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5. In triangle $\mathrm{ABC}$, the sides $A B=4, B C=6$. Point $M$ lies on the perpendicular bisector of segment $A B$, and lines $A M$ and $A C$ are perpendicular. Find $M A$, if the radius of the circumscribed circle around triangle $A B C$ is 9.
|
Solution. Introduce a coordinate system with the origin at point A such that point C lies on the x-axis. From the problem statement, point M lies on the y-axis. Let's introduce notations for the unknown coordinates: $\mathrm{A}(0,0), \mathrm{B}\left(\mathrm{x}_{\mathrm{B}}, \mathrm{y}_{\mathrm{B}}\right), \mathrm{C}\left(\mathrm{x}_{\mathrm{C}}, 0\right), \mathrm{M}\left(0, \mathrm{y}_{\mathrm{M}}\right)$. Let $\mathrm{N}$ be the midpoint of $\mathrm{AB}$ and 0 be the center of the circumscribed circle, then $N\left(\frac{x_{B}}{2}, \frac{y_{B}}{2}\right)$ and $O\left(\frac{x_{C}}{2}, y_{0}\right)$. From the perpendicularity of vectors $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{MN}}$, it follows that $\mathrm{x}_{\mathrm{B}} \frac{\mathrm{x}_{\mathrm{B}}}{2}+\mathrm{y}_{\mathrm{B}}\left(\mathrm{y}_{\mathrm{M}}-\frac{\mathrm{y}_{\mathrm{B}}}{2}\right)=0$. From this, considering $\mathrm{AB}^{2}=\mathrm{x}_{\mathrm{B}}^{2}+\mathrm{y}_{\mathrm{B}}^{2}=16$, we get $\mathrm{y}_{\mathrm{M}}=\frac{8}{\mathrm{y}_{\mathrm{b}}}$. From the perpendicularity of vectors $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{MO}}$, it follows that $\mathrm{x}_{\mathrm{B}} \frac{\mathrm{x}_{C}}{2}+\mathrm{y}_{\mathrm{B}}\left(\mathrm{y}_{\mathrm{M}}-\mathrm{y}_{0}\right)=0$. Additionally, $\mathrm{BC}^{2}=\left(\mathrm{x}_{\mathrm{B}}-\mathrm{x}_{C}\right)^{2}+\mathrm{y}_{\mathrm{B}}^{2}=36$ and $\mathrm{AO}^{2}=\left(\frac{\mathrm{x}_{\mathrm{C}}}{2}\right)^{2}+\mathrm{y}_{0}^{2}=81$. These equations are sufficient to obtain $\mathrm{MA}=\left|\mathrm{y}_{\mathrm{M}}\right|=6$.
Answer: 6.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find all prime numbers whose representation in base 14 has the form 101010 ... 101 (ones and zeros alternate).
|
Solution: Let $2n+1$ be the number of digits in the number $A=101010 \ldots 101$. Let $q=$ 14 - the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2 n}=\frac{q^{2 n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2 k \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k+1}-1}{q-1} \cdot \frac{q^{2 k+1}+1}{q+1}$. Thus, the number $A$ is represented as the product of two integer factors (by the theorem of Bezout, the polynomial $q^{2 k+1} \pm 1$ is divisible without remainder by the polynomial $q \pm 1$), each of which is different from 1. Therefore, when $n$ is even, the number $A$ is not prime.
- $n=2 k-1 \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k}-1}{q^{2}-1} \cdot\left(q^{2 k}+1\right)$. For $k>1$, both factors are integers and different from 1; therefore, the number $A$ is composite. It remains to verify that for $k=1$, the number $A=q^{0}+$ $q^{2}=197$ is prime.
Answer: 197.
|
197
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. How many solutions of the equation $x^{2}-2 x \cdot \sin (x \cdot y)+1=0$ fall within the circle $x^{2}+y^{2} \leq 100$?
|
Solution. We will interpret the left side of the equation as a quadratic trinomial in terms of $x$. For the roots to exist, the discriminant must be non-negative, i.e., $D=4 \sin ^{2}(x \cdot y)-4 \geq 0 \Leftrightarrow \sin ^{2}(x \cdot y)=1 \Leftrightarrow \cos 2 x y=-1 \Leftrightarrow x y=\frac{\pi}{2}+\pi n, n \in \mathbb{Z}$. For even $n$, we get the equation $x^{2}-2 x+1=0 \Leftrightarrow x=1$, and for odd $n$, we find $x=-1$. The left side of the equation is an even function of $x$, so for $x=1$ and $x=-1$, the corresponding values of $y$ will be the same. The solution has the form $(x, y)=\left( \pm 1, \frac{\pi}{2}+2 \pi k\right), k \in \mathbb{Z}$. Only 6 solutions fall within the circle $x^{2}+y^{2} \leq 100$.
Answer: 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. It is known that the polynomial $f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4}$ has 4 distinct real roots $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$. The polynomial of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coefficient $b_{1}$ of the polynomial $g(x)$.
|
Solution: Let the coefficients of the given polynomial (except the leading one) be denoted by $a_{0}, a_{1}, a_{2}, a_{3}$
$$
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}
$$
Then, according to the problem, we have
$$
f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)
$$
Together with the polynomial $f(x)$, consider the polynomial $h(x)$, which has roots $\left\{-x_{1},-x_{2},-x_{3},-x_{4}\right\}$
$$
h(x)=\left(x+x_{1}\right)\left(x+x_{2}\right)\left(x+x_{3}\right)\left(x+x_{4}\right)=a_{0}-a_{1} x+a_{2} x^{2}-a_{3} x^{3}+x^{4}
$$
Consider the polynomial $G(x)=f(x) h(x)$:
$$
\begin{aligned}
& G(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)\left(x+x_{1}\right)\left(x+x_{2}\right)\left(x+x_{3}\right)\left(x+x_{4}\right)= \\
&=\left(x^{2}-x_{1}^{2}\right)\left(x^{2}-x_{2}^{2}\right)\left(x^{2}-x_{3}{ }^{2}\right)\left(x^{2}-x_{4}^{2}\right)
\end{aligned}
$$
By substituting the variable $y=x^{2}$, we obtain the required polynomial $g(y)$, since
$$
g(y)=\left(y-x_{1}{ }^{2}\right)\left(y-x_{2}{ }^{2}\right)\left(y-x_{3}{ }^{2}\right)\left(y-x_{4}{ }^{2}\right)
$$
In our case
$$
\begin{gathered}
f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4} \\
h(x)=8-32 x-12 x^{2}+4 x^{3}+x^{4} \\
g(x)=f(x) h(x)=64-1216 x^{2}+416 x^{4}-40 x^{6}+x^{8}
\end{gathered}
$$
Interregional Olympiad for Schoolchildren based on Departmental Educational Organizations
$$
g(y)=64-1216 y+416 y^{2}-40 y^{3}+y^{4}
$$
Answer: -1216.
|
-1216
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A robot is located in one of the cells of an infinite grid and can be given the following commands:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent cell to the right).
For example, if the robot executes a sequence of four commands (up, right, down, left), it will obviously return to its initial position, i.e., it will end up in the same cell from which it started. How many different sequences of 4 commands exist that return the robot to its initial position?
|
Solution. For brevity, let's denote the command to the left as L, to the right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be the number of L commands in the sequence. We will calculate the number $N_{k}$ of the desired sequences for $k$ from 0 to 2.
- $\boldsymbol{k}=\mathbf{0}$. The sequence consists only of U and D commands. Since they are equal in number, U commands should be placed on 2 out of 4 positions, and D commands on the remaining 2 positions. The number of ways to choose 2 positions out of 4 is $C_{4}^{2}$. Therefore, $N_{0}=C_{4}^{2}=6$.
- $\boldsymbol{k}=\mathbf{1}$. Each of the commands L, R, U, and D appears in the sequence exactly once. The number of permutations of 4 elements is $4!$. Therefore, $N_{1}=4!=24$;
- $\boldsymbol{k}=\mathbf{2}$. Here, there are two L, two R, and no U or D commands. Two L commands can be placed in $C_{4}^{2}$ ways. Thus, $N_{2}=C_{4}^{2}=6$.
Therefore, the total number of sequences is $N_{0}+N_{1}+N_{2}=36$.
Answer: 36.
|
36
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find all prime numbers whose decimal representation has the form 101010 ... 01.
|
Solution. Let $2 n+1$ be the number of digits in the number $A=101010 \ldots 101$ under investigation. Let $q=$ 10 be the base of the number system. Then $A=q^{0}+q^{2}+\cdots q^{2 n}=\frac{q^{2 n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2 k \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k+1}-1}{q-1} \cdot \frac{q^{2 k+1}+1}{q+1}$. Thus, the number $A$ is represented as the product of two integer factors (by the theorem of Bezout, the polynomial $q^{2 k+1} \pm 1$ is divisible without remainder by the polynomial $q \pm 1$), each of which is different from 1. Therefore, when $n$ is even, the number $A$ is not prime.
- $n=2 k-1 \Rightarrow A=\frac{q^{2 n+2}-1}{q^{2}-1}=\frac{q^{2 k}-1}{q^{2}-1} \cdot\left(q^{2 k}+1\right)$. For $k>1$, both factors are integers and different from 1; therefore, the number $A$ is composite. It remains to verify that for $k=1$, the prime number $A=q^{0}+$ $q^{2}=101$ is obtained.
Answer: 101.
|
101
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. The ordinary fraction $\frac{1}{221}$ is represented as a periodic decimal fraction. Find the length of the period. (For example, the length of the period of the fraction $\frac{25687}{99900}=0.25712712712 \ldots=0.25$ (712) is 3.)
|
Solution. Let's consider an example. We will convert the common fraction $\frac{34}{275}$ to a decimal. For this, we will perform long division (fig.). As a result, we find $\frac{34}{275}=0.123636363 \ldots=0.123(63)$. We obtain the non-repeating part 123 and the repeating part 63. Let's discuss why the non-repeating part appears here, and show that the fraction $\frac{1}{221}$ does not have a non-repeating part. The matter is that in the decimal representation of the fraction $\frac{34}{275}$, the digit 3 appears every time the remainder is 100 when dividing by 275. We see (and this is the key point!), that the same remainder 100 is given by different numbers: 650 and 1750. Where, in turn, did these 650 and 1750 come from? The number 650 was obtained by appending a zero to the number $r_{1}=65$ (the remainder from dividing 340 by 275). That is, $10 r_{1}=650$. Similarly, $10 r_{2}=1750$, where $r_{2}=175$. The numbers 650 and 1750 give the same remainder when divided by 275 because their difference is divisible by 275: $1750-650=10\left(r_{2}-r_{1}\right): 275$. This is possible only because the numbers 10 and 275 are not coprime. Now it is clear why the fraction $\frac{1}{221}$ does not have a non-repeating part: if $r_{1}$ and $r_{2}$ are different remainders from dividing by 221, then the product $10\left(r_{2}-r_{1}\right)$ is not divisible by 221 (the number 221, unlike 275, is coprime with 10 - the base of the number system, so there is no non-repeating part).
Thus, the decimal representation of the fraction $\frac{1}{221}$ has the form $\frac{1}{221}=0,\left(a_{1} a_{2} \ldots a_{n}\right)$. Let's find $n$. Denote $A=a_{1} a_{2} \ldots a_{n}$. Then $\frac{1}{221}=10^{-n} \cdot A+10^{-2 n} \cdot A+\cdots$. By the formula for
825
1750
1650
$-\boxed{1000}$ $\frac{825}{-1750}$ the sum of an infinite geometric series $\frac{1}{221}=\frac{A}{10^{n}-1}$. From this, $A=\frac{10^{n}-1}{221}$. Since $A$ is a natural number, we need to find (the smallest) natural $n$ for which the number $10^{n}$ gives a remainder of 1 when divided by 221.
Note that $221=13 \cdot 17$. In general, an integer $B$ (in our case $B=10^{n}$) gives a remainder of 1 when divided by 221 if and only if $B$ gives a remainder of 1 when divided by both 13 and 17. The necessity is obvious. Sufficiency: if $B=13 k_{1}+1$ and $B=17 k_{2}+1$, then $13 k_{1}=17 k_{2}$, which means that $k_{1}$ is divisible by 17, i.e., $k_{1}=17 m$. Therefore, $B=13 \cdot 17 m+1$, and when divided by 221, the remainder is indeed 1. Now let's find such $n$ that the number $10^{n}$ gives a remainder of 1 when divided by 13. Consider the sequence $b_{n}=10^{n}$. Replace its terms with remainders when divided by 13. We get this: $b_{1}=10, b_{2}=9, b_{3}=13, b_{4}=3, b_{5}=4, b_{6}=1, \ldots$ Each subsequent term is uniquely determined by the previous one. Therefore, $\left\{b_{n}\right\}$ is a periodic sequence, where every sixth term is 1. We will do the same for 17. There, 1 will be equal to every 16th term. Thus, the remainder 1 when divided by both 13 and 17 will be obtained at $n=\operatorname{LCM}(6,16)=48$.
Answer: 48.
|
48
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Anya and Borya are playing "battleship" with the following rules: 29 different points are chosen on a circle, numbered clockwise with natural numbers from 1 to 29. Anya draws a ship - an arbitrary triangle with vertices at these points. We will call a "shot" the selection of two different natural numbers $k$ and $m$ from 1 to 29. If the segment with endpoints at points numbered $k$ and $m$ has at least one common point with Anya's triangle, the ship is considered "wounded." Borya fires a "salvo" - several shots simultaneously. Anya drew a ship and showed it to Borya. And then they noticed that any "salvo" of K different shots will definitely wound Anya's ship. Indicate any position of Anya's ship for which the value of $K$ will be minimal.
|
Solution. The vertices of Anya's triangle divide the circle into three arcs (see figure). Let $x, y$, and $26-x-y$ be the number of points on these arcs (see figure), excluding the vertices of the triangle itself. For a "shot" with endpoints at points $k$ and $m$ to not hit the ship, both these points must lie on one of the arcs. Clearly, two different points can be chosen on the arc containing $x$ points in $C_{x}^{2}$ ways. The same applies to the other arcs. Therefore, the number $N$ of "safe" shots is the sum $N=C_{x}^{2}+C_{y}^{2}+C_{26-x-y}^{2}$. Then the next shot will definitely "hit" the ship, so $K=N+1$.
Thus, we need to find such non-negative integers $x, y$ that satisfy the condition $x+y \leq 26$, for which the value of $N$ is minimal.
Let's write the expression for $N$ in expanded form:
$$
N=\frac{x(x-1)}{2}+\frac{y(y-1)}{2}+\frac{(26-x-y)(25-x-y)}{2}
$$
Expanding the brackets and combining like terms, we get
$$
N=x^{2}-x(26-y)+y^{2}-26 y+325
$$
www.v-olymp.ru
Interregional Olympiad for Schoolchildren Based on Departmental Educational Organizations in Mathematics
For each fixed $y$ from 0 to 26, we will find such a value of $x$ that satisfies the inequality
$$
0 \leq x \leq 26-y
$$
for which the value of $N$ is minimal. If $y$ is fixed, then the right-hand side (1) takes its minimum value at
$$
x=\frac{26-y}{2}
$$
(the vertex of the parabola, belonging to the interval (2)). This minimum value is $13 y+156$. It, in turn, is minimal at $y=\frac{26}{3} \approx 8.6$. From (3), we then find $x=\frac{26}{3}$. Among the points with integer coordinates $(8,8),(8,9),(9,8),(9,9)$ - the nearest integer neighbors of the point of minimum $\left(\frac{26}{3}, \frac{26}{3}\right)$ - we choose the one for which the value of $N$ is the smallest. These are the points $(8,9),(9,8),(9,9)$. For them, $N=100$.
Answer: The ship should be placed such that on the three arcs into which the vertices of the ship divide the circle, there are 8, 9, and 9 points (excluding the vertices of the ship itself).
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (4 points) Numbers are marked on a chessboard (see Fig. 1). How many arrangements of 8 rooks, none of which attack each other, exist such that the numbers on the squares occupied by the rooks include all numbers from 0 to 7? $\left(\begin{array}{llllllll}0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0\end{array}\right)$ Fig. 1
|
# Solution:
Let's number the rows of the board from 1 to 8 from top to bottom. On the first row, the position of the rook can be chosen in 8 ways. For example, we choose the first cell with the number 0. On the second row, there are 7 options left. Six of them (cells from the second to the seventh) have the property that under them in the lower half of the board, there are numbers different from the number occupied by the first rook (in our case - the number 0), and one cell (the last one, with the number 7) is such that under it in the lower half stands the number occupied by the first rook (in our case - 0).
Consider the first group of 6 options and make an arbitrary choice (for example, the second cell with the number 1). After such a choice, it turns out that in the lower half of the board, two different numbers are prohibited for choosing positions (in our case, 7 and 6). Therefore, the positions with these numbers need to be chosen in the upper half of the board, and there are exactly 2 such options (in our case, the 7th and 8th cells in rows 3 and 4). After this choice, in the lower half of the board, there will be 4 columns with different numbers in the columns. That is, there are $4! = 24$ options in the lower half. Thus, we get $8 \times 6 \times 2 \times 24$ options.
Consider the option of choosing a position on the second row when under the chosen cell in the lower half of the board stands the number chosen on the first row (in our case, the last cell with the number 7). After such a choice, in the third row, there are 6 options leading to a single option in the fourth row. After choosing options in the upper half of the board, in the lower half, there are again 4 free columns with four different numbers, i.e., 24 options. Thus, we get $8 \times 1 \times 6 \times 24$ options.
In total, we get:
$8 \times 6 \times 2 \times 24 + 8 \times 1 \times 6 \times 24 = 3456$.
Answer: 3456.
|
3456
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (5 points) From point $A$, lying on a circle of radius 3, chords $A B, A C$ and tangent $A D$ are drawn. The angle between the chords is $\frac{\pi}{4}$, and the angle between chord $A C$ and tangent $A D$, which does not contain chord $A B$, is $\frac{5 \pi}{12}$. Calculate the integer area of triangle $A B C$.
