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Problem 2. Masha wrote the numbers $4,5,6, \ldots, 16$ 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 4 to 16 is 130. If at least one number is erased, the sum of the remaining numbers does not exceed 126. Let's sequentially consider the options:
- if the sum is 126, then Masha could have erased only the number 4; then the remaining numbers can be divided into two groups with a sum of 63:
$$
16+15+14+13+5=12+11+10+9+8+7+6
$$
- if the sum is 125, then Masha could have erased only the number 5; then the remaining numbers can be divided into five groups with a sum of 25:
$$
16+9=15+10=14+11=13+12=8+7+6+4
$$
- if the sum is 124, then Masha could have erased only the number 6; then the remaining numbers can be divided into two groups with a sum of 62:
$$
16+15+14+13+4=12+11+10+9+8+7+5
$$
- if the sum is 123, then Masha could have erased only the number 7; then the remaining numbers can be divided into three groups with a sum of 41:
$$
16+15+10=14+13+9+5=12+11+8+6+4
$$
- if the sum is 122, then Masha could have erased only the number 8; then the remaining numbers can be divided into two groups with a sum of 61:
$$
16+15+14+12+4=13+11+10+9+7+6+5
$$
- if Masha erased the number 9, 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 16, so the sum in this group will be at least 16; 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
|
Problem 3. In a surgical department, there are 4 operating rooms: A, B, V, and G. In the morning, they were all empty. At some point, an operation began in operating room A, after some time - in operating room B, then after some more time - in V, and then in $\Gamma$.
All four operations ended simultaneously, and the total duration of all operations was 3 hours and 5 minutes. 36 minutes before the completion of all operations, the total duration of the ongoing operations was 46 minutes, and 10 minutes before that - 19 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 $\Gamma$ can be determined.
Solution. First, let's prove that the durations of operations in operating rooms A, B, and C cannot be determined uniquely. Indeed, it is easy to verify that if the durations of the operations are $65, 45, 44, 31$ or $56, 55, 43, 31$ minutes, then all conditions of the problem are satisfied. However, in these two cases, the durations of the operations in operating rooms A, B, and C are different.
Now let's prove that the duration of the operation in operating room Г can be uniquely determined. For this, let's note that the total duration of the operations 46 and 36 minutes before the end of the operations increased by 27 minutes. This means that 36 minutes before the end of the operations, the operations in operating rooms A, B, and C 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 $46 + 36 \cdot 3 = 154$ minutes. Therefore, the operation in operating room Г lasted $185 - 154 = 31$ minutes.
Comment. The durations of the operations can be ( $65 ; 44.5+s ; 44.5-s ; 31$ ), where $s \in [0 ; 1.5]$ or ( $55.5+t ; 55.5-t ; 43 ; 31$ ), where $t \in [0 ; 9.5]$.
|
31
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{43} \cdot \operatorname{tg} \frac{2 \pi}{43}+\operatorname{tg} \frac{2 \pi}{43} \cdot \operatorname{tg} \frac{3 \pi}{43}+\ldots+\operatorname{tg} \frac{k \pi}{43} \cdot \operatorname{tg} \frac{(k+1) \pi}{43}+\ldots+\operatorname{tg} \frac{2021 \pi}{43} \cdot \operatorname{tg} \frac{2022 \pi}{43}
$$
|
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 2021, we get that the expression in the condition equals
$$
\left(\operatorname{tg} \frac{2022 \pi}{43}-\operatorname{tg} \frac{\pi}{43}\right) \cdot\left(\operatorname{tg} \frac{\pi}{43}\right)^{-1}-2021
$$
Notice that
$$
\operatorname{tg} \frac{2022 \pi}{43}=\operatorname{tg}\left(47 \pi+\frac{\pi}{43}\right)=\operatorname{tg} \frac{\pi}{43}
$$
which means $(*)$ equals -2021.
|
-2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what least 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: -19.
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:
|
-19
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. Given an isosceles triangle $A B C(A B=B C)$ with the angle at the vertex equal to $102^{\circ}$. Point $O$ is located inside triangle $A B C$ such that $\angle O C A=30^{\circ}$, and $\angle O A C=21^{\circ}$. Find the measure of angle $\angle B O A$.
|
Answer: $81^{\circ}$.
Solution. Let $B H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $C O$ and segment $B H$. Note that $A S=S C$. For example, in triangle $A S C$, the median $S H$ coincides with the height.
Let's calculate the angles:
|
81
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Pasha wrote the numbers $4,5,6, \ldots, 14$ 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: 91.
Solution. The sum of the numbers from 4 to 14 is 99. If at least one number is erased, the sum of the remaining numbers does not exceed 95. Let's sequentially consider the options:
- if the sum is 95, then Pasha could have erased only the number 4; then the remaining numbers can be divided into five groups with a sum of 19:
$$
14+5=13+6=12+7=11+8=10+9
$$
- if the sum is 94, then Pasha could have erased only the number 5; then the remaining numbers can be divided into two groups with a sum of 47:
$$
14+13+12+8=11+10+9+7+6+4.
$$
- if the sum is 93, then Pasha could have erased only the number 6; then the remaining numbers can be divided into three groups with a sum of 31:
$$
14+13+4=12+11+8=10+9+7+5
$$
- if the sum is 92, then Pasha could have erased only the number 7; then the remaining numbers can be divided into two groups with a sum of 46:
$$
14+13+11+8=12+10+9+6+5+4.
$$
- if Pasha erased the number 8, then the numbers left on the board have a sum of 91: they could be divided either into 7 groups with a sum of 13, or into 13 groups with a sum of 7, or into 91 groups with a sum of 1; but some group will include the number 14, so the sum in this group will be at least 14; therefore, in this case, it is impossible to divide the numbers into groups with the same sum.
|
91
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. In a surgical department, there are 4 operating rooms: A, B, C, and D. In the morning, they were all empty. At some point, a surgery began in operating room A, after some time - in operating room B, then after some more time - in C, and finally in D.
All four surgeries ended simultaneously, and the total duration of all surgeries was 2 hours and 38 minutes. Twenty-four minutes before the completion of all surgeries, the total duration of the ongoing surgeries was 1 hour and 9 minutes, and fifteen minutes before that, it was 33 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 D can be determined.
Solution. First, let's prove that the durations of the operations in operating rooms A, B, and C cannot be determined uniquely. Indeed, it is not difficult to verify that if the durations of the operations are $72, 35, 34, 17$ or $56, 55, 30, 17$ minutes, then all conditions of the problem are satisfied. However, in these two cases, the durations of the operations in operating rooms A, B, and C are different.
Now let's prove that the duration of the operation in operating room D can be uniquely restored. For this, let's note that the total duration of the operations 39 and 24 minutes before the end of the operations increased by 36 minutes. This means that 24 minutes before the end of the operations, the operations in operating rooms A, B, and C 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 $69 + 24 \cdot 3 = 141$ minutes. Therefore, the operation in operating room D lasted $158 - 141 = 17$ minutes.
Comment. The durations of the operations can be $(72; 35.5 + s; 35.5 - s; 17)$, where $s \in [0; 3.5]$ or $(55.5 + t; 55.5 - t; 30; 17)$, where $t \in [0; 16.5]$.
|
17
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Calculate
$$
\operatorname{tg} \frac{\pi}{47} \cdot \operatorname{tg} \frac{2 \pi}{47}+\operatorname{tg} \frac{2 \pi}{47} \cdot \operatorname{tg} \frac{3 \pi}{47}+\ldots+\operatorname{tg} \frac{k \pi}{47} \cdot \operatorname{tg} \frac{(k+1) \pi}{47}+\ldots+\operatorname{tg} \frac{2019 \pi}{47} \cdot \operatorname{tg} \frac{2020 \pi}{47}
$$
|
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 2019, we get that the expression in the condition equals
$$
\left(\operatorname{tg} \frac{2020 \pi}{47}-\operatorname{tg} \frac{\pi}{47}\right) \cdot\left(\operatorname{tg} \frac{\pi}{47}\right)^{-1}-2019
$$
Notice that
$$
\operatorname{tg} \frac{2020 \pi}{47}=\operatorname{tg}\left(43 \pi-\frac{\pi}{47}\right)=-\operatorname{tg} \frac{\pi}{47}
$$
which means $(*)$ equals -2021.
|
-2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what least 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: -50.
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:
|
-50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. Given an isosceles triangle $X Y Z(X Y=Y Z)$ with the angle at the vertex equal to $96^{\circ}$. Point $O$ is located inside triangle $X Y Z$ such that $\angle O Z X=30^{\circ}$, and $\angle O X Z=18^{\circ}$. Find the measure of the angle $\angle Y O X$.
|
Answer: $78^{\circ}$.
Solution. Let $Y H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $Z O$ and segment $Y H$. Note that $X S=S Z$. For example, since in triangle $X S Z$ the median $S H$ coincides with the height.
Let's calculate the angles:
|
78
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. The number 27 is written on the board. Every minute, the number is erased from the board and replaced with the product of its digits, increased by 12. For example, after one minute, the number on the board will be $2 \cdot 7+12=26$. What will be on the board after an hour?
|
# Answer: 14
Solution. Let's find the next few numbers that will appear on the board. After 26, it will be 24, then $20, 12, 14, 16, 18$ and again 20. Notice that the sequence has looped and each subsequent number will coincide with the one that is 5 positions earlier. Therefore, after an hour, that is, 60 minutes, the number written will be the same as the one written 55 minutes earlier, which in turn coincides with the one written 50 minutes earlier, and so on, down to the one written 5 minutes earlier, which is the number 14.
|
14
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. Angelica wants to choose a three-digit code for her suitcase. To make it easier to remember, Angelica wants all the digits in her code to be in non-decreasing order. How many different options does Angelica have for choosing the code?
|
Problem 2. Angelica wants to choose a three-digit code for her suitcase. To make it easier to remember, Angelica wants all the digits in her code to be in non-decreasing order. How many different ways can Angelica choose her code? Answer. 220.
|
220
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (Option 1) The decimal representation of the natural number $n$ contains sixty-three digits. Among these digits, there are twos, threes, and fours. No other digits are present. The number of twos is 22 more than the number of fours. Find the remainder when $n$ is divided by 9.
|
# Answer. 5.
(Option 2) The decimal representation of a natural number $n$ contains sixty-one digits. Among these digits, there are threes, fours, and fives. No other digits are present. The number of threes is 11 more than the number of fives. Find the remainder when $n$ is divided by 9.
Answer. 8.
Criteria. "干" The correct answer is obtained by considering a specific case.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 200 mushrooms (it is possible that some of them did not bring any mushrooms home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
|
Answer: 21.
Solution. First, let's prove that when $n \leqslant 20$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $0,1, \ldots, n-2$ mushrooms, and the $n$-th collected all the rest. Since
$$
0+1+\ldots+(n-2) \leqslant 0+1+\ldots+18=171=200-29
$$
the last mushroom picker collected at least 29 mushrooms, i.e., more than each of the others. Thus, when $n \leqslant 20$, there exists an example where Petya could be wrong.
Let's show that when $n=21$, Petya will always be right. Suppose he is wrong and let the mushroom pickers have collected $a_{0}<a_{1}<\ldots<a_{20}$ mushrooms. It is easy to see that $a_{i} \geqslant i$, hence
$$
200=a_{0}+a_{1}+\ldots+a_{20} \geqslant 0+1+\ldots+20=210
$$
a contradiction.
|
21
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. A circle is inscribed in trapezoid $ABCD$, touching the lateral side $AD$ at point $K$. Find the area of the trapezoid if $AK=16, DK=4$ and $CD=6$.
|
Answer: 432.
Solution. Let $L, M, N$ be the points of tangency of the inscribed circle with the sides $BC, AB, CD$ respectively; let $I$ be the center of the inscribed circle. Denote the radius of the circle by $r$. Immediately note that $DN = DK = 4$ (the first equality follows from the equality of the segments of tangents), from which $CL = CN = CD - DN = 2$ (the first equality follows from the equality of the segments of tangents, the second is obvious).
Since $I$ is the point of intersection of the bisectors of the internal angles of the trapezoid, then $\angle IAD + \angle IDA = (\angle DAB + \angle ADC) / 2 = 180^\circ / 2 = 90^\circ$, where the penultimate equality follows from the parallelism of the lines $AB$ and $CD$. Therefore, triangle $AID$ is a right triangle and $\angle AID = 90^\circ$. Similarly, triangle $BIC$ is also a right triangle.
Next, since $IK$ and $IL$ are radii drawn to the points of tangency, then $\angle IKD = 90^\circ$ and $\angle ILB = 90^\circ$. Therefore, $IK$ and $IL$ are altitudes in triangles $AID$ and $BIC$ respectively. Using the known fact that in a right triangle, the square of the altitude dropped to the hypotenuse equals the product of the segments into which it divides the hypotenuse. Then
$$
IK^2 = AK \cdot KD = 16 \cdot 4 = 64 = 8^2
$$
i.e., $r = IK = 8$, and also $8^2 = IL^2 = CL \cdot LB = 2 \cdot LB$, i.e., $LB = 32$.[^0]
Now we have everything to find the area. Note that $MN$ is the height of the trapezoid and $MN = 2r = 16, AB + CD = (AM + MB) + 6 = (AK + BL) + 6 = 16 + 32 + 6 = 54$, from which the answer $\frac{16 \cdot 54}{2} = 432$.
|
432
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $40^{\circ}$. Find the other two angles of triangle $A B C$.
|
Answer: $60^{\circ}$ and $80^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}$. We aim to prove that triangle $H B C_{1}$ is equilateral. Then $\angle B A C = \angle B C_{1} C = 60^{\circ}$, from which the answer follows given the condition.
First, note that $H$ is the intersection of the angle bisectors of triangle $A_{1} B_{1} C_{1}$. Indeed, for example, $\angle B_{1} C_{1} C = \angle B_{1} B C = 90^{\circ} - \angle C = \angle C A A_{1} = \angle C C_{1} A_{1}$, so $C_{1} H$ is the angle bisector of $\angle A_{1} C_{1} B_{1}$; similarly, it can be shown that $A_{1} H$ and $B_{1} H$ are also angle bisectors of the corresponding angles. Therefore, $H$ is the incenter of triangle $A_{1} B_{1} C_{1}$ and $H K = H L$. Moreover, it has been shown that $\angle H C_{1} L = \angle H B K$, so the right triangles $H C_{1} L$ and $H B K$ are congruent by a leg and an acute angle. Therefore, $H C_{1} = H B$.
It remains to note that $H B = B C_{1}$. This fact is well-known and can be proven in various ways. Here, we will present just one of them. Note that triangles $A H B$ and $A C_{1} B$ are congruent by a side (common side $A B$) and two angles ($\angle H A B = \angle A_{1} B_{1} B = \angle C_{1} B_{1} B = \angle C_{1} A B$; similarly, it can be shown that $\angle H B A = \angle C_{1} B A$).
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. For what values of the parameter $a$ does the equation
$$
3^{x^{2}-2 a x+a^{2}}=a x^{2}-2 a^{2} x+a^{3}+a^{2}-4 a+4
$$
have exactly one solution?
|
Answer: Only when $a=1$.
Solution. Let's denote $x-a$ by $t$. Note that the number of solutions of the equation does not change with such a substitution. Then the original equation will take the form
$$
3^{t^{2}}=a t^{2}+a^{2}-4 a+4.
$$
Notice that the expressions on both sides do not change when $t$ is replaced by $-t$, so the equation can have an odd number of solutions (in particular, exactly one solution) only if $t=0$ is its root:
$$
3^{0}=a \cdot 0+a^{2}-4 a+4
$$
i.e., $a^{2}-4 a+3=0$, from which $a=3$ or $a=1$. Thus, besides these two numbers, no other values of the parameter $a$ can satisfy the condition.
