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# Problem 4. Maximum 20 points
Svetlana is convinced that any feline can be interesting to her for two purposes - catching mice and conducting cat therapy. Based on these considerations, Svetlana has decided to acquire 2 kittens, each of which has an equal probability of being either a male or a female. The weekly production possibility curve for males is $\mathrm{M}=80-4 \mathrm{~K}$, and for females $\mathrm{M}=16-0.25 \mathrm{~K}$, where $\mathrm{M}$ is the number of mice caught per week, and $\mathrm{K}$ is the number of hours devoted to cat therapy.
(a) Determine the maximum number of mice that 2 kittens can catch in a week with the optimal combination of males and females for this purpose.
(b) Determine the number of different variants of the combined production possibility curve Svetlana has when acquiring 2 kittens.
(c) Determine the analytical and graphical form of the combined production possibility curve for 2 kittens for each of the possible combinations of males and females.
(d) In the end, the kitten seller delivered the order, consisting of a male and a female, but also noted that the male has a twin brother, whom Svetlana can receive for free in her caring hands. Determine the analytical and graphical form of the combined production possibility curve if Svetlana agrees to the offer to take the third kitten.
|
# Solution:
(a) Males catch mice better than females. Two males can catch $40 \cdot 2=80$ mice in a week.
Answer: 160 (2 points).
(b) There are 3 options: one male and one female, two males, two females.
Answer: 3 (2 points).
(c) Two males with individual PPFs: $M=80-4K$. The opportunity costs for males are constant and equal (they can produce cat therapy in any order). The first produces $K_1$ units of cat therapy, and the second produces $K_2$ units of cat therapy. $K_{1}+K_{2}=K$. Joint PPF: $M=80-4 K_{1}+80-4 K_{2}$.
$$
M=80-4\left(K-K_{2}\right)+80-4 K_{2}=160-4 K
$$
Answer: two males $M=160-4 K, K \leq 40+$ graph (2 points).
Two females with individual PPFs: $M=16-0.25 K$. The solution is similar to the previous part: $M=32-0.25 K, K \leq 128$.
Answer: two females $M=32-0.25 K, K \leq 128+$ graph (2 points).
One male $M_{1}=80-4 K_{1}$ and one female $M_{2}=16-0.25 K_{2}$. The opportunity costs for the female are lower, so she produces cat therapy first.
$$
\begin{gathered}
K_{1}=0, K_{2}=K, K \leq 64 \\
K_{1}=K-64, K_{2}=64,64<K \leq 84
\end{gathered}
$$
The joint PPF can be obtained by substituting these values into the function $M=M_1 + M_2 = 80-4 K_{1} + 16-0.25 K_{2}$:
PPF:
$$
\begin{aligned}
& M=96-0.25 K, 0<K \leq 64 \\
& M=336-4 K, 64<K \leq 84
\end{aligned}
$$
Answer:
$$
M=96-0.25 K, 0<K \leq 64
$$
$M=336-4 K, 64<K \leq 84+$ graph (4 points)
(d) The problem reduces to finding the joint PPF for three individual functions:
$M=80-2 K, M=80-2 K, M=16-0.25 K$. Clearly, the problem can be simplified by combining the joint PPF of two males from the previous part $M=160-4 K$ with the PPF of one female $M=16-0.25 K$.
Similarly to the reasoning in the previous part, we get:
$$
\begin{gathered}
K_{1}=0, K_{2}=K, K \leq 64 \\
K_{1}=K-64, K_{2}=64,64<K \leq 104
\end{gathered}
$$
Substituting these values of $K_{1}$ and $K_{2}$ into $M=M_{1}+M_{2}=160-4 K_{1}+16-0.25 K_{2}$.
Joint PPF of two males and one female:
$$
\begin{gathered}
M=176-0.25 K, 0<K \leq 64 \\
M=416-4 K, 64<K \leq 104
\end{gathered}
$$
Answer:
$$
\begin{gathered}
M=176-0.25 K, 0<K \leq 64 \text { (4 points) } \\
M=416-4 K, 64<K \leq 104+\text { graph (4 points) }
\end{gathered}
$$
Arithmetic errors that did not lead to a distortion of the meaning of the problem - penalty 2 points.
Answers without solutions are not credited.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 6. Maximum 15 points
Find the values of the variable $x$ such that the four expressions: $2 x-6, x^{2}-4 x+5, 4 x-8, 3 x^{2}-12 x+11$ differ from each other by the same number. Find all possible integer values of the variable $x$ for any order of the expressions.
|
# Solution:
From the properties of numbers that differ from the following in a numerical sequence by the same number, we form two equations:
$$
\begin{gathered}
x^{2}-4 x+5-(2 x-6)=4 x-8-\left(x^{2}-4 x+5\right) \\
4 x-8-\left(x^{2}-4 x+5\right)=3 x^{2}-12 x+11-(4 x-8)
\end{gathered}
$$
Only \( x=4 \) satisfies both equations.
## Criteria.
- A correct solution to the problem and the correct answer are provided - 15 points (maximum),
- A correct solution to the problem is provided, but the wrong answer or no answer is given - 8 points (rounded to the nearest whole number, half the maximum),
- An error in the solution, but the correct answer is obtained - 7 points (rounded to the nearest whole number, half the maximum minus one),
- The solution is incorrect or missing, but the correct answer is obtained - 1 point (under any circumstances),
- The solution is incorrect or missing, and the wrong answer or no answer is given - 0 points.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 5. Maximum 15 points
The great-grandfather-banker left a legacy to his newborn great-grandson. According to the agreement with the bank, the amount in the great-grandson's account increases. Every year, on the day after the birthday, the current amount is increased by 1 million rubles more than in the previous year. Thus, if the initial amount was zero rubles, after one year +1 million rubles, after 2 years $1+2$ million, after 3 years $1+2+3$ and so on. According to the terms of the agreement, the process will stop, and the great-grandson will receive the money when the amount in the account becomes a three-digit number consisting of three identical digits.
How old will the great-grandson be when the terms of the agreement are met?
|
# Solution:
Since the number consists of identical digits, it can be represented as 111 multiplied by a. According to the problem, the same number should be obtained as the sum of an arithmetic progression. The first element of the progression is 1, the last is \( \mathrm{n} \), and the number of elements in the progression is \( \mathrm{n} \). Since \( 111 = 3 \times 37 \):
\[
n(n+1) = 2 \times 3 \times 37 \times a
\]
Since 37 is a prime number, and \( \frac{n(n+1)}{2} \) should be less than 1000, then either \( \mathrm{n}+1=37 \) or \( \mathrm{n}=37 \). By checking, we find that \( \mathrm{n}=36 \), and the required three-digit number consisting of identical digits is 666.
## Criteria:
Correctly formulated equation in integers: 5 points;
Correct answer: 5 points;
Analysis of the equation and proof of the absence of other options: 5 points.
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 1. Maximum 20 points
Let's say in country A there are only two firms engaged in research and development in a certain field. Currently, each of them independently decides whether to participate in the development of a new technology. It is known that if a firm develops a new technology, it will bring it $V$ monetary units of income in the future, provided the second firm is either unsuccessful or does not participate in the development. If both firms develop the new technology, each will receive $0.5 \mathrm{~V}$ monetary units. The investment costs for research and development for each firm will be IC monetary units regardless of the success of the development. Let the probability of success for each firm be $\alpha, 0<\alpha<1$. Then we can write the expected income of each firm as $\alpha(1-\alpha) V+0.5 \alpha^{2} V$, if both firms participate in the development, or $\alpha V$, if only the given firm participates in the development.
(a) Under what conditions will both firms participate in the development?
(b) How many firms will participate in the development if $V=16, \alpha=0.5$, and $I C=5$?
(c) Let's assume that social welfare can be assessed as the total profit of the firms. Will the number of firms found in part (b) be socially optimal? Comment on the result obtained.
|
# Solution and Grading Scheme:
(a) For each firm to decide to participate in the development, it is necessary that the expected profit from participation for each firm is greater than 0 (we use the formula for expected income in the case where both firms are involved in development):
$\alpha(1-\alpha) V+0.5 \alpha^{2} V-I C \geq 0$ (4 points)
(b) Substitute the known variables into the inequality from part (a) and check its validity:
$$
\begin{gathered}
0.5^{2} \cdot 16+0.5 \cdot 0.5^{2} \cdot 16-5 \geq 0 \\
4+2-5>0 \text { (4 points) }
\end{gathered}
$$
It makes no sense to check the equilibrium where only one firm is involved in development, as two inequalities must be satisfied simultaneously:
$$
\alpha V-\text { IC } \geq 0 \text { and } \alpha(1-\alpha) V+0.5 \alpha^{2} V-\text { IC }<0
$$
We know that the second one is not satisfied. (2 points)
(c) Calculate in which case the total profit will be greater:
1) both firms participate in development:
$\pi_{1}+\pi_{2}=2\left((1-\alpha) V+0.5 \alpha^{2} V-I C\right)=2(6-5)=2$ (3 points)
2) one firm participates in development.
$\pi_{\text {one }}=\alpha V-\mathrm{IC}=0.5 \cdot 16-5=8-5=3$. (3 points)
We see that it is more efficient for one firm to be involved in development. This is due to the fact that in the case of two firms participating, the duplication of research costs leads to a greater increase in costs than the increase in potential revenue for the firms due to the increased probability of discovery. (4 points)
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 3. Maximum 20 points
At the conference "Economics of the Present," an intellectual tournament was held, in which more than 198 but fewer than 230 scientists, including doctors and candidates of sciences, participated. Within one match, participants had to ask each other questions and record correct answers within a certain time. Each participant played against each other exactly once. The winner of the match received one point, the loser received no points; in the event of a draw, both participants received half a point. At the end of the tournament, it turned out that in matches against doctors of sciences, each participant scored half of all their points. How many candidates of sciences participated in the tournament? Provide the smallest of all possible answers.
|
Solution: Let there be $\mathrm{n}$ scientists participating in the tournament, of which $\mathrm{m}$ are doctors and $\mathrm{n}-\mathrm{m}$ are candidates of science. All participants conducted $\mathrm{n}(\mathrm{n}-1) / 2$ matches and scored $\mathrm{n}(\mathrm{n}-1) / 2$ points. Among them, the doctors of science competed in $\mathrm{m}(\mathrm{m}-1) / 2$ matches and scored the same number of points. The candidates conducted $(n-m)(n-m-1)/2$ matches among themselves and scored $(n\mathrm{m})(\mathrm{n}-\mathrm{m}-1) / 2$ points when playing against each other.
Since each participant scored half of their points against the doctors of science, the doctors of science scored half of their points against each other, meaning they scored twice as much in total: $\mathrm{m}(\mathrm{m}-1)$.
Since the candidates of science scored half of their points against the doctors of science, they also scored half of their points when competing among themselves, and in total they scored $(n-m)(n-m-1)$ points. The total number of points scored by all participants is: $m(m-1)+(n-m)(n-m-1)=n(n-1) / 2$.
After a series of transformations, we get:
$n^{2}-4 n m+4 m^{2}=n$
From which $(\mathrm{n}-2 \mathrm{~m})^{2}=\mathrm{n}$, that is, $\mathrm{n}$ is the square of a natural number.
Given $198<\mathrm{n}<230$, we find that $\mathrm{n}=225$, then $(225-2 \mathrm{~m}) 2=225$, from which $\mathrm{m}=$ 105 or $\mathrm{m}=120$.
The number of participants in the tournament is 225; the number of candidates of science is 105, or 120. The smallest value is 105.
## Criteria.
- A correct solution to the problem and the correct answer are provided - 20 points (maximum),
- A correct solution to the problem, but an incorrect answer or no answer - 8 points (rounded to the nearest whole number, half the maximum),
- An error in the solution, but the correct answer is provided - 7 points (rounded to the nearest whole number, half the maximum minus one),
- The solution is incorrect or absent, but the correct answer is provided - 2 points (under any circumstances),
- The solution is incorrect or absent, and an incorrect answer or no answer is provided - 0 points.
|
105
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. Maximum 15 points
In the treasury of the Magic Kingdom, they would like to replace all old banknotes with new ones. There are a total of 3,628,800 old banknotes in the treasury. Unfortunately, the machine that prints new banknotes requires major repairs and each day it can produce fewer banknotes: on the first day, it can only produce half of the banknotes that need to be replaced; on the second day, only a third of the remaining old banknotes in the treasury; on the third day, only a quarter, and so on. Each run of the machine in any state costs the treasury 90,000 monetary units (m.u.), and major repairs will cost 800,000 m.u. After major repairs, the machine can produce no more than one million banknotes per day. The kingdom has allocated no more than 1 million m.u. for the renewal of banknotes in the treasury.
(a) After how many days will 80% of the old banknotes be replaced?
(b) Will the kingdom be able to replace all the old banknotes in the treasury?
|
# Solution:
(a) If a major repair is to be carried out, it is most effective to do so on the second day, as this will allow the production of new banknotes to increase from 604,800 to 1 million on that day, and the production will also be 1 million in subsequent days, which is more than the possibilities without major repairs. It is not efficient to carry out the major repair on the first day, as the production without it is 1,814,400 banknotes. (9 points).
With the major repair, the budget will cover the repair itself and 2 launches of the machine, thus the maximum number of banknotes that can be replaced is 1,814,400 + 1,000,000 = 2,814,400 banknotes, which is less than 80% of the number of old banknotes (2 points).
Let's consider the option of not repairing the machine. We have enough funds for only 11 launches. In percentage terms, the remaining amount is:
After the first day: 50%.
After the second day: 200/3.
After the third day: 25%.
After the fourth day: 20%.
Thus, it will take 4 days (2,903,040 banknotes will be replaced), which is 80% of the banknotes. (2 points)
(b) After 11 launches, we will only be able to replace 3,326,400 banknotes, i.e., in any case, we will not be able to recast all the banknotes without repairing the machine.
It can also be noted that with any launch, there will always be a part of the banknotes that remain unreplaced. (2 points)
# Federal State Budget Educational Institution of Higher Education Russian Presidential Academy of National Economy and Public Administration
Olympiad of Schoolchildren of RANEPA in Economics 2022 - 2023 academic year 10-11 grade Final stage Variant 2.
#
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 1. Maximum 20 points
Let's say in country A there are only two firms engaged in research and development in a certain field. Currently, each of them independently decides whether to participate in the development of a new technology. It is known that if a firm develops a new technology, it will bring it $V$ monetary units of income in the future, provided the second firm is either unsuccessful or does not participate in the development. If both firms develop the new technology, each will receive $0.5 \mathrm{~V}$ monetary units. The investment costs for research and development for each firm will be IC monetary units regardless of the success of the development. Let the probability of success for each firm be $\alpha, 0<\alpha<1$. Then we can write the expected income of each firm as $\alpha(1-\alpha) V+0.5 \alpha^{2} V$, if both firms participate in the development, or $\alpha V$, if only the given firm participates in the development.
(a) Under what conditions will both firms participate in the development?
(b) How many firms will participate in the development if $V=24, \alpha=0.5$, and $I C=7$?
(c) Let's assume that social welfare can be assessed as the total profit of the firms. Will the number of firms found in part (b) be socially optimal? Comment on the result obtained.
|
# Solution and Grading Scheme:
(a) For each firm to decide to participate in the development, it is necessary that the expected profit from participation for each firm is greater than 0 (we use the formula for expected income in the case where both firms are involved in development):
$\alpha(1-\alpha) V+0.5 \alpha^{2} V-I C \geq 0$ (4 points)
(b) Substitute the known variables into the inequality from part (a) and check its validity:
$$
0.5^{2} \cdot 24+0.5 \cdot 0.5^{2} \cdot 24-7 \geq 0
$$
$6+3-7>0$ (4 points)
It makes no sense to check the equilibrium where only one firm is involved in development, as two inequalities must be satisfied simultaneously:
$\alpha V - IC \geq 0$ and $\alpha(1-\alpha) V+0.5 \alpha^{2} V - IC < 0$.
We know that the second one is not satisfied. (2 points)
(c) Calculate in which case the total profit will be greater:
1) both firms participate in development:
$\pi_{1}+\pi_{2}=2\left((1-\alpha) V+0.5 \alpha^{2} V-I C\right)=2(9-7)=4$ (3 points)
2) one firm participates in development.
$\pi_{\text {one }}=\alpha V-I C=0.5 \cdot 24-7=12-7=5$. (3 points)
We see that it is more efficient for one firm to be involved in development. This is due to the fact that in the case of two firms participating, the duplication of research costs leads to a greater increase in costs than the increase in potential income for the firms due to the increased probability of discovery. (4 points)
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 3. Maximum 20 points
At the "Economics and Law" congress, a "Tournament of the Best" was held, in which more than 220 but fewer than 254 delegates—economists and lawyers—participated. Within one match, participants had to ask each other questions within a limited time and record the correct answers. Each participant played against each other exactly once. The winner of the match received one point, the loser received no points; in the event of a draw, both participants received half a point. At the end of the tournament, it turned out that in matches against economists, each participant scored half of all their points. How many lawyers participated in the tournament? Provide the smallest of all possible answers.
|
Solution: Let there be $\mathrm{n}$ delegates participating in the tournament, of which $\mathrm{m}$ are economists and $\mathrm{n}-\mathrm{m}$ are lawyers. All participants conducted $\mathrm{n}(\mathrm{n}-1) / 2$ matches and scored $\mathrm{n}(\mathrm{n}-1) / 2$ points. Among them, the economists competed in $\mathrm{m}(\mathrm{m}-1) / 2$ matches and scored the same number of points. The lawyers conducted $(n-m)(n-m-1)/2$ matches among themselves and scored $(n\mathrm{m})(\mathrm{n}-\mathrm{m}-1) / 2$ points while playing against each other.
Since each participant scored half of their points against economists, the economists scored half of their points against each other, meaning they scored twice as many points in total: $\mathrm{m}(\mathrm{m}-1)$.
Since the lawyers scored half of their points against economists, they also scored half of their points against each other, meaning they scored $(n\mathrm{m})(\mathrm{n}-\mathrm{m}-1)$ points in total. The total number of points scored by all participants is:
$m(m-1)+(n-m)(n-m-1)=n(n-1) / 2$.
After a series of transformations, we get:
$n^{2}-4 n m+4 m^{2}=n$
From which $(\mathrm{n}-2 \mathrm{~m})^{2}=\mathrm{n}$, meaning $\mathrm{n}$ is a perfect square of a natural number.
Given $220<\mathrm{n}<254$, we find that $\mathrm{n}=225$, then $(225-2 \mathrm{~m})^{2}=225$, from which $\mathrm{m}=$ 105 or $\mathrm{m}=120$.
The number of participants in the tournament is 225; the number of lawyers is 105, or 120. The smallest value is 105.
## Criteria.
- Correct solution and the right answer - 20 points (maximum),
- Correct solution but the wrong answer or no answer -
8 points (rounded to the nearest whole number, half the maximum),
- Error in the solution but the right answer - 7 points (rounded to the nearest whole number, half the maximum minus one),
- Incorrect or no solution, but the right answer - 2 points (under any circumstances),
- Incorrect or no solution, and the wrong answer or no answer - 0 points.
|
105
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the minimum loss, which is EXPENSE - INCOME, where the letters $\boldsymbol{P}, \boldsymbol{A}, \boldsymbol{C}, \boldsymbol{X}, \boldsymbol{O}, \boldsymbol{D}$ represent digits forming an arithmetic progression in the given order. (2 points).
#
|
# Solution:
The difference in the progression is 1; otherwise, 6 digits will not fit (if the "step" is 2, then 6 digits will exceed the field of digits). The smaller the first digit, the smaller the loss. Therefore, $\boldsymbol{P}=1, \boldsymbol{A}=2, \boldsymbol{C}=3, \boldsymbol{X}=4, \boldsymbol{O}=5, D=6$. Thus, 123456-65456=58000.
Answer: 58000.
| Points | Criteria for evaluating the performance of task № 1 |
| :---: | :--- |
| $\mathbf{9}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| $\mathbf{6}$ | A correct sequence of all steps of the solution is provided. There are gaps in the justification for the choice of correspondence between letters and digits or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| $\mathbf{3}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, the difference in the progression is found); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the choice of correspondence between letters and digits is incorrect). The solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not meet the above criteria for scoring 3 and 6 points. |
|
58000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In a certain state, only liars and economists live (liars always lie, while economists tell the truth). At a certain moment, the state decided to carry out monetary and credit, as well as budgetary and tax reforms. Since it was unknown what the residents expected, everyone was asked several questions (with only "yes" and "no" answers). The questions were answered affirmatively as follows:
"Will taxes be raised?" $40 \%$,
"Will there be an increase in the money supply?" $30 \%$,
"Will bonds be issued?" $50 \%$,
"Will the gold and foreign exchange reserves be squandered?" $\mathbf{0} \%$.
What percentage of the population actually expects taxes to be raised, if each of them realistically expects only one measure? (14 points).
|
# Solution:
Let $\boldsymbol{x}$ be the proportion of liars in the country, then (1-x) is the proportion of economists. Each economist answers affirmatively to one question, and each liar answers affirmatively to three. Therefore, we can set up the equation:
$3 x+1-x=0.4+0.3+0.5+0 ; 2 x=0.2 ; x=0.1$. Thus, in the country, $10\%$ are liars and $90\%$ are economists. Since no one said they expect a depletion of gold and foreign exchange reserves, this means that all liars are expecting it. Therefore, the actual percentage of the population expecting a tax increase is $40-10=30\%$.
Answer: $30\%$.
| Points | Criteria for evaluating the performance of task № $\mathbf{3}$ |
| :---: | :--- |
| $\mathbf{14}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are correctly performed. The correct answer is obtained. |
| $\mathbf{10}$ | A correct sequence of all steps of the solution is provided. There are gaps in justifying the percentage of the population that actually expects a tax increase, or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| $\mathbf{5}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is correctly performed, possibly inaccurately (for example, an equation is set up); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the percentage of the population that actually expects a tax increase is found incorrectly). The solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not meet the criteria for scoring 5 and 10 points. |
|
30
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. There are 2015 coins on the table. Two players play the following game: they take turns; on a turn, the first player can take any odd number of coins from 1 to 99, and the second player can take any even number of coins from 2 to 100. The player who cannot make a move loses. How many coins should the first player take on the first move to then guarantee a win with an unchanging strategy? (20 points)
#
|
# Solution:
The strategy of the first player: he takes 95 coins, and then on each move, he takes (101-x) coins, where $\boldsymbol{x}$ is the number of coins taken by the second player. Since $\boldsymbol{x}$ is even (by the condition), 101-x is odd. Then $2015-95=1920$, since 101-x+x=101 coins will be taken per move, 1920/101=19 (remainder 1). In total, there will be 19 moves, in which 101 coins will be taken each time. The second player will be left with 1 coin and will not be able to make a move.
Answer: 95.
| Points | Criteria for evaluating the solution to problem № 4 |
| :---: | :--- |
| $\mathbf{2 0}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| $\mathbf{1 4}$ | A correct sequence of all steps of the solution is provided. There are gaps in the justification of the first player's strategy or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| $\mathbf{7}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, the first move of the first player is indicated); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the first player's strategy is incorrectly determined). At the same time, the solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not meet the criteria for scoring 7 and 14 points. |
|
95
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. The bank issued a loan to citizen $N$ on September 9 in the amount of 200 mln rubles. The repayment date is November 22 of the same year. The interest rate on the loan is $25 \%$ per annum. Determine the amount (in thousands of rubles) that citizen N will have to return to the bank. Assume that there are 365 days in a year, the bank accrues interest daily on the loan amount, and no interest is charged on the days of loan issuance and repayment. (14 points).
|
# Solution:
Number of days of the loan: September - 21 days, October - 31 days, November - 21 days, i.e., $21+31+21=73$ days. The accrued debt amount is calculated using the formula:
$\boldsymbol{F V}=\boldsymbol{P V} \cdot(\boldsymbol{1}+\boldsymbol{t} \cdot \mathbf{Y}$ ), where $\boldsymbol{F} \boldsymbol{V}$ - the accrued (future - future value) amount of money after a certain period; $\boldsymbol{P} \boldsymbol{V}$ - the initial (present - present value) amount of money, $\boldsymbol{t}$ - the term of the operation in days, $\mathbf{Y}$ - the duration of the year in days, $\boldsymbol{i}$ - the interest rate for the period. We have $t=73, Y=365, i=0.25 F V=200 *(1+73 * 0.25 / 365)=1.05 * 200=210$ thousand rubles.
Answer: 210 thousand rubles.
## Grading Criteria:
1) The number of days of the loan is correctly calculated - 3 points.
2) The correct formula for calculating the debt amount is obtained - 7 points
3) The debt amount is correctly calculated - 4 points. If an arithmetic error is made while using the correct formula for calculating the debt amount, 3 points are given.
|
210
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 3. Maximum 20 points
Consider a consumer who plans to use a coffee machine. Let's assume the consumer lives for two periods and values the benefit of using the coffee machine at 10 monetary units in each period. Coffee machines can be produced in two types: durable, working for two periods, and low-quality, completely breaking down at the end of each period. It is known that the technology for producing durable machines allows them to be produced with constant average costs of 6 monetary units. Low-quality coffee machines can be produced with constant average costs of C monetary units.
(a) Suppose coffee machines are produced only by a monopoly. For what average costs C will the monopoly produce only durable coffee machines?
(b) How would your answer to part (a) change if coffee machines of both types are produced in a perfectly competitive market?
