The sides $\overline{B C}$, $\overline{C D}$, and $\overline{D A}$ of the isosceles trapezoid $A B C D$ are congruent to each other. Squares $A D E F$ and $C B G H$ are drawn on its legs from the outside. If the obtuse angle of the trapezoid is seven times larger than its acute angle, determine the measure of the angle formed by the lines $E F$ and $G H$.

Solution:
## First Solution.

If $\alpha$ is the measure of the acute angle and $\beta$ is the measure of the obtuse angle of the trapezoid $ABCD$, according to the problem, we have $\beta=7\alpha$. The sum of the interior angles of a quadrilateral is $360^{\circ}$, and in an isosceles trapezoid, the acute angles are congruent to each other and the obtuse angles are congruent to each other, so we have:

$$
2\alpha + 2\beta = 360^{\circ} \quad \alpha + \beta = 180^{\circ} \quad \alpha + 7\alpha = 180^{\circ} \quad 8\alpha = 180^{\circ}
$$

from which we get $\alpha = 22.5^{\circ}$ and $\beta = 157.5^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_ff246218007c389a8ad8g-3.jpg?height=408&width=1556&top_left_y=767&top_left_x=250)

Let the lines $EF$ and $GH$ intersect at point $S$. Let the line on which the base $\overline{AB}$ lies intersect the line $EF$ at point $K$ and the line $GH$ at point $L$. We want to determine the angle $|\angle KSL|$.

In $\triangle AFK$, we have $|\angle KFA| = 90^{\circ}$ and $|\angle FAK| = 180^{\circ} - |\angle DAF| - |\angle BAD| = 180^{\circ} - 90^{\circ} - 22.5^{\circ} = 67.5^{\circ}$.

Therefore, we conclude $|\angle AKF| = 180^{\circ} - |\angle KFA| - |\angle FAK| = 22.5^{\circ}$.

Similarly, we calculate $|\angle GBL| = 22.5^{\circ}$. Thus, $|\angle LKS| = |\angle SLK| = 22.5^{\circ}$.

It follows that $|\angle KSL| = 180^{\circ} - 22.5^{\circ} - 22.5^{\circ} = 135^{\circ}$. The desired angle between the lines is $45^{\circ}$.

Note. Instead of calculating the angles in triangle $AFK$, we can observe that the lines $AD$ and $KS$ are parallel (since the line $KS$ contains the side $\overline{EF}$ which is opposite to the side $\overline{AD}$), and conclude that $|\angle AKF| = |\angle BAD| = 22.5^{\circ}$ due to the equality of the angles on the transversal.

Answer: 45


Find (1) the smallest positive period of $f(x)=\sin \frac{1}{2} x+\cos \frac{1}{3} x$;
(2) the smallest positive period of $g(x)=\sin 3 x \cdot \cos 4 x$.

Solution:
(1) Let $l$ be the period of $f(x)$, then
$$
\begin{aligned}
& \sin \frac{1}{2}(x+l)+\cos \frac{1}{3}(x+l) \\
& =\sin \frac{1}{2} x+\cos \frac{1}{3} x, \\
\therefore \quad & \sin \frac{1}{2}(x+l)-\sin \frac{1}{2} x=\cos \frac{1}{3} x \\
& -\cos \frac{1}{3}(x+l) .
\end{aligned}
$$

Using the sum-to-product identities, we get
$$
\begin{array}{l}
2 \cos \left(\frac{1}{2} x+\frac{l}{4}\right) \sin \frac{l}{4} \\
\equiv-2 \sin \left(\frac{1}{3} x+\frac{l}{6}\right) \sin \left(-\frac{l}{6}\right) \\
\therefore \sin \frac{l}{4} \cdot \cos \left(\frac{1}{2} x+\frac{l}{4}\right) \\
-\sin \frac{l}{6} \sin \left(\frac{1}{3} x+\frac{l}{6}\right)=0
\end{array}
$$

This holds for any real number $x$,
$\therefore \sin \frac{l}{4}=0$, and $\sin \frac{l}{6}=0$. (This means that $\cos \left(\frac{1}{2} x+\frac{l}{4}\right)$ and $\sin \left(\frac{1}{3} x+\frac{l}{6}\right)$ are linearly independent)
$\therefore l=4 k \pi$ and $l=6 k \pi$ (where $k$ is a natural number),
$\therefore$ the smallest value of $l$ is $12 \pi$. Therefore, the smallest positive period of $f(x)=\sin \frac{1}{2} x+\cos \frac{1}{3} x$ is $12 \pi$.
(2) $\because g(x)=\sin 3 x \cdot \cos 4 x=\frac{1}{2}(\sin 7 x - \sin x)$, following the solution in (1), we know that the smallest positive periods of $\sin 7 x$ and $\sin x$ are $\frac{2 \pi}{7}$ and $2 \pi$ respectively. The least common multiple of these periods is $2 \pi$, which is the smallest positive period of $g(x)$.

In general, for a function of the form $\sum_{i=1}^{k} a_{i} \sin n_{i} x + b_{i} \cos n_{i} x$ (where $n_{1}, n_{2}, \cdots, n_{k}$ are distinct positive rational numbers and $a_{i}, b_{i}$ are not both zero), the smallest positive period is the least common multiple of the periods of the individual terms.

Answer: 2 \pi


Diameter $AB$ and chord $CD$ intersect at point $M, \angle CMB=73^{\circ}$, the angular magnitude of arc $BC$ is $110^{\circ}$. Find the magnitude of arc $BD$.

Solution:
$C M B$ - the external angle of triangle $M D B$.

## Solution

Since $C M B$ is the external angle of triangle $M D B$, then

$$
\angle A B D=\angle C M B-\angle C D B=73^{\circ}-55^{\circ}=18^{\circ} .
$$

Therefore,

$$
\cup A D=36^{\circ}, \cup B D=180^{\circ}-36^{\circ}=144^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-04.jpg?height=472&width=548&top_left_y=667&top_left_x=755)

Answer

$144^{\circ}$.

Answer: 144


Let $n$ be a natural number. Determine all positive real numbers $x$ for which

$$
\frac{2^{2}}{x+1}+\frac{3^{2}}{x+2}+\cdots+\frac{(n+1)^{2}}{x+n}+n x^{2}=n x+\frac{n(n+3)}{2}
$$

Solution:
## First Solution.

Since $n x=x+x+\cdots+x$, and $\frac{n(n+3)}{2}=(1+2+3+\cdots+n)+n$, the original equality is equivalent to

$$
\sum_{k=1}^{n} \frac{(k+1)^{2}}{x+k}+n x^{2}-\sum_{k=1}^{n} x-\sum_{k=1}^{n} k-n=0
$$

or equivalently,

$$
\sum_{k=1}^{n}\left(\frac{(k+1)^{2}}{x+k}-(x+k)\right)+n x^{2}-n=0
$$

By simplifying, we get

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

Thus, the above equality is equivalent to

$$
(1-x)\left(n+\sum_{k=1}^{n} \frac{k+1}{x+k}\right)+n\left(x^{2}-1\right)=0
$$

or equivalently,

$$
(1-x)\left(n+\left(\frac{2}{x+1}+\frac{3}{x+2}+\cdots+\frac{n+1}{x+n}\right)-n(1+x)\right)=0
$$

For $x=1$, the equality clearly holds. For $x \neq 1$, it must hold that

$$
n+\left(\frac{2}{x+1}+\frac{3}{x+2}+\cdots+\frac{n+1}{x+n}\right)=n(1+x)
$$

or equivalently,

$$
\frac{2}{x+1}+\frac{3}{x+2}+\cdots+\frac{n+1}{x+n}=n x
$$

If $0 < x < 1$, the left side is strictly less than $n$, while the right side is strictly greater than $n$. If $x > 1$, the opposite is true: the left side is strictly less than $n$, and the right side is strictly greater than $n$. Therefore, the only solution is $x=1$.

Answer: 1


The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution:
We can use [complementary counting](https://artofproblemsolving.com/wiki/index.php/Complementary_counting), counting all of the colorings that have at least one red $2\times 2$ square.

For at least one red $2 \times 2$ square:
There are four $2 \times 2$ squares to choose which one will be red. Then there are $2^5$ ways to color the rest of the squares. $4*32=128$
For at least two $2 \times 2$ squares:
There are two cases: those with two red squares on one side and those without red squares on one side.
The first case is easy: 4 ways to choose which the side the squares will be on, and $2^3$ ways to color the rest of the squares, so 32 ways to do that. For the second case, there will be only two ways to pick two squares, and $2^2$ ways to color the other squares. $32+8=40$
For at least three $2 \times 2$ squares:
Choosing three such squares leaves only one square left, with four places to place it. This is $2 \cdot 4 = 8$ ways.
For at least four $2 \times 2$ squares, we clearly only have one way.
By the [Principle of Inclusion-Exclusion](https://artofproblemsolving.com/wiki/index.php/Principle_of_Inclusion-Exclusion), there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \times 2$ square.
There are $2^9=512$ ways to paint the $3 \times 3$ square with no restrictions, so there are $512-95=417$ ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a $2 \times 2$ red square is $\frac{417}{512}$, and $417+512=\boxed{929}$.

Answer: 929


Find all integers $p$ such that $p, p+2$ and $p+4$ are all three prime?

A prime number is an integer $\geqslant 2$ that is divisible only by 1 and itself.

Solution:
Solution to Exercise 1 First, we can expect there to be very few. We wish to obtain information about these prime numbers.

Note that a prime number divisible by 3 is necessarily equal to 3. We consider 3 cases based on the remainder of the division of $p$ by 3 (modulo 3):

- $p=3k: p$ is divisible by 3, so $p=3$. $\{3,5,7\}$ is indeed a triplet of prime numbers.
- $p=3k+1: p+2$ is divisible by 3 and prime, so $p=1$ (impossible).
- $p=3k+2: p+4$ is divisible by 3 and prime, so $p=-1$ (impossible).

Hence $p=3$.

Comment from the graders The exercise is generally very well handled, except for some attempts modulo 10, the argument modulo 3 was well understood. Be careful not to forget the case $p=2$.

Answer: 3


At what moment between 13:00 and 14:00 do the hour and minute hands of a clock form a straight angle $\left(180^{\circ}\right)$?

Solution:
First solution.

Let this happen $x$ minutes after 13:00.

The large (minute) hand moves $\frac{1}{60}$ of a full circle, or $6^{\circ}$, in one minute.

The small (hour) hand moves $30^{\circ}$ in one hour, so it moves $0.5^{\circ}$ in one minute. (3 points) After $x$ minutes, the large hand has moved $6x$ degrees, and the small hand has moved $\frac{1}{2}x$ degrees.

The hands will form a straight angle $\left(180^{\circ}\right)$ if the angle moved by the large hand is $210^{\circ}$ greater than the angle moved by the small hand (the small hand starts from the position "1", and the large hand starts from the position "12").

(6 points)

![](https://cdn.mathpix.com/cropped/2024_05_30_20b7f5a96a45568d6d08g-3.jpg?height=532&width=481&top_left_y=288&top_left_x=755)

Therefore, we must have

$$
6x = 210 + \frac{1}{2}x
$$

or $x = \frac{420}{11} = 38 \frac{2}{11}$.

The hands will form a straight angle at 13 hours and $38 \frac{2}{11}$ minutes.

Answer: 13:38\frac{2}{11}


The perimeter of a triangle is $34 \mathrm{~cm}$. If one of the sides is increased by $3 \mathrm{~cm}$, another is increased by $2 \mathrm{~cm}$, and the third is decreased by $3 \mathrm{~cm}$, a new triangle is obtained which is equilateral. Determine the length of the side of the newly obtained triangle.

Solution:
Solution. First method. Since the perimeter of a triangle increases or decreases proportionally to the increase or decrease in the length of one of its sides, the perimeter of the new triangle will be equal to $34+3+2-3=36 \mathrm{~cm}$. However, this triangle is equilateral, so the length of its side is equal to $36: 3=12 \mathrm{~cm}$.

Second method. Let the length of the side of the resulting equilateral triangle be $x$. Then, for the lengths of the sides $a, b, c$ of the initial triangle, whose perimeter is $a+b+c=34$, it holds that $a+3=x, b+2=x, c-3=x$. This means that

$$
\begin{aligned}
& 3 x=a+3+b+2+c-3 \\
& 3 x=(a+b+c)+3+2-3 \\
& 3 x=34+3+2-3 \\
& 3 x=36 \\
& x=12 \mathrm{~cm}
\end{aligned}
$$

Answer: 12\mathrm{~}


Five Bunchkins sit in a horizontal field. No three of the Bunchkins are sitting in a straight line. Each Bunchkin knows the four distances between her and each of the others. Each Bunchkin calculates and then announces the total of these distances. These totals are $17, 43, 56, 66$ and 76. A straight line is painted joining each pair of Bunchkins. What is the total length of paint required?

Solution:
SOLUTION
129

Each line's length will be announced twice; once by each of the two Bunchkins at its ends. By adding up the total of the numbers announced we will therefore include the length of each line exactly twice.
The total length of paint required is $\frac{1}{2} \times(17+43+56+66+76)=\frac{258}{2}=129$.

Answer: 129


Arrange $1,2, \cdots, n^{2}$ in a clockwise spiral pattern into an $n$ by $n$ table $T_{n}$, with the first row being $1,2, \cdots, n$. For example, $T_{3}=\left[\begin{array}{lll}1 & 2 & 3 \\ 8 & 9 & 4 \\ 7 & 6 & 5\end{array}\right]$. Let 2018 be in the $i$-th row and $j$-th column of $T_{100}$. Then $(i, j)=$ $\qquad$

Solution:
8. $(34,95)$.

Let $1 \leqslant k \leqslant 50$.
Then the element in the $k$-th row and $k$-th column of $T_{100}$ is
$$
\begin{array}{l}
1+4 \sum_{i=1}^{k-1}(101-2 i) \\
=1+4(101-k)(k-1) .
\end{array}
$$

Therefore, 1901 is in the 6th row and 6th column, 1990 is in the 6th row and 95th column, 2018 is in the 34th row and 95th column.

Answer: (34,95)


Write down your birthday. Double the number of the day and add the number of the month to the result.

(Example: Birthday 27.8., number of the day 27, number of the month 8, calculation: 2 * 27 + 8 = 62.) Multiply this new result by 5 and add 400. Subtract the tenfold of the day number from this. From the double of this result, subtract the tenfold of the month number.

If you multiply the number thus obtained by 2.5 and finally subtract 38, you get the number 1962. How is this possible with such different initial values?

Solution:
If we denote the day number as $a$ and the month number as $b$, the following calculation follows from the instructions in the problem:

1) Doubling the day number: $2 a$
2) Adding the month number: $2 a + b$
3) Multiplying by 5: $5 \cdot (2 a + b) = 10 a + 5 b$
4) Adding 400: $10 a + 5 b + 400$
5) Subtracting the tenfold of the day number: $10 a + 5 b + 400 - 10 a = 5 b + 400$

We can see that part of the date "falls out" of the calculation.

