries. Thus, $|A|=9 \mathrm{C}_{6}^{3}$.

To determine the number of elements in $A \cap B$, we can choose a chair for the rightmost representative from the first country. The two chairs to the left of this chair are also occupied by representatives from the first country. From the remaining 6 chairs, the rightmost representative from the second country can choose from 4 chairs, and all positions of the representatives are then determined. Thus, $|A \cap B|=9 \times 4$.

To determine the number of elements in $A \cap B \cap C$, we can choose a chair for the rightmost representative from the first country. The two chairs to the left of this chair are also occupied by representatives from the first country. From the remaining 6 chairs, the seating arrangements for the representatives from the other two countries have 2 possibilities. Thus, $|A \cap B \cap C|=9 \times 2$.
In summary, $|A \cup B \cup C|=3\left(9 \mathrm{C}_{6}^{3}\right)-3(9 \times 4)+9 \times 2=450$.
The required probability is $1-\frac{|A \cup B \cup C|}{1680}=1-\frac{450}{1680}=\frac{41}{56}$.

Answer: \frac{41}{56}


Given a parallelogram $ABCD$. A line passing through vertex $C$ intersects lines $AB$ and $AD$ at points $K$ and $L$. The areas of triangles $KBC$ and $CDL$ are $p$ and $q$. Find the area of parallelogram $ABCD$.

Solution:
The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.

## Solution

Consider the case when point $L$ lies on side $A D$. Let $S$ be the area of triangle $AKL$. Then the similarity coefficient of triangles $A K L$ and $B K C$ is $\frac{\sqrt{S}}{\sqrt{p}}$, and for triangles $A K L$ and $D C L$ it is $\frac{\sqrt{S}}{\sqrt{p}-\sqrt{S}}$. Therefore,

$$
S=q\left(\frac{\sqrt{S}}{\sqrt{p}-\sqrt{S}}\right)^{2}
$$

From this, we find that

$$
q=(\sqrt{p}-\sqrt{S})^{2}, \sqrt{S}=\sqrt{p}-\sqrt{q}
$$

Thus,

$$
S_{\mathrm{ABCD}}=p-S+q=p-(\sqrt{p}-\sqrt{q})^{2}+q=2 \sqrt{p q}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-26.jpg?height=352&width=879&top_left_y=744&top_left_x=592)

## Answer

## $2 \sqrt{p q}$.

Answer: 2\sqrt{pq}


For the equation $(1984 x)^{2}-1983 \times 1985 x-1$ $=0$, the larger root is $r$, and for the equation $x^{2}+1983 x-1984=0$, the smaller root is $s$. What is $r-s$?

Solution:
Solve $\because 1984^{2}+(-1983) \times 1985$ $+(-1)=0$, according to property 1, we can get the roots of the first equation $x_{1}=1, \quad x_{2}=-\frac{1}{1984^{2}}$.

Also $\because 1+1983-1984=0$, we can get the roots of the second equation $x_{1}^{\prime}=1, x_{2}^{\prime}=-1984$.
$$
\text { Therefore } r-s=1985 \text {. }
$$

Answer: 1985


Given $\boldsymbol{a}=\left(\cos \frac{2}{3} \pi, \sin \frac{2}{3} \pi\right), \overrightarrow{O A}=\boldsymbol{a}-\boldsymbol{b}, \overrightarrow{O B}=\boldsymbol{a}+\boldsymbol{b}$, if $\triangle O A B$ is an isosceles right triangle with $O$ as the right-angle vertex, then the area of $\triangle O A B$ is $\qquad$

Solution:
4. 1 Detailed Explanation: Let vector $\boldsymbol{b}=(x, y)$, then $\left\{\begin{array}{l}(\boldsymbol{a}+\boldsymbol{b})(\boldsymbol{a}-\boldsymbol{b})=0 \\ |\boldsymbol{a}+\boldsymbol{b}|=|\boldsymbol{a}-\boldsymbol{b}|\end{array}\right.$,
$$
\begin{array}{l}
\text { i.e., }\left\{\begin{array} { l }
{ ( x - \frac { 1 } { 2 } , y + \frac { \sqrt { 3 } } { 2 } ) \cdot ( - x - \frac { 1 } { 2 } , - y + \frac { \sqrt { 3 } } { 2 } ) = 0 } \\
{ ( x - \frac { 1 } { 2 } ) ^ { 2 } + ( y + \frac { \sqrt { 3 } } { 2 } ) ^ { 2 } = ( x + \frac { 1 } { 2 } ) ^ { 2 } + ( y - \frac { \sqrt { 3 } } { 2 } ) ^ { 2 } }
\end{array} , \text { i.e., } \left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
x= \pm \sqrt{3} y
\end{array}\right.\right. \\
\therefore \boldsymbol{b}=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \text { or }\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right), \therefore S_{\triangle A O B}=\frac{1}{2}|\boldsymbol{a}+\boldsymbol{b}||\boldsymbol{a}-\boldsymbol{b}|=1 .
\end{array}
$$

Answer: 1


A sequence of integers $a_{1}, a_{2}, a_{3}, \ldots$ is defined by
$$
\begin{array}{c}
a_{1}=k, \\
a_{n+1}=a_{n}+8 n \text { for all integers } n \geq 1 .
\end{array}
$$

Find all values of $k$ such that every term in the sequence is a square.

Solution:
SolUtion
First solution
We have $a_{2}=a_{1}+8$ and the first two terms are squares, so we have $8=b^{2}-a^{2}=(b+a)(b-a)$ where $a$ and $b$ are positive integers.
Hence either $b-a=1$ and $b+a=8$, which is impossible for integers, or $b-a=2$ and $b+a=4$, so $a=1$ and $b=3$. Hence the only possible value of $k$ is 1 .
Alternatively, suppose that all the terms are squares, with the first two differing by 8 . Note that 16 and 25 differ by 9 and the gaps between successive squares beyond that are shown to be greater than 9 . It follows that the first two terms could only be $1,4,9$ or 16 . It is easy to observe that this forces $a_{1}=1$ and $a_{2}=9$, and so the only possible value of $k$ is 1 .
Evaluating, we obtain the sequence $1,9,25,49, \ldots$ which are all perfect squares, and we must show that this pattern continues.
From the definition of the sequence, we have $a_{n}=1+8(1+2+\cdots+(n-1))$, but we need a simpler expression for the sum $1+2+\cdots+(n-1)$. One method is to consider a rectangle of dots divided into two triangles.
The grey dots represent the numbers 1 to 7 , and the black dots represent the numbers 7 to 1 . Together they form a rectangle with $7 \times 8$ dots.
Hence $1+2+3+4+5+6+7=\frac{1}{2}(7 \times 8)$ and in general $1+2+\cdots+(n-1)=\frac{1}{2}(n-1) n$
As a consequence we have $a_{n}=1+4(n-1) n=(2 n-1)^{2}$, which is a perfect square.
Second solution
We have $a_{n}=k+(1+2+\cdots+(n-1))$ and so $a_{n}=k+4(n-1) n=4 n^{2}-4 n+k$, as above, but for general $k$. Rearrange this to obtain $a_{n}=(2 n-1)^{2}+k-1$
If $a_{n}$ is always square, then we have an infinite number of pairs of squares which differ by a fixed value $k-1$. However this cannot happen if $k-1>0$, since the gap between consecutive squares $N^{2}$ and $(N+1)^{2}$ is $2 N+1$, which will eventually be larger than $k-1$ for a fixed $k \geq 2$
Hence $k=1$ and $a_{n}=(2 n-1)^{2}$

Answer: 1


For a convex polyhedron, the internal dihedral angle at each edge is acute. How many faces can the polyhedron have?

Solution:
5. For each of the faces, consider the vector of the external normal, i.e., a vector perpendicular to this face and directed outward from the polyhedron.

$1^{\circ}$. Let's prove that the angle between any two external normals is obtuse or straight. Suppose this is not the case, and there exist two faces $\Gamma_{1}$ and $\Gamma_{2}$, the external normals of which form an angle not greater than $\pi / 2$. Then the faces $\Gamma_{1}$ and $\Gamma_{2}$ belong to half-planes $\Pi_{1}$ and $\Pi_{2}$, which form a dihedral angle of at least $\pi / 2$.

Take a point $P$ on the face $\Gamma_{2}$. Let $P^{\prime}$ be the projection of the point $P$ onto the plane of the face $\Gamma_{1}$. Since the angle between

![](https://cdn.mathpix.com/cropped/2024_05_21_6455f48598646f41a890g-335.jpg?height=412&width=504&top_left_y=1217&top_left_x=741)

Fig. 167 the external normals to the faces $\Gamma_{1}$ and $\Gamma_{2}$ is not greater than $\pi / 2$, and our polyhedron is convex, the point $P^{\prime}$ lies outside the polygon $\Gamma_{1}$. Therefore, there exists a line containing some edge $r$ of the face $\Gamma_{1}$ and separating $P^{\prime}$ from $\Gamma_{1}$. The polyhedron lies inside the acute dihedral angle corresponding to the edge $r$, but $P$ lies outside this dihedral angle (Fig. 167). Contradiction.

$2^{\circ}$. It remains to show that in space, there do not exist more than four vectors, the pairwise angles between which are obtuse or straight.

Suppose this is not the case, and $\vec{u}_{0}, \vec{u}_{1}, \vec{u}_{2}, \vec{u}_{3}, \vec{u}_{4}$ are five vectors, the pairwise angles between which are obtuse or straight. Introduce a rectangular coordinate system such that the $O z$ axis is directed along $\vec{u}_{0}$.

Denote by $\vec{v}_{i}$ the projection of the vector $\vec{u}_{i}$ onto the plane $O x y$. Let $\vec{u}_{i}=\left(x_{i}, y_{i}, z_{i}\right)$, then $x_{0}=y_{0}=0, z_{0}>0$ and $v_{0}=0$.

The condition that all angles are obtuse or straight is equivalent to all scalar products being negative. In particular,

$$
z_{0} z_{i}=\left(\vec{u}_{0}, \vec{u}_{i}\right) < 0 \quad \text{for} \quad 1 \leqslant i \leqslant 4.
$$

Since $z_{0} > 0$, it follows that $z_{i} < 0$ for $1 \leqslant i \leqslant 4$. Now consider the scalar products between the projections:

$$
\left(\vec{v}_{i}, \vec{v}_{j}\right)=\left(\vec{u}_{i}, \vec{u}_{j}\right)-z_{i} z_{j} < 0 \quad \text{for} \quad 1 \leqslant i, j \leqslant 4.
$$

Thus, $\vec{v}_{1}, \vec{v}_{2}, \vec{v}_{3}, \vec{v}_{4}$ are four vectors on the plane, the angles between which are obtuse or straight. But this is impossible.

Therefore, a polyhedron has four faces. Note that a polyhedron with four faces can only be a tetrahedron.

Idea of another solution. First, prove that at each vertex of the polyhedron, three faces meet. For this, it is sufficient to prove that the sum of the dihedral angles of any $n$-hedral angle is greater than $\pi(n-2)$. Since an $n$-hedral angle can be cut into $n-2$ trihedral angles, it is sufficient to prove this statement for a trihedral angle. Now, it can be shown that all planar angles at each vertex are acute. From this, it is not difficult to deduce that all faces are triangles. Now, it is not difficult to verify that the polyhedron is a tetrahedron.

Answer: 4


For the function $\mathrm{f}(\mathrm{x})$, the condition $\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x})))+3 \mathrm{f}(\mathrm{f}(\mathrm{x}))+9 \mathrm{f}(\mathrm{x})+27 \mathrm{x}=0$ is satisfied. Find $\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{f}(2))))$.

Solution:
Answer: 162

## Examples of how to write answers:

14

$1 / 4$

$-0.25$

#

Answer: 162


Planar vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy: $|\vec{a}|=|\vec{b}| \neq 0, \vec{a} \perp \vec{b},|\vec{c}|=2 \sqrt{2},|\vec{c}-\vec{a}|=1$, try to find the maximum possible value of $|\vec{a}+\vec{b}-\vec{c}|$.

Solution:
Question 93, Solution: As shown in the figure, let $\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\overline{\mathrm{OD}}$, then quadrilateral $0 \mathrm{ADB}$ is a square, $|\vec{c}-\vec{a}|=|A C|=1$, and $|\vec{a}+\vec{b}-\vec{c}|=|C D|$. Let the side length of the square $O A D B$ be $x$, then $|O D|=$ $\sqrt{2} x$. According to Ptolemy's theorem, we have:
$$
\begin{array}{l}
|C D| \cdot|O A| \leq|O D| \cdot|A C|+|A D| \cdot|O C| \\
\Rightarrow x \cdot|C D| \leq \sqrt{2} x \cdot 1+x \cdot 2 \sqrt{2} \\
\Rightarrow|C D| \leq 3 \sqrt{2}
\end{array}
$$

Therefore, the maximum possible value of $|\vec{a}+\vec{b}-\vec{c}|$ is $3 \sqrt{2}$.

Answer: 3\sqrt{2}


A cube-shaped container with an edge length of $12 \mathrm{~cm}$ was filled to $\frac{5}{8}$ of its capacity with a liquid, and then it was slightly tilted along one of its edges. The diagram shows the cross-section of the container with the horizontal level of the liquid inside. We know that the length of segment $L C$ is exactly twice the length of segment $K B$. Determine the length of segment $L C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_2ef57db81a3ef472104fg-1.jpg?height=265&width=271&top_left_y=331&top_left_x=922)

Solution:
Solution. Before the impact, the fluid level was at $\frac{5}{8}$ of the 12 cm height, i.e., at $7.5 \mathrm{~cm}$. The area of this rectangle's cross-section is $12 \cdot 7.5=90 \mathrm{~cm}^{2}$. After the impact, the area of the trapezoid $B C L K$ will be $90 \mathrm{~cm}^{2}$. Let $L C=2 x$, then according to the text, $K B=x$, so we can write the area of the trapezoid as:

$$
\frac{(2 x+x) \cdot 12}{2}=90
$$

from which $x=5$. The length of the segment $L C$ is $10 \mathrm{~cm}$.