Let $a=1$. Then the equation will take the form $3^{t^{2}}=t^{2}+1$. Note that $3^{x}>x \ln 3+1$ for $x>0$ (which can be proven, for example, by taking the derivatives of both sides). Then for $t \neq 0$ we get $3^{t^{2}}>t^{2} \ln 3+1>t^{2}+1$. Therefore, when $a=1$, the equation has a unique solution.
Let $a=3$. Then the equation will take the form $3^{t^{2}}=3 t^{2}+1$. Note that $3^{1}=3,3 \cdot 1+1=4$, but $3^{2^{2}}=81$, $3 \cdot 2^{2}+1=13$, i.e., at $t=1$ the left side is less than the right, and at $t=2$ the opposite is true. Therefore, by the intermediate value theorem, the equation has at least one more root in the interval $(1,2)$. Therefore, $a=3$ does not satisfy the condition.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. In the decimal representation of an even number $M$, only the digits $0, 2, 4, 5, 7$, and 9 are used, and digits can repeat. It is known that the sum of the digits of the number $2M$ is 35, and the sum of the digits of the number $M / 2$ is 29. What values can the sum of the digits of the number $M$ take? List all possible answers.
|
# Answer: 31.
Solution. Let's denote the sum of the digits of a natural number $n$ by $S(n)$. Notice the following facts, each of which is easy to verify if you add numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding $n$ and $n$.
Lemma 2. Let $n$ be a natural number. Then the number of digits in the number $n$ that are greater than or equal to 5 is equal to the number of carries when adding $n$ and $n$.
Lemma 3. Let $n$ and $m$ be natural numbers. Then $S(n+m)=S(n)+S(m)-9k$, where $k$ is the number of carries when adding $n$ and $m$.
Let $N$ be the number of odd digits in the number $M$; according to the condition, $N$ is the number of digits in the number $M$ that are greater than or equal to 5. Notice that then, by Lemma 1, when adding $M/2$ and $M/2$, there were exactly $N$ carries, from which, by Lemma 3, we have $S(M)=2S(M/2)-9N=58-9N$. By Lemma 2, when adding $M$ and $M$, there were also $N$ carries, from which, again by Lemma 3, we have $35=S(2M)=2S(M)-9N$.
Thus, $S(M)=58-9N, 2S(M)=35+9N$, from which $3S(M)=93, S(M)=31$.
|
31
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=1: 2, B N: B B_{1}=1: 3, C K: C C_{1}=1: 4$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K P$, if the volume of the prism is 16.
|
Answer: 4.
Solution. Suppose we have found the position of point $P$ such that the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, respectively. Note that
$V_{M N K P}=\frac{1}{3} V_{M N K M_{1} N_{1} K_{1}}$. Draw planes $\beta$ and $\beta_{1}$ through points $M$ and $M_{1}$, parallel to the plane $A B C$, and call $R$ and $R_{1}$ the points of intersection with edge $B B_{1}$, and $S$ and $S_{1}$ with edge $C C_{1}$. Note that the figures $M N K R S$ and $M_{1} N_{1} K_{1} R_{1} S_{1}$ are obtained from each other by a parallel translation, and thus are equal, and their volumes are also equal. Therefore, the volumes of prisms $M N K M_{1} N_{1} K_{1}$ and $M R S M_{1} R_{1} S_{1}$ are also equal. But $V_{M R S M_{1} R_{1} S_{1}}=\frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$, from which we get that $V_{M N K P}=\frac{1}{3} \frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$.
We need to find the position of plane $\alpha$ such that $M M_{1}$ is maximized. Note that at least one of the points $M_{1}, N_{1}, K_{1}$ lies within the original prism, from which $M M_{1}=N N_{1}=K K_{1} \leq \max \left\{A M, A_{1} M, B N, B_{1} N, C K, C_{1} K\right\}$. Substituting the given ratios in the problem, we finally get that $M M_{1}=N N_{1}=K K_{1}=C_{1} K=\frac{3}{4} C C_{1}$, from which $V_{M N K P}=\frac{1}{3} \frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}=\frac{1}{3} \frac{3}{4} 16=4$.
## Variant II
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 338 mushrooms (it is possible that some of them did not bring any mushrooms home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
|
Answer: 27.
Solution. First, let's prove that when $n \leqslant 26$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $0,1, \ldots, n-2$ mushrooms, and the $n$-th collected all the rest. Since
$$
0+1+\ldots+(n-2) \leqslant 0+1+\ldots+24=300=338-38
$$
the last mushroom picker collected at least 38 mushrooms, i.e., more than each of the others. Thus, when $n \leqslant 26$, there exists an example where Petya could be wrong.
Let's show that when $n=27$, Petya will always be right. Suppose he is wrong and let the mushroom pickers have collected $a_{0}<a_{1}<\ldots<a_{26}$ mushrooms. It is easy to see that $a_{i} \geqslant i$, hence
$$
338=a_{0}+a_{1}+\ldots+a_{25} \geqslant 0+1+\ldots+25=351
$$
a contradiction
|
27
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. A circle with radius 4 is inscribed in trapezoid $ABCD$, touching the base $AB$ at point $M$. Find the area of the trapezoid if $BM=16$ and $CD=3$.
|
Answer: 108.
Solution. Let $K, L, N$ be the points of tangency of the inscribed circle with the sides $AD, BC, CD$ respectively; let $I$ be the center of the inscribed circle. Immediately note that $BL = BM = 16$.
Since $I$ is the intersection point of the angle bisectors of the internal angles of the trapezoid, then $\angle IAD + \angle IDA = (\angle DAB + \angle ADC) / 2 = 180^\circ / 2 = 90^\circ$, where the penultimate equality follows from the parallelism of lines $AB$ and $CD$. Therefore, triangle $AID$ is a right triangle and $\angle AID = 90^\circ$. Similarly, triangle $BIC$ is also a right triangle.
Next, since $IK$ and $IL$ are radii drawn to the points of tangency, then $\angle IKD = 90^\circ$ and $\angle ILB = 90^\circ$. Therefore, $IK$ and $IL$ are altitudes in triangles $AID$ and $BIC$ respectively. Using the known fact that in a right triangle, the square of the altitude dropped to the hypotenuse equals the product of the segments into which it divides the hypotenuse. Then
$$
4^2 = IL^2 = CL \cdot LB = CL \cdot 16
$$
i.e., $CL = 1$. By the equality of tangent segments, we have $CN = CL = 1$, hence $DK = DN = CD - CN = 3 - 1 = 2$. In the right triangle $AID$, we get $4^2 = IK^2 = AK \cdot KD = AK \cdot 2$, i.e., $AK = 8$.
Now we have everything to find the area. Note that $LM$ is the height of the trapezoid and $LM = 2r = 8, AB + CD = AM + MB + CD = 8 + 16 + 3 = 27$, hence the answer $\frac{8 \cdot 27}{2} = 108$.
|
108
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $50^{\circ}$. Find the other two angles of triangle $A B C$.
|
Answer: $60^{\circ}$ and $70^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}$. We aim to prove that triangle $H B C_{1}$ is equilateral. Then $\angle B A C = \angle B C_{1} C = 60^{\circ}$, from which the answer follows given the condition.
First, note that $H$ is the intersection of the angle bisectors of triangle $A_{1} B_{1} C_{1}$. Indeed, for example, $\angle B_{1} C_{1} C = \angle B_{1} B C = 90^{\circ} - \angle C = \angle C A A_{1} = \angle C C_{1} A_{1}$, so $C_{1} H$ is the angle bisector of $\angle A_{1} C_{1} B_{1}$; similarly, it can be shown that $A_{1} H$ and $B_{1} H$ are also angle bisectors of the corresponding angles. Therefore, $H$ is the incenter of triangle $A_{1} B_{1} C_{1}$ and $H K = H L$. Moreover, it has been shown that $\angle H C_{1} L = \angle H B K$, so the right triangles $H C_{1} L$ and $H B K$ are congruent by a leg and an acute angle. Therefore, $H C_{1} = H B$.
It remains to note that $H B = B C_{1}$. This fact is well-known and can be proven in various ways. Here, we will present just one of them. Note that triangles $A H B$ and $A C_{1} B$ are congruent by a side (common side $A B$) and two angles ($\angle H A B = \angle A_{1} B_{1} B = \angle C_{1} B_{1} B = \angle C_{1} A B$; similarly, it can be shown that $\angle H B A = \angle C_{1} B A$.
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. For what values of the parameter $a$ does the equation
$$
3^{x^{2}+6 a x+9 a^{2}}=a x^{2}+6 a^{2} x+9 a^{3}+a^{2}-4 a+4
$$
have exactly one solution?
|
Answer: Only when $a=1$.
Solution. Let's denote $x+3a$ by $t$. Note that the number of solutions to the equation does not change with this substitution. Then the original equation will take the form
$$
3^{t^{2}}=a t^{2}+a^{2}-4 a+4
$$
Notice that the expressions on both sides do not change when $t$ is replaced by $-t$, so the equation can have an odd number of solutions (in particular, exactly one solution) only if $t=0$ is a root:
$$
3^{0}=a \cdot 0+a^{2}-4 a+4
$$
i.e., $a^{2}-4 a+3=0$, from which $a=3$ or $a=1$. Thus, besides these two numbers, no other values of the parameter $a$ can satisfy the condition.
Let $a=1$. Then the equation will take the form $3^{t^{2}}=t^{2}+1$. Note that $3^{x}>x \ln 3+1$ for $x>0$ (which can be proven, for example, by taking the derivatives of both sides). Then for $t \neq 0$ we get $3^{t^{2}}>t^{2} \ln 3+1>t^{2}+1$. Therefore, when $a=1$, the equation has a unique solution.
Let $a=3$. Then the equation will take the form $3^{t^{2}}=3 t^{2}+1$. Note that $3^{1}=3, 3 \cdot 1+1=4$, but $3^{2^{2}}=81$, $3 \cdot 2^{2}+1=13$, i.e., at $t=1$ the left side is less than the right, and at $t=2$ the opposite is true. Therefore, by the intermediate value theorem, the equation has at least one more root in the interval $(1,2)$. Therefore, $a=3$ does not satisfy the condition.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. In the decimal representation of an even number $M$, only the digits $0, 2, 4, 5, 7$, and 9 are used, and digits may repeat. It is known that the sum of the digits of the number $2M$ is 31, and the sum of the digits of the number $M / 2$ is 28. What values can the sum of the digits of the number $M$ take? List all possible answers.
|
Answer: 29.
Solution. Let $S(n)$ denote the sum of the digits of a natural number $n$. Notice the following facts, each of which is easy to verify by adding numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding $n$ and $n$.
Lemma 2. Let $n$ be a natural number. Then the number of digits in the number $n$ that are greater than or equal to 5 is equal to the number of carries when adding $n$ and $n$.
Lemma 3. Let $n$ and $m$ be natural numbers. Then $S(n+m)=S(n)+S(m)-9k$, where $k$ is the number of carries when adding $n$ and $m$.
Let $N$ be the number of odd digits in the number $M$; according to the condition, $N$ is the number of digits in the number $M$ that are greater than or equal to 5. Notice that then, by Lemma 1, when adding $M/2$ and $M/2$, there were exactly $N$ carries, from which, by Lemma 3, we have $S(M)=2S(M/2)-9N=56-9N$. By Lemma 2, when adding $M$ and $M$, there were also $N$ carries, from which, again by Lemma 3, we have $31=S(2M)=2S(M)-9N$.
Thus, $S(M)=56-9N, 2S(M)=31+9N$, from which $3S(M)=87, S(M)=29$.
|
29
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=2: 3, B N: B B_{1}=3: 5, C K: C C_{1}=4: 7$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K P$, if the volume of the prism is 27.
|
# Answer: 6.
Solution. Suppose we have found the position of point $P$ such that the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, respectively. Note that $V_{M N K P}=\frac{1}{3} V_{M N K M_{1} N_{1} K_{1}}$. Draw planes $\beta$ and $\beta_{1}$ through points $M$ and $M_{1}$, parallel to the plane $A B C$, and call $R$ and $R_{1}$ the points of intersection with edge $B B_{1}$, and $S$ and $S_{1}$ with edge $C C_{1}$. Note that the figures $M N K R S$ and $M_{1} N_{1} K_{1} R_{1} S_{1}$ are obtained from each other by a parallel translation, and thus are equal, and their volumes are also equal. Therefore, the volumes of prisms $M N K M_{1} N_{1} K_{1}$ and $M R S M_{1} R_{1} S_{1}$ are also equal. But $V_{M R S M_{1} R_{1} S_{1}}=\frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$, from which we get that $V_{M N K P}=\frac{1}{3} \frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$.
We need to find the position of plane $\alpha$ such that $M M_{1}$ is maximized. Note that at least one of the points $M_{1}, N_{1}, K_{1}$ lies within the original prism, from which $M M_{1}=N N_{1}=K K_{1} \leq \max \left\{A M, A_{1} M, B N, B_{1} N, C K, C_{1} K\right\}$. Substituting the given ratios in the problem, we finally get that $M M_{1}=N N_{1}=K K_{1}=A M=\frac{2}{3} A A_{1}$, from which $V_{M N K P}=\frac{1}{3} \frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}=\frac{12}{3} \frac{2}{3} 27=6$.
## Variant III
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. $n$ mushroom pickers went to the forest and brought a total of 450 mushrooms (each brought at least one mushroom home). Boy Petya, upon learning this, said: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
|
Answer: 30.
Solution. First, let's prove that when $n \leqslant 29$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $1, \ldots, n-1$ mushrooms, and the $n$-th collected all the rest. Since
$$
1+\ldots+(n-1) \leqslant 1+\ldots+28=406=450-44
$$
the last mushroom picker collected at least 44 mushrooms, i.e., more than each of the others. Thus, when $n \leqslant 29$, there is an example where Petya could be wrong.
Let's show that when $n=30$, Petya will always be right. Suppose he is wrong and let the mushroom pickers have collected $a_{1}<\ldots<a_{30}$ mushrooms. It is easy to see that $a_{i} \geqslant i$, hence
$$
450=a_{1}+\ldots+a_{30} \geqslant 1+\ldots+30=465
$$
a contradiction
|
30
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $70^{\circ}$. Find the other two angles of triangle $A B C$.
|
Answer: $60^{\circ}$ and $50^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}$. We aim to prove that triangle $H B C_{1}$ is equilateral. Then $\angle B A C = \angle B C_{1} C = 60^{\circ}$, from which the answer follows given the condition.
First, note that $H$ is the intersection of the angle bisectors of triangle $A_{1} B_{1} C_{1}$. Indeed, for example, $\angle B_{1} C_{1} C = \angle B_{1} B C = 90^{\circ} - \angle C = \angle C A A_{1} = \angle C C_{1} A_{1}$, i.e., $C_{1} H$ is the angle bisector of $\angle A_{1} C_{1} B_{1}$; similarly, it can be shown that $A_{1} H$ and $B_{1} H$ are also angle bisectors of the corresponding angles. Therefore, $H$ is the incenter of triangle $A_{1} B_{1} C_{1}$ and $H K = H L$. Moreover, it has been shown that $\angle H C_{1} L = \angle H B K$, i.e., right triangles $H C_{1} L$ and $H B K$ are congruent by a leg and an acute angle. Therefore, $H C_{1} = H B$.
It remains to note that $H B = B C_{1}$. This fact is well-known and can be proven in various ways. Here, we will present just one of them. Note that triangles $A H B$ and $A C_{1} B$ are congruent by a side (common side $A B$) and two angles ($\angle H A B = \angle A_{1} B_{1} B = \angle C_{1} B_{1} B = \angle C_{1} A B$; similarly, it can be shown that $\angle H B A = \angle C_{1} B A$).