#
|
# Solution:
Let $\angle D A C=\alpha, \angle B A C=\beta$, and the radius of the circle be $R$. It is known that $\angle A C B=\angle D A C=\alpha$.
By the Law of Sines, $\quad \frac{|A B|}{\sin \alpha}=\frac{|B C|}{\sin \beta}=2 R$. Therefore, $\quad|A B|=2 R \sin \alpha$, and $|B C|=2 R \sin \beta$.
$S_{\triangle A B C}=\frac{1}{2}|A B| \cdot|B C| \cdot \sin (\angle A B C)$.
Thus,
$S_{\triangle A B C}=2 R^{2} \sin \alpha \sin \beta \sin (\alpha+\beta)$
To calculate the area, we need to know the values of $\sin \alpha$ and $\sin (\alpha+\beta)$. We will compute these values under the conditions of the problem.
For this, we will calculate $\sin \alpha$ and $\cos \alpha$.
$$
\begin{aligned}
& \sin \alpha=\sin \frac{5 \pi}{12}=\sin \left(\frac{\pi}{2}-\frac{\pi}{12}\right)=\cos \frac{\pi}{12}=\sqrt{\frac{1}{2}\left(1+\cos \frac{\pi}{6}\right)}=\frac{1}{2} \sqrt{2+\sqrt{3}}=\frac{1+\sqrt{3}}{2 \sqrt{2}} \\
& \cos \alpha=\cos \frac{5 \pi}{12}=\cos \left(\frac{\pi}{2}-\frac{\pi}{12}\right)=\sin \frac{\pi}{12}=\sqrt{\frac{1}{2}\left(1-\cos \frac{\pi}{6}\right)}=\frac{1}{2} \sqrt{2-\sqrt{3}}=\frac{-1+\sqrt{3}}{2 \sqrt{2}} \\
& \sin (\alpha+\beta)=\sin \left(\frac{\pi}{4}+\frac{5 \pi}{12}\right)=\sin \frac{2 \pi}{3}=\frac{\sqrt{3}}{2}
\end{aligned}
$$
Substituting the obtained values into (2), it is not difficult to get that $S_{\triangle A B C}=\frac{9}{4}(3+\sqrt{3})$.
It remains to compute the answer.
Answer: 10.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (5 points) From point $A$, lying on a circle, chords $A B$, $A C$, and tangent $A D$ are drawn. The angle between the chords is $\frac{\pi}{6}$, and the angle between chord $A C$ and tangent $A D$, which does not contain chord $A B$, is $\frac{5 \pi}{12}$. Calculate the integer part of the radius of the circle if the area of triangle $A B C$ is 32.
|
# Solution:
Let $\angle D A C=\alpha, \angle B A C=\beta$, and the radius of the circle be $R$. It is known that $\angle A C B=\angle D A C=\alpha$.
By the Law of Sines, $\quad \frac{|A B|}{\sin \alpha}=\frac{|B C|}{\sin \beta}=2 R$. Therefore, $\quad|A B|=2 R \sin \alpha$, and $|B C|=2 R \sin \beta$. $S_{\square A B C}=\frac{1}{2}|A B| \cdot|B C| \cdot \sin (\angle A B C)$.
Thus, $S_{\triangle A B C}=2 R^{2} \sin \alpha \sin \beta \sin (\alpha+\beta)$.
$$
R=\sqrt{\frac{S_{\triangle A B C}}{2 \sin \alpha \sin \beta \sin (\alpha+\beta)}}
$$
To calculate the radius, we need to know the values of $\sin \alpha$ and $\sin (\alpha+\beta)$. We will compute these values under the conditions of the problem. First, we will calculate $\sin \alpha$ and $\cos \alpha$.
$$
\begin{aligned}
& \sin \alpha=\sin \frac{5 \pi}{12}=\sin \left(\frac{\pi}{2}-\frac{\pi}{12}\right)=\cos \frac{\pi}{12}=\sqrt{\frac{1}{2}\left(1+\cos \frac{\pi}{6}\right)}=\frac{1}{2} \sqrt{2+\sqrt{3}}=\frac{1+\sqrt{3}}{2 \sqrt{2}} \\
& \cos \alpha=\cos \frac{5 \pi}{12}=\cos \left(\frac{\pi}{2}-\frac{\pi}{12}\right)=\sin \frac{\pi}{12}=\sqrt{\frac{1}{2}\left(1-\cos \frac{\pi}{6}\right)}=\frac{1}{2} \sqrt{2-\sqrt{3}}=\frac{-1+\sqrt{3}}{2 \sqrt{2}} \\
& \sin (\alpha+\beta)=\sin \left(\frac{\pi}{6}+\frac{5 \pi}{12}\right)=\sin \left(\frac{7 \pi}{12}\right)=\sin \left(\frac{\pi}{2}+\frac{\pi}{12}\right)=\cos \frac{\pi}{12}=\frac{1+\sqrt{3}}{2 \sqrt{2}}
\end{aligned}
$$
Substituting the obtained values into (2), it is not difficult to get that $R=\frac{16}{1+\sqrt{3}}$. It remains to compute the answer.
Answer: 5.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. There is an unlimited number of test tubes of three types - A, B, and C. Each test tube contains one gram of a solution of the same substance. Test tubes of type A contain a $10\%$ solution of this substance, type B $-20\%$ solution, and type C $-90\%$ solution. Sequentially, one after another, the contents of the test tubes are poured into a certain container. In this process, two consecutive pourings cannot use test tubes of the same type. It is known that a $20.17\%$ solution was obtained in the container, performing the minimum number of pourings. What is the maximum number of test tubes of type C that can be used in this process?
|
Solution: Let the number of test tubes of types A, B, and C be $a$, $b$, and $c$ respectively. According to the problem, $0.1a + 0.2b + 0.9c = 0.2017 \cdot (a + b + c) \Leftrightarrow 1000 \cdot (a + 2b + 9c) = 2017 \cdot (a + b + c)$. The left side of the last equation is divisible by 1000, so the right side must also be divisible by 1000. Therefore, the smallest possible value of the sum $a + b + c$ is 1000. We will show that this estimate is achievable. That is, we will prove that there exist non-negative integers $a$, $b$, and $c$ such that
$$
\left\{\begin{array}{c}
a + b + c = 1000 \\
a + 2b + 9c = 2017 \\
a \leq 500, b \leq 500, c \leq 500
\end{array}\right.
$$
The last three inequalities are necessary and sufficient conditions to ensure that test tubes of the same type are not used in two consecutive transfers. From the first two equations of the system (1), we find
$$
a = 7c - 17, \quad b = 1017 - 8c
$$
Substituting these expressions into the last three inequalities of the system (1), we get
$$
7c \leq 517, \quad 8c \geq 518, \quad c \leq 500
$$
From these, the largest value of $c$ is 73. The corresponding values of $a$ and $b$ can be found from (2). They obviously satisfy the inequalities of the system (1). Thus, the solvability of the system (1) in non-negative integers is proven.
## Answer: 73
|
73
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $1^{2}+2^{2}+4^{2}=21$ ).
|
Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Conversely, multiplying a divisor of $a$ by a divisor of $b$ yields a divisor of the product $a b$. This is clearly true for the squares of divisors as well (the square of a divisor of the product is equal to the product of the squares of the divisors of the factors and vice versa).
Consider the prime factorization of the number $N$: $N=p_{1}^{k_{1}} \cdot \ldots \cdot p_{n}^{k_{n}}$. Here, $p_{i}$ are distinct prime numbers, and all $k_{i} \in N$. Then $\sigma(N)=\sigma\left(p_{1}^{k_{1}}\right) \cdot \ldots \cdot \sigma\left(p_{n}^{k_{n}}\right)$ and $\sigma\left(p^{k}\right)=1+p^{2}+p^{4}+\ldots p^{2 k}$. Since $1800=2^{3} \cdot 3^{2} \cdot 5^{2}$, we have $\sigma(1800)=\left(1+2^{2}+2^{4}+2^{6}\right) \cdot\left(1+3^{2}+3^{4}\right) \cdot\left(1+5^{2}+5^{4}\right)=5035485$.
Answer: 5035485.
|
5035485
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. In a triangle with sides $a, b, c$ and angles $\alpha, \beta, \gamma$, the equality $3 \alpha + 2 \beta = 180^{0}$ is satisfied. The sides $a, b, c$ lie opposite the angles $\alpha, \beta, \gamma$ respectively. Find the length of side $c$ when $a=2, b=3$.
|
Solution: From the condition, it follows that $c > b$. Let's find a point $D$ on the segment $AB$ such that $AC = AD$. Then the triangle $ACD$ is isosceles and $\angle ACD = \angle ADC = 90^\circ - \alpha / 2$. The angle $\angle ADC$ is the external angle of the triangle $CBD$. Therefore, $\angle BCD + \beta = \angle ADC = 90^\circ - \frac{\alpha}{2} = \alpha + \beta$. Hence, $\angle BCD = \alpha$, and the triangles $CDB$ and $ABC$ are similar. We have $\frac{BD}{BC} = \frac{BC}{AB}$ or $\frac{c - b}{a} = \frac{a}{c'}$, from which it follows that $a^2 + bc - c^2 = 0$. The quadratic equation $c^2 - 3c - 4 = 0$ has a unique positive root $c = 4$.
Answer: 4.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. Find the number of matrices that satisfy two conditions:
1) the matrix has the form $\left(\begin{array}{lll}1 & * & * \\ * & 1 & * \\ * & * & 1\end{array}\right)$, where each * can take the value 0 or 1
2) the rows of the matrix do not repeat.
|
Solution: Let $A$ be the set of matrices satisfying condition 1) and $B$ be the subset of $A$ consisting of matrices satisfying condition 2). We need to find the number of elements in the set $B$. Let $A_{ij}, i, j \in\{1,2,3\}, i \neq j$, be the subset of $A$ consisting of matrices in which row $i$ and $j$ are the same. Then $B=A \backslash\left(A_{12} \cup A_{23} \cup A_{13}\right)$ and $|B|=|A|-\left|A_{12} \cup A_{23} \cup A_{13}\right|$. The cardinality $\left|A_{12} \cup A_{23} \cup A_{13}\right|$ is conveniently calculated using the inclusion-exclusion principle:
$\left|A_{12} \cup A_{23} \cup A_{13}\right|=\left|A_{12}\right|+\left|A_{23}\right|+\left|A_{13}\right|-\left|A_{12} \cap A_{23}\right|-\left|A_{13} \cap A_{23}\right|-\left|A_{12} \cap A_{13}\right|+\left|A_{12} \cap A_{23} \cap A_{13}\right|$.
It is easy to calculate the cardinalities of the sets involved in this expression:
$$
\left|A_{12}\right|=\left|A_{23}\right|=\left|A_{13}\right|=2^{3},\left|A_{12} \cap A_{23}\right|=\left|A_{13} \cap A_{23}\right|=\left|A_{12} \cap A_{13}\right|=\left|A_{12} \cap A_{23} \cap A_{13}\right|=1 .
$$
We get $|B|=2^{6}-3 \cdot 2^{3}+3-1=42$.
Answer: 42
|
42
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A motorist set off from point A to point B, the distance between which is 100 km. At the moment when the navigator showed that 30 minutes of travel remained, the motorist first reduced the speed by 10 km/h, and at the moment when the navigator showed that 20 km remained to travel, the motorist reduced the speed by the same 10 km/h again. (The navigator determines the remaining time and distance based on the current speed of movement.) Determine the initial speed of the car, given that the car traveled 5 minutes longer at the reduced speed than at the twice-reduced speed.
NOTE. In the 9th grade, the same problem was given in a slightly different version.
|
Solution. According to the condition, the distance from C to D is 20 km. Let the distance from A to B be denoted as
$x$ (km), then the distance from B to C will be $(80 - x)$ km. Let $v \frac{\text{km}}{4}$ be the initial speed of the car. Then on the segments BC and CD, the speed is $(v-10) \frac{\text{km}}{4}$ and $(v-20) \frac{\text{km}}{4}$, respectively. According to the condition, the journey from B to D would have taken half an hour if the car had continued to move at a speed of $v \frac{\text{km}}{4}$, that is,
$$
100 - x = \frac{v}{2}
$$
Furthermore, the time spent on the journey from B to C is 5 minutes more than the time spent on the journey from C to D:
$$
\frac{80 - x}{v - 10} - \frac{20}{v - 20} = \frac{1}{12}
$$
Expressing $x$ from the first equation and substituting it into the second, we get an equation to determine $v$:
$$
\frac{\frac{v}{2} - 20}{v - 10} - \frac{20}{v - 20} = \frac{1}{12}
$$
which has roots 14 and 100. The root 14 is obviously extraneous, as $v > 20$.
Answer: $100 \frac{\text{km}}{4}$.
|
100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find all roots of the equation $\frac{1}{\cos ^{3} x}-\frac{1}{\sin ^{3} x}=4 \sqrt{2}$, lying in the interval $\left(-\frac{\pi}{2}, 0\right)$. Write the answer in degrees.
|
# Solution:
$\sin ^{3} x-\cos ^{3} x=4 \sqrt{2} \sin ^{3} x \cos ^{3} x \Leftrightarrow(\sin x-\cos x)\left(\sin ^{2} x+\sin x \cos x+\cos ^{2} x\right)=4 \sqrt{2} \sin ^{3} x \cos ^{3} x$. Substitution: $\sin x-\cos x=t, \sin x \cos x=\frac{1-t^{2}}{2}$. Then $t\left(3-t^{2}\right)=\sqrt{2}\left(1-t^{2}\right)^{3}$. Substitution: $t=z \sqrt{2}$. The equation will take the form $z\left(3-2 z^{2}\right)-\left(1-2 z^{2}\right)^{3}=0$. There is a root $z=-1$, and the left side can be factored
$$
(z+1)\left(8 z^{5}-8 z^{4}-4 z^{3}+2 z^{2}+4 z-1\right)=0
$$
Since $x \in\left(-\frac{\pi}{2}, 0\right)$, then $t=\sin x-\cos x=\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)\left|4 z^{3}\right|$ and $\left|8 z^{4}\right|>\left|2 z^{2}\right|$. Therefore, $z=-1$ is the only root of equation (1). It is easy to find that $\sin \left(x-\frac{\pi}{4}\right)=-1$, and $x=-\frac{\pi}{4}$.
Answer: -45
|
-45
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find the sum of the squares of the natural divisors of the number 1800. (For example, the sum of the squares of the natural divisors of the number 4 is $\left.1^{2}+2^{2}+4^{2}=21\right)$.
|
Solution: Let $\sigma(N)$ be the sum of the squares of the natural divisors of a natural number $N$. Note that for any two coprime natural numbers $a$ and $b$, the equality $\sigma(a b)=\sigma(a) \cdot \sigma(b)$ holds. Indeed, any divisor of the product $a b$ is the product of a divisor of $a$ and a divisor of $b$. Conversely, multiplying a divisor of $a$ by a divisor of $b$ yields a divisor of the product $a b$. This is clearly true for the squares of divisors as well (the square of a divisor of the product is equal to the product of the squares of the divisors of the factors and vice versa).
Consider the prime factorization of the number $N$: $N=p_{1}^{k_{1}} \cdot \ldots \cdot p_{n}^{k_{n}}$. Here, $p_{i}$ are distinct prime numbers, and all $k_{i} \in N$. Then $\sigma(N)=\sigma\left(p_{1}^{k_{1}}\right) \cdot \ldots \cdot \sigma\left(p_{n}^{k_{n}}\right)$ and $\sigma\left(p^{k}\right)=1+p^{2}+p^{4}+\ldots p^{2 k}$. Since $1800=2^{3} \cdot 3^{2} \cdot 5^{2}$, we have $\sigma(1800)=\left(1+2^{2}+2^{4}+2^{6}\right) \cdot\left(1+3^{2}+3^{4}\right) \cdot\left(1+5^{2}+5^{4}\right)=5035485$.
Answer: 5035485.
|
5035485
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A circle touches the sides of an angle at points $A$ and $B$. A point $M$ is chosen on the circle. The distances from $M$ to the sides of the angle are 24 and 6. Find the distance from $M$ to the line $A B$.
|
Solution: $\angle \mathrm{XBM}=\angle \mathrm{ZAM}=\frac{1}{2} \overline{\mathrm{BM}}$, therefore, triangles ВМХ and ZAM are similar, so $\frac{\mathrm{XM}}{\mathrm{ZM}}=\frac{\mathrm{BM}}{\mathrm{AM}} \cdot \angle \mathrm{ABM}=\angle \mathrm{YAM}=\frac{1}{2} \overline{\mathrm{AM}}$, therefore, triangles $\mathrm{AMY}$ and $\mathrm{BMZ}$ are similar, so $\frac{\mathrm{YM}}{\mathrm{ZM}}=$ $\frac{\mathrm{AM}}{\mathrm{BM}} \cdot$ Therefore,
$$
\mathrm{ZM}^{2}=\mathrm{XM} \cdot \mathrm{YM}=24 \cdot 6=144, \quad \mathrm{ZM}=12
$$
Answer: 12
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. It is known that the polynomial $f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4}$ has 4 distinct real roots $\left\{x_{1}, x_{2}, x_{3}, x_{4}\right\}$. The polynomial of the form $g(x)=b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3}+x^{4}$ has roots $\left\{x_{1}^{2}, x_{2}^{2}, x_{3}^{2}, x_{4}^{2}\right\}$. Find the coefficient $b_{1}$ of the polynomial $g(x)$.
|
Solution: Let the coefficients of the given polynomial (except the leading one) be denoted by $a_{0}, a_{1}, a_{2}, a_{3}$: $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}$.
Then, according to the problem statement, we have:
$f(\quad)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+x^{4}=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)$.