(c) What conditions for choosing the lifespan of a coffee machine have we not considered? Come up with at least one and explain how its inclusion in the model would affect the choice in the conditions of part (a) and part (b).
|
# Solution and Grading Scheme:
(a) When coffee machines are produced by a monopoly, it sets a selling price that will extract all consumer surplus:
1) when producing a durable machine, the price equals the consumer's benefit from using the coffee machine for 2 periods $\mathrm{p}_{\mathrm{L}}=2 \cdot 10=20$ (1 point)
2) when producing a low-quality coffee machine, the price will be equal to the benefit from using it for one period $\mathrm{p}_{\mathrm{s}}=10 .(\mathbf{1}$ point)
Calculate the profit from producing a durable machine: $\pi_{\mathrm{L}}=20-6=14$. ( $\mathbf{1}$ point)
Profit from producing a low-quality coffee machine: $\pi_{\mathrm{S}}=10$ - c. ( $\mathbf{1}$ point)
For the monopolist to find it profitable to produce a durable machine, the profit from its production must be greater than the profit from producing two low-quality machines (since the service life of the durable one is twice as long):
$$
14>2(10 \text { - c) (2 points) }
$$
From this, it follows that c $>3$. ( $\mathbf{1}$ point)
(b) In this case, coffee machines will be sold at prices equal to the costs of their production: $\mathrm{q}_{\mathrm{L}}=6$ and $\mathrm{q}_{\mathrm{s}}=\mathrm{c}$. (2 points) For only durable machines to be produced, it is necessary that consumers prefer to buy only them.
The net consumer surplus from buying a durable machine will be $2 \cdot 10-\mathrm{q}_{\mathrm{L}}=20-$ $6=14$. ( 2 points)
The net consumer surplus from buying two low-quality coffee machines will be $2\left(10-\mathrm{q}_{\mathrm{s}}\right)=2(10-\mathrm{c}) .(\mathbf{2}$ points)
Thus, the decision to buy only durable machines will be made if the inequality $14>2(10-$ c) and c $>3$ is satisfied, i.e., the answer does not change. ( $\mathbf{1}$ point)
(c) costs of purchasing (going to the store, searching online - can vary over two periods), costs of disposing of a broken coffee machine (need to recycle it), etc. You need to demonstrate how the inequalities change with the addition of this new condition. (6 points)
|
3
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. Maximum 20 points
In Moscow, a tennis tournament is being held. Each team consists of 3 players. Each team plays against every other team, with each participant of one team playing against each participant of the other exactly one match. Due to time constraints, a maximum of 150 matches can be played in the tournament. How many teams can participate in the tournament so that everyone gets to play?
|
# Solution:
Assume there are two teams - then each of the three members of one team plays against each of the other - i.e., $3 * 3 = 9$ games. The number of team pairs can be $150: 9 = 16.6 \ldots$ a maximum of 16. Two teams form only one pair; three teams form three pairs. If there is a fourth team, add 3 more pairs, so in total 6. With each subsequent team, the number of pairs increases by one more: (2 teams - 1 pair, 3 teams $-1+2$ pairs, 4 teams $-1+2+3$ pairs, etc.) Thus, if there are 6 teams, $1+2+3+4+5=15$ pairs. On the seventh day, it is already $15+6=21$, which exceeds the maximum of 16. Therefore, the answer is: 6 teams.
## Criteria:
Correct answer obtained with sound reasoning - 20 points Correct chain of reasoning, but an arithmetic error leads to an incorrect answer - 5 points.
## Federal State Budget Educational Institution of Higher Education Russian Presidential Academy of National Economy and Public Administration
## RANEPA School Olympiad in Economics 2022 - 2023 Academic Year 8-9 grade Final Stage Variant 2.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 3. Maximum 20 points
Consider a consumer who plans to use a coffee machine. Let's assume the consumer lives for two periods and evaluates the benefit from using the coffee machine at 20 monetary units in each period. Coffee machines can be produced in two types: durable, working for two periods, and low-quality, completely breaking down at the end of each period. It is known that the technology for producing durable machines allows them to be produced with constant average costs of 12 monetary units. Low-quality coffee machines can be produced with constant average costs of C monetary units.
(a) Suppose coffee machines are produced only by a monopoly. For what average costs C will the monopoly produce only durable coffee machines?
(b) How would your answer to part (a) change if coffee machines of both types are produced in a perfectly competitive market?
(c) What conditions for choosing the lifespan of a coffee machine have we not considered? Come up with at least one and explain how its inclusion in the model would affect the choice in the conditions of part (a) and part (b).
|
# Solution and Grading Scheme:
(a) When coffee machines are produced by a monopoly, it sets a selling price that will extract all consumer surplus:
1) when producing a durable machine, the price equals the consumer's benefit from using the coffee machine for 2 periods $\mathrm{p}_{\mathrm{L}}=2 \cdot 20=40$ (1 point)
2) when producing a low-quality coffee machine, the price will be equal to the benefit from using it for one period $\mathrm{p}_{\mathrm{s}}=20$. (1 point)
Let's calculate the profit from producing a durable machine: $\pi_{\mathrm{L}}=40-12=28$. (1 point)
Profit from producing a low-quality coffee machine: $\pi_{\mathrm{s}}=20$ - c. (1 point)
For the monopolist to find it profitable to produce a durable machine, the profit from its production must be greater than the profit from producing two low-quality machines (since the service life of the durable one is twice as long):
$$
28>2(20-c) (2 \text{ points) }
$$
From this, it follows that c $>6$. (1 point)
(b) In this case, coffee machines will be sold at prices equal to the costs of their production: $\mathrm{q}_{\mathrm{L}}=12$ and $\mathrm{q}_{\mathrm{S}}=\mathrm{c}$. (2 points) For only durable machines to be produced, it is necessary that consumers prefer to buy only them.
The net consumer surplus from buying a durable machine will be $2 \cdot 20-\mathrm{q}_{\mathrm{L}}=40-$ $12=28. (2$ points)
The net consumer surplus from buying two low-quality coffee machines will be $2\left(20-\mathrm{q}_{s}\right)=2(20-\mathrm{c}). (2$ points)
Thus, the decision to buy only durable machines will be made if the inequality $28>2(20-$ c) and c $>6$ is satisfied, i.e., the answer does not change. (1 point)
(c) costs of purchasing (going to the store, searching online - can vary over two periods), costs of disposing of a broken coffee machine (need to recycle it), etc. It is necessary to demonstrate how the inequalities change with the addition of this new condition. (6 points)
|
6
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. Maximum 20 points
In Moscow, a tennis tournament is being held. Each team consists of 3 players. Each team plays against every other team, with each participant of one team playing against each participant of the other exactly one match. Due to time constraints, a maximum of 200 matches can be played in the tournament. How many teams can participate in the tournament so that everyone gets to play?
#
|
# Solution:
Assume there are two teams - then each of the three members of one team plays against each of the other - i.e., $3 * 3=9$ games. The number of team pairs can be $200: 9=22.2 \ldots$ a maximum of 22. Two teams form only one pair; three teams form three pairs. If there is a fourth team, add 3 more pairs, so in total 6. With each subsequent team, one more is added to the number of pairs from the previous number: (2 teams - 1 pair, 3 teams $-1+2$ pairs, 4 teams $-1+2+3$ pairs, etc.). Thus, on the seventh day, it is possible to form $1+2+3+4+5+6=21$ pairs. On the eighth day, it is already $21+7=28$, which exceeds the maximum of 22. Therefore, the answer is: 7 teams.
## Criteria:
Correctly derived the correct answer - 20 points The chain of reasoning is correct, but due to an arithmetic error, an incorrect answer is obtained - 5 points.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. In the Tumba-Yumba tribe with a population of 30 people, a trader arrives. After studying the customs of the tribe, the trader proposes to play a game. For each natural exchange of goods conducted in the market by two tribespeople, the trader gives each participant one gold coin. If at the end of the day, two different tribespeople end up with the same number of coins, all goods and coins go to the trader. A tribesperson trades goods only with another known tribesperson, and the cunning trader knows this. By the end of the day, the trader finds a group of tribespeople with the same number of coins. The tribespeople are shocked, the trader is happy, and the offending tribespeople are expelled from the market. Then the Chief appears and offers the trader a rematch. Let the trader distribute 270 coins among the remaining tribespeople so that no one has the same number of coins (i.e., each tribesperson should have a unique number of coins). If the trader cannot do this, all the coins and goods remain in the tribe.
- Is it true that in the original game, the tribespeople had no chance?
- Is it true that the Chief can return all the goods and gold?
- Is it true that the Chief knew how many tribespeople were expelled from the market?
- What is the maximum number of tribespeople that needed to be expelled for the Chief to win?
- Is it true that if the trader had lost the first game, the tribespeople could have lost the second game?
Explain your answers.
|
Solution: In any company, there are at least two people who have the same number of acquaintances. Therefore, the natives had no chance. The chief proposed to distribute 270 coins, which means he knew that some natives would be removed. Let's say x people were removed. To distribute a different number of coins to everyone, you need to take $\frac{0+30-x-1}{2}(30-x)$ coins, and the Chief should make sure that 270 coins are not enough. Then, at least 24 natives should remain, so $24 * 23 / 2=276$ coins are needed to avoid repetition. Since $23 * 11=253$ contains an arithmetic error, which does not affect the result. The correct answer to 2, 3, or 4 questions | 5 |
| The correct answer and justification for only one of the questions in the problem | 2 |
| The answer is either incorrect or guessed, with no justification | 0 |
|
24
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In a certain kingdom, the workforce consists only of the clan of dwarves and the clan of elves. Historically, in this kingdom, dwarves and elves have always worked separately, and no enterprise has ever hired both at the same time. The aggregate supply of labor resources of the dwarves is represented by the function $w_{\text {dwarves }}^{S}=1+L / 3$, and the aggregate supply of labor resources of the elves is $w_{\text {elves }}^{S}=3+L$. The inverse function of the aggregate demand for dwarf labor is $w_{\text {dwarves }}^{D}=10-2 L / 3$, and the inverse function of the aggregate demand for elf labor is $w_{\text {elves }}^{D}=18-2 L$. The king, who recently ascended to the throne, is very concerned that the wage rates of his subjects are different, so he issued a law stipulating that the wages of elves and dwarves must be the same and that workers in the kingdom should not be discriminated against based on clan affiliation. At the same time, the king believes that regulatory wage rate control will have a negative impact on the economy of his kingdom and requires all his subjects to behave in a perfectly competitive manner. By what factor will the wage of the group of workers whose wage was lower before the king's intervention increase, if firms in the kingdom are indifferent to hiring elves or dwarves?
|
# Solution:
Before the law was introduced, the wages of elves and gnomes were determined by the condition of equality of supply and demand in each market:
$w_{\text {gnome }}^{S}=w_{\text {gnome }}^{D}$, from which $1+\frac{L}{3}=10-2 L / 3$ and $L_{\text {gnome }}=9$ and $w_{\text {gnome }}=4$
$w_{\text {elf }}^{S}=w_{\text {elf }}^{D}$, from which $3+L=18-2 L$ and $L_{\text {elf }}=5 w_{\text {elf }}=8$
After the law was introduced, all market participants must behave in a perfectly competitive manner, and the total demand for labor resources must equal the total supply of labor resources. Therefore,
$L_{\text {gnome }}^{S}+L_{\text {elf }}^{S}=L_{\text {elf }}^{D}+L_{\text {gnome }}^{D}$, where
$L_{\text {gnome }}^{S}+L_{\text {elf }}^{S}=\left\{\begin{array}{l}3 w-3, \text { if } w \leq 3 \\ 4 w-6, \text { if } w>3\end{array}\right.$
$L_{\text {elf }}^{D}+L_{\text {gnome }}^{D}=\left\{\begin{array}{l}9-w / 2, \text { if } w \geq 10 \\ 24-2 w, \text { if } w solutions. All transformations and calculations are correct. The correct answer is obtained. | $\mathbf{1 0}$ |
| The correct sequence of all steps in the solution is provided. There are gaps in the justification of the choice of number combinations (for example, one extra solution (0+4) is not considered, pairs of digits are not divided by 2, etc.) or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. | |
| The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, part of the combinations is considered and calculated correctly); | |
| - the other part is either not performed or performed incorrectly, possibly | |
| even with logical errors (for example, most combinations are not considered | |
| or extra combinations are considered, combinations are calculated incorrectly, cases of erasing the same digit (2+2, | |
| 5+5, 8+8) are not considered, etc.); | |
| In this case, the solution may be incomplete. | |
All cases of the solution that do not meet the above criteria for scoring 3 and 6 points.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Given a 2015-digit number divisible by 9. Let the sum of its digits be $\boldsymbol{a}$, the sum of the digits of $\boldsymbol{a}$ be $\boldsymbol{b}$, and the sum of the digits of $\boldsymbol{b}$ be $\boldsymbol{c}$. Find the number $\boldsymbol{c}$. (14 points).
|
Solution:
The sum of the digits of any number gives the same remainder when divided by 9 as the number itself.
The largest 2015-digit number consists of 2015 nines. The sum of its digits is $2015 * 9 = 18135$, i.e., it has 5 digits.
The sum of the digits of the largest 5-digit number is 45 (b).
Numbers less than 45 and divisible by 9 are 9, 18, 27, 36, and 45. The sum of their digits is 9.
## Answer: 9.
| Points | Criteria for evaluating the performance of task № 3 |
| :---: | :--- |
| $\mathbf{1 4}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| $\mathbf{1 0}$ | A correct sequence of all steps of the solution is provided. There are gaps in the justification of upper estimates or possible values of the numbers $\boldsymbol{a}$ and (or) $\boldsymbol{b}$, or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| $\mathbf{5}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, a correct upper estimate is obtained only for the number $\boldsymbol{a}$ or only for the number $\boldsymbol{b}$); - the other part is either not performed or performed incorrectly, possibly even with logical errors (for example, finding another upper estimate). In this case, the solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not correspond to the above criteria for scoring 5 and 10 points. |
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In how many ways can two knights, two bishops, two rooks, a queen, and a king be arranged on the first row of a chessboard so that the following conditions are met:
1) The bishops stand on squares of the same color;
2) The queen and the king stand on adjacent squares. (20 points).
|
# Solution:
Let's number the cells of the first row of the chessboard in order from left to right with numbers from **1** to **8** ( **1** - the first white cell, **8** - the last black cell). Since the queen and king are standing next to each other, they can occupy one of 7 positions: 1-2, 2-3, ..., 7-8. Additionally, on each position, we can swap their places (for example, on position 1-2, we can place the king first, then the queen, and vice versa). In total, we have 2$\cdot$7=14 ways to place the queen and king on the first row of the chessboard so that they stand next to each other. Now let's see how we can place the bishops. Suppose we have placed the queen and king somewhere (for example, on positions 4-5). Notice that no matter how we place the queen and king, we will have 3 cells of each color left: 3 white and 3 black. To place the bishops, we need to choose the color of the cell - this gives us 2 options. Then we have 3 positions on which we want to place two bishops (the number of combinations of 3 taken 2 at a time). This can be done in 3 ways. Therefore, we have $2 \cdot 3=6$ ways to place the two bishops on cells of the same color given a fixed position of the king and queen. We have 4 cells left, on which we need to place two knights and two rooks. That is, we need to choose 2 positions out of 4 for the rooks, and the knights will take the remaining 2 positions. We can choose 2 cells out of 4 in 6 ways (the number of combinations of 4 taken 2 at a time). Therefore, we have 6 ways to place the two rooks and two knights on 4 positions in any order. Since we fixed the positions sequentially, the multiplication rule applies. The total number of ways to place the figures is $N=14 \cdot 6 \cdot 6=504$.
## Answer: 504.
| Points | Criteria for evaluating the completion of task № 4 |
| :---: | :---: |
| 20 | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| 14 | A correct sequence of all steps of the solution is provided. There are gaps in the justification of the number of ways to arrange the figures (for example, rooks and knights), or the distinguishability of figures in one pair, or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| 7 | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, the number of ways to arrange the king, queen, and either only the bishops or only the rooks and knights; the problem is solved separately only with condition №1 and only with condition №2); - the other part is either not performed or performed incorrectly, possibly even with logical errors (for example, the number of ways to arrange only the rooks and knights or only the bishops). The solution may be incomplete. |
| 0 | All cases of the solution that do not meet the above criteria for scoring 7 and 14 points. |
|
504
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. In his country of Milnlandia, Winnie-the-Pooh decided to open a company that produces honey. Winnie-the-Pooh sells honey only in pots, and it costs him 10 milnovs (the monetary units of Milnlandia) to produce any pot of honey. The inverse demand function for honey is given by $\boldsymbol{P}=310-3 \boldsymbol{Q}$ (where $\boldsymbol{Q}$ is the number of pots, and $\boldsymbol{P}$ is the price of one pot). There are no other suppliers of honey.
a) How many pots of honey will Winnie-the-Pooh produce if his main goal is to maximize profit?
b) The government of Milnlandia decided to introduce a tax $\boldsymbol{t}$ milnovs on each pot of honey sold by Winnie-the-Pooh's company. What should be the amount of the tax to maximize the tax revenue for the budget of Milnlandia? (17 points).
|
# Solution:
a) Profit $=P(Q) \cdot Q - TC(Q) = (310 - 3Q) \cdot Q - 10 \cdot Q = 310Q - 3Q^2 - 10Q = 300Q - 3Q^2$.
Since the graph of the function $300Q - 3Q^2$ is a parabola opening downwards, its maximum is achieved at the vertex: $Q = -b / 2a = -300 / (-6) = 50$.
b) Winnie-the-Pooh maximizes the quantity $P(Q) \cdot Q - TC(Q) = (310 - 3Q) \cdot Q - 10 \cdot Q - tQ = (300 - t)Q - 3Q^2$, $Q = (300 - t) / 6$. The state receives a monetary amount in the form of tax revenue $Q \cdot t = (300 - t) / 6 \cdot t = (50 - 1/6 \cdot t) \cdot t = -1/6 \cdot t^2 + 50t$. The graph of this function is a parabola opening downwards, its maximum is achieved at $t = -50 / (-2/6) = 50 \cdot 6 / 2 = 50 \cdot 3 = 150$.
Answer: a) 50; b) 150.
## Grading Criteria:
a)
1) The problem of profit maximization for a single producer is correctly formulated: 3 points
2) The solution to the profit maximization problem is correctly found and justified: 4 points. The absence of justification for the maximum value at the extremum point results in a 1-point deduction.
b)
1) The consequences of introducing a profit tax for a single producer are correctly accounted for: 3 points
2) The tax amount is correctly found: 7 points. If there is an arithmetic error in calculating the tax, 6 points are given.
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. To climb from the valley to the mountain top, one must walk 4 hours on the road, and then -4 hours on the path. On the mountain top, two fire-breathing dragons live. The first dragon spews fire for 1 hour, then sleeps for 17 hours, then spews fire for 1 hour again, and so on. The second dragon spews fire for 1 hour, then sleeps for 9 hours, then spews fire for 1 hour again, and so on. It is dangerous to walk on both the road and the path during the first dragon's fire-spewing, while it is dangerous to walk only on the path during the second dragon's fire-spewing. Both dragons start spewing fire simultaneously at midnight. Is it possible to safely climb from the valley to the mountain top and return back? (8 points)
|
# Solution:
The path along the road and the trail (there and back) takes 16 hours. Therefore, if you start immediately after the first dragon's eruption, this dragon will not be dangerous. The path along the trail (there and back) takes 8 hours. Therefore, if you start moving along the trail immediately after the second dragon's eruption, this dragon will not be dangerous. For a safe ascent, it is sufficient for the first dragon to stop erupting by the start of the journey along the road, and for the second dragon to stop erupting 4 hours later, by the start of the journey along the trail. The first dragon erupts in hours numbered $18 x+1$, and the second dragon in hours numbered $10 y+1$, where $x, y$ are natural numbers. We need them to erupt with a 4-hour shift. We get the following Diophantine equation:
$$
10 y+1=(18 x+1)+4,5 y=9 x+2 .
$$
The smallest solution in natural numbers is $x=2, y=4$. Therefore, for a safe ascent, you need to start at the beginning of the 38th hour ( $18 \cdot 2+1=37)$.
Answer: You need to start at the beginning of the 38th hour.
| llы | Я№ 1 |
| :---: | :---: |
| 8 | Contains all steps of the solution. and explanation. |
| 5 | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is correctly performed, possibly inaccurately (for example, a Diophantine equation is formulated, or a graph of eruptions is constructed, or the possibility of a safe path is demonstrated in principle); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the solution to the Diophantine equation is not found, or there are errors in the graph of eruptions, or a specific moment for the start of a safe path is not indicated, or the possibility of stopping and waiting is allowed). The solution may be incomplete. |
| 0 | All cases of the solution that do not correspond to the above criteria for scoring 5 and 8 points. |
|
38
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. On the extensions of sides $\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C D}$ and $\boldsymbol{A}$ of the convex quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$, points $\boldsymbol{B}_{1}, \boldsymbol{C}_{1}, \boldsymbol{D}_{1}$ and $\boldsymbol{A}_{1}$ are taken such that $\boldsymbol{B} \boldsymbol{B}_{1}=\boldsymbol{A B}, \boldsymbol{C} \boldsymbol{C}_{1}=\boldsymbol{B C}, \boldsymbol{D} \boldsymbol{D}_{1}=\boldsymbol{C D}$ and $\boldsymbol{A} \boldsymbol{A}_{1}=\boldsymbol{D} \boldsymbol{A}$. How many times larger is the area of quadrilateral $\boldsymbol{A}_{1} \boldsymbol{B}_{1} \boldsymbol{C}_{1} \boldsymbol{D}_{1}$ compared to the area of quadrilateral $A B C D .(10$ points)
|
# Solution:
Let the area of quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$ be $\boldsymbol{S}$. The median divides the area of a triangle in half. Therefore, $S_{A B C}=S_{C B B 1}=S_{C B 1 C 1}$. Consequently, $S_{B B 1 C 1}=2 S_{A B C}$. Similarly, we have $S_{C C 1 D 1}=2 S_{B C D}$, $S_{D D 1 A 1}=2 S_{C D A}$, and $S_{A A 1 B 1}=2 S_{D A B}$. Adding these four equations, we get: $S_{B B 1 C 1}+S_{C C 1 D 1}+S_{D D 1 A 1}+S_{A A 1 B 1}=2\left(S_{A B C}+S_{B C D}+S_{C D A}+S_{D A B}\right)=4 S$. Therefore, $S_{A I B I C I D 1}=4 S+S=5 S$.
Answer: 5.
| Points | Criteria for evaluating the performance of task № 3 |
| :---: | :--- |
| $\mathbf{1 0}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| $\mathbf{8}$ | A correct sequence of all steps of the solution is provided. There are gaps in the justification of the area relationships of triangles or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| $\mathbf{5}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, the property of the median of a triangle is given or the areas of right triangles are correctly calculated); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the area relationships of triangles are obtained incorrectly or only a special case of a quadrilateral is considered). The solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not meet the criteria for scoring 5, 8, and 10 points. |
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The power in the kingdom of gnomes was seized by giants. The giants decided to get rid of the gnomes and told them the following: "Tomorrow we will line you up so that each of you will see those who stand after and not see those who stand before (i.e., the 1st sees everyone, the last sees no one). We will put either a black or a white hat on each of you (equally likely, each will have either a black or a white hat) and ask what color it is. Those who answer correctly will be released, and those who answer incorrectly will be executed." How many gnomes can be risked at a minimum with certain agreements before the execution, if there are p gnomes in the kingdom and p $<\infty$. Justify your answer. (12 points)
#
|
# Solution:
The gnomes agree as follows: they risk the first gnome, telling him the following: "Denote a white hat as 1 and a black hat as 0, and count the sum of the remaining n-1 gnomes. If the sum is even, say 'white'; if it is odd, say 'black'. Under this condition, the first gnome dies with a probability of $1 / 2$. The second gnome counts the sum of the (n-2) gnomes. If the sum has changed, it means he has a white hat. The others hear him and adjust the parity of the remaining gnomes, and so on. Answer: 1 gnome.
Answer: 1.
| Points | Criteria for evaluating the completion of task № 5 |
| :---: | :---: |
| 12 | A correct and justified sequence of all steps of the solution is provided. The correct answer is obtained. |
| 10 | A correct sequence of all steps of the solution is provided. There are gaps in the justification of the parity of the sum, but they do not affect the further course of the solution. |
| 8 | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, it is proposed to denote a white hat as 1 and a black hat as 0); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the necessity of adjusting the parity of the sum is not proven). The solution may be incomplete. |
| 5 | The problem is not solved, but there is some progress in its solution, i.e.: - the initial part of the solution is performed correctly (for example, the assumption about the probable death of the first gnome is correctly stated or the cases of even and odd numbers of gnomes are considered); - the main part is either not performed or performed incorrectly, possibly even with logical errors (for example, the correspondence between the color of the hat and the numbers is not shown, or the probability of the first gnome's death is insufficiently justified, or an incorrect conclusion is drawn from the parity and non-parity of the number of gnomes, or the assumption is made that the gnomes can exchange additional information or that execution occurs immediately upon an incorrect answer and the others see it). The solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not meet the criteria for scoring 5, 8, 10, and 12 points. |
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. To climb from the valley to the mountain top, one must walk 6 hours on the road, and then - 6 hours on the path. On the mountain top, two fire-breathing dragons live. The first dragon spews fire for 1 hour, then sleeps for 25 hours, then spews fire for 1 hour again, and so on. The second dragon spews fire for 1 hour, then sleeps for 13 hours, then spews fire for 1 hour again, and so on. During the first dragon's fire-spewing, it is dangerous to walk both on the road and on the path, while during the second dragon's fire-spewing, it is dangerous to walk only on the path. At midnight, both dragons start spewing fire simultaneously. Is it possible to safely climb from the valley to the mountain top and return back? (6 points)
|
# Solution:
The path along the road and the trail (there and back) takes 24 hours. Therefore, if you start immediately after the first dragon's eruption, this dragon will not be dangerous. The path along the trail (there and back) takes 12 hours. Therefore, if you start moving along the trail immediately after the second dragon's eruption, this dragon will not be dangerous. For a safe ascent, it is necessary that by the start of movement along the road, the first dragon has stopped erupting, and 6 hours later, by the start of movement along the trail, the second dragon has stopped erupting. The first dragon erupts in hours numbered $26 x+1$, and the second dragon in hours numbered $14 y+1$, where $x, y$ are natural numbers. We need them to erupt with a 6-hour shift. We get the following Diophantine equation:
$$
14 y+1=(26 x+1)+6, \quad 7 y=13 x+3
$$
The smallest solution in natural numbers is $x=3, y=6$. Therefore, for a safe ascent, you need to start at the beginning of the 80th hour $(26 \cdot 3+1=79)$.