Answer: 10\mathrm{~}


What are the integers $k$ such that for all real numbers $a, b, c$,

$$
(a+b+c)(a b+b c+c a)+k a b c=(a+b)(b+c)(c+a)
$$

Solution:
Suppose the equation is verified for an integer $k$. By taking $a=b=c=1$, we get $3 \times 3 + k = 2^{3}$, so $k = 8 - 9 = -1$.

It remains to verify that $(a+b+c)(ab+bc+ca) - abc = (a+b)(b+c)(c+a)$ for all real numbers $a, b, c$. Now,

$$
(a+b+c)(ab+bc+ca) - abc = a^2b + a^2c + abc + b^2c + b^2a + abc + c^2a + c^2b + abc - abc
$$

so

$$
(a+b+c)(ab+bc+ca) - abc = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc
$$

and

$$
(a+b)(b+c)(a+c) = (ab + b^2 + ac + bc)(a+c) = a^2b + ab^2 + a^2c + abc + abc + b^2c + ac^2 + bc^2
$$

so

$$
(a+b)(b+c)(a+c) = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc
$$

For $k = -1$, we indeed have for all real numbers $a, b, c$,

$$
(a+b+c)(ab+bc+ca) + kabc = (a+b)(b+c)(c+a)
$$

The unique solution is therefore $k = -1$.

Answer: -1


Calculate the area of the common part of two rhombuses, the lengths of the diagonals of the first of which are 4 and $6 \mathrm{~cm}$, and the second is obtained by rotating the first by $90^{\circ}$ around its center.

Solution:
Solution.

The required area $S$ is equal to $4\left(S_{\triangle A O B}-S_{\triangle A F K}\right)$ (Fig. 10.127). We find $S_{\triangle A O B}=0.5 \cdot 3 \cdot 2=3 \, \text{cm}^{2}$. The side of the rhombus is $\sqrt{2^{2}+3^{2}}=\sqrt{13}$ (cm). In $\triangle A O B$, the segment $O K$ is the angle bisector; then, using the formula

$$
l_{c}=\frac{\sqrt{a b(a+b+c)(a+b-c)}}{a+b}, \text{ we have } O K=\frac{\sqrt{6(5+\sqrt{13})(5-\sqrt{13})}}{5}=\frac{6 \sqrt{2}}{5} \, \text{cm.}
$$

Next, $S_{\triangle A F K}=\frac{1}{2} A F \cdot K H$, where $K H=\frac{O K}{\sqrt{2}}=\frac{6}{5}$ cm, $A F=3-2=1$ cm; therefore, $S_{\triangle A F K}=0.6 \, \text{cm}^{2}$. Finally, $S=4(3-0.6)=9.6 \, \text{cm}^{2}$.

Answer: $9.6 \, \text{cm}^{2}$.

Answer: 9.6\,


Find the derivative of the specified order.

$y=\left(1+x^{2}\right) \operatorname{arctg} x, y^{\prime \prime \prime}=?$

Solution:
## Solution

$y^{\prime}=\left(\left(1+x^{2}\right) \operatorname{arctg} x\right)^{\prime}=2 x \cdot \operatorname{arctg} x+\left(1+x^{2}\right) \cdot \frac{1}{1+x^{2}}=$

$=2 x \cdot \operatorname{arctg} x+1$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=(2 x \cdot \operatorname{arctg} x+1)^{\prime}=2 \operatorname{arctg} x+2 x \cdot \frac{1}{1+x^{2}}=$

$=2 \operatorname{arctg} x+\frac{2 x}{1+x^{2}}$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(2 \operatorname{arctg} x+\frac{2 x}{1+x^{2}}\right)^{\prime}=\frac{2}{1+x^{2}}+\frac{2 \cdot\left(1+x^{2}\right)-2 x \cdot 2 x}{\left(1+x^{2}\right)^{2}}=$

$=\frac{2\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}+\frac{2+2 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{4+4 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{4}{\left(1+x^{2}\right)^{2}}$

Problem Kuznetsov Differentiation $19-10$

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


If $a$, $b$, $c$ are all real numbers, and $a+b+c=0$, $abc=2$, then the minimum value that $|a|+|b|+|c|$ can reach is $\qquad$ .

Solution: (Tip: Refer to Example 3. Answer: 4. )

Answer: 4


Along the road connecting Masha's and Sasha's houses, there are 17 apple trees and 18 poplars. When Masha was going to visit Sasha, she took photos of all the trees. Right after the tenth apple tree, Masha's phone memory ran out, and she couldn't photograph the remaining 13 trees. The next day, when Sasha was going to visit Masha, starting from the eighth apple tree, he picked one leaf from each tree. How many leaves did Sasha pick?

Solution:
Answer: 22.

Solution. Note that the tenth apple tree, counting from Masha's house, is the eighth apple tree, counting from Sasha's house. Therefore, Sasha will not pick a leaf from exactly 13 trees. We get that he will pick a total of $17+18-13=22$ leaves.

Answer: 22


Given an isosceles triangle \(ABC\) with base \(BC\). On the extension of side \(AC\) beyond point \(C\), point \(K\) is marked, and a circle is inscribed in triangle \(ABK\) with center at point \(I\). A circle passing through points \(B\) and \(I\) is tangent to line \(AB\) at point \(B\). This circle intersects segment \(BK\) again at point \(L\). Find the angle between lines \(IK\) and \(CL\). Answer: \(90^{\circ}\).

Solution:
Solution. Let $I M$ and $I N$ be the perpendiculars dropped from point $I$ to $A K$ and $B K$ respectively. Note that $\angle A B I = \angle B L I$, since line $A B$ is tangent to the circumcircle of triangle $B L I$. Line $A I$ is the angle bisector of angle $A$, and $A B = A C$ by the given condition. Therefore, triangles $A B I$ and $A C I$ are congruent by two sides and the included angle. Hence,

$$
\angle A C I = \angle A B I = \angle B L I \quad \text{and} \quad I M = I N
$$

which means that right triangles $C M I$ and $L N I$ are congruent. Consequently,

$$
K C = K M - C M = K N - L N = K L
$$

Line $K I$ is the angle bisector of isosceles triangle $C L K$, from which it follows that $K I \perp C L$.

Answer: 90


Given that $a$ is a real number, the equation about $x$
$$
27 x^{2}+2 a^{2} x+a=0
$$

has real roots. Then the maximum value of $x$ is ( ).
(A) 0
(B) $-\frac{3}{2}$
(C) $-\frac{1}{3}$
(D) $\frac{1}{6}$

Solution:
4.D.

Consider the original equation as an equation in terms of $a$, and transform it into $2 x a^{2}+a+27 x^{2}=0$.
When $x=0$, $a=0$;
When $x \neq 0$, $\Delta=1-216 x^{3} \geqslant 0$, solving this yields $x \leqslant \frac{1}{6}$.
When $a=-\frac{3}{2}$, $x=\frac{1}{6}$.
Therefore, the maximum value of $x$ is $\frac{1}{6}$.

Answer: D


Find the total number of different integers the function

\[ f(x) = \left[x \right] + \left[2 \cdot x \right] + \left[\frac{5 \cdot x}{3} \right] + \left[3 \cdot x \right] + \left[4 \cdot x \right] \]

takes for $0 \leq x \leq 100.$

Solution:
1. **Identify the function and the range:**
   The function given is:
   \[
   f(x) = \left[x \right] + \left[2 \cdot x \right] + \left[\frac{5 \cdot x}{3} \right] + \left[3 \cdot x \right] + \left[4 \cdot x \right]
   \]
   We need to find the total number of different integers this function takes for \(0 \leq x \leq 100\).

2. **Analyze the intervals:**
   The function \(f(x)\) changes its value when any of the floor functions \(\left[x\right]\), \(\left[2x\right]\), \(\left[\frac{5x}{3}\right]\), \(\left[3x\right]\), or \(\left[4x\right]\) changes its value. This happens at specific points where \(x\) is a rational number that causes any of these expressions to be an integer.

3. **Determine the critical points in the interval \([0, 3)\):**
   The critical points in \([0, 3)\) are:
   \[
   0, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}, 1, \frac{6}{5}, \frac{5}{4}, \frac{4}{3}, \frac{3}{2}, \frac{5}{3}, \frac{7}{4}, \frac{9}{5}, 2, \frac{9}{4}, \frac{7}{3}, \frac{12}{5}, \frac{5}{2}, \frac{8}{3}, \frac{11}{4}
   \]
   These points are where the value of \(f(x)\) changes. There are 22 such points.

4. **Calculate the number of different integers in \([0, 3)\):**
   Since there are 22 critical points, there are 22 different integers generated by \(f(x)\) in the interval \([0, 3)\).

5. **Extend the interval to \([0, 99)\):**
   The interval \([0, 99)\) can be divided into 33 subintervals of length 3 each: \([0, 3), [3, 6), [6, 9), \ldots, [96, 99)\). Each of these subintervals will generate 22 different integers.

   Therefore, the total number of different integers generated in \([0, 99)\) is:
   \[
   33 \times 22 = 726
   \]

6. **Consider the interval \([99, 100]\):**
   The interval \([99, 100]\) is analogous to the interval \([0, 1]\) because the floor functions will behave similarly. We need to determine the number of different integers generated in \([0, 1]\).

   The critical points in \([0, 1]\) are:
   \[
   0, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}, 1
   \]
   There are 8 such points.

7. **Calculate the total number of different integers:**
   Adding the number of different integers generated in \([0, 99)\) and \([99, 100]\):
   \[
   726 + 8 = 734
   \]

The final answer is \(\boxed{734}\)

Answer: 734


The line $l$ passes through the focus of the parabola $y^{2}=a(x+1)(a>0)$ and is perpendicular to the $x$-axis. If the segment cut off by $l$ on the parabola is 4, then $a=$ $\qquad$ .

Solution:
$=.1 .4$.
Since the parabolas $y^{2}=a(x+1)$ and $y^{2}=a x$ have the same length of the focal chord with respect to their directrix, the general equation $y^{2}=a(x+1)$ can be replaced by the standard equation $y^{2}=a x$ for solving, and the value of $a$ remains unchanged. Using the formula for the length of the latus rectum, we get $a=4$.

Answer: 4


The sum of the greatest integer less than or equal to $x$ and the least integer greater than or equal to $x$ is $5$. The solution set for $x$ is
$\textbf{(A)}\ \Big\{\frac{5}{2}\Big\}\qquad \textbf{(B)}\ \big\{x\ |\ 2 \le x \le 3\big\}\qquad \textbf{(C)}\ \big\{x\ |\ 2\le x < 3\big\}\qquad\\  \textbf{(D)}\ \Big\{x\ |\ 2 < x\le 3\Big\}\qquad \textbf{(E)}\ \Big\{x\ |\ 2 < x < 3\Big\}$

Solution: If $x \leq 2$, then $\lfloor x \rfloor + \lceil x \rceil \leq 2+2 < 5$, so there are no solutions with $x \leq 2$. If $x \geq 3$, then $\lfloor x \rfloor + \lceil x \rceil \geq 3+3$, so there are also no solutions here. Finally, if $2<x<3$, then $\lfloor x \rfloor + \lceil x \rceil = 2 + 3 = 5$, so the solution set is $\boxed{E}$.

Answer: E


In the following figure, $AD=AB, \angle DAB=\angle DCB=\angle AEC=90^{\circ}$ and $AE=5$. Find the area of the quadrangle $ABCD$.

Solution:
31 Ans: 25.
Upon rotating $A E D$ anticlock wise $90^{\circ}$ through $A$, we get a square. Thus the area is $5^{2}=25$.

Answer: 25


In how many ways can you partition the set $\{1,2, \ldots, 12\}$ into six mutually disjoint two-element sets in such a way that the two elements in any set are coprime?

Solution:
Solution. No two even numbers can be in the same set (pair). Let us call partitions of $\{1,2, \ldots, 12\}$ with this property, that is one even and one odd number in each pair, even-odd partitions. The only further limitations are, that 6 nor 12 cannot be paired with 3 or 9 , and 10 cannot be paired with 5 .

That means, that odd numbers 1,7 and 11 can be paired with numbers $2,4,6$, 8, 10, 12, numbers 3 and 9 with 2, 4, 8, 10 and number 5 can be paired with 2, 4, 6, 8,12 . We cannot use the product rule directly, we distinguish two cases: 5 is paired with 6 or 12 , in the second one 5 is paired with one of 2,4 , and 8 . The possible pairings are $2 \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 1=144$ in the first case, $3 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1=108$ in the second case. Together $144+108=252$ pairings.

Answer: 252


In triangle $\mathrm{ABC}$ with sides $a, b$ and $c$, it is known that $b=50 \mathrm{~cm}, c=48 \mathrm{~cm}$ and the internal angle at vertex A is twice the internal angle at vertex B. Determine the side $a$.