6) Doubling the result: $10 b + 800$
7) Subtracting the tenfold of the month number: $10 b + 800 - 10 b = 800$

Now the rest of the date has "fallen out." For all possible initial values, the same intermediate result is now present.

8) Multiplying by 2.5: $2.5 \cdot 800 = 2000$
9) Subtracting 38: $2000 - 38 = 1962$

Answer: 1962


In a right-angled triangle $A B C$, with a right angle at vertex $C$, the length of the hypotenuse is 12. Squares $A B D E$ and $A C G F$ are constructed outward on sides $\overline{A B}$ and $\overline{A C}$. If points $D, E$, $F$ and $G$ lie on the same circle, calculate the perimeter of triangle $A B C$.

Solution:
## Solution.

Let $a=\overline{B C}, b=\overline{A C}$, and $c=\overline{A B}$.

The segments $\overline{D E}$ and $\overline{F G}$ are chords of the circle, so their perpendicular bisectors intersect at the center of the given circle. The perpendicular bisector of $\overline{D E}$ intersects the hypotenuse $\overline{A B}$ at point $S$. The perpendicular bisector of $\overline{F G}$ is parallel to the side $\overline{B C}$ of triangle $A B C$ and intersects the leg $\overline{A C}$ at its midpoint $R$.

Then, by Thales' theorem on proportional segments (or the similarity of triangles $A S R$ and $A B C$), the hypotenuse is intersected at its midpoint, i.e., at point $S$.

(This conclusion can also be derived from the fact that the perpendicular bisectors of the sides $\overline{D E}$ and $\overline{F G}$ are also the perpendicular bisectors of the sides $\overline{A B}$ and $\overline{A C}$, and the perpendicular bisectors of the sides in a right triangle intersect at the midpoint of the hypotenuse.)

We conclude that the center of the circle is at the midpoint of the hypotenuse and that $|R S|=\frac{a}{2}$. Then $|S E|=|S G|=r$. Applying the Pythagorean theorem to triangle $S A E$ gives us

$$
\begin{aligned}
& |S A|^{2}+|A E|^{2}=|S E|^{2} \\
& \left(\frac{c}{2}\right)^{2}+c^{2}=r^{2} \\
& r^{2}=36+144=180
\end{aligned}
$$

In triangle $G Q S$, we have $|S Q|^{2}+|Q G|^{2}=|G S|^{2}$, or

$$
\begin{aligned}
& \left(\frac{a}{2}+b\right)^{2}+\left(\frac{b}{2}\right)^{2}=r^{2}=180 \\
& \frac{a^{2}}{4}+a b+b^{2}+\frac{b^{2}}{4}=180 \\
& a b+b^{2}=144
\end{aligned}
$$

Since $a^{2}+b^{2}=144$ (Pythagorean theorem in triangle $A B C$), it follows that $a b-a^{2}=0$, or $a(b-a)=0$.

This is only possible if $a=b$. Then $2 a^{2}=144$ and $a=6 \sqrt{2}$.

The perimeter of triangle $A B C$ is $o=a+b+c=2 a+c=12 \sqrt{2}+12$.

## NATIONAL MATHEMATICS COMPETITION

2nd grade - high school - B variant

Poreč, March 29, 2019.

Answer: 12\sqrt{2}+12


There are 10 students standing in a row, and their birthdays are in different months. There are $n$ teachers who will select these students to join $n$ interest groups. Each student is selected by exactly one teacher, and the order of the students is maintained. Each teacher must select students whose birthdays are in months that are either strictly increasing or strictly decreasing (selecting one or two students is also considered strictly increasing or decreasing). Each teacher should select as many students as possible. For all possible orderings of the students, find the minimum value of $n$.

Solution:
15. If $n \leqslant 3$, let's assume the birth months of these 10 students are $1,2, \cdots, 10$.

When the students are sorted by their birthdays as $4,3,2,1,7,6,5,9, 8,10$, there exists at least one teacher who must select two students from the first four. Since the birth months of these two students are decreasing, and the birth months of the last six students are all greater than those of the first four, this teacher cannot select any of the last six students; among the remaining no more than two teachers, there must be one who has to select two students from the fifth to the seventh. Similarly, this teacher cannot select any of the last three students; the remaining no more than one teacher also cannot select any of the last three students, leading to a contradiction.

Below is the proof: For any distinct ordered sequence of real numbers $a_{1}, a_{2}, \cdots, a_{m}$, when $m \geqslant 5$, there must exist three numbers $a_{i}, a_{j}, a_{k}(i<a_{j}>a_{k}$.
Let the maximum and minimum numbers be $a_{s}$ and $a_{t}$. Without loss of generality, assume $s<a_{s+1}>a_{t} ; s+1=t$.
Since $m \geqslant 5$, there must be at least two numbers either before $a_{s}, a_{s+1}$ or after $a_{s}, a_{s+1}$.

Without loss of generality, assume there are two numbers $a_{s+2}$ and $a_{s+3}$ after $a_{s}, a_{s+1}$. Thus,
$$
a_{s}>a_{s+2}>a_{s+3} \text { or } a_{s+1}<a_{s+2}<a_{s+3}
$$

must hold.
Using the above conclusion, when $n=4$, the first teacher can select at least 3 students; if the remaining students are greater than or equal to 5, the second teacher can also select at least 3 students; at this point, the number of remaining students does not exceed 4, and can be selected by the two teachers.
Therefore, the minimum value of $n$ is 4.

Answer: 4


Find all functions $f: \mathbf{Q} \rightarrow \mathbf{Q}$, satisfying the condition $f(1)=2$ and the identity

$$
f(x y) \equiv f(x) f(y)-f(x+y)+1, \quad x, y \in \mathbf{Q}
$$

Solution:
20.6. Putting \( y=1 \) in the original identity, we get the identity

\[
f(x) \equiv f(x) f(1)-f(x+1)+1 \quad (x \in \mathbb{Q})
\]

i.e.

\[
f(x+1) \equiv f(x)+1
\]

From this, for all \( x \in \mathbb{Q}, n \in \mathbb{Z} \) we have

\[
f(x+n)=f(x)+n
\]

therefore

\[
f(n)=f(1)+n-1=n+1
\]

Next, substituting into the original identity the values \( x=1 / n, y=n \) for \( n \in \mathbb{Z} \), we get

\[
f\left(\frac{1}{n} \cdot n\right)=f\left(\frac{1}{n}\right) \cdot f(n)-f\left(\frac{1}{n}+n\right)+1
\]

from which

\[
2=f(1 / n)(n+1)-f(1 / n)-n+1, \text{ i.e. } f(1 / n)=1+1 / n
\]

Finally, substituting the values \( x=p, y=1 / q \) for \( p \in \mathbb{Z}, q \in \mathbb{N}_{1} \) we have

\[
f\left(p \cdot \frac{1}{q}\right)=f(p) \cdot f\left(\frac{1}{q}\right)-f\left(p+\frac{1}{q}\right)+1
\]

from which

\[
f(p / q)=(p+1)(1 / q+1)-1 / q-p=p / q+1 .
\]

Thus, we have only one function \( f(x)=x+1 \), which indeed satisfies all the conditions of the problem. Then

Answer: f(x)=x+1


In a Cartesian coordinate system, there are 10 different points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots, P_{10}\left(x_{10}, y_{10}\right)$. If $x_{i}=x_{j}$ or $y_{i}=y_{j}$, then $P_{i}$ and $P_{j}$ are called a "coordinate pair" (the order of $P_{i}$ and $P_{j}$ does not matter). If the 10 different points satisfy: the number of points that form a "coordinate pair" with each point does not exceed $m$; and in any case, they can be exactly divided into 5 pairs, none of which are "coordinate pairs". What is the maximum value of $m$?

Solution:
Given 10 different points as $P_{1}(1,1), P_{2}(1,2), P_{3}(1,3), P_{4}(1,4), P_{5}(1,5), P_{6}(1,6)$, $P_{7}(2,2), P_{8}(2,3), P_{9}(2,4), P_{10}(2,5)$. Since $P_{1}(1,1), P_{2}(1,2), P_{3}(1,3), P_{4}(1,4), P_{5}(1,5), P_{6}(1,6)$ form "same-label point pairs" with each other, it is impossible to divide the 10 different points into 5 point pairs such that no point pair is a "same-label point pair". Therefore, $m \leqslant 4$. If $m \leqslant 4$, first randomly divide the 10 points into 5 point pairs: $\left(Q_{1}, R_{1}\right),\left(Q_{2}, R_{2}\right),\left(Q_{3}, R_{3}\right),\left(Q_{4}, R_{4}\right)$, $\left(Q_{5}, R_{5}\right)$. If there exists a "same-label point pair", such as $\left(Q_{1}, R_{1}\right)$, consider the following adjustments:
(1) Adjust $\left(Q_{1}, R_{1}\right),\left(Q_{2}, R_{2}\right)$ to $\left(Q_{1}, Q_{2}\right),\left(R_{1}, R_{2}\right)$, leaving the rest unchanged;
(2) Adjust $\left(Q_{1}, R_{1}\right),\left(Q_{2}, R_{2}\right)$ to $\left(Q_{1}, R_{2}\right),\left(R_{1}, Q_{2}\right)$, leaving the rest unchanged;
(3) Adjust $\left(Q_{1}, R_{1}\right),\left(Q_{3}, R_{3}\right)$ to $\left(Q_{1}, Q_{3}\right),\left(R_{1}, R_{3}\right)$, leaving the rest unchanged;
(4) Adjust $\left(Q_{1}, R_{1}\right),\left(Q_{3}, R_{3}\right)$ to $\left(Q_{1}, R_{3}\right),\left(R_{1}, Q_{3}\right)$, leaving the rest unchanged;
(5) Adjust $\left(Q_{1}, R_{1}\right),\left(Q_{4}, R_{4}\right)$ to $\left(Q_{1}, Q_{4}\right),\left(R_{1}, R_{4}\right)$, leaving the rest unchanged;
(6) Adjust $\left(Q_{1}, R_{1}\right),\left(Q_{4}, R_{4}\right)$ to $\left(Q_{1}, R_{4}\right),\left(R_{1}, Q_{4}\right)$, leaving the rest unchanged;
(7) Adjust $\left(Q_{1}, R_{1}\right),\left(Q_{5}, R_{5}\right)$ to $\left(Q_{1}, Q_{5}\right),\left(R_{1}, R_{5}\right)$, leaving the rest unchanged;
(8) Adjust $\left(Q_{1}, R_{1}\right),\left(Q_{5}, R_{5}\right)$ to $\left(Q_{1}, R_{5}\right),\left(R_{1}, Q_{5}\right)$, leaving the rest unchanged.

Since $Q_{1}, R_{1}$, except for $R_{1}, Q_{1}$, can form "same-label point pairs" with at most 3 other points each, one of the 8 methods above must result in two point pairs that are not "same-label point pairs". Thus, after one adjustment, the number of "same-label point pairs" is reduced by at least one. If there are still "same-label point pairs" after the adjustment, continue the adjustment. After a finite number of adjustments, it is certain that no point pair will be a "same-label point pair". In summary, the maximum value of $m$ is 4.

Answer: 4


Find the smallest positive integer $n$ such that
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}-\sqrt{\frac{n-2012}{2011}} \\
<\sqrt[3]{\frac{n-2013}{2011}}-\sqrt[3]{\frac{n-2011}{2013}} .
\end{array}
$$

Solution:
6. From the known, we must have $n \geqslant 2$ 013. At this time,
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}4023, \\
\sqrt[3]{\frac{n-2013}{2011}} \geqslant \sqrt[3]{\frac{n-2011}{2013}} \\
\Leftrightarrow 2013(n-2013) \geqslant 2011(n-2011) \\
\Leftrightarrow n \geqslant 4024 .
\end{array}
$$

From equations (1) and (2), when $n \geqslant 4024$,
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}-\sqrt{\frac{n-2012}{2011}}\sqrt[3]{\frac{n-2013}{2011}}-\sqrt[3]{\frac{n-2011}{2013}} .
\end{array}
$$

In summary, the smallest positive integer $n$ that satisfies the condition is
4024.

Answer: 4024


Given that $n$ is a positive integer. If in any permutation of $1,2, \cdots, n$, there always exists a sum of six consecutive numbers greater than 2013, then $n$ is called a "sufficient number". Find the smallest sufficient number.

Solution:
Three, the smallest well-fed number is 671.
First, prove: 671 is a well-fed number.
For any permutation $b_{1}, b_{2}, \cdots, b_{671}$ of $1,2, \cdots, 671$, divide it into 112 groups, where each group contains six numbers, and the last two groups share one number, i.e.,
$$
\begin{array}{l}
A_{1}=\left(b_{1}, b_{2}, \cdots, b_{6}\right), A_{2}=\left(b_{7}, b_{8}, \cdots, b_{12}\right), \cdots, \\
A_{111}=\left(b_{661}, b_{662}, \cdots, b_{666}\right), \\
A_{112}=\left(b_{666}, b_{667}, \cdots, b_{671}\right) .
\end{array}
$$

Then the average value of the sums of these 112 groups is greater than
$$
\frac{1+2+\cdots+671}{112}=\frac{671 \times 672}{2 \times 112}=2013 \text {. }
$$

Therefore, there must be at least one group whose sum is greater than 2013. At this point, it is always possible to find a sequence of six consecutive numbers whose sum is greater than 2013.
Thus, 671 is a well-fed number.
Next, construct a permutation of $1,2, \cdots, 670$ such that the sum of any six consecutive numbers is less than or equal to 2013.

Construct a $112 \times 6$ number table: place $670,669, \cdots, 335$ sequentially in the left $112 \times 3$ number table (larger numbers on the left (top) side of smaller numbers), and place $1,2, \cdots, 334$ sequentially in the right $112 \times 3$ number table (smaller numbers on the left (top) side of larger numbers), resulting in the number table shown in Figure 8 (with two empty positions).
\begin{tabular}{cccccc}
670 & 669 & 668 & 1 & 2 & 3 \\
667 & 666 & 665 & 4 & 5 & 6 \\
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\
340 & 339 & 338 & 331 & 332 & 333 \\
337 & 336 & 335 & 334 & &
\end{tabular}

Figure 8
Arrange the numbers in this table from left to right and from top to bottom in a single row, which will be the permutation that meets the condition (the sum of any six consecutive numbers is $2013-3k$ for $k=0,1,2,3$).
Thus, 670 is not a well-fed number.
Therefore, the smallest well-fed number is 671.
(Li Zhang, Tianjin Foreign Studies University Affiliated Foreign Languages School, 300230)

Answer: 671


In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, the degree of the dihedral angle $A-BD_{1}-A_{1}$ is $\qquad$ .