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 8. For what values of the parameter $a$ does the equation
$$
5^{x^{2}-6 a x+9 a^{2}}=a x^{2}-6 a^{2} x+9 a^{3}+a^{2}-6 a+6
$$
have exactly one solution?
|
Answer: Only when $a=1$.
Solution. Let's denote $x-3a$ by $t$. Note that the number of solutions to the equation does not change with this substitution. Then the original equation will take the form
$$
5^{t^{2}}=a t^{2}+a^{2}-6a+6
$$
Notice that the expressions on both sides do not change when $t$ is replaced by $-t$, so the equation can have an odd number of solutions (in particular, exactly one solution) only if $t=0$ is a root:
$$
5^{0}=a \cdot 0+a^{2}-6a+6
$$
i.e., $a^{2}-6a+5=0$, from which $a=5$ or $a=1$. Thus, besides these two numbers, no other values of the parameter $a$ can satisfy the condition.
Let $a=1$. Then the equation will take the form $5^{t^{2}}=t^{2}+1$. Note that $5^{x}>x \ln 5+1$ for $x>0$ (which can be proven, for example, by taking the derivatives of both sides). Then for $t \neq 0$ we get $5^{t^{2}}>t^{2} \ln 5+1>t^{2}+1$. Therefore, when $a=1$, the equation has a unique solution.
Let $a=5$. Then the equation will take the form $5^{t^{2}}=5t^{2}+1$. Note that $5^{1}=5, 5 \cdot 1+1=6$, but $5^{2^{2}}=625$, $5 \cdot 2^{2}+1=21$, i.e., at $t=1$ the left side is less than the right, and at $t=2$ the opposite is true. Therefore, by the Intermediate Value Theorem, the equation has at least one more root in the interval $(1,2)$. Therefore, $a=5$ does not satisfy the condition.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. In the decimal representation of an even number $M$, only the digits $0, 2, 4, 5, 7$, and 9 are used, and digits can repeat. It is known that the sum of the digits of the number $2M$ is 43, and the sum of the digits of the number $M / 2$ is 31. What values can the sum of the digits of the number $M$ take? List all possible answers.
|
# Answer: 35.
Solution. Let $S(n)$ denote the sum of the digits of a natural number $n$. Notice the following facts, each of which is easy to verify by adding numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding $n$ and $n$.
Lemma 2. Let $n$ be a natural number. Then the number of digits in the number $n$ that are greater than or equal to 5 is equal to the number of carries when adding $n$ and $n$.
Lemma 3. Let $n$ and $m$ be natural numbers. Then $S(n+m) = S(n) + S(m) - 9k$, where $k$ is the number of carries when adding $n$ and $m$.
Let $N$ be the number of odd digits in the number $M$; according to the condition, $N$ is the number of digits in the number $M$ that are greater than or equal to 5. Notice that then, by Lemma 1, when adding $M/2$ and $M/2$, there were exactly $N$ carries, from which, by Lemma 3, we have $S(M) = 2S(M/2) - 9N = 62 - 9N$. By Lemma 2, when adding $M$ and $M$, there were also $N$ carries, from which, again by Lemma 3, we have $43 = S(2M) = 2S(M) - 9N$.
Thus, $S(M) = 62 - 9N, 2S(M) = 43 + 9N$, from which $3S(M) = 105, S(M) = 35$.
|
35
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=3: 7, B N: B B_{1}=2: 5, C K: C C_{1}=4: 9$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K P$, if the volume of the prism is 40.
|
Answer: 8.
Solution. Suppose we have found the position of point $P$ such that the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, respectively. Note that
$V_{M N K P}=\frac{1}{3} V_{M N K M_{1} N_{1} K_{1}}$. Draw planes $\beta$ and $\beta_{1}$ through points $M$ and $M_{1}$, parallel to the plane $A B C$, and call $R$ and $R_{1}$ the points of intersection with edge $B B_{1}$, and $S$ and $S_{1}$ with edge $C C_{1}$. Note that the figures $M N K R S$ and $M_{1} N_{1} K_{1} R_{1} S_{1}$ are obtained from each other by a parallel translation, and thus are equal, and their volumes are also equal. Therefore, the volumes of prisms $M N K M_{1} N_{1} K_{1}$ and $M R S M_{1} R_{1} S_{1}$ are also equal. But $V_{M R S M_{1} R_{1} S_{1}}=\frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$, from which we get that $V_{M N K P}=\frac{1}{3} \frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$.
We need to find the position of plane $\alpha$ such that $M M_{1}$ is maximized. Note that at least one of the points $M_{1}, N_{1}, K_{1}$ lies within the original prism, from which $M M_{1}=N N_{1}=K K_{1} \leq \max \left\{A M, A_{1} M, B N, B_{1} N, C K, C_{1} K\right\}$. Substituting the given ratios in the problem, we finally get that $M M_{1}=N N_{1}=K K_{1}=B_{1} N=\frac{3}{5} B B_{1}$, from which $V_{M N K P}=\frac{1}{3} \frac{N N_{1}}{B B_{1}} V_{A B C A_{1} B_{1} C_{1}}=\frac{1}{3} \frac{3}{5} 40=8$.
## Variant IV
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. $n$ mushroom pickers went to the forest and brought a total of 162 mushrooms (each brought at least one mushroom home). Boy Petya, upon learning this, declared: "Some two of them must have brought the same number of mushrooms!" For what smallest $n$ will Petya definitely be right? Don't forget to justify your answer.
|
Answer: 18.
Solution. First, let's prove that when $n \leqslant 17$, Petya can be wrong. Suppose the first $n-1$ mushroom pickers collected $1, \ldots, n-1$ mushrooms, and the $n$-th collected all the rest. Since
$$
1+\ldots+(n-1) \leqslant 1+\ldots+16=136=162-26
$$
the last mushroom picker collected at least 26 mushrooms, i.e., more than each of the others. Thus, when $n \leqslant 17$, there is an example where Petya could be wrong.
Let's show that when $n=18$, Petya will always be right. Suppose he is wrong and let the mushroom pickers have collected $a_{1}<\ldots<a_{18}$ mushrooms. It is easy to see that $a_{i} \geqslant i$, hence
$$
162=a_{1}+\ldots+a_{18} \geqslant 1+\ldots+18=171
$$
a contradiction
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. A circle with radius 2 is inscribed in trapezoid $ABCD$, touching the base $CD$ at point $N$. Find the area of the trapezoid if $DN=1$ and $AB=12$.
|
Answer: 27.
Solution. Let $K, L, M$ be the points of tangency of the inscribed circle with the sides $AD, BC, AB$ respectively; let $I$ be the center of the inscribed circle. Immediately note that $DK = DN = 1$.
Since $I$ is the intersection of the angle bisectors of the internal angles of the trapezoid, $\angle IAD + \angle IDA = (\angle DAB + \angle ADC) / 2 = 180^\circ / 2 = 90^\circ$, where the penultimate equality follows from the parallelism of lines $AB$ and $CD$. Therefore, triangle $AID$ is a right triangle and $\angle AID = 90^\circ$. Similarly, triangle $BIC$ is also a right triangle.
Next, since $IK$ and $IL$ are radii drawn to the points of tangency, $\angle IKD = 90^\circ$ and $\angle ILB = 90^\circ$. Therefore, $IK$ and $IL$ are altitudes in triangles $AID$ and $BIC$ respectively. Using the known fact that in a right triangle, the square of the altitude dropped to the hypotenuse equals the product of the segments into which it divides the hypotenuse. Then
$$
4 = IK^2 = AK \cdot KD = 1 \cdot AK
$$
i.e., $AK = 4$. By the equality of tangent segments, we have $AM = AK = 4$, hence $BL = BM = AB - AM = 12 - 4 = 8$. In the right triangle $BIC$, we get $4 = IL^2 = CL \cdot LB = 8 \cdot CL$, i.e., $CL = 0.5$.
Now we have everything to find the area. Note that $LM$ is the height of the trapezoid and $LM = 2r = 4, AB + CD = AB + DN + CN = 12 + 1 + 0.5 = 13.5$, hence the answer $\frac{4 \cdot 13.5}{2} = 27$.
|
27
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. Points $A_{1}, B_{1}, C_{1}$ are the points of intersection of the extensions of the altitudes of an acute-angled triangle $A B C$ with the circumcircle of $A B C$. The incircle of triangle $A_{1} B_{1} C_{1}$ touches one of the sides of $A B C$, and one of the angles of triangle $A B C$ is $80^{\circ}$. Find the other two angles of triangle $A B C$.
|
Answer: $60^{\circ}$ and $40^{\circ}$.
Solution. Without loss of generality, let the circle $\omega$, inscribed in $A_{1} B_{1} C_{1}$, touch the side $B C$. Let $H$ be the orthocenter of triangle $A B C$, $K$ be the point of tangency of $\omega$ and $B C$, and $L$ be the point of tangency of $\omega$ and $A_{1} C_{1}$. We aim to prove that triangle $H B C_{1}$ is equilateral. Then $\angle B A C = \angle B C_{1} C = 60^{\circ}$, from which the answer follows given the condition.
First, note that $H$ is the intersection of the angle bisectors of triangle $A_{1} B_{1} C_{1}$. Indeed, for example, $\angle B_{1} C_{1} C = \angle B_{1} B C = 90^{\circ} - \angle C = \angle C A A_{1} = \angle C C_{1} A_{1}$, i.e., $C_{1} H$ is the angle bisector of $\angle A_{1} C_{1} B_{1}$; similarly, it can be shown that $A_{1} H$ and $B_{1} H$ are also angle bisectors of the corresponding angles. Therefore, $H$ is the incenter of triangle $A_{1} B_{1} C_{1}$ and $H K = H L$. Moreover, it has been shown that $\angle H C_{1} L = \angle H B K$, i.e., the right triangles $H C_{1} L$ and $H B K$ are congruent by a leg and an acute angle. Therefore, $H C_{1} = H B$.
It remains to note that $H B = B C_{1}$. This fact is well-known and can be proven in various ways. Here, we will present just one of them. Note that triangles $A H B$ and $A C_{1} B$ are congruent by a side (common side $A B$) and two angles ($\angle H A B = \angle A_{1} B_{1} B = \angle C_{1} B_{1} B = \angle C_{1} A B$; similarly, it can be shown that $\angle H B A = \angle C_{1} B A$.
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. For what values of the parameter $a$ does the equation
$$
5^{x^{2}+2 a x+a^{2}}=a x^{2}+2 a^{2} x+a^{3}+a^{2}-6 a+6
$$
have exactly one solution?
|
Answer: Only when $a=1$.
Solution. Let's denote $x+a$ by $t$. Note that the number of solutions of the equation does not change with such a substitution. Then the original equation will take the form
$$
5^{t^{2}}=a t^{2}+a^{2}-6 a+6
$$
Notice that the expressions on both sides do not change when $t$ is replaced by $-t$, so the equation can have an odd number of solutions (in particular, exactly one solution) only if $t=0$ is its root:
$$
5^{0}=a \cdot 0+a^{2}-6 a+6
$$
i.e., $a^{2}-6 a+5=0$, from which $a=5$ or $a=1$. Thus, apart from these two numbers, no other values of the parameter $a$ can satisfy the condition.
Let $a=1$. Then the equation will take the form $5^{t^{2}}=t^{2}+1$. Note that $5^{x}>x \ln 5+1$ for $x>0$ (which can be proven, for example, by taking the derivatives of both sides). Then for $t \neq 0$ we get $5^{t^{2}}>$ $t^{2} \ln 5+1>t^{2}+1$. Therefore, when $a=1$, the equation has a unique solution.
Let $a=5$. Then the equation will take the form $5^{t^{2}}=5 t^{2}+1$. Note that $5^{1}=5,5 \cdot 1+1=6$, but $5^{2^{2}}=625$, $5 \cdot 2^{2}+1=21$, i.e., at $t=1$ the left side is less than the right, and at $t=2$ the opposite is true. Therefore, by the intermediate value theorem, the equation has at least one more root in the interval $(1,2)$. Therefore, $a=5$ does not satisfy the condition.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. In the decimal representation of an even number $M$, only the digits $0, 2, 4, 5, 7$, and 9 are used, and digits may repeat. It is known that the sum of the digits of the number $2M$ is 39, and the sum of the digits of the number $M / 2$ is 30. What values can the sum of the digits of the number $M$ take? List all possible answers.
|
Answer: 33.
Solution. Let $S(n)$ denote the sum of the digits of a natural number $n$. Notice the following facts, each of which is easy to verify by adding numbers in a column.
Lemma 1. Let $n$ be a natural number. Then the number of odd digits in the number $2n$ is equal to the number of carries when adding $n$ and $n$.
Lemma 2. Let $n$ be a natural number. Then the number of digits in the number $n$ that are greater than or equal to 5 is equal to the number of carries when adding $n$ and $n$.
Lemma 3. Let $n$ and $m$ be natural numbers. Then $S(n+m) = S(n) + S(m) - 9k$, where $k$ is the number of carries when adding $n$ and $m$.
Let $N$ be the number of odd digits in the number $M$; according to the condition, $N$ is the number of digits in the number $M$ that are greater than or equal to 5. Notice that then, by Lemma 1, when adding $M/2$ and $M/2$, there were exactly $N$ carries, from which, by Lemma 3, we have $S(M) = 2S(M/2) - 9N = 60 - 9N$. By Lemma 2, when adding $M$ and $M$, there were also $N$ carries, from which, again by Lemma 3, we have $39 = S(2M) = 2S(M) - 9N$.
Thus, $S(M) = 60 - 9N, 2S(M) = 39 + 9N$, from which $3S(M) = 99, S(M) = 33$.
|
33
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10. Points $M, N$, and $K$ are located on the lateral edges $A A_{1}, B B_{1}$, and $C C_{1}$ of the triangular prism $A B C A_{1} B_{1} C_{1}$ such that $A M: A A_{1}=5: 6, B N: B B_{1}=6: 7, C K: C C_{1}=2: 3$. Point $P$ belongs to the prism. Find the maximum possible value of the volume of the pyramid $M N K P$, if the volume of the prism is 35.
|
Answer: 10.
Solution. Suppose we have found the position of point $P$ at which the volume of pyramid $M N K P$ is maximized. Draw a plane $\alpha$ through it, parallel to the plane $M N K$, and call $M_{1}, N_{1}$, and $K_{1}$ the points of intersection of this plane with the edges $A A_{1}, B B_{1}$, and $C C_{1}$, respectively. Note that $V_{M N K P}=\frac{1}{3} V_{M N K M_{1} N_{1} K_{1}}$. Draw planes $\beta$ and $\beta_{1}$ through points $M$ and $M_{1}$, parallel to the plane $A B C$, and call $R$ and $R_{1}$ the points of intersection with edge $B B_{1}$, and $S$ and $S_{1}$ with edge $C C_{1}$. Note that the figures $M N K R S$ and $M_{1} N_{1} K_{1} R_{1} S_{1}$ are obtained from each other by a parallel translation, and therefore are equal, and their volumes are also equal. Then the volumes of prisms $M N K M_{1} N_{1} K_{1}$ and $M R S M_{1} R_{1} S_{1}$ are also equal. But $V_{M R S M_{1} R_{1} S_{1}}=\frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$, from which we get that $V_{M N K P}=\frac{1}{3} \frac{M M_{1}}{A A_{1}} V_{A B C A_{1} B_{1} C_{1}}$.