Answer: You need to start at the beginning of the 80th hour.
| Points | Criteria for evaluating the performance of task No. 1 |
| :---: | :--- |
| $\mathbf{6}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| $\mathbf{3}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, a Diophantine equation is formulated); - the other part is either not performed or performed incorrectly, possibly even with logical errors (for example, the solution to the Diophantine equation is not found). The solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not meet the above criteria for scoring 3 and 6 points. |
|
80
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. On the extensions of sides $\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C D}$ and $\boldsymbol{A}$ of the convex quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$, points $\boldsymbol{B}_{1}, \boldsymbol{C}_{1}, \boldsymbol{D}_{1}$ and $\boldsymbol{A}_{1}$ are taken such that $\boldsymbol{B} \boldsymbol{B}_{1}=\boldsymbol{A B}, \boldsymbol{C} \boldsymbol{C}_{1}=\boldsymbol{B C}, \boldsymbol{D D}_{1}=\boldsymbol{C D}$ and $\boldsymbol{B} \boldsymbol{B}_{1}=\boldsymbol{A B}$ and $\boldsymbol{A} \boldsymbol{A}_{1}=\boldsymbol{A}$. How many times smaller is the area of quadrilateral $\boldsymbol{A} \boldsymbol{B} \boldsymbol{C D}$ compared to the area of quadrilateral $\boldsymbol{A}_{1} \boldsymbol{B}_{1} C_{1} \boldsymbol{D}_{1}$. (10 points)
|
# Solution:
Let the area of quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$ be $\boldsymbol{S}$. A median divides the area of a triangle in half. Therefore, $S_{A B C}=S_{C B B 1}=S_{C B 1 C 1}$. Consequently, $S_{B B 1 C 1}=2 S_{A B C}$. Similarly, we have $S_{C C 1 D 1}=2 S_{B C D}$, $S_{D D 1 A 1}=2 S_{C D A}$, and $S_{A A 1 B 1}=2 S_{D A B}$. Adding these four equalities, we get: $S_{B B 1 C 1}+S_{C C 1 D 1}+S_{D D 1 A 1}+S_{A A 1 B 1}=2\left(S_{A B C}+S_{B C D}+S_{C D A}+S_{D A B}\right)=4 S$. Therefore, $S_{A I B I C I D 1}=4 S+S=5 S$.
## Answer: 5.
| Points | Criteria for evaluating the performance of task No. 3 |
| :---: | :--- |
| $\mathbf{9}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| $\mathbf{6}$ | A correct sequence of all steps of the solution is provided. There are gaps in the justification of the area relationships of the triangles or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| $\mathbf{3}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, the property of the median of a triangle is given); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the area relationships of the triangles are obtained incorrectly). The solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not correspond to the above criteria for scoring 9, 6, and 3 points. |
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In all other cases - o points.
## Task 2
Maximum 15 points
Solve the equation $2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$.
How many solutions of this equation satisfy the condition:
$0 \leq x \leq 2020 ?$
|
Solution. Let $t=\frac{\pi x}{4}$. Then the equation takes the form
$2 \sqrt{2} \sin ^{3} t=\cos \left(\frac{\pi}{4}-t\right)$.
$2 \sqrt{2} \sin ^{3} t=\cos \frac{\pi}{4} \cos t+\sin \frac{\pi}{4} \sin t$
$2 \sqrt{2} \sin ^{3} t=\frac{\sqrt{2}}{2} \cos t+\frac{\sqrt{2}}{2} \sin t$
$4 \sin ^{3} t=\cos t+\sin t ; 4 \sin ^{3} t-\sin t=\cos t$
Case 1: $\sin t=0 \rightarrow \cos t=0 \rightarrow \emptyset$
Case 2: $\sin t \neq 0$. Dividing both sides of the equation by $\sin t$, we get:
$4 \sin ^{2} t-1=\cot t$. Considering that $\sin ^{2} t=\frac{1}{1+\cot ^{2} t}$ and denoting $y=\cot t$, we obtain
$\frac{4}{1+y^{2}}-1=y$, or after transformation $y^{3}+y^{2}+y-3=0$
$(y-1)\left(y^{2}+2 y+3\right)=0$, the only root of which is $y=1$.
Making reverse substitutions: $\cot t=1 \rightarrow t=\frac{\pi}{4}+\pi n, n \in \mathbb{Z}$.
$\frac{\pi x}{4}=\frac{\pi}{4}+\pi n, n \in \mathbb{Z}$, from which $x=1+4 n, n \in \mathbb{Z}$.
$0 \leq x \leq 2020 ; 0 \leq 1+4 n \leq 2020, -0.25 \leq n \leq 504.75$.
Since $n$ is an integer, $0 \leq n \leq 504$. The number of integers $n$ is 505.
Answer: 505.
## Evaluation Criteria:
|
505
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In all other cases - $\mathbf{0}$ points.
## Task 2
## Maximum 15 points
Solve the equation $2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\sin \left(\frac{\pi}{4}(1+x)\right)$.
How many solutions of this equation satisfy the condition: $2000 \leq x \leq 3000$?
|
# Solution:
$\sin \left(\frac{\pi}{4}(1+x)\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$. The equation becomes
$2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$, i.e., we get problem 2 from option 1, the solution of which is:
$x=1+4 n, n \in Z . \quad 2000 \leq x \leq 3000,2000 \leq 1+4 n \leq 3000,499.75 \leq n \leq 749.75$.
Since $n$ is an integer, $500 \leq n \leq 749$. The number of integers $n$ is 250.
Answer: 250.
## Grading Criteria:
|
250
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In all other cases, $-\mathbf{0}$ points.
## Task 4
Maxim Andreevich, a former university lecturer, gives math lessons to groups of schoolchildren. The cost of one hour (60 minutes) of Maxim Andreevich's lesson with a group of schoolchildren is 3 thousand rubles (after all taxes). In addition to income from tutoring, he has a steady income from investment activities, which amounts to 14 thousand rubles per month. Maxim Andreevich dedicates his weekends entirely to his family and does not work. Maxim Andreevich is very disciplined and strictly monitors the distribution of his time and income. On working days, of which there are exactly 21 in each month, he sleeps for eight hours a day, works for some time, devotes some time to his hobby - creating metal engravings, and spends the rest of the time on rest and household chores. He considers his hobby an important part of his life, so if he works L hours in a day, he will definitely spend 2L hours on creating engravings. He does not sell the engravings but gives them to friends and acquaintances. In addition, Maxim Andreevich is an active philanthropist and donates money to a fund for helping sick children. Every working day, he sets aside exactly $\mathrm{k} / 3$ thousand rubles for charity if he spends a total of $\mathrm{k}$ hours on rest and household chores that day. He spends 70 thousand rubles per month on all household expenses, hobbies, and household management (excluding charitable donations), and he has no other expenses. Of course, Maxim Andreevich loves to rest and do household chores, but he has to work, and every working day he decides for himself how many lessons with students he will conduct that day.
(a) What is the maximum amount of money Maxim Andreevich can donate to charity each month?
(b) Since Maxim Andreevich is a very good math teacher, the parents of the students he teaches are willing to offer him a higher rate for additional lessons (beyond those he already conducts daily). Maxim Andreevich has long taken care of the days when he will no longer be able to work, so he confidently spends all the money he receives each month. Will Maxim Andreevich agree to their offer? If yes, what should the pay for the extra lessons be for him to agree to conduct additional lessons? If no, explain why he would refuse such an offer.
|
# Solution and Grading Scheme:
(a) Maxim Andreevich works $\left(24-8-2 L_{i}-k_{i}\right)=16-2 L_{i}-k_{i}$ hours a day, where $L_{i}$ is the number of hours he works as a tutor on the $i$-th working day, and $k_{i}$ is the number of hours he spends on rest and household chores on the $i$-th working day. The number of working hours in one working day can differ from the number of working hours in another working day.
Then $16-2 L_{i}-k_{i}=L_{i}$, or $L_{i}=\left(16-k_{i}\right) / 3$
His daily earnings from tutoring are $3 L_{i}=3\left(16-k_{i}\right) / 3=16-k_{i}$.
In a month, Maxim Andreevich earns from tutoring
$$
\begin{gathered}
3 L_{1}+3 L_{2}+\cdots+3 L_{20}=\left(16-k_{1}\right)+\left(16-k_{2}\right)+\cdots+\left(16-k_{20}\right) \\
3\left(L_{1}+L_{2}+\cdots+L_{20}\right)=21 * 16-\left(k_{1}+k_{2}+\cdots+k_{20}\right)
\end{gathered}
$$
Or $3 L^{*}=336-K^{*}$,
where $L^{*}$ is the total number of working hours in a month, and $K^{*}$ is the total number of hours in a month that Maxim Andreevich spends on rest and household chores.
His monthly expenses amount to $70+\left(k_{1}+k_{2}+\cdots+k_{20}\right) / 3=70+K^{*} / 3$. Since he spends all his earnings, including income from investment activities, then
$14+336-K^{*}=70+K^{*} / 3$, from which we find the maximum number of hours in a month that Maxim Andreevich spends on rest and household chores: $K^{*}=210$.
Then in a month, Maxim Andreevich will spend a maximum of $\frac{K^{*}}{3}=\frac{210}{3}=$ 70 thousand rubles on charity.
Full solution - 10 points.
For a partial solution assuming that the number of working hours, leisure, etc., are the same every day of the month - 6 points.
If only the equations for the number of hours are written - 2 points.
(b) If Maxim Andreevich declines the offer to work more, then in a month he will work $L^{*}=(336-70) / 3=42$ hours.
Suppose Maria Ivanovna would agree to work extra for a higher hourly wage A>3. Then her monthly earnings from tutoring would be
$3 * 42+A(\tilde{L}-42)=126-42 \mathrm{~A}+\mathrm{A} \tilde{L}$
Since $L_{i}=\left(16-k_{i}\right) / 3$, then with the new distribution of working time, as before
$\tilde{L}=L_{1}+L_{2}+\cdots+L_{20}=\left(16-k_{1}\right) / 3+\left(16-k_{2}\right) / 3+\cdots+\left(16-k_{20}\right) / 3=16 * 21 / 3-$ $\widetilde{K} / 3$
Then in a month, including investments, Maxim Andreevich will earn
$126-42 A+A(112-\widetilde{K} / 3)+14$ thousand rubles.
Since the maximum expenses will equal the income with the proportions in the distribution of time between work and rest, then
$$
\begin{gathered}
126-42 A+A(112-\widetilde{K} / 3)+14=70+\widetilde{K} / 3 \\
70(A+1)=\frac{\widetilde{K}}{3}(A+1)
\end{gathered}
$$
And, as in part (a), we get: $K^{*}=210$. That is, Maria Ivanovna will not agree to work more than the hours she works at a rate of 2 thousand rubles/hour, regardless of the increased wage rate.
Full justification - 10 points,
Partial justification - 5 points
## Task 5
Company "Zolotnik" is engaged in gold and precious stone mining. "Zolotnik" has signed a contract to process a square plot of land with an area of 10 km². The workers of "Zolotnik" suffer from chronic fatigue due to the very labor-intensive work, which leads to a 2-fold decrease in their productivity each day. The amount of land that any worker can process on the first day of work is $1 / 2$ km². The salary of one worker for the entire period of work is 3 monetary units (mon. units), and the daily rental cost of equipment for processing the entire plot of land is 7 mon. units. The contract with a team of workers, which can consist of any number of workers, can only be signed on the first day the contract for processing the plot of land comes into effect.
(a) Find the minimum number of workers required to process the specified plot of land.
(b) Find the number of workers who can process the plot of land in one day.
(c) If "Zolotnik" minimizes its costs for processing the plot of land, how many workers will the company hire and how long will the processing take? What will be the total costs of the company for renting equipment and paying workers?
(a) Answer: 11.
We will prove by contradiction that no smaller number of workers can complete the task. Suppose $n$ workers processed the entire plot of land in $k$ days, where $n \leq 10$. Then each of them processed no more than
$\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{k}}=1-\frac{1}{2^{k}}$
square kilometers of land, which is strictly less than 1.
Thus, in total, all workers processed no more than
$$
n\left(1-\frac{1}{2^{k}}\right)=n-\frac{n}{2^{k}} \leq 10-\frac{10}{2^{k}}10
$$
Thus, 11 is the minimum number of workers required to process the given plot of land.
## Full justification - 5 points
(b) Answer: 20.
Indeed, on the first day, each worker processes $\frac{1}{2}$ square kilometers of land, so to process the entire plot, $10: \frac{1}{2}=20$ workers are required.
## Full justification - 5 points
(c) Answer: the company will hire 14 workers for 2 days;
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3.
## Maximum 10 points
In the Country of Wonders, a pre-election campaign is being held for the position of the best tea lover, in which the Mad Hatter, March Hare, and Dormouse are participating. According to a survey, $20 \%$ of the residents plan to vote for the Mad Hatter, $25 \%$ for the March Hare, and $30 \%$ for the Dormouse. The rest of the residents are undecided. Determine the smallest percentage of the undecided voters that the Mad Hatter must attract to ensure he does not lose to the March Hare and the Dormouse (under any distribution of votes), knowing that each of the undecided voters will vote for one of the candidates. The winner is determined by a simple majority of votes. Justify your answer.
|
# Solution:
Let the number of residents in Wonderland be $N$, then $0.2 N$ residents are going to vote for Dum, $0.25 N$ residents for the Rabbit, and $0.3 N$ residents for Sonya. The undecided voters are $0.25 N$ residents. Let $\alpha$ be the fraction of the undecided voters who are going to vote for Dum. Dum will not lose the election if the number of votes for him is not less than the number of votes for his opponents. He will receive $0.2 N + 0.25 N \alpha$ votes. In the "worst" case, the remaining fraction $1 - \alpha$ of the undecided voters will vote for the Rabbit or Sonya. Therefore, for the condition of the problem to be met, the following must hold:
$$
\left\{\begin{array} { c }
{ 0.2 N + 0.25 N \alpha \geq 0.25 N + 0.25 N (1 - \alpha), } \\
{ 0.2 N + 0.25 N \alpha \geq 0.3 N + 0.25 N (1 - \alpha). }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
0.2 + 0.25 \alpha \geq 0.25 + 0.25(1 - \alpha) \\
0.2 + 0.25 \alpha \geq 0.3 + 0.25(1 - \alpha)
\end{array}\right.\right.
$$
It is clear that it is sufficient to solve the second inequality of the system:
$$
0.2 + 0.25 \alpha \geq 0.3 + 0.25 - 0.25 \alpha \Leftrightarrow 0.5 \alpha \geq 0.35 \Leftrightarrow \alpha \geq 0.7
$$
The smallest value of $\alpha = 0.7$, which corresponds to $70\%$ of the undecided voters.
Answer: $70\%$.
## Grading Criteria:
- 10 points - the correct answer is obtained with proper justification.
- 5 points - there is a correct sequence of reasoning leading to the right answer, but an arithmetic error has been made.
- 0 points - the solution does not meet any of the criteria provided.
|
70
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4.
## Maximum 15 points
Find the value of the expression under the condition that the summation is performed infinitely
$$
\sqrt{20+\sqrt{20+\sqrt{20+\cdots}}}
$$
#
|
# Solution:
Let $x=\sqrt{20+\sqrt{20+\sqrt{20+\cdots}}}, x>0$.
Square both sides of the obtained equation. We get:
$x^{2}=20+x$
$x^{2}-x-20=0 ;$
$D=1+80=81 ; x_{1}=\frac{1-9}{2}=-4-$ does not satisfy the condition $x>0$;
$x_{1}=\frac{1+9}{2}=5-$ fits.
Answer: 5.
## Criteria
0 points - If calculated approximately, then $0 \%$.
8 points - If both roots of the equation are given: -4 and 5, then $50 \%$.
15 points - If the answer is correct, and a valid and justified solution is provided without using a calculator, then $100 \%$.
#
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Assignment 7.
## Maximum 10 points
In the modern world, every consumer often has to make decisions about replacing old equipment with more energy-efficient alternatives. Consider a city dweller who uses a 60 W incandescent lamp for 100 hours each month. The electricity tariff is 5 rubles/kWh.
The city dweller can buy a more energy-efficient lamp with a power of 12 W, which costs 120 rubles and provides the same light output as the aforementioned incandescent lamp. Alternatively, he can approach an investor (an energy service company) that will install such an energy-saving lamp itself, but in return, the city dweller must pay the company 75% of the resulting cost savings on electricity for 10 months (the tariff payment is made regardless of the method of lamp installation).
(a) If the city dweller plans his expenses only for the next 10 months, will he install the energy-saving lamp himself or approach the investor company?
(b) If the city dweller plans his expenses for the entire service life of a typical energy-saving lamp, what decision will he make?
|
# Solution and Grading Scheme:
a) Expenses for 10 months when installing an energy-saving lamp independently:
$$
120 \text { rub. }+12 \text { (W) * } 100 \text { (hours) / } 1000 \text { * } 5 \text { (rub./kW*hour) * } 10 \text { (months) = } 180 \text { rub. }
$$
Expenses for 10 months when turning to an energy service company:
$$
(12+(60-12) * 0.75)(W) * 100 \text { (hours) / } 1000 \text { * } 5 \text { (rub./kW*hour) * } 10 \text { (months) = } 240 \text { rub. }
$$
Thus, the most cost-effective of the alternative options considered is to install the energy-saving lamp independently.
Full correct solution - 6 points. Only one option correctly calculated - 3 points.
b) The expected service life of a typical lamp is more than 10 months. Since after 10 months, you only need to pay for electricity consumption, as in part a), it will be more cost-effective to install the lamp independently.
Justification (including calculations) 4 points
#
|
180
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 4. 25 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. In the cryptogram, the amount of tax revenue to be collected was specified. It was also emphasized that a larger amount of tax revenue could not be collected on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the amount of tax revenue were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit commodity tax on the consumer in the amount of 30 monetary units per unit of the good. It is known that the market supply is given by $Q_s = 6P - 312$, and the market demand is linear. In the situation where no taxes are imposed, at the equilibrium point, the price elasticity of market supply is 1.5 times higher than the absolute value of the price elasticity of the market demand function. After the tax was introduced, the consumer price increased to 118 monetary units.
1) Restore the market demand function.
2) Determine the amount of tax revenue collected at the chosen rate.
3) Determine the rate of the quantity tax that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler indicated to collect?
|
# Solution:
1) Let the demand function be linear $Q_{d}=a-b P$. It is known that $1.5 \cdot\left|E_{p}^{d}\right|=E_{p}^{s}$. For linear demand functions, using the definition of elasticity, we get: $1.5 \cdot \frac{b P_{e}}{Q_{e}}=\frac{6 P_{e}}{Q_{e}}$. From this, we find that $b=4$.
If a per-unit tax $t=30$ is introduced, then $P_{d}=118 . a-4 P_{d}=6\left(P_{d}-30\right)-312 ; 0.1 a+$ $49.2=P_{d}=118$; from which $a=688$.
The market demand function is $Q_{d}=688-4 P$.
2) It is known that $P_{d}(t=30)=118$. Therefore, $Q_{d}=688-472=216, T=216 \cdot 30=6480$.
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenues are $\quad T=Q \cdot t=288 t-2.4 t^{2}$. The graph of the function is a parabola opening downwards, the maximum of the function is achieved at $t^{*}=60$.
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$.
## Grading Criteria:
1) Correctly determined the slope of the demand curve - 3 points.
Correctly determined the selling price with the incorrect tax - 1 point.
Correctly determined the sales volume with the incorrect tax - 2 points.
Correctly determined the intercept of the demand curve - 3 points.
Correctly found the demand function - 1 point.
If an error/typo is made in the solution that leads to results contradicting economic logic, only 3 points are awarded for this item even if the rest of the solution is correct.
2) Correctly found the tax revenues - 3 points.
3) Explained how to find the rate that maximizes tax revenues - 4 points.
Found the rate that maximizes tax revenues - 4 points. If the rate found contradicts logic, even if the value of $t$ is correct, these 4 points are not awarded.
4) Found the quantity of goods that will be sold at the rate that maximizes tax revenues - 2 points. If the quantity found contradicts logic, even if the value of $t$ is correct, these 2 points are not awarded.
Found the tax revenues at this tax rate - 2 points. If the tax revenues are found but their value does not align with the condition in the problem about the mixed-up digits, these 2 points are not awarded.
## Penalties:
An arithmetic error was made in calculations that did not lead to results contradicting economic logic - 1 point.
The fact that the tax rate found in item 3) gives a maximum, not a minimum of the objective function, is not justified.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4. 25 points
A novice economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. In the cryptogram, the amount of tax revenue to be collected was specified. It was also emphasized that it was impossible to collect a larger amount of tax revenue on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the amount of tax revenue were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit tax on the producer in the amount of 90 monetary units per unit of the product. It is known that the market demand is given by $Q d=688-4 P$, and the market supply is linear. In the situation where there are no taxes, the price elasticity of market supply at the equilibrium point is 1.5 times higher than the absolute value of the price elasticity of the market demand function. After the tax was introduced, the producer's price decreased to 64 monetary units.
1) Restore the market supply function.
2) Determine the amount of tax revenue collected at the chosen rate.
3) Determine the rate of the quantity tax that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler indicated to collect?
|
# Solution:
1) Let the supply function be linear $Q_{s}=c+d P$. It is known that $1.5 \cdot\left|E_{p}^{d}\right|=E_{p}^{s}$. Using the definition of price elasticity for linear demand functions, $1.5 \cdot$ $\frac{4 P_{e}}{Q_{e}}=\frac{d P_{e}}{Q_{e}}$. We find that $d=6$. If a per-unit tax $t=90$ is introduced, then $P_{s}=64$. Therefore, $688-4\left(P_{s}+90\right)=6 P_{s}+c ; 0.1 c+32.8=P_{s}=64 ; c=-312$.
The market supply function is $Q_{s}=6 P-312$.
2) It is known that $P_{s}(t=90)=64$. Therefore, $Q_{s}=6 P-312=72$. Then the government revenue is $T=72 \cdot 90=6480$.
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenue will be $T=Q \cdot t=288 t-2.4 t^{2}$. The graph of the function is a parabola opening downwards, and the maximum of the function is achieved at $t^{*}=60$.
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$.
## Grading Criteria:
1) Correctly determined the slope of the supply curve - 3 points.
Correctly determined the purchase price with the incorrect tax - 1 point.
Correctly determined the sales volume with the incorrect tax - 2 points.
Correctly determined the intercept of the supply curve - 3 points.
Correctly found the supply function - 1 point.
If an error/typo is made in the solution that leads to results contradicting economic logic, only 3 points are awarded for this section even if the rest of the solution is correct.
2) Correctly found the tax revenue - 3 points.
3) Explained how to find the rate that maximizes tax revenue - 4 points.
Found the rate that maximizes tax revenue - 4 points. If the rate found contradicts logic, even if the value of $t$ is correct, these 4 points are not awarded.
4) Found the quantity of goods sold at the rate that maximizes tax revenue - 2 points. If the quantity found contradicts logic, even if the value of $t$ is correct, these 2 points are not awarded. Found the tax revenue at this tax rate - 2 points. If the tax revenue is found but does not align with the condition in the problem about mixed-up digits, these 2 points are not awarded.
## Penalties:
An arithmetic error was made in calculations that did not lead to results contradicting economic logic -1 point.
The fact that the tax rate found in section 3) gives a maximum, not a minimum, of the objective function is not justified.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 2. Maximum 16 points
Settlements $A, B$, and $C$ are connected by straight roads. The distance from settlement $A$ to the road connecting settlements $B$ and $C$ is 100 km, and the sum of the distances from point $B$ to the road connecting $A$ and $C$, and from point $C$ to the road connecting $A$ and $B$ is 300 km. It is known that point $D$ is equidistant from the roads connecting points $A, B, C$ and lies within the area bounded by these roads. Any resident of all settlements spends no more than 1 liter of fuel for every 10 km of road. What is the maximum amount of fuel that would be needed by a motorist who needs to get from settlement $D$ to any of the roads connecting the other settlements?
#
|
# Solution
The settlements form a triangle $\mathrm{ABC}$, and point $\mathrm{D}$, being equidistant from the sides of the triangle, is the incenter of the triangle (i.e., the center of the inscribed circle). Note that the fuel consumption will be maximal when the distance from point $\mathrm{D}$ to the sides of triangle $\mathrm{ABC}$ is maximal, or, equivalently, when the radius $r$ of the inscribed circle in the given triangle is maximal under the conditions of the problem. The conditions of the problem can be easily reformulated as follows (see the figure):
$\mathrm{ha}=100, \mathrm{hb}+\mathrm{hc}=300$,
where $\mathrm{ha}, \mathrm{hb}, \mathrm{hc}$ are the heights of triangle ABC. It is not difficult to establish that the following equality holds:
1ha+1hb+1hc=1r.