Together with the polynomial $f(x)$, consider the polynomial $h(x)$, which has roots $\left\{-x_{1},-x_{2},-x_{3},-x_{4}\right\}$:
$$
h(x)=\left(x+x_{1}\right)\left(x+x_{2}\right)\left(x+x_{3}\right)\left(x+x_{4}\right)=a_{0}-a_{1} x+a_{2} x^{2}-a_{3} x^{3}+x^{4}
$$
Consider the polynomial $G($ 可 $)=f(x) h(x)$:
$$
\begin{aligned}
& G(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)\left(x+x_{1}\right)\left(x+x_{2}\right)\left(x+x_{3}\right)\left(x+x_{4}\right)= \\
&=\left(x^{2}-x_{1}{ }^{2}\right)\left(x^{2}-x_{2}{ }^{2}\right)\left(x^{2}-x_{3}{ }^{2}\right)\left(x^{2}-x_{4}{ }^{2}\right) .
\end{aligned}
$$
By substituting the variable $y=x^{2}$, we obtain the required polynomial $g(y)$, since
$$
g(y)=\left(y-x_{1}{ }^{2}\right)\left(y-x_{2}{ }^{2}\right)\left(y-x_{3}{ }^{2}\right)\left(y-x_{4}{ }^{2}\right)
$$
In our case:
$$
\begin{gathered}
f(x)=8+32 x-12 x^{2}-4 x^{3}+x^{4} \\
h(x)=8-32 x-12 x^{2}+4 x^{3}+x^{4} \\
f(x)=f(x) h(x)=64-1216 x^{2}+416 x^{4}-40 x^{6}+x^{8} \\
g(y)=64-1216 y+416 y^{2}-40 y^{3}+y^{4}
\end{gathered}
$$
Answer: -1216
|
-1216
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Find the number of matrices that satisfy two conditions:
3) the matrix has the form $\left(\begin{array}{lll}1 & * & * \\ * & 1 & * \\ * & * & 1\end{array}\right)$, where each * can take the value 0 or 1 and the rows of the matrix do not repeat.
|
Solution: Let $A$ be the set of matrices satisfying condition 1) and $B$ be the subset of $A$ consisting of matrices satisfying condition 2). We need to find the number of elements in the set $B$. Let $A_{ij}, i, j \in\{1,2,3\}, i \neq j$, be the subset of $A$ consisting of matrices in which row $i$ and $j$ are the same. Then $B=A \backslash\left(A_{12} \cup A_{23} \cup A_{13}\right)$ and $|B|=|A|-\left|A_{12} \cup A_{23} \cup A_{13}\right|$. The cardinality $\left|A_{12} \cup A_{23} \cup A_{13}\right|$ is conveniently calculated using the inclusion-exclusion principle:
$\left|A_{12} \cup A_{23} \cup A_{13}\right|=\left|A_{12}\right|+\left|A_{23}\right|+\left|A_{13}\right|-\left|A_{12} \cap A_{23}\right|-\left|A_{13} \cap A_{23}\right|-\left|A_{12} \cap A_{13}\right|+\left|A_{12} \cap A_{23} \cap A_{13}\right|$.
It is easy to calculate the cardinalities of the sets involved in this expression:
$$
\left|A_{12}\right|=\left|A_{23}\right|=\left|A_{13}\right|=2^{3},\left|A_{12} \cap A_{23}\right|=\left|A_{13} \cap A_{23}\right|=\left|A_{12} \cap A_{13}\right|=\left|A_{12} \cap A_{23} \cap A_{13}\right|=1
$$
We get $|B|=2^{6}-3 \cdot 2^{3}+3-1=42$.
|
42
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Non-zero numbers $a$ and $b$ are roots of the quadratic equation $x^{2}-5 p x+2 p^{3}=0$. The equation $x^{2}-a x+b=0$ has a unique root. Find $p$.
|
Solution. Since the equation $x^{2}-a x+b=0$ has a unique root, then $b=\frac{a^{2}}{4}$. By Vieta's theorem, we have the equalities: $a+b=5 p ; a b=2 p^{3}$. Substituting $b=\frac{a^{2}}{4}$ into the last equality, we get: $a=2 p$. Considering that $a$ and $b$ are non-zero, we find $p=3$.
Answer: 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. On the sides $B C$ and $C D$ of the square $A B C D$, points $E$ and $F$ are chosen such that the angle $E A F$ is $45^{\circ}$. The length of the side of the square is 1. Find the perimeter of triangle $C E F$.
|
Solution. If point $D$ is reflected across line $A F$, and then across line $A E$, it will transition to point $B$. Indeed, the composition of two axial symmetries relative to intersecting lines is a rotation by twice the angle between the lines. In our case, these two symmetries are equivalent to a $90^{\circ}$ rotation around point $A$.
This means that the image of point $D$ under the symmetry relative to $A F$ and the image of point $B$ under the symmetry relative to $A E$ are the same
point; on the diagram, it is denoted by $K$. From point $K$, segments $A E$ and $A F$ are seen at a $90^{\circ}$ angle (since angles are preserved under symmetry, for example, angles $A B E$ and $A K E$ are equal). Therefore, point $K$ is the foot of the perpendicular dropped from point $A$ to line $E F$.
Finally, since $B E = E K$ and $D F = F K$ (lengths of segments are preserved under symmetry), we see that the perimeter of triangle $C E F$ is equal to the sum of the lengths of sides $B C$ and $C D$ of the square.
Answer: 2.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (5 points) The sides of a parallelogram are 5 and 13, and the angle between them is $\arccos \frac{6}{13}$. Two mutually perpendicular lines divide this parallelogram into four equal-area quadrilaterals. Find the lengths of the segments into which these lines divide the sides of the parallelogram.
#
|
# Solution:
Let in parallelogram $ABCD$, $|AB| = |CD| = a$, $|AD| = |BC| = b$, $\angle BAC = \alpha = \arccos c$. Clearly, $\cos \alpha = c$, and $\cos (\pi - \alpha) = -c$.
Denote $|DN| = x$, $|LC| = y$.
By contradiction, it can be shown that the mutually perpendicular lines $MN$ and $KL$, specified in the problem, must pass through the intersection point of the diagonals of parallelogram $ABCD$. It is evident that $\square KON = \square LON$.
From this and the equality of areas, $S_{\text{KOND}} = S_{\text{LONC}}$, it follows that the areas coincide, $S_{\square KND} = S_{\square LNC}$. This fact implies the validity of the equality
$$
\frac{1}{2}(a-x) y \cdot \sin \alpha = \frac{1}{2}(b-y) x \cdot \sin (\pi - \alpha),
$$
from which the relation follows
$$
y = \frac{b}{a} x
$$
Based on the equality $|KN| = |LN|$ and the cosine theorem, we can write the equation
$$
(a-x)^2 + y^2 - 2(a-x) y \cdot \cos \alpha = (b-y)^2 + x^2 - 2(b-y) x \cdot \cos (\pi - \alpha)
$$
from which and from (1), in turn, follows the equation
$$
\frac{4 b c}{a} x^2 - \left(2a - \frac{2b^2}{a} + 4bc\right) x + a^2 - b^2 = 0
$$
For $a = 5$, $b = 13$, and $c = \frac{6}{13}$, the solutions to (2) are $x = -10$ and $x = 3$. The first is discarded for geometric reasons. The corresponding value of $y = \frac{39}{5}$.
Answer: 3 and $2, \frac{39}{5}$ and $\frac{26}{5}$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (2 points) For a natural number $x$, five statements are made:
$$
3 x>91
$$
$$
\begin{aligned}
& x37 \\
& 2 x \geq 21 \\
& x>7
\end{aligned}
$$
It is known that only three of them are true, and two are false. Find $x$.
|
Solution. Let's transform the original system of inequalities
$$
\begin{aligned}
& x>\frac{91}{3} \\
& x \geq \frac{21}{2} \\
& x>\frac{37}{4} \\
& x>7 \\
& x<120
\end{aligned}
$$
The first inequality does not hold, since otherwise, four inequalities would be satisfied immediately. Therefore, $x \leq \frac{91}{3}$, and the fifth inequality is satisfied.
The second inequality also does not hold, since otherwise, four inequalities would be satisfied: the second, third, fourth, and fifth. Therefore, $x<\frac{21}{2}$.
Now it is immediately clear that the desired number $x=10$.
Answer: $x=10$
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the value of $f(2019)$, given that $f(x)$ simultaneously satisfies the following three conditions:
1) $f(x)>0$ for any $x>0$
2) $f(1)=1$;
3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}$ for any $a, b \in \mathbb{R}$.
|
Solution. In the identity given in the problem
$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$
let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find
$$
f(0)=0
$$
Next, by setting $b=-a$ in (1), we get, taking (2) into account, that
$$
f(a) \cdot f(-a)=-a^{2}
$$
Finally, when $b=0$, identity (1) (taking (2) into account) becomes $f(a) \cdot f(a)=a^{2}$. Therefore, it is necessary that $f(a)=a$ for $a>0$, since by condition $f(x)>0$ for $x>0$. Furthermore, according to (3), $f(a)=a$ for $a<0$ as well. Ultimately, $f(x)=x$ for any $x \in \mathbb{R}$. It is easy to verify that such an $f(x)$ indeed satisfies the requirements 1$), 2), 3$) from the problem statement. Thus, $f(x)=x$.
Answer: 2019.
|
2019
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent cell to the right).
For example, if the robot executes a sequence of four commands (up, right, down, left), it will obviously return to the initial position, i.e., it will end up in the same cell from which it started. How many different sequences of 8 commands exist that return the robot to the initial position?
|
Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be the number of L commands in the sequence. We will count the number $N_{k}$ of the desired sequences for $k$ from 0 to 4.
- $\boldsymbol{k}=\mathbf{0}$. The sequence consists only of U and D commands. Since their numbers are equal, on 4 out of 8 positions there should be a U, and on the remaining positions, a D. The number of ways to choose 4 positions out of 8 is $C_{8}^{4}$. Therefore, $N_{0}=C_{8}^{4}=70$;
- $\boldsymbol{k}=\mathbf{1}$. The sequence consists of one L command, one R command, and three U and three D commands. The number of ways to place the two commands L and R on 8 positions is $C_{8}^{2} \cdot C_{2}^{1}$: $C_{8}^{2}$ is the number of ways to choose 2 positions out of 8, and $C_{2}^{1}=2$ is the number of ways to place the commands L and R on these two positions. On the remaining 6 positions, 3 U commands can be placed in $C_{6}^{3}$ ways. Therefore, $N_{1}=C_{8}^{2} \cdot C_{2}^{1} \cdot C_{6}^{3}=1120$;
- $\boldsymbol{k}=\mathbf{2}$. Here, there are two L commands, two R commands, and two U and two D commands. For L and R, there are $C_{8}^{4} \cdot C_{4}^{2}$ ways to place them. On the remaining 4 positions, 2 U commands can be placed in $C_{4}^{2}$ ways. Thus, $N_{2}=C_{8}^{4} \cdot C_{4}^{2} \cdot C_{4}^{2}=2520$.
By reasoning similarly, it can be shown that $N_{3}=N_{1}$ and $N_{4}=N_{0}$. Therefore, the desired number of sequences is $2 \cdot\left(N_{0}+N_{1}\right)+N_{2}=4900$.
Answer: 4900.
|
4900
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A robot is located in one of the cells of an infinite grid paper, to which the following commands can be given:
- up (the robot moves to the adjacent cell above);
- down (the robot moves to the adjacent cell below);
- left (the robot moves to the adjacent cell to the left);
- right (the robot moves to the adjacent cell to the right).
For example, if the robot executes a sequence of four commands (up, right, down, left), it will obviously return to the initial position, i.e., it will end up in the same cell from which it started. How many different sequences of 8 commands exist that return the robot to the initial position?
|
Solution. For brevity, let's denote the command to move left as L, right as R, up as U, and down as D. For the robot to return to its initial position, it is necessary and sufficient that the number of L commands equals the number of R commands, and the number of U commands equals the number of D commands. Let $k$ be the number of L commands in the sequence. We will count the number $N_{k}$ of the desired sequences for $k$ from 0 to 4.
- $\boldsymbol{k}=\mathbf{0}$. The sequence consists only of U and D commands. Since they are equal in number, 4 out of 8 positions should be U, and the rest should be D. The number of ways to choose 4 positions out of 8 is $C_{8}^{4}$. Therefore, $N_{0}=C_{8}^{4}=70$;
## Interregional School Olympiad in Mathematics - $\boldsymbol{k}=\mathbf{1}$. The sequence consists of one L command, one R command, and three U and three D commands. The number of ways to place two L and R commands on 8 positions is $C_{8}^{2} \cdot C_{2}^{1}$: $C_{8}^{2}$ is the number of ways to choose 2 positions out of 8, and $C_{2}^{1}=2$ is the number of ways to place the L and R commands on these two positions. The number of ways to place 3 U commands on the remaining 6 positions is $C_{6}^{3}$. Therefore, $N_{1}=C_{8}^{2} \cdot C_{2}^{1} \cdot C_{6}^{3}=1120$;
- $\boldsymbol{k}=\mathbf{2}$. Here, there are two L commands, two R commands, and two U and two D commands. The number of ways to place L and R commands is $C_{8}^{4} \cdot C_{4}^{2}$. The number of ways to place 2 U commands on the remaining 4 positions is $C_{4}^{2}$. Therefore, $N_{2}=$ $C_{8}^{4} \cdot C_{4}^{2} \cdot C_{4}^{2}=2520$.
By reasoning similarly, it can be shown that $N_{3}=N_{1}$ and $N_{4}=N_{0}$. Thus, the desired number of sequences is $2 \cdot\left(N_{0}+N_{1}\right)+N_{2}=4900$.
Answer: 4900.
|
4900
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find all prime numbers whose decimal representation has the form 101010 ... 101 (ones and zeros alternate).
|
Solution: Let $2n+1$ be the number of digits in the number $A=101010$...101 under investigation. Let $q=$ 10 be the base of the numeral system. Then $A=q^{0}+q^{2}+\cdots q^{2n}=\frac{q^{2n+2}-1}{q^{2}-1}$. Consider the cases of even and odd $n$.
- $n=2k \Rightarrow A=\frac{q^{2n+2}-1}{q^{2}-1}=\frac{q^{2k+1}-1}{q-1} \cdot \frac{q^{2k+1}+1}{q+1}$. Thus, the number $A$ is represented as the product of two integer factors (by the theorem of Bezout, the polynomial $q^{2k+1} \pm 1$ is divisible without remainder by the polynomial $q \pm 1$), each of which is different from 1. Therefore, when $n$ is even, the number $A$ is not prime.
- $n=2k-1 \Rightarrow A=\frac{q^{2n+2}-1}{q^{2}-1}=\frac{q^{2k}-1}{q^{2}-1} \cdot (q^{2k}+1)$. For $k>1$, both factors are integers and different from 1; thus, the number $A$ is composite. It remains to verify that for $k=1$, the resulting number $A=q^{0}+q^{2}=101$ is prime.
Answer: 101.
|
101
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Anya and Borya are playing "battleship" with the following rules: 29 different points are chosen on a circle, numbered clockwise with natural numbers from 1 to 29. Anya draws a ship - an arbitrary triangle with vertices at these points. We will call a "shot" the selection of two different natural numbers $k$ and $m$ from 1 to 29. If the segment with endpoints at points numbered $k$ and $m$ has at least one common point with Anya's triangle, the ship is considered "wounded." Borya fires a "salvo" - several shots simultaneously. Anya drew a ship and showed it to Borya. And then they noticed that any "salvo" of K different shots will definitely wound Anya's ship. Indicate some placement of Anya's ship for which the value of $K$ will be minimal.
Interregional School Olympiad in Mathematics for Students of Departmental Educational Organizations
|
Solution. The vertices of Anya's triangle divide the circle into three arcs. Let $x, y$ and $26-x-y$ be the number of points on these arcs (see figure), excluding the vertices of the triangle itself. For a "shot" with endpoints at points $k$ and $m$ to not hit the ship, both these points must lie on one of the arcs. Clearly, the number of ways to choose two different points on an arc containing $x$ points is $C_{x}^{2}$. The same applies to the other arcs. Therefore, the number $N$ of "safe" shots is the sum $N=C_{x}^{2}+C_{y}^{2}+C_{26-x-y}^{2}$. Then the next shot will definitely "hit" the ship, so $K=N+1$.
Thus, we need to find such non-negative integers $x, y$ that satisfy the condition $x+y \leq 26$, for which the value of $N$ is minimal. Let's write the expression for $N$ as follows:
$$
N=\frac{x(x-1)}{2}+\frac{y(y-1)}{2}+\frac{(26-x-y)(25-x-y)}{2}
$$
Expanding the brackets and combining like terms, we get
$$
N=x^{2}-x(26-y)+y^{2}-26 y+325
$$
For each fixed $y$ from 0 to 26, we will find such a value of $x$ that satisfies the inequality
$$
0 \leq x \leq 26-y
$$
for which the value of $N$ is minimal. If $y$ is fixed, then the right-hand side (1) takes its minimum value at
$$
x=\frac{26-y}{2}
$$
(the vertex of the parabola, belonging to the interval (2)).
This minimum value is $\frac{3}{4} y^{2}-13 y+156$. It, in turn, is minimal at $y=\frac{26}{3} \approx 8.6$. From (3), we then find $x=\frac{26}{3}$. Among the points with integer coordinates $(8,8),(8,9),(9,8),(9,9)$ - the nearest integer neighbors of the point of minimum $\left(\frac{26}{3}, \frac{26}{3}\right)$ - we choose the one for which the value of $N$ is the smallest. These are the points $(8,9),(9,8),(9,9)$. For them, $N=100$.
Answer: The ship should be placed such that on the three arcs into which the vertices of the ship divide the circle, there are 8, 9, and 9 points (excluding the vertices of the ship itself).
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A square table consists of 2014 rows and 2014 columns. In each cell at the intersection of the row with number $i$ and the column with number $j$, the number $a_{i, j}=(-1)^{i}(2015-i-j)^{2}$ is written. Find the sum of all the numbers in the table
|
Problem 3. Answer: $\mathbf{0 .}$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{l}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$.