We need to find the position of plane $\alpha$ such that $M M_{1}$ is maximized. Note that at least one of the points $M_{1}, N_{1}, K_{1}$ lies within the original prism, from which $M M_{1}=N N_{1}=K K_{1} \leq \max \left\{A M, A_{1} M, B N, B_{1} N, C K, C_{1} K\right\}$. Substituting the given ratios in the problem, we finally get that $M M_{1}=N N_{1}=K K_{1}=B N=\frac{6}{7} B B_{1}$, from which $V_{M N K P}=\frac{1}{3} \frac{N N_{1}}{B B_{1}} V_{A B C A_{1} B_{1} C_{1}}=\frac{16}{3} \frac{6}{7} 35=10$.
[^0]: ${ }^{1}$ see https://ru.wikipedia.org/wiki/Малая_теорема_Ферма\#Альтернативная_формулировка
[^1]: ${ }^{2}$ see https://ru.wikipedia.org/wiki/Малая_теорема_Ферма\#Альтернативная_формулировка
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. In a football championship, 20 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other?
|
Answer: 90 games.
Solution. We will consider the games that have not been played. The condition means that the unplayed games do not form triangles. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ (the estimate is obvious).
Inductive step: Suppose it is proven for $k$, we will prove for $k+1$. If there are no unplayed games, then everything is proven. Otherwise, select any teams $A$ and $B$ that have not played against each other. Note that the number of unplayed games involving teams $A$ or $B$ is no more than $2k$ (not counting the game between $A$ and $B$), since for any team $C$, at least one of the games $AC$ and $BC$ has been played. Now consider all teams except $A$ and $B$ and apply the inductive hypothesis - among them, no more than $k^2$ games have not been played. Therefore, the total number of unplayed games is no more than $k^2 + (2k + 1) = (k+1)^2$, which is what we needed to prove.
Substituting $k=10$ gives that the number of unplayed games is no more than 100, and the total number of possible games is $\frac{20 \cdot 19}{2} = 190$, from which the number of played games is no less than $190 - 100 = 90$.
The estimate is achieved if the teams are divided into two equal groups, all matches are played within each group, and no matches are played between the groups.
|
90
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. In a football championship, 16 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other?
#
|
# Answer: 56 games.
Solution. We will consider the games that have not been played. The condition means that there will be no three teams that have not played against each other at all. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ (the estimate is obvious).
Inductive step: Suppose it is proven for $k$, we will prove for $k+1$. If there are no unplayed games, then everything is proven. Otherwise, select any teams $A$ and $B$ that have not played against each other. Note that the number of unplayed games involving teams $A$ or $B$ is no more than $2k$ (not counting the game between $A$ and $B$), because for any team $C$, at least one of the games $AC$ or $BC$ has been played. Now consider all teams except $A$ and $B$ and apply the inductive hypothesis - among them, no more than $k^2$ games have not been played. Therefore, the total number of unplayed games is no more than $k^2 + (2k + 1) = (k+1)^2$, which is what we needed to prove.
Substituting $k=8$ gives that the number of unplayed games is no more than 64, and the total number of possible games is $\frac{16 \cdot 15}{2} = 120$, so the number of played games is no less than $120 - 64 = 56$.
The estimate is achieved if the teams are divided into two equal groups, all matches are played within each group, and no matches are played between the groups.
|
56
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. In a football championship, 16 teams participate, each playing against each other once. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other?
|
Answer: 56 games.
Solution. We will consider the games that have not been played. The condition means that there will be no three teams that have not played with each other at all. We will prove by induction on $k$ that for $2k$ teams, the maximum number of unplayed games is no more than $k^2$.
Base case: $k=1$ (the estimate is obvious).
Inductive step: Suppose it is proven for $k$, we will prove for $k+1$. If there are no unplayed games, then everything is proven. Otherwise, select any teams $A$ and $B$ that have not played against each other. Note that the number of unplayed games involving teams $A$ or $B$ is no more than $2k$ (not counting the game between $A$ and $B$), because for any team $C$, at least one of the games $AC$ or $BC$ has been played. Now consider all teams except $A$ and $B$ and apply the inductive hypothesis - among them, no more than $k^2$ games have not been played. Therefore, the total number of unplayed games is no more than $k^2 + (2k + 1) = (k+1)^2$, which is what we needed to prove.
Substituting $k=8$ gives that the number of unplayed games is no more than 64, and the total number of possible games is $\frac{16 \cdot 15}{2} = 120$, so the number of played games is no less than $120 - 64 = 56$.
The estimate is achieved if the teams are divided into two equal groups, all matches are played within each group, and no matches are played between the groups.
|
56
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 30?
|
Answer: 32.
Solution. First, let's provide an example of 32 numbers whose sum is 0 and the sum of their squares is 30. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{16}=\sqrt{\frac{15}{16}}, x_{17}=x_{18}=\ldots=x_{32}=-\sqrt{\frac{15}{16}}$ will work.
Now, we will prove that fewer than 32 numbers will not suffice. Suppose the contrary. Then, among all the numbers, either positive or negative, there are no more than 15. By multiplying all the numbers by -1 if necessary, we can assume that there are no more than 15 negative numbers.
Let $y_{1}, y_{2}, \ldots, y_{k}$ be all the negative numbers, and $y_{k+1}, \ldots, y_{n}$ be all the non-negative numbers. Then $y_{1}^{2}+y_{2}^{2}+\ldots+y_{k}^{2}<k \leqslant 15$, and
$$
y_{k+1}^{2}+y_{k+2}^{2}+\ldots+y_{n}^{2} \leqslant y_{k+1}+y_{k+2}+\ldots+y_{n}=-y_{1}-y_{2}-\ldots-y_{k}<k \leqslant 15
$$
Adding the obtained inequalities, we get, $30=y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}<30$. Contradiction.
|
32
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. In a right triangle $ABC$, a circle is constructed on the leg $AC$ as its diameter, which intersects the hypotenuse $AB$ at point $E$. A tangent to the circle is drawn through point $E$, which intersects the leg $CB$ at point $D$. Find the length of $DB$, if $AE=6$, and $BE=2$.
|
Answer: 2.
Solution. The solution is based on two simple observations. First, $\angle A E C=90^{\circ}$, since it subtends the diameter. Second, $D E$ and $D C$ are tangents to the circle from the condition, so $D E=D C$. Therefore, in the right triangle $C E B$, a point $D$ is marked on the hypotenuse $B C$ such that $D E=D C$; it is well known that then $D$ is the midpoint of this hypotenuse, $B D=B C / 2$.
The solution can be completed in several ways; we will present two of them.
First method, height of a right triangle and the Pythagorean theorem. In the right triangle $A B C$, the height $C E$ divides the hypotenuse $A B$ into segments $A E=6$ and $B E=2$. We get that $C E^{2}=A E \cdot B E=6 \cdot 2$; by the Pythagorean theorem $B C^{2}=B E^{2}+C E^{2}=2^{2}+2 \cdot 6=16, B D=B C / 2=2$.
Second method, the tangent-secant theorem. According to the tangent-secant theorem for the tangent $B C$ and the secant $B A$, we have: $B C^{2}=B E \cdot B A=2 \cdot(2+6)=16, B D=B C / 2=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-11 x^{2}+a x-8=0$ have three distinct real roots that form a geometric progression?
|
Answer: only 22.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-11 x^{2}+a x-8=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a polynomial of the third degree:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=11 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=8
\end{array}\right.
$$
Since $x_{1}, x_{2}, x_{3}$ form a geometric progression (let's assume in this order), there exist $b$ and $q$ such that $x_{1}=b, x_{2}=b q, x_{3}=b q^{2}$. Then from the equality $x_{1} x_{2} x_{3}=8$, we have $b^{3} q^{3}=8$, from which $x_{2}=b q=2, x_{1}=2 / q, x_{3}=2 q$.
Then $2\left(\frac{1}{q}+1+q\right)=11(*)$, after transformations $2 q^{2}-9 q+2=0$. The discriminant of this expression is $D=9^{2}-4 \cdot 2 \cdot 2>0$, so such $q$, and with it $x_{1}, x_{2}, x_{3}$, will exist. Then
$$
a=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=x_{1} x_{2} x_{3}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=8 \cdot\left(\frac{q}{2}+\frac{1}{2}+\frac{1}{2 q}\right)=4\left(q+1+\frac{1}{q}\right)=2 \cdot 11=22
$$
(In the penultimate transition, we used the equality $(*)$).
Comment. Naturally, $q, x_{1}, x_{2}, x_{3}$ can be calculated explicitly:
$$
q_{1,2}=\frac{9 \pm \sqrt{65}}{4}
$$
Choose $q$ with the “+” (if you choose with the “-”, then $x_{1}$ and $x_{3}$ will swap places, which will not affect the answer); then
$$
x_{1}=\frac{2 \cdot 4}{9+\sqrt{65}}=\frac{9-\sqrt{65}}{2}, x_{2}=2, x_{3}=\frac{9+\sqrt{65}}{2}
$$
$a$ could have been calculated by substituting the obtained numbers into the expression $x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}$.
|
22
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 42?
|
Answer: 44.
Solution. First, let's provide an example of 44 numbers whose sum is 0 and the sum of their squares is 42. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{22}=\sqrt{\frac{21}{22}}, x_{23}=x_{24}=\ldots=x_{44}=-\sqrt{\frac{21}{22}}$ will work.
Now, we will prove that it is impossible to achieve this with fewer than 44 numbers. Suppose the contrary. Then, among all the numbers, either the positive or the negative ones do not exceed 21. By multiplying all the numbers by -1 if necessary, we can assume that there are no more than 21 negative numbers.
Let $y_{1}, y_{2}, \ldots, y_{k}$ be all the negative numbers, and $y_{k+1}, \ldots, y_{n}$ be all the non-negative numbers. Then $y_{1}^{2}+y_{2}^{2}+\ldots+y_{k}^{2}<k \leqslant 21$, and
$$
y_{k+1}^{2}+y_{k+2}^{2}+\ldots+y_{n}^{2} \leqslant y_{k+1}+y_{k+2}+\ldots+y_{n}=-y_{1}-y_{2}-\ldots-y_{k}<k \leqslant 21
$$
Adding the obtained inequalities, we get, $42=y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}<42$. Contradiction.
|
44
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. In a right triangle $P Q R$, a circle is constructed on the leg $P R$ as its diameter, which intersects the hypotenuse $P Q$ at point $T$. A tangent to the circle is drawn through point $T$, which intersects the leg $R Q$ at point $S$. Find the length of $S Q$, if $P T=15$, and $Q T=5$.
|
Answer: 5.
Solution. The solution is based on two simple observations. First, $\angle P T R=90^{\circ}$, since it subtends the diameter. Second, $S T$ and $S R$ are tangents to the circle from the given conditions, so $S T=S R$. Therefore, in the right triangle $R T Q$, a point $S$ is marked on the hypotenuse $Q R$ such that $S T=S R$; it is well known that then $S$ is the midpoint of this hypotenuse, $Q S=Q R / 2$.
The solution can be completed in several ways; we will present two of them.
First method, height of a right triangle and the Pythagorean theorem. In the right triangle $P Q R$, the height $R T$ divides the hypotenuse $P Q$ into segments $P T=15$ and $Q T=5$. We get that $R T^{2}=P T \cdot Q T=15 \cdot 5$; by the Pythagorean theorem $Q R^{2}=Q T^{2}+R T^{2}=5^{2}+15 \cdot 5=100, Q S=Q R / 2=5$.
Second method, the tangent-secant theorem. According to the tangent-secant theorem for the tangent $Q R$ and the secant $Q P$, we have: $Q R^{2}=Q T \cdot Q P=5 \cdot(5+15)=100, Q S=Q R / 2=5$.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-14 x^{2}+a x-27=0$ have three distinct real roots that form a geometric progression?
|
Answer: only 42.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-14 x^{2}+a x-27=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a cubic polynomial:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=14 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=27
\end{array}\right.
$$
Since $x_{1}, x_{2}, x_{3}$ form a geometric progression (let's assume in this order), there exist $b$ and $q$ such that $x_{1}=b, x_{2}=b q, x_{3}=b q^{2}$. Then from the equality $x_{1} x_{2} x_{3}=27$, we have $b^{3} q^{3}=27$, hence $x_{2}=b q=3, x_{1}=3 / q, x_{3}=3 q$.
Then $3\left(\frac{1}{q}+1+q\right)=14(*)$, after transformations $3 q^{2}-11 q+3=0$. The discriminant of this expression is $D=11^{2}-4 \cdot 3 \cdot 3>0$, so such $q$, and with it $x_{1}, x_{2}, x_{3}$, exist. Then $a=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=x_{1} x_{2} x_{3}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=27 \cdot\left(\frac{q}{3}+\frac{1}{3}+\frac{1}{3 q}\right)=9\left(q+1+\frac{1}{q}\right)=3 \cdot 14=42$. (In the penultimate transition, we used the equality $(*)$ ).
Comment. Naturally, $q, x_{1}, x_{2}, x_{3}$ can be calculated explicitly:
$$
q_{1,2}=\frac{11 \pm \sqrt{85}}{6}
$$
Choose $q$ with the “+” (if you choose with the “-”, then $x_{1}$ and $x_{3}$ will swap places, which will not affect the answer); then
$$
x_{1}=\frac{3 \cdot 6}{11+\sqrt{85}}=\frac{11-\sqrt{85}}{2}, x_{2}=3, x_{3}=\frac{11+\sqrt{85}}{2} .
$$
$a$ could have been calculated by substituting the obtained numbers into the expression $x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}$.
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 36?
#
|
# Answer: 38.
Solution. First, let's provide an example of 38 numbers whose sum is 0 and the sum of their squares is 36. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{19}=\sqrt{\frac{18}{19}}, x_{20}=x_{21}=\ldots=x_{38}=-\sqrt{\frac{18}{19}}$ will work.
Now, we will prove that fewer than 38 numbers will not suffice. Suppose the contrary. Then, among all the numbers, either the positive or the negative ones do not exceed 18. By multiplying all the numbers by -1 if necessary, we can assume that there are no more than 18 negative numbers.
Let $y_{1}, y_{2}, \ldots, y_{k}$ be all the negative numbers, and $y_{k+1}, \ldots, y_{n}$ be all the non-negative numbers. Then $y_{1}^{2}+y_{2}^{2}+\ldots+y_{k}^{2}<k \leqslant 18$, and
$$
y_{k+1}^{2}+y_{k+2}^{2}+\ldots+y_{n}^{2} \leqslant y_{k+1}+y_{k+2}+\ldots+y_{n}=-y_{1}-y_{2}-\ldots-y_{k}<k \leqslant 18
$$
Adding the obtained inequalities, we get, $36=y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}<36$. Contradiction.
|
38
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. In a right triangle $X Y Z$, a circle is constructed on the leg $X Z$ as its diameter, which intersects the hypotenuse $X Y$ at point $W$. A tangent to the circle is drawn through point $W$, which intersects the leg $Z Y$ at point $V$. Find the length of $V Y$, if $X W=12$, and $Y W=4$.
|
Answer: 4.
Solution. The solution is based on two simple observations. First, $\angle X W Z=90^{\circ}$, since it subtends the diameter. Second, $V W$ and $V Z$ are tangents to the circle from the condition, so $V W=V Z$. Therefore, in the right triangle $Z W Y$, a point $V$ is marked on the hypotenuse $Y Z$ such that $V W=V Z$; it is well known that then $V$ is the midpoint of this hypotenuse, $Y V=Y Z / 2$.
The solution can be completed in several ways; we will present two of them.
First method, height of a right triangle and the Pythagorean theorem. In the right triangle $X Y Z$, the height $Z W$ divides the hypotenuse $X Y$ into segments $X W=12$ and $Y W=4$. We get that $Z W^{2}=X W \cdot Y W=12 \cdot 4$; by the Pythagorean theorem, $Y Z^{2}=Y W^{2}+Z W^{2}=4^{2}+12 \cdot 4=64, Y V=Y Z / 2=4$.