Indeed, let $S$ be the area of the triangle, then
$$
\mathrm{S}=12 \mathrm{aha}=12 \mathrm{bhb}=12 \mathrm{chc}=\text { pr. }
$$
From this, we have:
$$
h a=2 S a, h b=2 S b, h c=2 S c, r=S p .
$$
And we get:
$$
1 \mathrm{ha+}+1 \mathrm{hb}+1 \mathrm{hc}=\mathrm{a} 2 \mathrm{~S}+\mathrm{b} 2 \mathrm{~S}+\mathrm{c} 2 \mathrm{~S}=\mathrm{a}+\mathrm{b}+\mathrm{c} 2 \mathrm{~S}=\mathrm{pS}=1 \mathrm{r} \text {. }
$$
From the condition, it follows that
$$
1 \mathrm{r}=1100+1 \mathrm{hb}+1 \mathrm{hc}=1100+1 \mathrm{hb}+1300-\mathrm{hb}=1100+300 \mathrm{hb} 300-\mathrm{hb} .
$$
The radius $\mathrm{r}$ is maximal when the sum on the right side of the equation is minimal. But this is equivalent to the maximality of the expression
hb300-hb.
This is a quadratic function, and its maximum is achieved at its vertex, i.e., as it is not difficult to understand, when $\mathrm{hb}=150$. Therefore, the maximum value of the radius of the inscribed circle is found from the equation
$$
1 \mathrm{r}=1100+300150300-150=1100+175=7300 \Leftrightarrow \mathrm{r}=3007 \text { km. }
$$
From this, the maximum fuel consumption is:
$$
\text { r10=307 l. }
$$
Answer: 307 l.
|
307
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 5. Maximum 20 points
The commander of a tank battalion, in celebration of being awarded a new military rank, decided to invite soldiers to a tank festival, where the main delicacy is buckwheat porridge. The commander discovered that if the soldiers are lined up by height, there is a certain pattern in the change of their demand functions. The demand for buckwheat porridge of the shortest soldier is given by $Q d=510-5.1 P$, the next tallest soldier's demand is $Q d=520-5.2 P$, the next one's is $Q d=530-5.3 P$, and so on. The commander's individual demand for buckwheat porridge is $Q d=500-5 P$. The battalion commander always tries to feed as many soldiers as possible. Apart from the commander and the invited soldiers, no one else demands buckwheat porridge. At the festival, 25 perfectly competitive firms offer buckwheat porridge, with each firm's supply function being $Q s=302 P$. The battalion commander, after consulting with his family, decided to make a surprise - initially, the guests and the commander will make decisions about consuming buckwheat porridge, assuming that everyone will pay for it themselves, but at the end of the celebration, the commander's family will pay the total bill from their savings, which amount to 2,525,000 monetary units. If this amount is insufficient, the commander will ask all guests to equally share the remaining bill. It is known that the equilibrium price is set at 20 monetary units. Assume that if guests, like the commander, have to make a certain fixed payment, this payment does not affect their demand for buckwheat porridge.
(a) How is the individual demand of the 60th soldier invited to the celebration described? How will the aggregate demand for buckwheat porridge be described if the commander plans to invite 40 soldiers and does not refuse the porridge himself? Explain the construction of the demand function in this case.
(b) How many soldiers did the commander invite to the celebration? Did the guests have to pay part of the bill? If so, how much did each of them pay?
|
# Solution
(a) The commander tries to feed as many soldiers as possible, which means he will invite relatively short soldiers first - all other things being equal, their consumption of porridge is less.
Note that the individual demand of soldiers is determined by the formula $Q_{d}=500+10 n- (5+0.1 n) P$, where $n$ is the ordinal number of the soldier. Also note that all consumers are willing to purchase the product at $P \in [0 ; 100]$, so to find the market demand, it is sufficient to sum the individual demand functions.
The individual demand of the 60th soldier is $Q_{d}=1100-11 P$. To find the demand of 40 soldiers, we can use the formula for the arithmetic progression:
$Q_{d}=\frac{510-5.1 P+500+10 n-(5+0.1 n) P)}{2} \cdot n$
$Q_{d}=\frac{1410-14.1 P}{2} \cdot 40$
$Q_{d}=28200-282 P$
Adding the commander's demand, we get the market demand: $Q_{d}=28700-287 P$.
Answer: $Q_{d}=1100-11 P, Q_{d}=28700-287 P$.
(b) Knowing the individual supply of one firm, we can find the market supply:
$Q_{s}=302 P \cdot 25=7550 P$
It is known that $Q_{s}(20)=Q_{d}(20)$, so $Q_{d}=7550 \cdot 20=151000$.
From the previous part, we know that
$Q_{d}=5 n^{2}+505 n+500-\left(0.05 n^{2}+5.05 n+5\right) P$
$Q_{d}(20)=4 n^{2}+404 n+400=151000$
$(4 n+1004)(n-150)=0$
$n_{1}=-251, n_{2}=150$.
Obviously, $n>0$. The commander invited 150 guests.
$T R=151000 \cdot 20=3020000>2525000$ - the expenses for buckwheat porridge exceed the family budget, which means the soldiers will help pay part of the bill. Each soldier will contribute an amount of $\frac{495000}{150}=3300$ monetary units.
Answer: the commander invited 150 guests, each of whom paid 3300 monetary units.
## Grading Criteria
(a) Individual demand is formulated - (3p), market demand is formulated - (5p).
(b) Market supply function is formulated - (2p), equilibrium quantity is determined - (2p), market demand function in terms of the number of guests is formulated - (2p), maximum number of guests is determined - (3p), behavior of guests and their payment of the bill is determined - (3p).
A penalty of 1 point for arithmetic errors that did not lead to significant distortion of the results. Alternative solutions may be awarded full points if they contain a correct and justified sequence of actions.
|
150
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The number 2458710411 was written 98 times in a row, resulting in a 980-digit number. From this number, it is required to erase 4 digits. What is the number of ways this can be done so that the newly obtained 976-digit number is divisible by 6?
|
Solution. If a number is divisible by 6, then it is divisible by 3 and 2. A number is divisible by 2 if and only if its last digit is even. The 980-digit number given in the condition ends with 2458710411, i.e., it has the form ...2458710411. For the number to be divisible by 2, it is necessary to strike out the last two ones so that the number ends with an even digit. The resulting number will have the form 2458710411...245871041124587104. Now it is necessary for this number to be divisible by 3. A number is divisible by 3 if and only if the sum of its digits is divisible by 3. The sum of the digits of the resulting number is:
$(2+4+5+8+7+1+0+4+1+1) * 97 + (2+4+5+8+7+1+0+4) = 33 * 97 + 31 = 3232$.
The remainder of the division of the number 3232 by 3 is 1. This means that the numbers must be struck out in such a way that their sum is divisible by 3 and leaves a remainder of 1. Then, upon subtraction, a number will be obtained whose sum of digits will be exactly divisible by 3. The maximum sum that can be obtained by striking out digits is $16 (8+8)$. Therefore, it is necessary to consider all numbers from 0 to 16 that are divisible by 3 and leave a remainder of 1, and all possible combinations of their formation. These numbers are: $1, 4, 7, 10, 13, 16$. The combinations by which they can be formed are as follows:
$1 (0+1); 4 (0+4, 2+2); 7 (2+5, 7+0); 10 (5+5, 2+8); 13 (8+5); 16 (8+8)$.
It remains to calculate the number of ways in which each combination can be obtained. The first combination can be obtained by striking out one 1 and one 0. There are $97 * 3 + 1 = 292$ ones in our number, and 98 zeros. Therefore, the number of ways in which this can be done is $(97 * 3 + 1) * 98 = 28616$. For subsequent combinations, we get:
$(0+4): 98 * 98 = 9604$
$(2+2): 98 * 97 / 2 = 4753$
$(2+5): 98 * 98 = 9604$
$(7+0): 98 * 98 = 9604$
$(5+5): 98 * 97 / 2 = 4753$
$(2+8): 98 * 98 = 9604$
$(5+8): 98 * 98 = 9604$
$(8+8): 98 * 97 / 2 = 4753$.
Summing all the obtained combinations, we get the number of ways 90895. However, it should be noted that one of the proposed ways is not suitable. If we strike out the last digits 0 and 4, then our number will end with 1, i.e., it will not be divisible by 2, and therefore, it will not be divisible by 6. Therefore, the number of combinations is $90895 - 1 = 90894$ (ways).
Answer: 90894 ways.
| Criteria for evaluating the performance of the task | Points |
| :--- | :---: |
| A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are correctly performed. The correct answer is obtained. | **10** |
| A correct sequence of all steps of the solution is provided. Gaps in the justification of the choice of number combinations (for example, one extra solution (0+4) is not taken into account, pairs of digits are not divided by 2, etc.) or a computational error or typo that does not affect the further course of the solution are allowed. As a result of this error or typo, an incorrect answer may be obtained. | **6** |
| The task is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, part of the combinations is taken into account and calculated correctly); |
| - the other part is either not performed or performed incorrectly, possibly even with logical errors (for example, most combinations are not taken into account or extra combinations are taken into account, combinations are calculated incorrectly, cases of striking out the same digit (2+2, |
| 5+5, 8+8) are not taken into account, etc.); |
All cases of solutions that do not meet the above criteria for scoring 3 and 6 points.
|
90894
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that a larger amount of tax revenue could not be collected on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the tax revenue amount were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit tax on the consumer in the amount of 30 monetary units per unit of the good. The market supply is given by \( Q_s = 6P - 312 \), and the market demand is linear. Additionally, it is known that a change in price by one unit results in a change in the quantity demanded that is 1.5 times smaller than the change in the quantity supplied. After the tax was introduced, the consumer price increased to 118 monetary units.
1) Restore the market demand function.
2) Determine the amount of tax revenue collected at the chosen tax rate.
3) Determine the per-unit tax rate that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler specified to be collected?
|
# Solution:
1) Let the demand function be linear $Q_{d}=a-b P$. It is known that $1.5 b=6$. We find that $b=$ 4. If a per-unit tax $t=30$ is introduced, then $P_{d}=118 . a-4 P_{d}=6\left(P_{d}-30\right)-312 ; 0.1 a+$ $49.2=P_{d}=118 ; a=688$. The market demand function is $Q_{d}=688-4 P$. (8 points).
2) It is known that $P_{d}(t=30)=118$. Therefore, $Q_{d}=688-472=216, T=216 \cdot 30=6480$. (4 points).
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenues are $T=Q \cdot t=288 t-2.4 t^{2}$. This is a downward-opening parabola, the maximum of the function is reached at $t^{*}=60$. (4 points).
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$. (4 points).
Penalties: an arithmetic error was made - 2 points, lack of justification for the maximum - 5 points.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that a larger amount of tax revenue could not be collected on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the tax revenue amount were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit tax on the producer in the amount of 90 monetary units per unit of the product. It is known that the market demand is given by $Q_d = 688 - 4P$, and the market supply is linear. Additionally, it is known that a change in price by one unit results in a change in the quantity demanded that is 1.5 times smaller than the change in the quantity supplied. After the tax was introduced, the producer's price decreased to 64 monetary units.
1) Restore the market supply function.
2) Determine the amount of tax revenue collected at the chosen rate.
3) Determine the rate of the quantity tax that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler indicated to collect?
|
# Solution:
1) Let the supply function be linear $Q_{s}=c+d P$. It is known that $1.5 \cdot 4=d$. We find that $d=6$. If a per-unit tax $t=90$ is introduced, then $P_{s}=64.688-4\left(P_{s}+90\right)=6 P_{s}+c$; $0.1 c+32.8=P_{s}=64 ; c=-312$. The market supply function is $Q_{s}=6 P-312$. (8 points).
2) It is known that $P_{s}(t=90)=64$. Therefore, $Q_{s}=6 P-312=72, T=72 \cdot 90=6480$. (4 points).
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenues are $T=Q \cdot t=288 t-2.4 t^{2}$. This is a downward-opening parabola, and the maximum of the function is achieved at $t^{*}=60$. (4 points).
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$. (4 points).
Penalties: An arithmetic error was made -2 points, lack of justification for the maximum -5 points.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Consider the vectors $\vec{a}=(3, x), \vec{b}=(\sqrt{x-2}, 4)$, then $\vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|$, which is equivalent to
$$
3 \sqrt{x-2}+4 x \leq \sqrt{\left(9+x^{2}\right)(x+14)}
$$
Equality is possible if and only if the vectors are collinear
$$
\frac{\sqrt{x-2}}{3}=\frac{4}{x}>0 \Leftrightarrow x \sqrt{x-2}=12
$$
The left side of the considered equation is a monotonically increasing function on the domain of definition, so the equation has no more than one root. By trial, we find $x=6$.
|
Answer: 1. 6. 2.6 3.74. 7.
## Checking Criteria:
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In all other cases - 0 points
## *Important: the numerical assessment of the free area (solution) is not the only possible one, for example, the "gap" can be more than 10 m.
## Assignment 2 (12 points)
Crocodile Gena and Old Lady Shapoklyak entered into a futures contract, according to which Gena agreed to invest in the 1st project an amount equal to $p_{1}>0$ thousand rubles, and after a year receive $x_{1}$ thousand rubles such that
$$
4 x_{1}-3 p_{1}-44=0
$$
and at the same time Gena agreed to invest in the 2nd project an amount equal to $p_{2}>0$ thousand rubles, and after a year receive $x_{2}$ thousand rubles such that
$$
p_{2}^{2}-12 p_{2}+x_{2}^{2}-8 x_{2}+43=0
$$
According to the terms of the contract, the parameters $p_{1}, x_{1}, p_{2}, x_{2}$ were chosen so that the value
$$
d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(p_{1}-p_{2}\right)^{2}}
$$
was minimized. Determine:
1) the minimum value of the quantity $d$;
2) is the deal profitable for Gena? Find the amount of profit or loss for Gena if he agrees to this deal.
|
# Solution:
1) It is clear that the expression $4 x_{1}-3 p_{1}-44=0 \Leftrightarrow x=\frac{3}{4} p+11$ defines the equation of a certain line $l$ in the plane $x O p$. Consider the expression
$$
\begin{aligned}
p^{2}-12 p+x^{2}-8 x+4 & =0 \Leftrightarrow(p-6)^{2}+(x-4)^{2}-36-16+43=0 \\
\Leftrightarrow & (p-6)^{2}+(x-4)^{2}=3^{2}
\end{aligned}
$$
It defines in the plane $x O p$ the equation of a circle $\Phi$ with center at the point $(6 ; 4)$ and radius 3. Then the expression of the form
$$
\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(p_{1}-p_{2}\right)^{2}}
$$
defines the distance between points lying on $l$ and $\Phi$. Therefore, according to the problem's condition, we need to find the minimum distance between points lying on
the line and the circle. Let $O_{1} H$ be the perpendicular from the center of the circle $O_{1}$ to the line $l$, then $O_{1} P$ is the radius of the circle. We will show that $H P=O_{1} H-O_{1} P$ is the desired value. Indeed, let $H_{1} \in l, F \in \Phi$. Draw the tangent $a$ to the circle at point $P$. The radius $O_{1} P$, drawn to the point of tangency, is perpendicular to $a$. Therefore, $a \| l$. And thus, $HP$ is equal to the distance between the parallel lines. Now we have:
$$
H_{1} F=H_{1} P_{1}+P_{1} F \geq H_{1} P_{1} \geq H P .
$$
Thus, $H P$ is the shortest distance between the points of the circle and the line $l$. Let's find $O_{1} H$ using the formula for the distance from a point to a line (one of the ways to find it):
$$
O_{1} H=\frac{|4 \cdot 4-3 \cdot 6-44|}{\sqrt{3^{2}+4^{2}}}=\frac{46}{5}=9.2 .
$$
Therefore, $H P=9.2-3=6.2$.
2) Let's define the equation of the line $O_{1} H$. Since it is perpendicular to the line $l$ and passes through the point $O_{1}(6 ; 4)$, its equation in the plane $x O p$ has the form
$$
x=-\frac{4}{3}(p-6)+4 \Leftrightarrow x=-\frac{4}{3} p+12
$$
Let's find the coordinates of the point $H$ as the coordinates of the intersection of the lines $l$ and $O_{1} H$. Form the equation:
$$
-\frac{4}{3} p+12=\frac{3}{4} p+11 \Leftrightarrow \mathrm{p}=\frac{12}{25}=0.48
$$
Thus, $p_{1}=0.48$ and $x_{1}=\frac{3}{4} p_{1}+11=11.36$.
Let's find the coordinates of the point $P$ as the coordinates of the intersection of the line $O_{1} H$ and the circle $\Phi$ :
$$
(p-6)^{2}+\left(-\frac{4}{3} p+12-4\right)^{2}=3^{2}
$$
We get two solutions $p=\frac{39}{5}$ and $p=\frac{21}{5}$. Since the point $P$ lies to the left of the point $O_{1}$, we finally have $p_{2}=\frac{21}{5}=4.2$. Thus, $x_{2}=6.4$.
The advantage of the deal will be:
$$
x_{1}+x_{2}-p_{1}-p_{2}=11.36+6.4-0.48-4.2=13.08
$$
Thus, the deal is profitable for Crocodile Gena, and the profit will be 13080 rubles.
Answer: 1) 6.2. 2) profitable, profit 13080 rubles.
## Criteria:
|
13080
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 3. Maximum 14 points
Settlements $A, B$, and $C$ are connected by straight roads. The distance from settlement $A$ to the road connecting settlements $B$ and $C$ is 100 km, and the sum of the distances from settlement $B$ to the road connecting $A$ and $C$, and from settlement $C$ to the road connecting $A$ and $B$ is 300 km. It is known that settlement $D$ is equidistant from the roads connecting settlements $A, B, C$ and lies within the area bounded by these roads. Any resident of all settlements spends no more than 1 liter of fuel for every 10 km of road. What is the maximum amount of fuel that would be needed by a motorist who needs to get from settlement $D$ to any of the roads connecting the other settlements?
#
|
# Solution
The settlements form a triangle $\mathrm{ABC}$, and point $\mathrm{D}$, being equidistant from the sides of the triangle, is the incenter of the triangle (i.e., the center of the inscribed circle). Note that the fuel consumption will be maximal when the distance from point $\mathrm{D}$ to the sides of triangle $\mathrm{ABC}$ is maximal, or, equivalently, when the radius $\mathrm{r}$ of the inscribed circle in the triangle is maximal under the conditions of the problem. The conditions of the problem can be easily reformulated as follows (see the figure):
$$
\text { ha=100, hb+hc=300, }
$$
where $\mathrm{ha}, \mathrm{hb}, \mathrm{hc}$ are the heights of triangle ABC. It is not difficult to establish that the following equality holds:
$$
1 \mathrm{ha} + 1 \mathrm{hb} + 1 \mathrm{hc} = 1 \mathrm{r} \text {. }
$$
Indeed, let $S$ be the area of the triangle, then
$$
\text { S=12aha=12bhb=12chc=pr. }
$$
From this, we have:
$$
\text { ha=2Sa, hb=2Sb, hc=2Sc, r=Sp. }
$$
And we get:
$$
1 \mathrm{ha} + 1 \mathrm{hb} + 1 \mathrm{hc} = \mathrm{a} / 2 \mathrm{~S} + \mathrm{b} / 2 \mathrm{~S} + \mathrm{c} / 2 \mathrm{~S} = (\mathrm{a} + \mathrm{b} + \mathrm{c}) / 2 \mathrm{~S} = \mathrm{p} / \mathrm{S} = 1 \mathrm{r} \text {. }
$$
From the condition, it follows that
$$
1 \mathrm{r} = 1 / 100 + 1 / \mathrm{hb} + 1 / \mathrm{hc} = 1 / 100 + 1 / \mathrm{hb} + 1 / (300 - \mathrm{hb}) = 1 / 100 + 300 / (\mathrm{hb} (300 - \mathrm{hb})).
$$
The radius $\mathrm{r}$ is maximal when the sum on the right side of the equation is minimal. This is equivalent to the maximality of the expression
$$
\mathrm{hb} / (300 - \mathrm{hb}).
$$
This is a quadratic function and its maximum is achieved at its vertex, i.e., as it is easy to understand, when $\mathrm{hb} = 150$. Therefore, the maximum value of the radius of the inscribed circle is found from the equation
$$
1 \mathrm{r} = 1 / 100 + 300 / (150 * (300 - 150)) = 1 / 100 + 1 / 75 = 7 / 300 \Leftrightarrow \mathrm{r} = 300 / 7 \text { km. }
$$
From this, the maximum fuel consumption is:
$$
\text { r10 = 30 / 7 = 307 liters. }
$$
Answer: 307 liters.
|
307
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 5. Maximum 20 points
The commander of a tank battalion, in celebration of being awarded a new military rank, decided to organize a mass celebration, inviting subordinate soldiers. Only the soldiers whom the commander personally invites can attend. The main delicacy at the celebration is buckwheat porridge. However, after consulting with his family, the tank battalion commander decided to make a surprise - initially, guests, as well as the commander himself, will make decisions about consuming buckwheat porridge, assuming that they will pay for it themselves, but at the end of the celebration, the commander will pay the total bill from his family's savings. He discovered that if the soldiers are lined up by height, there is a certain pattern in the changes in their demand functions. The demand for buckwheat porridge of the shortest soldier is given by \( Q_d = 510 - 5.1P \), the next tallest \( Q_d = 520 - 5.2P \), the next \( Q_d = 530 - 5.3P \), and so on. The commander will first invite the shortest tankers. The individual demand of the commander for buckwheat porridge is given by \( Q_d = 500 - 5P \).
(a) How is the individual demand of the 45th soldier invited to the celebration defined? How will the aggregate demand for buckwheat porridge be defined if the commander invites 45 soldiers and does not refuse the porridge himself?
(b) The commander can invite the soldiers to a local cafe, the only one in the region. The marginal costs of producing buckwheat porridge are 0. The commander's family is willing to spend no more than 2,525,000 monetary units on the celebration. What is the maximum number of guests whose demand for buckwheat porridge will be satisfied at the celebration if the commander eats the porridge with everyone?
(c) After estimating how many soldiers he can feed in the cafe, the battalion commander decided to invite the military personnel to a tank festival. In addition to the commander and the soldiers he invites, no one else demands porridge. Buckwheat porridge at the festival is offered by 25 perfectly competitive firms, each with a supply function \( Q_s = 302P \). The commander's family is willing to spend the entire amount allocated for the celebration, but if it is not enough, the commander will ask all guests to equally share the remaining bill. It is known that the equilibrium price was set at 20 monetary units. How many soldiers did the commander invite to the festival? Did the guests have to pay part of the bill? If so, how much did each of them pay? Assume that if the guests, like the commander, have to make a certain fixed payment, the payment does not affect their demand for buckwheat porridge.
(d) Which option (b or c) would the soldiers prefer if each of them made a decision about the amount of buckwheat porridge to consume according to their demand function as described in the problem? Justify your answer.
|
# Solution
(a) The commander tries to feed as many soldiers as possible, which means he will primarily invite relatively short soldiers - all other things being equal, their porridge consumption is less.
Note that the individual demand of soldiers is determined by the formula $Q_{d}=500+10 n-(5+0.1 n) P$, where $n$ is the ordinal number of the soldier. Also note that all consumers are willing to purchase the product at $P \in [0; 100]$, which means that to find the market demand, it is sufficient to sum the individual demand functions.
The individual demand of the 45th soldier is $Q_{d}=950-9.5 P$. To find the demand of 45 soldiers, we can use the formula for the arithmetic progression:
$Q_{d}=\frac{(510-5.1 P + 500 + 10 n - (5 + 0.1 n) P)}{2} \cdot n$
If $n=45$, then:
$Q_{d}=32850-328.5 P$
Adding the commander's demand, we get the market demand: $Q_{d}=33350-333.5 P$.
Answer: $Q_{d}=950-9.5 P, Q_{d}=33350-333.5 P$.
(b) For the canteen, the profit maximization problem coincides with the revenue maximization problem, since marginal costs are zero. If the commander invites $n$ soldiers, then the market demand will be:
$Q_{d}=500-5 P + \frac{1010 n + 10 n^2 - (10.1 n + 0.1 n^2) P}{2}$
$Q_{d}=5 n^2 + 505 n + 500 - (0.05 n^2 + 5.05 n + 5) P$
$T R=(5 n^2 + 505 n + 500) P - (0.05 n^2 + 5.05 n + 5) P^2$
The graph of $T R$ is a parabola opening downwards, since for $n \geq 0$ it holds that $0.05 n^2 + 5.05 n + 5 > 0$. The maximum value is at the vertex of the parabola:
$$
\begin{aligned}
& P^{*}=\frac{-5 n^2 - 505 n - 500}{-2(0.05 n^2 + 5.05 n + 5)}=50 \\
& T R_{\max }=T R(50)=125 n^2 + 12625 n + 12500 \\
& T R_{\max } \leq 2525000 \\
& 125 n^2 + 12625 n + 12500 - 2525000 \leq 0 \\
& n^2 + 101 n - 20100 \leq 0 \\
& (n + 201)(n - 100) \leq 0 \\
& n_{1}=-201, n_{2}=100
\end{aligned}
$$
Obviously, the number of guests cannot be negative or a non-integer, so the maximum number of guests the commander can invite is 100.
Answer: 100.
(c) Knowing the individual supply of one firm, we can find the market supply: $Q_{s}=302 P \cdot 25=7550 P$
It is known that $Q_{s}(20)=Q_{d}(20)$, which means $Q_{d}=7550 \cdot 20=151000$.
From the previous part, we know that
$Q_{d}=5 n^2 + 505 n + 500 - (0.05 n^2 + 5.05 n + 5) P$
$Q_{d}(20)=4 n^2 + 404 n + 400=151000$
$(4 n + 1004)(n - 150)=0$
$n_{1}=-251, n_{2}=150$
The commander invited 150 guests.
$T R=151000 \cdot 20=3020000>2525000$ - the expenses for buckwheat porridge exceed the family budget, which means the soldiers will help pay part of the bill. Each soldier will contribute an amount of $\frac{495000}{150}=3300$ monetary units.