Interregional Olympiad for Schoolchildren based on Departmental Educational Organizations
|
Solution. We will prove that ${ }^{1}$
Multiply the equation (a) of the original system
$$
a_{2} b_{3}=a_{3} b_{2}
$$
$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & (\mathrm{a}) \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1 & \text { (d) }\end{cases}
$$
by $b_{2}$ and subtract from it the equation (b), multiplied by $b_{1}$. As a result, we get
$$
a_{2} \cdot \Delta=b_{2}
$$
Here $\boldsymbol{\Delta}=\mathbf{b}_{\mathbf{2}} \boldsymbol{b}_{\mathbf{3}}-\boldsymbol{b}_{\mathbf{1}} \boldsymbol{b}_{\mathbf{4}}$. Similarly, from (c) and (d) we find that
$$
a_{3} \cdot \Delta=b_{3}
$$
Note that $\Delta \neq 0$, otherwise from (3) it would follow that $b_{3}=0$, and thus $a_{2} b_{3}=0$, which contradicts the condition of the problem. It remains to express $a_{2}$ and $a_{3}$ from (2) and (3) and substitute the obtained expressions into (1). The validity of the relation (1) will thereby be proven.
Further, from equation (d) and equality (1), it follows that $a_{4} b_{4}=1-a_{3} b_{2}=1-a_{2} b_{3}=-6$.
Answer: $a_{4} b_{4}=-6$.
Comment. ${ }^{1}$ The system of equations in the problem is the component-wise recording of the matrix equality $\mathbf{A} \cdot \mathbf{B}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$, where $\mathbf{A}=\left(\begin{array}{ll}a_{1} & a_{2} \\ a_{3} & a_{4}\end{array}\right)$ and $\mathbf{B}=\left(\begin{array}{ll}b_{1} & b_{2} \\ b_{3} & b_{4}\end{array}\right)$. It is well known that if the product of two matrices is equal to the identity matrix, then such matrices commute, which means that the system of equations in the problem will remain valid if all $a_{i}$ are replaced by $b_{i}$ and vice versa. From this observation, equality (1) follows immediately.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. It is known that there exists a natural number $N$ such that
$$
(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}
$$
Find $N$.
|
Solution. Suppose that raising the number $a+b \sqrt{3}$ to the power $N$ results in the number $A+B \sqrt{3}$ (here $a, b, A, B$ are integers). Expanding the expression $(a+b \sqrt{3})^{N}$, we get a sum of monomials (with integer coefficients that are not important to us now) of the form $a^{N-n}(b \sqrt{3})^{n}$. The coefficient $B$ will be contributed by those monomials where the exponent $n$ is odd. Therefore, if $(a+b \sqrt{3})^{N}=A+B \sqrt{3}$, then $(a-b \sqrt{3})^{N}=A-B \sqrt{3}$. Multiplying the equations $(\sqrt{3}-1)^{N}=4817152-2781184 \cdot \sqrt{3}$ and $(-\sqrt{3}-1)^{N}=4817152+2781184 \cdot \sqrt{3}$, we get $(-2)^{N}=4817152^{2}-3 \cdot 2781184^{2}$. The exponent $N$ can be found by successively dividing both sides by 2 (for example, each term on the right can be immediately divided by 256).
Answer: $N=16$.
|
16
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (2 points) It is known about the numbers $x_{1}$ and $x_{2}$ that $x_{1}+x_{2}=2 \sqrt{1703}$ and $\left|x_{1}-x_{2}\right|=90$. Find $x_{1} \cdot x_{2}$.
|
Solution: Let $A=x_{1}+x_{2}, B=x_{1} \cdot x_{2}, C=\left|x_{1}-x_{2}\right|$. Since $C^{2}=A^{2}-4 \cdot B$, we find $B=-322$.
Answer: -322 .
|
-322
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (3 points) Find the number of integers from 1 to 1000 inclusive that give the same remainder when divided by 11 and by 12.
|
Solution. Let $r_{n}(a)$ be the remainder of the division of the number $a$ by the number $n$. Suppose $a \in [1 ; 1000]$ and $r_{11}(a)=r_{12}(a)=t$. Then $t \in \{0, \ldots, 10\}$ and the following equality holds:
$$
t+11 k=t+12 m=a, \quad k, m \in N_{0}
$$
From the last equality, it follows that $k$ is divisible by 12, and $m$ is divisible by 11. Therefore,
$$
t+11 \cdot 12 \cdot s=t+132 \cdot s=a, \quad s \in N_{0}
$$
It remains to take into account the condition $a=t+132 \cdot s \leq 1000, \quad t \in \{0, \ldots, 10\}$:
$$
\begin{aligned}
& t=0 \Rightarrow 132 s \leq 1000 \Rightarrow s \leq 7.6 \Rightarrow 7 \text{ numbers } \quad(s \neq 0) \\
& t=1 \Rightarrow 1+132 s \leq 1000 \Rightarrow s \leq 7.57 \Rightarrow 8 \text{ numbers } \\
& t=2 \Rightarrow 2+132 s \leq 1000 \Rightarrow s \leq 7.56 \Rightarrow 8 \text{ numbers } \\
& t=3 \Rightarrow 3+132 s \leq 1000 \Rightarrow s \leq 7.55 \Rightarrow 8 \text{ numbers } \\
& t=4 \Rightarrow 4+132 s \leq 1000 \Rightarrow s \leq 7.55 \Rightarrow 8 \text{ numbers } \\
& t=5 \Rightarrow 5+132 s \leq 1000 \Rightarrow s \leq 7.54 \Rightarrow 8 \text{ numbers } \\
& t=6 \Rightarrow 6+132 s \leq 1000 \Rightarrow s \leq 7.54 \Rightarrow 8 \text{ numbers } \\
& t=7 \Rightarrow 7+132 s \leq 1000 \Rightarrow s \leq 7.53 \Rightarrow 8 \text{ numbers } \\
& t=8 \Rightarrow 8+132 s \leq 1000 \Rightarrow s \leq 7.52 \Rightarrow 8 \text{ numbers } \\
& t=9 \Rightarrow 9+132 s \leq 1000 \Rightarrow s \leq 7.51 \Rightarrow 8 \text{ numbers } \\
& t=10 \Rightarrow 10+132 s \leq 1000 \Rightarrow s \leq 7.5 \Rightarrow 8 \text{ numbers }
\end{aligned}
$$
In total, we get 87 numbers. (obviously, the last enumeration could have been shortened)
Answer: 87 numbers.
P.S. An alternative approach consists in enumerating $s$, where $s \leq \frac{1000}{132} \approx 7.57$, and finding those $t \in \{0, \ldots, 10\}$ for which the inequality $a=t+132 \cdot s \leq 1000$ holds.
|
87
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the sum of all natural numbers $n$ that are multiples of three and for which the number of divisors (including 1 and $n$ itself) is equal to $\frac{n}{3}$. (For example, the number 12 has 6 divisors: $1,2,3,4,6,12$.)
|
Solution. Let the canonical decomposition of the number $n$ be: $n=2^{t_{1}} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdot p^{t_{k}}$. Then the number of divisors of the number $n$ is $\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)$. From the condition of the problem, we have the equality:
$\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)=2^{t_{1}} \cdot 3^{t_{2}-1} \cdot 5^{t_{3}} \cdots \cdots \cdot p^{t_{k}} \cdot\left({ }^{*}\right)$
Note that $2^{t_{1}} 3^{t_{2}-1}>\left(t_{1}+1\right)\left(t_{2}+1\right)$ for $t_{1} \geq 4$ and $t_{2} \geq 3, \quad \ldots, \quad p^{t_{k}}>t_{k}+1$ for $t_{k} \geq 1$. Therefore, $t_{1}$ can take the values $0,1,2$ or 3 and $t_{2}$ can take the values 1 or 2. Substituting the specified values into the equality (*), we find that $n=9, n=18$ or $n=24$.
Answer: 51.
|
51
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (4 points) Find the area of a triangle if two of its medians are equal to $\frac{15}{7}$ and $\sqrt{21}$, and the cosine of the angle between them is $\frac{2}{5}$.
#
|
# Solution:
Let $|A D|=a$ and $|B E|=b$ be the medians of the triangle, and $\cos (\angle A O F)=c$.
It is known that the area
$$
\begin{aligned}
& S_{A O E}=\frac{1}{6} \cdot S_{A B C} \\
& S_{A O E}=\frac{1}{2} \cdot\left(\frac{2}{3} \cdot a\right) \cdot\left(\frac{1}{3} \cdot b\right) \cdot \sin (\angle A O F)
\end{aligned}
$$
or, equivalently,
$S_{A O E}=\frac{a b}{9} \sqrt{1-c^{2}}$.
Therefore, $S_{A B C}=\frac{2 a b}{3} \sqrt{1-c^{2}}$.
Substituting the specific values of the parameters, we get the answer.
Answer: 6.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (5 points) Numbers are marked on a chessboard (see Fig. 1). How many arrangements of 8 rooks are there such that no rook attacks another and the numbers on the squares occupied by the rooks include all numbers from 0 to 7?
\(\left(\begin{array}{llllllll}0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0\end{array}\right)\)
Fig. 1.
Let's number the rows of the board from 1 to 8 from top to bottom. On the first row, the position of the rook can be chosen in 8 ways. For example, we choose the first cell with the number 0. On the second row, there are 7 options left. Six of them (cells from the second to the seventh) have the property that under them in the lower half of the board are numbers different from the number occupied by the first rook (in our case, 0), and one cell (the last one, with the number 7) is such that under it in the lower half stands the number occupied by the first rook (in our case, 0).
Consider the first group of 6 options and make an arbitrary choice (for example, the second cell with the number 1). After such a choice, it turns out that in the lower half of the board, positions with two different numbers are prohibited (in our case, 7 and 6). Therefore, positions with these numbers need to be chosen in the upper half of the board, and there are exactly 2 such options (in our case, the 7th and 8th cells in rows 3 and 4). After this choice, in the lower half of the board, there will be 4 columns with different numbers in the columns. That is, in the lower half, there are \(4! = 24\) options. Thus, we get \(8 \times 6 \times 2 \times 24\) options.
Consider the option of choosing a position on the second row when under the chosen cell in the lower half of the board stands the number chosen on the first row (in our case, the last cell with the number 7). After such a choice, in the third row, there are 6 options leading to a single option in the fourth row. After choosing options in the upper half of the board, in the lower half, there are again 4 free columns with four different numbers, i.e., 24 options. Thus, we get \(8 \times 1 \times 6 \times 24\) options.
In total, we get:
\(8 \times 6 \times 2 \times 24 + 8 \times 1 \times 6 \times 24 = 3456\).
|
Answer: 3456.
## Variant 2
|
3456
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (4 points) Find the area of a triangle if two of its medians are equal to 3 and $2 \sqrt{7}$, and the cosine of the angle between them is $-\frac{3}{4}$.
#
|
# Solution:
Let $|A D|=a$ and $|B E|=b$ be the medians of the triangle, and $\cos (\angle A O F)=c$.
It is known that the area
$$
\begin{aligned}
& S_{A O E}=\frac{1}{6} \cdot S_{A B C} \cdot \\
& S_{A O E}=\frac{1}{2} \cdot\left(\frac{2}{3} \cdot a\right) \cdot\left(\frac{1}{3} \cdot b\right) \cdot \sin (\angle A O F)
\end{aligned}
$$
or, equivalently,
$S_{A O E}=\frac{a b}{9} \sqrt{1-c^{2}}$.
Therefore, $S_{A B C}=\frac{2 a b}{3} \sqrt{1-c^{2}}$.
Substituting the specific values of the parameters, we get the answer.
Answer: 7.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Eight numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $b_{1}, b_{2}, b_{3}, b_{4}$ satisfy the relations
$$
\left\{\begin{array}{l}
a_{1} b_{1}+a_{2} b_{3}=1 \\
a_{1} b_{2}+a_{2} b_{4}=0 \\
a_{3} b_{1}+a_{4} b_{3}=0 \\
a_{3} b_{2}+a_{4} b_{4}=1
\end{array}\right.
$$
It is known that $a_{2} b_{3}=7$. Find $a_{4} b_{4}$.
|
Solution. We will prove that ${ }^{1}$
$$
a_{2} b_{3}=a_{3} b_{2}
$$
Multiply equation (a) of the original system
$$
\begin{cases}a_{1} b_{1}+a_{2} b_{3}=1 & (\mathrm{a}) \\ a_{1} b_{2}+a_{2} b_{4}=0 & \text { (b) } \\ a_{3} b_{1}+a_{4} b_{3}=0 & \text { (c) } \\ a_{3} b_{2}+a_{4} b_{4}=1 & \text { (d) }\end{cases}
$$
by $b_{2}$ and subtract from it equation (b), multiplied by $b_{1}$. The result is
$$
a_{2} \cdot \Delta=b_{2}
$$
Here $\Delta=b_{2} b_{3}-b_{1} b_{4}$. Similarly, from (c) and (d) we find that
$$
a_{3} \cdot \Delta=b_{3}
$$
Note that $\Delta \neq 0$, otherwise from (3) it would follow that $b_{3}=0$, and thus $a_{2} b_{3}=0$, which contradicts the problem's condition. It remains to express $a_{2}$ and $a_{3}$ from (2) and (3) and substitute the obtained expressions into (1). The validity of relation (1) will thereby be proven.
Further, from equation (d) and equality (1), it follows that $a_{4} b_{4}=1-a_{3} b_{2}=1-a_{2} b_{3}=-6$.
Answer: $a_{4} b_{4}=-6$.
Comment. ${ }^{1}$ The system of equations in the problem is the component-wise recording of the matrix equality $\mathbf{A} \cdot \mathbf{B}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$, where $\mathbf{A}=\left(\begin{array}{ll}a_{1} & a_{2} \\ a_{3} & a_{4}\end{array}\right)$ and $\mathbf{B}=\left(\begin{array}{ll}b_{1} & b_{2} \\ b_{3} & b_{4}\end{array}\right)$. It is well known that if the product of two matrices is the identity matrix, then these matrices commute, which means the system of equations in the problem will remain valid if all $a_{i}$ are replaced by $b_{i}$ and vice versa. From this observation, equality (1) follows immediately.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. [6] On an island, there live knights, liars, and yes-men; each knows who everyone else is. All 2018 residents were lined up and asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?". The residents answered in turn, and everyone could hear their answers. Knights answered truthfully, liars lied. Each yes-man answered the same way as the majority of those who answered before him, and if the number of "Yes" and "No" answers was equal, he gave either of these answers. It turned out that there were exactly 1009 "Yes" answers. What is the maximum number of yes-men that could be among the residents?
|
Answer: 1009 yes-men. Solution: Let's call knights and liars principled people.
Estimate. First method. (Buchaev Abdulqadyr) We will track the balance - the difference between the number of "Yes" and "No" answers. At the beginning and at the end, the balance is zero, and with each answer, it changes by 1. Zero values of the balance divide the row of residents into groups. Within each group, the balance retains its sign. Yes-men always increase the absolute value of the balance. Therefore, to make the balance zero, there must be no fewer principled people than yes-men. This is true for each group, and therefore for all the island's residents.
Second method. A yes-man could not increase the minimum of the current number of "Yes" and "No" answers. Since this minimum increased from 0 to 1009, and with each answer it changed by no more than one unit, there must be at least 1009 principled people.
Example. First, all 1009 yes-men said "No," and then 1009 knights said "Yes."
Note. Any sequence of 1009 principled people can be thinned out by yes-men so that the condition is met.
|
1009
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. [8] A $7 \times 7$ board is either empty or has an invisible $2 \times 2$ ship placed on it "by cells." It is allowed to place detectors in some cells of the board and then turn them on simultaneously. An activated detector signals if its cell is occupied by the ship. What is the smallest number of detectors needed to guarantee that, based on their signals, it can be determined whether there is a ship on the board, and if so, which cells it occupies?
(R. Zhinodarov)
|
Answer: 16 detectors. Solution: Evaluation. In each $2 \times 3$ rectangle, there should be at least two detectors: the rectangle consists of three $1 \times 2$ dominoes, and if a detector is in the outer domino, we cannot determine whether there is a ship on the other two dominoes, and if a detector is in the middle domino, we cannot determine which of the outer dominoes the ship occupies along with the middle domino. Therefore, there should be no fewer than 16 detectors (see the left image).
Example. On the right image, the black squares indicate the placement of 16 detectors. Any ship intersects with exactly one black $2 \times 2$ square in one cell, two adjacent cells, or all four cells. In any case, the position of the ship or its absence is uniquely determined.
Note. 16 detectors can be placed in only three fundamentally different ways. One is shown in the solution, and the other two are shown in the image below.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [5] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly 4 chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order?
(Egor Bakaev)
|
Answer: For 61 rubles. Solution: Let's number the chips and cells in order from 0 to 99. The free operation does not change the remainder of the cell number when divided by 5.
Estimate. Let's mentally arrange the piles of chips in a circle. First, the pile of chips with a remainder of 0, then with 1, and so on up to 4. The paid operation swaps a pair of chips from adjacent piles. Chips from the zero pile must participate in at least one such swap to reach the fourth pile. Similarly for chips from the fourth pile. Chips from the first pile must participate in at least two swaps to reach the third pile. Similarly for the third pile. Therefore, at least $(20+20+40+40): 2=60$ rubles are required. But if only 60 rubles are spent, the chips from the first pile will have to go through the second pile, so at least one chip from the second pile will participate in swaps. Therefore, more than 60 rubles are necessary.
Algorithm. It is clear that the free operations can arrange the chips within a pile in any order. Therefore, the correct arrangement of all chips from the zero and fourth piles can be done for 20 rubles. Consider the remaining three piles. Let's mentally leave only one chip $A$ in the second pile. We will swap it with a chip from the first pile. Each time we will move the chip that has come to the second pile further using a new chip. Then for 40 rubles, we will move all the chips from the first pile to the third, and from the third to the first except one: it will remain in the second pile, not reaching the first. We will swap it with $A$, and all the chips will be in the correct piles.
|
61
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given a triangular pyramid $S A B C$, the base of which is an equilateral triangle $A B C$, and all plane angles at vertex $S$ are equal to $\alpha$. For what least $\alpha$ can we assert that this pyramid is regular?