Solution:
Solution: Extend side CA through vertex A to point D such that $\overline{\mathrm{AD}}=\overline{\mathrm{AB}}$. Then, triangle DBA is isosceles, so $\triangle D=\triangle D B A$. For this triangle, $\triangle B A C=2 \beta$ is an exterior angle. It is equal to the sum of the base angles

![](https://cdn.mathpix.com/cropped/2024_06_05_e7266f2e82216ac5d245g-5.jpg?height=644&width=411&top_left_y=252&top_left_x=1049)
$D B$, i.e., $\triangle D+\triangle D B A=2 \beta$. Therefore, $\triangle D=\triangle D B A=\beta$. However, then $\triangle C B D=2 \beta$, so triangles $\mathrm{ABC}$ and BDC are similar. We have $\overline{\mathrm{BC}}: \overline{\mathrm{AC}}=\overline{\mathrm{CD}}: \overline{\mathrm{BC}}$ or $a: 50=(50+48): a ; a^{2}=50 \cdot 48$; i.e., $a=70 \mathrm{~cm}$.

Answer: 70


A two-digit number has the following property: if it is added to the number with the same digits but in reverse order, the result is a perfect square.

Find all such two-digit numbers.

Solution:
Solution. If $x$ is the unit digit and $y$ is the tens digit of the desired two-digit number, then according to the condition of the problem, we have

$$
(10 y + x) + (10 x + y) = m^{2}
$$

i.e.

$$
11(x + y) = m^{2}
$$

According to this, 11 is a divisor of $m$; if $m = 11k$, we get

$$
x + y = 11k^{2}
$$

From $1 \leq x \leq 9$ and $1 \leq y \leq 9$, it follows that $2 \leq x + y \leq 18$, which means that

$$
x + y = 11
$$

From (3) it follows that the desired numbers are: $29, 38, 47, 56, 65, 74, 83$ and 92.

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


Calculate the volumes of bodies bounded by the surfaces.

$$
\frac{x^{2}}{4}+\frac{y^{2}}{9}-\frac{z^{2}}{36}=-1, z=12
$$

Solution:
## Solution

In the section of the given figure by the plane $z=$ const, there is an ellipse:

$$
\frac{x^{2}}{4}+\frac{y^{2}}{9}=\frac{z^{2}}{36}-1
$$

The area of the ellipse described by the formula: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $\pi \cdot a \cdot b$

Let's find the radii of the ellipse:

![](https://cdn.mathpix.com/cropped/2024_05_22_348f4289c7b5f46bc246g-25.jpg?height=768&width=1057&top_left_y=998&top_left_x=888)

$$
\begin{aligned}
& \frac{x^{2}}{4 \cdot \frac{z^{2}-36}{36}}+\frac{y^{2}}{9 \cdot \frac{z^{2}-36}{36}}=1 \rightarrow a=\frac{1}{3} \sqrt{z^{2}-36} ; b=\frac{1}{2} \sqrt{z^{2}-36} \\
& \Rightarrow S=\pi a b=\pi \cdot \frac{1}{3} \sqrt{z^{2}-36} \cdot \frac{1}{2} \sqrt{z^{2}-36}=\frac{\pi}{6} \cdot\left(z^{2}-36\right)
\end{aligned}
$$

$$
\begin{aligned}
& V=\int_{6}^{12} S(z) d z=\frac{\pi}{6} \int_{6}^{12}\left(z^{2}-36\right) d z=\left.\frac{\pi}{6}\left(\frac{z^{3}}{3}-36 z\right)\right|_{6} ^{12}= \\
& =\frac{\pi}{6}\left(\frac{12^{3}}{3}-36 \cdot 12-\frac{6^{3}}{3}+36 \cdot 6\right)=\frac{\pi}{6}(576-432-72+216)=48 \pi
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 84 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+20-14$ »

Categories: Kuznetsov's Problem Book Integrals Problem 20 | Integrals | Problems for Checking

- Last modified: 06:35, 23 June 2010.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 20-15

## Material from PlusPi

Answer: 48\pi


Given $a, b \in \mathbf{R}, h>0$, then â$|a-b|<2 h$â is â$|a-1|<h$ and $|b-1|<h$â ( ).
(A) A sufficient but not necessary condition
(B) A necessary but not sufficient condition
(C) A necessary and sufficient condition
(D) Neither a sufficient nor a necessary condition

Solution:
Solution: Since $a, b \in \mathbf{R}, h>0$, if we take $a=b=1+h$, then $|a-b|=0<2h$, but $|a-1|=h$ and $|b-1|=h$, meaning $|a-1|<h$ and $|b-1|<h$ do not hold, so the conclusion cannot be derived from the condition;

Conversely, if $|a-1|<h$ and $|b-1|<h$, then $|a-b|=|(a-1)-(b-1)| \leqslant |a-1|+|b-1|<2h$, which means the condition can be derived from the conclusion.
Therefore, the condition is a necessary but not sufficient condition for the conclusion to hold. Choose B.
Note: A correct proposition requires a rigorous proof process; while negating a proposition, only one counterexample is needed. The above two questions are typical examples that need to be understood. Example 2: If considered on the number line, it is more intuitive. $|a-1|<h$ and $|b-1|<h$ indicate points on the number line whose "distance" from 1 is less than $h$, naturally the "distance" between $a$ and $b$ is less than $2h$; conversely, no matter how small the "distance" between $a$ and $b$ is, the "distance" between $a$ or $b$ and 1 may not be less than $h$.

Answer: B


Write down all three-term arithmetic sequences of prime numbers with a common difference of 8.

Solution:
Let these three numbers be $a$, $a+8$, $a+16$.
Since the remainders when $a$, $a+8$, $a+16$ are divided by 3 are all different, one of these three numbers must be 3. And since $a+8$, $a+16$ are prime numbers greater than 3, it follows that $a=3$.
Therefore, the sequence of numbers is $3,11,19$.

Answer: 3,11,19


For any positive integers $n$ and $k(k \leqslant n)$, $f(n, k)$ represents the number of positive integers not exceeding $\left[\frac{n}{k}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$) that are coprime with $n$. Then $f(100,3)=(\quad)$.
(A) 11
(B) 13
(C) 14
(D) 19

Solution:
4. C.

From $\left[\frac{100}{3}\right]=33$, we know that the required number is the count of numbers from 1 to 33 that are coprime with 100.

First, remove all even numbers, leaving 17 odd numbers. Then, remove multiples of 5 (a total of three), thus, the required number is 14.

Answer: C


Find the number of rational numbers $r$, $0<r<1$, such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution: Let the numerator and denominator $x,y$ with $\gcd(x,y)=1$ and $x+y = 1000.$ Now if $\gcd(x,y) = 1$ then $\gcd(x,y) =\gcd(x,1000-x)= \gcd(x,1000-x-(-1)x)=\gcd(x,1000)=1.$ Therefore any pair that works satisfies $\gcd(x,1000)= 1.$ By Euler's totient theorem, there are $\phi(1000) = 400$ numbers relatively prime to 1000 from 1 to 1000. Recall that $r=\frac{x}{y}<1$ and note by Euclidean algorithm $\gcd(1000,1000-x)=1$, so we want $x<y=1000-x.$ Thus the $400$ relatively prime numbers can generate $\boxed{200}$ desired fractions.

Answer: 200


A line passing through point $A$, located outside the circle at a distance of 7 from its center, intersects the circle at points $B$ and $C$. Find the radius of the circle, given that $AB=3$, $BC=5$.

Solution:
The product of the entire secant and its external part for a given point and a given circle is constant.

## Solution

Since $A B<B C$, point $B$ lies between points $A$ and $C$. Let $O$ be the center of the circle, $R$ its radius, and the line $A O$ intersects the circle at points $D$ and $E$ (with $D$ between $A$ and $O$). Then

$$
A D \cdot A E=A B \cdot A C, (7-R)(7+R)=3 \cdot 8, 49-R^{2}=24
$$

$R^{2}=25$. Therefore, $R=5$.

## Answer

5.

Answer: 5


A tournament of dodgeball was held at school. In each game, two teams competed. 15 points were awarded for a win, 11 for a draw, and no points for a loss. Each team played against each other once. By the end of the tournament, the total number of points scored was 1151. How many teams were there?

Solution:
5. Answer: 12 teams.

Solution. Let there be $\mathrm{N}$ teams. Then the number of games was $\mathrm{N}(\mathrm{N}-1) / 2$. For each game, a total of 15 or 22 points are scored. Therefore, the number of games was no less than 53 $(1151 / 22)$ and no more than $76(1151 / 15)$.

Note that if there were no more than 10 teams, then the number of games would not exceed 45. And if there were no fewer than 13 teams, then the number of games would be no less than 78.

Thus, the number of teams was 11 (55 games) or 12 (66 games). In each game, teams score exactly 15 points in total! And 7 additional points if there was a draw. Therefore, the total number of draws is (1151-55*15)/7 (a non-integer - this could not have been the case) or (1151-66*15)/7=23. Therefore, the only option is that there were 12 teams.

## Grading Criteria.

Correct and justified solution - 7 points.

Justified that the number of teams is 11 or 12 without further progress - 3 points. Justified that the number of games was no less than 53 and no more than 76 without further progress - $\mathbf{2}$ points. Correct answer with a constructed example - $\mathbf{2}$ points. Just the correct answer - $\mathbf{0}$ points.

Answer: 12


Find all natural numbers that are 36 times the sum of their digits. ANSWER: $324 ; 648$.

Solution:
Solution. Let $S(x)$ denote the sum of the digits of the number $x$. Then the equation $x=36 \cdot S(x)$ has no solutions for $n \geq 5$, since $x \geq 10^{n-1}$, while $36 \cdot S(x) \leq 36 \cdot 9 \cdot n=324 n36(a+b+c+d) \Leftrightarrow 964 a+64 b>26 \cdot 9+35 \cdot 9 \geq 26 c+35 d$.

For $n=3: a \cdot 100+b \cdot 10+c=36(a+b+c) \Leftrightarrow 64 a=26 b+35 c$. After this, we perform a check. As a result, the numbers 324 and 648 are obtained.

For $n=2: a \cdot 10+b=36(a+b) \Leftrightarrow 26 a=35 b-$ there are no solutions.

Answer: 324;648


Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms
$$
S_{n}=2 a_{n}-1,(n=1,2, \cdots)
$$

The sequence $\left\{b_{n}\right\}$ satisfies
$$
b_{1}=3, b_{k+1}=a_{k}+b_{k},(k=1,2, \cdots) .
$$

Find the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$.

Solution:
[Solution] From $s_{n}=2 a_{n}-1$, we know,
$$
\begin{array}{l}
a_{1}=2 a_{1}-1, \\
a_{1}=1 .
\end{array}
$$

Also, $a_{k}=s_{k}-s_{k-1}=\left(2 a_{k}-1\right)-\left(2 a_{k-1}-1\right)=2 a_{k}-2 a_{k-1}$,
so $a_{k}=2 a_{k-1}$.
Thus, the sequence $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and common ratio 2.
From $b_{k+1}=a_{k}+b_{k}$, we get
$$
\left\{\begin{array}{l}
b_{2}=a_{1}+b_{1}, \\
b_{3}=a_{2}+b_{2}, \\
\cdots \cdots \cdots \cdots \\
b_{n}=a_{n-1}+b_{n-1} .
\end{array}\right.
$$

Adding these equations,
$$
b_{n}=s_{n-1}+b_{1}=\frac{2^{n-1}-1}{2-1}+3=2^{n-1}+2 .
$$

Therefore, the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is
$$
S_{n}^{\prime}=1+2+\cdots+2^{n-1}+2 n=2^{n}+2 n-1 .
$$

Answer: 2^{n}+2n-1


Solution:
8.57 Noticing that

$$
\begin{aligned}
& \log _{\frac{1}{4}} \frac{x+1}{4 x-1}=\log _{4} \frac{4 x-1}{x+1} \\
& \log _{\frac{1}{3}} \log _{4} \frac{4 x-1}{x+1}=-\log _{3} \log _{4} \frac{4 x-1}{x+1}
\end{aligned}
$$

we transform the given inequality:

$$
\begin{aligned}
& \log _{3} \log _{4} \frac{4 x-1}{x+1}+\log _{3} \log _{4} \frac{4 x-1}{x+1}>\frac{2}{3}.
\end{aligned}
$$

Answer: $\left(\frac{2}{3}, \infty\right)$.

Answer: (\frac{2}{3},\infty)


The amount of electricity flowing through a conductor, starting from the moment of time $t=0$, is given by the formula $Q=3 t^{2}-3 t+4$. Find the current strength at the end of the 6th second.

Solution: Solution. The current is the derivative of the quantity of electricity with respect to time: therefore, we need to find the derivative of the function $Q=3 t^{2}-3 t+4$ and calculate its value at $t=6 \mathrm{c}$. We have $I=Q^{\prime}=$ $=6 t-3$, from which at $t=6$ we get $I=33$ (A).

Answer: 33


Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant 180^{\circ}\right)$, and the complex numbers $z, (1+i)z, 2\bar{z}$ correspond to three points $P, Q, R$ on the complex plane. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by segments $PQ, PR$ is $S$. Then the maximum distance from point $S$ to the origin is

Solution:
5.3.

Let the complex number $w$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have
$$
\begin{aligned}
w+z & =2 \bar{z}+(1+\mathrm{i}) z, \text { i.e., } w=2 \bar{z}+\mathrm{i} z . \\
\therefore|w|^{2} & =(2 \bar{z}+\mathrm{i} z)(2 z-\mathrm{i} \bar{z}) \\
& =4+1+2 \mathrm{i}\left(z^{2}-\bar{z}^{2}\right)=5-4 \sin 2 \theta \\
& \leqslant 5+4=9(\text { when } \theta=135 \text {, the equality holds }) .
\end{aligned}
$$

Thus, $|w|_{\text{max}}=3$.

Answer: 3


The formation for the radio calisthenics competition is a rectangle, with a total of 128 participants. Xiao Ming's position is the 7th from the left, 1st from the front, and 8th from the back. Therefore, Xiao Ming's position is the __th from the right.