Answer: the commander invited 150 guests, each of whom paid 3300 monetary units.
(d) Let's find the consumer surplus (CS) for the soldier with number $i$ in each case. For a linear demand function, $C S=\frac{(P_{\max }-P_{e}) Q}{2}$. The demand function of each soldier is linear, and $P_{\max }=100$.
In part (b), consumers do not pay for the buckwheat porridge, i.e., $P_{e}=0$.
$Q_{i}^{1}=500+10 i-(5+0.1 i) \cdot 50=250+5 i$
$C S_{i}^{1}=(P_{\max }-P_{e}) \cdot 0.5 \cdot Q_{1}^{i}=(100-0) \cdot 0.5(250+5 i)=12500+250 i$.
In part (c), consumers do not pay for the buckwheat porridge directly, but spend 3300 monetary units.
$Q_{i}^{2}=500+10 i-(5+0.1 i) \cdot 20=400+8 i$
$C S_{i}^{2}=(100-0) \cdot 0.5(400+8 i)-3300=16700+400 i$
It is clear that even with the expenses for partial payment of the bill, for any $i \geq 0$ it holds that $C S_{i}^{2}>C S_{i}^{1}$.
Answer: all soldiers prefer the option (c).
## Grading Criteria
(a) Individual demand is formulated - (1p), market demand is formulated - (2p).
(b) The firm's problem is formulated - (1p), the optimal price is determined - (2p), the maximum number of guests is determined - (3p).
(c) The market supply function is formulated - (1p), the equilibrium quantity is determined - (1p), the market demand function in terms of the number of guests is formulated - (2p), the maximum number of guests is determined - (2p), the behavior of guests and their payment of the bill is determined - (2p).
(d) A justified comparison of benefits is formulated, from which a conclusion is drawn about the preference of a certain option - (3p).
A penalty of 1 point for arithmetic errors that did not lead to significant distortion of the results. Alternative solutions may be evaluated for the full number of points if they contain a correct and justified sequence of actions.
|
150
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that it was impossible to collect a larger amount of tax revenue on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the amount of tax revenue were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit tax on the consumer in the amount of 30 monetary units per unit of the good. The market supply is given by $Q_s = 6P - 312$, and the market demand is linear. Additionally, it is known that a change in price by one unit results in a change in the quantity demanded that is 1.5 times smaller than the change in the quantity supplied. After the tax was introduced, the consumer price increased to 118 monetary units.
1) Restore the market demand function.
2) Determine the amount of tax revenue collected at the chosen tax rate.
3) Determine the per-unit tax rate that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler specified to be collected?
|
# Solution:
1) Let the demand function be linear $Q_{d}=a-b P$. It is known that $1.5 b=6$. We find that $b=$ 4. If a per-unit tax $t=30$ is introduced, then $P_{d}=118 . a-4 P_{d}=6\left(P_{d}-30\right)-312 ; 0.1 a+$ $49.2=P_{d}=118 ; a=688$. The market demand function is $Q_{d}=688-4 P$. (8 points).
2) It is known that $P_{d}(t=30)=118$. Therefore, $Q_{d}=688-472=216, T=216 \cdot 30=6480$. (4 points).
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenues are $T=Q \cdot t=288 t-2.4 t^{2}$. This is a downward-opening parabola, the maximum of the function is reached at $t^{*}=60$. (4 points).
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$. (4 points).
Penalties: an arithmetic error was made - 2 points, lack of justification for the maximum - 5 points.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that a larger amount of tax revenue could not be collected on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the tax revenue amount were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit tax on the producer in the amount of 90 monetary units per unit of the product. It is known that the market demand is given by $Q_d = 688 - 4P$, and the market supply is linear. Additionally, it is known that a change in price by one unit results in a change in the quantity demanded that is 1.5 times smaller than the change in the quantity supplied. After the tax was introduced, the producer's price decreased to 64 monetary units.
1) Restore the market supply function.
2) Determine the amount of tax revenue collected at the chosen tax rate.
3) Determine the per-unit tax rate that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler indicated to collect?
|
# Solution:
1) Let the supply function be linear $Q_{s}=c+d P$. It is known that $1.5 \cdot 4=d$. We find that $d=6$. If a per-unit tax $t=90$ is introduced, then $P_{s}=64.688-4\left(P_{s}+90\right)=6 P_{s}+c$; $0.1 c+32.8=P_{s}=64 ; c=-312$. The market supply function is $Q_{s}=6 P-312$. (8 points).
2) It is known that $P_{s}(t=90)=64$. Therefore, $Q_{s}=6 P-312=72, T=72 \cdot 90=6480$. (4 points).
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenues are $T=Q \cdot t=288 t-2.4 t^{2}$. This is a downward-opening parabola, and the maximum of the function is achieved at $t^{*}=60$. (4 points).
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$. (4 points).
Penalties: an arithmetic error is made -2 points, lack of justification for the maximum -5 points.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Maximum 15 points. On side AB of an equilateral triangle $\mathrm{ABC}$, a right triangle $\mathrm{A} H \mathrm{~B}$ is constructed ( $\mathrm{H}$ - the vertex of the right angle), such that $\angle \mathrm{HBA}=60^{\circ}$. Let point K lie on ray $\mathrm{BC}$ beyond point $\mathrm{C}$ and $\angle \mathrm{CAK}=15^{\circ}$. Find the angle between line HK and the median of triangle $\mathrm{AHB}$, drawn from vertex $\mathrm{H}$.
|
# Solution:
Extend NB and NA beyond points B and A respectively (H-B-B1, H-A-A1)
$\angle \mathrm{B} 1 \mathrm{BC}=60^{\circ}$
$\angle$ KAA1 $=75^{\circ}$, so BK is the bisector of $\angle \mathrm{ABB} 1$
AK is the bisector of $\angle \mathrm{A} 1 \mathrm{AC}$, therefore, HK is the bisector of $\angle \mathrm{AHB}$, so we need to find the angle between the bisector and the median of triangle AHB, drawn from the vertex of the right angle. Answer: $15^{\circ}$
| Criteria | Score |
| :--- | :---: |
| The solution algorithm is provided, the drawing is correct, and the correct answer is obtained | 15 |
| The solution algorithm is provided, the drawing is correct, but due to an arithmetic error, the answer is incorrectly calculated or the answer is correct, but the logic of the proof is violated | 6 |
| The solution algorithm has significant logical errors, but some points of the proof are correct | 2 |
| The solution does not meet any of the presented criteria | 0 |
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Maximum 15 points. Masha was given a chest with multicolored beads (each bead has a unique color, there are a total of $\mathrm{n}$ beads in the chest). Masha chose seven beads for her dress and decided to try all possible combinations of them on the dress (thus, Masha selects from a set of options to sew one, two, three, four, five, six, or seven beads, and the order of the beads does not matter to her). Then she counted how many options she got and was very surprised that the number turned out to be odd.
1) What number did Masha get?
2) Is it true that if Masha had chosen from an even number of beads, she could have gotten an even number of options?
3) Is it true that if the order of the beads sewn on the dress mattered to Masha, she could have gotten both even and odd numbers?
|
# Solution:
1) Consider one bead. Before Masha sews it onto the dress, there are two options: to take the bead or not. If we now choose two beads, the number of options becomes four, which can be obtained by multiplying the first option by two. By increasing the number of beads, we conclude that the total number of all possible options for p beads is $2^{\text {n }}$. According to the problem, Masha chose 7 beads, but the option of not sewing anything is not considered, so the number of options is $2^{7}-1=127$.
2) The number of options is odd (see point 1).
3) This is correct; it is sufficient to consider the case of one and two beads. In the first case, the number of options is 1, and in the second case, it is 4.
| Criterion | Score |
| :--- | :---: |
| The correct answer is obtained in points 1), 2), and 3) through a logically justified algorithm | 15 |
| The correct answer is obtained in only one of the points through a logically justified algorithm | 2 |
| The solution does not meet any of the presented criteria | 0 |
|
127
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Maximum 15 points. The company "Intelligence, Inc" has developed a robot with artificial intelligence. To manufacture it, a special machine is required, which can produce 1 robot in 1 hour. The company owns a large number of such machines, but the created robot is so intelligent that it can produce an exact copy of itself in exactly 2 hours. "Intelligence, Inc" has received an urgent order for 3250 robots, which need to be manufactured in just 10 hours. The company's expenses for creating one robot (regardless of the time of its production and whether it was created by another robot or a special machine) are constant and amount to 100 monetary units, and the maintenance of one special machine during its operation to produce a robot costs the company 70 monetary units. "Intelligence, Inc" does not incur any other expenses for creating robots. What are the minimum costs the company will incur in fulfilling the order?
|
# Solution:
To minimize the company's costs, it is necessary to find the minimum number of machines that will allow the company to complete the order within the specified time frame.
Let $x$ be the number of machines. Then, in the first hour of operation, they will produce $x$ robots. These $x$ robots will start manufacturing new robots, and after 3 hours, there will be $2x$. Meanwhile, the machines continue their work and will produce a new batch of robots by the end of the 2nd hour, which will also start manufacturing their own kind. Thus, after 3 hours of the company's operation, there will be $2x + x$ robots. If we consider that after 9 hours from the start of the work on the order, the robots manufacturing their own kind will not have time to produce new ones by the deadline, then after 10 hours of operation, $62x$ robots will be produced.
The calculation of the number of robots is also presented in the table:
| 1 hour | 2 hour | 3 hour | 4 hour | 5 hour | 6 hour | 7 hour | 8 hour | 9 hour | 10 hour | Total |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{x}$ | - | $2x + x$ | - | $6x + x$ | - | $14x + x$ | - | $30x + x$ | - | $60x + 2x$ |
| - | $\mathrm{x}$ | - | $2x + x$ | - | $6x + x$ | - | $14x + x$ | - | $30x + x$ | |
It remains to find $x$ and round it up to the nearest whole number (otherwise, there won't be enough machines to produce the required number of robots within the specified time).
$$
x = \frac{3250}{62} \approx 52.42
$$
Thus, the company's costs for fulfilling the order will be $100 * 3250 + 70 * 53 = 328710$ monetary units.
## Grading Scheme:
Full correct solution 15 points.
Correctly found the number of robots - 10 points out of 15.
Correctly found the company's expenses for fulfilling the order - 5 points out of 15.
The number of robots is found incorrectly, but there is a correct logic for finding the costs - 2 points out of 15.
There is progress in finding the number of robots, but there are minor logical errors and the number of robots is found incorrectly - 5 points out of 15.
|
328710
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Maximum 15 points. A school economics teacher believes it is extremely important to know the basics of both microeconomics and macroeconomics. Therefore, for his subject, he has introduced the following final grading system. First, the student's study of the basics of microeconomics ( $O_{\text {micro }}$ ) is assessed out of 5 points, and then the basics of macroeconomics ( $O_{\text {macro }}$ ) are assessed out of 5 points. The final grade in economics is formed as the minimum value of two quantities: $0.25 * O_{\text {micro }}+0.75 * O_{\text {macro }}$ and $0.75 * O_{\text {micro }}+0.25 * O_{\text {macro }}$ (the final value is rounded to the nearest integer according to the rules of mathematical rounding). Student Ivanov seriously approached the planning of time for studying each school subject and decided that he would spend no more than 4.6 time units on studying economics. However, studying the basics of one area of economics is more difficult for Ivanov than the other. Thus, each unit of time spent on studying macroeconomics increases his final score (in points) in this area by exactly 1.5 points, while the same unit of time invested in preparing for microeconomics increases the final score in this area by 2.5 points.
(a) What is the maximum grade that student Ivanov can get in economics?
(b) Explain why the final grade in economics reflects the teacher's opinion that it is extremely important to know both microeconomics and macroeconomics. Will this always be the case with such a grading system?
|
# Solution:
(a) If the minimum value of the two is determined by the expression $0.75 * O_{\text {мИкро }} + 0.25 * O_{\text {мАкро }}$, i.e., $O_{\text {мИкро }}O_{\text {мАкро }}$, then it is more advantageous for Ivanov to spend an additional unit of time studying macroeconomics.
Thus, to achieve the highest grade, Ivanov should not spend time studying only one area of economics. How should he correctly allocate his time? As long as the expressions $0.25 * O_{\text {мИкро }} + 0.75 * O_{\text {мАкро }}$ and $0.75 * O_{\text {мИкро }} + 0.25 * O_{\text {мАкро }}$ are not equal, as shown above, it will always be more advantageous to switch to studying one of the areas, increasing the resulting grade. Therefore, the maximum final grade can only be achieved if the time is allocated such that $0.25 * O_{\text {мИкро }} + 0.75 * O_{\text {мАкро }} = 0.75 * O_{\text {мИкро }} + 0.25 * O_{\text {мАкро }}$. From this, $O_{\text {мИкро }} = O_{\text {мАкро }}$. To achieve this, with the full use of the time resource, it is necessary to spend $5 / 8$ of the total time on studying macroeconomics, and the rest on studying microeconomics. Then the calculated final grade in economics will be $O_{\text {мИкро }} = O_{\text {мАкро }} = 3 * 2.5 * 4.6 / 8 = 4.3125$, and after mathematical rounding, it will be 4 points.
It should be noted that since the grade was rounded down, Ivanov can achieve a grade of 4 without using the entire planned time resource of 4.6 time units, and can spend the remaining time on another activity.
(b) Using this formula, it turns out that the student will know both microeconomics and macroeconomics. However, if the coefficients for the grades for knowledge of microeconomics and/or macroeconomics, and the coefficients of the labor intensity of studying both areas of economics were different, it could result in a boundary solution where the student would spend time preparing only for one area of economics (for example, if the labor intensity ratio was $1: 5$ with the given formula for the final grade).
## Grading Scheme:
(a) Total 10 points:
- 5 points for a complete justification that the solution will be internal (that $O_{\text {мИкро }} = O_{\text {мАкро }}$);
- 5 points for finding the maximum grade.
(b) Total 5 points:
- 2 points for explaining the internal solution;
- 3 points for explaining the possibility of a boundary solution.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4. Maximum 20 points
## Option 1
At a school, the remote stage of a team geometry tournament is taking place, where participants' results are evaluated based on the number of points earned for a fully solved problem. A complete solution to a planimetry problem is worth 7 points, and a problem in stereometry is worth 12 points. The tournament winner is the team that scores the highest number of points. Andrey is organizing his team of 3 people, where he will be the captain. He is considering whether to invite Volodya and Zhanna or Petya and Galina. Therefore, he asked all the candidates to honestly provide information about their capabilities in solving problems within the allotted time for this stage of the tournament. It is known that the opportunity costs for each student in solving planimetry and stereometry problems are always constant.
| Name | Maximum number of stereometry problems if solving only them | Maximum number of planimetry problems if solving only them |
| :--- | :---: | :---: |
| Andrey | 7 | 7 |
| Volodya | 6 | 3 |
| Zhanna | 3 | 18 |
| Petya | 12 | 3 |
| Galina | 7 | 14 |
Help Andrey decide which pair of students to take into his team if the team's only goal is to win the tournament.
|
# Solution:
Let's find out what the maximum result the team of Andrey, Volodya, and Zhanna could achieve.
Andrey, instead of solving 1 problem in planimetry, can solve 1 problem in stereometry. Since a problem in stereometry is more valuable, he should specialize in stereometry problems, earning $12 * 7 = 84$ points for the team.
Volodya, instead of solving 1 problem in planimetry, can solve 2 problems in stereometry. Since 1 problem in planimetry is less valuable than 2 problems in stereometry, he should specialize in stereometry problems, earning $12 * 6 = 72$ points for the team.
Zhanna, instead of solving 1 problem in stereometry, can solve 6 problems in planimetry. Since 1 problem in stereometry is less valuable than 6 problems in planimetry, she should specialize in planimetry problems, earning $7 * 18 = 126$ points for the team. In total, the team of Andrey, Volodya, and Zhanna can earn a maximum of 84 + 72 + 126 = 282 points.
Let's find out what the maximum result the team of Andrey, Petya, and Galina could achieve.
Petya, instead of solving 1 problem in planimetry, can solve 4 problems in stereometry. Since 1 problem in planimetry is less valuable than 4 problems in stereometry, he should specialize in stereometry problems, earning $12 * 12 = 144$ points for the team. Galina, instead of solving 1 problem in stereometry, can solve 2 problems in planimetry. Since 1 problem in stereometry is less valuable than 2 problems in planimetry, she should specialize in planimetry problems, earning $7 * 14 = 98$ points for the team.
In total, the team of Andrey, Petya, and Galina can earn a maximum of $84 + 144 + 98 = 326$ points.
Thus, Andrey should invite Petya and Galina to his team.
## Grading Criteria:
To compare the benefits of forming each team, it is sufficient to compare the contributions of only four participants: Volodya, Zhanna, Petya, and Galina.
A correct calculation of the points each of them can bring to the team, conducted in any correct way - 4 points.
A correct calculation of the total number of points (for the team as a whole or for the pair of participants that Andrey will invite to the team) - 2 points.
A comparison of the benefits of each pair of participants and the correct answer - 2 points.
Any arithmetic error that did not lead to a significant distortion of the results is penalized by 1 point.
Any arithmetic error that led to a significant distortion of the results is penalized by 5 points.
## Variant 2
A school is holding the remote stage of a team tournament in physics, where the results of the participants are evaluated based on the number of points earned for a fully solved problem. A fully solved problem in kinematics is worth 14 points, and a problem in thermodynamics is worth 24 points. The team that scores the most points wins the tournament. Volodya is forming his team of 3 people, in which he will be the captain. He is considering whether to invite Andrey and Tatyana, or Semyon and Maria. Therefore, he asked all the candidates to honestly indicate in the table their capabilities for solving problems during the allocated time for this stage of the tournament. It is known that the opportunity costs of each student for solving problems in kinematics and thermodynamics are always constant.
| Name | Maximum number of kinematics problems if solving only them | Maximum number of thermodynamics problems if solving only them |
| :--- | :---: | :---: |
| Volodya | 7 | 7 |
| Semyon | 4 | 8 |
| Maria | 8 | 2 |
| Andrey | 2 | 12 |
| Tatyana | 2 | 1 |
Help Volodya decide which pair of students to invite to his team if the team's only goal is to win the tournament.
## Solution:
Let's find out what the maximum result the team of Volodya, Semyon, and Maria could achieve.
Volodya, instead of solving 1 problem in thermodynamics, can solve 1 problem in kinematics. Since a problem in thermodynamics is more valuable, he should specialize in thermodynamics problems, earning $24 * 7 = 168$ points for the team.
Semyon, instead of solving 1 problem in kinematics, can solve 2 problems in thermodynamics. Since 1 problem in kinematics is less valuable than 2 problems in thermodynamics, he should specialize in thermodynamics problems, earning $24 * 8 = 192$ points for the team. Maria, instead of solving 1 problem in thermodynamics, can solve 4 problems in kinematics. Since 1 problem in thermodynamics is less valuable than 4 problems in kinematics, she should specialize in kinematics problems, earning $14 * 8 = 112$ points for the team. In total, the team of Volodya, Semyon, and Maria can earn a maximum of $168 + 192 + 112 = 472$ points.
Let's find out what the maximum result the team of Volodya, Andrey, and Tatyana could achieve.
Andrey, instead of solving 1 problem in kinematics, can solve 6 problems in thermodynamics. Since 1 problem in kinematics is less valuable than 6 problems in thermodynamics, he should specialize in thermodynamics problems, earning $24 * 12 = 288$ points for the team.
Tatyana, instead of solving 1 problem in thermodynamics, can solve 2 problems in kinematics. Since 1 problem in thermodynamics is less valuable than 2 problems in kinematics, she should specialize in kinematics problems, earning $14 * 2 = 28$ points for the team.
In total, the team of Volodya, Andrey, and Tatyana can earn a maximum of $168 + 288 + 28 = 484$ points.
Thus, Volodya should invite Andrey and Tatyana to his team.
## Grading Criteria:
To compare the benefits of forming each team, it is sufficient to compare the contributions of only four participants: Semyon, Maria, Andrey, and Tatyana.
A correct calculation of the points each of them can bring to the team, conducted in any correct way - 4 points.
A correct calculation of the total number of points (for the team as a whole or for the pair of participants that Volodya will invite to the team) - 2 points.
A comparison of the benefits of each pair of participants and the correct answer - 2 points.
Any arithmetic error that did not lead to a significant distortion of the results is penalized by 1 point.
Any arithmetic error that led to a significant distortion of the results is penalized by 5 points.
|
326
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. Maximum 20 points
In the city of Eifyadl, runic stones are sold. It is known that the first merchant offers a fixed discount of $\mathrm{n} \%$ for every 5th stone purchased, while the second merchant increases the discount by $1 \%$ for each subsequent stone purchased (0% for the 1st stone, 3% for the 4th stone, and so on), but not exceeding $20 \%$. An order for 100 runic stones has been placed. Assume that all stones must be purchased from only one merchant.
a) Find the minimum discount rate for the first merchant so that it is more advantageous for the buyer to purchase all the stones from him.
b) What is the maximum number of runic stones that would be more advantageous to buy from the first merchant with the discount found in part a)?
c) Suppose now that the buyer has increased the order to 175 stones. Upon learning this, the second merchant increased the discount limit to $30 \%$ and the discount increment to $2 \%$. Can the first merchant compete with the second?
|
# Solution and Evaluation Criteria:
a) Let's assume the cost of one rune stone without a discount is 1 unit of currency. We find the average cost of a rune stone from the first merchant:
$(20(1-n)+80) / 100=(100-20 n) / 100$
For the second merchant:
$(1+0.99+0.98+\ldots+0.81+0.8 * 80) / 100=82.1 / 100$
Then:
$100-20 n>82.1$
$17.9 / 20 \quad n>0.895$
Thus, $n=90\%$
Answer: $90\%$.
## Evaluation Criteria:
Logically justified solution - 5 points, logically justified solution with arithmetic error - 0 points
b) We find the average cost of a rune stone from the second merchant:
$(1+0.98+\ldots+0.72+0.7 * 159) / 175=125.06 / 175$
And from the first merchant:
$(35 *(1-n)+140) / 175=(175-35 n) / 175$
Then:
$125.06>175-35 n$
$n>(175-125.06) / 35$
$n>49.94 / 35$
$n>1.427$, which is impossible, as the maximum discount is $100\%$, and paying extra for the purchase of a stone significantly affects the merchant's costs. Answer - no, $n>100\%$
## Evaluation Criteria:
Logically justified solution - 7 points,
justified solution with arithmetic error - 3.5 points, incorrect answer - 0 points
c) Let the number of stones be X. We write the general equation for the average cost of a stone from the first and second merchants:
$1: (x / 5 * (1-0.9) + (x - x / 5)) / x$
$2: (18.1 + 0.8 * (x - 20)) / x$
We get:
$0.1 x / 5 + x - x / 5 < 18.1 + 0.8 * (x - 20)$
$x - 0.18 x - 0.8 x < 2.1$
$0.02 x < 2.1$
$x < 105$
Then: $x=104$ stones.
Answer: 104.
Evaluation Criteria:
Logically justified solution - 8 points,
solution with arithmetic error - 0 points, unsolved part a) - 0 points.
|
104
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4. Maximum 20 points
## Option 1
At a school, the remote stage of a team geometry tournament is taking place, where participants' results are evaluated based on the number of points earned for a fully solved problem. A complete solution to a planimetry problem is worth 7 points, and a problem in stereometry is worth 12 points. The team that scores the highest number of points wins the tournament. Andrey is organizing his team of 3 people, where he will be the captain. He is considering whether to invite Volodya and Zhanna or Petya and Galina. Therefore, he asked all the candidates to honestly provide information about their capabilities in solving problems within the allotted time for this stage of the tournament. It is known that the opportunity costs for each student in solving planimetry and stereometry problems are always constant.
| Name | Maximum number of stereometry problems if solving only them | Maximum number of planimetry problems if solving only them |
| :--- | :---: | :---: |
| Andrey | 7 | 7 |
| Volodya | 6 | 3 |
| Zhanna | 3 | 18 |
| Petya | 12 | 3 |
| Galina | 7 | 14 |
Help Andrey decide which pair of students to take into his team if the team's only goal is to win the tournament.
|
# Solution:
Let's find out what the maximum result the team of Andrey, Volodya, and Zhanna could achieve.
Andrey, instead of solving 1 problem in planimetry, can solve 1 problem in stereometry. Since a problem in stereometry is more valuable, he should specialize in stereometry problems, earning $12 * 7 = 84$ points for the team.
Volodya, instead of solving 1 problem in planimetry, can solve 2 problems in stereometry. Since 1 problem in planimetry is less valuable than 2 problems in stereometry, he should specialize in stereometry problems, earning $12 * 6 = 72$ points for the team.
Zhanna, instead of solving 1 problem in stereometry, can solve 6 problems in planimetry. Since 1 problem in stereometry is less valuable than 6 problems in planimetry, she should specialize in planimetry problems, earning $7 * 18 = 126$ points for the team. In total, the team of Andrey, Volodya, and Zhanna can earn a maximum of 84 + 72 + 126 = 282 points.
Let's find out what the maximum result the team of Andrey, Petya, and Galina could achieve.
Petya, instead of solving 1 problem in planimetry, can solve 4 problems in stereometry. Since 1 problem in planimetry is less valuable than 4 problems in stereometry, he should specialize in stereometry problems, earning $12 * 12 = 144$ points for the team. Galina, instead of solving 1 problem in stereometry, can solve 2 problems in planimetry. Since 1 problem in stereometry is less valuable than 2 problems in planimetry, she should specialize in planimetry problems, earning $7 * 14 = 98$ points for the team.
In total, the team of Andrey, Petya, and Galina can earn a maximum of $84 + 144 + 98 = 326$ points.