M. Malkin
|
Answer: $60^{\circ}$.
We will prove that when $\alpha=60^{\circ}$, the pyramid is regular. Let the sides of the triangle $ABC$ be equal to 1. Note that in any triangle with an angle of $60^{\circ}$, the side opposite this angle is the middle-length side (and if it is strictly less than one of the sides, it is strictly greater than the other). Suppose one of the lateral edges of the pyramid is not equal to 1, for example, $SA>1$. Then in the faces $SAB$ and $SAC$, the edges $SB$ and $SC$ will be less than 1, and thus in the face $SBC$, the edge $BC$ will not be the middle side, which is a contradiction.
Now we will show how to construct an irregular pyramid with dihedral angles $\alpha<60^{\circ}$ at the vertex $S$.
First method. First, we remove the lateral edge $SA$, and rotate the remaining lateral face $SBC$ around its base edge $BC$ until this face lies in the plane of the base and contains the base triangle. During this motion, there will be "current" pyramids, in which two equal dihedral angles are initially equal to $\alpha$, then greater than $\alpha$ (at the moment when the vertex projects onto the vertex of the base - since at this moment the sine of the angle at the vertex $S$ in the lateral faces $SAC$ and $SAB$ is $1: SB$, and in the face $SBC$ it is the ratio of the lateral height of this triangle to $SB$, which is less than 1), and at the end, in the "degenerate" pyramid, they are equal to $\alpha / 2$. Therefore, by continuity, there will be $\alpha$ again.
Second method. Consider the triangle $SAB$ with $SA=SB$ and $\angle S=\alpha$. Since $AB<SB$, there exists a point $C$ on the side $SA$ such that $BC=AB$. Now take a trihedral angle in which all dihedral angles are equal to $\alpha$, and mark segments $SA, SB, SC$ on its edges. The pyramid $SABC$ is the desired one.
Third method. (Sergey Komarov, 11th grade, MSU SUNT) Let $S'ABC$ be a regular pyramid with a dihedral angle $\alpha$ at the vertex $S'$. Let $X, Y$ be points such that $XY=AB$, and construct triangles $XYZ$ and $XYT$ on $XY$ such that $\angle ZXY=\angle ZYX=90^{\circ}-\frac{\alpha}{2}$, $\angle TXY=90^{\circ}+\frac{\alpha}{2}$, $\angle TXY=90^{\circ}-\frac{3\alpha}{2}$, then it is clear that $\angle XTY=\angle XZY=\alpha$. By the Law of Sines for these triangles, we have
$$
\frac{TY}{\sin \left(90^{\circ}+\frac{\alpha}{2}\right)}=\frac{XY}{\sin \alpha}=\frac{ZX}{\sin \left(90^{\circ}-\frac{\alpha}{2}\right)}=\frac{ZY}{\sin \left(90^{\circ}-\frac{\alpha}{2}\right)}
$$
which implies $TY=ZX=ZY$, since the sines of adjacent angles are equal.
Now let $S$ be a point on the extension of the segment $S'B$ beyond point $B$ such that $SB=TX$. Then $\angle SBC=\angle SBA=90^{\circ}+\frac{\alpha}{2}=\angle TXY, BC=BA=XY$, and $SB=SB=TX$, so $\triangle SBC=\triangle SBA=\triangle TXY$, from which $SC=SA=TY=ZX=ZY$, and thus $\triangle SCA=\triangle ZXY$ (by three sides). From all the indicated triangle equalities, it follows that the pyramid $SABC$ has all dihedral angles at the vertex $S$ equal to $\alpha$, so it satisfies the conditions of the problem. But it is not regular, since the line $S'B$ is not perpendicular to $ABC$, and on it only one point projects to the center of the triangle $ABC$, which is the point $S'$, and not the point $S$.
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. [4] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles required to rearrange the chips in reverse order?
(Egor Bakaev)
|
Answer. 50 rubles.
Solution. Evaluation. Each chip must change the parity of its number. A free operation does not change the parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Example. Let's number the chips in order from 0 to 99. We will color the cells in four colors: $a b c d a b c d . . . d$. A free operation changes the chips in adjacent cells of the same color. Therefore, in cells of the same color, the chips can be freely rearranged in any order. We will swap the chips in all pairs $b c$ and $d a$ - this is 49 paid operations. In cells of color $b$ and $c$, the chips can already be arranged as needed for free. In cells of color $a$ and $d$, we will arrange the chips so that chips 0 and 99 are next to each other. We will swap them with the last paid operation and then arrange all the chips in the required order.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. [9] Inside an isosceles triangle $A B C$, a point $K$ is marked such that $C K=A B=B C$ and $\angle K A C=30^{\circ}$. Find the angle $A K B$.
E. Bakayev
|
Answer: $150^{\circ}$. Solution 1. Construct an equilateral triangle $B C L$ (see the figure; points $A$ and $L$ are on the same side of the line $B C$). Points $A, C$, and $L$ lie on a circle with radius $B A$ and center at point $B$. Since $K$ lies inside triangle $A B C$, angle $A B C$ is greater than $60^{\circ}$, hence $L$ and $B$ lie on opposite sides of $A C$ and $L$ lies on the smaller arc $A C$. Therefore, the inscribed angle $C A L$ is half of the central angle $C B L$, which is $30^{\circ}$.
Obviously, the point $K$ that satisfies the conditions of the problem is unique, so it coincides with the point symmetric to $L$ with respect to side $A C$. Thus, triangle $A K L$ is equilateral, and point $K$, like point $B$, lies on the perpendicular bisector of segment $A L$, from which $\angle A K B = \angle L K B = 180^{\circ} - 1 / 2 \angle A K L = 150^{\circ}$.
Solution 2. Let the height $B M$ of triangle $A B C$ intersect line $A K$ at point $O$ (see the figure). Then $\angle C O M = \angle A O M = 60^{\circ}$. Therefore, $\angle A O C = 120^{\circ}$ and $\angle C O B = 120^{\circ}$. Consequently, triangles $B O C$ and $K O C$ are congruent by two sides and the angle opposite the larger side (the so-called fourth criterion of triangle congruence). Therefore, $O B = O K$, meaning triangle $B O K$ is isosceles with an angle of $120^{\circ}$ at vertex $O$. Therefore, $\angle O K B = 30^{\circ}$, and $\angle A K B = 150^{\circ}$.
Solution 3. Construct an equilateral triangle $A C L$ on $A C$ such that points $L$ and $B$ lie on the same side of $A C$ (see the figure).
Draw the height $B M$ in triangle $A B C$, which is also the perpendicular bisector of side $A C$. Since $A L C$ is equilateral, point $L$ also lies on line $B M$. Additionally, draw the height $A N$ in triangle $A L C$. Since $A N$ is the angle bisector of $\angle L A C$, point $K$ lies on this line. Note also that $K$ lies on the same side of $B M$ as $A$, because due to $C K = C B$, it cannot lie inside triangle $B M C$; thus, $K$ lies on segment $A N$.
Notice that right triangles $B M C$ and $K N C$ are congruent by a leg and the hypotenuse (since $M C = A C / 2 = L C / 2 = N C$ and $B C = K C$). Therefore, first, $B M = K N$, and second, $B$ lies on segment $L M$ (since $B M = K N 2 \beta > \beta$, then $\angle A K C = 180^{\circ} - \beta$. Thus, $\angle A C K = \beta - 30^{\circ}$, from which $\angle K C B = \angle C - \angle A C K = (90^{\circ} - \beta) - (\beta - 30^{\circ}) = 120^{\circ} - 2 \beta$, $\angle B K C = 1 / 2 (180^{\circ} - \angle K C B) = 30^{\circ} + \beta$, $\angle A K B = 360^{\circ} - \angle B K C - \angle A K C = 360^{\circ} - (30^{\circ} + \beta) - (180^{\circ} - \beta) = 150^{\circ}$.
|
150
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. [12] There are 100 piles, each with 400 stones. In one move, Petya selects two piles, removes one stone from each, and earns as many points as the absolute difference in the number of stones in these two piles. Petya must remove all the stones. What is the maximum total number of points he can earn?
M. Diden
|
Answer: 3920000. Solution. Estimation. We will assume that the stones in the piles are stacked on top of each other, and Petya takes the top (at the moment) stones from the selected piles. We will number the stones in each pile from bottom to top with numbers from 1 to 400. Then the number of points Petya gets on each move is equal to the difference of the numbers of the removed stones. As a result, he will get a sum of the form $\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|+\ldots+\mid a_{39999}-a_{40000}$, where $a_{i}$ are the numbers of the corresponding stones.
Notice that after expanding the brackets, we get an algebraic sum $S$ of one hundred numbers 400, one hundred numbers 399, ..., one hundred twos, and one hundred ones, with a minus sign in front of exactly half of these numbers.
We will call the numbers from 1 to 200 small, and the rest - large. If it were allowed to take any stones from the piles, then the maximum value of $S$ is obviously achieved when all large numbers enter $S$ with a plus sign, and all small numbers - with a minus sign. Such a sum is equal to $100(400+399+\ldots+201-200-199-\ldots-1)=100((400-200)+(399-199)+\ldots+(201-1))=100 \cdot 200^{2}$.
However, note that each large number will enter the sum with a minus sign at least once: this will happen, for example, the first time Petya removes a stone with this number. Similarly, each of the 200 small numbers will enter the sum with a plus sign at least once (the moment Petya removes the last stone with this number). Therefore, the maximum result Petya can achieve does not exceed $99 \cdot(400+399+\ldots+201)-99 \cdot(200+199+\ldots+1)-(400+399+\ldots+201)+(200+199+\ldots+1)=98 \cdot 200^{2}$.
Example. The specified result can be achieved, for example, as follows. In the first 200 moves, Petya takes 200 stones from the first two piles (at this point, 200 large numbers each get a minus sign once). In the next 200 moves, he removes 200 top stones from the third pile and 200 bottom stones from the first pile, then 200 stones from the second and fourth piles, the third and sixth piles, ..., the 98th and 100th piles (at this point, all numbers enter with the "correct" signs). Finally, there are 200 bottom stones left in the last two piles, which are removed in the last 200 moves (and 200 plus signs appear before the numbers from 200 to 1).
|
3920000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [5] Several natural numbers are written in a row with a sum of 2019. No number and no sum of several consecutive numbers equals 40. What is the maximum number of numbers that could have been written?
A. Shapovalov
|
Answer: 1019 numbers.
Lemma. The sum of any forty consecutive numbers is not less than 80.
Proof. Let the numbers $a_{1}, \ldots, a_{40}$ be written consecutively. Among the numbers $b_{0}=0, b_{1}=a_{1}$, $b_{2}=a_{1}+a_{2}, \ldots, b_{40}=a_{1}+a_{2}+\ldots+a_{40}$, there will be two $b_{i}$ and $b_{j}(i1019$ numbers. By the lemma, the sum of the first $1000=25 \cdot 40$ of them is not less than $25 \cdot 80=2000$. The sum of the remaining numbers (at least 20) is not less than 20. Therefore, the total sum is not less than 2020. Contradiction.
Example. 25 groups 1, $, ., 1,41$ (in each group 39 ones and the number 41) and then 19 ones.
|
1019
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On the first day, $2^{n}$ schoolchildren played ping-pong in a "knockout" format: first, two played, then the winner played with the third, the winner of this pair played with the fourth, and so on until the last schoolchild (there are no draws in ping-pong). On the second day, the same schoolchildren played for the cup: first, they arbitrarily split into pairs and played, the losers were eliminated, and the winners again arbitrarily split into pairs and played, and so on. It turned out that the sets of pairs that played on the first and second days were the same (possibly with different winners). Find the largest possible value of $n$.
B. Frenkin
|
Answer: 3.
Solution. Let's construct the following graph: vertices are players, edges are played matches. According to the condition, for both tournaments, this graph is the same. Consider the first tournament and select the matches where the winners had not won before (for example, the first match). Then the corresponding edges form a path, and the other edges are connected to this path by one end. In particular, if we remove all pendant vertices, only our path without the end vertices will remain.
Now consider the same graph as the graph of the cup tournament. If we remove the pendant vertices from it, a graph of the tournament on $2^{n-1}$ winners of the first stage will remain. It is obviously a path only for $n \leq 3$, otherwise, the tournament winner will have a degree of at least 3. Therefore, $n \leq 3$.
It remains to provide an example for $n=3$. Let the participants be numbered from 1 to 8 and the pairs in the cup be as follows (the loser is indicated first, the winner second): $1-2,3-4,5-6,7-8,2-4,6-8$, 4-8. Then in the knockout stage, the pairs could have been as follows (the winner is again indicated second): $1-2$, $2-4,3-4,4-8,7-8,8-6,6-5$.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. There is a 1001-digit natural number $A$. The 1001-digit number $Z$ is the same number $A$, written from end to beginning (for example, for four-digit numbers, these could be 7432 and 2347). It is known that $A > Z$. For which $A$ will the quotient $A / Z$ be the smallest (but strictly greater than 1)?
|
Answer. For $A$, whose notation (from left to right) is: 501 nines, an eight, and 499 nines.
Solution 1. Let $A=\overline{a_{1000} a_{999 \ldots a_{0}}}$. Since $A>Z$, among the digits $a_{0}, a_{1}, \ldots, a_{499}$ there is at least one non-nine. Therefore, $Z \leqslant Z_{0}=\underbrace{99 \ldots 9}_{499} 8 \underbrace{99 \ldots 9}_{501}$.
We will show that $A-Z \geqslant 10^{501}-10^{499}$. From this it will follow that
$$
\frac{A}{Z}-1 \geqslant \frac{10^{501}-10^{499}}{Z_{0}}
$$
this estimate is achieved when $Z=Z_{0}$, which gives the answer.
We have
$$
\begin{aligned}
A-Z & =\left(a_{1000}-a_{0}\right)\left(10^{1000}-1\right)+\left(a_{999}-a_{1}\right)\left(10^{999}-10\right)+\cdots+\left(a_{501}-a_{499}\right)\left(10^{501}-10^{499}\right)= \\
& =\varphi_{499} \Delta_{499}+\varphi_{498} \Delta_{498}+\cdots+\varphi_{0} \Delta_{0}
\end{aligned}
$$
where $\varphi_{i}=a_{501+i}-a_{499-i}$ and $\Delta_{i}=10^{501+i}-10^{499-i}$ for $i=0,1, \ldots, 499$. Note that $\Delta_{i+1}>10 \Delta_{i}$.
Let $j$ be the largest index for which $\varphi_{j} \neq 0$. Then
$$
\begin{aligned}
\left|\varphi_{j} \Delta_{j}+\varphi_{j-1} \Delta_{j-1}+\cdots+\varphi_{0} \Delta_{0}\right| & \geqslant\left|\varphi_{j} \Delta_{j}\right|-\left|\varphi_{j-1} \Delta_{j-1}\right|-\cdots-\left|\varphi_{0} \Delta_{0}\right| \geqslant \\
& \geqslant \Delta_{j}\left(1-\frac{9}{10}-\frac{9}{100}-\cdots-\frac{9}{10^{j}}\right)=\frac{\Delta_{j}}{10^{j}} \geqslant \Delta_{0}
\end{aligned}
$$
which is what we needed.
Solution 2. It is clear that we can minimize (the positive) number $\frac{A}{Z}-1=\frac{A-Z}{Z}$.
Number the digits in $A$ from left to right as $a_{1}, a_{2}, \ldots, a_{1001}$. Let $k$ be the smallest index for which $a_{k} \neq a_{1002-k}$ (then $k \leqslant 500$ and $a_{k}>a_{1002-k}$, since $A>Z$).
Consider an arbitrary optimal example. Replace the first and last $k-1$ digits with nines. $A-Z$ will not change, and $Z$ will not decrease, so our fraction will not increase. For the same reason, $a_{501}$ can be replaced with 9.
Replace $a_{k}$ with 9 and $a_{1002-k}$ with 8. In this case, $A-Z$ will not increase, and $Z$ will not decrease.
Replace all digits $a_{k+1}, \ldots, a_{500}$ with zeros, and $a_{502}, \ldots, a_{1001-k}$ with nines. Then $A-Z$ will not increase, and $Z$ will not decrease (this will only happen if the second half was already nines!). Since in the optimal example $A-Z<Z$ (simply fewer digits), it is clear that $\frac{A-Z}{Z}$ will not increase.
Thus, we can assume that $A$ has the form
$$
\underbrace{99 \ldots 9}_{k} \underbrace{00 \ldots 0}_{500-k} 9 \underbrace{99 \ldots 9}_{500-k} 8 \underbrace{99 \ldots 9}_{k-1} .
$$
In this case,
$$
A-Z=10^{501}+10^{500}-10^{k}-10^{k-1}
$$
This expression reaches its minimum when $k=500$, and for this same $k$, the maximum value of the considered $Z$ is achieved. Therefore, this is the answer.
|
501
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. [4] In each cell of a strip of length 100, there is a chip. You can swap any two adjacent chips for 1 ruble, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order?
(Egor Bakaev)
|
Answer. 50 rubles.
Solution. Evaluation. Each chip must change the parity of its number. A free operation does not change the parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Example. Let's number the chips in order from 0 to 99. We will color the cells in four colors: $a b c d a b c d . . . d$. A free operation changes the chips in adjacent cells of the same color. Therefore, in cells of the same color, the chips can be freely rearranged in any order. We will swap the chips in all pairs $b c$ and $d a$ - this is 49 paid operations. In cells of color $b$ and $c$, the chips can already be arranged as needed for free. In cells of color $a$ and $d$, we will arrange the chips so that chips 0 and 99 are next to each other. We will swap them with the last paid operation and then arrange all the chips in the required order.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. [5] On an island, there live knights, liars, and yes-men; each knows who everyone else is. All 2018 inhabitants were lined up and asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?". The inhabitants answered in turn, and everyone could hear their answers. Knights answered truthfully, liars lied. Each yes-man answered the same way as the majority of those who answered before him, and if the number of "Yes" and "No" answers was equal, he gave either of these answers. It turned out that there were exactly 1009 "Yes" answers. What is the maximum number of yes-men that could be among the inhabitants of the island?