Solution: answer: 10

Answer: 10


Solve the inequality:

$$
\sqrt{2 x+8}+\sqrt{10-2 x} \geq \log _{2}\left(4 x^{2}-4 x+65\right)
$$

Solution:
# Solution:

It is not difficult to understand that

$$
\log _{2}\left(4 x^{2}-4 x+65\right)=\log _{2}\left((2 x-1)^{2}+64\right) \geq \log _{2} 64=6
$$

Let's estimate the expression $\sqrt{2 x+8}+\sqrt{10-2 x}$. Consider the vectors $\overrightarrow{\boldsymbol{a}}(1,1)$ and $\vec{b}(\sqrt{2 x+8}, \sqrt{10-2 x})$. We have

$$
\begin{gathered}
|\vec{a} \cdot \vec{b}| \leq|\vec{a}| \cdot|\vec{b}| \\
\Leftrightarrow \sqrt{2 x+8}+\sqrt{10-2 x} \leq \sqrt{1+1} \cdot \sqrt{2 x+8+10-2 x}=6
\end{gathered}
$$

Therefore, the inequality is equivalent to the system

$$
\left\{\begin{array} { l }
{ \operatorname { l o g } _ { 2 } ( 4 x ^ { 2 } - 4 x + 6 5 ) = 6 } \\
{ \sqrt { 2 x + 8 } + \sqrt { 1 0 - 2 x } = 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
2 x-1=0 \\
\sqrt{2 x+8}+\sqrt{10-2 x}=6
\end{array}\right.\right.
$$

Answer: $\frac{1}{2}$.

## Evaluation Criteria:

- 15 points - the correct answer is obtained with proper justification.
- 10 points - the solution contains a correct sequence of steps leading to the correct answer, but one arithmetic error is made during the solution.
- 5 points - the solution contains generally correct steps, the correct answer is obtained, but the solution contains inaccuracies and/or is not fully justified.
- 0 points - the solution does not meet any of the criteria provided.

Answer: \frac{1}{2}


The sequence $\left\{a_{n}\right\}$ has 9 terms, where $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$. Find the number of such sequences.

Solution: Prompt: Categorized Count. Answer: 491.

Answer: 491


Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x \geqslant 0$, $f(x)=2^{x}+2 x+b$ ( $b$ is a constant). Then $f(-10)=$ $\qquad$ .

Solution:
2. -1043 .

From the given condition, we easily know that
$$
f(0)=2^{0}+2 \times 0+b=0 \text {. }
$$

Solving for $b$ yields $b=-1$.
By the property of odd functions $f(-x)=-f(x)$, we have
$$
\begin{array}{l}
f(-10)=-f(10)=-2^{10}-2 \times 10+1 \\
=-1043 .
\end{array}
$$

Answer: -1043


Find the natural number $n$ from the equation

$$
3^{2} \cdot 3^{5} \cdot 3^{8} \cdots 3^{3 n-1}=27^{5}
$$

Solution:
## Solution.

$3^{2+5+8+\ldots+3 n-1}=3^{15}, 2+5+8+\ldots+3 n-1=15$.

On the left side of the equation, we have the sum of the terms of an arithmetic progression $S_{k}$, where $a_{1}=2, d=3, a_{k}=3 n-1, k=\frac{a_{k}-a_{1}}{d}+1=\frac{3 n-1-2}{3}+1=n$.

Then $S_{k}=\frac{a_{1}+a_{k}}{2} \cdot k=\frac{2+3 n-1}{2} \cdot n=\frac{3 n^{2}+n}{2}$, and the equation becomes $\frac{3 n^{2}+n}{2}=15,3 n^{2}+n-30=0$, from which $n=3$.

Answer: 3.

Solve the equations (7.048 - 7.127):

Answer: 3


Given a real-coefficient polynomial function $y=ax^2+bx+c$, for any $|x| \leqslant 1$, it is known that $|y| \leqslant 1$. Try to find the maximum value of $|a|+|b|+|c|$.

Solution:
Proof: First, we prove the following auxiliary proposition.
Proposition: Let real numbers $A, B$ satisfy $|A| \leqslant 2, |B| \leqslant 2$. Then $|A+B| + |A-B| \leqslant 4$.
In fact, without loss of generality, let $|A| \geqslant |B|$.
From $A^2 \geqslant B^2$, we have $(A+B)(A-B) \geqslant 0$,
thus $|A+B| + |A-B| = |(A+B) + (A-B)| = |2A| \leqslant 4$.

Now, let's solve the original problem.
When $x=0$, $|y| \leqslant 1$,
i.e., $|c| \leqslant 1$.
For any $|x| \leqslant 1$, we have
\[
|a x^2 + b x| = |y - c| \leqslant |y| + |c| \leqslant 2.
\]

Taking $x=1$, we get $|a + b| \leqslant 2$,
taking $x=-1$, we get $|a - b| \leqslant 2$.
According to the auxiliary proposition, we have
\[
\begin{aligned}
|a| + |b| &= \frac{1}{2}(|2a| + |2b|) \\
&= \frac{1}{2}(|(a+b) + (a-b)| + |(a+b) - (a-b)|) \\
&\leqslant \frac{1}{2} \times 4 = 2.
\end{aligned}
\]

From (1) and (2), we know $|a| + |b| + |c| \leqslant 3$.
Next, we show that $|a| + |b| + |c|$ can achieve the maximum value of 3.
Take $a=2, b=0, c=-1$. In this case, $y = 2x^2 - 1$. It is easy to verify that when $|x| \leqslant 1$, $|y| \leqslant 1$.
Therefore, the maximum value of $|a| + |b| + |c|$ is 3.
(Chen Kuanhong, Nengshi High School, Yueyang County, Hunan Province, 414113)

Answer: 3


Solution:
5. $\frac{1005}{3017}$.

Let $f(x)=f_{0}(x)=\frac{1+x}{1-3 x}$. Then
$f_{1}(x)=f(f(x))=\frac{1+\frac{1+x}{1-3 x}}{1-3 \cdot \frac{1+x}{1-3 x}}=-\frac{1-x}{1+3 x}$,
$f_{2}(x)=f\left(f_{1}(x)\right)=\frac{1-\frac{1-x}{1+3 x}}{1+3 \cdot \frac{1-x}{1+3 x}}=x$,
$f_{3}(x)=f(x)=f_{0}(x)$.
Thus, the expression of $f_{n}(x)$ repeats in a cycle of 3.
$$
\begin{array}{l}
\text { Then } f_{2011}(x)=f_{3 \times 670+1}(x) \\
=f_{1}(x)=\frac{x-1}{1+3 x} .
\end{array}
$$

Therefore, $f_{2011}(2011)=\frac{2011-1}{1+3 \times 2011}=\frac{1005}{3017}$.

Answer: \frac{1005}{3017}


As shown in Figure $1, M_{1}$, $M_{2}$, $M_{3}$, $M_{4}$ are points on the sides $AB$, $BC$, $CD$, $DA$ of quadrilateral $ABCD$, respectively, and $AM_{1}: M_{1}B = AM_{4}: M_{4}D = CM_{2}: M_{2}B = CM_{3}: M_{3}D = 1:2$. $P$ is a point inside quadrilateral $ABCD$, and the sum of the areas of quadrilaterals $PM_{1}AM_{4}$ and $PM_{2}CM_{3}$ is 8. Then the area $S$ of quadrilateral $ABCD$ is ( ).
(A) 16
(B) 24
(C) 20
(D) 32

Solution:
2. B.

As shown in Figure 5, connect $P A$,
$P B, P C, P D$.
Given $A M_{1}: M_{1} B=$
$1: 2$, it is easy to see that
$$
A B=3 A M_{1} .
$$

Thus, $S_{\triangle P A B}=3 S_{\triangle P A M_{1}}$.
Similarly, $S_{\triangle P B C}=3 S_{\triangle P C M_{2}}, S_{\triangle P C D}=3 S_{\triangle P C M_{3}}$,
$$
S_{\triangle P A D}=3 S_{\triangle P A M_{4}} \text {. }
$$

Adding the above four equations, we get

Answer: B


Let $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ non-negative real numbers $\left(n>2, n \in \mathbf{N}^{*}\right)$, and $\sum_{i=1}^{n} x_{i}=n, \sum_{i=1}^{n} i x_{i}=2 n-2$, find the maximum value of the expression $S=\sum_{i=1}^{n} i^{2} x_{i}$.

Solution:
10. Solution: Let $y_{i}=\sum_{j=i}^{n} x_{j}, i=1,2, \cdots, n$ then $y_{1}=n$ and $\sum_{i=1}^{n} y_{i}=2 n-2$. $S=\sum_{k=1}^{n} k^{2} x_{k}=\sum_{k=1}^{n-1} k^{2}\left(y_{k} -y_{k-1}\right)+n^{2} y_{n}=\sum_{k=1}^{n-1} k^{2} y_{k}-\sum_{k=2}^{n}(k-1)^{2} y_{k}+n^{2} y_{n}=y_{1}+\sum_{k=2}^{n-1}(2 k-1) y_{k}-(n-1)^{2} y_{n}+n^{2} y_{n}=\sum_{k=1}^{n}(2 k -1) y_{k}$.

Notice that, $y_{1}=n$, and $y_{2} \geqslant y_{3} \geqslant \cdots \geqslant y_{n}$. If there exists $i \in\{2,3, \cdots, n-1\}$ such that $y_{i}>y_{n}$ (and assume $i$ is the largest index for which this holds), then by the expression of $S$, the coefficient of $y_{i}$ is less than the coefficients of $y_{i-1}, \cdots, y_{n}$. Therefore, under the condition of keeping $y_{i}+y_{i-1}+\cdots+y_{n}$ unchanged, replacing each number with their arithmetic mean increases $S$. Hence, when $y_{2}=y_{3}=\cdots=y_{n}$, $S$ is maximized, at which point $S_{\max }=n+\frac{n-2}{n-1} \sum_{k=2}^{n}(2 k-1)=n^{2}-2$.

Note: This problem can also be solved by letting $y_{k}=\sum_{i=1}^{k} x_{2}$, and directly using Abel's transformation, we get $S=\sum_{k=1}^{n}(2 k-1) y_{k}$, at this time $y_{n}=n, y_{n-1} \geqslant y_{n-2} \geqslant \cdots \geqslant y_{1}$.

Answer: n^2-2


Given the polynomial $p(n)=n^{3}-n^{2}-5 n+2$, find all integers $n$ such that $p^{2}(n)$ is the square of a prime number.

Solution:
Solve: Since $p(n)=(n+2)\left(n^{2}-3 n+1\right)$. Let $p$ be a prime, then the necessary and sufficient condition for $p^{2}(n)=p^{2}$ is $p(n)= \pm p$, i.e., $(n+2)\left(n^{2}-3 n+1\right)= \pm p$.
There are only two possibilities:
(1) $n+2= \pm 1, n^{2}-3 n+1= \pm p$,
(2) $n+2= \pm p, n^{2}-3 n+1= \pm 1$.
In case (1), $n=-1$ or -3. When $n=-1$, $n^{2}-3 n+1=5$ is a prime; when $n=-3$, $n^{2}-3 n+1=19$ is a prime.

In case (2), from $n^{2}-3 n+1=1$ we get $n=0$ or $n=3$, and from $n^{2}-3 n+1=-1$ we get $n=1$ or $n=2$.

When $n=0$, $n+2$ is a prime; when $n=3$, $n+2=5$ is a prime; when $n=1$, $n+2=3$ is a prime; when $n=2$, $n+2=4$ is not a prime.
In summary, all possible values of $n$ are 5: $-1,-3,0,1,3$.

Answer: -1,-3,0,1,3


Consider all functions f from the set of all positive integers into itself satisfying \( f(t^2 f(s)) = s f(t)^2 \) for all s and t. Determine the least possible value of f(1998).