Thus, Andrey should invite Petya and Galina to his team.
## Grading Criteria:
To compare the benefits of forming each team, it is sufficient to compare the contributions of only four participants: Volodya, Zhanna, Petya, and Galina.
A correct calculation of the points each of them can bring to the team, conducted in any correct way - 4 points.
A correct calculation of the total number of points (for the team as a whole or for the pair of participants that Andrey will invite to the team) - 2 points.
A comparison of the benefits of each pair of participants and the correct answer - 2 points.
Any arithmetic error that did not lead to a significant distortion of the results is penalized by 1 point.
Any arithmetic error that led to a significant distortion of the results is penalized by 5 points.
## Variant 2
A school is holding the remote stage of a team tournament in physics, where the results of the participants are evaluated based on the number of points earned for a fully solved problem. A fully solved problem in kinematics is worth 14 points, and a problem in thermodynamics is worth 24 points. The team that scores the most points wins the tournament. Volodya is forming his team of 3 people, in which he will be the captain. He is considering whether to invite Andrey and Tatyana, or Semyon and Maria. Therefore, he asked all the candidates to honestly indicate in the table their capabilities for solving problems during the allocated time for this stage of the tournament. It is known that the opportunity costs of each student for solving problems in kinematics and thermodynamics are always constant.
| Name | Maximum number of kinematics problems if solving only them | Maximum number of thermodynamics problems if solving only them |
| :--- | :---: | :---: |
| Volodya | 7 | 7 |
| Semyon | 4 | 8 |
| Maria | 8 | 2 |
| Andrey | 2 | 12 |
| Tatyana | 2 | 1 |
Help Volodya decide which pair of students to invite to his team if the team's only goal is to win the tournament.
## Solution:
Let's find out what the maximum result the team of Volodya, Semyon, and Maria could achieve.
Volodya, instead of solving 1 problem in thermodynamics, can solve 1 problem in kinematics. Since a problem in thermodynamics is more valuable, he should specialize in thermodynamics problems, earning $24 * 7 = 168$ points for the team.
Semyon, instead of solving 1 problem in kinematics, can solve 2 problems in thermodynamics. Since 1 problem in kinematics is less valuable than 2 problems in thermodynamics, he should specialize in thermodynamics problems, earning $24 * 8 = 192$ points for the team. Maria, instead of solving 1 problem in thermodynamics, can solve 4 problems in kinematics. Since 1 problem in thermodynamics is less valuable than 4 problems in kinematics, she should specialize in kinematics problems, earning $14 * 8 = 112$ points for the team. In total, the team of Volodya, Semyon, and Maria can earn a maximum of $168 + 192 + 112 = 472$ points.
Let's find out what the maximum result the team of Volodya, Andrey, and Tatyana could achieve.
Andrey, instead of solving 1 problem in kinematics, can solve 6 problems in thermodynamics. Since 1 problem in kinematics is less valuable than 6 problems in thermodynamics, he should specialize in thermodynamics problems, earning $24 * 12 = 288$ points for the team.
Tatyana, instead of solving 1 problem in thermodynamics, can solve 2 problems in kinematics. Since 1 problem in thermodynamics is less valuable than 2 problems in kinematics, she should specialize in kinematics problems, earning $14 * 2 = 28$ points for the team.
In total, the team of Volodya, Andrey, and Tatyana can earn a maximum of $168 + 288 + 28 = 484$ points.
Thus, Volodya should invite Andrey and Tatyana to his team.
## Grading Criteria:
To compare the benefits of forming each team, it is sufficient to compare the contributions of only four participants: Semyon, Maria, Andrey, and Tatyana.
A correct calculation of the points each of them can bring to the team, conducted in any correct way - 4 points.
A correct calculation of the total number of points (for the team as a whole or for the pair of participants that Volodya will invite to the team) - 2 points.
A comparison of the benefits of each pair of participants and the correct answer - 2 points.
Any arithmetic error that did not lead to a significant distortion of the results is penalized by 1 point.
Any arithmetic error that led to a significant distortion of the results is penalized by 5 points.
|
326
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. Maximum 20 points
In the city of Eifyadl, runic stones are sold. It is known that the first merchant offers a fixed discount of $\mathrm{n} \%$ for every 5th stone purchased, while the second merchant increases the discount by $1 \%$ for each subsequent stone purchased (0% for the 1st stone, 3% for the 4th stone, and so on), but not exceeding $20 \%$. An order for 100 runic stones has been placed. Assume that all stones must be purchased from only one merchant.
a) Find the minimum discount rate for the first merchant so that it is more advantageous for the buyer to purchase all the stones from him.
b) What is the maximum number of runic stones that it would be more advantageous to buy from the first merchant with the discount found in part a)?
c) Suppose now that the buyer has increased the order to 175 stones. Upon learning this, the second merchant increased the discount limit to $30 \%$ and the discount increment to $2 \%$. Can the first merchant compete with the second?
|
# Solution and Evaluation Criteria:
a) Let's assume the cost of one rune stone without a discount is 1 unit of currency. We find the average cost of a rune stone from the first merchant:
$(20(1-n)+80) / 100=(100-20 n) / 100$
For the second merchant:
$(1+0.99+0.98+\ldots+0.81+0.8 * 80) / 100=82.1 / 100$
Then:
$100-20 n>82.1$
$17.9 / 20 \quad n>0.895$
Thus, $n=90\%$
Answer: $90\%$.
## Evaluation Criteria:
Logically justified solution - 5 points, logically justified solution with arithmetic error - 0 points
b) We find the average cost of a rune stone from the second merchant:
$(1+0.98+\ldots+0.72+0.7 * 159) / 175=125.06 / 175$
And from the first merchant:
$(35 *(1-n)+140) / 175=(175-35 n) / 175$
Then:
$125.06>175-35 n$
$n>(175-125.06) / 35$
$n>49.94 / 35$
$n>1.427$, which is impossible, as the maximum discount is $100\%$, and paying extra for the purchase of a stone significantly affects the merchant's costs. Answer - no, $n>100\%$
## Evaluation Criteria:
Logically justified solution - 7 points,
justified solution with arithmetic error - 3.5 points, incorrect answer - 0 points
c) Let the number of stones be X. We write the general equation for the average cost of a stone from the first and second merchants:
$1: (x / 5 * (1-0.9) + (x - x / 5)) / x$
$2: (18.1 + 0.8 * (x - 20)) / x$
We get:
$0.1 x / 5 + x - x / 5 < 18.1 + 0.8 * (x - 20)$
$x - 0.18 x - 0.8 x < 2.1$
$0.02 x < 2.1$
$x < 105$
Then: $x=104$ stones.
Answer: 104.
Evaluation Criteria:
Logically justified solution - 8 points,
solution with arithmetic error - 0 points, unsolved part a) - 0 points.
|
104
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A necklace consists of 30 blue and a certain number of red beads. It is known that on both sides of each blue bead there are beads of different colors, and one bead away from each red bead there are also beads of different colors. How many red beads can be in this necklace? (The beads in the necklace are arranged cyclically, that is, the last one is adjacent to the first.)
|
Answer: 60.
Solution. It is obvious that blue beads appear in the necklace in pairs, separated by at least one red bead. Let there be $n$ red beads between two nearest pairs of blue beads. We will prove that $n=4$. Clearly, $n \leqslant 4$, since the middle one of five consecutive red beads does not satisfy the condition of the problem. For $n<4$, there are three possible situations:
CCKKKCC, CCKKCC, CCKCC.
In the first case, the middle red bead does not satisfy the condition, and in the other cases, all red beads do not satisfy the condition.
Thus, pairs of blue beads must be separated by four red beads. It is clear that such a necklace satisfies the condition of the problem. In it, the number of red beads is twice the number of blue beads, that is, there are 60.
|
60
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On the segment $A B$ of length 10, a circle $\omega$ is constructed with $A B$ as its diameter. A tangent to $\omega$ is drawn through point $A$, and a point $K$ is chosen on this tangent. A line through point $K$, different from $A K$, is tangent to the circle $\omega$ at point $C$. The height $C H$ of triangle $A B C$ intersects segment $B K$ at point $L$. Find the area of triangle $C K L$, given that $B H: A H=1: 4$.
|
Answer: 8.
Solution. Let $O$ be the center of $\omega$. Note that
$$
B H=\frac{1}{5} A B=2, \quad A H=8, \quad O H=\frac{1}{2} A B-B H=3, \quad C H=\sqrt{O C^{2}-O H^{2}}=4
$$
Right triangles $B H C$ and $O A K$ are similar because
$$
\angle A B C=\frac{1}{2} \angle A O C=\frac{1}{2}\left(180^{\circ}-\angle A K C\right)=90^{\circ}-\angle A K O=\angle A O K
$$
Then
$$
\frac{A B}{A K}=2 \cdot \frac{A O}{O K}=2 \cdot \frac{B H}{C H}=1
$$
from which $\angle A B K=45^{\circ}$ and $L H=B H=2$. Therefore,
$$
S_{C K L}=\frac{1}{2} \cdot C L \cdot A H=\frac{1}{2} \cdot(C H-L H) \cdot A H=\frac{1}{2} \cdot 2 \cdot 8=8
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A knight is placed in each cell of a chessboard. What is the smallest number of knights that can be removed from the board so that no knight remains that attacks exactly three other knights? (A knight attacks the squares that are one square away horizontally and two squares away vertically, or vice versa.)
|
Answer: 8 knights.
Solution 1. We will say that a knight controls a square on the board if it attacks this square or stands on it. First, we will prove that it is impossible to remove fewer than 8 knights. It is sufficient to check that at least 4 knights must be removed from each half of the board. Consider, for definiteness, the upper half and mark six knights on it as shown in the figure:
(for convenience, they are highlighted in different colors). We will call the squares marked with a circle on the figure multiple, and the other squares simple. We will divide the figure into two $4 \times 4$ squares and fix one of them. The black knights standing in the square attack exactly three squares. Therefore, one of the following actions must be performed.
1) Remove two knights standing on simple squares controlled by black knights (they can be the black knights themselves).
2) Remove a knight standing on a multiple square. As a result, the white knight from this square will attack exactly three other knights. Therefore, it will be necessary to remove another knight from a simple square controlled by the white knight (possibly the white knight itself).
The same actions must be performed for the other square. Thus, each square determines a pair of squares in the upper half of the board from which knights need to be removed. These pairs do not intersect, since no two marked knights from different squares control the same squares. In other words, the actions with the squares are performed independently of each other. Therefore, at least four knights will have to be removed from the upper half of the board.
We will provide an example showing that 8 knights are sufficient. The knights that need to be removed from the board are marked on the figure.
Solution 2. We will first show that it is necessary to remove at least 8 knights. Mark 12 knights on the board as shown in the left figure:
(for convenience, they are highlighted in different colors). We will call a square useful if it is occupied or attacked by at least one black knight, and useless otherwise. If a square is attacked by two black knights, it is called key (on the left figure, key squares are marked with circles). We will divide the board into four $4 \times 4$ squares. Since black knights attack exactly three squares on the board, the following two facts are true.
1) In each square, at least one useful square must be cleared.
2) If in some square exactly one useful square is cleared, then it must be a key square. Suppose that no more than 7 knights can be removed from the board. Then in some square, exactly one knight will be removed. Let this be, for definiteness, the top-left square. By 2), the knight must be removed from a key square, after which the white knights from this square will attack exactly three knights. Therefore, for each white knight, it will be necessary to clear another square. By assumption, these squares do not lie in the top-left square (and, therefore, they are different). Thus, one of the squares will lie in the bottom-left square, and the other in the top-right square. Obviously, both of them are useless.
By 1), there is at least one cleared square in the top-left and bottom-right squares. Therefore, in the top-right and bottom-left squares, a total of no more than five squares will be cleared. Then in some square (for example, the top-right square), no more than two squares will be cleared. One of them, as shown above, is useless. Therefore, by 1) and 2), another square will be cleared, and it will be a key square. After this, for the white knights standing in the top-right square, it will be necessary to clear two more useless squares - one in the top-left square, and the other in the bottom-right square. In the end, we will clear 4 useless squares (they are different since they lie in different squares), and at least 4 useful squares by 1). Thus, at least 8 knights will be removed, which is impossible.
We will now provide an example showing that 8 knights are sufficient. The knights that need to be removed from the board are marked on the right figure.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A necklace consists of 175 beads of red, blue, and green colors. It is known that each red bead has neighbors of different colors, and on any segment of the necklace between two green beads, there is at least one blue bead. What is the minimum number of blue beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is adjacent to the first.)
|
Answer: 30.
Solution 1. We will show that any block of six consecutive beads contains a blue bead. We can assume that there is no more than one green bead in it, otherwise there is nothing to prove. If the block contains 5 red beads, then at least 3 of them are consecutive, and the middle one does not satisfy the problem's condition. Therefore, the block contains no more than four red beads, and thus, there is a blue one.
Fix a blue bead in the necklace, and divide the rest into 29 blocks of 6 beads each. By the proven fact, each block contains at least one blue bead. Therefore, there are at least 30 blue beads in the necklace.
It remains to provide an example of a necklace with exactly 30 blue beads:
$$
\text {RRGRBS; RRGRBS; ..; RRGRBS (29 times); B. }
$$
Solution 2. Fix any green or blue bead, and divide the rest into 58 consecutive triplets. Each triplet contains no more than two red beads, otherwise the middle red bead would have same-colored neighbors. Therefore, there are no more than $2 \cdot 58 = 116$ red beads in the necklace, and thus, there are at least $175 - 116 = 59$ blue and green beads together. If we mentally remove all the red beads, we notice that the blue beads now make up at least half, as there is a blue bead between any two green beads. Therefore, there are at least $\frac{59}{2} = 29 \frac{1}{2}$ blue beads, meaning there are at least 30.
It remains to provide an example of a necklace with exactly 30 blue beads:
$$
\text {RRGRBS; RRGRBS; ...; RRGRBS (29 times); B. }
$$
|
30
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given a right triangle $ABC$ with a right angle at $C$. On its leg $BC$ of length 26, a circle is constructed with $BC$ as its diameter. A tangent $AP$ is drawn from point $A$ to this circle, different from $AC$. The perpendicular $PH$, dropped from point $P$ to segment $BC$, intersects segment $AB$ at point $Q$. Find the area of triangle $BPQ$, given that $BH: CH = 4: 9$.
|
Answer: 24.
Solution. Let $O$ be the center of $\omega$. Note that
$$
B H=\frac{4}{13} B C=8, \quad C H=18, \quad O H=\frac{1}{2} B C-B H=5, \quad P H=\sqrt{O P^{2}-O H^{2}}=12
$$
Right triangles $B H P$ and $O C A$ are similar because
$$
\angle C B P=\frac{1}{2} \angle C O P=\frac{1}{2}\left(180^{\circ}-\angle C A P\right)=90^{\circ}-\angle C A O=\angle C O A
$$
Then
$$
\frac{A C}{B C}=\frac{1}{2} \cdot \frac{A C}{O C}=\frac{1}{2} \cdot \frac{P H}{B H}=\frac{3}{4}
$$
From the similarity of triangles $B H Q$ and $B C A$, we get $Q H=\frac{3}{4} B H=6$. Therefore,
$$
S_{B P Q}=\frac{1}{2} \cdot P Q \cdot B H=\frac{1}{2} \cdot(P H-Q H) \cdot B H=\frac{1}{2} \cdot 6 \cdot 8=24
$$
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A knight is placed in each cell of a chessboard. What is the minimum number of knights that can be removed from the board so that no knight remains that attacks exactly four other knights? (A knight attacks the squares that are one square away horizontally and two squares away vertically, or vice versa.)
|
Answer: 8 knights.
Solution. First, we will show that no fewer than 8 knights need to be removed. On the left diagram, all knights that attack exactly 4 squares of the board are marked (for convenience, they are highlighted in different colors). Let's call such knights bad. To stop a knight from attacking four others, one must remove either this knight or one of those it attacks. We will show that even to get rid of the bad black knights, one must free no fewer than 8 squares. On the middle diagram, the circles mark the squares under attack by the bad black knights. Three bad black knights in the upper left corner attack four squares marked with the symbol $\odot$. If only one of these squares is freed, one of the black knights will remain bad. Therefore, for this trio of knights, at least two squares need to be cleared. Thus, for all four trios, at least $4 \cdot 2 = 8$ squares need to be freed.
Now, let's provide an example showing that 8 knights are sufficient. On the right diagram, the knights that need to be removed from the board are marked.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A necklace consists of 100 beads of red, blue, and green colors. It is known that among any five consecutive beads, there is at least one blue one, and among any seven consecutive beads, there is at least one red one. What is the maximum number of green beads that can be in this necklace? (The beads in the necklace are arranged cyclically, that is, the last one is adjacent to the first.)
|
Answer: 65.
Solution. Let there be a set of beads $A$ such that in every set of $n$ consecutive beads, there is at least one from $A$. We will show that $A$ contains no fewer than $\frac{100}{n}$ elements. Indeed, between any two adjacent beads from $A$, there are no more than $n-1$ beads. If the set $A$ contains $m$ elements, then
$$
100 \leqslant m+m(n-1)=m n, \quad \text { hence } \quad m \geqslant \frac{100}{n}
$$
By the proven result, the number of blue beads is no less than $\frac{100}{5}=20$, and the number of red beads is no less than $\frac{100}{7}=14 \frac{2}{7}$, which means at least 15. Therefore, the number of green beads is no more than $100-20-15=65$.
Let's provide an example of a necklace containing 65 green beads. Place the blue beads at positions that are multiples of 5, and the red beads at positions with the following numbers:
$$
1,8,14 ; 21,28,34 ; 41,48,54 ; 61,68,74 ; 81,88,94
$$
Fill the remaining positions with green beads.
|
65
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. For a natural number ending not in zero, one of its digits (not the most significant) was erased. As a result, the number decreased by 9 times. How many numbers exist for which this is possible?
|
Answer: 28.
Solution. Let's represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, and $k, m, n$ are non-negative integers, with $m>0$. By erasing the digit $a$, we get the number $m+10^{k} n$. According to the condition,
$$
m+10^{k} a+10^{k+1} n=9\left(m+10^{k} n\right) \Longleftrightarrow 8 m=10^{k}(a+n)
$$
Note that $k>0$, otherwise $m=0$ and $a=n=0$. Then the number $8 m$ is divisible by 10 and thus ends in 0. Given the condition, the number $m$ does not end in 0. Therefore, the last digit of $m$ is 5 and the number $m$ is odd. Therefore, $8 m$ is not divisible by 16, which implies $k \leqslant 3$. Let's consider three cases.
1) Suppose $k=3$. Then $m=125(a+n)$. Since the number $m$ is odd and less than $1000, a+n$ can take the values $1,3,5,7$. Note that the pair $(a, n)$ uniquely determines the original number, and each value of $a+n$ gives $a+n$ different pairs. Thus, we get $1+3+5+7=16$ options.
2) Suppose $k=2$. Then $m=25 \cdot \frac{a+n}{2}$. Since the number $m$ is odd and less than $100, a+n$ equals 2 or 6. These values give us $2+6=8$ options.
3) Suppose $k=1$. Then $m=5 \cdot \frac{a+n}{4}$. Since the number $m$ is odd and less than 10, we get $a+n=4$, which gives us 4 options.
Note that in 1) we get four-digit numbers, in 2) — three-digit numbers, in 3) — two-digit numbers. Therefore, each number satisfying the condition of the problem falls into exactly one of the sets 1) - 3). Thus, the total number of options is $16+8+4=28$.
|
28
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A square $4 \times 4$ is divided into 16 squares $1 \times 1$. We will call a path a movement along the sides of the unit squares, in which no side is traversed more than once. What is the maximum length that a path connecting two opposite vertices of the large square can have?
|
Answer: 32.
Solution. Let's call the sides of the $1 \times 1$ squares edges, the vertices of these squares nodes, and the number of edges adjacent to a node the multiplicity of the node. Notice that the $1 \times 1$ squares generate 40 distinct edges. If a path passes through a node of multiplicity 3, it enters the node along one edge and exits along another, while the third edge adjacent to the node cannot belong to the path. Consider a triplet of nodes of multiplicity 3 lying on one side of the $4 \times 4$ square. Each node has an edge adjacent to it that is free from the path (regardless of whether the path passes through the node or not), and for non-adjacent nodes, these edges are different. Therefore, there are at least $2 \cdot 4 = 8$ edges free from the path, and the length of the path does not exceed $40 - 8 = 32$. An example of a path of length 32 is shown in the figure.
|
32
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A string is threaded with 150 beads of red, blue, and green. It is known that among any six consecutive beads, there is at least one green, and among any eleven consecutive beads, there is at least one blue. What is the maximum number of red beads that can be on the string?
|
Answer: 112.
Solution. We can choose $\left[\frac{150}{11}\right]=13$ consecutive blocks of 11 beads each. Since each block contains at least one blue bead, there are at least 13 blue beads on the string. In addition, we can group all the beads into 25 consecutive blocks of 6 beads each. Each block contains at least one green bead, so there are at least 25 of them on the string. Therefore, the number of red beads is no more than $150-25-13=112$.
Let's provide an example where the string contains exactly 112 red beads. Place the green beads at positions that are multiples of 6, and the blue beads at positions
$$
11,22,33,44,55 ; 65,76,87,98,109 ; 119,130,141
$$
Fill the remaining positions with red beads.
|
112
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. For a natural number ending not in zero, one of its digits was replaced by zero (if it is the leading digit, it was simply erased). As a result, the number decreased by 9 times. How many numbers exist for which this is possible?
|
Answer: 7.
Solution. Let's represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, and $k, m, n$ are non-negative integers, with $m>0$, otherwise $m=a=0$. Then the number $8 m$ is a multiple of 10 and therefore ends in 0. By the condition, the number $m$ does not end in 0. Thus, the last digit of $m$ is 5 and the number $m$ is odd. Therefore, $8 m$ is not divisible by 16, which implies $k \leqslant 3$. Let's consider three cases.
1) Suppose $k=3$. Then $m=125 a$. Since the number $m$ is odd and less than 1000, the digit $a$ can take the values $1,3,5,7$, which gives us 4 options.
2) Suppose $k=2$. Then $m=\frac{25 a}{2}$. Since the number $m$ is odd and less than 100, the digit $a$ is 2 or 6. These values give us another 2 options.
3) Suppose $k=1$. Then $m=\frac{5 a}{4}$. Since the number $m$ is odd and less than 10, we get $a=4$.
Note that in 1) we get four-digit numbers, in 2) - three-digit numbers, in 3) - two-digit numbers. Therefore, each number satisfying the condition of the problem falls into exactly one of the sets 1) - 3). Thus, the total number of options is $4+2+1=7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A rectangle $3 \times 5$ is divided into 15 squares $1 \times 1$. We will call a path a movement along the sides of the unit squares, such that no side is traversed more than once. What is the maximum length that a path connecting two opposite vertices of the rectangle can have?
|
Answer: 30.
Solution. Let the rectangle be denoted as $A B C D$, and let the path connect its vertices $A$ and $C$. We will call the sides of the $1 \times 1$ squares edges, the vertices of these squares - nodes, and the number of edges adjacent to a node - the multiplicity of the node. Note that the $1 \times 1$ squares generate 38 different edges. If the path passes through a node of multiplicity 3, then it enters the node along one edge and exits along another, while the third edge adjacent to the node cannot belong to the path. Let $X$ be the set of nodes lying on the broken line $B A D$, excluding points $B$ and $D$. It consists of point $A$ and six nodes of multiplicity 3. Note that point $A$ has multiplicity 2, and the path does not enter it. Therefore, for each node in $X$, there is an edge free from the path (regardless of whether the path passes through the given node or not). For non-adjacent nodes, these edges are certainly different, and there cannot be more than 3 pairs of adjacent nodes. Thus, the set $X$ provides at least 4 edges free from the path. The same reasoning applies to the broken line $B C D$. Therefore, there are at least $2 \cdot 4 = 8$ edges free from the path, and the length of the path does not exceed $38 - 8 = 30$. An example of a path of length 30 is shown in the figure.
|
30
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A necklace consists of 50 blue and a certain number of red beads. It is known that in any segment of the necklace containing 8 blue beads, there are at least 4 red ones. What is the minimum number of red beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is adjacent to the first.)
|
Answer: 29.
Solution. Note that any segment of the necklace consisting of 11 beads contains no more than 7 blue and no fewer than 4 red beads (otherwise, it would contain 8 blue beads and no more than 3 red ones). Fix a red bead in the necklace. The 7 consecutive segments of 11 beads adjacent to it do not cover the entire necklace, as these segments contain no more than 49 blue beads out of 50. Therefore, the total number of red beads is no less than $7 \cdot 4 + 1 = 29$.
Let's provide an example of a necklace containing exactly 29 red beads. Consider the block
$$
\text{B = RC RC RC RC; CCC,}
$$
consisting of 4 red and 7 blue beads. Then the desired necklace has the form
$$
\text{B, B, .., B (7 times); RC.}
$$
|
29
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 6, 24, and 24. A truncated cone is placed on the table with its smaller base down, and it shares a generatrix with each of the other cones. Find the radius of the smaller base of the truncated cone.
|
Answer: 2.