(M. Kuznev)
|
Answer: 1009 yes-men. Solution: Let's call knights and liars principled people.
Estimate. First method. (Buchaev Abdulqadir) We will track the balance - the difference between the number of "Yes" and "No" answers. At the beginning and at the end, the balance is zero, and with each answer, it changes by 1. Zero values of the balance divide the row of residents into groups. Within each group, the balance retains its sign. Yes-men always increase the absolute value of the balance. Therefore, to make the balance zero, there must be no fewer principled people than yes-men. This is true for each group, and therefore for all the island's residents.
Second method. A yes-man could not increase the minimum of the current number of "Yes" and "No" answers. Since this minimum increased from 0 to 1009, and with each answer, it changed by no more than one, there must be at least 1009 principled people.
Example. First, all 1009 yes-men said "No," and then 1009 knights said "Yes."
Note. Any sequence of 1009 principled people can be thinned out by yes-men so that the condition is met.
|
1009
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. [8] The three medians of a triangle divide its angle into six angles, of which exactly $k$ are greater than $30^{\circ}$. What is the greatest possible value of $k$?
(N. Sedrakyan)
|
Answer. $k=3$.
Evaluation. First method. Let $h_{1} \leq h_{2} \leq h_{3}$ be the heights of the triangle, and $m_{1}, m_{2}, m_{3}$ be the medians from the corresponding vertices (in fact, $m_{1} \leq m_{2} \leq m_{3}$, but this will not be used). Then we have $m_{3} \geq h_{i}$ for $i=1,2$, 3. Dropping perpendiculars from the end of $m_{3}$ to the adjacent sides, each of them equals half of the corresponding height and, therefore, is not greater than $m_{3} / 2$. Thus, the angles adjacent to $m_{3}$ are not greater than $30^{\circ}$. Similarly, one of the angles at $m_{2}$ is not greater than $30^{\circ}$.
Second method. Let there be a triangle $A B C$ with medians $A X, B Y, C Z$. Both angles $B A X$ and $B C Z$ cannot be greater than $30^{\circ}$: otherwise, points $A$ and $C$ lie inside a circle with chord $Z X$ and a diameter twice as large as $Z X$, but then $A C=2 Z X$, meaning a segment equal to the diameter lies strictly inside the circle - a contradiction. Similarly, in each of the other two pairs, at most one angle is greater than $30^{\circ}$.
Example 1. Consider first a triangle $A B C$ where $\angle C=90^{\circ}, \angle A=30^{\circ}$. Let $B K$ and $C L$ be its medians. Then $\angle A C L=30^{\circ}, \angle B C L=60^{\circ}, \angle C B K>30^{\circ}$ (since the angle bisector of angle $B$ lies between the leg and the median). Now slightly decrease the leg $A C$. In this case, the angle $A C L$ will slightly increase (i.e., become greater than $30^{\circ}$), while the angles $B C L$ and $C B K$ will slightly decrease but remain greater than $30^{\circ}$.
Example 2. Consider a right triangle with legs 4 and 6 (see figure). We have $\operatorname{tg} 30^{\circ}<\operatorname{tg} \alpha<\operatorname{tg} \beta<\operatorname{tg} \gamma$, since $\frac{1}{\sqrt{3}}<\frac{2}{3}<\frac{3}{4}<\frac{3}{2}$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. [5] Pentagon $A B C D E$ is circumscribed around a circle. The angles at its vertices $A, C$ and $E$ are $100^{\circ}$. Find the angle $A C E$.
|
Answer: $40^{\circ}$. It is not hard to understand that such a pentagon exists.
Solution 1. The lines connecting the vertices with the center $O$ of the inscribed circle $\omega$ are the angle bisectors of the pentagon. Therefore, $\angle O A E=\angle O E A=50^{\circ}, \angle A O E=80^{\circ}$. Let $\omega$ touch the sides $B C$ and $A E$ at points $K$ and $M$ respectively. Then $\angle O C K=50^{\circ}$, and the right triangles $O M A, O K C$ and $O M E$ are equal by the leg and the opposite angle. Thus, $O A=O C=O E$, i.e., points $A, C$ and $E$ lie on a circle with center $O$. Therefore, $\angle A C E=\frac{1}{2} \angle A O E=40^{\circ}$. Solution 2. Since the angles $A, C, E$ are equal, all tangents to the circle from these vertices are equal. Since the tangents from vertex $B$ are also equal, triangle $A B C$ is isosceles, and triangle $C D E$ is also (analogously). The sum of angles $B$ and $D$ is $540^{\circ}-3 \cdot 100^{\circ}=240^{\circ}$, so $\angle A C B+\angle E C D=\left(2 \cdot 180^{\circ}-240^{\circ}\right): 2=$ $=60^{\circ}, \mathrm{a} \angle A C E=100^{\circ}-60^{\circ}=40^{\circ}$.
|
40
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.
It is known that three numbers $a_{1}, a_{2}, a_{3}$ were obtained as follows. First, a natural number $A$ was chosen and the numbers $A_{1}=[A]_{16}, A_{2}=$ $[A / 2]_{16}, A_{3}=[A / 4]_{16}$ were found, where $[X]_{16}$ is the remainder of the integer part of the number $X$ divided by 16 (for example, $[53 / 2]_{16}=10$). Then an integer $B$ was chosen such that 0 $\leq \mathrm{B} \leq 15$. The numbers $A_{1}, A_{2}, A_{3}$ and $B$ are written in binary, i.e., each is represented as a string of 0s and 1s of length 4, adding the necessary number of zeros on the left. We agree to add such strings symbol by symbol "in a column" without carrying to the next digit according to the rule: $1+1=0+0=0$ and $0+1=1+0=0$, and denote the operation of symbol-by-symbol addition by the symbol $\diamond$. For example, $3 \diamond 14=(0011) \diamond$ $(1110)=(1101)=13$. Let $a_{1}=A_{1} \diamond B, a_{2}=A_{2} \diamond B, a_{3}=A_{3} \diamond B$. Find the sum of all possible values of the number $a_{3}$, given that $a_{1}=2, a_{2}=$ 9. Choose the correct answer:
#
|
# Answer: 16.
## Conditions and answers to the problems of the final stage of the 2012-13 academic year
#
|
16
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
Task 2.
In a table consisting of $n$ rows and $m$ columns, numbers (not necessarily integers) were written such that the sum of the elements in each row is 408, and the sum of the elements in each column is 340. After that, $k$ columns were added to the table, the sum of the elements in each of which is 476, and a column, the sum of the elements in which is 272. The resulting table has a sum of elements in each row equal to 544. Find the numbers $n, m$, and $k$ for which the expression $2n - 3m + 6k$ takes the smallest possible natural value. For the found parameters $n, m$, and $k$, provide an example of the specified table.
|
Answer: 4 when $n=65, m=78, k=18$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $ik$. The user enters digits from the left column with the left hand, and the rest with the right hand. Restore the lock code if it is known that when entering the code, the user entered the digits as follows:
when $a_{3}=3$ - right, right, left, right;
when $a_{3}=9$ - left, left, left, left;
when $a_{3}=1$ - right, right, right, right;
when $a_{3}=7$ - left, right, right, right.
In your answer, specify the obtained code.
| $\mathrm{a}_{1}$ | $\mathrm{a}_{2}$ | $\mathrm{a}_{3}$ | | |
| :--- | :--- | :--- | :---: | :---: |
| $\mathrm{a}_{4}$ | $\mathrm{a}_{5}$ | $\mathrm{a}_{6}$ | | |
| $\mathrm{a}_{7}$ | $\mathrm{a}_{8}$ | $\mathrm{a}_{9}$ | | |
| | | | | $\mathrm{a}_{0}$ |
| | | | | |
| | | | | |
|
5. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The layout of the digits on the keyboard changes after entering the code based on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The user enters digits from the left column with the left hand, and the rest with the right hand. Restore the lock code if it is known that when entering the code, the user entered the digits as follows:
when $a_{3}=3$ - right, right, left, right;
when $a_{3}=9$ - left, left, left, left;
when $a_{3}=1$ - right, right, right, right;
when $a_{3}=7$ - left, right, right, right.
In the answer, specify the obtained code.
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 0 | | |
| | | |
| $a_{1}$ | $a_{2}$ | $\mathrm{a}_{3}$ |
| :---: | :---: | :---: |
| $\mathrm{a}_{4}$ | $a_{5}$ | $\mathrm{a}_{6}$ |
| $a_{7}$ | $a_{8}$ | $\mathrm{a}_{9}$ |
| | $\mathrm{a}_{0}$ | |
Answer: 3212
|
3212
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. At the robot running competition, a certain number of mechanisms were presented. The robots were released on the same distance in pairs. The protocol recorded the differences in the finish times of the winner and the loser in each of the races. All of them turned out to be different: 1 sec., 2 sec., 3 sec., 4 sec., 5 sec., 7 sec. It is known that during the races, each robot competed with each other exactly once, and that each robot always ran at the same speed. In your answer, indicate the time in seconds of the slowest mechanism, if the best time to complete the distance was 30 seconds.
|
3. At the robot running competition, a certain number of mechanisms were presented. The robots were released on the same distance in pairs. The protocol recorded the differences in the finish times of the winner and the loser in each of the races. All of them turned out to be different: 1 sec., 2 sec., 3 sec., 4 sec., 5 sec., 7 sec. It is known that during the races, each robot competed with each other exactly once, and that each robot always ran at the same speed. In the answer, indicate the time in seconds of the slowest mechanism, if the best time to complete the distance was 30 seconds.
Answer: 37.
|
37
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The user enters digits from the left column with the left hand, and the rest with the right hand. Restore the lock code if it is known that when entering the code, the user entered the digits as follows:
when $a_{3}=3$ - right, right, left, right;
when $a_{3}=9$ - left, left, left, left;
when $a_{3}=1$ - right, right, right, right;
when $a_{3}=7$ - left, right, right, right.
In your answer, specify the obtained code.
| 1 | 2 | 3 | |
| :--- | :--- | :--- | :---: |
| 4 | 5 | 6 | |
| 7 | 8 | 9 | |
| 0 | | | |
| | | | |
| $\mathrm{a}_{1}$ | $\mathrm{a}_{2}$ | $\mathrm{a}_{3}$ |
| :---: | :---: | :---: |
| $\mathrm{a}_{4}$ | $\mathrm{a}_{5}$ | $\mathrm{a}_{6}$ |
| $\mathrm{a}_{7}$ | $\mathrm{a}_{8}$ | $\mathrm{a}_{9}$ |
| | | $\mathrm{a}_{0}$ |
| | | |
| | | |
|
6. Unlocking the communicator is done by entering a 4-digit numerical code on the touch screen. The arrangement of digits on the keyboard changes after entering the code depending on a random prime number $k$ from 7 to 2017, and the digit $i$ is displayed as $a_{i}$, which is the last digit of the number $i k$. The user enters digits from the left column with the left hand, and the rest with the right hand. Restore the lock code if it is known that when entering the code, the user entered the digits as follows:
when $a_{3}=3$ - right, right, left, right;
when $a_{3}=9$ - left, left, left, left;
when $a_{3}=1$ - right, right, right, right;
when $a_{3}=7$ - left, right, right, right.
In the answer, specify the obtained code.
| 1 | 2 | 3 | |
| :--- | :--- | :--- | :---: |
| 4 | 5 | 6 | |
| 7 | 8 | 9 | |
| 0 | | | |
| | | | |
| $\mathrm{a}_{1}$ | $\mathrm{a}_{2}$ | $\mathrm{a}_{3}$ | | |
| :--- | :--- | :--- | :---: | :---: |
| $\mathrm{a}_{4}$ | $\mathrm{a}_{5}$ | $\mathrm{a}_{6}$ | | |
| $\mathrm{a}_{7}$ | $\mathrm{a}_{8}$ | $\mathrm{a}_{9}$ | | |
| | | | | $\mathrm{a}_{0}$ |
| | | | | |
Answer: 3212.
|
3212
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Workers at an aluminum plant played the following game during their break. They drew cards from a deck (36 cards) until they had cards of all 4 suits among the drawn cards. What is the minimum number of cards that must be drawn to ensure that all 4 suits are among them?
a) 4
b) 9
c) 18
d) 28
e) 36
|
8. answer: g. There are 9 cards of each suit. If you draw 27 cards, they could all be from only three suits. Suppose you draw 28 cards and among the first 27 cards, they are only from three suits, then in the deck, only cards of the fourth suit will remain. And the 28th card will already be of the 4th suit.
|
28
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
12. The director of an aluminum plant wants the advertisement of his plant, which lasts 1 min. 30 sec., to be shown exactly every 18 min. 30 sec. after the end of the previous advertising break. How much time will the advertisement of the aluminum plant be shown in a week?
a) 12 h.
b) 12 h. 12 min.
c) 12 h. 24 min.
d) 12 h. 36 min.
e) 12 h. 48 min.
|
12. answer: g. In a week, there are $7 \cdot 24=168$ hours. For every hour, advertisements are shown for $1.5 \cdot 3=4.5$ minutes. Therefore, in a week, advertisements make up $4.5 \cdot 168=756=12$ hours and 36 minutes of airtime. In a week, there are $7 \cdot 24=168$ hours. For every hour, advertisements are shown for $1.5 \cdot 3=4.5$ minutes. Therefore, in a week, advertisements make up $4.5 \cdot 168=756=12$ hours and 36 minutes of airtime.
|
12
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
8. During the lunch break on the day of the "13th Element. ALchemy of the Future" competition, a cafeteria worker mixed 2 liters of one juice with a $10 \%$ sugar content and 3 liters of another juice with a $15 \%$ sugar content. What is the sugar content of the resulting mixture?
a) $5 \%$
б) $12.5 \%$
в) $12.75 \%$
г) $13 \%$
д) $25 \%$
|
8. Answer g. In the first juice, there is $0.1 \cdot 2=0.2$ liters of sugar, and in the second juice, there is $0.15 \cdot 3=0.45$ liters of sugar. Then, in the resulting mixture, there will be $0.2+0.45=0.65$ liters of sugar. The total volume is 5 liters. Then the percentage of sugar in the mixture is $\frac{0.65}{5} \cdot 100 \% = 13 \%$.
|
13
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
15. Hydrogen was passed over a heated powder (X1). The resulting red substance (X2) was dissolved in concentrated sulfuric acid. The resulting solution of the substance blue (X3) was neutralized with potassium hydroxide - a blue precipitate (X4) formed, which upon heating turned into a black powder (X1). What substances are involved in the described process? Indicate the molar mass of the initial and final substance (X1).
|
15. 80 g/mol; $\mathrm{m}(\mathrm{CuO})=80$ g/mol; $\mathrm{X}_{1}-\mathrm{CuO} ; \mathrm{X}_{2}-\mathrm{Cu} ; \mathrm{X}_{3}-\mathrm{CuSO}_{4 ;} \mathrm{X}_{4}-\mathrm{Cu}(\mathrm{OH})_{2}$
|
80
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
16. At the quiz in the Museum of Entertaining Sciences of SFU, 10 schoolchildren are participating. In each round, the students are divided into pairs. Each participant meets every other participant exactly once. A win in a match earns 1 point, a draw earns 0.5 points, and a loss earns 0 points. What is the minimum number of rounds after which an early winner can appear?
## Answers and solutions:
## Translation of the question and answers into English:
16. At the quiz in the Museum of Entertaining Sciences of SFU, 10 schoolchildren are participating. In each round, the students are divided into pairs. Each participant meets every other participant exactly once. A win in a match earns 1 point, a draw earns 0.5 points, and a loss earns 0 points. What is the minimum number of rounds after which an early winner can appear?
## Answers and solutions:
|
16. 7. Evaluation. After the sixth round, 30 points have been played, and the leader has no more than 6 points, while the other nine participants have collectively scored no less than 24 points. Therefore, among them, there is at least one participant with more than three points. Since there are still 3 rounds ahead, the winner is still unknown.
Example. Suppose in the first 7 rounds, the leader won all their matches, and all other matches ended in draws. Then two students who have not yet faced the leader have 3.5 points each, while the others have 3 points each. Since only 2 rounds remain until the end of the tournament, the winner is already determined.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Young researcher Petya noticed that the iron at home heats up by $9^{\circ} \mathrm{C}$ every 20 seconds, and after being turned off, it cools down by $15^{\circ} \mathrm{C}$ every 30 seconds. Last evening, after turning off the iron, it cooled down for exactly 3 minutes. How long was the iron turned on?
a) 3 min 20 sec
b) 3 min 40 sec
c) 3 min
d) 4 min
e) 5 min
|
8. answer: a. 3 minutes $=180$ seconds. In 180 seconds, the iron cools down by $180 / 30 \cdot 15=90$ degrees. To heat up by 90 degrees, the iron takes $90 / 9 \cdot 20=200$ seconds. 200 seconds $=3$ minutes 20 seconds.