Solution:
Let the cube be a unit cube, and establish a spatial rectangular coordinate system as shown in Figure 20-10. Then, $D(0,0,0), A(1,0,0), D_{1}(0,0,1), B(1,1,0), A_{1}(1,0,1)$.

Thus, $\overrightarrow{A D_{1}}=(-1,0,1), \overrightarrow{A B}=(0,1,0), \overrightarrow{A_{1} B}=(0,1,-1), \overrightarrow{A_{1} D_{1}}=(-1,0,0)$.

Let the normal vectors of plane $A B D_{1}$ and plane $A_{1} B D_{1}$ be $\boldsymbol{n}_{1}=\left(x_{1}, y_{1}, z_{1}\right), \boldsymbol{n}_{2}=\left(x_{2}, y_{2}, z_{2}\right)$. From $\boldsymbol{n}_{1} \perp \overrightarrow{A B}, \boldsymbol{n}_{1} \perp \overrightarrow{A D_{1}}, \boldsymbol{n}_{2} \perp \overrightarrow{A_{1} B}, \boldsymbol{n}_{2} \perp \overrightarrow{A_{1} D_{1}}$, we get
$$
y_{1}=0,-x_{1}+z_{1}=0, y_{2}-z_{2}=0,-x_{2}=0 \text {. }
$$

Let $x_{1}=a$, then $z_{1}=a$; let $y_{2}=b$, then $z_{2}=b$.
Thus, $\boldsymbol{n}_{1}=(a, 0, a), \boldsymbol{n}_{2}=(0, b, b)$, and let the angle between $\boldsymbol{n}_{1}$ and $\boldsymbol{n}_{2}$ be $\alpha$.
Therefore, $\cos \alpha=\frac{\boldsymbol{n}_{1} \cdot \boldsymbol{n}_{2}}{\left|\boldsymbol{n}_{1}\right| \cdot\left|\boldsymbol{n}_{2}\right|}=\frac{(a, 0, a) \cdot(0, b, b)}{\sqrt{2} a \cdot \sqrt{2} b}=\frac{1}{2}$.
So, $\alpha=60^{\circ}$. It is also easy to see that the dihedral angle $A-B D_{1}-A_{1}$ has a plane angle $\theta=\alpha$.
Therefore, the dihedral angle $A-B D_{1}-A_{1}$ is $60^{\circ}$.

Answer: 60


Find the range of the function $y=x^{2}+x \sqrt{x^{2}-1}$.

Solution:
11. The domain of the function is $\{x \mid x \geqslant 1$ or $x \leqslant-1\}$.
(1) It is easy to see that the function $y=x^{2}+x \sqrt{x^{2}-1}$ is an increasing function on $[1,+\infty)$. Therefore, when $x \geqslant 1$, $y \geqslant 1$.
(2) When $x \leqslant-1$,
$$
y=x\left(x+\sqrt{x^{2}-1}\right)=\frac{x}{x-\sqrt{x^{2}-1}}=\frac{1}{1+\sqrt{1-\frac{1}{x^{2}}}} .
$$

Given $x \leqslant-1$, we know
$$
0 \leqslant 1-\frac{1}{x^{2}} < \frac{1}{2}.
$$
Thus, the range of the function $y=x^{2}+x \sqrt{x^{2}-1}$ is $\left(\frac{1}{2},+\infty\right)$.

Answer: \left(\frac{1}{2},+\infty\right)


Let $ABCDEF$ be a regular hexagon, $M$ a point on the diagonal $AC$ and $N$ a point on $CE$ such that

$$
\frac{AM}{AC}=\frac{CN}{CE}=r
$$

What can be said about $r$ if $B, M$ and $N$ are collinear?

Solution:
Let's start by drawing the figure, adding G as the intersection point of $A C$ and $B E$ (by the symmetries of the hexagon, $G$ is the midpoint of $A C$):

![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-110.jpg?height=440&width=380&top_left_y=211&top_left_x=840)

Consider the triangle CEG: the points $M, N$, and $B$ are on $G C, C E$, and $E G$ respectively, and are collinear, so we can use Menelaus' theorem:

$$
\frac{\mathrm{GM}}{\mathrm{MC}} \cdot \frac{\mathrm{CN}}{\mathrm{NE}} \cdot \frac{\mathrm{EB}}{\mathrm{BG}}=-1
$$

Let's calculate the three ratios:

$$
\begin{gathered}
\left.\mathrm{GM}=\mathrm{AM}-\mathrm{AG}=\mathrm{r} A C-\frac{1}{2} A C \text { so } \frac{\mathrm{GM}}{\mathrm{MC}}=\frac{(\mathrm{r}-1 / 2) A C}{(1-r) A C}\right)=\frac{2 \mathrm{r}-1}{2-2 r} \\
\frac{\mathrm{CN}}{\mathrm{NE}}=\frac{\mathrm{rCE}}{\mathrm{CE}-\mathrm{rCE}}=\frac{\mathrm{r}}{1-\mathrm{r}}
\end{gathered}
$$

For the last ratio, we will calculate $E B$ and $B G$ using the properties of the regular hexagon. Let $c$ be the side length of the hexagon, $EB$ is a major diagonal and measures $2 c$ (we divide the hexagon into 6 equilateral triangles and it is obvious), and $B G=c / 2$ (since $A G$ is the axis of symmetry of one of the small equilateral triangles), $\frac{E B}{B G}=-4$. Therefore, $r$ satisfies the following equation:

$$
\begin{aligned}
& -4 \cdot \frac{2 r-1}{2-2 r} \cdot \frac{r}{1-r}=-1 \\
& \text { if and only if } \quad 2(2 r-1) r=(1-r)^{2} \\
& \text { if and only if } \quad 4 r^{2}-2 r=r^{2}-2 r+1 \\
& \text { if and only if } \\
& r= \pm \frac{\sqrt{3}}{3} .
\end{aligned}
$$

The real number $r$ is either $\sqrt{3} / 3$ or $-\sqrt{3} / 3$.

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


3. Find all integer solutions $a, b, c$ that satisfy $\frac{1}{2}(a+b)(b+c)(c+a)+(a+b+c)^{3}=1-a b c$, and provide necessary proof.

Solution:
3. Solution: Let $s=a+b+c, p(x)=(x-a)(x-b)(x-c)=x^{3}-s x^{2}+(a b+b c+c a) x-a b c$.

Then $(a+b)(b+c)(c+a)=p(s)=(a b+b c+c a) s-a b c$.
Thus, the given equation can be rearranged as $(a b+b c+c a) s-a b c=-2 s^{3}+2-2 a b c$,
which simplifies to $p(-s)+2=0$.
This indicates that $(2 a+b+c)(a+2 b+c)(a+b+2 c)=2$.
It is easy to see that one of the factors on the left side is 2, and the other two factors are 1,1 (or -1,-1), or one factor is -2, and the other two factors are 1,-1 (or -1,1).
Solving these, we get $a=1, b=0, c=0 ; a=2, b=-1, c=-1$.
When one of the factors is -2, the corresponding two equations can lead to $4(a+b+c)=-2$, which clearly has no integer solutions $(a, b, c)$.

Since the equation is symmetric in $a, b, c$, the complete set of possible values for $(a, b, c)$ is $(1,0,0),(0,1,0)$, $(0,0,1),(2,-1,-1),(-1,2,-1),(-1,-1,2)$.

Answer: (1,0,0),(0,1,0),(0,0,1),(2,-1,-1),(-1,2,-1),(-1,-1,2)


Find the maximum value of the expression $\sin x \sin y \sin z + \cos x \cos y \cos z$.

Solution:
What is the greatest value of the expression $\sin x \sin y+\cos x \cos y$?

## Solution

When $x=y=z=0$, the expression $\sin x \sin y \sin z+\cos x \cos y \cos z$ equals 1. We will show that a value greater than 1 cannot be achieved. The expression $\sin x \sin y \sin z+\cos x \cos y \cos z_{\text {does not }}$ exceed

$|\sin x\|\sin y|+| \cos x\| \cos y| \cdot$ There exist angles $x^{\prime}, y^{\prime}$ from the interval $[0, \pi / 2]$ such that $\sin x^{\prime}=|\sin x|_{\text {and }} \sin y^{\prime}=|\sin y|$. Then $|\sin x\|\sin y|+| \cos x\| \cos y|=\sin x^{\prime} \sin y^{\prime}+\cos x^{\prime} \cos y^{\prime}=\cos \left(x^{\prime}-y^{\prime}\right)$, which obviously does not exceed 1.

## Answer

1.00

Answer: 1


In the sequence $\left\{a_{n}\right\}$, if $a_{1}, a_{2}$ are positive integers, and $a_{n}=\left|a_{n-1}-a_{n-2}\right|$, $n=3,4,5, \cdots$, then $\left\{a_{n}\right\}$ is called an "absolute difference sequence".
(1) Give an example of an "absolute difference sequence" whose first five terms are not zero (only the first ten terms are required to be written);
(2) If in the "absolute difference sequence" $\left\{a_{n}\right\}$, $a_{20}=3, a_{21}=$ 0, and the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=a_{n}+a_{n+1}+a_{n+2}, n=1,2$, $3, \cdots$, determine whether the limits of $a_{n}$ and $b_{n}$ exist as $n \rightarrow \infty$. If they exist, find their limit values;
(3) Prove: Any "absolute difference sequence" contains infinitely many zero terms.

Solution:
(1) $a_{1}=3, a_{2}=1, a_{3}=2, a_{4}=1, a_{5}=1$, $a_{6}=0, a_{7}=1, a_{8}=1, a_{9}=0, a_{10}=1$. (The answer is not unique.)
(2) Since in the absolute difference sequence $\left\{a_{n}\right\}$, $a_{20}=3, a_{21}=$ 0, starting from the 20th term, the sequence is $a_{20}=3, a_{21}=$ $0, a_{22}=3, a_{23}=3, a_{24}=0, a_{25}=3, a_{26}=3, a_{27}=$ $0, \cdots$.

That is, starting from the 20th term, every three adjacent terms cyclically take the values $3,0,3$, so when $n \rightarrow \infty$, the limit of $a_{n}$ does not exist.

When $n \geqslant 20$, $b_{n}=a_{n}+a_{n+1}+a_{n+2}=6$, so $\lim _{n \rightarrow \infty} b_{n}=6$.
(3) Proof: According to the definition, the sequence $\left\{a_{n}\right\}$ must have a zero term after a finite number of terms, as proven below:

Assume that $\left\{a_{n}\right\}$ has no zero terms. Since $a_{n}=\mid a_{n-1}-$ $a_{n-2} \mid$, for any $n$, we have $a_{n} \geqslant 1$, thus

When $a_{n-1} \geqslant a_{n-2}$, $a_{n}=a_{n-1}-a_{n-2} \leqslant a_{n-1}-1$ $(n \geqslant 3)$;

When $a_{n-1}a_{2 n}\right)$, \\
$a_{2 n}\left(a_{2 n-1}0(n=1,2,3, \cdots)$, which contradicts the assumption, thus $\left\{a_{n}\right\}$ must have a zero term.

If the first zero term appears at the $n$-th term, i.e., $a_{n}=0$, let $a_{n-1}=A(A \neq 0)$, then $a_{n+1}=A, a_{n+2}=0, \cdots$, hence starting from the $n$-th term, every three adjacent terms cyclically take the values 0, and the absolute difference sequence $\left|a_{n}\right|$ has infinitely many zero terms.

Answer: 6


Each student in class 7B ate the same number of chocolate bars during math lessons over the week. Nine of them together ate fewer than 288 chocolate bars in a week, while ten of them together ate more than 300 chocolate bars. How many chocolate bars did each student in class 7B eat? Explain your answer.

Solution:
Answer: 31 chocolates each

Solution 1: Since ten students ate more than 300 chocolates, nine students ate more than (310:10)$\cdot$9 = 270 chocolates. It is also known that nine students together ate less than 288 chocolates. The only number divisible by 9 in the range from 271 to 287 is 279. Therefore, nine students ate 279 chocolates, and one student ate 31 chocolates.

Solution 2: Since nine students together ate less than 288 chocolates in a week, one student ate less than $288: 9=32$ chocolates in a week. Since ten students together ate more than 300 chocolates in a week, one student ate more than 300:10 = 30 chocolates in a week. Ultimately, each student ate more than 30 but less than 32, i.e., 31 chocolates.

Criteria: Only the answer - 0 points. 7 points for the correct answer with correct reasoning. 3-5 points for partial progress in the solution.

Answer: 31


Drop perpendiculars $D D_{1}, D D_{2}, D D_{3}$ from point $D$ to the planes $S B C$, $S A C$, and $S A B$ respectively. Let $D D_{1}=x, D D_{2}=y, D D_{3}=z$. According to the condition, we form the system of equations

$$
\left\{\begin{array}{l}
y^{2}+z^{2}=5 \\
x^{2}+z^{2}=13 \\
x^{2}+y^{2}=10
\end{array}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_778d240aea7dd323243eg-9.jpg?height=526&width=630&top_left_y=95&top_left_x=1848)

Fig. 6:

From here, we find $x=3, y=1, z=2$. Let the lengths of the edges $S A, S B$, and $S C$ be $a, b$, and $c$ respectively. Since points $A, B, C$, and $D$ lie in the same plane, the relation $\frac{3}{a}+\frac{1}{b}+\frac{2}{c}=1$ holds.

Using the inequality between the arithmetic mean and the geometric mean for three variables, we get:

$$
\begin{aligned}
& \frac{\frac{3}{a}+\frac{1}{b}+\frac{2}{c}}{3} \geqslant \sqrt[3]{\frac{3}{a} \cdot \frac{1}{b} \cdot \frac{2}{c}}=\sqrt[3]{\frac{6}{a b c}} \Longleftrightarrow \\
& \Longleftrightarrow 1=\left(\frac{3}{a}+\frac{1}{b}+\frac{2}{c}\right)^{3} \geqslant \frac{6 \cdot 27}{a b c} \Longleftrightarrow a b c \geqslant 6 \cdot 27
\end{aligned}
$$

with equality holding when $\frac{3}{a}=\frac{1}{b}=\frac{2}{c}=\frac{1}{3}$. The volume of the pyramid $V=\frac{a b c}{6}$, so $V \geqslant 27$. Equality holds when $a=9, b=3, c=6$.

Solution:
Answer: 27.

Answer to option 17-2: 108.

Answer to option $17-3: 27$.

Answer to option $17-4: 108$.

[^0]:    ${ }^{1}$ This equality can be proven by expressing $B C^{2}$ from two triangles $B A C$ and $B D C$ using the planimetric cosine theorem.

Answer: 27


Let the sequence of positive integers $a_{1}, a_{2}, a_{3}, a_{4}$ be a geometric sequence, with the common ratio $r$ not being an integer, and $r>1$. In such a sequence, the smallest value that $a_{4}$ can take is $\qquad$ .