Solution: 120 Solution Let f(1) = k. Then f(kt 2 ) = f(t) 2 and f(f(t)) = k 2 t. Also f(kt) 2 = 1·f(kt) 2 = f(k 3 t 2 ) = f(1 2 f(f(kt 2 ))) = k 2 f(kt 2 ) = k 2 f(t) 2 . Hence f(kt) = k f(t). By an easy induction k n f(t n+1 ) = f(t) n+1 . But this implies that k divides f(t). For suppose the highest power of a prime p dividing k is a > b, the highest power of p dividing f(t). Then a > b(1 + 1/n) for some integer n. But then na > (n + 1)b, so k n does not divide f(t) n+1 . Contradiction. Let g(t) = f(t)/k. Then f(t 2 f(s)) = f(t 2 kg(s)) = k f(t 2 g(s) = k 2 g(t 2 g(s)), whilst s f(t) 2 = k 2 s f(t) 2 . So g(t 2 g(s)) = s g(t) 2 . Hence g is also a function satisfying the conditions which evidently has smaller values than f (for k > 1). It also satisfies g(1) = 1. Since we want the smallest possible value of f(1998) we may restrict attention to functions f satisfying f(1) = 1. Thus we have f(f(t) = t and f(t 2 ) = f(t) 2 . Hence f(st) 2 = f(s 2 t 2 ) = f(s 2 f(f(t 2 ))) = f(s) 2 f(t 2 ) = f(s) 2 f(t) 2 . So f(st) = f(s) f(t). Suppose p is a prime and f(p) = m·n. Then f(m)f(n) = f(mn) = f(f(p)) = p, so one of f(m), f(n) = 1. But if f(m) = 1, then m = f(f(m)) = f(1) = 1. So f(p) is prime. If f(p) = q, then f(q) = p. Now we may define f arbitarily on the primes subject only to the conditions that each f(prime) is prime and that if f(p) = q, then f(q) = p. For suppose that s = p 1 a 1 ...p r a r and that f(p i ) = q i . If t has any additional prime factors not included in the q i , then we may add additional p's to the expression for s so that they are included (taking the additional a's to be zero). So suppose t = q 1 b 1 ...q r b r .Then t 2 f(s) = q 1 2b 1 +a 1 ...q r 2b r +a r and hence f(t 2 f(s) = p 1 2b 1 +a 1 ...p r 2b r +a r = s f(t) 2 . We want the minimum possible value of f(1998). Now 1998 = 2.3 3 .37, so we achieve the minimum value by taking f(2) = 3, f(3) = 2, f(37) = 5 (and f(37) = 5). This gives f(1998) = 3·2 3 5 = 120. 39th IMO 1998 © John Scholes jscholes@kalva.demon.co.uk 26 Oct 1998 Last corrected/updated 20 Aug 03

Answer: 120


Try to find all natural number triples $(A, B, C)$, such that
$$
A^{2}+B-C=100, A+B^{2}-C=124 .
$$

Solution:
Solution: From the problem,
$A^{2}+B-C=100$,
$A+B^{2}-C=124$.
(2) - (1), $A+B^{2}-A^{2}-B=24$,

$\quad(B-A)(A+B-1)=24$.
Since $A, B, C$ are natural numbers,
we can obtain the following system of equations
(1) $\left\{\begin{array}{l}B-A=1, \\ A+B-1=24\end{array}\right.$
(2) $\left\{\begin{array}{l}B-A=24, \\ A+B-1=1\end{array}\right.$;
(3) $\left\{\begin{array}{l}B-A=2, \\ A+B-1=12 ;\end{array}\right.$
(4) $\left\{\begin{array}{l}B-A=12, \\ A+B-1=2 ;\end{array}\right.$
(5) $\left\{\begin{array}{l}B-A=3, \\ A+B-1=8 ;\end{array}\right.$
(6) $\left\{\begin{array}{l}B-A=8, \\ A+B-1=3\end{array}\right.$;
(7) $\left\{\begin{array}{l}B-A=4, \\ A+B-1=6\end{array}\right.$
(8) $\left\{\begin{array}{l}B-A=6 \\ A+B-1=4\end{array}\right.$
Solving (1) and (5) yields $\left\{\begin{array}{l}A=12, \\ B=13 ;\end{array}\left\{\begin{array}{l}A=3, \\ B=6 .\end{array}\right.\right.$
(2), (3), (4), (6), (7), (8) have no solutions in the natural number range.
Substituting $A=12, B=13$ into (1) gives $C=57$;
Substituting $A=3, B=6$ into (1) gives $C=-35$.
Therefore, the natural number solution set that meets the problem's requirements is $A=12, B=$ $13, C=57$.

Answer: A=12, B=13, C=57


Which number is greater,

$$
\log _{3} 2019 \text { or } \quad 4+\sqrt{\log _{3} 18171} \quad ?
$$

Solution: 1. We have $18171=9 \cdot 2019=3^{2} \cdot 2019$, so it follows that $\log _{3} 18171=\log _{3} 3^{2}+\log _{3} 2019=$ $2+\log _{3}$ 2019. Let's introduce the substitution $t=\log _{3}$ 2019. Then we need to compare the numbers $t$ and $4+\sqrt{2+t}$. From $3^{6}=729<2019<2187=3^{7}$ it follows that $6<t<7$. We will prove the inequality $t<4+\sqrt{2+t}$, i.e., $t-4<\sqrt{2+t}$. After squaring, it remains to prove $t^{2}-8 t+16<2+t$, i.e., $t^{2}-9 t+14<0$. The quadratic function on the left side has zeros at points $\frac{9 \pm \sqrt{81-56}}{2}=\frac{9 \pm 5}{2}$, so it is negative for $t \in(2,7)$; since $6<t<7$, the desired function is indeed negative for $t=\log _{3} 2019$, so among the two observed numbers, the larger one is $4+\sqrt{\log _{3} 18171}$.

Answer: 4+\sqrt{\log_{3}18171}


In the morning, 5 foreign cars were parked along the road. By noon, 2 domestic cars were parked between each pair of foreign cars. And by evening, a motorcycle was parked between each pair of adjacent cars. How many motorcycles were parked in total $?$

Answer: 12 .

Solution: Solution. Between 5 foreign cars there are 4 gaps, so there were $4 \cdot 2=8$ domestic cars parked there; that is, a total of $5+8=13$ cars were parked. Between them, 12 motorcycles were parked.

Answer: 12


Compute the largest possible number of distinct real solutions for $x$ to the equation \[x^6+ax^5+60x^4-159x^3+240x^2+bx+c=0,\] where $a$, $b$, and $c$ are real numbers.

Solution:
1. **Understanding the Polynomial and Its Degree:**
   The given polynomial is of degree 6:
   \[
   P(x) = x^6 + ax^5 + 60x^4 - 159x^3 + 240x^2 + bx + c
   \]
   A polynomial of degree 6 can have at most 6 real roots.

2. **Application of Newton's Inequalities:**
   Newton's inequalities provide conditions on the coefficients of a polynomial that can restrict the number of real roots. However, in this context, we need to consider the nature of the roots (real vs. complex).

3. **Complex Roots and Their Pairs:**
   Complex roots of polynomials with real coefficients always come in conjugate pairs. Therefore, the number of complex roots must be even.

4. **Analyzing Possible Numbers of Real Roots:**
   - **6 Real Roots:** If all roots were real, we would have 6 real roots. However, Newton's inequalities or other constraints might prevent this.
   - **5 Real Roots:** This is impossible because the remaining root would have to be complex, and complex roots must come in pairs.
   - **4 Real Roots:** This is possible if the remaining 2 roots are a pair of complex conjugates.
   - **3 Real Roots:** This is possible if the remaining 3 roots are complex, but they must form 1 real root and 1 pair of complex conjugates, which is not possible.
   - **2 Real Roots:** This is possible if the remaining 4 roots are two pairs of complex conjugates.
   - **1 Real Root:** This is possible if the remaining 5 roots are complex, but they must form 2 pairs of complex conjugates and 1 real root, which is not possible.
   - **0 Real Roots:** This is possible if all 6 roots are complex, forming 3 pairs of complex conjugates.

5. **Constructing a Polynomial with 4 Real Roots:**
   To construct a polynomial with exactly 4 real roots, we can consider:
   \[
   (x - r_1)(x - r_2)(x - r_3)(x - r_4)(x^2 + px + q)
   \]
   Here, \(r_1, r_2, r_3, r_4\) are real roots, and \(x^2 + px + q\) has complex conjugate roots. This construction is feasible and satisfies the conditions.

6. **Conclusion:**
   The largest possible number of distinct real solutions for the given polynomial is 4.

The final answer is $\boxed{4}$

Answer: 4


Try to determine the largest integer not exceeding $\frac{\sqrt{14}+2}{\sqrt{14}-2}$

Solution:
4. 3 .

Notice that,
$$
\begin{array}{l}
\frac{\sqrt{14}+2}{\sqrt{14}-2}=\frac{(\sqrt{14}+2)^{2}}{(\sqrt{14}-2)(\sqrt{14}+2)} \\
=\frac{14+4+4 \sqrt{14}}{14-4}=\frac{18+4 \sqrt{14}}{10} .
\end{array}
$$

And $3<\sqrt{14}<4$, thus, $30<18+4 \sqrt{14}<34$.
Therefore, $3<\frac{18+4 \sqrt{14}}{10}<3.4$.
Hence, the largest integer not exceeding $\frac{\sqrt{14}+2}{\sqrt{14}-2}$ is $=3$.

Answer: 3


On a line, there are blue and red points, with no fewer than 5 red points. It is known that on any segment with endpoints at red points, containing a red point inside, there are at least 4 blue points. And on any segment with endpoints at blue points, containing 3 blue points inside, there are at least 2 red points. What is the maximum number of blue points that can be on a segment with endpoints at red points, not containing other red points?

Solution:
Answer: 4.

Solution: Note that on a segment with endpoints at red points, not containing other red points, there cannot be 5 blue points. Indeed, in this case, between the outermost blue points, there would be 3 blue points, which means there would be at least 2 more red points. Therefore, on such a segment, there are no more than 4 blue points.

We will show that 4 blue points can lie on a segment with endpoints at red points, not containing other red points. Suppose the points are arranged on a line in the following order: 2 red - 4 blue - 2 red - 4 blue - 2 red. Then all conditions are satisfied, and there is a segment with 4 blue points.

Comment: It is proven that there are no more than 4 blue points between neighboring red points - 4 points.

A correct example with 4 blue points is provided - 3 points.

Answer: 4


How many ways are there to place eight non-attacking rooks on the remaining squares of an $8 \times 8$ chessboard after deleting the four corners?

Solution:
1. **Label the Columns:**
   Label the columns of the chessboard from 1 through 8, from left to right. Notice that there must be exactly one rook in each of these columns.

2. **Consider the Corners:**
   The four corners of the chessboard are removed. These corners are the squares (1,1), (1,8), (8,1), and (8,8). This means that the rooks in columns 1 and 8 cannot be placed in rows 1 or 8.

3. **Placing Rooks in Columns 1 and 8:**
   For columns 1 and 8, we need to place the rooks in the remaining 6 rows (2 through 7). We need to choose 2 out of these 6 rows for the rooks in columns 1 and 8. The number of ways to choose 2 rows out of 6 is given by the binomial coefficient:
   \[
   \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
   \]
   Since there are 2 columns (1 and 8) and each can be placed in any of the chosen rows, we have:
   \[
   2 \cdot \binom{6}{2} = 2 \cdot 15 = 30
   \]

4. **Placing Rooks in Columns 2 through 7:**
   For columns 2 through 7, we have 6 columns and 6 rows (2 through 7) available. We need to place one rook in each column such that no two rooks are in the same row. The number of ways to arrange 6 rooks in 6 rows is given by the factorial:
   \[
   6! = 720
   \]

5. **Combining the Results:**
   The total number of ways to place the rooks is the product of the number of ways to place the rooks in columns 1 and 8 and the number of ways to place the rooks in columns 2 through 7:
   \[
   2 \cdot \binom{6}{2} \cdot 6! = 30 \cdot 720 = 21600
   \]

The final answer is \(\boxed{21600}\).

Answer: 21600


Fill the circles in the figure with the numbers $2, 4, 6, 8, 10, 12, 14$ so that the sum of the three numbers in each row is equal. What is this equal sum? $\qquad$ (Write down all possible answers.)

Solution: 【Analysis】 $3 A=56+2 X, X=2,8,14 \Rightarrow A=20,24,28$. Examples omitted.

Answer: 20,24,28


As shown in the figure, in the rectangular coordinate system $x O y$, the coordinates of point $A$ are $(10,0)$, and point $B$ moves on the positive half-axis of the $y$-axis. If the lengths of the three sides of $\triangle O A B$ are all integers, then the ordinate of point $B$ is $\qquad$ .

Answer: 24


In a school, 35% are girls, and there are 252 more boys than girls. How many boys and how many girls are there in the school?

Solution: Solution. In the school, there are $65\%$ boys, which means there are $30\%$ more boys than girls. If $30\%$ is 252 children, then $10\%$ is 84 children. Therefore, we conclude that there are 840 students in the school, of which 294 are female, and 546 are male.

Answer: 294


Compute the integral

$$
\int_{-\infty}^{+\infty} \frac{x+1}{x^{2}-2 x+2} \cos x \, dx
$$

Solution:
Solution. To solve the problem, it is sufficient to compute the improper integral

$$
\int_{-\infty}^{+\infty} \frac{x+1}{x^{2}-2 x+2} e^{i x} d x
$$

and use the formula

$$
\int_{-\infty}^{+\infty} \frac{x+1}{x^{2}-2 x+2} \cos x d x=\operatorname{Re} \int_{-\infty}^{+\infty} \frac{x+1}{x^{2}-2 x+2} e^{i x} d x
$$

1. To apply the residue theorem, we introduce the function of a complex variable

$$
f(z)=\frac{z+1}{z^{2}-2 z+2} e^{i z}
$$

and construct a contour consisting of the segment of the real axis $[-\varrho, \varrho]$ and the semicircle $C_{\varrho}=\{|z|=\varrho, \operatorname{Im} z \geqslant 0\}$, choosing $\varrho$ so that all singular points $z_{k}(k=1,2, \ldots, n)$ of the function $f(z)$, lying in the upper half-plane, are inside the contour. Then, by the residue theorem,

$$
\int_{-\varrho}^{\varrho} \frac{x+1}{x^{2}-2 x+2} e^{i x} d x+\int_{C_{\varrho}} f(z) d z=2 \pi i \sum_{k=1}^{n} \operatorname{res}_{z=z_{k}} f(z)
$$

We take the limit as $\varrho \rightarrow+\infty$. Since in our case

$$
g(z)=\frac{z+1}{z^{2}-2 z+2}
$$

is a proper rational fraction and $\lambda=1>0$, the conditions of Jordan's lemma are satisfied, and therefore,

$$
\lim _{\varrho \rightarrow+\infty} \int_{C_{\varrho}} f(z) d z=0
$$

Since the right-hand side in (5) does not depend on $\varrho$, we have

$$
\int_{-\infty}^{+\infty} \frac{(x+1)}{x^{2}-2 x+2} e^{i x} d x=2 \pi i \sum_{k=1}^{n} \operatorname{res}_{z_{k}} \frac{z+1}{z^{2}-2 z+2} e^{i z}
$$

where $z_{k}$ are the singular points of the function

$$
f(z)=\frac{z+1}{z^{2}-2 z+2} e^{i z}
$$

lying in the upper half-plane.