Second method, the tangent-secant theorem. According to the tangent-secant theorem for the tangent $Y Z$ and the secant $Y X$, we have: $Y Z^{2}=Y W \cdot Y X=4 \cdot(4+12)=64, Y V=Y Z / 2=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what values of the parameter $a$ does the equation $x^{3}-15 x^{2}+a x-64=0$ have three distinct real roots that form a geometric progression?
|
Answer: only 60.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-15 x^{2}+a x-64=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a polynomial of the third degree:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=15 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=64
\end{array}\right.
$$
Since $x_{1}, x_{2}, x_{3}$ form a geometric progression (let's assume in this order), there exist $b$ and $q$ such that $x_{1}=b, x_{2}=b q, x_{3}=b q^{2}$. Then from the equality $x_{1} x_{2} x_{3}=64$, we have $b^{3} q^{3}=64$, from which $x_{2}=b q=4, x_{1}=4 / q, x_{3}=4 q$.
Then $4\left(\frac{1}{q}+1+q\right)=15(*)$, after transformations $4 q^{2}-11 q+4=0$. The discriminant of this expression is $D=11^{2}-4 \cdot 4 \cdot 4>0$, so such $q$, and with it $x_{1}, x_{2}, x_{3}$, will exist. Then
$a=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=x_{1} x_{2} x_{3}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=64 \cdot\left(\frac{q}{4}+\frac{1}{4}+\frac{1}{4 q}\right)=16\left(q+1+\frac{1}{q}\right)=4 \cdot 15=60$. (In the penultimate transition, we used the equality $(*)$ ).
Comment. Naturally, $q, x_{1}, x_{2}, x_{3}$ can be calculated explicitly:
$$
q_{1,2}=\frac{11 \pm \sqrt{57}}{8}
$$
Choose $q$ with “+” (if you choose with “-”, then $x_{1}$ and $x_{3}$ will swap places, which will not affect the answer); then
$$
x_{1}=\frac{4 \cdot 8}{11+\sqrt{57}}=\frac{11-\sqrt{57}}{2}, x_{2}=4, x_{3}=\frac{11+\sqrt{57}}{2} .
$$
$a$ could have been calculated by substituting the obtained numbers into the expression $x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 40?
|
Answer: 42.
Solution. First, let's provide an example of 42 numbers whose sum is 0 and the sum of their squares is 40. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{21}=\sqrt{\frac{20}{21}}, x_{22}=x_{23}=\ldots=x_{42}=-\sqrt{\frac{20}{21}}$ will work.
Now, we will prove that fewer than 42 numbers will not suffice. Suppose the contrary. Then, among all the numbers, either the positive or the negative ones do not exceed 20. By multiplying all the numbers by -1 if necessary, we can assume that there are no more than 20 negative numbers.
Let $y_{1}, y_{2}, \ldots, y_{k}$ be all the negative numbers, and $y_{k+1}, \ldots, y_{n}$ be all the non-negative numbers. Then $y_{1}^{2}+y_{2}^{2}+\ldots+y_{k}^{2}<k \leqslant 20$, and
$$
y_{k+1}^{2}+y_{k+2}^{2}+\ldots+y_{n}^{2} \leqslant y_{k+1}+y_{k+2}+\ldots+y_{n}=-y_{1}-y_{2}-\ldots-y_{k}<k \leqslant 20
$$
Adding the obtained inequalities, we get, $40=y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}<40$. Contradiction.
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. In a right triangle $KLM$, a circle is constructed on the leg $KM$ as its diameter, which intersects the hypotenuse $KL$ at point $G$. A tangent to the circle is drawn through point $G$, intersecting the leg $ML$ at point $F$. Find the length of $FL$, if $KG=5$ and $LG=4$.
|
Answer: 3.
Solution. The solution is based on two simple observations. First, $\angle K G M=90^{\circ}$, since it subtends the diameter. Second, $F G$ and $F M$ are tangents to the circle from the condition, so $F G=F M$. Therefore, in the right triangle $M G L$, a point $F$ is marked on the hypotenuse $L M$ such that $F G=F M$; it is well known that then $F$ is the midpoint of this hypotenuse, $L F=L M / 2$.
The solution can be completed in several ways; we will present two of them.
First method, height of a right triangle and the Pythagorean theorem. In the right triangle $K L M$, the height $M G$ divides the hypotenuse $K L$ into segments $K G=5$ and $L G=4$. We get that $M G^{2}=K G \cdot L G=5 \cdot 4$; by the Pythagorean theorem $L M^{2}=L G^{2}+M G^{2}=4^{2}+5 \cdot 4=36, L F=L M / 2=3$.
Second method, the tangent-secant theorem. According to the tangent-secant theorem for the tangent $L M$ and the secant $L K$, we have: $L M^{2}=L G \cdot L K=4 \cdot(4+5)=36, L F=L M / 2=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what values of the parameter $a$ does the equation $x^{3}+16 x^{2}+a x+64=0$ have three distinct real roots that form a geometric progression?
|
Answer: only 64.
Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}+16 x^{2}+a x+64=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a cubic polynomial:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=-16 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=-64
\end{array}\right.
$$
Since $x_{1}, x_{2}, x_{3}$ form a geometric progression (let's assume in this order), there exist $b$ and $q$ such that $x_{1}=b, x_{2}=b q, x_{3}=b q^{2}$. Then from the equality $x_{1} x_{2} x_{3}=-64$, we have $b^{3} q^{3}=-64$, from which $x_{2}=b q=-4, x_{1}=-4 / q, x_{3}=-4 q$.
Then $-4\left(\frac{1}{q}+1+q\right)=-16(*)$, after transformations $q^{2}-3 q+1=0$. The discriminant of this expression is $D=3^{2}-4 \cdot 1 \cdot 1>0$, so such a $q$, and with it $x_{1}, x_{2}, x_{3}$, will exist. Then
$a=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=x_{1} x_{2} x_{3}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=-64 \cdot\left(-\frac{q}{4}-\frac{1}{4}-\frac{1}{4 q}\right)=16\left(q+1+\frac{1}{q}\right)=4 \cdot 16=64$.
(In the penultimate transition, we used the equality $\left({ }^{*}\right)$ ).
Comment. Naturally, $q, x_{1}, x_{2}, x_{3}$ can be calculated explicitly:
$$
q_{1,2}=\frac{3 \pm \sqrt{5}}{2}
$$
Choose $q$ with the “+” (if you choose with the “-”, then $x_{1}$ and $x_{3}$ will swap places, which will not affect the answer); then
$$
x_{1}=\frac{-4 \cdot 2}{3+\sqrt{5}}=-2(3-\sqrt{5}), x_{2}=-4, x_{3}=-2(3+\sqrt{5})
$$
$a$ could have been calculated by substituting the obtained numbers into the expression $x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}$.
|
64
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. Each knight gives one affirmative answer to four questions, while a liar gives three. In total, there were $105+45+85+65=300$ affirmative answers. If all the residents of the city were knights, the total number of affirmative answers would be 200. The 100 extra "yes" answers come from the lies of the liars. Thus, the number of liars is $\frac{100}{2}=50$. Let $k$ be the number of knights living in district B, then $45-k$ is the number of affirmative answers to the second question given by the liars. Therefore, the number of liars living in district B is $50-(45-k)=k+5$. In the other districts, the number of liars is less than the number of knights.
|
Answer: in Block B, on 5.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10. The sum of the surface areas of the polyhedra into which a parallelepiped is divided by sections is equal to the sum of the surface area of the parallelepiped and the areas of the internal surfaces. The sum of the areas of the internal surfaces is equal to twice the sum of the areas of the sections.
Let's find the maximum possible area of a section passing through the diagonal $XY$ of an arbitrary parallelepiped with edges $a \leq b \leq c$. The section is a parallelogram $ZXTY$, whose vertices lie on opposite edges of the parallelepiped. The area of the parallelogram is the product of the length of the diagonal $XY$ and the distance from point $Z$ to $XY$.
Consider the projection of the parallelepiped onto a plane perpendicular to the diagonal $XY$. From the diagram, it is clear that the distance from point $Z$ to the broken line $ABC$, and thus to the diagonal $XY$, is greatest if
$Z$ coincides with one of the vertices $A, B$, or $C$.
This means that the section passes through one of the edges of the parallelepiped. Therefore, the section with the largest area is one of the diagonal sections. All these sections are rectangles. Let's find the largest of their areas:
$$
S_{1}=a \sqrt{b^{2}+c^{2}}, \quad S_{2}=b \sqrt{a^{2}+c^{2}}, \quad \text{and} \quad S_{3}=c \sqrt{b^{2}+a^{2}}
$$
From the condition $a \leq b \leq c$, it follows that
$a^{2} b^{2}+a^{2} c^{2} \leq c^{2} b^{2}+a^{2} c^{2}$, and $a^{2} b^{2}+c^{2} b^{2} \leq c^{2} b^{2}+a^{2} c^{2}$. Therefore, $S_{1} \leq S_{3}$ and $S_{2} \leq S_{3}$. This means that the section with the largest area passes through the largest edge. According to the condition, the largest length is the edge $AA_{1}$, so the sections $AA_{1}C_{1}C$ and $BB_{1}D_{1}D$ have the largest area:
$5 \sqrt{4^{2}+3^{2}}=25$. The sum of the surface areas of the polyhedra into which the parallelepiped is divided by these sections (see the diagram) is
$$
2\left(A A_{1} \cdot A B + A A_{1} \cdot A D + A B \cdot A D\right) + 4 \cdot 25 = 194
$$
|
Answer: 194.
[^0]: ${ }^{1}$ This statement is a particular case of Cauchy's inequality.
|
194
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. Each of the two workers was assigned to process the same number of parts. The first completed the work in 8 hours. The second spent more than 2 hours on setting up the equipment and with its help finished the work 3 hours earlier than the first. It is known that the second worker processed as many parts in 1 hour after the equipment started as the first did by that time. By what factor does the equipment increase labor productivity?
|
Answer: 4 times.
Solution: Let $x$ be the time spent on equipment setup. Then the second worker worked (on the equipment) $8-3-x=5-x$ hours, producing as much per hour as the first worker in $x+1$ hours. Therefore, $\frac{8}{5-x}=\frac{x+1}{1}$. We get $x^{2}-4 x+3=0$. But by the condition $x>2$, so $x=3$, and the required ratio is $\frac{x+1}{1}=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$.
|
Answer: 4.
Solution. By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. The function $f$ is such that $f(2 x-3 y)-f(x+y)=-2 x+8 y$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$.
|
Answer: 3.
Solution. By substituting $y=-x$, we get that $f(5 x)=-10 x+f(0)$, i.e., $f(t)=-2 t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(5 t)-f(t)}{f(4 t)-f(3 t)}$.
|
Answer: 4.
Solution. Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. The function $f$ is such that $f(x+2 y)-f(3 x-2 y)=2 y-x$ for all $x, y$. Find all possible values of the expression $\frac{f(4 t)-f(t)}{f(3 t)-f(2 t)}$.
|
Answer: 3.
Solution. Substituting $x=-2 y$, we get that $f(0)-f(-8 y)=-4 y$, i.e., $f(t)=\frac{1}{2} t+c$ (where $c$ is some constant). Therefore, the desired expression is always (when it is defined) equal to 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 13 adventurers have rubies; exactly 9 have emeralds; exactly 15 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has sapphires, then they have either emeralds or diamonds (but not both at the same time);
- if an adventurer has emeralds, then they have either rubies or sapphires (but not both at the same time).
What is the minimum number of adventurers that can be in such a group?
|
Answer: 22.
Solution. Note that the number of adventurers who have sapphires is equal to the total number of adventurers who have emeralds or diamonds. Then, from the first condition, it follows that 9 adventurers have sapphires and emeralds, and 6 have sapphires and diamonds. That is, every adventurer who has emeralds must also have sapphires. Then, from the second condition, there cannot be an adventurer who has both emeralds and rubies. Therefore, there must be at least $13+9=22$ adventurers.
This number of adventurers is indeed possible: let's say we have 9 adventurers who have sapphires and emeralds, 6 adventurers who have sapphires, diamonds, and rubies, and 7 adventurers who have only rubies. One can verify that this example meets all the conditions.
|
22
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. A team of workers was working on pouring the rink on the large and small fields, with the area of the large field being twice the area of the small field. In the part of the team that worked on the large field, there were 4 more workers than in the part that worked on the small field. When the pouring of the large rink was completed, the part of the team that was on the small field was still working. What is the maximum number of workers that could have been in the team
|
Answer: 10.
Solution. Let the number of workers on the smaller field be $n$, then the number of workers on the larger field is $n+4$, and the total number of people in the team is $2n+4$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivities of each part of the team are $an$ and $a(n+4)$. If the area of the smaller field is $S$, then the area of the larger field is $2S$. The time spent on completing the entire work by each team is $\frac{S}{an}$ and $\frac{2S}{a(n+4)}$, respectively. According to the problem, $\frac{S}{an} > \frac{2S}{a(n+4)}$. Due to the positivity of all variables, this inequality is equivalent to the inequality $n+4 > 2n$, or $n < 4$. Therefore, $n \leqslant 3$, and $2n+4 \leqslant 10$. The equality situation is clearly possible: just take any positive $S$ and $a$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $BC=8$ and $AC=4$. Find the length of side $AB$ if the length of the vector $4 \overrightarrow{OA} - \overrightarrow{OB} - 3 \overrightarrow{OC}$ is 10.
|
Answer: 5.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cdot \overrightarrow{O B})+2 y z(\overrightarrow{O B} \cdot \overrightarrow{O C})+2 x z(\overrightarrow{O A} \cdot \overrightarrow{O C})= \\
& =\left(x^{2}+y^{2}+z^{2}\right) R^{2}+x y\left(O A^{2}+O B^{2}-A B^{2}\right)+y z\left(O B^{2}+O C^{2}-B C^{2}\right)+x z\left(O A^{2}+O C^{2}-A C^{2}\right)= \\
& =\left(x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right) R^{2}-x y A B^{2}-y z B C^{2}-x z A C^{2}= \\
& =(x+y+z)^{2} R^{2}-x y A B^{2}-y z B C^{2}-x z A C^{2}
\end{aligned}
$$
Then, for $x=4, y=-1, z=-3$, we obtain the equality $10^{2}=0 \cdot R^{2}+4 A B^{2}-3 B C^{2}+12 A C^{2}$, from which $A B^{2}=\frac{1}{4}\left(100-12 A C^{2}+3 B C^{2}\right)=25$, i.e., $A B=5$.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. For what values of the parameter $a$ does the equation
$$
\log _{2}^{2} x+(a-6) \log _{2} x+9-3 a=0
$$
have exactly two roots, one of which is four times the other?
|
Answer: $-2.2$
Solution. Let $t=\log _{2} x$, then the equation becomes $t^{2}+(a-6) t+(9-3 a)=0$. Notice that $3 \cdot(3-a)=9-3 a, 3+(3-a)=6-a$, from which, by the theorem converse to Vieta's theorem, the roots of this equation are -3 and $3-a$. We make the reverse substitution: $\log _{2} x=3$ or $\log _{2} x=3-a$, i.e., either $x=8$, or $x=2^{3-a}$. We get two cases: $8=4 \cdot 2^{3-a}$ or $2^{3-a}=4 \cdot 8$. In the first case $a=2$, in the second $a=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. In triangle $A B C$, side $A C=42$. The bisector $C L$ is divided by the point of intersection of the bisectors of the triangle in the ratio $2: 1$, counting from the vertex. Find the length of side $A B$, if the radius of the circle inscribed in triangle $A B C$ is 14.
|
Answer: 56.
Answer: Let $I$ be the center of the inscribed circle in triangle $ABC$ (i.e., the point of intersection of the angle bisectors). Noting that $AI$ is the angle bisector in triangle $ALC$, by the angle bisector theorem, we have: $AC: AL = CI: IL = 2: 1$, from which $AL = AC / 2 = 21$.