Solution:
10. 27.

Brief solution: Given $r \in \mathbf{Q}$, there exist coprime positive integers $p, q(p>q \geqslant 2)$ such that $r=\frac{q}{p}$, then $a_{4}=a_{1} r^{3}=$ $\frac{a_{1} q^{3}}{p^{3}}$
Since $a_{4} \in \mathbf{Z}$, $a_{1}$ must be a multiple of $p^{3}$. Therefore, we can set $a_{1}=k p^{3}$ ($k$ is a positive integer), thus $a_{4}=k q^{3}$. Since $q>p \geqslant 2$, to make $a_{4}$ the smallest, we must have $k=1, q=3$, at this point $a_{4}=27$, and the sequence is $8,12,18$, 27.

Answer: 27


Find the angle between the planes:

$x-3 y-2 z-8=0$

$x+y-z+3=0$

Solution:
## Solution

The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes are:

$\overrightarrow{n_{1}}=\{1 ;-3 ;-2\}$

$\overrightarrow{n_{2}}=\{1 ; 1 ;-1\}$

The angle $\phi$ between the planes is determined by the formula:

$\cos \phi=\frac{\left(\overrightarrow{n_{1}}, \overrightarrow{n_{2}}\right)}{\left|\overrightarrow{n_{1}}\right| \cdot\left|\overrightarrow{n_{2}}\right|}=\frac{1 \cdot 1+(-3) \cdot 1+(-2) \cdot(-1)}{\sqrt{1^{2}+(-3)^{2}+(-2)^{2}} \cdot \sqrt{1^{2}+1^{2}+(-1)^{2}}}=$

$=\frac{1-3+2}{\sqrt{1+9+4} \cdot \sqrt{1+1+1}}=\frac{0}{\sqrt{14} \cdot \sqrt{3}}=0$
$\phi=\arccos 0 \approx \frac{\pi}{2}$

## Problem Kuznetsov Analytic Geometry 10-17

Answer: \frac{\pi}{2}


Three points $X, Y,Z$ are on a straight line such that $XY = 10$ and $XZ = 3$. What is the product of all possible values of $YZ$?

Solution:
To solve this problem, we need to consider the relative positions of the points \(X\), \(Y\), and \(Z\) on a straight line. We are given that \(XY = 10\) and \(XZ = 3\). We need to find the product of all possible values of \(YZ\).

1. **Case 1: \(X\) lies between \(Y\) and \(Z\)**

   In this case, the distance \(YZ\) is the sum of \(XY\) and \(XZ\):
   \[
   YZ = XY + XZ = 10 + 3 = 13
   \]

2. **Case 2: \(Y\) lies between \(X\) and \(Z\)**

   In this case, the distance \(YZ\) is the difference between \(XY\) and \(XZ\):
   \[
   YZ = XY - XZ = 10 - 3 = 7
   \]

3. **Case 3: \(Z\) lies between \(X\) and \(Y\)**

   This case is essentially the same as Case 2, where \(YZ = 7\). Therefore, we do not need to consider it separately.

Now, we have two possible values for \(YZ\): 13 and 7. The product of these values is:
\[
13 \times 7 = 91
\]

The final answer is \(\boxed{91}\)

Answer: 91


Select several numbers from $1, 2, 3, \cdots, 9, 10$ such that each of the 20 numbers $1, 2, 3, \cdots, 19, 20$ is equal to one of the selected numbers or the sum of two selected numbers (which can be the same). How many numbers at least need to be selected?

Solution:
【Solution】The solution is as follows:
$$
\begin{array}{l}
1=1 ; 2=2 ; 3=1+2 ; 4=2+2 ; 5=5 ; 6=1+5 ; 7=2+5 ; 8=8 ; 9=9 ; 10=10 ; 11=1+10 ; \\
12=2+10 ; 13=5+8 ; 14=7+7 ; 15=5+10 ; 16=8+8 ; 17=8+9 ; 18=8+10 ; 19=9+10 ;
\end{array}
$$

By observation, it can be seen that selecting several numbers from $1, 2, 3, \cdots, 9, 10$ as $\{1,2,5,8,9,10\}$; can make each number in the 20 numbers $1, 2, 3, \cdots, 19, 20$ equal to a certain selected number or the sum of two selected numbers (which can be equal). Therefore, at least 6 numbers need to be selected. The answer is 6.

Answer: 6


A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a$ and $b$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution:
1. Given the function \( f(z) = (a + bi)z \), where \( a \) and \( b \) are positive numbers, and the property that the image of each point in the complex plane is equidistant from that point and the origin, we have:
   \[
   |f(z)| = |z - f(z)|
   \]
   This implies:
   \[
   |(a + bi)z| = |z - (a + bi)z|
   \]

2. Simplify the right-hand side:
   \[
   |(a + bi)z| = |z(1 - (a + bi))|
   \]
   Since \( |kz| = |k||z| \) for any complex number \( k \):
   \[
   |a + bi||z| = |1 - (a + bi)||z|
   \]
   Dividing both sides by \( |z| \) (assuming \( z \neq 0 \)):
   \[
   |a + bi| = |1 - (a + bi)|
   \]

3. Let \( a + bi = c \). Then:
   \[
   |c| = |1 - c|
   \]
   Given \( |c| = 8 \):
   \[
   8 = |1 - c|
   \]

4. Let \( c = a + bi \). Then:
   \[
   |1 - (a + bi)| = 8
   \]
   This can be written as:
   \[
   |(1 - a) - bi| = 8
   \]

5. The magnitude of a complex number \( x + yi \) is given by \( \sqrt{x^2 + y^2} \):
   \[
   \sqrt{(1 - a)^2 + b^2} = 8
   \]

6. Squaring both sides:
   \[
   (1 - a)^2 + b^2 = 64
   \]

7. We also know that:
   \[
   a^2 + b^2 = 64
   \]

8. We now have two equations:
   \[
   (1 - a)^2 + b^2 = 64
   \]
   \[
   a^2 + b^2 = 64
   \]

9. Subtract the second equation from the first:
   \[
   (1 - a)^2 + b^2 - (a^2 + b^2) = 64 - 64
   \]
   \[
   (1 - a)^2 - a^2 = 0
   \]
   \[
   1 - 2a + a^2 - a^2 = 0
   \]
   \[
   1 - 2a = 0
   \]
   \[
   a = \frac{1}{2}
   \]

10. Substitute \( a = \frac{1}{2} \) into \( a^2 + b^2 = 64 \):
    \[
    \left(\frac{1}{2}\right)^2 + b^2 = 64
    \]
    \[
    \frac{1}{4} + b^2 = 64
    \]
    \[
    b^2 = 64 - \frac{1}{4}
    \]
    \[
    b^2 = \frac{256}{4} - \frac{1}{4}
    \]
    \[
    b^2 = \frac{255}{4}
    \]

11. Given \( b^2 = \frac{m}{n} \) where \( m \) and \( n \) are relatively prime positive integers, we have \( m = 255 \) and \( n = 4 \). Therefore:
    \[
    m + n = 255 + 4 = 259
    \]

The final answer is \(\boxed{259}\)

Answer: 259


a) Determine the parameters $p, q$ such that the expressions

$$
(x+3)^{2}+(7 x+p)^{2} \text { and }(3 x+5)^{2}+(p x+q)^{2}
$$

are equal to the square of a first-degree polynomial.

b) Let $a, b, c, A, B, C$ be given, non-zero numbers such that the polynomials

$$
(a x+b)^{2}+(A x+B)^{2} \text { and }(b x+c)^{2}+(B x+C)^{2}
$$

are equal to the square of a first-degree polynomial. Show that the same holds for the following polynomial:

$$
(c x+a)^{2}+(C x+A)^{2}
$$

Solution:
I. solution. It is generally useful to determine under what conditions the sum of the squares of two linear expressions is equal to the square of a linear expression, because we can use this in the answers to each question. Let $\alpha, \beta, \gamma, \delta$ be any four numbers, and consider the polynomial

$$
(\alpha x+\beta)^{2}+(\gamma x+\delta)^{2}=\left(\alpha^{2}+\gamma^{2}\right) x^{2}+2(\alpha \beta+\gamma \delta) x+\beta^{2}+\delta^{2}
$$

If here $\alpha=\gamma=0$, then we are dealing with a non-negative constant, which is indeed the square of a constant. If this is not the case, then complete the square for the expression:

$$
\begin{gathered}
\left(\sqrt{\alpha^{2}+\gamma^{2}} x+\frac{\alpha \beta+\gamma \delta}{\sqrt{\alpha^{2}+\gamma^{2}}}\right)^{2}+\beta^{2}+\delta^{2}-\frac{(\alpha \beta+\gamma \delta)^{2}}{\alpha^{2}+\gamma^{2}}=\left(\sqrt{\alpha^{2}+\gamma^{2}} x+\frac{\alpha \beta+\gamma \delta}{\sqrt{\alpha^{2}+\gamma^{2}}}\right)^{2}+ \\
\frac{\alpha^{2} \delta^{2}+\beta^{2} \gamma^{2}-2 \alpha \beta \gamma \delta}{\alpha^{2}+\gamma^{2}}
\end{gathered}
$$

The expression is the square of a linear expression if and only if the last fraction is 0. The numerator of this fraction is $(\alpha \delta-\beta \gamma)^{2}$, so our quadratic polynomial is the square of a linear expression if and only if

$$
\alpha \delta=\beta \gamma
$$

(This is also true in the case $\alpha=\gamma=0$, so more precisely, we should say "at most quadratic" and "at most linear" polynomials.)
a) For the first expression under (1),

$$
\alpha=1, \quad \beta=3, \quad \gamma=7, \quad \delta=p
$$

thus (4) gives the value $p=21$. For the second expression,

$$
\alpha=3, \quad \beta=5, \quad \gamma=p, \quad \delta=q
$$

so from (4), $3 q=5 p=105$, i.e., $q=35$. Indeed,

$$
\begin{aligned}
& (x+3)^{2}+(7 x+21)^{2}=(\sqrt{50} x+3 \sqrt{50})^{2} \quad \text { and } \\
& (3 x+5)^{2}+(21 x+35)^{2}=(3 \sqrt{50} x+5 \sqrt{50})^{2}
\end{aligned}
$$

b) The two expressions under (2) are, according to (4), the square of at most a linear expression if and only if

$$
a B=b A \quad \text { and } \quad b C=c B
$$

Multiplying these two conditions,

$$
a b B C=b c A B, \quad \text { i.e., } \quad b B(a C-c A)=0
$$

Since by condition $b$ and $B$ are not 0, then

$$
a C-c A=0, \quad a C=c A
$$

and from this, according to (4), it follows that the expression (3) is also the square of at most a linear expression.

András Csikvári (Budapest, Berzsenyi D. Gymnasium, 2nd year)

II. solution. Examine the function generated by the individual expressions for $x$, and how they can become 0. The sum of two squares can only be 0 if both squares are 0, and the square of a linear expression disappears for the value of $x$ that makes the base 0. The problem thus wants to find, in each case, the value of $x$ for which the bases of the two added squares are zero.

a) For the first and second expressions under (1), the corresponding $x$ values are

$$
x_{1}=-3=-\frac{p}{7} \quad \text { and } \quad x_{2}=-\frac{5}{3}=-\frac{q}{p}
$$

and from this, as in the first solution, the values

$$
p=21, \quad q=\frac{5}{3} p=35
$$

are obtained.

b) For the expressions under (2), these values are

$$
x_{3}=-\frac{b}{a}=-\frac{B}{A} \quad \text { and } \quad x_{4}=-\frac{c}{b}=-\frac{C}{B}
$$

If these relationships hold, then by multiplying them, we get

$$
\frac{c}{a}=\frac{C}{A}
$$

Denoting the common value of these two fractions by $\lambda$, and expressing $c$ and $C$ from this, the polynomial (3) becomes:

$$
(\lambda a x+a)^{2}+(\lambda A x+A)^{2}=\left[\sqrt{a^{2}+A^{2}}(\lambda x+1)\right]^{2}
$$

which is indeed the square of a linear expression.

Using the work of István Dettai (Pannonhalma, Bencés Gymnasium, 2nd year)

Answer: p=21,q=35


Determine the natural number $\overline{a b c d}$ such that for its digits the following conditions hold:

$$
a \cdot d + c \cdot d = 72, \quad a \cdot c + c \cdot d = 56 \text{ and } a \cdot b \cdot c \cdot d = 0
$$

Solution:
## First Solution.

From the condition $a \cdot b \cdot c \cdot d=0$, it follows that at least one of the factors is equal to 0. If $d=0$, then the first equation would read $0=72$, which is impossible. If $c=0$, then the second given equation would read $0=56$, which is also impossible. If $a=0$, then $c \cdot d=72$ and $c \cdot d=56$ would both hold, which is also impossible. This means that it is necessary that $b=0$.

Let's write the expressions on the left sides of the first two given equations in the form of a product:

$$
d \cdot(a+c)=72 \quad \text{and} \quad c \cdot(a+d)=56
$$

Let's list all possibilities for $c$, $a$, and $d$ that satisfy the second equation, and check which of these possibilities also satisfy the first equation.

| $c$ | $a+d$ | $a$ | $d$ | $d \cdot(a+c)$ |
| :---: | :---: | :---: | :---: | :---: |
| 1 | 56 | $/$ | $/$ | $/$ |
| 2 | 28 | $/$ | $/$ | $/$ |
| 4 | 14 | 5 | 9 | 81 |
| 4 | 14 | 6 | 8 | 80 |
| 4 | 14 | 7 | 7 | 77 |
| 4 | 14 | 8 | 6 | 72 |
| 4 | 14 | 9 | 5 | 65 |
| 7 | 8 | 1 | 7 | 56 |
| 7 | 8 | 2 | 6 | 54 |
| 7 | 8 | 3 | 5 | 50 |

| $c$ | $a+d$ | $a$ | $d$ | $d \cdot(a+c)$ |
| :---: | :---: | :---: | :---: | :---: |
| 7 | 8 | 4 | 4 | 44 |
| 7 | 8 | 5 | 3 | 36 |
| 7 | 8 | 6 | 2 | 26 |
| 7 | 8 | 7 | 1 | 14 |
| 8 | 7 | 1 | 6 | 54 |
| 8 | 7 | 2 | 5 | 50 |
| 8 | 7 | 3 | 4 | 44 |
| 8 | 7 | 4 | 3 | 36 |
| 8 | 7 | 5 | 2 | 26 |
| 8 | 7 | 6 | 1 | 14 |

The only possibility that satisfies both conditions is $a=8, c=4$, and $d=6$, i.e., the desired number is 8046.