2. We find the singular points of the function

$$
f(z)=\frac{z+1}{z^{2}-2 z+2} e^{i z}=\frac{z+1}{(z-1-i)(z-1+i)} e^{i z}
$$

as the zeros (of the first order) of its denominator: $z=1+i$ and $z=1-i$. Thus, the points $z=1+i$ and $z=1-i$ are poles of the first order. In the upper half-plane, there is only one point $z=1+i$.

3. We compute the residue at the simple pole $z=1+i$ using the formula

$$
\operatorname{res}_{z=z_{0}} \frac{\varphi(z)}{\psi(z)}=\frac{\varphi\left(z_{0}\right)}{\psi^{\prime}\left(z_{0}\right)}
$$

where $\varphi(z)=(z+1) e^{i z}$ and $\psi(z)=z^{2}-2 z+2$. We get

$$
\operatorname{res}_{z=1+i} \frac{(z+1) e^{i z}}{z^{2}-2 z+2}=\left.\frac{(z+1) e^{i z}}{2 z-2}\right|_{z=1+i}=\frac{(2+i) e^{-1+i}}{2 i}
$$

4. We compute the improper integral using formula (6):

$$
\int_{-\infty}^{+\infty} \frac{x+1}{x^{2}-2 x+2} e^{i x} d x=2 \pi i \frac{(2+i) e^{-1+i}}{2 i}=\pi e^{-1}(2+i)(\cos 1+i \sin 1)
$$

5. Using formula (4), we compute the desired integral

$$
\int_{-\infty}^{+\infty} \frac{x+1}{x^{2}-2 x+2} \cos x d x=\operatorname{Re}\left[\int_{-\infty}^{+\infty} \frac{x+1}{x^{2}-2 x+2} e^{i x} d x\right]=
$$

$$
=\operatorname{Re}\left(\pi e^{-1}(2+i)(\cos 1+i \sin 1)\right)=\pi e^{-1}(2 \cos 1-\sin 1)
$$

Answer. $\int_{-\infty}^{+\infty} \frac{x+1}{x^{2}-2 x+2} \cos x d x=\pi e^{-1}(2 \cos 1-\sin 1)$.

### Conditions of the Problems. Compute the integrals using residues.

1. $\int_{-\infty}^{+\infty} \frac{x \cos x}{x^{2}-2 x+10} d x$.
2. $\int_{-\infty}^{+\infty} \frac{\cos 2 x}{x^{2}-x+1} d x$
3. $\int_{-\infty}^{+\infty} \frac{(x-1) \cos x}{x^{2}-2 x+2} d x$
4. $\int_{-\infty}^{+\infty} \frac{x \sin 6 x}{x^{2}+4 x+13} d x$.
5. $\int_{-\infty}^{+\infty} \frac{(x+1) \cos x}{x^{2}-4 x+6} d x$
6. $\int_{0}^{\infty} \frac{x \sin x}{x^{4}+1} d x$.
7. $\int_{0}^{\infty} \frac{\cos x}{\left(x^{2}+4\right)^{2}} d x$.
8. $\int_{0}^{\infty} \frac{\cos x}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x$.
9. $\int_{0}^{\infty} \frac{x \sin 3 x}{x^{2}+16} d x$.
10. $\int_{0}^{\infty} \frac{x \sin x}{\left(x^{2}+1\right)^{2}} d x$.

### Answers.

1. $\frac{\pi}{3} e^{-3}(\cos 1-3 \sin 1)$.
2. $\frac{2 \pi}{\sqrt{3}} e^{-\sqrt{3}} \cos 1$.
3. $-\pi e^{-1} \sin 1$.
4. $\frac{\pi}{3} e^{-18}(3 \cos 12+2 \sin 12)$.
5. $\frac{\pi}{\sqrt{2}} e^{-4 \sqrt{2}}(3 \cos 8-\sqrt{2} \sin 8)$.
6. $\frac{\pi}{2} e^{-1 / \sqrt{2}} \sin \frac{1}{\sqrt{2}}$. 7. $\frac{3 \pi}{32} e^{-2}$.
7. $\frac{\pi}{10}\left(\frac{e^{-2}}{2}-\frac{e^{-3}}{3}\right)$.
8. $\frac{\pi}{2} e^{-12}$.
9. $\frac{\pi}{4} e^{-1}$.

### Chapter 2

### OPERATIONAL CALCULUS

When studying the topic of OPERATIONAL CALCULUS, you will become familiar with the concepts of the original function and its image (by Laplace), study the properties of originals and images, and learn to apply them to solve ordinary differential equations and systems using the operational method.

Using the RESHEBNIK VM package, you will be able to compute integrals, find the partial fraction decomposition of rational functions, compute derivatives, and perform all other actions necessary for studying the topic. Once you are proficient in it, you will be able to compute images and restore originals from their images with simple keystrokes on the computer keyboard, using the capabilities of the STEM Plus module, which is part of the RESHEBNIK VM package.

### 2.1. Concepts of Original and Image

### Problem Statement. Prove that the function $f(t)$ is an original and find its image (by Laplace).

### Plan of Solution. A complex-valued function $f(t)$ of a real variable $t$ is called an original if it satisfies three conditions:

a) $f(t) \equiv 0$ for all $t<0$ and $s \geqslant 0$ such that

$$
|f(t)|<M e^{s t} \quad \text{for all } t>0
$$

The smallest number $s$ for which this inequality holds is called the growth indicator of the function $f(t)$.

If $f(t)$ is an original, then its Laplace transform $F(p)$ (where $p$ is a complex variable) is defined by the formula

$$
F(p)=\int_{0}^{\infty} e^{-p t} f(t) d t
$$

The function $F(p)$ of the complex variable $p$ is also called the image (by Laplace) of the function $f(t)$. The connection between the original and the image is denoted by the symbol

$$
f(t) \longleftrightarrow F(p) .
$$

Remark. In the half-plane $\operatorname{Re} p>s$ (where $s$ is the growth indicator of the original $f(t)$), the integral (1) converges absolutely and defines an analytic function $F(p)$.

1. We prove that the function $f(t)$ is an original by verifying the conditions a)-c). We determine the growth indicator $s$ of the function $f(t)$.
2. We find the

Answer: \pie^{-1}(2\cos1-\sin1)


Make all permutations of 5 distinct elements $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$, where $a_{1}$ is not in the 1st or 2nd position, $a_{2}$ is not in the 2nd or 3rd position, $a_{3}$ is not in the 5th position, $a_{4}$ is not in the 4th or 5th position, and $a_{5}$ is not in the 3rd or 4th position. How many different permutations can be made?

Solution:
Solution: This is a permutation problem with restrictions, and the corresponding $5 \times 5$ chessboard $R_{5}$ with forbidden cells is shown in Figure 4.5. The rook polynomial of the forbidden cell chessboard $C$ of $R_{5}$ is
$$
\begin{array}{l}
=t \cdot(\times \times) \cdot\left(\begin{array}{cc}
& \times \\
\times & \times \\
\times &
\end{array}\right)+\left(\begin{array}{ll}
\times & \times \\
& \times
\end{array}\right) \cdot\left(\begin{array}{ll}
\times \\
\times & \times \\
\times & \times
\end{array}\right) \\
=t(1+2 t)\left(1+4 t+3 t^{2}\right)+ \\
\left(1+3 t+t^{2}\right)\left(1+5 t+6 t^{2}+t^{3}\right) \\
=\left(t+6 t^{2}+11 t^{3}+6 t^{4}\right)+ \\
\left(1+8 t+22 t^{2}+24 t^{3}+9 t^{4}+t^{5}\right) \\
=1+9 t+28 t^{2}+35 t^{3}+15 t^{4}+t^{5} \text {, } \\
\end{array}
$$

Therefore, the hit polynomial of $R_{5}$ is
$$
\begin{aligned}
E(t)= & 5!+9 \cdot 4!(t-1)+28 \cdot 3!(t-1)^{2}+ \\
& 35 \cdot 2!(t-1)^{3}+15(t-1)^{4}+(t-1)^{5},
\end{aligned}
$$

Thus, the number of all permutations is
$$
\begin{aligned}
N & =E(0) \\
& =5!-9 \cdot 4!+28 \cdot 3!-35 \cdot 2!+15-1 \\
& =120-216+168-70+15-1 \\
& =16 .
\end{aligned}
$$

Answer: 16


For nonnegative integers $a$ and $b$ with $a + b \leq 6$, let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$. Let $S$ denote the sum of all $T(a, b)$, where $a$ and $b$ are nonnegative integers with $a + b \leq 6$. Find the remainder when $S$ is divided by $1000$.

Solution:
Let $c=6-(a+b)$, and note that $\binom{6}{a + b}=\binom{6}{c}$. The problem thus asks for the sum $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$. Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately, which is equal to $\binom{18}{6}=18564$. Therefore, the answer is $\boxed{564}$.
-rocketscience

Answer: 564


Solution:
4. (A).

Let $y=-x$, then $y \in\left(0, \frac{1}{2}\right)$, and $a_{1}=$ $\cos (\sin y \pi), a_{2}=\sin (\cos y \pi), a_{3}=\cos (1-y) \pi<0$.
Since $\sin y \pi+\cos y \pi$
$$
=\sqrt{2} \sin \left(y \pi+\frac{\pi}{4}\right) \leqslant \sqrt{2}<\frac{\pi}{2},
$$

therefore, $0<\cos y \pi<\frac{\pi}{2}-\sin y \pi<\frac{\pi}{2}$.
Thus, $0<\sin (\cos y \pi)<\cos (\sin y \pi)$.

Answer: A


Find all integer pairs $(x, y)$ such that
$$
1+2^{x}+2^{2 x+1}=y^{2} .
$$

Solution:
4. If $(x, y)$ is a solution, then $x \geqslant 0, (x, -y)$ is also a solution.

When $x=0$, the solutions are $(0,2), (0,-2)$.
Assume $(x, y)$ is a solution, $x>0$. Without loss of generality, let $y>0$.
Thus, the original equation is equivalent to
$$
2^{x}\left(1+2^{x+1}\right)=(y-1)(y+1) \text {. }
$$

Therefore, $y-1$ and $y+1$ are even numbers, with exactly one being divisible by 4. Hence, $x \geqslant 3$, and one of the factors is divisible by $2^{x-1}$ but not by $2^{x}$.
So, $y=2^{x-1} m+\varepsilon$, where $m$ is an odd number, and $\varepsilon= \pm 1$.
Substituting into the original equation, we get
$$
\begin{array}{l}
2^{x}\left(1+2^{x+1}\right)=\left(2^{x-1} m+\varepsilon\right)^{2}-1 \\
=2^{2 x-2} m^{2}+2^{x} m \varepsilon,
\end{array}
$$

which simplifies to $1+2^{x+1}=2^{x-2} m^{2}+m \varepsilon$.
Thus, $1-\varepsilon m=2^{x-2}\left(m^{2}-8\right)$.
When $\varepsilon=1$, we have $m^{2}-8 \leqslant 0$, i.e., $m=1$, which does not satisfy the equation.
When $\varepsilon=-1$, we have
$$
1+m=2^{x-2}\left(m^{2}-8\right) \geqslant 2\left(m^{2}-8\right),
$$

which implies
$$
2 m^{2}-m-17 \leqslant 0 \text {. }
$$

Therefore, $m \leqslant 3$.
On the other hand, $m \neq 1$, and since $m$ is odd, we have $m=3$.
Thus, $x=4, y=23$.
Therefore, all solutions are
$$
(0,2),(0,-2),(4,23),(4,-23) \text {. }
$$

Answer: (0,2),(0,-2),(4,23),(4,-23)


Fix an integer $n \geq 2$. A fairy chess piece leopard may move one cell up, or one cell to the right, or one cell diagonally down-left. A leopard is placed onto some cell of a $3 n \times 3 n$ chequer board. The leopard makes several moves, never visiting a cell twice, and comes back to the starting cell. Determine the largest possible number of moves the leopard could have made.

Solution:
Solution. The required maximum is $9 n^{2}-3$. We first show that the number of visited cells is divisible by 3 . To this end, refer to a coordinate frame whose axes point upward and rightward, respectively, so that the centres of the cells have integral coordinates in the range 1 through $3 n$. Assign each cell $(i, j)$ the residue of $i+j$ modulo 3 . Notice that equally many cells are assigned each residue, since this is the case for every $3 \times 1$ rectangle. Any move of the leopard is made from a cell assigned residue $r$ to one assigned residue $r+1$ modulo 3. Hence, the residues along any path of the leopard are $\cdots \rightarrow 0 \rightarrow 1 \rightarrow 2 \rightarrow 0 \rightarrow 1 \rightarrow \cdots$. Consequently, every cyclic path contains equally many cells assigned each residue, so the total number of cells along the cycle is divisible by 3 .

Next, we show that the leopard cannot visit all cells. Argue indirectly. The leopard's cycle forms a closed non-self-intersecting polyline enclosing some region along whose boundary the leopard moves either clockwise or counter-clockwise. On the other hand, the cell $(3 n, 1)$ may be visited only as $(3 n-1,1) \rightarrow(3 n, 1) \rightarrow(3 n, 2)$, while the cell $(1,3 n)$ may be visited only as $(1,3 n-1) \rightarrow(1,3 n) \rightarrow(2,3 n)$ - see the left figure below. The region bounded by the leopard's cycle lies to the left of the path in the former case, but to the right of the path in the latter case. This is a contradiction.

The argument above shows that the number of visited cells does not exceed $9 n^{2}-3$. The figure in the middle below shows that this bound is indeed achieved.