Solution. Let $C$ be the center of the smaller base of the truncated cone, and $r$ be the radius of this base. Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other cones. Denote by $\mathcal{K}_{0}$ the cone that completes the truncated cone to a regular cone, and by $\mathcal{K}_{1}$ the cone with the base center at $O_{1}$. On the left diagram, the section of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ by the plane $\Pi$ passing through points $O_{1}$ and $C$ and perpendicular to the table is shown. By the problem's condition, $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ have a common generatrix, which lies in $\Pi$ since it passes through the vertices of the cones. Let $B$ be the point of intersection of this generatrix with the table. Then $B$ lies on the boundaries of the bases of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$, as well as on the segment $C O_{1}$ connecting the centers of the bases. Hence, the bases of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ touch each other at point $B$, i.e., $B C = r$. Similarly, it can be verified that the distance from $C$ to the bases of the other two cones is also $r$. Therefore, the following equalities hold:
$$
r = C O_{1} - 6 = C O_{2} - 24 = C O_{3} - 24
$$
Then $C O_{2} = C O_{3}$, which means that point $C$ lies on the common tangent $A O_{1}$ to the bases of the larger cones (see the right diagram). Note that
$$
A O_{1} = \sqrt{O_{1} O_{2}^{2} - A O_{2}^{2}} = \sqrt{30^{2} - 24^{2}} = 18
$$
Therefore,
$$
(24 + r)^{2} = C O_{2}^{2} = 24^{2} + A C^{2} = 24^{2} + (A O_{1} - r - 6)^{2} = 24^{2} + (12 - r)^{2}, \quad \text{from which } r = 2
$$
## Variant 6
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A necklace consists of 100 red and a certain number of blue beads. It is known that in any segment of the necklace containing 10 red beads, there are at least 7 blue ones. What is the minimum number of blue beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is adjacent to the first.)
|
Answer: 78.
Solution. Note that any segment of the necklace containing 16 beads has no more than 9 red and no fewer than 7 blue beads (otherwise, it would contain 10 red beads and no more than 6 blue ones). Fix a blue bead in the necklace. The 11 consecutive segments of 16 beads adjacent to it do not cover the entire necklace, as these segments contain no more than 99 red beads out of 100. Therefore, the total number of blue beads is no less than $11 \cdot 7+1=78$.
Let's provide an example of a necklace containing exactly 78 blue beads. Consider the block
$$
\text { B = CB CB CB CB CB CB CB; RR, }
$$
consisting of 7 blue and 9 red beads. Then the desired necklace has the form
$$
\text { B, B, .., B (11 times); CR. }
$$
|
78
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A necklace consists of 50 blue, 100 red, and 100 green beads. We will call a sequence of four consecutive beads good if it contains exactly 2 blue beads and one each of red and green. What is the maximum number of good quartets that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last bead is adjacent to the first.)
|
Answer: 99.
Solution. Blue beads make up one fifth of the total. Therefore, there will be two consecutive blue beads (let's call them $a$ and $b$), separated by at least three beads. Note that $a$ and $b$ are part of no more than three good quartets, while the other blue beads are part of no more than four. If we add these inequalities, the right side will be $48 \cdot 4 + 2 \cdot 3 = 198$, and the left side will be twice the number of good quartets, since each will be counted twice. Therefore, there can be no more than 99 good quartets.
Let's provide an example of a necklace that gives exactly 99 good quartets:
$$
\text{RBYB; RBYB; ..; RBYB (25 times); RZ ...}
$$
(the ellipsis at the end means any combination of green and red beads). The good quartets will start at positions $1, 2, \ldots, 99$, and only these.
|
99
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 23, 46, and 69. A truncated cone is placed on the table with its smaller base down, and it shares a generatrix with each of the other cones. Find the radius of the smaller base of the truncated cone.
#
|
# Answer: 6.
Solution. Let $C$ be the center of the smaller base of the truncated cone, $R$ be the radius of this base, and $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other cones. Denote by $\mathcal{K}_{0}$ the cone that completes the truncated cone to a regular cone, and by $\mathcal{K}_{1}$ the cone with the base center at $O_{1}$. On the left diagram, the section of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ by the plane $\Pi$ passing through points $O_{1}$ and $C$ and perpendicular to the table is shown. By the problem's condition, $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ have a common generatrix, which lies in $\Pi$ since it passes through the vertices of the cones. Let $B$ be the point of intersection of this generatrix with the table. Then $B$ lies on the boundaries of the bases of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$, as well as on the segment $C O_{1}$ connecting the centers of the bases. Hence, the bases of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ touch each other at point $B$, i.e., $B C=R$. Similarly, it can be verified that the distance from $C$ to the bases of the other two cones is also $R$. Let $r=23$ and note that
$$
O_{1} O_{2}=3 r, \quad O_{1} O_{3}=4 r, \quad O_{2} O_{3}=5 r
$$
which means that triangle $O_{1} O_{2} O_{3}$ is a right triangle. Direct the coordinate axes along the rays $O_{1} O_{3}$ and $O_{1} O_{2}$ (see the right diagram). Let the coordinates of point $C$ be $(x, y)$. Since
$$
C O_{1}=B O_{1}+B C=r+R, \quad C O_{2}=2 r+R, \quad C O_{3}=3 r+R
$$
the following equalities hold:
$$
(r+R)^{2}-x^{2}=y^{2}=(3 r+R)^{2}-(4 r-x)^{2} \Longleftrightarrow r^{2}+2 r R=6 r R+8 r x-7 r^{2} \Longleftrightarrow 2 x=2 r-R
$$
and also
$$
(r+R)^{2}-y^{2}=x^{2}=(2 r+R)^{2}-(3 r-y)^{2} \Longleftrightarrow r^{2}+2 r R=4 r R+6 r y-5 r^{2} \Longleftrightarrow 3 y=3 r-R
$$
Since $x^{2}+y^{2}=C O_{1}^{2}=(r+R)^{2}$, we get
$$
36(r+R)^{2}=(6 r-3 R)^{2}+(6 r-2 R)^{2} \Longleftrightarrow 23 R^{2}+132 r R-36 r^{2}=0
$$
This equation in terms of $R$ has a unique positive solution $R=\frac{6}{23} r=6$.
## Variant 8
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A thread is strung with 75 blue, 75 red, and 75 green beads. We will call a sequence of five consecutive beads good if it contains exactly 3 green beads and one each of red and blue. What is the maximum number of good quintets that can be on this thread?
|
Answer: 123.
Solution. Note that the first and last green beads are included in no more than three good fives, the second and second-to-last - in no more than four fives, and the rest - in no more than five fives. If we add these inequalities, we get $2 \cdot 3 + 2 \cdot 4 + 71 \cdot 5 = 369$ on the right side, and three times the number of good fives on the left side, since each will be counted three times. Therefore, there can be no more than 123 good fives.
Let's provide an example of bead placement that gives exactly 123 good fives:
$$
\text { KSGGG; KSGGG; ...; KSGGG (25 times); KS ... }
$$
(the ellipsis at the end means an arbitrary combination of blue and red beads). The good fives will be those starting at positions $1, 2, \ldots, 123$, and only those.
|
123
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are $2 r, 3 r$, and $10 r$. A frustum of a cone is placed on the table with its smaller base down, and it shares a common generatrix with each of the other cones. Find $r$ if the radius of the smaller base of the frustum is 15.
|
Answer: 29.
Solution. Let $C$ be the center of the smaller base of the truncated cone, and $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other cones, with $R=15$. Denote by $\mathcal{K}_{0}$ the cone that completes the truncated cone to a regular cone, and by $\mathcal{K}_{1}$ the cone with the center of its base at $O_{1}$. On the left diagram, the section of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ by the plane $\Pi$ passing through points $O_{1}$ and $C$ and perpendicular to the table is shown. By the problem's condition, $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ have a common generatrix, which lies in $\Pi$ since it passes through the vertices of the cones. Let $B$ be the point of intersection of this generatrix with the table. Then $B$ lies on the boundaries of the bases of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$, as well as on the segment $C O_{1}$ connecting the centers of the bases. Hence, the bases of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ touch each other at point $B$, i.e., $B C=R$. Similarly, it can be verified that the distance from $C$ to the bases of the other two cones is also $R$. Note that
$$
O_{1} O_{2}=5 r, \quad O_{1} O_{3}=12 r, \quad O_{2} O_{3}=13 r
$$
which means that triangle $O_{1} O_{2} O_{3}$ is a right triangle. Direct the coordinate axes along the rays $O_{1} O_{3}$ and $O_{1} O_{2}$ (see the right diagram). Let the coordinates of point $C$ be $(x, y)$. Since
$$
C O_{1}=B O_{1}+B C=2 r+R, \quad C O_{2}=3 r+R, \quad C O_{3}=10 r+R
$$
the following equalities hold:
$$
(2 r+R)^{2}-x^{2}=y^{2}=(10 r+R)^{2}-(12 r-x)^{2} \Longleftrightarrow 4 r^{2}+4 r R=20 r R+24 r x-44 r^{2} \Longleftrightarrow 3 x=6 r-2 R
$$
and also
$$
(2 r+R)^{2}-y^{2}=x^{2}=(3 r+R)^{2}-(5 r-y)^{2} \Longleftrightarrow 4 r^{2}+4 r R=6 r R+10 r y-16 r^{2} \Longleftrightarrow 5 y=10 r-R
$$
Since $x^{2}+y^{2}=C O_{1}^{2}=(2 r+R)^{2}$, we get
$$
225(2 r+R)^{2}=(30 r-10 R)^{2}+(30 r-3 R)^{2} \Longleftrightarrow 225 r^{2}-420 R r-29 R^{2}=0
$$
This equation in terms of $r$ has a unique positive solution $r=\frac{29}{15} R=29$.
## Variant 9
|
29
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A necklace consists of 80 beads of red, blue, and green colors. It is known that on any segment of the necklace between two blue beads, there is at least one red bead, and on any segment of the necklace between two red beads, there is at least one green bead. What is the minimum number of green beads that can be in this necklace? (The beads in the necklace are arranged cyclically, that is, the last one is adjacent to the first.)
|
Answer: 27.
Solution. If the blue beads are arranged in a circle, the number of pairs of adjacent beads is equal to the number of beads. Since there is a red bead between any two blue beads, there are no fewer red beads in the necklace than blue ones. Similarly, it can be proven that there are no fewer green beads than red ones. Therefore, there are no fewer than $\frac{80}{3}=26 \frac{2}{3}$ green beads, which means there are at least 27. An example of a necklace containing exactly 27 green beads is as follows:
GKR 3 GKR ... ; GKR (26 times); GK.
|
27
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In the cells of an $80 \times 80$ table, pairwise distinct natural numbers are placed. Each of them is either a prime number or a product of two prime numbers (possibly the same). It is known that for any number $a$ in the table, there is a number $b$ in the same row or column such that $a$ and $b$ are not coprime. What is the maximum number of prime numbers that can be in the table?
|
Answer: 4266.
Solution. We will say that a composite number $a$ serves a prime number $p$ if $a$ and $p$ are not coprime (i.e., $a$ is divisible by $p$). For each prime number in the table, there is a composite number that serves it. Since each composite number has no more than two distinct prime divisors, it serves no more than two prime numbers. Thus, if the table contains $n$ composite numbers, then the number of primes does not exceed $2n$. Therefore, the total number of numbers in the table does not exceed $3n$. Then
$$
3n \geqslant 80^{2} \Longrightarrow n \geqslant \frac{80^{2}}{3}=2133 \frac{1}{3} \Longrightarrow n \geqslant 2134 \Longrightarrow 80^{2}-n \leqslant 80^{2}-2134=4266
$$
Thus, the number of prime numbers in the table does not exceed 4266.
Now, let's show how 4266 prime numbers can be placed in the table. We will use the following algorithm for filling rows and columns.
1) The first 52 positions are filled with different prime numbers $p_{1}, p_{2}, \ldots, p_{52}$. These numbers must be new, i.e., not used before in the table.
2) In the next 26 cells, we place the numbers $p_{1} p_{2}, p_{3} p_{4}, \ldots, p_{51} p_{52}$.
3) The last two positions are left unfilled.
We apply this algorithm sequentially to the rows $1,2, \ldots, 80$, and then to the last two columns. Thus, we will place $80 \cdot 52 + 2 \cdot 52 = 4264$ prime numbers. It remains to fill the $2 \times 2$ square in the bottom right corner. In it, on one diagonal, we will place a pair of new prime numbers, and on the other diagonal - their squares. In the end, we will place 4266 different prime numbers.
|
4266
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. On the table, there are three cones standing on their bases, touching each other. The heights of the cones are the same, and the radii of their bases are 1, 2, and 3. A sphere is placed on the table, touching all the cones. It turns out that the center of the sphere is equidistant from all points of contact with the cones. Find the radius of the sphere.
|
Answer: 1.
Solution: Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the cones, $O$ be the center of the sphere, $R$ be the radius of the sphere, $C$ be the point of contact of the sphere with the table, $2 \alpha$ and $2 \beta$ be the angles at the vertices of the first and second cones, $A$ and $B$ be the points of intersection of the segments $C O_{1}$ and $C O_{2}$ with the bases of the cones. The left diagram shows the section of the first cone by the plane $C O O_{1}$. The contact of the sphere with the first cone means that the perpendicular $O E$, dropped from point $O$ to the generatrix $P A$, is equal to $R$. Indeed, in this case, the sphere and the cone touch the plane passing through the generatrix $P A$ perpendicular to the section, and lie on opposite sides of it. Then the sphere touches the lines $A C$ and $A E$ at points $C$ and $E$, from which
$$
O C=O E=R, \quad A C=A E \quad \text { and } \quad \angle A O C=\frac{1}{2} \angle E O C=\frac{\pi}{4}-\frac{\alpha}{2} .
$$
Therefore, $A C=R \cdot \operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$ and, similarly, $B C=R \cdot \operatorname{tg}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)$. Let
$$
\lambda=\operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right), \quad \mu=\operatorname{tg}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)
$$
Since $O_{1} O_{2}=3, O_{1} O_{3}=4, O_{2} O_{3}=5$, the triangle $O_{1} O_{2} O_{3}$ is a right triangle. By the condition, point $C$ is equidistant from the points of contact of the bases of the cones. Then $C$ lies at the intersection of the angle bisectors of triangle $O_{1} O_{2} O_{3}$ and, therefore, is the center of its inscribed circle, the radius of which is 1. By calculating $C O_{1}^{2}$ and $C O_{2}^{2}$ in two ways, we get
$$
\left\{\begin{array} { l }
{ ( 1 + \lambda R ) ^ { 2 } = 2 , } \\
{ ( 2 + \mu R ) ^ { 2 } = 5 }
\end{array} \Longleftrightarrow \left\{\begin{array} { l }
{ \lambda ^ { 2 } R ^ { 2 } + 2 \lambda R = 1 , } \\
{ \mu ^ { 2 } R ^ { 2 } + 4 \mu R = 1 }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
\lambda^{2} R^{2}+2 \lambda R-1=0, \\
\left(\lambda^{2}-\mu^{2}\right) R^{2}=2 R(2 \mu-\lambda)
\end{array}\right.\right.\right.
$$
(we subtracted the second equation from the first). Note that
$$
\frac{1}{\lambda}-\lambda=\frac{1+\operatorname{tg} \frac{\alpha}{2}}{1-\operatorname{tg} \frac{\alpha}{2}}-\frac{1-\operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg} \frac{\alpha}{2}}=\frac{4 \operatorname{tg} \frac{\alpha}{2}}{1-\operatorname{tg}^{2} \frac{\alpha}{2}}=2 \operatorname{tg} \alpha \quad \text { and, similarly, } \quad \frac{1}{\mu}-\mu=2 \operatorname{tg} \beta
$$
By the equality of the heights of the cones, the relation $\operatorname{tg} \beta=2 \operatorname{tg} \alpha$ holds. Therefore,
$$
1-\mu^{2}=\frac{2 \mu}{\lambda}\left(1-\lambda^{2}\right) \quad \text { and } \quad \lambda^{2}-\mu^{2}=\lambda^{2}-1+1-\mu^{2}=\frac{\left(1-\lambda^{2}\right)(2 \mu-\lambda)}{\lambda}
$$
The second equation of $(*)$ gives
$$
R=\frac{2(2 \mu-\lambda)}{\left(\lambda^{2}-\mu^{2}\right)}=\frac{2 \lambda}{\left(1-\lambda^{2}\right)} \Longleftrightarrow R-\lambda^{2} R=2 \lambda
$$
From the first equation of the system $(*)$, it follows that $\lambda R=\sqrt{2}-1$. Excluding $\lambda$, we get
$$
R-\frac{(\sqrt{2}-1)^{2}}{R}=\frac{2(\sqrt{2}-1)}{R} \Longleftrightarrow R^{2}=(\sqrt{2}-1)(\sqrt{2}+1)=1 \Longleftrightarrow R=1
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Around a round table, 50 schoolchildren are sitting: blondes, brunettes, and redheads. It is known that in any group of schoolchildren sitting in a row, between any two blondes there is at least one brunette, and between any two brunettes - at least one redhead. What is the minimum number of redheads that can sit at this table?
|
Answer: 17.
Solution. If only blondes were sitting at the table, the number of pairs of neighbors would be equal to the number of blondes. Since there is a brunette between any two blondes, there are no fewer brunettes than blondes sitting at the table. Similarly, it can be proven that there are no fewer redheads than brunettes. Therefore, there are no fewer than $\frac{50}{3}=16 \frac{2}{3}$ redheads sitting at the table, which means there are at least 17. Let's provide an example of seating with exactly 17 redheads:
$$
\text { RCB; RCB; ... RCB; (16 times); RC }
$$
(The letters R, C, and B denote redheads, brunettes, and blondes, respectively).
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Inside an angle of $30^{\circ}$ with vertex $A$, a point $K$ is chosen, the distances from which to the sides of the angle are 1 and 2. Through point $K$, all possible lines are drawn, intersecting the sides of the angle. Find the minimum area of the triangle cut off by the line from the angle.
|
Answer: 8.
Solution 1. Choose points $P$ and $Q$ on the sides of the angle such that point $K$ is the midpoint of segment $P Q$. We will show that line $P Q$ cuts off a triangle of the smallest area. Take another line passing through point $K$. Let it intersect the sides of the angle at points $X$ and $Y$. By swapping $P$ with $Q$ and $X$ with $Y$ if necessary, we can assume that $X$ lies on segment $A P$. It is sufficient to prove that $K X < K Y$. Suppose $K X \geqslant K Y$. Mark a point $Z$ on segment $K X$ such that $K Z = K Y$ (see the left figure). Triangles $K Q Y$ and $K P Z$ are equal by two sides and the angle, hence
$$
\angle K Q Y = \angle K P Z \quad \text{and} \quad \angle Z P Q + \angle A Q P = 180^{\circ}.
$$
But this is impossible, since point $Z$ lies inside triangle $A P Q$ and
$$
\angle Z P Q + \angle A Q P \leqslant \angle A P Q + \angle A Q P = 150^{\circ}.
$$
2. The area of triangle $A X Y$ is expressed by the function
$$
S(x) = \frac{x^2}{x-2}, \quad \text{hence} \quad S'(x) = \frac{x(x-4)}{(x-2)^2}
$$
Thus, the derivative $S$ is negative for $x \in (2,4)$ and positive for $x > 4$. Therefore, the minimum value of $S$ is achieved at $x = 4$ and is equal to 8.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In the cells of a $75 \times 75$ table, pairwise distinct natural numbers are placed. Each of them has no more than three distinct prime divisors. It is known that for any number $a$ in the table, there is a number $b$ in the same row or column such that $a$ and $b$ are not coprime. What is the maximum number of prime numbers that can be in the table?
|
Answer: 4218.
Solution. We will say that a composite number $a$ serves a prime number $p$ if the numbers $a$ and $p$ are not coprime (that is, $a$ is divisible by $p$). For each prime number in the table, there is a composite number that serves it. Since each composite number has no more than three distinct prime divisors, it serves no more than three prime numbers. Thus, if the table contains $n$ composite numbers, then the number of primes does not exceed $3n$. Therefore, the total number of numbers in the table does not exceed $4n$. Then
$$
4n \geqslant 75^{2} \Longrightarrow n \geqslant \frac{75^{2}}{4}=1406 \frac{1}{4} \Longrightarrow n \geqslant 1407 \Longrightarrow 75^{2}-n \leqslant 75^{2}-1407=4218
$$
Thus, the number of prime numbers in the table does not exceed 4218.
Now let's show how 4218 prime numbers can be placed in the table. We will use the following algorithm for filling rows and columns.
1) The first 54 positions are filled with different prime numbers $p_{1}, p_{2}, \ldots, p_{54}$. These numbers must be new, that is, not used before in the table.
2) In the next 18 cells, we place the numbers $p_{1} p_{2} p_{3}, p_{4} p_{5} p_{6}, \ldots, p_{52} p_{53} p_{54}$.
3) The last three positions are left unfilled.
We apply this algorithm sequentially to the rows $1,2, \ldots, 75$, and then to the last three columns. Thus, we will place $75 \cdot 54 + 3 \cdot 54 = 4212$ prime numbers. It remains to fill the cells of the $3 \times 3$ square in the lower right corner. We choose new prime numbers $q_{1}, \ldots, q_{6}$ and place them as follows:
$$
\left(\begin{array}{ccc}
q_{1} q_{2} & q_{3} q_{4} & q_{5} q_{6} \\
q_{1} & q_{3} & q_{5} \\
q_{2} & q_{4} & q_{6}
\end{array}\right)
$$
In the end, we will place 4218 different prime numbers.
|
4218
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. On the table, there are three cones standing on their bases, touching each other. The heights of the cones are the same, and the radii of their bases are $2r$, $3r$, and $10r$. A sphere with a radius of 2 is placed on the table, touching all the cones. It turns out that the center of the sphere is equidistant from all points of contact with the cones. Find $r$.
|
Answer: 1.
Solution: Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the cones, $O$ be the center of the sphere, $R$ be the radius of the sphere, $C$ be the point of contact of the sphere with the table, $2 \alpha$ and $2 \beta$ be the angles at the vertices of the first and second cones, $A$ and $B$ be the points of intersection of the segments $C O_{1}$ and $C O_{2}$ with the bases of the cones. The left diagram shows the section of the first cone by the plane $C O O_{1}$. The contact of the sphere with the first cone means that the perpendicular $O E$, dropped from point $O$ to the generatrix $P A$, is equal to $R$. Indeed, in this case, the sphere and the cone touch the plane passing through the generatrix $P A$ perpendicular to the section, and lie on opposite sides of it. Then the sphere touches the lines $A C$ and $A E$ at points $C$ and $E$, from which
$$
O C=O E=2, \quad A C=A E \quad \text { and } \quad \angle A O C=\frac{1}{2} \angle E O C=\frac{\pi}{4}-\frac{\alpha}{2} .
$$
Therefore, $A C=2 \operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$ and, similarly, $B C=2 \operatorname{tg}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)$. Let
$$
\lambda=\operatorname{tg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right), \quad \mu=\operatorname{tg}\left(\frac{\pi}{4}-\frac{\beta}{2}\right)
$$
Since $O_{1} O_{2}=5 r, O_{1} O_{3}=12 r, O_{2} O_{3}=13 r$, the triangle $O_{1} O_{2} O_{3}$ is a right triangle. By the condition, point $C$ is equidistant from the points of contact of the bases of the cones. Then $C$ lies at the intersection of the angle bisectors of triangle $O_{1} O_{2} O_{3}$ and, therefore, is the center of its inscribed circle, the radius of which is $2 r$. Calculating $C O_{1}^{2}$ and $C O_{2}^{2}$ in two ways, we get
$$
\left\{\begin{array} { l }
{ ( 2 r + 2 \lambda ) ^ { 2 } = 8 r ^ { 2 } , } \\
{ ( 3 r + 2 \mu ) ^ { 2 } = 1 3 r ^ { 2 } }
\end{array} \Longleftrightarrow \left\{\begin{array} { l }
{ 4 \lambda ^ { 2 } + 8 \lambda r = 4 r ^ { 2 } , } \\
{ 4 \mu ^ { 2 } + 1 2 \mu r = 4 r ^ { 2 } }
\end{array} \Longleftrightarrow \left\{\begin{array}{l}
\lambda^{2}+2 \lambda r-r^{2}=0 \\
\lambda^{2}-\mu^{2}=r(3 \mu-2 \lambda)
\end{array}\right.\right.\right.
$$
(we subtracted the second equation from the first). Note that
$$
\frac{1}{\lambda}-\lambda=\frac{1+\operatorname{tg} \frac{\alpha}{2}}{1-\operatorname{tg} \frac{\alpha}{2}}-\frac{1-\operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg} \frac{\alpha}{2}}=\frac{4 \operatorname{tg} \frac{\alpha}{2}}{1-\operatorname{tg}^{2} \frac{\alpha}{2}}=2 \operatorname{tg} \alpha \quad \text { and, similarly, } \quad \frac{1}{\mu}-\mu=2 \operatorname{tg} \beta
$$
By the equality of the heights of the cones, the relation $\operatorname{tg} \beta=\frac{3}{2} \operatorname{tg} \alpha$ holds. Therefore,
$$
1-\mu^{2}=\frac{3 \mu}{2 \lambda}\left(1-\lambda^{2}\right) \quad \text { and } \quad \lambda^{2}-\mu^{2}=\lambda^{2}-1+1-\mu^{2}=\frac{\left(1-\lambda^{2}\right)(3 \mu-2 \lambda)}{2 \lambda}
$$
The second equation of $(*)$ gives
$$
r=\frac{\lambda^{2}-\mu^{2}}{3 \mu-2 \lambda}=\frac{1-\lambda^{2}}{2 \lambda}=\frac{1}{2}\left(\frac{1}{\lambda}-\lambda\right)
$$
From the first equation of the system $(*)$, it follows that $\lambda=r(\sqrt{2}-1)$. Excluding $\lambda$, we get
$$
2 r=\frac{\sqrt{2}+1}{r}-(\sqrt{2}-1) r \Longleftrightarrow \sqrt{2}+1=(\sqrt{2}+1) r^{2} \Longleftrightarrow r=1
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second must be different. For example, 101 and 292 have this form, while 222 and 123 do not. Similarly, we define the form of the number $\overline{a b c a b d}$. How many odd numbers of the form $\overline{\text { adabcd }}$ are divisible by 5?
|
Answer: 448.