Next, $AC \cdot AL \cdot \sin \angle A = 2 S_{\triangle ALC} = 2 S_{\triangle AIC} + 2 S_{\triangle AIL} = AC \cdot r + AL \cdot r = (AC + AL) \cdot r$, where $r$ is the radius of the inscribed circle in triangle $ABC$. Thus, $42 \cdot 21 \cdot \sin \angle A = (42 + 21) \cdot 14$, i.e., $\sin \angle A = 1, \angle A = 90^{\circ}$.
By the angle bisector theorem for $BI$ in triangle $CLB$, we have $BC: BL = CI: IL = 2: 1$. Letting $BL = x$, we have $BC = 2x$. By the Pythagorean theorem: $AC^2 + AB^2 = BC^2$, i.e., $42^2 + (21 + x)^2 = (2x)^2$, from which $x = 35$, and $AB = x + 21 = 56$.
|
56
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. The function $F$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $F(n a, n b, n c)=n \cdot F(a, b, c)$, $F(a+n, b+n, c+n)=F(a, b, c)+n$, $F(a, b, c)=F(c, b, a)$ hold. Find $F(58,59,60)$.
|
Answer: 59.
Solution. Note that $F(-1,0,1)=F(1,0,-1)=(-1) \cdot F(-1,0,1)$, from which $F(-1,0,1)=0$. Then $F(58,59,60)=F(-1,0,1)+59=59$.
Comment. The function $F$ cannot be uniquely determined. For example, the functions $F(a, b, c)=(a+b+c) / 3$, $F(a, b, c)=b$, and $F(a, b, c)=$ median of the numbers $\{a, b, c\}$ all fit.
|
59
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 5 adventurers have rubies; exactly 11 have emeralds; exactly 10 have sapphires; exactly 6 have diamonds. In addition, it is known that
- if an adventurer has diamonds, then they have either emeralds or sapphires (but not both at the same time);
- if an adventurer has emeralds, then they have either rubies or diamonds (but not both at the same time).
What is the minimum number of adventurers that can be in such a group?
|
Answer: 16.
Solution. Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or diamonds. Then, from the second condition, it follows that 5 adventurers have rubies and emeralds, and 6 have emeralds and diamonds. That is, every adventurer who has diamonds must also have emeralds. Then, from the first condition, there cannot be an adventurer who has both sapphires and diamonds. Therefore, there are at least $10+6=16$ adventurers.
Indeed, there can be this many adventurers: let's say we have 6 adventurers who have emeralds and diamonds, 5 adventurers who have rubies, emeralds, and sapphires, and 5 adventurers who have only sapphires. One can verify that this example meets all the conditions.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. A landscaping team worked on a large and a small football field, with the area of the large field being twice the area of the small field. In the part of the team that worked on the large field, there were 6 more workers than in the part that worked on the small field. When the landscaping of the large field was completed, the part of the team that was on the small field was still working. What is the maximum number of workers that could have been in the team
|
Answer: 16.
Solution. Let the number of workers on the smaller field be $n$, then the number of workers on the larger field is $n+6$, and the total number of people in the team is $2n+6$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivities of each part of the team are $an$ and $a(n+6)$. If the area of the smaller field is $S$, then the area of the larger field is $2S$. The time spent on completing the entire work by each team is $\frac{S}{an}$ and $\frac{2S}{a(n+6)}$, respectively. According to the problem, $\frac{S}{an} > \frac{2S}{a(n+6)}$. Due to the positivity of all variables, this inequality is equivalent to the inequality $n+6 > 2n$, or $n < 6$. Therefore, $n \leqslant 5$, and $2n+6 \leqslant 16$. The equality situation is clearly possible: just take any positive $S$ and $a$.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $AB=8$ and $AC=5$. Find the length of side $BC$ if the length of the vector $\overrightarrow{OA}+3 \overrightarrow{OB}-4 \overrightarrow{OC}$ is 10.
|
Answer: 4.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cdot \overrightarrow{O B})+2 y z(\overrightarrow{O B} \cdot \overrightarrow{O C})+2 x z(\overrightarrow{O A} \cdot \overrightarrow{O C})= \\
& =\left(x^{2}+y^{2}+z^{2}\right) R^{2}+x y\left(O A^{2}+O B^{2}-A B^{2}\right)+y z\left(O B^{2}+O C^{2}-B C^{2}\right)+x z\left(O A^{2}+O C^{2}-A C^{2}\right)= \\
& =\left(x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right) R^{2}-x y A B^{2}-y z B C^{2}-x z A C^{2}= \\
& =(x+y+z)^{2} R^{2}-x y A B^{2}-y z B C^{2}-x z A C^{2}
\end{aligned}
$$
Then, for $x=1, y=3, z=-4$, we obtain the equality $10^{2}=0 \cdot R^{2}-3 A B^{2}+12 B C^{2}+4 A C^{2}$, from which $B C^{2}=\frac{1}{12}\left(100+3 A B^{2}-4 A C^{2}\right)=16$, i.e., $B C=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. The function $f$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $f(n a, n b, n c)=n \cdot f(a, b, c)$, $f(a+n, b+n, c+n)=f(a, b, c)+n$, $f(a, b, c)=f(c, b, a)$ hold. Find $f(24,25,26)$.
|
Answer: 25.
Solution. Note that $f(-1,0,1)=f(1,0,-1)=(-1) \cdot f(-1,0,1)$, from which $f(-1,0,1)=0$. Then $f(24,25,26)=f(-1,0,1)+25=25$.
Comment. The function $f$ cannot be uniquely determined. For example, the functions $f(a, b, c)=(a+b+c) / 3$, $f(a, b, c)=b$, and $f(a, b, c)=$ median of the numbers $\{a, b, c\}$ all fit.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 4 adventurers have rubies; exactly 10 have emeralds; exactly 6 have sapphires; exactly 14 have diamonds. Moreover, it is known that
- if an adventurer has rubies, then they have either emeralds or diamonds (but not both at the same time)
- if an adventurer has emeralds, then they have either rubies or sapphires (but not both at the same time).
What is the minimum number of adventurers that can be in such a group?
|
Answer: 18.
Solution. Note that the number of adventurers who have emeralds is equal to the total number of adventurers who have rubies or sapphires. Then, from the second condition, it follows that 4 adventurers have rubies and emeralds, and 6 have emeralds and sapphires. That is, every adventurer who has rubies must also have emeralds. Then, from the first condition, there cannot be an adventurer who has both rubies and diamonds. Therefore, there must be at least $4+14=18$ adventurers.
Indeed, there can be this many adventurers: suppose we have 4 adventurers who have emeralds and rubies, 6 adventurers who have diamonds, emeralds, and sapphires, and 10 adventurers who have only sapphires. One can verify that this example meets all the conditions.
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. A team of workers was laying linoleum in a store's warehouse and in the cash hall, with the warehouse area being 3 times larger than the cash hall area. In the part of the team working in the warehouse, there were 5 more workers than in the part working in the cash hall. When the work in the warehouse was completed, the part of the team working in the cash hall was still working. What is the maximum number of workers that could have been in the team?
|
Answer: 9.
Solution. Let the number of workers in the cash hall be denoted as $n$, then the number of workers in the warehouse is $n+5$, and the total number of people in the team is $2n+5$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivity of each part of the team is $an$ and $a(n+5)$. If the area of the cash hall is $S$, then the area of the warehouse is $3S$. The time spent on completing the entire work by each part of the team is $\frac{S}{an}$ and $\frac{3S}{a(n+5)}$, respectively. According to the problem, $\frac{S}{an} > \frac{3S}{a(n+5)}$. Due to the positivity of all variables, this inequality is equivalent to $n+5 > 3n$, or $2n < 5$. Therefore, $n \leqslant 2$, and $2n+5 \leqslant 9$. The equality situation is clearly possible: just take any positive $S$ and $a$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $BC=5$ and $AB=4$. Find the length of side $AC$ if the length of the vector $3 \overrightarrow{OA}-4 \overrightarrow{OB}+\overrightarrow{OC}$ is 10.
|
Answer: 8.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cdot \overrightarrow{O B})+2 y z(\overrightarrow{O B} \cdot \overrightarrow{O C})+2 x z(\overrightarrow{O A} \cdot \overrightarrow{O C})= \\
& =\left(x^{2}+y^{2}+z^{2}\right) R^{2}+x y\left(O A^{2}+O B^{2}-A B^{2}\right)+y z\left(O B^{2}+O C^{2}-B C^{2}\right)+x z\left(O A^{2}+O C^{2}-A C^{2}\right)= \\
& =\left(x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right) R^{2}-x y A B^{2}-y z B C^{2}-x z A C^{2}= \\
& =(x+y+z)^{2} R^{2}-x y A B^{2}-y z B C^{2}-x z A C^{2}
\end{aligned}
$$
Then, for $x=3, y=-4, z=1$, we obtain the equality $10^{2}=0 \cdot R^{2}+12 A B^{2}+4 B C^{2}-3 A C^{2}$, from which $A C^{2}=\frac{1}{3}\left(12 A C^{2}+4 B C^{2}-100\right)=64$, i.e., $A C=8$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. The function $G$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$, the equalities $G(n a, n b, n c)=n \cdot G(a, b, c)$, $G(a+n, b+n, c+n)=G(a, b, c)+n$, $G(a, b, c)=G(c, b, a)$ hold. Find $G(89,90,91)$.
|
Answer: 90.
Solution. Note that $G(-1,0,1)=G(1,0,-1)=(-1) \cdot G(-1,0,1)$, from which $G(-1,0,1)=0$. Then $G(89,90,91)=G(-1,0,1)+90=90$.
Comment. The function $G$ cannot be uniquely determined. For example, the functions $G(a, b, c)=(a+b+c) / 3, G(a, b, c)=b$ and $G(a, b, c)=$ median of the numbers $\{a, b, c\}$ are suitable.
|
90
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. A group of adventurers is showing off their loot. It is known that exactly 9 adventurers have rubies; exactly 8 have emeralds; exactly 2 have sapphires; exactly 11 have diamonds. Moreover, it is known that
- if an adventurer has diamonds, then they have either rubies or sapphires (but not both at the same time);
- if an adventurer has rubies, then they have either emeralds or diamonds (but not both at the same time).
What is the smallest number of adventurers that can be in such a group?
|
Answer: 17.
Solution. Note that the number of adventurers who have diamonds is equal to the total number of adventurers who have rubies or sapphires. Then, from the first condition, it follows that 9 adventurers have rubies and diamonds, and 2 have sapphires and diamonds. That is, every adventurer who has rubies must also have diamonds. Then, from the second condition, there cannot be an adventurer who has both rubies and emeralds. Therefore, there must be at least $9+8=17$ adventurers.
Indeed, there can be this many adventurers: let's say we have 9 adventurers who have rubies and diamonds, 2 adventurers who have emeralds, sapphires, and diamonds, and 6 adventurers who have only emeralds. One can verify that this example meets all the conditions.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. A team of lumberjacks was cutting trees on a large and a small plot, with the area of the small plot being 3 times less than that of the large plot. In the part of the team that worked on the large plot, there were 8 more lumberjacks than in the part that worked on the small plot. When the tree harvesting on the large plot was completed, the part of the team that was on the small plot was still working. What is the maximum number of lumberjacks that could have been in the team?
|
Answer: 14.
Solution. Let the number of workers on the smaller plot be denoted as $n$, then the number of workers on the larger plot is $n+8$, and the total number of workers in the team is $2n+8$. The problem "implicitly assumes" that the productivity of each worker is the same, denoted as $a$. Therefore, the productivity of each part of the team is $an$ and $a(n+8)$. If the area of the smaller plot is $S$, then the area of the larger plot is $3S$. The time spent on completing the entire work by each part of the team is $\frac{S}{an}$ and $\frac{3S}{a(n+8)}$, respectively. According to the problem, $\frac{S}{an} > \frac{3S}{a(n+8)}$. Given the positivity of all variables, this inequality is equivalent to $n+8 > 3n$, or $n < 4$. Therefore, $n \leqslant 3$, and $2n+8 \leqslant 14$. The equality situation is clearly possible: just take any positive $S$ and $a$.
|
14
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $AB=5, AC=8$, and $BC=4$. Find the length of the vector $\overrightarrow{O A}-4 \overrightarrow{O B}+3 \overrightarrow{O C}$.
|
Answer: 10.
Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:
$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cdot \overrightarrow{O B})+2 y z(\overrightarrow{O B} \cdot \overrightarrow{O C})+2 x z(\overrightarrow{O A} \cdot \overrightarrow{O C})= \\
& =\left(x^{2}+y^{2}+z^{2}\right) R^{2}+x y\left(O A^{2}+O B^{2}-A B^{2}\right)+y z\left(O B^{2}+O C^{2}-B C^{2}\right)+x z\left(O A^{2}+O C^{2}-A C^{2}\right)= \\
& =\left(x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right) R^{2}-x y A B^{2}-y z B C^{2}-x z A C^{2}= \\
& =(x+y+z)^{2} R^{2}-x y A B^{2}-y z B C^{2}-x z A C^{2}
\end{aligned}
$$
Then, for $x=1, y=-4, z=3$, we obtain the equality $(\overrightarrow{O A}-4 \overrightarrow{O B}+3 \overrightarrow{O C})^{2}=0 \cdot R^{2}+4 A B^{2}+12 B C^{2}-$ $3 A C^{2}=100$, i.e., the answer is 10.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. In triangle $A B C$, side $B C=28$. The bisector $B L$ is divided by the point of intersection of the bisectors of the triangle in the ratio $4: 3$, counting from the vertex. Find the radius of the circumscribed circle around triangle $A B C$, if the radius of the inscribed circle in it is 12.
|
Answer: 50.
Solution. Let $I$ be the center of the inscribed circle in triangle $ABC$ (i.e., the point of intersection of the angle bisectors). Noting that $CI$ is the angle bisector in triangle $BLC$, by the angle bisector theorem, we have: $BC: CL = BI: IL = 4: 3$, from which $CL = 3BC / 4 = 21$.
Next, $BC \cdot CL \cdot \sin \angle C = 2 S_{\triangle BCL} = 2 S_{\triangle BIC} + 2 S_{\triangle CIL} = BC \cdot r + CL \cdot r = (BC + CL) \cdot r$, where $r$ is the radius of the inscribed circle in triangle $ABC$. Thus, $28 \cdot 21 \cdot \sin \angle C = (28 + 21) \cdot 12$, i.e., $\sin \angle C = 1, \angle C = 90^\circ$.
By the angle bisector theorem for $AI$ in triangle $ALB$, we have $BA: AL = BI: IL = 4: 3$. Letting $AL = 3x$, we have $AB = 4x$. By the Pythagorean theorem: $AC^2 + BC^2 = AB^2$, i.e., $28^2 + (21 + 3x)^2 = (4x)^2$, from which $x = 25$, and $R = AB / 2 = 4 \cdot 25 / 2 = 50$.
|
50
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. The function $g$ is defined on the set of triples of integers and takes real values. It is known that for any four integers $a, b, c$ and $n$ the equalities $g(n a, n b, n c)=$ $n \cdot g(a, b, c), g(a+n, b+n, c+n)=g(a, b, c)+n, g(a, b, c)=g(c, b, a)$ hold. Find $g(14,15,16)$.
|
Answer: 15.
Solution. Note that $g(-1,0,1)=g(1,0,-1)=(-1) \cdot g(-1,0,1)$, from which $g(-1,0,1)=0$. Then $g(14,15,16)=g(-1,0,1)+15=15$.