![](https://cdn.mathpix.com/cropped/2024_06_04_1105870bb241d4c1b2cbg-11.jpg?height=485&width=1591&top_left_y=1395&top_left_x=234)

Remarks. (1) Variations of the argument in the second paragraph above may equally well work. For instance, consider the situation around the cell $(1,1)$. The cycle comes to that cell from $(2,2)$, and leaves it to, say, $(1,2)$. The previous part of the cycle should then look like $(i, 2) \rightarrow(i-1,1) \rightarrow(i-1,2) \rightarrow(i-2,1) \rightarrow \cdots \rightarrow(2,2) \rightarrow(1,1) \rightarrow(1,2)$, say with a maximal $i$. Then the cell $(i, 1)$ could only be visited as $(i+1,2) \rightarrow(i, 1) \rightarrow(i+1,1)$; again, the two parts have opposite orientations with respect to the interior region.

(2) The problem becomes more symmetric under an affine transformation mapping the square lattice to the honeycomb illustrated in the right figure above.

Answer: 9n^2-3


Let $F=\left\{(x, y, z) \in \mathbb{R}^{3}: x+y+z=0\right\}$ be a subspace of $\mathbb{R}^{3}$. Determine a basis of $F$ and then the dimension of $F$.

Solution: . We write: $F=\left\{(x, y, z) \in \mathbb{R}^{3}: x=-y-z, y=y, z=z\right\}=$ $\{(-y-z, y, z), y, z \in \mathbb{R}\}=\{(-y, y, 0)+(-z, 0, z), y, z \in \mathbb{R}\}=\operatorname{Vect}\{(-1,1,0),(-1,0,1)\}$. This writing allows us to deduce that the family $\{(-1,1,0),(-1,0,1)\}$ is a generating set of $F$. We also notice that this family is linearly independent. Thus, it is a basis of $F$. Since it contains 2 elements, we conclude that $F$ is of dimension 2 (it is then a vector plane).

Answer: 2


On an island, there live red, yellow, green, and blue chameleons.

- On a cloudy day, either one red chameleon changes its color to yellow, or one green chameleon changes its color to blue.
- On a sunny day, either one red chameleon changes its color to green, or one yellow chameleon changes its color to blue.

In September, there were 18 sunny and 12 cloudy days. As a result, the number of yellow chameleons increased by 5. By how much did the number of green chameleons increase?

Solution:
Answer: 11.

Solution. Let $A$ be the number of green chameleons on the island, and $B$ be the number of yellow chameleons. Consider the quantity $A-B$. Note that each cloudy day it decreases by 1, and each sunny day it increases by 1. Since there were $18-12=6$ more sunny days than cloudy days in September, the quantity $A-B$ increased by 6 over this period. Since $B$ increased by 5, $A$ must have increased by $5+6=11$.

Answer: 11


Divide the number 80 into two parts such that $30\%$ of one part is 10 more than $20\%$ of the other part. Write the smaller of the two parts in your answer.

Solution:
Answer: 28

Solution. Let one part of the number be $x$, then the other part will be $80-x$. We get the equation $0.3 \cdot x = 0.2 \cdot (80 - x) + 10$, solving it we get $x = 52$, and the other part of the number is 28.

Answer: 28


What is the probability that the chords $AB$ and $CD$ intersect when points $A, B, C$, and $D$ are randomly placed on the circumference of a circle, independently of each other?

Solution: Solution. The chords intersect if the points on the circumference are in the order $A C B D$ in one of the traversal directions. Place the points at arbitrary positions on the circumference. We can do this because the placement does not affect the intersection, only the names we give to the points. Let one of the placed points be $A$. Which one we choose still does not affect the intersection. If we now label one of the neighbors of $A$ as $B$, then the segments $A B$ and $C D$ do not intersect if we label the "opposite" one, otherwise there will be an intersection point. The selection of $B$ can be made randomly, so there is a $\frac{2}{3}$ probability of choosing a point adjacent to $A$. According to this, the probability that the chords intersect is $\frac{1}{3}$.

Answer: \frac{1}{3}


Let $ABCD$ be a right trapezoid with bases $AB$ and $CD$, and right angles at $A$ and $D$. Given that the shorter diagonal $BD$ is perpendicular to the side $BC$, determine the smallest possible value for the ratio $\frac{CD}{AD}$.

Solution:
Let $A \widehat{B} D=B \widehat{D} C=\alpha$. Then we have that $D C=\frac{B D}{\cos \alpha}$ and $A D=B D \sin \alpha$, hence

$\frac{D C}{A D}=\frac{\frac{B D}{\cos \alpha}}{B D \sin \alpha}=\frac{1}{\sin \alpha \cos \alpha}=\frac{2}{\sin 2 \alpha} \geq 2$.

Equality occurs when $\sin 2 \alpha=1$, that is, when $\alpha=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-42.jpg?height=326&width=576&top_left_y=1935&top_left_x=1143)

Answer: 2


In quadrilateral $ABCD$, the angle bisectors of $\angle A$, $\angle B$, $\angle C$, and $\angle D$ intersect at a single point $P$. Let the areas of $\triangle APD$, $\triangle APB$, $\triangle BPC$, and $\triangle DPC$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$, respectively. Then ( . ).
(A) $S_{1}+S_{3}=S_{2}+S_{4}$
(B) $S_{1}+S_{2}=S_{3}+S_{4}$
(C) $S_{1}+S_{4}=S_{2}+S_{3}$
(D) $S_{1}+S_{3} \neq S_{2}+S_{4}$

Solution:
6. A.

From the point on the angle bisector to the two sides of the angle, the distances are equal, so the distances from point $P$ to each side of the quadrilateral are equal, denoted as $h$ (as shown in Figure 11). It is easy to see that
$\triangle A P E \cong \triangle A P F$,
$\triangle B P E \cong \triangle B P H$,
$\triangle C P G \cong \triangle C P H$,
$\triangle D P G \cong \triangle D P F$.

Adding them up, we get $S_{\triangle A B P}+S_{\triangle C D P}=S_{\triangle B C P}+S_{\triangle D P}$.

Answer: A


In the Rhind Papyrus (Ancient Egypt), among other information, there are decompositions of fractions into the sum of fractions with a numerator of 1, for example,

$\frac{2}{73}=\frac{1}{60}+\frac{1}{219}+\frac{1}{292}+\frac{1}{x}$

One of the denominators here is replaced by the letter $x$. Find this denominator.

Solution:
# Solution:

First, find $\stackrel{1}{-.}$ from the equation:

$\frac{2}{73}=\frac{1}{60}+\frac{1}{219}+\frac{1}{292}+\frac{1}{x}$

$\frac{2}{73}-\frac{1}{219}-\frac{1}{292}-\frac{1}{60}=\frac{1}{x}$

$\frac{2}{73}-\frac{1}{73 \cdot 3}-\frac{1}{73 \cdot 4}-\frac{1}{60}=\frac{1}{x}$

$\frac{2 \cdot 3 \cdot 4-4-3}{73 \cdot 3 \cdot 4}-\frac{1}{60}=\frac{1}{x}$
$\frac{17}{73 \cdot 3 \cdot 4}-\frac{1}{5 \cdot 3 \cdot 4}=\frac{1}{x}$

$\frac{17 \cdot 5-73}{73 \cdot 3 \cdot 4 \cdot 5}=\frac{1}{x}$

$\frac{12}{73 \cdot 3 \cdot 4 \cdot 5}=\frac{1}{x}$

$\frac{1}{73 \cdot 5}=\frac{1}{x}$

$\frac{1}{365}=\frac{1}{x}$

Thus, $x=365$.

Answer: $x=365$.

Answer: 365


In the relay race (4 x 100 meters), 8-man teams compete in the final. How many runners are competing in total for the victory?

Solution: $8 \cdot 4=32$ To fight for the victory, 32 runners are competing.

Answer: 32


Find all pairs of natural numbers $a$ and $b$, such that their greatest common divisor is equal to both numbers 30 - $a$ and 42 - $b$.

Solution:
Solution. Let $d$ be the greatest common divisor of numbers $a$ and $b$. Since $d$ divides both $a$ and $b$ and also $d=30-a=42-b$, $d$ divides both numbers 30 and 42, and thus also their greatest common divisor 6. For each of the possibilities $d \in\{1,2,3,6\}$, we can easily calculate the values of $a=30-d$ and $b=42-d$ and verify whether $d$ is indeed their greatest common divisor $(a, b)$:

| $d$ | $a$ | $b$ | $(a, b)$ |
| :--- | :--- | :--- | :---: |
| $\mathbf{1}$ | $\mathbf{2 9}$ | $\mathbf{4 1}$ | $\mathbf{1}$ |
| 2 | $28=2^{2} \cdot 7$ | $40=2^{3} \cdot 5$ | 4 |
| $\mathbf{3}$ | $\mathbf{2 7}=3^{3}$ | $\mathbf{3 9}=3 \cdot 13$ | $\mathbf{3}$ |
| 6 | $24=2^{3} \cdot 3$ | $36=2^{2} \cdot 3^{2}$ | 12 |

The values in the first and last columns of the table match only for $d=1$ and $d=3$, so the solutions to the problem are the pairs $a=29, b=41$ and the pairs $a=27, b=39$.

Answer: (29,41),(27,39)


A bouncy ball falls from point $A$ to the ground, bounces up to point $B$, then falls to a platform $20 \mathrm{~cm}$ high, bounces up to point $C$, and finally falls to the ground. Each time it bounces, the height it reaches is $80 \%$ of the height from which it fell. It is known that point $A$ is $68 \mathrm{~cm}$ higher above the ground than point $C$. Find the height of point $C$ above the ground.

Solution:
Three, 15. Let the height of point $C$ above the ground be $x \mathrm{~cm}$, then $\frac{\frac{x-20}{80 \%}+20}{80 \%}-68=x$. Solving this, we get $x=132$. Answer:略.

Note: The word "略" at the end is not translated as it seems to be a placeholder or abbreviation in the original text, possibly indicating that the detailed steps or further explanation are omitted. If you need a specific translation for "略", please provide more context.

Answer: 132


In an arm wrestling tournament, 510 athletes are participating. 1 point is awarded for a win, and 0 points for a loss. If the winner initially had fewer points than the opponent, the winner additionally receives one point from the loser. In each round, participants with a difference of no more than 1 point in their scores compete. The tournament ends as soon as a sole leader is determined. What is the minimum number of rounds that need to be played?

Solution:
# Answer: 9.

Solution. Let there be $N=2 m+k$ athletes in the leading group before the next round, with $2 m$ of them meeting each other, and $k$ meeting participants who have one point less. Note that if athletes with $n$ and $n+1$ points meet, they will end up with $n$ and $n+2$ points regardless of the outcome. Therefore, after the round, there will be $m+k$ participants in the leading group. The minimum value of this sum is $\frac{N}{2}$ for even $N$ and $\frac{N+1}{2}$ for odd $N$. According to the problem, after the first round, there will be no fewer than 255 leaders, after the second round - no fewer than 128, and after each subsequent round, the number of leaders will decrease by no more than half. Therefore, it will take no fewer than $2+\log _{2} 128=9$ rounds.

We will show that the tournament can be completed in 9 rounds. We will prove two statements.

1) In a tournament with $2^{n}$ participants, the winner can be determined in $n$ rounds. We will use induction on $n$. For $n=1$, this is obvious. Suppose the statement is true for some $n$. Then, in a tournament with $2^{n+1}$ participants, after the first round, $2^{n}$ athletes will have 0 points and $2^{n}$ will have 1 point. We will conduct two independent tournaments in these groups. By the induction hypothesis, both can be completed in $n$ rounds. The winner of the second tournament will become the overall winner, and he will be determined in $n+1$ rounds.
2) For $n \geqslant 3$, in a tournament with $2^{n}-2$ participants, the winner can be determined in $n$ rounds. After the first round, $2^{n-1}-1$ athletes will have 0 points and $2^{n-1}-1$ will have 1 point. If in the second round, only one pair of opponents has a different number of points, then after the round, $2^{n-2}$ participants will have 0 points and 2 points, and $2^{n-1}-2$ participants will have 1 point. We will now use induction on $n$. For $n=3$, we will pair athletes with the same number of points, after which the tournament will end. Suppose $n>3$ and the statement is true for $n-1$. By statement 1) and the induction hypothesis, independent tournaments can be organized in the groups with 0, 1, and 2 points, and the first and third groups will end in $n-2$ rounds, while the second group will end in $n-1$ rounds (but the last round will not be needed). The winner of the third group will win the entire tournament, as he will have $n+1$ points, while participants from other groups will have no more than $n$ points. In total, the tournament will last $n$ rounds.

It remains to apply statement 2) for $n=9$.

Answer: 9


Simplify the expression $\frac{2 \cos ^{2} x}{\sin x-\sin ^{-1} x}-\sin x$. For which values of $x$ does the expression have no meaning?

Solution:
2. We consider $\sin ^{-1} x=\frac{1}{\sin x}$ and rearrange the denominator $\sin x-\sin ^{-1} x=\sin x-\frac{1}{\sin x}=\frac{\sin ^{2} x-1}{\sin x}$. We use the relationship between sine and cosine $\sin ^{2} x+\cos ^{2} x=1$, from which we express $\sin ^{2} x-1=-\cos ^{2} x$. We transform the fraction into $\mathrm{v} \frac{2 \cos ^{2} x}{\frac{-\cos 2 x}{\sin x}}=-2 \sin x, \cos x \neq 0$ and $\sin x \neq 0$ or $x \neq \frac{\pi}{2}+k \pi$ and $x \neq k \pi, k \in \mathbb{Z}$. We simplify the expression $-2 \sin x-\sin x=-3 \sin x$. The expression is undefined for $x=\frac{k \pi}{2}, k \in \mathbb{Z}$.