Solution. Odd numbers divisible by 5 are numbers ending in 5, so for $d$ we have only one option. For $a$ we have 8 options, as the number cannot start with zero, and $a$ cannot be equal to $d$. The digit $b$ cannot be equal to $a$ or $d$, and there are no other restrictions on it - we get 8 possible values. Similarly, for the digit $c$ - 7 options. Therefore, the total number of such numbers is $1 \cdot 8 \cdot 8 \cdot 7=448$.
|
448
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. (40 points) The numbers $s_{1}, s_{2}, \ldots, s_{1008}$ are such that their sum is $2016^{2}$. It is known that
$$
\frac{s_{1}}{s_{1}+1}=\frac{s_{2}}{s_{2}+3}=\frac{s_{3}}{s_{3}+5}=\ldots=\frac{s_{1008}}{s_{1008}+2015}
$$
Find $s_{17}$.
|
Answer: 132.
Solution. Note that none of the $s_{i}$ is equal to zero (otherwise, all the fractions $\frac{s_{i}}{s_{i}+2 i-1}$ would be equal to zero, and, consequently, all $s_{i}$ would have to be equal to zero, which contradicts the fact that their sum is $2016^{2}$). Therefore, the original condition is equivalent to the condition
$$
\frac{s_{1}+1}{s_{1}}=\frac{s_{2}+3}{s_{2}}=\frac{s_{3}+5}{s_{3}}=\ldots=\frac{s_{1008}+2015}{s_{1008}} \Leftrightarrow \frac{1}{s_{1}}=\frac{3}{s_{2}}=\frac{5}{s_{3}}=\ldots=\frac{2015}{s_{1008}}
$$
From the first equality, express $s_{2}$ in terms of $s_{1}: s_{2}=3 s_{1}$; then, similarly, $s_{3}$ in terms of $s_{1}$: $s_{3}=5 s_{1}; \ldots s_{1008}=2015 s_{1}$. Now we have
$$
\sum_{i=1}^{1008} s_{i}=\sum_{i=1}^{1008}(2 i-1) s_{1}=s_{1} \cdot \frac{(1+2015) \cdot 1008}{2}=1008^{2} s_{1}=2016^{2}
$$
From this, $s_{1}=4$ and, consequently, $s_{17}=(2 \cdot 17-1) \cdot 4=132$.
|
132
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Cells of an infinite grid paper are painted in $k$ colors (each cell is painted entirely in one color). What is the largest $k$ such that in every grid rectangle with sides 3 and 4, cells of all these colors will be found?
|
Answer: 10.
Solution. Divide the infinite grid paper into ten-cell figures as shown in the figure.
The required coloring with 10 colors can be achieved by coloring one such figure in 10 colors and coloring the rest in exactly the same way.
It is obvious that the sheet cannot be painted in more than 12 colors in the required manner. Suppose there is a coloring in 11 or 12 colors. Assume that there is a coloring in 11 colors such that in each $3 \times 3$ square all colors are different. Consider one such square. In the containing $3 \times 4$ rectangle, there are only 11 different colors. Therefore, in the additional $3 \times 1$ strip, some two colors coincide. Then, in the $3 \times 3$ square containing this strip, some two colors coincide.
Suppose we have a coloring in $n$ colors, where $n$ is 11 or 12. Choose a $3 \times 3$ square in which there are no more than $n-3$ different colors. Extend it to rectangles $3 \times 4$ and $4 \times 4$ as shown in the figure. Then, in the strips $3 \times 1$ and $1 \times 3$, the same triplets of colors are represented. Cover the cell marked with a cross with a horizontal rectangle $3 \times 4$. Then this rectangle will contain 5 cells from the strips, meaning it has at least two pairs of cells of the same color, so the total number of colors does not exceed 10. Contradiction.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. What is the minimum number of cells that need to be marked in a $50 \times 50$ table so that each vertical or horizontal strip of $1 \times 6$ contains at least one marked cell.
|
Answer: 416.
Solution. A $50 \times 50$ square can easily be cut into four rectangles of $24 \times 26$ and a central square of $2 \times 2$. Each rectangle can be cut into $4 \cdot 26=104$ strips of $1 \times 6$. Each such strip must have its own marked cell, so there will be no fewer than 416 such cells.
We will show how to mark 416 cells in the required manner. Mark all parallel diagonals with lengths of $5, 11, 17, 23, 29, 35, 41$, and $47$. In total, there will be $2 \cdot (5+11+17+23+29+35+41+47)=416$ marked cells.
|
416
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In some cells of a $1 \times 2021$ strip, one chip is placed in each. For each empty cell, the number equal to the absolute difference between the number of chips to the left and to the right of this cell is written. It is known that all the written numbers are distinct and non-zero. What is the minimum number of chips that can be placed in the cells?
|
Answer: 1347.
Solution. Let $n$ be the number of chips placed. Note that the numbers in the empty cells lie in the range from 1 to $n$ and have the same parity. Therefore, there can be no more than $\left[\frac{n+1}{2}\right]$ such numbers. This means that the number of empty cells does not exceed $\left[\frac{n+1}{2}\right]$, otherwise the numbers placed in them would repeat. Then
$$
2021-n \leqslant\left[\frac{n+1}{2}\right] \leqslant \frac{n+1}{2}, \quad \text { hence } \quad n \geqslant \frac{4041}{3}=1347
$$
We will show that the value $n=1347$ is achievable. A suitable arrangement is
$$
\underbrace{0101 \ldots 01}_{674 \text { pairs }} \underbrace{111 \ldots 1}_{673 \text { numbers }}
$$
where ones represent chips, and zeros represent empty cells. In this case, the numbers in the positions of the zeros will be:
$$
1347,1345,1343, \ldots, 3,1
$$
Remark. The implementation provided in the solution is not unique. For example, the following also works:
$$
1011011 \ldots 0110
$$
The corresponding numbers are:
$$
1345,1341, \ldots, 5,1,3,7, \ldots, 1343,1347
$$
|
1347
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In some cells of a $1 \times 2100$ strip, one chip is placed. In each of the empty cells, a number is written that is equal to the absolute difference between the number of chips to the left and to the right of this cell. It is known that all the written numbers are distinct and non-zero. What is the minimum number of chips that can be placed in the cells?
|
Answer: 1400.
Solution. Let $n$ be the number of chips placed. Note that the numbers in the empty cells lie in the range from 1 to $n$ and have the same parity. Therefore, there can be no more than $\left[\frac{n+1}{2}\right]$ such numbers. This means that the number of empty cells does not exceed $\left[\frac{n+1}{2}\right]$, otherwise the numbers placed in them would repeat. Then
$$
2100-n \leqslant\left[\frac{n+1}{2}\right] \leqslant \frac{n+1}{2}, \quad \text { hence } n \geqslant \frac{4199}{3}=1399 \frac{2}{3} \text { and } n \geqslant 1400
$$
We will show that the value $n=1400$ is achievable. The arrangement
$$
\underbrace{0101 \ldots 01}_{700 \text { pairs }} \underbrace{111 \ldots 1}_{700 \text { numbers }}
$$
where 1s represent chips and 0s represent empty cells, will work. In this case, the numbers in the positions of the zeros will be:
$$
1400,1398,1396, \ldots, 4,2
$$
Remark. The implementation provided in the solution is not unique. For example, the following arrangement also works:
$$
\underbrace{101101 \ldots 101}_{350 \text { triples }} \underbrace{110110 \ldots 110}_{350 \text { triples }}
$$
The corresponding numbers are:
$$
1398,1394, \ldots, 6,2,4,8, \ldots, 1396,1400
$$
|
1400
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Positive numbers $a, b, c$ are such that $a^{2} b+b^{2} c+c^{2} a=3$. Find the minimum value of the expression
$$
A=a^{7} b+b^{7} c+c^{7} a+a b^{3}+b c^{3}+c a^{3}
$$
|
Answer: 6.
Solution. By the Cauchy-Schwarz inequality
$$
A=\left(a^{7} b+a b^{3}\right)+\left(b^{7} c+b c^{3}\right)+\left(c^{7} a+c a^{3}\right) \geqslant 2\left(a^{4} b^{2}+b^{4} c^{2}+c^{4} a^{2}\right) \geqslant \frac{2}{3}\left(a^{2} b+b^{2} c+c^{2} a\right)^{2}=6
$$
Equality is achieved when $a=b=c=1$.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Given an acute scalene triangle $A B C$. In it, the altitudes $B B_{1}$ and $C C_{1}$ are drawn, intersecting at point $H$. Circles $\omega_{1}$ and $\omega_{2}$ with centers $H$ and $C$ respectively touch the line $A B$. From point $A$ to $\omega_{1}$ and $\omega_{2}$, tangents other than $A B$ are drawn. Denote the points of tangency with these circles by $D$ and $E$ respectively. Find the angle $B_{1} D E$.
|
Answer: $180^{\circ}$.
Solution. Since right triangles $A C C_{1}$ and $A C E$ are equal by leg and hypotenuse, line $C A$ is the bisector of isosceles triangle $C_{1} C E$. Therefore, it is the perpendicular bisector of segment $C_{1} E$, meaning points $C_{1}$ and $E$ are symmetric with respect to line $A C$. From the equality of triangles $A H D$ and $A H C_{1}$, it follows that $\angle A H D = \angle A H C_{1}$. The quadrilateral $A B_{1} H C_{1}$ has two opposite right angles. Therefore, it is cyclic, from which $\angle A B_{1} C_{1} = \angle A H C_{1}$. By the symmetry of points $C_{1}$ and $E$,
$$
\angle A B_{1} E = \angle A B_{1} C_{1} = \angle A H C_{1}.
$$
Moreover, $\angle A D H = 90^{\circ} = \angle A B_{1} H$. Therefore, quadrilateral $A D B_{1} H$ is also cyclic, and
$$
\angle A B_{1} D = \angle A H D = \angle A H C_{1}
$$
Thus, $\angle A B_{1} D = \angle A B_{1} E$, which means $\angle B_{1} D E = 180^{\circ}$.
|
180
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A circle is circumscribed around an acute-angled triangle $A B C$. Point $K$ is the midpoint of the smaller arc $A C$ of this circle, and point $L$ is the midpoint of the smaller arc $A K$ of this circle. Segments $B K$ and $A C$ intersect at point $P$. Find the angle between the lines $B C$ and $L P$, given that $B K = B C$.
|
Answer: $90^{\circ}$.
Solution. Let $\angle A B L=\varphi$. Then $\angle K B L=\varphi$ and $\angle K B C=2 \varphi$. Since $B K=B C$, we get $\angle B K C=\angle B C K=90^{\circ}-\varphi$. Moreover,
$$
\angle L C K+\angle B K C=\angle L B K+\angle B K C=\varphi+\left(90^{\circ}-\varphi\right)=90^{\circ},
$$
which means that lines $B K$ and $C L$ are perpendicular. On the other hand,
$$
\angle B L C+\angle A C L=\angle B K C+\angle A B L=\left(90^{\circ}-\varphi\right)+\varphi=90^{\circ},
$$
so lines $B L$ and $A C$ are also perpendicular. Therefore, point $P$ is the orthocenter of triangle $B C L$, from which it follows that $L P \perp B C$.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{3}+b^{3}+c^{3}}{(a+b+c)^{3}-26 a b c}
$$
|
Answer: 3.
Solution. Note that $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6 a b c+B$, where
$$
B=3\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right) \geqslant 18 \sqrt[6]{a^{6} b^{6} c^{6}}=18 a b c
$$
(we used the Cauchy inequality). Then
$$
a^{3}+b^{3}+c^{3} \leqslant(a+b+c)^{3}-24 a b c
$$
Let $t=\frac{(a+b+c)^{3}}{a b c}$. By the Cauchy inequality, $t \geqslant 27$, hence
$$
A \leqslant \frac{(a+b+c)^{3}-24 a b c}{(a+b+c)^{3}-26 a b c}=\frac{t-24}{t-26}=1+\frac{2}{t-26} \leqslant 3
$$
Equality is achieved when $a=b=c$.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$, and a circle with center at point $O$ encompasses circles $\omega_{1}$ and $\omega_{2}$, touching them at points $C$ and $D$ respectively. It turns out that points $A, C$, and $D$ lie on the same line. Find the angle $A B O$.
|
Answer: $90^{\circ}$.
Solution. Let $O_{1}$ and $O_{2}$ be the centers of circles $\omega_{1}$ and $\omega_{2}$. Triangles $C O D$, $C O_{1} A$, and $A O_{2} D$ are isosceles. Since points $C, A, D$ lie on the same line, we get
$$
\angle O_{1} A C = \angle O_{1} C A = \angle O C D = \angle O D C
$$
Therefore, $O_{1} A \| O O_{2}$ and, similarly, $O_{2} A \| O O_{1}$. Thus, $O O_{1} A O_{2}$ is a parallelogram, from which we have
$$
O_{1} A = O_{2} O \quad \text{and} \quad \angle A O_{1} O_{2} = \angle O_{1} O_{2} O
$$
Since triangles $A O_{1} B$ and $A O_{2} B$ are isosceles, segment $O_{1} O_{2}$ is the perpendicular bisector of $A B$. Then
$$
O_{1} B = O_{1} A = O_{2} O \quad \text{and} \quad \angle B O_{1} O_{2} = \angle A O_{1} O_{2} = \angle O_{1} O_{2} O
$$
Therefore, $O_{1} O_{2} O B$ is an isosceles trapezoid, from which $B O \| O_{1} O_{2}$ and, hence, $B O \perp A B$.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{4}+b^{4}+c^{4}}{(a+b+c)^{4}-80(a b c)^{4 / 3}}
$$
|
Answer: 3.
Solution. Note that $(a+b+c)^{4}=a^{4}+b^{4}+c^{4}+B$, where
\[
\begin{aligned}
& B=4\left(a^{3} b+a b^{3}+b^{3} c+b c^{3}+c^{3} a+c a^{3}\right)+6\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+12\left(a b c^{2}+b c a^{2}+c a b^{2}\right) \geqslant \\
& \geqslant 24 \sqrt[6]{(a b c)^{8}}+18 \sqrt[3]{(a b c)^{4}}+36 \sqrt[3]{(a b c)^{4}}=78(a b c)^{4 / 3}
\end{aligned}
\]
(we used the Cauchy inequality). Then
\[
a^{4}+b^{4}+c^{4} \leqslant(a+b+c)^{4}-78(a b c)^{4 / 3}
\]
Let $t=\frac{(a+b+c)^{4}}{(a b c)^{4 / 3}}$. By the Cauchy inequality, $t \geqslant 81$, hence
\[
A \leqslant \frac{(a+b+c)^{4}-78(a b c)^{4 / 3}}{(a+b+c)^{4}-80(a b c)^{4 / 3}}=\frac{t-78}{t-80}=1+\frac{2}{t-80} \leqslant 3
\]
Equality is achieved when $a=b=c$.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)}{a^{3}+b^{3}+c^{3}-2 a b c}
$$
|
Answer: 6.
Solution. By the Cauchy-Schwarz inequality,
$$
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)=\frac{1}{3}\left((a+b+c)^{3}-\left(a^{3}+b^{3}+c^{3}+6 a b c\right)\right) \leqslant \frac{1}{3}\left((a+b+c)^{3}-9 a b c\right)
$$
Note that
$$
a^{3}+b^{3}+c^{3}-2 a b c \geqslant \frac{1}{9}(a+b+c)^{3}-2 a b c=\frac{1}{9}\left((a+b+c)^{3}-18 a b c\right)
$$
Let $t=\frac{(a+b+c)^{3}}{a b c}$. By the Cauchy-Schwarz inequality, $t \geqslant 27$, hence
$$
A \leqslant \frac{3\left((a+b+c)^{3}-9 a b c\right)}{(a+b+c)^{3}-18 a b c}=\frac{3(t-9)}{t-18}=3\left(1+\frac{9}{t-18}\right) \leqslant 6
$$
Equality is achieved when $a=b=c$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Circles $\omega_{1}$ and $\omega_{2}$ with centers $O_{1}$ and $O_{2}$ respectively intersect at point $A$. Segment $O_{2} A$ intersects circle $\omega_{1}$ again at point $K$, and segment $O_{1} A$ intersects circle $\omega_{2}$ again at point $L$. The line passing through point $A$ parallel to $K L$ intersects circles $\omega_{1}$ and $\omega_{2}$ again at points $C$ and $D$ respectively. Segments $C K$ and $D L$ intersect at point $N$. Find the angle between the lines $O_{1} A$ and $O_{2} N$.
|
Answer: $90^{\circ}$.
Solution. Triangles $O_{1} A K$ and $O_{2} A L$ are isosceles and have a common base angle. Therefore, their vertex angles are equal. Let the common value of these angles be $2 \alpha$. Since an inscribed angle is half the corresponding central angle, we have $\angle A D L = \angle A C K = \alpha$. Since $K L \| C D$, we also get $\angle K L N = \angle L K N = \alpha$. Then
$$
\angle L N K = 180^{\circ} - 2 \angle K L N = 180^{\circ} - 2 \alpha = 180^{\circ} - \angle A O_{2} L = 180^{\circ} - \angle K O_{2} L
$$
Therefore, the quadrilateral $L N K O_{2}$ is cyclic, from which we get
$$
\angle K O_{2} N = \angle K L N = \alpha = \frac{1}{2} \angle A O_{2} L
$$
This means that the line $O_{2} N$ is the angle bisector of the vertex angle in the isosceles triangle $A O_{2} L$, and therefore also the altitude. Thus, the lines $O_{2} N$ and $O_{1} A$ are perpendicular.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b)}{(a+b+c)^{4}-79(a b c)^{4 / 3}}
$$
|
Answer: 3.
Solution. By the Cauchy-Schwarz inequality,
$a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b)=(a+b+c)\left(a^{3}+b^{3}+c^{3}\right)-\left(a^{4}+b^{4}+c^{4}\right) \leqslant(a+b+c)\left(a^{3}+b^{3}+c^{3}\right)-3(a b c)^{4 / 3}$.
Note that $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6 a b c+B$, where
$$
B=3\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right) \geqslant 18 \sqrt[6]{a^{6} b^{6} c^{6}}=18 a b c
$$
(we used the Cauchy-Schwarz inequality). Then $a^{3}+b^{3}+c^{3} \leqslant(a+b+c)^{3}-24 a b c$ and
$(a+b+c)\left(a^{3}+b^{3}+c^{3}\right)-3(a b c)^{4 / 3} \leqslant(a+b+c)^{4}-24 a b c(a+b+c)-3(a b c)^{4 / 3} \leqslant(a+b+c)^{4}-75(a b c)^{4 / 3}$.
Let $t=\frac{(a+b+c)^{4}}{(a b c)^{4 / 3}}$. By the Cauchy-Schwarz inequality, $t \geqslant 81$, hence
$$
A \leqslant \frac{(a+b+c)^{4}-75(a b c)^{4 / 3}}{(a+b+c)^{4}-79(a b c)^{4 / 3}}=\frac{t-75}{t-79}=1+\frac{4}{t-79} \leqslant 3
$$
Equality is achieved when $a=b=c$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Circles $\omega_{1}$ and $\omega_{2}$ with centers $O_{1}$ and $O_{2}$ respectively intersect at point $B$. The extension of segment $O_{2} B$ beyond point $B$ intersects circle $\omega_{1}$ at point $K$, and the extension of segment $O_{1} B$ beyond point $B$ intersects circle $\omega_{2}$ at point $L$. The line passing through point $B$ parallel to $K L$ intersects circles $\omega_{1}$ and $\omega_{2}$ again at points $A$ and $C$ respectively. The rays $A K$ and $C L$ intersect at point $N$. Find the angle between the lines $O_{1} N$ and $O_{2} B$.
|
Answer: $90^{\circ}$.
Solution. Since $\angle O_{1} B K = \angle O_{2} B L$, the isosceles triangles $O_{1} B K$ and $O_{2} B L$ are similar. Therefore, their vertex angles are equal. Let the common value of these angles be $2 \alpha$. Since an inscribed angle is half the corresponding central angle, we have $\angle B C L = \angle B A K = \alpha$. Since $K L \| A C$, we also get $\angle K L N = \angle L K N = \alpha$. Then,
$$
\angle L N K = 180^{\circ} - 2 \angle L K N = 180^{\circ} - 2 \alpha = 180^{\circ} - \angle B O_{1} K = 180^{\circ} - \angle L O_{1} K
$$
Therefore, the quadrilateral $L N K O_{1}$ is cyclic, from which we get
$$
\angle L O_{1} N = \angle L K N = \alpha = \frac{1}{2} \angle B O_{1} K
$$
Thus, the line $O_{1} N$ is the angle bisector of the vertex angle in the isosceles triangle $B O_{1} K$, and therefore also the altitude. Hence, the lines $O_{1} N$ and $O_{2} B$ are perpendicular.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. For what smallest $k$ can $k$ cells be marked on a $10 \times 11$ board such that any placement of a three-cell corner on the board touches at least one marked cell?
|
Answer: 50.
Solution. It is not hard to notice that in any $2 \times 2$ square, there are at least two marked cells. Since 25 such squares can be cut out from a $10 \times 11$ board, there must be no fewer than 50 marked cells in it. An example with 50 marked cells is obtained if the cells with an even first coordinate are marked.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A warehouse stores 400 tons of cargo, with the weight of each being a multiple of a centner and not exceeding 10 tons. It is known that any two cargos have different weights. What is the minimum number of trips that need to be made with a 10-ton truck to guarantee the transportation of these cargos from the warehouse?
|
Answer: 51.
Solution. We will show that it is always possible to transport the goods in 51 trips, even if the warehouse contains all weights from 1 to 100 tons. Indeed, we can divide all the goods, except for the 50-ton and 100-ton ones, into 49 pairs as follows:
$$
(1,99), \quad(2,98), \quad(3,97), \quad \ldots, \quad(49,51)
$$
For their transportation, 49 trips are sufficient, since each pair fits into the truck. Two more trips are needed to transport the 50-ton and 100-ton goods.
Now we will show that it is not always possible to transport the goods in 50 trips. Suppose the warehouse stores goods weighing 31 and $47,48, \ldots, 100$ tons. Their total weight is $31+\frac{54 \cdot(47+100)}{2}=4000$ tons. No two
goods weighing from 50 to 100 tons can be transported together in one truck. But there are 51 such goods, so at least 51 trips are required.
|
51
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. For what smallest $k$ can $k$ cells be marked on a $9 \times 9$ board such that any placement of a three-cell corner piece will touch at least two marked cells?
|
Answer: 56.
Solution. It is not hard to notice that in any $2 \times 2$ square, at least three cells must be marked, and in each $1 \times 2$ rectangle, at least two cells must be marked. Since from a $9 \times 9$ board, 16 $2 \times 2$ squares and 8 $1 \times 2$ rectangles can be cut out, at least $16 \cdot 3 + 8 = 56$ cells must be marked in total. An example with 56 marked cells is shown in the figure.
|
56
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A cinema was visited by 50 viewers, the total age of whom is 1555 years, and among them, there are no viewers of the same age. For what largest $k$ can we guarantee to select 16 viewers whose total age is not less than $k$ years?
|
Answer: 776.
Solution. We will show that $k \geqslant 776$, that is, the total age of the 16 oldest viewers is always not less than 776. Arrange the viewers in order of increasing age, and let $a_{i}$ be the age of the $i$-th viewer. Since there are no viewers of the same age, we get
$$
a_{1} \leqslant a_{2}-1 \leqslant a_{3}-2 \leqslant \ldots \leqslant a_{50}-49
$$
that is, the numbers $b_{i}=a_{i}-(i-1)$ are increasing. Moreover,
$$
b_{1}+b_{2}+\ldots+b_{50}=a_{1}+a_{2}+\ldots+a_{50}-(0+1+2+\ldots+49)=1555-\frac{49 \cdot 50}{2}=330
$$
from which the arithmetic mean of the numbers $b_{k}$ is $\frac{330}{50}=6.6$. Due to the increasing nature of the numbers $b_{i}$, there exists an index $m$ such that $b_{i} \leqslant 6$ for $i \leqslant m$ and $b_{i} \geqslant 7$ for $i>m$. If $m \leqslant 34$, then
$$
b_{35}+b_{36}+\ldots+b_{50} \geqslant 16 \cdot 7=112
$$
and if $m>34$
$$
b_{1}+b_{2}+\ldots+b_{34} \leqslant 34 \cdot 6=204 \quad \text { and } \quad b_{35}+b_{36}+\ldots+b_{50} \geqslant 330-204=126>112
$$
In both cases
$$
a_{35}+a_{36}+\ldots+a_{50}=b_{35}+b_{36}+\ldots+b_{50}+(34+35+\ldots+49) \geqslant 112+\frac{16 \cdot(34+49)}{2}=776
$$
Now we will show that $k \leqslant 776$. We need to provide an example where the total age of the 16 oldest viewers is 776. Suppose people aged $6,7, \ldots, 25$ and $27,28, \ldots, 56$ years came to the cinema. Their total age is $\frac{51 \cdot(6+56)}{2}-26=1555$, and the total age of the 16 oldest of them is $41+42+\ldots+56=\frac{16 \cdot(41+56)}{2}=776$.
|
776
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. For what smallest $k$ can $k$ cells be marked on a $12 \times 12$ board such that any placement of a four-cell figure $\square \square$ on the board touches at least one marked cell? (The figure can be rotated and flipped.)
|
Answer: 48.
Solution. It is not hard to notice that in any $2 \times 3$ rectangle, there are at least two marked cells. Since the $12 \times 12$ board is divided into 24 such rectangles, there must be no fewer than 48 marked cells. An example with 48 marked cells can be obtained by marking the cells where the sum of the coordinates is divisible by 4.
|
48
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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