Comment. The function $g$ cannot be uniquely determined. For example, the functions $g(a, b, c)=(a+b+c) / 3$, $g(a, b, c)=b$, and $g(a, b, c)=$ median of the numbers $\{a, b, c\}$ all fit.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. In the $1^{\text{st}}$ grade class, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 2. Among the answers received were: $(13,11),(17,11),(14,14)$. How many boys and how many girls are in the class?
|
Answer: 15 boys and 12 girls.
Solution. First solution. Let's denote the children who gave the answers (13,11), (17,11), (14,14) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the answers of the girls has the same parity as $m$, and in the answers of the boys - the opposite. Therefore, children A and B are of the same gender, and C is of the other gender.
The first numbers in the answers of A and B differ by 4, so they are both incorrect. Thus, the number of classmates for A and B is 15, and the number of female classmates is 11.
If A and B are boys, then there are 16 boys and 11 girls in the class. In this case, girl C would have 16 classmates and 10 female classmates, and her answer (14,14) would contradict the condition. Therefore, A and B are girls, and there are 15 boys and 12 girls in the class.
Second solution. Suppose a child wrote the numbers $(m, d)$. If he wrote both numbers correctly, he would have written one of the four options: $(m-2, d),(m+2, d),(m, d-2),(m, d+2)$. Then, if this child is a boy, there are four possible options for the number of boys and girls in the class: $(m-1, d),(m+3, d),(m+1, d-2)$, and $(m+1, d+2)$. Similarly, if this child is a girl, the possible options in this case are: $(m-2, d+1),(m+2, d+1),(m, d-1),(m, d+3)$.
Thus, each answer gives us eight possible options for how many boys and girls could be in the class, one of which must be correct:
for $(13,11)$ these are $(12,11),(16,11),(14,9),(14,13),(11,12),(15,12),(13,10),(13,14)$;
for $(17,11)$ these are $(16,11),(20,11),(18,9),(18,13),(15,12),(19,12),(17,10),(17,14)$;
for $(14,14)$ these are $(13,14),(17,14),(15,12),(15,16),(12,15),(16,15),(14,13),(14,17)$.
It remains to note that only the option $(15,12)$ appears in all three rows.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=5: 4$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $4 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C}$ is 2016.
|
Answer: 224.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=5: 4$, from which $\overrightarrow{B L}=5 / 9 \cdot \overrightarrow{B C}$. Now
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{5}{9} \cdot \overrightarrow{B C}=\overrightarrow{A B}+\frac{5}{9}(\overrightarrow{A C}-\overrightarrow{A B})=\frac{4}{9} \cdot \overrightarrow{A B}+\frac{5}{9} \cdot \overrightarrow{A C}=\frac{1}{9}(4 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C})
$$
Therefore, $|A L|=1 / 9 \cdot|4 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C}|=1 / 9 \cdot 2016=224$.
|
224
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 9. The Federation of Wrestling has assigned each participant in the competition a qualification number. It is known that in matches between wrestlers whose qualification numbers differ by more than 2, the wrestler with the lower number always wins. The tournament for 256 wrestlers is held on an Olympic system: at the beginning of each day, the wrestlers are paired, the loser is eliminated from the competition (there are no draws). What is the highest qualification number that the winner can have?
|
Answer: 16
Solution. Note that a wrestler with number $k$ can only lose to a wrestler with number $k+1$ or $k+2$, so after each round, the smallest number cannot increase by more than 2 numbers. In a tournament with 256 participants, there are 8 rounds, so the number of the tournament winner does not exceed $1+2 \cdot 8=17$.
Suppose the wrestler with number 17 can win. Then in the first round, the wrestlers with numbers 1 and 2 must be eliminated. This is only possible if the wrestler with number 1 loses to the wrestler with number 3, and the wrestler with number 2 loses to the wrestler with number 4. Thus, after the first round, the wrestlers with numbers 3 and 4 will remain. Similarly, after the second round, the wrestlers with numbers 5 and 6 will remain, after the third round - 7 and $8, \ldots$, after the seventh round - 15 and 16. This means that in the final, decisive match, the wrestlers with numbers 15 and 16 will meet. Contradiction with the assumption that the wrestler with number 17 can win.
We will show that the wrestler with number 16 can win. Let's call the wrestlers with numbers greater than 16 weak. Suppose in the round with number $k \leqslant 7$ the wrestler with number $2 k-1$ loses to the wrestler with number $2 k+1$, the wrestler with number $2 k$ loses to the wrestler with number $2 k+2$, the wrestlers with numbers $2 k+3, \ldots, 16$ win against some weak wrestlers, and the remaining weak wrestlers play among themselves. Then after 7 rounds, the wrestlers with numbers 15 and 16 will remain, and in the final match, the wrestler with number 16 will win.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. In $1^{st}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 4. Among the answers received were: $(15,18),(15,10),(12,13)$. How many boys and how many girls are in the class?
|
Answer: 16 boys and 14 girls.
Solution. First solution. Let's denote the children who gave the answers (15,18), (15,10), (12,13) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the answers of girls has the same parity as $m$, while in the answers of boys, it has the opposite parity. Therefore, children A and B are of the same gender, and C is of the opposite gender.
The second numbers in the answers of A and B differ by 8, which means they are both incorrect. Thus, the number of classmates for A and B is 15, and the number of female classmates is 14.
If A and B are girls, then there are 15 boys and 15 girls in the class. In this case, boy C would have 14 classmates and 15 female classmates, and his answer (12,13) would contradict the condition. Therefore, A and B are boys, and there are 16 boys and 14 girls in the class.
Second solution. Suppose a child wrote the numbers $(m, d)$. If he wrote both numbers correctly, he would have written one of the four options: $(m-4, d), (m+4, d), (m, d-4), (m, d+4)$. Then, if this child is a boy, there are four possible combinations of the number of boys and girls in the class: $(m-3, d), (m+5, d), (m+1, d-4)$, and $(m+1, d+4)$. Similarly, if this child is a girl, the possible combinations in this case are: $(m-4, d+1), (m+4, d+1), (m, d-3), (m, d+5)$.
Thus, each answer gives us eight possible combinations of the number of boys and girls in the class, one of which must be correct:
for $(15,18)$, these are $(12,18), (20,18), (16,14), (16,22), (11,19), (19,19), (15,15), (15,23)$;
for $(15,10)$, these are $(12,10), (20,12), (16,6), (16,14), (11,11), (19,11), (15,7), (15,15)$;
for (12,13), these are (9,13), (17,15), (13,9), (13,17), (8,14), (16,14), (12,10), (12,18).
It remains to note that only the option $(16,14)$ appears in all three rows.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=4: 3$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $3 \cdot \overrightarrow{A B}+4 \cdot \overrightarrow{A C}$ is 2016.
|
Answer: 288.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=4: 3$, from which $\overrightarrow{B L}=4 / 7 \cdot \overrightarrow{B C}$. Now
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{4}{7} \cdot \overrightarrow{B C}=\overrightarrow{A B}+\frac{4}{7}(\overrightarrow{A C}-\overrightarrow{A B})=\frac{3}{7} \cdot \overrightarrow{A B}+\frac{4}{7} \cdot \overrightarrow{A C}=\frac{1}{7}(3 \cdot \overrightarrow{A B}+4 \cdot \overrightarrow{A C})
$$
Therefore, $|A L|=1 / 7 \cdot|3 \cdot \overrightarrow{A B}+4 \cdot \overrightarrow{A C}|=1 / 7 \cdot 2016=288$.
|
288
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 9. The Federation of Wrestling has assigned each participant in the competition a qualification number. It is known that in matches between wrestlers whose qualification numbers differ by more than 2, the wrestler with the lower number always wins. The tournament for 512 wrestlers is held on an Olympic system: at the beginning of each day, the wrestlers are divided into pairs, the loser is eliminated from the competition (there are no draws). What is the highest qualification number that the winner can have?
|
Answer: 18.
Solution. Note that a wrestler with number $k$ can only lose to a wrestler with number $k+1$ or $k+2$, so after each round, the smallest number cannot increase by more than 2 numbers. In a tournament with 512 participants, there are 9 rounds, so the number of the tournament winner does not exceed $1+2 \cdot 9=19$.
Suppose the wrestler with number 19 can win. Then in the first round, the wrestlers with numbers 1 and 2 must be eliminated. This is only possible if the wrestler with number 1 loses to the wrestler with number 3, and the wrestler with number 2 loses to the wrestler with number 4. Thus, after the first round, the wrestlers with numbers 3 and 4 will remain. Similarly, after the second round, the wrestlers with numbers 5 and 6 will remain, after the third round - 7 and 8, ..., after the eighth round - 17 and 18. This means that in the final, decisive battle, the wrestlers with numbers 17 and 18 will meet. Contradiction with the assumption that the wrestler with number 19 can win.
We will show that the wrestler with number 18 can win. Let's call the wrestlers with numbers greater than 18 weak. Suppose in the round with number $k \leqslant 8$ the wrestler with number $2 k-1$ loses to the wrestler with number $2 k+1$, the wrestler with number $2 k$ loses to the wrestler with number $2 k+2$, the wrestlers with numbers $2 k+3, \ldots, 18$ win against some weak wrestlers, and the remaining weak wrestlers play among themselves. Then after 8 rounds, the wrestlers with numbers 17 and 18 will remain, and in the final battle, the wrestler with number 18 will win.
|
18
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. In $1^{\text {st }}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 2. Among the answers received were: $(12,18),(15,15),(11,15)$. How many boys and how many girls are in the class?
|
Answer: 13 boys and 16 girls.
Solution. First solution. Let's denote the children who gave the answers (12,18), (15,15), (11,15) as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the girls' answers has the same parity as $m$, and in the boys' answers - the opposite. Therefore, children B and C are of the same gender, and A is of the other gender.
The first numbers in the answers of B and C differ by 4, so they are both incorrect. Thus, the number of classmates for B and C is 13, and the number of female classmates is 15.
If B and C are boys, then there are 14 boys and 15 girls in the class. In this case, girl A would have 14 classmates and 14 female classmates, and her answer (12,18) would contradict the condition. Therefore, B and C are girls, and there are 13 boys and 16 girls in the class.
Second solution. Suppose a child wrote the numbers $(m, d)$. If he wrote both numbers correctly, then he would have written one of the four options: $(m-2, d),(m+2, d),(m, d-2),(m, d+2)$. Then, if this child is a boy, there are four possible options for the number of boys and girls in the class: $(m-1, d),(m+3, d),(m+1, d-2)$, and $(m+1, d+2)$. Similarly, if this child is a girl, the possible options in this case are: $(m-2, d+1),(m+2, d+1),(m, d-1),(m, d+3)$.
Thus, each answer gives us eight options for how many boys and girls could be in the class, one of which must be correct:
for (12,18) these are (11,18),(15,18),(13,16),(13,20),(10,19),(14,19),(12,17),(12,21);
for (15,15) these are (14,15),(18,15),(16,13),(16,17),(13,16),(17,16),(15,14),(15,18);
for (11,15) these are (10,15),(14,15),(12,13),(12,17),(9,16),(13,16),(11,14),(11,18).
It remains to note that only the option (13,16) appears in all three rows.
|
13
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=5: 2$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $2 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C}$ is 2016.
|
Answer: 288.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=5: 2$, from which $\overrightarrow{B L}=5 / 7 \cdot \overrightarrow{B C}$. Now,
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{4}{7} \cdot \overrightarrow{B C}=\overrightarrow{A B}+\frac{5}{7}(\overrightarrow{A C}-\overrightarrow{A B})=\frac{2}{7} \cdot \overrightarrow{A B}+\frac{5}{7} \cdot \overrightarrow{A C}=\frac{1}{7}(2 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C})
$$
Therefore, $|A L|=1 / 7 \cdot|2 \cdot \overrightarrow{A B}+5 \cdot \overrightarrow{A C}|=1 / 7 \cdot 2016=288$.
|
288
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. In the $1^{\text{st}}$ grade, each child was asked to write down two numbers: the number of their classmates and the number of their female classmates (in that exact order; the child does not count themselves). Each child wrote one number correctly, and the other number was off by exactly 4. Among the answers received were: $(10,14),(13,11),(13,19)$. How many boys and how many girls are in the class?
|
Answer: 14 boys and 15 girls.
Solution. First solution. Let's denote the children who gave the answers $(10,14),(13,11),(13,19)$ as A, B, and C, respectively. Note that if there are $m$ boys in the class, then the first number in the girls' answers has the same parity as $m$, and in the boys' answers - the opposite. Therefore, children B and C are of the same gender, and A is of the other gender.
The second numbers in the answers of B and C differ by 8, so they are both incorrect. Thus, the number of classmates for B and C is 13, and the number of female classmates is 15.
If B and C are girls, then there are 13 boys and 16 girls in the class. In this case, boy A would then have 12 classmates and 16 female classmates, and his answer $(10,14)$ would contradict the condition. Therefore, B and C are boys, and there are 14 boys and 15 girls in the class.
Second solution. Suppose a child wrote the numbers $(m, d)$. If he wrote both numbers correctly, he would have written one of the four options: $(m-4, d),(m+4, d),(m, d-4),(m, d+4)$. Then, if this child is a boy, there are four possible combinations of the number of boys and girls in the class: $(m-3, d),(m+5, d),(m+1, d-4)$, and $(m+1, d+4)$. Similarly, if this child is a girl, the possible combinations in this case are: $(m-4, d+1),(m+4, d+1),(m, d-3),(m, d+5)$.
Thus, each answer gives us eight possible combinations of the number of boys and girls in the class, one of which must be correct:
for $(10,14)$ these are $(7,14),(15,14),(11,10),(11,18),(6,15),(14,15),(10,11),(10,19)$;
for $(13,11)$ these are $(10,11),(18,11),(14,7),(14,15),(9,12),(17,12),(13,8),(13,16)$;
for $(13,19)$ these are $(10,19),(18,19),(14,15),(14,23),(9,20),(17,20),(13,16),(13,24)$.
It remains to note that only the variant $(14,15)$ appears in all three rows.
|
14
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. How many natural numbers $n>1$ exist, for which there are $n$ consecutive natural numbers, the sum of which is equal to 2016?
|
Answer: 5.
Solution. Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a+(n-1) n / 2=2016$, or, after algebraic transformations, $n(2 a+n-1)=4032=2^{6} \cdot 3^{2} \cdot 7$.
Note that $n$ and $2 a+n-1$ have different parity. Therefore, if $n$ is even, then $n$ must be divisible by $2^{6}=64$, from which $n \geqslant 64, 2 a+n-1 \leqslant 4032 / 64=63$, which contradicts $2 a+n-1>n$.
Thus, $n$ is an odd divisor of the number 4032, i.e., a divisor of the number 63. Let's check that each of them fits. If $n=3: 2 a+3-1=1344, a=671$; if $n=7: 2 a+7-1=576, a=285$; if $n=9$ : $2 a+9-1=448, a=220$; if $n=21: 2 a+21-1=192, a=86$; if $n=63: 2 a+63-1=64, a=1$. In total, there are five suitable $n$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. In triangle $A B C$ with the ratio of sides $A B: A C=7: 2$, the bisector of angle $B A C$ intersects side $B C$ at point $L$. Find the length of segment $A L$, if the length of the vector $2 \cdot \overrightarrow{A B}+7 \cdot \overrightarrow{A C}$ is 2016.
|
Answer: 224.