Answer: 8046


Determine all pairs $(x ; y)$ of real numbers $x, y$ that satisfy the following system of equations:

$$
\begin{aligned}
x + xy + xy^2 & = -21 \\
y + xy + x^2 y & = 14 \\
x + y & = -1
\end{aligned}
$$

Solution:
Adding (1) and (2) results in the new equation

$$
(x+y)+2 x y+x y(x+y)=-7
$$

from which $x y=-6$ is immediately obtained using (3).

By substituting this into (1), one obtains the additional linear equation

$$
x-6 y=-15
$$

which, in conjunction with (3), ultimately leads to the unique solution $x=-3, y=2$.

Answer: -3,2


Given that $p$ is a prime number, $r$ is the remainder when $p$ is divided by 210. If $r$ is a composite number and can be expressed as the sum of two perfect squares, find $r$.

Solution:
If let $p=210 n+r, 0<\sqrt{210}$, then by $r=q m>\sqrt{210}$. $\sqrt{210}=210$, which is impossible.
Therefore, the prime factors of the composite number $r$ can only be $2,3,5,7,11,13$.
So we should classify and discuss based on the prime factor $q$ of $r$.
(1) When $q=2,3,5,7$, since
$$
p=210 n+q m,
$$

then at this time $p$ has a divisor of $2,3,5,7$, which contradicts that $p$ is a prime number.
(2) When $q=11$,
$\because$ if $r$ can be expressed as the sum of two squares, let $r=a^{2}+b^{2}$, where $a, b$ are positive integers, at this time $a^{2}+b^{2}$ can be divisible by 11.
Since $a^{2} \equiv 0,1,4,9,5,3(\bmod 11)$,
if $a^{2}+b^{2} \equiv 0 \quad(\bmod 11)$,

then $a$ and $b$ must both be divisible by 11, at this time $a^{2}+b^{2}$ can be divided by $121+121=242>210$, leading to a contradiction.
(3) When $q=13$,
since $r=q m<210$, then $m<\frac{210}{q}<17$, i.e., $m \leqslant 16$.
Also, since $m \geqslant q$, so $m=13$.
At this time $r=13^{2}=169=12^{2}+5^{2}$.
From the above, only $r=169$ meets the requirements.

Answer: 169


As shown in the figure, $P, Q$ are points on the side $AD$ and the diagonal $AC$ of the square $ABCD$, respectively, and $PD: AP=4: 1$, $QC: AQ=2: 3$. If the area of the square $ABCD$ is 25, then the area of triangle $PBQ$ is $\qquad$ .

Solution:
【Analysis】Draw $E F / / A B$, intersecting $A D$ at $F$ and $B C$ at $E$, and $Q G \perp D C$ at $G$. According to the similarity ratio, the lengths of each segment can be calculated. Then, subtract the areas of the other parts from the area of the square to get the final result.

【Solution】Solution: Connect $Q D$, draw $E F / / A B$, intersecting $A D$ at $F$ and $B C$ at $E$, and $Q G \perp D C$ at $G$. Since the area of the square $A B C D$ is 25,

so $A D=E F=5$;
Since $Q C: A Q=2: 3$, according to the symmetry of the square, $Q E=Q G=2, Q F=3$.
Since $P D: A P=4: 1$, so $A P=1, P D=4$.
$$
\begin{array}{l}
S \triangle P Q D=S_{\text {square } A B C D}-S_{\triangle C Q B}-S \triangle D Q C-S_{\triangle P Q D}-S_{\triangle P A B} \\
=25-2 \times 5 \div 2-2 \times 5 \div 2-4 \times 3 \div 2-1 \times 5 \div 2 \\
=25-5-5-6-2.5=6.5 .
\end{array}
$$

Answer: The area of triangle $P B Q$ is 6.5.
The answer is: 6.5.

Answer: 6.5


Consider the sequence defined as follows: $x_{1}=1$,

$$
x_{n+1}=x_{n}+\frac{1}{2 x_{n}}
$$

for any $n$.

a) Find the limit of the sequence.

b) Calculate

$$
\lim _{n \rightarrow \infty} \frac{x_{n}}{\sqrt{n}}
$$

Solution:
Solution. a) The sequence is strictly increasing. If it were bounded, it would converge to a real number $L$, and we would obtain by passing to the limit $L=L+\frac{1}{2 L}$, which is absurd, so the limit is $\infty$

b) We calculate, using Stolz-Cesaro,

$$
\lim _{n \rightarrow \infty} \frac{x_{n}^{2}}{n}=\lim _{n \rightarrow \infty} \frac{x_{n+1}^{2}-x_{n}^{2}}{n+1-n}=\lim _{n \rightarrow \infty}\left(\left(x_{n}+\frac{1}{2 x_{n}}\right)^{2}-x_{n}^{2}\right)=\lim _{n \rightarrow \infty}\left(1+\frac{1}{4 x_{n}^{2}}\right)=1
$$

so the required limit is 1

Answer: 1


If the function $f(x)=x^{2}+a|x-1|$ is monotonically increasing on $[0,+\infty)$, then the range of the real number $a$ is $\qquad$ .

Solution:
3. $[-2,0]$.

On the interval $[1,+\infty)$, $f(x)=x^{2}+a x-a$ is monotonically increasing, which is equivalent to $-\frac{a}{2} \leqslant 1 \Rightarrow a \geqslant-2$.

On the interval $[0,1]$, $f(x)=x^{2}-a x+a$ is monotonically increasing, which is equivalent to $\frac{a}{2} \leqslant 0 \Rightarrow a \leqslant 0$.
Therefore, the real number $a \in[-2,0]$.

Answer: [-2,0]


If 7 is subtracted from the thought three-digit number, the resulting difference will be divisible by 7; if 8 is subtracted, the resulting difference will be divisible by 8; if 9 is subtracted, the resulting difference will be divisible by 9. What is the smallest possible number thought of?

Solution: 5. The intended number is divisible by $7, 8, 9$. The smallest number divisible by 7, 8, and 9 is $7 \cdot 8 \cdot 9=504$. This number meets the requirement of the problem.

Answer: 504


Let $n$ be an even natural number. We partition the numbers $1,2, \ldots, n^{2}$ into two sets $A$ and $B$ of equal size, such that each of the $n^{2}$ numbers belongs to exactly one of the two sets. Let $S_{A}$ and $S_{B}$ be the sum of all elements in $A$ and $B$ respectively. Determine all $n$ for which there exists a partition such that

$$
\frac{S_{A}}{S_{B}}=\frac{39}{64}
$$

Answer: The natural numbers $n$ sought are all multiples of 206.

Solution:
Let $k$ be the integer such that $S_{A}=39 k$ and $S_{B}=64 k$. We then have that the sum of all numbers from 1 to $n^{2}$ is

$$
\frac{n^{2}\left(n^{2}+1\right)}{2}=S_{A}+S_{B}=39 k+64 k=103 k
$$

Thus, $103 \left\lvert\, \frac{n^{2}\left(n^{2}+1\right)}{2}\right.$, and since 103 is a prime number, it must divide either $\frac{n^{2}}{2}$ or $n^{2}+1$. However, since $n^{2}$ and $1=1^{2}$ are coprime, all prime divisors of $n^{2}+1$ must be congruent to $1 \bmod 4$. Since $103 \equiv 3 \bmod 4$, it cannot divide $n^{2}+1$ and thus must be a divisor of $\frac{n^{2}}{2}$. This can only be the case if $103 \mid n$, so since $n$ must be even, there can only be a solution if $n$ is a multiple of 206.

Now let's show that if $n=206 m$ for some natural number $m$, such a partition exists. In this case we have

$$
k=\frac{S_{A}+S_{B}}{103}=\frac{n^{2}\left(n^{2}+1\right)}{206}
$$

and it suffices to show that we can choose half of the numbers so that their sum is $39 k$. In fact, in this case, the sum of the remaining numbers would necessarily be $64 k$. We prove here a stronger property, namely that all sums from $\frac{n^{2}\left(n^{2}+2\right)}{8}$ to $\frac{3 n^{4}+2 n^{2}}{8}$ are achievable. Since $\frac{n^{2}\left(n^{2}+2\right)}{8}=\frac{103 k}{4}\frac{n^{2}\left(n^{2}+2\right)}{8}$, the set $A$ must contain a number greater than $\frac{n^{2}}{2}$ and thus $B$ must contain a number less than or equal to $\frac{n^{2}}{2}$. In particular, this means that there is a number in $A$ that is strictly greater than a number in $B$. Therefore, there exists a number $x$ such that $x \in B$ and $x+1 \in A$. By swapping $x$ and $x+1$, the sum in $A$ decreases by 1, which proves that it is indeed possible to achieve all sums.

Answer: 206


A number when divided by 3 leaves a remainder of 2, when divided by 4 leaves a remainder of 3, and when divided by 5 leaves a remainder of 4. This number is $\qquad$

Solution:
【Analysis】Interpreting "divided by 3 leaves a remainder of 2, divided by 4 leaves a remainder of 3, divided by 5 leaves a remainder of 4" as being 1 less than when divided by 3, 1 less than when divided by 4, and 1 less than when divided by 5, means this number is at least 1 less than the least common multiple (LCM) of 3, 4, and 5. Since 3, 4, and 5 are pairwise coprime, the LCM of these three numbers is their product. Find the LCM of 3, 4, and 5, then subtract 1.

【Solution】Solution: $3 \times 4 \times 5-1$,
$$
\begin{array}{l}
=60-1, \\
=59 ;
\end{array}
$$

Answer: The number is 59.
Therefore, the answer is: 59.
【Comment】This problem mainly examines the method of finding the least common multiple (LCM) when three numbers are pairwise coprime: If three numbers are pairwise coprime, their LCM is the product of these three numbers.

Answer: 59


Find all values of the real number $k$ for which the expression

$$
\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)
$$

does not depend on $x$.

Solution:
Solution. We have

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

The expression does not depend on $x$ if $3+2 k=0$, i.e., for $k=-\frac{3}{2}$.

Answer: -\frac{3}{2}


The 8th layer contains how many regular triangular tiles.

Solution: answer: 90

Answer: 90


Isosceles triangle $\triangle{ABC}$ has $\angle{ABC}=\angle{ACB}=72^\circ$ and $BC=1$. If the angle bisector of $\angle{ABC}$ meets $AC$ at $D$, what is the positive difference between the perimeters of $\triangle{ABD}$ and $\triangle{BCD}$?

Solution:
1. Given that $\triangle{ABC}$ is an isosceles triangle with $\angle{ABC} = \angle{ACB} = 72^\circ$ and $BC = 1$. We need to find the positive difference between the perimeters of $\triangle{ABD}$ and $\triangle{BCD}$, where $D$ is the point where the angle bisector of $\angle{ABC}$ meets $AC$.

2. Let $AB = AC = x$. Since $\angle{ABC} = \angle{ACB} = 72^\circ$, $\angle{BAC} = 180^\circ - 2 \times 72^\circ = 36^\circ$.

3. The angle bisector of $\angle{ABC}$ splits $\angle{ABC}$ into two $36^\circ$ angles. Therefore, $\angle{ABD} = 36^\circ$ and $\angle{DBC} = 36^\circ$.

4. By the Angle Bisector Theorem, the ratio of the segments $AD$ and $DC$ is equal to the ratio of the sides $AB$ and $BC$. Since $AB = AC = x$ and $BC = 1$, we have:
   \[
   \frac{AD}{DC} = \frac{AB}{BC} = \frac{x}{1}
   \]
   Let $AD = k$ and $DC = \frac{k}{x}$.

5. Since $AD + DC = AC = x$, we have:
   \[
   k + \frac{k}{x} = x
   \]
   Solving for $k$, we get:
   \[
   k \left(1 + \frac{1}{x}\right) = x \implies k = \frac{x^2}{x+1}
   \]

6. Now, we calculate the perimeters of $\triangle{ABD}$ and $\triangle{BCD}$:
   - Perimeter of $\triangle{ABD}$:
     \[
     AB + BD + AD = x + BD + \frac{x^2}{x+1}
     \]
   - Perimeter of $\triangle{BCD}$:
     \[
     BC + BD + DC = 1 + BD + \frac{x^2}{x+1}
     \]

7. To find $BD$, we use the Law of Cosines in $\triangle{ABD}$ and $\triangle{BCD}$:
   - In $\triangle{ABD}$:
     \[
     BD^2 = AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos(36^\circ)
     \]
     \[
     BD^2 = x^2 + \left(\frac{x^2}{x+1}\right)^2 - 2 \cdot x \cdot \frac{x^2}{x+1} \cdot \cos(36^\circ)
     \]
   - In $\triangle{BCD}$:
     \[
     BD^2 = BC^2 + DC^2 - 2 \cdot BC \cdot DC \cdot \cos(36^\circ)
     \]
     \[
     BD^2 = 1 + \left(\frac{x^2}{x+1}\right)^2 - 2 \cdot 1 \cdot \frac{x^2}{x+1} \cdot \cos(36^\circ)
     \]

8. Since $BD$ is common in both triangles, we can simplify the calculations by noting that the positive difference between the perimeters is:
   \[
   \left(x + BD + \frac{x^2}{x+1}\right) - \left(1 + BD + \frac{x^2}{x+1}\right) = x - 1
   \]

The final answer is $\boxed{1}$.

Answer: 1


In a two-story house, which is inhabited not only on both floors but also in the basement, 35 people live above someone and 45 people live below someone. At the same time, one third of all the people living in the house live on the 1st floor.

How many people live in the house in total?

Solution:
People who live above someone are residents of the 2nd and 1st floors. People who live below someone are residents of the 1st floor and the ground floor. In the sum $35+45=80$, the residents of the 1st floor are counted twice.

If we denote the number of residents on the 1st floor by $p$, then the total number of residents in the building can be expressed as either $80-p$ or $3p$. From this, we get the equation, which we can easily solve:

$$
\begin{aligned}
3 p & =80-p, \\
4 p & =80, \\
p & =20 .
\end{aligned}
$$

There are a total of 60 people living in the building.

Evaluation. 2 points for the observation that the residents of the 1st floor are counted twice in the sum $35+45$; 2 points for setting up and solving the equation; 2 points for the total number of people in the building.

Note. If $d, p$, and $z$ represent the number of residents on the 2nd floor, the 1st floor, and the ground floor, respectively, then from the problem statement we have

$$
d+p=35, \quad p+z=45, \quad d+p+z=3 p .
$$

From this, we can express all unknowns in various ways: $d=15, p=20$, and $z=25$. Such solutions should be evaluated with 2 points for setting up the equations, 2 points for solving them, and 2 points for the conclusion.

Answer: 60


Solve in $\mathbb{R}$ the equation $\sqrt{x^{2}+9}+\sqrt{x^{2}-6 x+10}=5$.