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
12. The security guard of the aluminum plant works on Tuesdays, Fridays, and on odd-numbered days. What is the maximum number of consecutive days the security guard can work?
a) 3
b) 4
c) 5
d) 6
e) 7
|
12. answer: g. 6 days. An example of such a situation: 29th (odd), 30th (Tuesday), 31st (odd), 1st (odd), 2nd (Friday), 3rd (odd).
|
6
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
16. A worker at an aluminum plant can produce 16 blanks or 10 parts from blanks in one shift. It is known that exactly one part is made from each blank. What is the maximum number of blanks the worker can produce in one shift to make parts from them in the same shift?
|
16. 6. Let's denote by $x$ the duration of the worker's working day in hours. Then the worker makes one part from a blank in $\frac{x}{10}$ hours, one blank in $\frac{x}{16}$ hours, and a blank and a part from it in $\frac{x}{10}+\frac{x}{16}=\frac{13 x}{80}$ hours. Since $x: \frac{13 x}{80}=6 \frac{2}{13}$, the maximum number of blanks that the worker can make in a day so that parts can be made from all of them on the same day is 6.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. What is the mass fraction (%) of oxygen in aluminum oxide $\mathrm{Al}_{2} \mathrm{O}_{3}$?
a) $53 \%$
b) $27 \%$
c) $16 \%$
d) $102 \%$
|
7. $53 \% . \operatorname{Mr}(\mathrm{Al} 2 \mathrm{O} 3)=27 * 2+16 * 3=102 \quad \mathrm{~W}(\mathrm{Al})=27 * 2 / 102 * 100 \%=53 \%$
7. $53 \% . \operatorname{Mr}(\mathrm{Al} 2 \mathrm{O} 3)=27 * 2+16 * 3=102 \quad \mathrm{~W}(\mathrm{Al})=27 * 2 / 102 * 100 \%=53 \%$
The text is already in a format that is a mix of mathematical notation and chemical formula, which is the same in English. Therefore, the translation is identical to the original text.
|
53
|
Other
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
16. Every sixth bus in the bus park of the aluminum plant is equipped with an air conditioner. After the plant director ordered to install it on 5 more buses, a quarter of the buses had an air conditioner. How many buses are there in the park of the plant, if each bus is equipped with only one air conditioner?
|
16. 60. Let the number of buses in the park be denoted by $x$. Then, $\frac{1}{6} x + 5$ buses are equipped with air conditioning. This constitutes a quarter of all the buses in the park. We have the equation: $\left(\frac{1}{6} x + 5\right) \cdot 4 = x$ or $\frac{1}{3} x = 20$. From this, $x = 60$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three is to the sum of all terms except the last three as $4: 3$. Find the number of terms in this progression.
|
Answer: 20.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all terms except the first three is $-\frac{(n-3) \cdot(a+3 d+a+(n-1) d)}{2}$, and the sum of all terms except the last three is $-\frac{(n-3) \cdot(a+a+(n-4) d)}{2}$. From the condition, we have the system:
$$
\left\{\begin{aligned}
2 \cdot 13(a+6 d) & =13(a+(n-7) d) \\
3 \cdot(n-3) \cdot(2 a+(n+2) d) & =4 \cdot(n-3) \cdot(2 a+(n-4) d)
\end{aligned}\right.
$$
or, after transformations,
$$
\left\{\begin{aligned}
a & =(n-19) d \\
-2 a & =(n-22) d
\end{aligned}\right.
$$
(since otherwise the sum of all terms except the first three would equal the sum of all terms except the last three), we get $n=20$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number written with the same digits in reverse order is equal to $N$. It turns out that the number $N$ is divisible by 100. Find $N$.
|
Answer: 11000.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{d c b a}=1000 d+100 c+10 b+a$, where $a$, $b, c, d$ are digits and $a \neq 0$.
According to the condition, $X+Y$ is divisible by 100, i.e., $1001(a+d)+110(b+c) \vdots 100$.
We have $1001(a+d) \vdots 10$, i.e., $a+d \vdots 10$, from which, since $a$ and $d$ are digits and $a \neq 0, 1 \leq a+d \leq 18$, thus $a+d=10$. Further, $1001 \cdot 10+110(b+c) \vdots 100$, i.e., $b+c+1 \vdots 10$, from which, since $b$ and $c$ are digits, $1 \leq b+c+1 \leq 19$, thus, $b+c=9$.
Therefore, $N=X+Y=1001 \cdot 10+110 \cdot 9=11000$.
|
11000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 65 and 31, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
|
Answer: 2015.
Solution. Let $O$ be the intersection point of diagonals $AC$ and $BD$. From the similarity of triangles $AOB$ and $COD$, it follows that $\overrightarrow{OC} = \frac{31}{65} \overrightarrow{AO}$, and $\overrightarrow{OD} = \frac{31}{65} \overrightarrow{BO}$.
Let the vector $\overrightarrow{AO}$ be denoted by $\vec{a}$, and the vector $\overrightarrow{BO}$ by $\vec{b}$. Then, from the condition, it follows that $(\vec{a}, \vec{b}) = 0$ and
$$
\overrightarrow{AD} = \overrightarrow{AO} + \overrightarrow{OD} = \vec{a} + \frac{31}{65} \vec{b}, \quad \overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = \vec{b} + \frac{31}{65} \vec{a}
$$
From which
$$
(\overrightarrow{AD}, \overrightarrow{BC}) = \left(\vec{a} + \frac{31}{65} \vec{b}, \vec{b} + \frac{31}{65} \vec{a}\right) = \frac{31}{65} \left(|\vec{a}|^2 + |\vec{b}|^2\right) + (\ldots) \cdot (\vec{a}, \vec{b}) = \frac{31}{65} |AB|^2 = 2015
$$
where the penultimate equality follows from the fact that triangle $AOB$ is a right triangle.
|
2015
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10. A cylinder of volume 9 is inscribed in a cone. The plane of the upper base of this cylinder cuts off a frustum of volume 63 from the original cone. Find the volume of the original cone.
|
Answer: 64.
Solution. Let the height and radius of the original cone be $H$ and $R$, and the height and radius of the cylinder be $h$ and $r$. We use the formula for the volume of a frustum of a cone: $\frac{1}{3} \pi\left(R^{2}+R r+r^{2}\right) h=63$. We also know that $\pi r^{2} h=9$. Dividing the corresponding parts of the equations, we get $\left(\frac{R}{r}\right)^{2}+\left(\frac{R}{r}\right)+1=$ $\frac{63 \cdot 3}{9}=21$. Solving the quadratic equation, we get the roots 4 and -5, only the positive root has geometric meaning. $R / r=4, \frac{H-h}{H}=4, \frac{h}{H}=\frac{3}{4}$, from which we obtain for the original cone:
$$
V=\frac{1}{3} \pi R^{2} H=\frac{1}{3}\left(\pi r^{2} h\right)\left(\frac{R}{r}\right)^{2} \frac{H}{h}=\frac{1}{3} \cdot 9 \cdot 4^{2} \cdot \frac{4}{3}=64
$$
## Variant II
|
64
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is to the sum of all terms except the last three as $5: 4$. Find the number of terms in this progression.
|
Answer: 22.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all terms except the first three is $-\frac{(n-3) \cdot(a+3 d+a+(n-1) d)}{2}$, and the sum of all terms except the last three is $-\frac{(n-3) \cdot(a+a+(n-4) d)}{2}$. From the condition, we have the system:
$$
\left\{\begin{aligned}
2 \cdot 13(a+6 d) & =13(a+(n-7) d) \\
4 \cdot(n-3) \cdot(2 a+(n+2) d) & =5 \cdot(n-3) \cdot(2 a+(n-4) d)
\end{aligned}\right.
$$
or, after transformations,
$$
\left\{\begin{aligned}
a & =(n-19) d \\
-2 a & =(n-28) d
\end{aligned}\right.
$$
Multiplying the first equation by 2 and adding it to the second, we get $(3 n-66) d=0$. Since $d \neq 0$ (otherwise, the sum of all terms except the first three would equal the sum of all terms except the last three), we obtain $n=22$.
|
22
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number obtained from $X$ by swapping its second and third digits is divisible by 900. Find the remainder when the number $X$ is divided by 90.
|
Answer: 45.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{a c b d}=1000 a+100 c+10 b+d$, where $a$, $b, c, d$ are digits and $a \neq 0, d \neq 0$.
According to the condition, $X+Y$ is divisible by 900, i.e., $2000 a+110(b+c)+2 d \vdots 900$.
We have, $2 d \vdots 10$, i.e., $d \vdots 5$, so since $d \neq 0$ and $d$ is a digit, $d=5$. Next, $110(b+c)+10 \vdots 100$, i.e., $b+c+1 \vdots 10$, from which, since $b$ and $c$ are digits, $1 \leq b+c+1 \leq 19, b+c=9$. Finally, $2000 a+110 \cdot 9+10 \vdots 9$, i.e., $2 a+1 \vdots 9$, from which, since $a$ is a digit, $a=4$.
Thus, $X=4000+90 b+90+5=90 q+45$.
|
45
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 155 and 13, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
|
Answer: 2015.
Solution. Let $O$ be the intersection point of diagonals $AC$ and $BD$. From the similarity of triangles $AOB$ and $COD$, it follows that $\overrightarrow{OC} = \frac{13}{155} \overrightarrow{AO}$, and $\overrightarrow{OD} = \frac{13}{155} \overrightarrow{BO}$.
Let the vector $\overrightarrow{AO}$ be denoted by $\vec{a}$, and the vector $\overrightarrow{BO}$ by $\vec{b}$. Then, from the condition, it follows that $(\vec{a}, \vec{b}) = 0$ and
$$
\overrightarrow{AD} = \overrightarrow{AO} + \overrightarrow{OD} = \vec{a} + \frac{13}{155} \vec{b}, \quad \overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = \vec{b} + \frac{13}{155} \vec{a}
$$
From which
$$
(\overrightarrow{AD}, \overrightarrow{BC}) = \left(\vec{a} + \frac{13}{155} \vec{b}, \vec{b} + \frac{13}{155} \vec{a}\right) = \frac{13}{155} \left(|\vec{a}|^2 + |\vec{b}|^2\right) + (\ldots) \cdot (\vec{a}, \vec{b}) = \frac{13}{155} |AB|^2 = 2015
$$
where the penultimate equality follows from the fact that triangle $AOB$ is a right triangle.
|
2015
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is to the sum of all terms except the last three as $3: 2$. Find the number of terms in this progression.
|
Answer: 18.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all terms except the first three is $-\frac{(n-3) \cdot(a+3 d+a+(n-1) d)}{2}$, and the sum of all terms except the last three is $-\frac{(n-3) \cdot(a+a+(n-4) d)}{2}$. From the condition, we have the system:
$$
\left\{\begin{aligned}
2 \cdot 13(a+6 d) & =13(a+(n-7) d) \\
2 \cdot(n-3) \cdot(2 a+(n+2) d) & =3 \cdot(n-3) \cdot(2 a+(n-4) d)
\end{aligned}\right.
$$
or, after transformations,
$$
\left\{\begin{aligned}
a & =(n-19) d \\
-2 a & =(n-16) d
\end{aligned}\right.
$$
(since otherwise the sum of all terms except the first three would equal the sum of all terms except the last three), we get $n=18$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 65 and 31, respectively, and its lateral sides are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$.
|
Answer: -2015.
Solution. Let $O$ be the point of intersection of the lines containing the lateral sides $AD$ and $BC$. From the similarity of triangles $AOB$ and $DOC$, it follows that $\overrightarrow{OC} = -\frac{31}{65} \overrightarrow{BO}$, and $\overrightarrow{OD} = -\frac{31}{65} \overrightarrow{AO}$.
Let the vector $\overrightarrow{AO}$ be denoted by $\vec{a}$, and the vector $\overrightarrow{BO}$ by $\vec{b}$. Then, from the condition, it follows that $(\vec{a}, \vec{b}) = 0$ and
$$
\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = \vec{a} - \frac{31}{65} \vec{b}, \quad \overrightarrow{BD} = \overrightarrow{BO} + \overrightarrow{OD} = \vec{b} - \frac{31}{65} \vec{a}
$$
From which
$$
(\overrightarrow{AC}, \overrightarrow{BD}) = \left(\vec{a} - \frac{31}{65} \vec{b}, \vec{b} - \frac{31}{65} \vec{a}\right) = -\frac{31}{65}\left(|\vec{a}|^2 + |\vec{b}|^2\right) + (\ldots) \cdot (\vec{a}, \vec{b}) = -\frac{31}{65}|AB|^2 = -2015
$$
where the penultimate equality follows from the fact that triangle $AOB$ is a right triangle.
|
-2015
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. The sum of the first thirteen terms of a certain arithmetic progression is $50 \%$ of the sum of the last thirteen terms of this progression. The sum of all terms of this progression except the first three terms is in the ratio $6: 5$ to the sum of all terms except the last three. Find the number of terms in this progression.
|
Answer: 24.
Solution. Let $a$ be the first term of the arithmetic progression, $d$ be its common difference, and $n$ be the number of terms. Then, the sum of the first thirteen terms is $\frac{13 \cdot(a+(a+12 d))}{2}$, the sum of the last thirteen terms is $-\frac{13 \cdot(a+(n-13) d+a+(n-1) d)}{2}$, the sum of all terms except the first three is $-\frac{(n-3) \cdot(a+3 d+a+(n-1) d)}{2}$, and the sum of all terms except the last three is $-\frac{(n-3) \cdot(a+a+(n-4) d)}{2}$. From the condition, we have the system:
$$
\left\{\begin{aligned}
2 \cdot 13(a+6 d) & =13(a+(n-7) d) \\
5 \cdot(n-3) \cdot(2 a+(n+2) d) & =6 \cdot(n-3) \cdot(2 a+(n-4) d)
\end{aligned}\right.
$$
or, after transformations,
$$
\left\{\begin{aligned}
a & =(n-19) d \\
-2 a & =(n-34) d
\end{aligned}\right.
$$
(since otherwise the sum of all terms except the first three would equal the sum of all terms except the last three), we get $n=24$.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. If the sum of the digits of a four-digit number $X$ is subtracted from $X$, the result is a natural number $N=K^{2}$, where $K$ is a natural number that gives a remainder of 5 when divided by 20 and a remainder of 3 when divided by 21. Find the number $N$.
|
Answer: 2025.
Solution. Let $X=\overline{a b c d}=1000 a+100 b+10 c+d$, where $a, b, c, d$ are digits and $a \neq 0$.
According to the condition, $X-a-b-c-d=999 a+99 b+9 c=K^{2}$, where $K=20 u+5, K=21 v+3$.
Notice that $999 a+99 b+9 c \vdots 9$, i.e., $K^{2} \vdots 9$, which means $K=3 M$, and $M^{2}=111 a+11 b+c \leq$ $111 \cdot 9+11 \cdot 9+9=1107, M \leq 33, K=3 M \leq 99$.
Thus, we need to find among the numbers from 1 to 99 all those numbers that are divisible by 3, give a remainder of 5 when divided by 20, and a remainder of 3 when divided by 21. This can be done in various ways, for example, by simple enumeration: list all numbers from 1 to 99 that give a remainder of 3 when divided by 21:
$$
3,24,45,66,87
$$
Among them, only one number - 45 gives a remainder of 5 when divided by 20. That is, $K=45$, and $N=2025$.
|
2025
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 155 and 13, respectively, and its lateral sides are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$.
|
Answer: -2015.
Solution. Let $O$ be the point of intersection of the lines containing the lateral sides $AD$ and $BC$. From the similarity of triangles $AOB$ and $DOC$, it follows that $\overrightarrow{OC} = -\frac{13}{155} \overrightarrow{BO}$, and $\overrightarrow{OD} = -\frac{13}{155} \overrightarrow{AO}$.
Let the vector $\overrightarrow{AO}$ be denoted by $\vec{a}$, and the vector $\overrightarrow{BO}$ by $\vec{b}$. Then, from the condition, it follows that $(\vec{a}, \vec{b}) = 0$ and
$$
\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = \vec{a} - \frac{13}{155} \vec{b}, \quad \overrightarrow{BD} = \overrightarrow{BO} + \overrightarrow{OD} = \vec{b} - \frac{13}{155} \vec{a}
$$
From which
$$
(\overrightarrow{AC}, \overrightarrow{BD}) = \left(\vec{a} - \frac{13}{155} \vec{b}, \vec{b} - \frac{13}{155} \vec{a}\right) = -\frac{13}{155}\left(|\vec{a}|^{2} + |\vec{b}|^{2}\right) + (\ldots) \cdot (\vec{a}, \vec{b}) = -\frac{13}{155}|AB|^{2} = -2015
$$
where the penultimate equality follows from the fact that triangle $AOB$ is a right triangle.
|
-2015
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. What is greater: 1 or $\frac{21}{64}+\frac{51}{154}+\frac{71}{214} ?$
|
Answer: One is greater.
## First solution.
$$
\frac{21}{64}+\frac{51}{154}+\frac{71}{214}<\frac{21}{63}+\frac{51}{153}+\frac{71}{213}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1
$$
## Second solution.
$$
\begin{aligned}
\frac{21}{64}+\frac{51}{154}+\frac{71}{214} & =\frac{21 \cdot 154 \cdot 214+64 \cdot 51 \cdot 214+64 \cdot 154 \cdot 71}{64 \cdot 154 \cdot 214}= \\
& =\frac{692076+698496+699776}{2109184}=\frac{2090348}{2109184}\left[=\frac{522587}{527296}\right]<1 .
\end{aligned}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. In a football tournament, seven teams played: each team played once with each other. In the next round, teams that scored thirteen or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round?
|
Answer: 4.
Solution. In total, the teams played $\frac{7 \cdot 6}{2}=21$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $21 \cdot 3=63$. Hence, the number of teams $n$ advancing to the next stage satisfies the inequality $n \cdot 13 \leqslant 63$, from which $n \leqslant 4$.
On the other hand, an example of a tournament table can be provided where 4 teams advance to the next round:
| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | Total |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | $\mathrm{X}$ | 0 | 1 | 3 | 3 | 3 | 3 | 13 |
| 2 | 3 | $\mathrm{X}$ | 0 | 1 | 3 | 3 | 3 | 13 |
| 3 | 1 | 3 | $\mathrm{X}$ | 0 | 3 | 3 | 3 | 13 |
| 4 | 0 | 1 | 3 | $\mathrm{X}$ | 3 | 3 | 3 | 13 |
| 5 | 0 | 0 | 0 | 0 | $\mathrm{X}$ | 1 | 1 | 2 |
| 6 | 0 | 0 | 0 | 0 | 1 | $\mathrm{X}$ | 1 | 2 |
| 7 | 0 | 0 | 0 | 0 | 1 | 1 | $\mathrm{X}$ | 2 |
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. For what least natural $k$ is the expression $2017 \cdot 2018 \cdot 2019 \cdot 2020+k$ a square of a natural number?
|
Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2018$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1\right)+1=\left(n^{2}+n-1\right)^{2}
\end{aligned}
$$
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. In a football tournament, seven teams played: each team played once with each other. In the next round, teams that scored twelve or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round?
|
Answer: 5.