The simplified denominator $\sin x-\sin ^{-1} x=\sin x-\frac{1}{\sin x}=\frac{\sin ^{2} x-1}{\sin x} \ldots \ldots \ldots \ldots 1+1^{*}$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_97c791425a4f6cf5fca5g-13.jpg?height=88&width=1630&top_left_y=1755&top_left_x=207)

![](https://cdn.mathpix.com/cropped/2024_06_07_97c791425a4f6cf5fca5g-13.jpg?height=54&width=1636&top_left_y=1829&top_left_x=210)

![](https://cdn.mathpix.com/cropped/2024_06_07_97c791425a4f6cf5fca5g-13.jpg?height=60&width=1627&top_left_y=1872&top_left_x=206)

![](https://cdn.mathpix.com/cropped/2024_06_07_97c791425a4f6cf5fca5g-13.jpg?height=54&width=1625&top_left_y=1926&top_left_x=210)

Answer: -3\sinx


What is the largest angle of the triangle if

$$
\frac{1}{V_{a}^{2}}=\frac{1}{V_{b}^{2}}+\frac{1}{V_{c}^{2}} ?
$$

Solution:
Let's denote the sides and altitudes of the triangle in the usual way as $a, b, c, m_{a}, m_{b}, m_{c}$. If we rotate the triangle around one of its sides, the resulting solid is either a cone, or the sum or difference of two cones, depending on whether there is no obtuse angle or there is one among the angles on the axis of rotation. The volume of the resulting solid in each case (using the notation in the diagram):
![](https://cdn.mathpix.com/cropped/2024_05_02_7d66de0a624e4f23811dg-1.jpg?height=1328&width=370&top_left_y=347&top_left_x=866)

\[
\begin{gathered}
V_{1}=\frac{\pi}{3} \cdot A C^{2} \cdot A B=\frac{\pi}{3} \cdot m_{c}^{2} \cdot c \\
V_{2}=\frac{\pi}{3} \cdot C T^{2} \cdot A T+\frac{\pi}{3} \cdot C T^{2} \cdot B T=\frac{\pi}{3} \cdot m_{c}^{2} \cdot c \\
V_{3}=\frac{\pi}{3} \cdot C T^{2} \cdot B T-\frac{\pi}{3} \cdot C T^{2} \cdot A T=\frac{\pi}{3} \cdot m_{c}^{2} \cdot c
\end{gathered}
\]

Thus, if we rotate the triangle around the side $c$, the volume of the resulting solid is $\frac{\pi}{3} \cdot m_{c}^{2} \cdot c$ in all cases. Similarly, $V_{a}=\frac{\pi}{3} \cdot m_{a}^{2} \cdot a$ and $V_{b}=\frac{\pi}{3} \cdot m_{b}^{2} \cdot b$. Substituting these into our original equation and rearranging, we get:

\[
\frac{1}{m_{a}^{4} \cdot a^{2}}=\frac{1}{m_{b}^{4} \cdot b^{2}}+\frac{1}{m_{c}^{4} \cdot c^{2}}
\]

If the area of the triangle is $t$, then $16 t^{4}=m_{a}^{4} \cdot a^{4}=m_{b}^{4} \cdot b^{4}=m_{c}^{4} \cdot c^{4}$. Using this, we can rewrite (1) as:

\[
\frac{1}{\frac{t^{4}}{a^{2}}}=\frac{1}{\frac{t^{4}}{b^{2}}}+\frac{1}{\frac{t^{4}}{c^{2}}}
\]

or:

\[
a^{2}=b^{2}+c^{2}
\]

From this, by the converse of the Pythagorean theorem, it follows that the angle opposite the side $a$ is a right angle; thus, the largest angle of the triangle is $90^{\circ}$.

Answer: 90


A beam of light with a diameter of $d_{1}=5 \mathrm{~cm}$ falls on a thin diverging lens with an optical power of $D_{p}=-6$ Diopters. On a screen positioned parallel to the lens, a bright spot with a diameter of $d_{2}=20 \mathrm{~cm}$ is observed. After replacing the thin diverging lens with a thin converging lens, the size of the spot on the screen remains unchanged. Determine the optical power $D_{c}$ of the converging lens.

Solution:
# Answer: 10 Dptr

Solution. The optical scheme corresponding to the condition:

(3 points)

![](https://cdn.mathpix.com/cropped/2024_05_06_be79a21ed2ca3a6294abg-10.jpg?height=825&width=1170&top_left_y=73&top_left_x=520)

The path of the rays after the diverging lens is shown in black, and after the converging lens in red. We have $C D=\left|\frac{1}{D_{p}}\right|=\frac{1}{6}$.

From the similarity of triangles, it follows that: $\frac{C D}{C F}=\frac{B K}{A L}=\frac{1}{4}$,

then $D F=C F-C D=4 C D-C D=3 C D=\frac{1}{2}$.

From the similarity of triangles, it follows that: $\frac{D E}{F E}=\frac{B K}{A L}$.

We get that: $D E=\frac{1}{5} D F=\frac{1}{10}$.

As a result, the optical power of the converging lens: $D_{c}=\frac{1}{D E}=10$ Dptr.

(2 points)

Answer: 10


Let $F_{1}$ and $F_{2}$ be the two foci of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, and $P$ be a point on the ellipse such that $\left|P F_{1}\right|:\left|P F_{2}\right|=2: 1$. Then the area of $\triangle P F_{1} F_{2}$ is $\mathrm{J}$

Solution:
From the definition of an ellipse, we know that $\left|P F_{1}\right|+\left|P F_{2}\right|=2 a=6$, and $\left|P F_{1}\right|$ : $\left|P F_{2}\right|=2: 1$. Therefore, $\left|P F_{1}\right|=4 \cdot\left|P F_{2}\right|=2$, and $\left|F_{1} F_{2}\right|=2 c=$ $2 \sqrt{5}$. Combining $4^{2}+2^{2}=(2 \sqrt{5})^{2}$, we know that $\triangle P F_{1} F_{2}$ is a right triangle, hence
$$
S_{\triangle P F_{1} F_{0}}=\frac{1}{2}\left|P F_{1}\right| \cdot\left|P F_{2}\right|=4 .
$$

Answer: 4


In the following drawing, the chords $DE$ and $BC$ are perpendicular, with $BC$ being a diameter of the circle with center at $A$. Additionally, $\angle CGF=40^{\circ}$ and $GH=2 \text{~cm}$.

a) Determine the value of the angle $\angle CHF$.
b) Find the length of $HJ$.

Solution:
Solution

a) Since $BC$ is a diameter, it follows that $\angle BFC=90^{\circ}$. Thus, as we also have $\angle CHG=90^{\circ}$, the circle $\Gamma$ with diameter $CG$ passes through $F$ and $H$. In this circle, the angles $\angle CGF$ and $\angle CHF$ are inscribed in the same arc $CF$, so $\angle CHF = \angle CGF = 40^{\circ}$.

b) Observing the circle $\Gamma$ again, we can conclude that $\angle HCG = \angle HFG$, as both are inscribed in the arc $GH$. Considering now the circle with diameter $BC$, we have $\angle ICB = \angle IFB$, because both are inscribed in the arc $IB$. Thus, $\angle ICH = \angle BFH = \angle HFG = \angle HCG$. Therefore, since the right triangles $CHG$ and $CHJ$ have the same angles and a common leg, they are congruent, resulting in $HJ = GH = 2 \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-33.jpg?height=746&width=667&top_left_y=259&top_left_x=782)

Answer: 2


Find the derivative of the function $f(x) = 3x^4 - 2userImageRelation not relevant to the problem was mistakenly included. Please disregard that part and focus on the math problem provided.

Solution:
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{2 x-1}{4} \cdot \sqrt{2+x-x^{2}}+\frac{9}{8} \cdot \arcsin \frac{2 x-1}{3}\right)^{\prime}= \\
& =\frac{2}{4} \cdot \sqrt{2+x-x^{2}}+\frac{2 x-1}{4} \cdot \frac{1}{2 \sqrt{2+x-x^{2}}} \cdot(1-2 x)+\frac{9}{8} \cdot \frac{1}{\sqrt{1-\left(\frac{2 x-1}{3}\right)^{2}}} \cdot \frac{2}{3}= \\
& =\frac{\sqrt{2+x-x^{2}}}{2}-\frac{(2 x-1)^{2}}{8 \sqrt{2+x-x^{2}}}+\frac{9}{4 \sqrt{9-(2 x-1)^{2}}}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{4\left(2+x-x^{2}\right)}{8 \sqrt{2+x-x^{2}}}-\frac{4 x^{2}-4 x+1}{8 \sqrt{2+x-x^{2}}}+\frac{9}{4 \sqrt{9-\left(4 x^{2}-4 x+1\right)}}= \\
& =\frac{8+4 x-4 x^{2}-4 x^{2}+4 x-1}{8 \sqrt{2+x-x^{2}}}+\frac{9}{4 \sqrt{8-4 x^{2}+4 x}}= \\
& =\frac{7+8 x-8 x^{2}}{8 \sqrt{2+x-x^{2}}}+\frac{9}{8 \sqrt{2+x-x^{2}}}= \\
& =\frac{16+8 x-8 x^{2}}{8 \sqrt{2+x-x^{2}}}=\frac{2+x-x^{2}}{\sqrt{2+x-x^{2}}}=\sqrt{2+x-x^{2}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 10-3

Answer: \sqrt{2+x-x^{2}}


Note that $x=0$ is not a root of the equation for any $a$. Transform the equation by dividing it by $x^{3}$:

$$
\left(a^{2}+1\right)\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)+2=12 a
$$

Thus, if $x_{0}$ is a root of the equation, then $\frac{1}{x_{0}}$ is also a root. Therefore, the only possible roots are $x=1$ or $x=-1$.

In the first case, we get $8 a^{2}+4=12 a$, from which $a=1$ or $a=\frac{1}{2}$, and in the second case, we have $8 a^{2}=-12 a$, from which $a=0$ or $a=-\frac{3}{2}$. Substituting the values $a=-3 / 2,1 / 2,1$ into the equation $\left(a^{2}+1\right) t^{3}-3 t+2-12 a=0$, where $t=x+\frac{1}{x}$, we verify that for each of them, it has a unique solution $t=2$ or $t=-2$, and the original equation has a unique solution $x=1$ or $x=-1$. In the case $a=0$, the equation $t^{3}-3 t+2=0$ has two solutions $t=-2$ and $t=1$. The root $t=-2$ gives the solution $x=-1$, while the root $t=1$ does not provide solutions for $x$.

Solution:
Answer: $a=-\frac{3}{2}, 0, \frac{1}{2}, 1$.

Answer to option: $5-2: a=-\frac{2}{3}, 1, 2$.

## Lomonosov Moscow State University

Olympiad "Conquer Sparrow Hills"

Variant 2-1 (Chelyabinsk)

Answer: -\frac{3}{2},0,\frac{1}{2},1


Given the parabola $C: y=a x^{2}(a>0)$, the line $y=x+2$ intersects the parabola at points $A$ and $B$, and $M$ is the midpoint of segment $AB$. A vertical line through $M$ intersects the parabola $C$ at point $N$.
(1) Prove: The tangent line $l$ to the parabola $C$ at point $N$ is parallel to $AB$.
(2) Does there exist a real number $a$ such that $\overrightarrow{N A} \cdot \overrightarrow{N B}=0$? If it exists, find the value of $a$; if not, explain the reason.

Solution:
13. (1) From $\left\{\begin{array}{l}y=x+2, \\ y=a x^{2}\end{array}\right.$ we get $a x^{2}-x-2=0$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then $x_{1}+x_{2}=\frac{1}{a}, x_{1} x_{2}=-\frac{2}{a}$.
Thus, $x_{N}=x_{M}=\frac{x_{1}+x_{2}}{2}=\frac{1}{2 a}, y_{N}=a x_{N}^{2}=\frac{1}{4 a}$.
From $y^{\prime}=\left(a x^{2}\right)^{\prime}=2 a x$, we know that the slope of the tangent line $l$ to the parabola $C$ at point $N$ is $k_{l}=2 a \cdot \frac{1}{2 a}=1$.

Therefore, the tangent line $l$ to the parabola $C$ at point $N$ is parallel to the line $A B$.
(2) Suppose there exists a real number $a$ such that $\overrightarrow{N A} \cdot \overrightarrow{N B}=0$. Since $M$ is the midpoint of segment $A B$, we have
$$
\overrightarrow{N A} \cdot \overrightarrow{N B}=0 \Leftrightarrow N A \perp N B \Leftrightarrow|M N|=\frac{1}{2}|A B| \text {. }
$$

Since $M N \perp x$-axis, we have
$$
\begin{array}{l}
|M N|=\left|\frac{1}{2 a}+2-\frac{1}{4 a}\right|=\frac{1}{4 a}+2 . \\
\text { Also, }|A B|=\sqrt{2} \cdot\left|x_{1}-x_{2}\right| \\
=\sqrt{2} \cdot \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{2} \cdot \sqrt{\frac{1}{a^{2}}+\frac{8}{a}},
\end{array}
$$

Then $\left(\frac{1}{4 a}+2\right)^{2}=\frac{1}{4} \times 2\left(\frac{1}{a^{2}}+\frac{8}{a}\right)$.
Solving this, we get $a=\frac{7}{8}$ or $a=-\frac{1}{8}$ (discard).
Therefore, there exists a real number $a=\frac{7}{8}$ such that $\overrightarrow{N A} \cdot \overrightarrow{N B}=0$.

Answer: a=\frac{7}{8}