Solution:
Solution

First Solution: We can eliminate radicals by squaring both sides of the equation below:

$$
\begin{aligned}
\sqrt{x^{2}+9}+\sqrt{x^{2}-6 x+10} & =5 \\
\left(\sqrt{x^{2}-6 x+10}\right)^{2} & =\left(5-\sqrt{x^{2}+9}\right)^{2} \\
x^{2}-6 x+10 & =25-10 \sqrt{x^{2}+9}+x^{2}+9 \\
10 \sqrt{x^{2}+9} & =6 x+24 \\
25\left(x^{2}+9\right) & =(3 x+12)^{2} \\
25 x^{2}+225 & =9 x^{2}+72 x+144 \\
16 x^{2}-72 x+81 & =0 \\
(4 x-9)^{2} & =0
\end{aligned}
$$

Consequently, $4 x-9=0$, or $x=\frac{9}{4}$. To verify that $x=\frac{9}{4}$ is the solution, we can write

$$
\begin{aligned}
\sqrt{x^{2}+9}+\sqrt{x^{2}-6 x+10} & =\sqrt{\frac{81}{16}+9}+\sqrt{\frac{81}{16}-6 \cdot \frac{9}{4}+10} \\
& =\frac{15}{4}+\frac{5}{4} \\
& =5
\end{aligned}
$$

Second Solution: Consider points $F, D$, and $E$ in the Cartesian plane with coordinates $(x, 0)$, $(0,3)$, and $(3,-1)$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-35.jpg?height=597&width=720&top_left_y=1632&top_left_x=768)

We can associate the distances between some points with the given radicals:

$$
\begin{aligned}
D F & =\sqrt{x^{2}+9} \\
F E & =\sqrt{(x-3)^{2}+1^{2}} \\
& =\sqrt{x^{2}-6 x+10} \\
D E & =\sqrt{9+16} \\
& =5
\end{aligned}
$$

By the triangle inequality,

$$
\begin{aligned}
\sqrt{x^{2}+9}+\sqrt{(x-3)^{2}+1^{2}} & =D F+F E \\
& \geq D E \\
& =5
\end{aligned}
$$

Equality holds only when $D, F$, and $E$ are collinear. Therefore, the point $F$ must coincide with the intersection $G$ between $D E$ and the $O x$ axis, i.e., $x=9 / 4$.

Observation: A related problem that could be approached similarly using Analytic Geometry is:

What is the minimum value of the real function

$$
f(x)=\sqrt{a^{2}+x^{2}}+\sqrt{(b-x)^{2}+c^{2}}
$$

where $a, b, c$ are positive real numbers?

Repeating the previous argument, if $E=(b,-c), D=(0, a)$, and $F=(x, 0)$, then $f(x)=F E+F D$ and, by the triangle inequality, $f(x) \geq D E$ where equality occurs only if $D, F$, and $E$ are collinear. The intersection of $D E$ and the $O x$ axis is $\left(\frac{a b}{a+c}, 0\right)$, and thus the minimum occurs at $x=\frac{a b}{a+c}$.

Answer: \frac{9}{4}


The main stained glass window of a modern cathedral is a circle with a diameter of 2 m, intersected by a cross formed by two perpendicular lines that intersect at a point 50 cm away from the center of the window. During the great mass, one slightly absent-minded Parisian decided to calculate the sum of the squares of the lengths of the two segments of this cross.

"Look at this! - he exclaimed. - The number of square meters is equal to the number of mortal sins ${ }^{2}$".

Why did this happen?

Solution:
15. Let the cross form segments $A B$ and $C D$, intersecting at point $M$; further, $O$ - the center of the stained glass, $H$ - its projection on $A B$, and finally, $\alpha$ - the measure of $O M H$. Consider the right triangle $O H B$:

$$
H B^{2}=O B^{2}-O H^{2}=O B^{2}-O M^{2} \cdot \sin ^{2} \alpha
$$

But $A B=2 H B$. Therefore, $A B^{2}=4 O B^{2}-4 O M^{2} \cdot \sin ^{2} \alpha=\left(4-\sin ^{2} \alpha\right) \mathrm{M}^{2}$. $C D^{2}$ can be calculated similarly, but instead of $\alpha$, we now take $90^{\circ}-\alpha$. In the end, we get

$C D^{2}=4 O C^{2}-4 O M^{2} \cdot \sin ^{2}\left(90^{\circ}-\alpha\right)=\left(4-\cos ^{2} \alpha\right) \mathrm{M}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-158.jpg?height=943&width=932&top_left_y=568&top_left_x=582)

$$
A B^{2}+C D^{2} \text { - the number of deadly sins }
$$

Thus,

$$
A B^{2}+C D^{2}=7 \mathrm{~m}^{2}
$$

but 7 is the number of deadly sins.

Answer: 7\mathrm{~}^{2}


In how many ways can we color the cells of a $2 \times 2016$ board in two colors so that there are no three cells of the same color that can be simultaneously covered by a tile of the shape shown in the figure? The tile can be rotated.

![](https://cdn.mathpix.com/cropped/2024_05_30_1a4376e76c5095e3db07g-06.jpg?height=149&width=140&top_left_y=1916&top_left_x=1552)

Solution:
## Solution.

If both fields in the first column are white, then in the second column, no field can be white. Thus, both fields in the second column are black. Similarly, we conclude that both fields in the third column are white, and so on. This is one valid coloring.

Analogously, if both fields in the first column are black, we get another valid coloring.

![](https://cdn.mathpix.com/cropped/2024_05_30_1a4376e76c5095e3db07g-06.jpg?height=149&width=605&top_left_y=2421&top_left_x=654)

If the fields in the first column are of different colors, then in the second column, we cannot have two fields of the same color. Thus, the fields in the second column must also be of different colors, and both possibilities (the black field is in the first or second row) are allowed. The total number of colorings in this case is \(2 \cdot 2 \cdot \ldots \cdot 2 = 2^{2016}\).

## COUNTY COMPETITION IN MATHEMATICS

2nd grade - high school - A variant
March 3, 2016.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.

Answer: 2^{2016}+2


The common difference of an arithmetic sequence with real terms is 4, and the square of the first term plus the sum of the remaining terms does not exceed 100. Such a sequence can have at most $\qquad$ terms.

Solution:
Let the sequence $\left\{a_{n}\right\}$ have the first term $a_{1}$, and the common difference be $4 \Rightarrow S_{n}=n a_{1}+2 n(n-1)$
$$
\begin{array}{l}
\Rightarrow a_{1}^{2}+S_{n}-a_{1} \leqslant 100 \Rightarrow a_{1}^{2}+(n-1) a_{1}+2 n^{2}-2 n-100 \leqslant 0 \\
\Rightarrow\left(a_{1}+\frac{n-1}{2}\right)^{2}+\frac{7}{4} n^{2}-\frac{3}{2} n-\frac{401}{4} \leqslant 0 \Rightarrow 7 n^{2}-6 n-401 \leqslant 0 \\
\Rightarrow n \leqslant\left[\frac{3+\sqrt{9+7 \cdot 401}}{7}\right]=8 . \text { Therefore, such a sequence can have at most } 8 \text { terms. }
\end{array}
$$

Answer: 8


Let $n$ be a positive integer,
$$
\begin{aligned}
S= & \{(x, y, z) \mid x, y, z \in\{0,1, \cdots, n\}, \\
& x+y+z>0\}
\end{aligned}
$$

is a set of $(n+1)^{3}-1$ points in three-dimensional space. Try to find the minimum number of planes whose union contains $S$ but does not contain $(0,0,0)$.

Solution:
【Analysis】Supplementary definition based on Definition 1:
For a function $f(x, y, z)$ on $\mathbf{R}$, the first-order difference is
$$
\begin{array}{l}
\Delta_{x} f=f(x+1, y, z)-f(x, y, z), \\
\Delta_{y} f=f(x, y+1, z)-f(x, y, z), \\
\Delta_{z} f=f(x, y, z+1)-f(x, y, z) .
\end{array}
$$

We call $\Delta \in\left\{\Delta_{x}, \Delta_{y}, \Delta_{z}\right\}$ a difference operator.
Returning to the original problem.
Suppose there exist $m$ planes
$$
a_{i} x+b_{i} y+c_{i} z-d_{i}=0\left(1 \leqslant i \leqslant m, d_{i} \neq 0\right)
$$

satisfying the conditions.
Construct an $m$-degree polynomial
$$
f(x, y, z)=\prod_{i=1}^{m}\left(a_{i} x+b_{i} y+c_{i} z-d_{i}\right) \text {. }
$$

By the problem statement, when $(x, y, z) \in S$,
$$
f(x, y, z) \equiv 0 \text {, but } f(0,0,0) \neq 0 \text {. }
$$

If $m<3 n$, by part (3) of Theorem 4, we easily get
$$
\Delta_{x}^{n} \Delta_{y}^{n} \Delta_{z}^{n} f \equiv 0 .
$$

By part (2) of Theorem 3, we have
$$
\Delta^{n} f(x)=\sum_{i=0}^{n}(-1)^{n-i} \mathrm{C}_{n}^{i} f(x+i) \text {. }
$$

Generalizing to three dimensions, we have
$$
\begin{array}{l}
\Delta_{x}^{n} \Delta_{y}^{n} \Delta_{z}^{n} f \\
=\sum_{(i, j, k) \in S U(0,0,0)!}(-1)^{3 n-i-j-k} \mathrm{C}_{n}^{i} \mathrm{C}_{n}^{j} \mathrm{C}_{n}^{k} \text {. } \\
f(x+i, y+j, z+k) \text {. } \\
\end{array}
$$

Let $x=y=z=0$, we get
$$
\begin{array}{l}
f(0,0,0) \\
=\sum_{(i, j, k) \in s}(-1)^{i+j+k+1} \mathrm{C}_{n}^{i} \mathrm{C}_{n}^{j} \mathrm{C}_{n}^{k} f(i, j, k)=0 .
\end{array}
$$

This contradicts the problem statement.
Therefore, $m \geqslant 3 n$.
Construct $3 n$ planes to include all points in $S$.
Notice that each point in $S$ has at least one coordinate as $1,2, \cdots, n$.
Thus, the $3 n$ planes
$$
\begin{array}{l}
x-i=0(1 \leqslant i \leqslant n), \\
y-j=0(1 \leqslant j \leqslant n), \\
z-k=0(1 \leqslant k \leqslant n)
\end{array}
$$

clearly meet the requirements.
Hence, the minimum number of planes satisfying the conditions is $3 n$.

Answer: 3n


Given numbers $x_{1}, \ldots, x_{n}$ from the interval $\left[0, \frac{\pi}{2}\right]$. Find the maximum value of the expression

$$
A=\left(\sqrt{\sin x_{1}}+\ldots+\sqrt{\sin x_{n}}\right) \cdot\left(\sqrt{\cos x_{1}}+\ldots+\sqrt{\cos x_{n}}\right) .
$$

Solution:
Answer: $\frac{n^{2}}{\sqrt{2}}$.

Solution. Note that for any $a_{1}, \ldots, a_{n}$

$$
\left(\sum_{k=1}^{n} a_{k}\right)^{4} \leqslant\left(n \sum_{k=1}^{n} a_{k}^{2}\right)^{2} \leqslant n^{3} \sum_{k=1}^{n} a_{k}^{4}
$$

From this, by the Cauchy-Schwarz inequality,

$$
A^{2} \leqslant \frac{1}{2}\left(\left(\sum_{k=1}^{n} \sqrt{\sin x_{k}}\right)^{4}+\left(\sum_{k=1}^{n} \sqrt{\cos x_{k}}\right)^{4}\right) \leqslant \frac{n^{3}}{2}\left(\sum_{k=1}^{n} \sin ^{2} x_{k}+\sum_{k=1}^{n} \cos ^{2} x_{k}\right)=\frac{n^{4}}{2}
$$

Therefore, $A \leqslant \frac{n^{2}}{\sqrt{2}}$. Equality is achieved when $x_{1}=\ldots=x_{n}=\frac{\pi}{4}$.

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


In a circle with radius $10 \mathrm{~cm}$, segment $A B$ is a diameter and segment $A C$ is a chord of $12 \mathrm{~cm}$. Determine the distance between points $B$ and $C$.

Solution:
Given that $AB$ is a diameter, the triangle $\triangle ABC$ is inscribed in a semicircle. This implies that this triangle is a right triangle at vertex $C$. By the Pythagorean Theorem,

$$
BC^{2}=AB^{2}-AC^{2}
$$

that is,

$$
BC^{2}=20^{2}-12^{2}=256=16^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-27.jpg?height=352&width=415&top_left_y=526&top_left_x=1321)

Thus, we obtain that $BC=16$.

Answer: 16


The height of the cylinder is $H$, and the radius of its base is $R$. Inside the cylinder, a pyramid is placed, the height of which coincides with the generatrix $A A_{1}$ of the cylinder, and the base is an isosceles triangle $A B C(A B=A C)$ inscribed in the base of the cylinder. Find the lateral surface area of the pyramid if $\angle A=120^{\circ}$.

Solution:
2.21. Draw $A D \perp B C$ and connect points $A_{1}$ and $D$ (Fig. 2.19). According to the theorem of three perpendiculars, we have $A_{1} D \perp B C$. Since arc $C A B$ contains $120^{\circ}$, and arcs $A C$ and $A B$ each contain $60^{\circ}$, then $B C=R \sqrt{3}$, $A B=R$. In $\triangle A B D$ we have $A D=\frac{R}{2}$. Further, from $\triangle A A_{1} D$ we get $A_{1} D=\sqrt{H^{2}+\frac{R^{2}}{4}}=\frac{1}{2} \sqrt{R^{2}+4 H^{2}}$. There-

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-466.jpg?height=469&width=388&top_left_y=327&top_left_x=752)

Fig. 2.19

fore,

$$
\begin{gathered}
S_{\triangle A_{1} A B}=\frac{1}{2} A B \cdot A A_{1}=\frac{1}{2} R H \\
S_{\triangle A_{1} B C}=\frac{1}{2} B C \cdot A_{1} D=\frac{1}{2} R \sqrt{3} \cdot \frac{1}{2} \sqrt{R^{2}+4 H^{2}}=\frac{1}{4} \sqrt{3 R^{2}+12 H^{2}}
\end{gathered}
$$

Finally, we find

$$
\begin{gathered}
S_{60 \mathrm{~K}}=2 S_{\triangle \mathrm{A}_{1} A B}+S_{\triangle \mathrm{A}_{1} B C}=R H+\frac{1}{4} R \sqrt{3 R^{2}+12 H^{2}}= \\
=\frac{R}{4}\left(4 H+\sqrt{3 R^{2}+12 H^{2}}\right)
\end{gathered}
$$

Answer: $0.25 R\left(4 H+\sqrt{3 R^{2}+12 H^{2}}\right)$.