Solution. Note that by the property of the angle bisector of a triangle $B L: L C=B A: A C=7: 2$, from which $\overrightarrow{B L}=7 / 9 \cdot \overrightarrow{B C}$. Now,
$$
\overrightarrow{A L}=\overrightarrow{A B}+\overrightarrow{B L}=\overrightarrow{A B}+\frac{7}{9} \cdot \overrightarrow{B C}=\overrightarrow{A B}+\frac{7}{9}(\overrightarrow{A C}-\overrightarrow{A B})=\frac{2}{9} \cdot \overrightarrow{A B}+\frac{7}{9} \cdot \overrightarrow{A C}=\frac{1}{9}(2 \cdot \overrightarrow{A B}+7 \cdot \overrightarrow{A C})
$$
Therefore, $|A L|=1 / 9 \cdot|2 \cdot \overrightarrow{A B}+7 \cdot \overrightarrow{A C}|=1 / 9 \cdot 2016=224$.
|
224
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. Egor borrowed 28 rubles from Nikita, and then returned them in four payments. It turned out that Egor always returned a whole number of rubles, and the amount of each payment increased and was divisible by the previous one. What was the last amount Egor paid?
|
Answer: 18 rubles
Solution:
1) If Egor paid $a$ rubles the first time, then the second time - not less than $2a$, the third - not less than $4a$, the fourth - not less than $8a$, and in total - not less than $15a$. Since $15a \leq 28$, we get that $a=1$.
2) The second time he paid 2 or 3 rubles (because if 4, then he paid at least $1+4+8+16=29>28$).
2.1) If he paid 2 rubles, then he had 25 rubles left to pay, and in this case, all subsequent payments would be even. This case is impossible.
2.2) Therefore, he paid 3 rubles, and for the last two times, he had 24 rubles left to pay. Let's say he paid $a$ times more the third time than the second, and $b$ times more the fourth time than the third. Then $3a + 3ab = 24$, which means $a + ab = a(b+1) = 8$. It follows that $a$ and $b+1$ are powers of two, and $b>1$, which is only possible with $a=2$ and $b=3$. Hence the answer.
## Criteria:
Only the answer $0-0$ points.
Only the answer with verification (all 4 sums are written out, showing that all conditions are met) - 1 point, it does not add up with anything.
Each of the points 1), 2), 2.1) is worth one point, these points are added up. A generally correct solution, but 2.2) is proven by incomplete enumeration - no more than 5 points.
|
18
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. On the surface of a pentagonal pyramid (see fig.), several gnomes live in pairwise distinct points, and they can live both inside the faces and on the edges or at the vertices. It turned out that on each face (including the vertices and edges that bound it) a different number of gnomes live. What is the minimum number of gnomes living on the pyramid?
|
Answer: 6
Solution: There are 6 faces in total, so at least 5 gnomes live on the most "populated" one. If there are exactly 5 gnomes in total, then all of them live on one face (let's call it face $A$), so the faces where 4 and 3 gnomes live (let's call them $B$ and $C$ respectively) are adjacent to it. Then, on the edge $A-B$ there should be 4, and on the edge $A-C$ there should be 3 gnomes. This means there are already at least 7 gnomes, and we could have counted only one of them twice - the one sitting at the vertex connecting faces $A$, $B$, and $C$. Therefore, there are at least 6 gnomes - a contradiction. An example with 6 gnomes is shown in the figure.
## Criteria:
Only the answer - 0 points.
Only a correct example - 2 points.
Only proof that there cannot be fewer than 5 - 1 point.
Only proof that there cannot be fewer than 6 - 3 points.
These points DO NOT add up.
Justification of the example is not required.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.4. Ten chess players over nine days played a full round-robin tournament, during which each of them played exactly one game with each other. Each day, exactly five games were played, with each chess player involved in exactly one of them. For what maximum $n \leq 9$ can it be claimed that, regardless of the schedule, at the end of some game day with number $k \leq 8$, there will definitely be $n$ chess players who have already played all their scheduled games with each other?
|
Answer. $n=5$.
Solution. By the end of the eighth day, each chess player has played 8 games, meaning they have not played one. The unplayed games divide the chess players into 5 non-intersecting pairs. By the Pigeonhole Principle, among any six chess players, there will always be two belonging to the same pair, that is, who have not played against each other even by the end of the penultimate day of the tournament. Therefore, $n \leq 5$.
We will show that by the end of the seventh day of the tournament, regardless of the schedule, there will always be five chess players who have already played all their scheduled games with each other, from which it follows that $n=5$. We will construct a chain of chess players: the first is any player, the second is the one who played the first on the eighth day, the third is the one who played the second on the ninth day, the fourth is the one who played the third on the eighth day, and so on. Since the number of chess players is finite, at some point in the chain, a repetition will occur for the first time. This cannot happen at any player except the first, otherwise, the repeated player would have played at least three games in two days. Therefore, the chain will form a cycle of even length, as the days of the games considered alternate when moving along it. If the constructed cycle includes all 10 chess players, we stop; if not, we choose any player not included in it and repeat the procedure for constructing the cycle. The new cycle will not intersect with the already constructed one. After several steps, we will construct several cycles of even length, not less than 4, including all 10 chess players (there are few options - either 10 or $6+4$). We select half of the participants from each cycle, every other one, totaling 5 chess players. No two of them are adjacent in the cycles, so they did not play against each other on the eighth and ninth days.
Therefore, all games between them were played in the first 7 days. Thus, $n \geq 5$, and hence $n=5$. In this case, $k=7$.
Grading criteria. ($\cdot$) Proof that $n \leq 5: 3$ points. ($\cdot$) Proof that $n \geq 5: 4$ points.
10.5.. Find the maximum value of the sum $x_{1}\left(1-x_{2}\right)+x_{2}\left(1-x_{3}\right)+\ldots+x_{6}\left(1-x_{7}\right)+x_{7}\left(1-x_{1}\right)$ for arbitrary numbers $x_{1}, x_{2}, \ldots, x_{7}$ from the interval $[0,1]$.
Answer. 3.
Solution. We isolate in the left-hand side of the inequality all terms containing the variable $x_{1}$, obtaining $x_{1}\left(1-x_{2}-x_{7}\right)$. For fixed other variables, this expression, and hence the entire left-hand side, reaches its maximum when $x_{1}=1$ if $x_{2}+x_{7}<1$ and when $x_{1}=0$ if $x_{2}+x_{7} \geq 1$. Therefore, if we replace each variable, starting from $x_{1}$ to $x_{7}$, with 1 or 0 according to the condition whether the sum of the neighboring variables is less than 1 or not less than 1, we will not decrease the value of the sum on the right-hand side of the inequality. Thus, the maximum of the right-hand side is achieved for a set of $x_{1}, x_{2}, \ldots, x_{7}$ where each variable is either 1 or 0. Moreover, in such a set, two adjacent variables cannot both be 1, as in this case, the second one could not have appeared according to the replacement rules. Therefore, the set contains no more than 3 ones, and the rest are zeros, meaning that no more than three terms in the right-hand side of the inequality are non-zero, each obviously not exceeding 1, and the entire sum is no more than 3. An example where the value 3 is achieved: $x_{1}=x_{3}=x_{5}=1$, the rest are 0.
Grading criteria. Proven that the maximum of the sum does not exceed 3: 5 points. Example where the value 3 is achieved: 2 points.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. At a round table, 410 deputies sat down, each of whom was either a knight, who always tells the truth, or a liar, who always lies. Each of the deputies said: "Among my twenty neighbors to the left and twenty neighbors to the right, there are exactly 20 liars in total." It is known that at least half of the people at the table are liars. How many knights are there at the table?
|
Answer: None.
Solution: Let's divide all the people sitting at the table into ten groups of 41 people each. Then, at least one group will have at least 21 liars. Otherwise, in each group, there would be a maximum of 20, i.e., no more than \(20 \times 10 = 200\) in total, which is less than half of the total number. Consider this group.
If a knight is sitting in the center, he tells the truth, and there are exactly 20 liars in this group. But we have already established that there are at least 21. Contradiction. Therefore, a liar is sitting in the center, and there are at least 22 liars in the group (there cannot be exactly 21, as otherwise, the central person would be telling the truth). Now consider the left neighbor of the central liar. Among his 40 neighbors, there are at least 21 liars, since we lost one extreme neighbor (who might be a liar) by shifting to the left, but gained a liar neighbor who was previously central. Repeating the reasoning for him, we get that he is also a liar. Shift left again and continue this process. Eventually, we will get that everyone sitting at the table is a liar.
## Criteria:
Only the answer, answer with verification - 0 points.
Proved that there will be a group with 21 liars - 2 points.
Proved that in this group, a liar is sitting in the center - another 1 point.
|
0
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. On a circular route, two buses operate with the same speed and a movement interval of 21 minutes. What will be the movement interval if 3 buses operate on this route at the same constant speed?
|
Answer: 14.
Solution: Since the interval of movement with two buses on the route is 21 minutes, the length of the route in "minutes" is 42 minutes. Therefore, the interval of movement with three buses on the route is $42: 3=14$ minutes.
Criteria: only answer, answer with verification - 3 points.
|
14
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. Natasha and Inna bought the same box of tea bags. It is known that one tea bag is enough for two or three cups of tea. Natasha's box was enough for only 41 cups of tea, while Inna's was enough for 58 cups. How many tea bags were in the box?
|
Answer: 20.
Solution: Let there be $n$ tea bags in the box. Then the number of brewings can vary from $2n$ to $3n$. Therefore, 58 is not greater than $3n$, which means $19<n$. Additionally, 41 is not less than $2n$, so $n<21$. Since the number of tea bags must be a natural number that is less than 21 but greater than 19, there are exactly 20 tea bags in the box.
Criteria: answer only - 0 points, answer with verification - 1 point. Correct system of inequalities without conclusions - 2 points, correct system of inequalities and correct answer - 4 points. Proof of one of the estimates for the number of packets in the pack (that there are not fewer than 20 or not more than 20) - 3 points regardless of whether the correct answer is given. These points DO NOT add up.
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. On a line, one hundred points are marked: green, blue, and red. It is known that between any two red points there is a blue one, and between any two blue points there is a green one. In addition, there are no fewer red points than blue ones, and no fewer blue points than green ones. How many points are painted blue?
|
Answer: 33.
Solution: Let the number of red points be $n$, then the number of blue points is not less than $n-1$ (the number of intervals between "adjacent" red points), and since by condition the number of red points is not less than the number of blue points, the number of blue points is either $n$ or $n-1$. Similarly, if the number of blue points is $m$, then the number of green points is either $m$ or $m-1$. Thus, there are 4 possible cases.
| red | blue | green | total |
| :--: | :--: | :--: | :--: |
| $n$ | $n$ | $n$ | $3n$ |
| $n$ | $n$ | $n-1$ | $3n-1$ |
| $n$ | $n-1$ | $n-1$ | $3n-2$ |
| $n$ | $n-1$ | $n-2$ | $3n-3$ |
In each case, it is easy to find the total number of points (see table), which by condition is 100. Thus, we get 4 equations, only one of which has an integer solution $100=3n-2$. From this, the number of red points is 34, blue - 33, and green - 33.
Criteria: only answer, answer with verification - 0 points. Obtaining the estimate that the number of blue points is not less than the number of red points minus 1, and the number of green points is not less than the number of blue points minus 1 - 3 points (for both estimates together, not for each).
## Solutions to the tasks of the first stage of the All-Siberian Open Olympiad for schoolchildren in mathematics 2014-15, November 9, 2014
## 8th grade
Each task is worth 7 points.
|
33
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. On a circular route, 12 trams run in one direction at the same speed and at equal intervals. How many trams need to be added so that at the same speed, the intervals between trams decrease by one fifth?
#
|
# Answer: 3.
Solution: Let's take the entire distance as 60 arbitrary units, which means the trams are currently 5 arbitrary units apart. We want this distance to be reduced by $1 / 5$, making it equal to 4 arbitrary units. For this, we need $60: 4=15$ trams, which is 3 more than the current number.
Criteria: only the answer - 0 points, answer with verification - 2 points, deduct no less than 2 points for arithmetic errors.
|
3
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. Each digit of the natural number $N$ is strictly greater than the one to its left. What is the sum of the digits of the number $9 N$?
|
Answer: 9.
Solution: Note that $9 N=10 N-N$. Let's perform this subtraction in a column. In the units place, there will be the difference between 10 and the last digit of the number $N$, in the tens place - the last and the second-to-last digit, decreased by 1. In all subsequent places, there will be the difference between two adjacent digits, as the smaller digit will always be subtracted from the larger one. Each digit will appear once as the minuend and once as the subtrahend. When finding the sum of the digits, all these minuends and subtrahends will cancel each other out. Only the aforementioned $10-1=9$ will remain.
Criteria: only the answer, the answer with verification on specific cases - 0 points, the scheme of proof by example -2 points, the transition $9 N=10 N-N-1$ point. These points $\mathrm{NE}$ are not summed.
## Solutions to the tasks of the first stage of the All-Siberian Open Olympiad for schoolchildren in mathematics 2014-15
## 9th grade
## Each task is worth 7 points
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. In the morning, a dandelion blooms, it flowers yellow for three days, on the fourth morning it turns white, and by the evening of the fifth day, it withers. On Monday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and on Wednesday - 15 yellow and 11 white. How many white dandelions will there be on the meadow on Saturday?
|
Answer. Six dandelions.
Solution. A blooming dandelion is white on the fourth and fifth day. This means that on Saturday, the dandelions that bloomed on Tuesday or Wednesday will be white. Let's determine how many there are. The dandelions that were white on Monday had already flown away by Wednesday, and 20 yellow ones definitely survived until Wednesday (possibly turning white).
On Wednesday, there were $15+11=26$ dandelions on the meadow. We know that 20 of them were on the meadow since Monday, and the remaining $26-20=6$ bloomed on Tuesday and Wednesday.
Criteria. The correct answer is guessed with verification: 2 points.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. Inside a semicircle of radius 12, there are a circle of radius 6 and a small semicircle, each touching the others pairwise, as shown in the figure. Find the radius of the small semicircle.
|
Answer: 4.
Solution. Let the radius of the small semicircle be $x$, the center of the large semicircle be $A$, the center of the circle be $B$, and the center of the small semicircle be $C$. The centers of the tangent circles and semicircle, and the corresponding points of tangency, lie on the same straight line, so $A B C$ is a right triangle with legs $A B=6, A C=12-x, B C=6+x$. By the Pythagorean theorem, we have $(x+6)^{2}=6^{2}+(12-x)^{2}$, from which $x=4$.
Criteria. It should be clearly explained why $A B=6, A C=12-x, B C=6+x$, otherwise 2 points will be deducted.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.4. In a semicircle with a radius of 18 cm, a semicircle with a radius of 9 cm is constructed on one of the halves of the diameter, and a circle is inscribed, touching the larger semicircle from the inside, the smaller semicircle from the outside, and the second half of the diameter. Find the radius of this circle.
#
|
# Answer: 8 cm.
Solution. Let O, O_1, O_2 be the centers of the large semicircle, the small semicircle, and the inscribed circle, respectively, and let P, Q, R be the points of tangency of the inscribed circle with the diameter of the large semicircle, the small semicircle, and the large semicircle, respectively. Then the points O_1, Q, O_2 lie on the same line, and the points O, O_2, R lie on the same line. Let the radius of the inscribed circle be x. From the Pythagorean theorem for triangle OO_2P, we have OP = \sqrt{(18-x)^2 - x^2} = \sqrt{324 - 36x}. From the Pythagorean theorem for triangle O_1O_2P, we have O_1P = \sqrt{(9 + x)^2 - x^2} = \sqrt{81 + 18x}. Therefore, O_1P = OP + 9 = \sqrt{324 - 36x} + 9 = \sqrt{81 + 18x}. Dividing by 3, we get \sqrt{36 - 4x} + 3 = \sqrt{9 + 2x}. From the last equation, x = 8.
Criteria. There is no clear justification for the calculation of OP and O_1P: deduct 2 points. The equation for x is correctly set up, but not solved: 3 points.
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8
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Geometry
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math-word-problem
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Yes
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Yes
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olympiads
| false
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