Solution. In total, the teams played $\frac{7 \cdot 6}{2}=21$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $21 \cdot 3=63$. Hence, the number of teams $n$ advancing to the next stage satisfies the inequality $n \cdot 12 \leqslant 63$, from which $n \leqslant 5$.
On the other hand, an example of a tournament table can be provided where 5 teams advance to the next round:
| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | Total |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | $\mathrm{X}$ | 0 | 0 | 3 | 3 | 3 | 3 | 12 |
| 2 | 3 | $\mathrm{X}$ | 0 | 0 | 3 | 3 | 3 | 12 |
| 3 | 3 | 3 | $\mathrm{X}$ | 0 | 0 | 3 | 3 | 12 |
| 4 | 0 | 3 | 3 | $\mathrm{X}$ | 0 | 3 | 3 | 12 |
| 5 | 0 | 0 | 3 | 3 | $\mathrm{X}$ | 3 | 3 | 12 |
| 6 | 0 | 0 | 0 | 0 | 0 | $\mathrm{X}$ | 1 | 1 |
| 7 | 0 | 0 | 0 | 0 | 0 | 1 | $\mathrm{X}$ | 1 |
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. For what least natural $k$ is the expression $2019 \cdot 2020 \cdot 2021 \cdot 2022 + k$ a square of a natural number?
|
Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2020$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1\right)+1=\left(n^{2}+n-1\right)^{2}
\end{aligned}
$$
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. What is greater: 1 or $\frac{23}{93}+\frac{41}{165}+\frac{71}{143}$?
|
Answer: One is greater.
## First solution.
$$
\frac{23}{93}+\frac{41}{165}+\frac{71}{143}<\frac{23}{92}+\frac{41}{164}+\frac{71}{142}=\frac{1}{4}+\frac{1}{4}+\frac{1}{2}=1 .
$$
## Second solution.
$$
\begin{aligned}
\frac{23}{93}+\frac{41}{165}+\frac{71}{143} & =\frac{23 \cdot 165 \cdot 143+93 \cdot 41 \cdot 143+93 \cdot 165 \cdot 71}{93 \cdot 165 \cdot 143}= \\
& =\frac{542685+545259+1089495}{2194335}=\frac{2177439}{2194335}\left[=\frac{65983}{66495}\right]<1
\end{aligned}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. In a football tournament, eight teams played: each team played once with each other. In the next round, teams that scored fifteen or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round?
|
Answer: 5.
Solution. In total, the teams played $\frac{8 \cdot 7}{2}=28$ games, in each of which 2 or 3 points were at stake. Therefore, the maximum total number of points that all teams can have is $28 \cdot 3=84$. Hence, the number of teams $n$ that advance to the next stage satisfies the inequality $n \cdot 15 \leqslant 84$, from which $n \leqslant 5$.
On the other hand, an example of a tournament table can be provided in which 5 teams qualify for the next round:
| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | Sum |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | $\mathrm{X}$ | 0 | 0 | 3 | 3 | 3 | 3 | 3 | 15 |
| 2 | 3 | $\mathrm{X}$ | 0 | 0 | 3 | 3 | 3 | 3 | 15 |
| 3 | 3 | 3 | $\mathrm{X}$ | 0 | 0 | 3 | 3 | 3 | 15 |
| 4 | 0 | 3 | 3 | $\mathrm{X}$ | 0 | 3 | 3 | 3 | 15 |
| 5 | 0 | 0 | 3 | 3 | $\mathrm{X}$ | 3 | 3 | 3 | 15 |
| 6 | 0 | 0 | 0 | 0 | 0 | $\mathrm{X}$ | 1 | 1 | 2 |
| 7 | 0 | 0 | 0 | 0 | 0 | 1 | $\mathrm{X}$ | 1 | 2 |
| 8 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | $\mathrm{X}$ | 2 |
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. For what least natural $k$ is the expression $2018 \cdot 2019 \cdot 2020 \cdot 2021+k$ a square of a natural number
|
Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2019$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1\right)+1=\left(n^{2}+n-1\right)^{2}
\end{aligned}
$$
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. In a football tournament, six teams played: each team played once with each other. In the next round, teams that scored twelve or more points qualify. Three points are awarded for a win, one point for a draw, and zero points for a loss. What is the maximum number of teams that can advance to the next round?
|
Answer: 3
Solution. In total, the teams played $\frac{6 \cdot 5}{2}=15$ games, each of which awarded 2 or 3 points. Therefore, the maximum total number of points that all teams can have is $15 \cdot 3=45$. Hence, the number of teams $n$ that advance to the next stage satisfies the inequality $n \cdot 12 \leqslant 45$, from which $n \leqslant 3$.
On the other hand, an example of a tournament table can be provided where 3 teams advance to the next round:
| | 1 | 2 | 3 | 4 | 5 | 6 | Sum |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | $\mathrm{X}$ | 0 | 3 | 3 | 3 | 3 | 12 |
| 2 | 3 | $\mathrm{X}$ | 0 | 3 | 3 | 3 | 12 |
| 3 | 0 | 3 | $\mathrm{X}$ | 3 | 3 | 3 | 12 |
| 4 | 0 | 0 | 0 | $\mathrm{X}$ | 1 | 1 | 2 |
| 5 | 0 | 0 | 0 | 1 | $\mathrm{X}$ | 1 | 2 |
| 6 | 0 | 0 | 0 | 1 | 1 | $\mathrm{X}$ | 2 |
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. For what least natural $k$ is the expression $2016 \cdot 2017 \cdot 2018 \cdot 2019 + k$ a square of a natural number?
|
Answer: 1.
Solution. We will prove that $k=1$ already works. Let $n=2017$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1\right)+1=\left(n^{2}+n-1\right)^{2}
\end{aligned}
$$
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Dasha wrote the numbers $9,10,11, \ldots, 22$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the remaining numbers on the board?
|
Answer: 203.
Solution. The sum of numbers from 9 to 22 is 217. If at least one number is erased, the sum of the remaining numbers does not exceed 208. Let's sequentially consider the options:
- if the sum is 208, then Dasha could have erased only the number 9; then the remaining numbers can be divided into two groups with a sum of 104:
$$
22+21+20+19+12+10=18+17+16+15+14+13+11
$$
- if the sum is 207, then Dasha could have erased only the number 10; then the remaining numbers can be divided into three groups with a sum of 69:
$$
22+21+17+9=20+19+18+12=16+15+14+13+11
$$
- if the sum is 206, then Dasha could have erased only the number 11; then the remaining numbers can be divided into two groups with a sum of 103:
$$
22+21+20+19+12+9=18+17+16+15+14+13+10
$$
- if the sum is 205, then Dasha could have erased only the number 12; then the remaining numbers can be divided into five groups with a sum of 41:
$$
22+19=21+20=18+13+10=17+15+9=16+14+11
$$
- if the sum is 204, then Dasha could have erased only the number 13; then the remaining numbers can be divided into two groups with a sum of 102:
$$
22+21+20+19+11+9=18+17+16+15+14+12+10
$$
- if Dasha erased the number 14, then the numbers left on the board have a sum of 203: they could be divided into 7 groups with a sum of 29, or 29 groups with a sum of 7, or 203 groups with a sum of 1; one of these groups will contain the number 22; since we have no options with a sum of 22, at least one more number will be added to this group; therefore, the sum in this group will be at least 31; thus, in this case, it is impossible to divide the numbers into groups with the same sum.
|
203
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. In the surgical department, there are 4 operating rooms: I, II, III, and IV. In the morning, they were all empty. At some point, an operation began in operating room I, then after some time in operating room II, then after some more time in operating room III, and finally in operating room IV.
All four operations ended simultaneously, and the total duration of all operations was 2 hours and 32 minutes. Thirty minutes before the completion of all operations, the total duration of the ongoing operations was 52 minutes, and ten minutes before that, it was 30 minutes. Which operating rooms' operation durations can be determined from these data, and which cannot?
|
Answer: Only the duration of the operation in Operating Room IV can be determined.
Solution. First, let's prove that the durations of the operations in Operating Rooms I, II, and III cannot be determined uniquely. Indeed, it is not difficult to verify that if the durations of the operations are 70, 39, 33, 10 or 56, 54, 32, 10 minutes, then all conditions of the problem are satisfied. However, in these two variants, the durations of the operations in Operating Rooms I, II, and III are different.
Now let's prove that the duration of the operation in Operating Room IV can be uniquely restored. For this, let's note that the total duration of the operations 40 and 30 minutes before the end of the operations increased by 22 minutes. This means that 30 minutes before the end of the operations, the operations in Operating Rooms I, II, and III were already in progress, otherwise, the total duration would have increased by no more than 20 minutes. Then, by the end of all operations, their total duration is $52 + 30 \cdot 3 = 142$ minutes. Therefore, the operation in Operating Room IV lasted $152 - 142 = 10$ minutes.
Comment. The durations of the operations can be (in minutes) $(70; 36+s; 36-s; 10)$, where $s \in [0; 4]$ or $(55+t; 55-t; 32; 10)$, where $t \in [0; 15]$.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{43} \cdot \operatorname{tg} \frac{2 \pi}{43}+\operatorname{tg} \frac{2 \pi}{43} \cdot \operatorname{tg} \frac{3 \pi}{43}+\ldots+\operatorname{tg} \frac{k \pi}{43} \cdot \operatorname{tg} \frac{(k+1) \pi}{43}+\ldots+\operatorname{tg} \frac{2019 \pi}{43} \cdot \operatorname{tg} \frac{2020 \pi}{43}
$$
|
Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{\operatorname{tg}(\alpha-\beta)}-1
$$
Then
$$
\operatorname{tg} \frac{k \pi}{43} \cdot \operatorname{tg} \frac{(k+1) \pi}{43}=\frac{\operatorname{tg} \frac{(k+1) \pi}{43}-\operatorname{tg} \frac{k \pi}{43}}{\operatorname{tg}\left(\frac{(k+1) \pi}{43}-\frac{k \pi}{43}\right)}-1=\left(\operatorname{tg} \frac{(k+1) \pi}{43}-\operatorname{tg} \frac{k \pi}{43}\right) \cdot\left(\operatorname{tg} \frac{\pi}{43}\right)^{-1}-1
$$
Adding these equalities for all $k$ from 1 to 2019, we get that the expression in the condition equals
$$
\left(\operatorname{tg} \frac{2020 \pi}{43}-\operatorname{tg} \frac{\pi}{43}\right) \cdot\left(\operatorname{tg} \frac{\pi}{43}\right)^{-1}-2019
$$
Notice that
$$
\operatorname{tg} \frac{2020 \pi}{43}=\operatorname{tg}\left(47 \pi-\frac{\pi}{43}\right)=-\operatorname{tg} \frac{\pi}{43}
$$
which means $(*)$ equals -2021.
|
-2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what greatest value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-2 x+a x^{2}\right)^{8}$ be equal to -1540?
|
Answer: -5.
Solution. Applying the polynomial formula, we get
$$
\left(1-2 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-2 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-2)^{n_{2}} \cdot a^{n_{3}} \cdot x^{n_{2}+2 n_{3}}
$$
To determine which terms in the sum contain $x^{4}$, we need to solve the system of equations in non-negative integers:
$$
\left\{\begin{array}{l}
n_{1}+n_{2}+n_{3}=8 \\
n_{2}+2 n_{3}=4
\end{array}\right.
$$
From the second equation, it follows that $n_{2}$ is even. Due to the non-negativity of the variables, $n_{2}$ can take the values 0, 2, and 4. Solving the system for each of these $n_{2}$, we will have three cases:
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. Given an isosceles triangle $K L M(K L=L M)$ with the angle at the vertex equal to $114^{\circ}$. Point $O$ is located inside triangle $K L M$ such that $\angle O M K=30^{\circ}$, and $\angle O K M=27^{\circ}$. Find the measure of angle $\angle L O M$.
|
Answer: $150^{\circ}$.
Solution. Let $L H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $M O$ and segment $L H$. Note that $K S=S M$. For example, since in triangle $K S M$ the median SH coincides with the height.
Let's calculate the angles:
|
150
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Sasha wrote the numbers $7, 8, 9, \ldots, 17$ on the board, and then erased one or several of them. It turned out that the remaining numbers on the board cannot be divided into several groups such that the sums of the numbers in the groups are equal. What is the greatest possible value of the sum of the remaining numbers on the board?
|
Answer: 121.
Solution. The sum of numbers from 7 to 17 is 132. If at least one number is erased, the sum of the remaining numbers does not exceed 125. Let's sequentially consider the options:
- if the sum is 125, then Sasha could have erased only the number 7; then the remaining numbers can be divided into five groups with a sum of 25:
$$
8+17=9+16=10+15=11+14=12+13
$$
- if the sum is 124, then Sasha could have erased only the number 8; then the remaining numbers can be divided into two groups with a sum of 62:
$$
17+16+15+14=13+12+11+10+9+7
$$
- if the sum is 123, then Sasha could have erased only the number 9; then the remaining numbers can be divided into three groups with a sum of 41:
$$
17+16+8=15+14+12=13+11+10+7
$$
- if the sum is 122, then Sasha could have erased only the number 10; then the remaining numbers can be divided into two groups with a sum of 61:
$$
17+16+15+13=14+12+11+9+8+7
$$
- if Sasha erased the number 11, then the numbers left on the board have a sum of 121: they could be divided either into 11 groups with a sum of 11, or into 121 groups with a sum of 1; but some group will include the number 17, so the sum in this group will be at least 17; therefore, in this case, it is impossible to divide the numbers into groups with the same sum.
|
121
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. In the surgical department, there are 4 operating rooms: 1, 2, 3, and 4. In the morning, they were all empty. At some point, a surgery began in operating room 1, after some time - in operating room 2, then after some more time - in operating room 3, and finally in operating room 4.
All four surgeries ended simultaneously, and the total duration of all surgeries was 2 hours and 7 minutes. 18 minutes before the completion of all surgeries, the total duration of the ongoing surgeries was 60 minutes, and 15 minutes before that - 25 minutes. Which operating rooms' surgery durations can be determined from these data, and which cannot?
|
Answer: Only the duration of the operation in operating room 4 can be determined.
Solution. First, let's prove that the durations of operations in operating rooms 1, 2, and 3 cannot be determined uniquely. Indeed, it is not difficult to verify that if the durations of the operations are 58, 29, 27, 13 minutes or 46, 45, 23, 13 minutes, then all conditions of the problem are satisfied. However, in these two cases, the durations of the operations in operating rooms 1, 2, and 3 are different.
Now let's prove that the duration of the operation in operating room 4 can be uniquely restored. For this, let's note that the total duration of the operations 33 and 18 minutes before the end of the operations increased by 35 minutes. This means that 18 minutes before the end of the operations, the operations in operating rooms 1, 2, and 3 were already in progress, otherwise, the total duration would have increased by no more than 30 minutes. Then, by the end of all operations, their total duration is \(60 + 18 \cdot 3 = 114\) minutes. Therefore, the operation in operating room 4 lasted \(127 - 114 = 13\) minutes.
Comment. The durations of the operations can be \(58; 28+s; 28-s; 13\) minutes, where \(s \in [0; 5]\) or \((45.5+t; 45.5-t; 23; 13)\), where \(t \in [0; 12.5]\).
|
13
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{47} \cdot \operatorname{tg} \frac{2 \pi}{47}+\operatorname{tg} \frac{2 \pi}{47} \cdot \operatorname{tg} \frac{3 \pi}{47}+\ldots+\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}+\ldots+\operatorname{tg} \frac{2021 \pi}{47} \cdot \operatorname{tg} \frac{2022 \pi}{47}
$$
|
Answer: -2021.
Solution. From the formula
$$
\operatorname{tg}(\alpha-\beta)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}
$$
we express the product of tangents:
$$
\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{\operatorname{tg}(\alpha-\beta)}-1
$$
Then
$$
\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}=\frac{\operatorname{tg} \frac{(k+1) \pi}{47}-\operatorname{tg} \frac{k \pi}{47}}{\operatorname{tg}\left(\frac{(k+1) \pi}{47}-\frac{k \pi}{47}\right)}-1=\left(\operatorname{tg} \frac{(k+1) \pi}{47}-\operatorname{tg} \frac{k \pi}{47}\right) \cdot\left(\operatorname{tg} \frac{\pi}{47}\right)^{-1}-1
$$
Adding these equalities for all $k$ from 1 to 2021, we get that the expression in the condition equals
$$
\left(\operatorname{tg} \frac{2022 \pi}{47}-\operatorname{tg} \frac{\pi}{47}\right) \cdot\left(\operatorname{tg} \frac{\pi}{47}\right)^{-1}-2021
$$
Notice that
$$
\operatorname{tg} \frac{2022 \pi}{47}=\operatorname{tg}\left(43 \pi+\frac{\pi}{47}\right)=\operatorname{tg} \frac{\pi}{47}
$$
which means ( ${ }^{*}$ ) equals -2021.
|
-2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what greatest value of the parameter $a$ will the coefficient of $x^{4}$ in the expansion of the polynomial $\left(1-3 x+a x^{2}\right)^{8}$ be equal to $70$?
|
Answer: -4.
Solution. Applying the polynomial formula, we get
$$
\left(1-3 x+a x^{2}\right)^{8}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot 1^{n_{1}} \cdot(-3 x)^{n_{2}} \cdot\left(a x^{2}\right)^{n_{3}}=\sum_{n_{1}+n_{2}+n_{3}=8} \frac{8!}{n_{1}!\cdot n_{2}!\cdot n_{3}!} \cdot(-3)^{n_{2}} \cdot a^{n_{3}} \cdot x^{n_{2}+2 n_{3}}
$$
To determine which terms in the sum contain $x^{4}$, we need to solve the system of equations in non-negative integers:
$$
\left\{\begin{array}{l}
n_{1}+n_{2}+n_{3}=8 \\
n_{2}+2 n_{3}=4
\end{array}\right.
$$
From the second equation, it follows that $n_{2}$ is even. Due to the non-negativity of the variables, $n_{2}$ can take the values 0, 2, and 4. Solving the system for each of these $n_{2}$, we will have three cases:
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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