Answer: 0.25R(4H+\sqrt{3R^{2}+12H^{2}})


Given the sequence $\left\{a_{n}\right\}$ with the general term $a_{n}=\left(\frac{2}{3}\right)^{n-1}\left[\left(\frac{2}{3}\right)^{n-1}-1\right]$, which of the following statements is correct? ( ).
(A) The maximum term is $a_{1}$, the minimum term is $a_{4}$
(B) The maximum term is $a_{1}$, the minimum term does not exist
(C) The maximum term does not exist, the minimum term is $a_{3}$
(D) The maximum term is $a_{1}$, the minimum term is $a_{3}$

Solution:
6 From the problem, we have $a_{1}=0, a_{2}=-\frac{2}{9}, a_{3}=-\frac{20}{81}$. When $n \geqslant 3$, $t=\left(\frac{2}{3}\right)^{n-1}$ is a decreasing function, and
$$
0a_{n}>a_{n-1}>\cdots>a_{4}>a_{3} .
$$

Therefore, $a_{1}$ is the largest, and $a_{3}$ is the smallest. Hence, the answer is D.

Answer: D


A Pretti number is a seven-digit positive integer with the following properties:

- The integer formed by its leftmost three digits is a perfect square.
- The integer formed by its rightmost four digits is a perfect cube.
- Its ten thousands digit and ones (units) digit are equal.
- Its thousands digit is not zero.

How many Pretti numbers are there?

Solution:
Since a Pretti number has 7 digits, it is of the form $a b c d$ efg.

From the given information, the integer with digits $a b c$ is a perfect square.

Since a Pretti number is a seven-digit positive integer, then $a>0$, which means that $a b c$ is between 100 and 999, inclusive.

Since $9^{2}=81$ (which has two digits) $10^{2}=100$ (which has three digits) and $31^{2}=961$ (which has three digits) and $32^{2}=1024$ (which has four digits), then $a b c$ (which has three digits) must be one of $10^{2}, 11^{2}, \ldots, 30^{2}, 31^{2}$, since $32^{2}$ has 4 digits.

From the given information, the integer with digits defg is a perfect cube.

Since the thousands digit of a Pretti number is not 0 , then $d>0$.

Since $9^{3}=729$ and $10^{3}=1000$ and $21^{3}=9261$ and $22^{3}=10648$, then defg (which has four digits) must be one of $10^{3}, 11^{3}, \ldots, 20^{3}, 21^{3}$, since $22^{3}$ has 5 digits.

Since the ten thousands digit and units digit of the original number are equal, then $c=g$. In other words, the units digits of $a b c$ and defg are equal.

The units digit of a perfect square depends only on the units digit of the integer being squared, since in the process of multiplication no digit to the left of this digit affects the resulting units digit.

The squares $0^{2}$ through $9^{2}$ are $0,1,4,9,16,25,36,49,64,81$.

This gives the following table:

| Units digit of $n^{2}$ | Possible units digits of $n$ |
| :---: | :---: |
| 0 | 0 |
| 1 | 1,9 |
| 4 | 2,8 |
| 5 | 5 |
| 6 | 4,6 |
| 9 | 3,7 |

Similarly, the units digit of a perfect cube depends only on the units digit of the integer being cubed.

The cubes $0^{3}$ through $9^{3}$ are $0,1,8,27,64,125,216,343,512,729$.

This gives the following table:

| Units digit of $m^{3}$ | Possible units digits of $m$ |
| :---: | :---: |
| 0 | 0 |
| 1 | 1 |
| 2 | 8 |
| 3 | 7 |
| 4 | 4 |
| 5 | 5 |
| 6 | 6 |
| 7 | 3 |
| 8 | 2 |
| 9 | 9 |

We combine this information to list the possible values of $c=g$ (from the first table, these must be $0,1,4,5,6,9)$, the squares between $10^{2}$ and $31^{2}$, inclusive, with this units digit, and the cubes between $10^{3}$ and $21^{3}$ with this units digit:

| Digit $c=g$ | Possible squares | Possible cubes | Pretti numbers |
| :---: | :---: | :---: | :---: |
| 0 | $10^{2}, 20^{2}, 30^{2}$ | $10^{3}, 20^{3}$ | $3 \times 2=6$ |
| 1 | $11^{2}, 19^{2}, 21^{2}, 29^{2}, 31^{2}$ | $11^{3}, 21^{3}$ | $5 \times 2=10$ |
| 4 | $12^{2}, 18^{2}, 22^{2}, 28^{2}$ | $14^{3}$ | $4 \times 1=4$ |
| 5 | $15^{2}, 25^{2}$ | $15^{3}$ | $2 \times 1=2$ |
| 6 | $14^{2}, 16^{2}, 24^{2}, 26^{2}$ | $16^{3}$ | $4 \times 1=4$ |
| 9 | $13^{2}, 17^{2}, 23^{2}, 27^{2}$ | $19^{3}$ | $4 \times 1=4$ |

For each square in the second column, each cube in the third column of the same row is possible. (For example, $19^{2}$ and $11^{3}$ give the Pretti number 3611331 while $19^{2}$ and $21^{3}$ give the Pretti number 3619 261.) In each case, the number of Pretti numbers is thus the product of the number of possible squares and the number of possible cubes.

Therefore, the number of Pretti numbers is $6+10+4+2+4+4=30$.

Answer: 30


In a carriage, any $m(m \geqslant 3)$ passengers have a unique common friend (if A is a friend of B, then B is also a friend of A, and no one is a friend of themselves), how many friends does the person with the most friends have in this carriage?

Solution:
19 Let the person with the most friends have $k$ friends, clearly $k \geqslant m$. If $k>m$, let $A$ have $k$ friends $B_{1}$, $B_{2}, \cdots, B_{k}$. And denote
$$
S=\left\{B_{1}, B_{2}, B_{3}, \cdots, B_{k}\right\},
$$

Let $B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}$ be any $m-1$ elements chosen from $S$. Then $A, B_{i_{1}}, \cdots, B_{i_{m-1}}$ these $m$ people have a common friend, denoted as $C_{i}$. Since $C_{i}$ is a friend of $A$, it follows that $C_{i} \in S$. If
$$
\left\{B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}\right\} \neq\left\{B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{m-1}}\right\},
$$

and the unique friends corresponding to $\left\{A, B_{i_{1}}, \cdots, B_{i_{m-1}}\right\}$ and $\left\{A, B_{j_{1}}, \cdots, B_{j_{m-1}}\right\}$ are $C_{i}$, $C_{j} \in S$, respectively, then it must be that $C_{i} \neq C_{j}$, otherwise
$$
\left\{B_{i_{1}}, B_{i_{2}}, \cdots, B_{i_{m-1}}\right\} \cup\left\{B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{m-1}}\right\}
$$

would have at least $m$ elements, and they would have at least two friends $A$ and $C_{i}$, which contradicts the problem statement. In this way, under the above correspondence, each element in $S$ corresponds to at most one $(m-1)$-element subset of $S$, thus we have
$$
C_{k}^{m-1} \leqslant k
$$

Since $m \geqslant 3, m-1 \geqslant 2$, we have
$$
C_{k}^{m-1}>C_{k}^{1}=k
$$

Clearly, (1) and (2) are contradictory. Therefore, the maximum value sought is $m$.

Answer: m


Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.

Solution:
Answer: 20.

Solution. Lines $C_{1} Y$ and $Z X$ lie in parallel planes $B B_{1} C_{1} C$ and $A A_{1} D_{1} D$, so they do not intersect. Since these two lines also lie in the same plane $C_{1} X Z Y$, they are parallel. Similarly, lines $Y Z$ and $C_{1} X$ are parallel. Therefore, quadrilateral $C_{1} X Z Y$ is a parallelogram.

![](https://cdn.mathpix.com/cropped/2024_05_06_0323bbf84409a1adeb34g-46.jpg?height=504&width=680&top_left_y=1143&top_left_x=385)

Fig. 15: to the solution of problem 11.8

Let us lay off segment $A T$ on $A D$, equal to $B Y$ (Fig. 15). Quadrilateral $B Y T A$ is a parallelogram, meaning segments $Y T, B A$, and $C_{1} D_{1}$ are parallel and equal. Additionally, segments $Y Z$ and $C_{1} X$ are parallel and equal; from the parallelism follows the equality of angles $Z Y T$ and $X C_{1} D_{1}$, which gives the equality of triangles $Z Y T$ and $X C_{1} D_{1}$.

Then

$$
\begin{aligned}
D Z & =Z T+A D-A T=X D_{1}+B_{1} C_{1}-B Y= \\
& =\left(A_{1} D_{1}-A_{1} X\right)+B_{1} C_{1}-B Y=(14-5)+14-3=20
\end{aligned}
$$

Another solution. As in the previous solution, we will use the fact that $C_{1} X Z Y$ is a parallelogram.

Consider the chain of vector equalities:

$$
\overrightarrow{D Z}=\overrightarrow{D C}+\overrightarrow{C Y}+\overrightarrow{Y Z}=\overrightarrow{D_{1} C_{1}}+\overrightarrow{C Y}+\overrightarrow{C_{1} X}=\overrightarrow{C Y}+\overrightarrow{D_{1} X}
$$

In the last part, all vectors are oriented the same as $\overrightarrow{D Z}$, so we can transition to the equality of segments and continue:

$$
D Z=C Y+D_{1} X=(C B-B Y)+\left(D_{1} A_{1}-A_{1} X\right)=2 B_{1} C_{1}-B Y-A_{1} X
$$

Substituting the lengths given in the condition, we get $D Z=2 \cdot 14-3-5=20$.

Answer: 20


Let the prime $p$ satisfy that there exist positive integers $x, y$ such that $p-1=2 x^{2}, p^{2}-1=2 y^{2}$.
Then the number of primes $p$ that meet the condition is ( ).
(A) 1
(B) 2
(C) 3
(D) 4

Solution:
6. A.

Obviously, $p \neq 2$.
When the prime $p \geqslant 3$, obviously,
$$
p>y>x \text {, }
$$

and $p(p-1)=2(y+x)(y-x)$.
From $p \mid 2(y+x)(y-x)$, we know $p \mid(y+x)$.
Then $p \leqslant x+y<2 p \Rightarrow p=x+y$
$$
\begin{array}{l}
\Rightarrow p-1=2(y-x) \Rightarrow p=4 x-1 \\
\Rightarrow 4 x-1-1=2 x^{2} \Rightarrow x=1 \Rightarrow p=3 .
\end{array}
$$

Upon verification, $p=3$ meets the conditions.

Answer: A


If the areas of rectangles POED, OMCE, AKOP are 2, 4, 6 respectively, determine the area of rectangle $A B C D$.

If the perimeters of rectangles POED, OMCE, AKOP are 6, 10, 10 respectively, calculate the perimeter of rectangle $A B C D$.

Solution:
Solution.1) Since the area of OMCE is twice the area of $P O E D$, it follows that side $O M$ is twice as large as side $O P$. Similarly, since the area of $A K O P$ is three times the area of $P O E D$, it follows that $O K$ is three times larger than $O E$. We conclude that the area of rectangle BMOK is $2 \cdot 3=6$ times larger than the area of rectangle $P O E D$, i.e., it is $6 \cdot 2=12$. The area of $A B C D$ is the sum of the areas of the four rectangles and is equal to $2+4+6+12=24$.

2) From the fact that the perimeter of $O M C E$ is 4 units larger than the perimeter of $P O E D$, it follows that $O M$ is $4: 2=2$ units larger than $O P$. Similarly, $O K$ is two units larger than $O E$. It follows that the perimeter of $K B M O$ is $2 \cdot 2+2 \cdot 2=8$ units larger than the perimeter of $P O E D$, and is $6+8=14$. To determine the perimeter of rectangle $A B C D$, the perimeters of the four rectangles need to be summed and the result divided by 2, i.e., the perimeter of $A B C D$ is $\frac{6+10+10+14}{2}=20$.

## VI section

Answer: 20


In a match without ties, the game ends when one person wins 2 more games than the other, and the one with more wins is the winner. It is known that in the odd-numbered games, the probability of A winning is $\frac{3}{5}$; in the even-numbered games, the probability of B winning is $\frac{3}{5}$. Then the expected number of games played when the match ends is

Solution:
6. $\frac{25}{6}$.

Let the expected number of games played until the end of the match be $E$.
The probability that player A wins the first game and player B wins the second, or player B wins the first and player A wins the second, is
$$
\frac{3}{5} \times \frac{3}{5}+\frac{2}{5} \times \frac{2}{5}=\frac{13}{25} \text {. }
$$

In this case, the expected number of additional games to be played is $E$, and the total expected number of games is $E+2$.

If either player A or player B wins both of the first two games, the probability is $1-\frac{13}{25}=\frac{12}{25}$, then the match ends exactly after two games. Therefore, $E=\frac{12}{25} \times 2+\frac{13}{25}(E+2) \Rightarrow E=\frac{25}{6}$.

Answer: \frac{25}{6}


From a rectangular sheet of paper with dimensions $54 \mathrm{~cm}$ and $15 \mathrm{~cm}$, Marko cuts out the largest possible square. From the remaining piece of paper, he again cuts out the largest possible square and continues the process until the entire sheet is cut into squares, i.e., until the remaining piece of paper is also a square. What is the sum of the perimeters of all the squares thus obtained?

Solution:
## Solution.

From a rectangular piece of paper with dimensions $54 \mathrm{~cm}$ and $15 \mathrm{~cm}$, Marko first cuts out three squares with a side length of $15 \mathrm{~cm}$.

|  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |

|  |
| :--- |
|  |
|  |
|  |

After that, from the rectangular piece of paper with dimensions $54-3 \cdot 15=9$, i.e., $9 \mathrm{~cm}$ and $15 \mathrm{~cm}$, Marko cuts out one square with a side length of $9 \mathrm{~cm}$.

After that, from the rectangular piece of paper with dimensions $15-9=6$,

i.e., $6 \mathrm{~cm}$ and $9 \mathrm{~cm}$, Marko cuts out one square with a side length of $6 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_e5ca9b3e8971fb159cf9g-1.jpg?height=200&width=282&top_left_y=1913&top_left_x=1595)

After that, from the rectangular piece of paper with dimensions $9-6=3$,

i.e., $3 \mathrm{~cm}$ and $6 \mathrm{~cm}$, Marko cuts out two squares with a side length of $3 \mathrm{~cm}$.

The sum of the perimeters of all the squares is $3 \cdot 60+36+24+2 \cdot 12=180+36+24+24=264 \mathrm{~cm}$.

Answer: 264\mathrm{~}


A three-digit number $A$ is such that the difference between the largest and smallest three-digit numbers formed by its digits is still $A$. What is this number $A$?

Solution:
【Solution】Solution: Let the three digits that form the three-digit number $A$ be $a, b, c$, and $a>b>c$, then the largest three-digit number is $a \times 100 + b \times 10 + c$, and the smallest three-digit number is $c \times 100 + b \times 10 + a$,

So the difference is $(a \times 100 + b \times 10 + c) - (c \times 100 + b \times 10 + a) = 99 \times (a - c)$.
Therefore, the original three-digit number is a multiple of 99, the possible values are $198, 297, 396, 495, 594, 693, 792$, 891,

Among them, only 495 meets the requirement, $954 - 459 = 495$.
Answer: The three-digit number $A$ is 495.

Answer: 495


Find the derivative of the function $y=9 x^{5}$.

Solution:
Solution. Using rule V and the formula $\left(x^{n}\right)^{\prime}=n x^{n-1}$, we get

$$
y^{\prime}=\left(9 x^{5}\right)^{\prime}=9 \cdot 5 x^{4}=45 x^{4}
$$

When skilled, intermediate steps can be skipped:

$$
\left(9 x^{5}\right)^{\prime}=45 x^{4}
$$

Answer: 45x^{4}


Evaluate $ \int_0^1 xe^{x^2}dx$.

Let $ I_n=\int_0^1 x^{2n-1}e^{x^2}dx$. Express $ I_{n+1}$ in terms of $ I_n$.

Solution:
### Part 1: Evaluate \( \int_0^1 xe^{x^2} \, dx \)

1. **Substitution**: Let \( x^2 = t \). Then \( 2x \, dx = dt \) or \( x \, dx = \frac{1}{2} \, dt \).

2. **Change of Limits**: When \( x = 0 \), \( t = 0 \). When \( x = 1 \), \( t = 1 \).

3. **Rewrite the Integral**:
   \[
   \int_0^1 xe^{x^2} \, dx = \int_0^1 e^t \cdot \frac{1}{2} \, dt = \frac{1}{2} \int_0^1 e^t \, dt
   \]

4. **Evaluate the Integral**:
   \[
   \frac{1}{2} \int_0^1 e^t \, dt = \frac{1}{2} \left[ e^t \right]_0^1 = \frac{1}{2} (e^1 - e^0) = \frac{1}{2} (e - 1)
   \]

### Part 2: Express \( I_{n+1} \) in terms of \( I_n \)

1. **Definition**: Let \( I_n = \int_0^1 x^{2n-1} e^{x^2} \, dx \).

2. **Expression for \( I_{n+1} \)**:
   \[
   I_{n+1} = \int_0^1 x^{2n+1} e^{x^2} \, dx = \int_0^1 x^{2n} \cdot x e^{x^2} \, dx
   \]

3. **Integration by Parts**: Let \( u = x^{2n} \) and \( dv = x e^{x^2} \, dx \). Then \( du = 2n x^{2n-1} \, dx \) and \( v = \frac{e^{x^2}}{2} \).

4. **Apply Integration by Parts**:
   \[
   \int u \, dv = uv - \int v \, du
   \]
   \[
   I_{n+1} = \left. x^{2n} \cdot \frac{e^{x^2}}{2} \right|_0^1 - \int_0^1 \frac{e^{x^2}}{2} \cdot 2n x^{2n-1} \, dx
   \]

5. **Evaluate the Boundary Term**:
   \[
   \left. x^{2n} \cdot \frac{e^{x^2}}{2} \right|_0^1 = \left( 1^{2n} \cdot \frac{e^1}{2} \right) - \left( 0^{2n} \cdot \frac{e^0}{2} \right) = \frac{e}{2} - 0 = \frac{e}{2}
   \]

6. **Simplify the Integral**:
   \[
   I_{n+1} = \frac{e}{2} - n \int_0^1 x^{2n-1} e^{x^2} \, dx = \frac{e}{2} - n I_n
   \]

The final answer is \( \boxed{\frac{1}{2}(e - 1)} \) for part 1 and \( \boxed{I_{n+1} = \frac{e}{2} - n I_n} \) for part 2.

Answer: I_{n+1} = \frac{e}{2} - n I_n


Danka had a paper flower with ten petals. On each petal, there was exactly one digit, and no digit was repeated on any other petal. Danka tore off two petals so that the sum of the numbers on the remaining petals was a multiple of nine. Then she tore off another two petals so that the sum of the numbers on the remaining petals was a multiple of eight. Finally, she tore off two more petals so that the sum of the numbers on the remaining petals was a multiple of ten.

Find the three sums that could have remained after each tearing. Determine all such triplets of sums.

Solution:
The sum of all numbers on the petals is $0+1+\ldots+9=45$. Danka could have torn off a sum of at least 1 (if she tore off petals 0 and 1) and at most 17 (if she tore off petals 8 and 9).

- After the first tearing, the sum remaining on the flower must be in the range of 28 to 44 ( $45-17=28$ and $45-1=44$ ). Among these numbers, there is only one multiple of nine, which is 36.
- After the second tearing, the sum remaining on the flower must be in the range of 19 to 35 ( $36-17=19$ and $36-1=35$ ). Among these numbers, there are two multiples of eight, namely 32 or 24.
- Similarly, we determine the possible sums after the third tearing:
- If the sum remaining after the second tearing is 32, the sum remaining after the third tearing must be in the range of 15 to 31 ( $32-17=15$ and $32-1=31$ ). Among these numbers, there are two multiples of ten, namely 30 or 20.
- If the sum remaining after the second tearing is 24, the sum remaining after the third tearing must be in the range of 7 to 23 ( $24-17=7$ and $24-1=23$ ). Among these numbers, there are two multiples of ten, namely 20 or 10.

After each tearing, the following triples of sums could remain:

$$
(36,32,20),(36,32,30),(36,24,20),(36,24,10).
$$

For each of these possibilities, we need to verify whether they can actually be realized, i.e., whether pairs of petals can be torn off in such a way that no digit is used twice. In the following scheme, we write above the arrows an example of which petals could have been torn off in each step:

$$
\begin{aligned}
& 45 \xrightarrow{0,9} 36 \xrightarrow{1,3} 32 \xrightarrow{4,8} 20 \\
& 45 \xrightarrow{4,5} 36 \xrightarrow{1,3} 32 \xrightarrow{0,2} 30 \\
& 45 \xrightarrow{2,7} 36 \xrightarrow{4,8} 24 \xrightarrow{1,3} 20 \\
& 45 \xrightarrow{2,7} 36 \xrightarrow{4,8} 24 \xrightarrow{5,9} 10
\end{aligned}
$$

We see that all the given possibilities can be realized, the problem has four solutions.

Evaluation. 2 points for finding the four possibilities; 2 points for verifying that each possibility can be realized; 2 points for a process that excludes the possibility of missing any other possibility, or for justifying that no other possibility exists.

Answer: (36,32,20),(36,32,30),(36,24,20),(36,24,10)


A and B take turns picking distinct numbers from $0,1, \cdots, 81$, with A starting first, and each person picks one number from the remaining numbers each time. After all 82 numbers are picked, let $A$ and $B$ be the sums of all numbers chosen by A and B, respectively. During the process of picking numbers, A hopes that the greatest common divisor (GCD) of $A$ and $B$ is as large as possible, while B hopes that the GCD of $A$ and $B$ is as small as possible. Under the best strategies for both A and B, find the GCD of $A$ and $B$ after all numbers are picked.

Solution:
【Analysis】Answer: The greatest common divisor of $A$ and $B$ is
41.
Let $d=(A, B)$ denote the greatest common divisor of $A$ and $B$.
Notice that, $d \perp (A+B)$.
And $A+B=0+1+\cdots+81=41 \times 81$, then
$d \mid 81 \times 41$.
First, prove: Player A can ensure $41 \mid d$.
Pair the numbers $0,1, \cdots, 81$ into 41 pairs:
$\{x, x+41\}(x=0,1, \cdots, 40)$.
Player A can ensure that exactly one number is chosen from each pair.
In fact, in the first round, Player A can choose any number $a$. If Player B chooses the paired number $a'$, then in the second round, Player A can choose any unselected number $b$; if Player B chooses a number $c \neq a'$ in the first round, then in the second round, Player A can choose the paired number $c'$ of $c$. This continues until the last round.
At this point, Player A has chosen exactly one number from each pair, so
$A \equiv 0+1+\cdots+40 \equiv 0(\bmod 41)$.
Thus, $41 \mid B \Rightarrow 41 \mid d$.
Next, prove: Player B can ensure $d$ is not a multiple of 3.
Pair the numbers $0,1, \cdots, 81$ into 41 pairs:
$$
\begin{array}{l}
\{x, x+39\}(x=0,1, \cdots, 38), \\
\{78,81\},\{79,80\} .
\end{array}
$$

Player B can ensure that exactly one number is chosen from each pair.
In fact, Player B only needs to choose the paired number of the number chosen by Player A in each round. Except for the last pair $\{79,80\}$, the numbers chosen by Player A and Player B are congruent modulo 3, while $79$ and $80$ are not congruent modulo 3, so $A-B$ is not a multiple of 3. Therefore, $d$ is not a multiple of 3.
Combining $d \mid 81 \times 41$, we know $d \mid 41$.
In summary, under the best strategies of both players, $d=41$.
【Summary】Notice that, $0+1+\cdots+81=3^{4} \times 41$, Player A can adopt an appropriate grouping to make the greatest common divisor $d$ a multiple of 41, while Player B can ensure $d$ is not a multiple of 3 through another grouping method, thus the balanced state of the game is $d=41$.

Answer: 41


In trapezoid $A B C D$, $A B / / D C$, $D C=2 A B=2 A D$. If $B D=6, B C=4$, then $S_{A B C D}=\ldots$ ($S_{A B C D}$ represents the area of quadrilateral $A B C D$, the same below.)

Solution:
7. 18

As shown in Figure 6, take the midpoint $M$ of $CD$, and connect $BM$, $AM$.

From the given conditions, it is easy to see that quadrilateral $ADMB$ is a rhombus, and quadrilateral $AMCB$ is a parallelogram. Therefore, $AM = BC = 4$.
Noting that $\triangle ADM \cong \triangle MAB \cong \triangle BMC$, we have $S_{ABCD} = \frac{3}{2} S_{\triangle DMB} = \frac{3}{2} \times \frac{1}{2} \times 6 \times 4 = 18$

Answer: 18


Grandma baked 21 batches of dumplings, with $N$ dumplings in each batch, $N>70$. Then she laid out all the dumplings on several trays, with 70 dumplings on each tray. What is the smallest possible value of $N$?

Solution:
Answer: 80.

Solution: The total number of baked buns is $21 \cdot N$. This number must be divisible by 70 to be able to distribute them into several trays of 70 each. $70=2 \cdot 5 \cdot 7$, and 21 is already divisible by 7. Therefore, $N$ must be divisible by 10, and the smallest such $N$ is 80.

Answer: 80


For how many positive integers $n \le 500$ is $n!$ divisible by $2^{n-2}$?

Solution:
To determine how many positive integers \( n \le 500 \) make \( n! \) divisible by \( 2^{n-2} \), we need to analyze the power of 2 in the factorial \( n! \). Specifically, we need to find the number of \( n \) such that the exponent of 2 in \( n! \) is at least \( n-2 \).

1. **Calculate the exponent of 2 in \( n! \):**
   The exponent of 2 in \( n! \) is given by:
   \[
   \nu_2(n!) = \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{4} \right\rfloor + \left\lfloor \frac{n}{8} \right\rfloor + \cdots
   \]
   We need this to be at least \( n-2 \):
   \[
   \nu_2(n!) \geq n-2
   \]

2. **Simplify the inequality:**
   Using the property of the sum of the digits in the binary representation of \( n \), denoted as \( s_2(n) \), we know:
   \[
   \nu_2(n!) = n - s_2(n)
   \]
   Therefore, the inequality becomes:
   \[
   n - s_2(n) \geq n - 2 \implies s_2(n) \leq 2
   \]

3. **Interpret \( s_2(n) \leq 2 \):**
   This means that \( n \) can have at most two 1's in its binary representation. Thus, \( n \) can be:
   - A power of 2 (which has exactly one 1 in its binary representation).
   - A sum of two distinct powers of 2 (which has exactly two 1's in its binary representation).

4. **Count the valid \( n \):**
   - **Powers of 2:** The powers of 2 less than or equal to 500 are \( 2^0, 2^1, 2^2, \ldots, 2^8 \). This gives us 9 numbers.
   - **Sum of two distinct powers of 2:** We need to count the sums \( 2^i + 2^j \) where \( i > j \) and both \( i \) and \( j \) are non-negative integers such that \( 2^i + 2^j \leq 500 \).

     Let's list these sums:
     - For \( i = 8 \): \( 2^8 + 2^7, 2^8 + 2^6, \ldots, 2^8 + 2^0 \) (8 sums)
     - For \( i = 7 \): \( 2^7 + 2^6, 2^7 + 2^5, \ldots, 2^7 + 2^0 \) (7 sums)
     - For \( i = 6 \): \( 2^6 + 2^5, 2^6 + 2^4, \ldots, 2^6 + 2^0 \) (6 sums)
     - For \( i = 5 \): \( 2^5 + 2^4, 2^5 + 2^3, \ldots, 2^5 + 2^0 \) (5 sums)
     - For \( i = 4 \): \( 2^4 + 2^3, 2^4 + 2^2, \ldots, 2^4 + 2^0 \) (4 sums)
     - For \( i = 3 \): \( 2^3 + 2^2, 2^3 + 2^1, 2^3 + 2^0 \) (3 sums)
     - For \( i = 2 \): \( 2^2 + 2^1, 2^2 + 2^0 \) (2 sums)
     - For \( i = 1 \): \( 2^1 + 2^0 \) (1 sum)

     Adding these, we get:
     \[
     8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36
     \]

5. **Total count:**
   Adding the counts of powers of 2 and sums of two distinct powers of 2, we get:
   \[
   9 + 36 = 45
   \]
   However, we need to exclude \( n = 1 \) because it does not satisfy \( n \leq 500 \) and \( n! \) divisible by \( 2^{n-2} \).

   Therefore, the final count is:
   \[
   45 - 1 = 44
   \]

The final answer is \(\boxed{44}\)

Answer: 44


How many three-digit numbers less than 200 can be expressed in the form $a^{2} \cdot b$, where $a$ and $b$ are prime numbers?

Solution:
4. If $a=2$, then $a^{2}=4$. For the product to be a three-digit number, $b$ must be a prime number greater than or equal to 25. For the product to be less than 200, $b$ must be a prime number less than 50.

Then $b \in\{29,31,37,41,43,47\}$.

1 POINT

If $a=3$, then $a^{2}=9$, so $b$ needs to be a prime number greater than 11 and less than or equal to 22.

Then $b \in\{13,17,19\}$.

1 POINT

For $a=5$, $a^{2}=25$, so $b \in\{5,7\}$.

1 POINT

For $a=7$, $a^{2}=49$, so $b \in\{3\}$.

1 POINT

For $a \geq 11$, $a^{2} \cdot b>200$ for any prime number $b$, so there are no more solutions.

Therefore, the total number of solutions is $6+3+2+1=12$.

Answer: 12