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6. Maximum 15 points. Mr. N, maximizing his profit, owns a patent for the production of a unique good and can sell his product only in countries A and B. The demand for his product in these countries is represented by the functions $q_{A}=40-2P$ and $q_{B}=26-P$, where $q_{A}$ and $q_{B}$ are the quantities of the prod... | # Solution:
Since resale between countries is impossible, the monopolist, maximizing its profit, will increase sales of its product in these countries until the marginal costs of its production become equal to the marginal revenue in each of the countries. That is, until $M R_{A}\left(q_{A}\right)=M R_{B}\left(q_{B}\r... | 133.7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,531 |
7. Maximum 10 points. Imagine that your family has decided to renovate the apartment: completely replace the electrical wiring, change the plumbing, level and plaster the walls, and finish the renovation with finishing work. For this, you are looking for a construction company or a private renovation team that is ready... | # Solution:
Repair services are a classic example of credence goods - these are goods and services whose utility is difficult or impossible for the consumer to determine. Unlike experience goods, the utility from using or the decline in consumer quality of goods is hard to measure even after consumption.
In this exam... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,532 |
4. Students could present another solution to this problem. In such a case, it is important to have a well-argued and clear proof that the point they found is indeed the one sought. Accordingly, the maximum number of points for finding (constructing) and justifying it, depending on the degree of argumentation, is up to... | # Solution:
The economist chose method 1. Let "E" represent the economist, and "L" represent the lawyer. Suppose L divides the coins into parts with a and b coins in each, such that a < b. Then E, pursuing their own interest, divides b into two parts - 1 coin and b-1 coins. These parts become the largest and smallest,... | Method1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,533 |
5. Maximum 15 points. Below is a fragment of a map of one of Moscow's districts (scale 1:50). Propose three different economically justified arguments explaining the reasons for the location of the "Perekrestok" stores. Explain what two disadvantages this strategy of store placement within one network might have. (Note... | # Solution:
The concentration of stores of one network in a relatively small area can be explained by several economically justified reasons.
1) The placement of "Perekrestoks" can prevent the emergence of other stores, which means that this can contribute to the monopolization of the supply of goods sold in such sup... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,535 |
# Task 1.
Maximum 10 points
In the Dark Dungeon, 1000 kind gnomes are held captive by 100 orcs. The chief orc has devised a scheme: orcs are assigned numbers from 00 to 99, and gnomes are assigned numbers from 000 to 999. An orc is assigned to watch a gnome if the orc's number can be obtained by deleting one digit fr... | # Solution:
a) Answer: The chief orc is right. Indeed, mathematically, the problem boils down to whether we can find 50 two-digit numbers from which, by adding one digit at the beginning, in the middle, or at the end, we can obtain all 1000 three-digit numbers. (Obviously, within the problem, there are $10 * 10$ two-d... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,536 |
# Task 2.
Maximum 15 points
Solve the equation:
$$
|3 x+3|^{3}+|3 x-5|^{3}=131
$$
# | # Solution:
Let's make a variable substitution:
$$
t=\frac{3 x+3+3 x-5}{2}=3 x-1
$$
The equation becomes:
$$
|t+4|^{3}+|t-4|^{3}=131
$$
It is not hard to see that the function $f(t)=|t+4|^{3}+|t-4|^{3}$ is even, so it is sufficient to solve the equation for $t \geq 0$. Consider the cases for removing the absolute ... | \frac{\\sqrt{2}+4}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,537 |
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 $3... | # 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 n... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,538 |
# 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 approximate... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,539 |
# Assignment 5.
## Maximum 15 points
In a small state called Konyr, the beloved good "ravot" is sold in a perfectly competitive market. The Ministry of Economic Development of this state is conducting a study on the state of the economy and is collecting data on various markets, including the market for "ravot." Anal... | # Solution and Evaluation Criteria:
Let the demand function in spring be: $Q_{\text {spring }}^{D}(p)=a-b p$, then the demand functions in winter and summer are, respectively: $Q_{\text {winter }}^{D}(p)=(a-b p) / 2$ and $Q_{\text {summer }}^{D}(p)=2(a-b p)$.
Let the supply function in spring, summer, and autumn be: ... | 10,172.5,19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,540 |
# Task 6.
## Maximum 10 points
In April this year, Rossiyskaya Gazeta in the Economics section published an article titled "With Their Own Policies" (https://rg.ru/2019/04/24/strahovat-zhivotnyh-stali-chashche.html). Based solely on the information provided by the authors of the article and providing economic justifi... | # Solution and Evaluation Criteria:
Let's consider possible arguments in answering the questions of the assignment, referring to the provided publication text. A maximum of 5 points is awarded for a reasoned answer to each item of the assignment.
1) According to the article, the ratio of insurance coverage to the cos... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,541 |
# 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 ... | # 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 s... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,542 |
11. It is necessary to clearly follow the requirements for the volume of work if it is specified in the task.
## Task 1 (Maximum 12 points)
Economist Sasha has a 2016-sided die. The probability of each face landing is the same. Sasha rolls this die 2016 times and counts the sum of the points that come up, but unfortu... | 2033138 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,544 | |
Task 1. 15 points
Find the minimum value of the expression
$$
\sqrt{x^{2}-\sqrt{3} \cdot|x|+1}+\sqrt{x^{2}+\sqrt{3} \cdot|x|+3}
$$
as well as the values of $x$ at which it is achieved. | # Solution:
Let $t=|x| \geq 0$. We need to find the minimum value of the function
$$
f(t)=\sqrt{t^{2}-\sqrt{3} \cdot t+1}+\sqrt{t^{2}+\sqrt{3} \cdot t+3}
$$
for $t \geq 0$. Transform the function $f(t)$ to the form:
$$
f(t)=\sqrt{\left(t-\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{4}}+\sqrt{\left(t+\frac{\sqrt{3}}{2}\ri... | \sqrt{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,545 |
# Task 3. 20 points
Retail purchases at wholesale prices are called group buying when people cooperate to buy goods from suppliers without a markup.
The practice of group buying became popular in Russia in the mid-2000s and is still used today. Buyers unite on specialized websites or social networks. In each purchase... | # Solution:
(a) Let's list several factors that can explain the profitability of such purchases
1) Joint purchases allow for significant savings on the acquisition of goods, as they are essentially made at wholesale prices, while overhead costs associated with delivering the goods to the buyer and paying for the serv... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,547 |
# Problem 4. 25 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. In the cryptogram, the amount of tax revenue to be collected was specified. It was also emphasized that a larger amount of ta... | # Solution:
1) Let the demand function be linear $Q_{d}=a-b P$. It is known that $1.5 \cdot\left|E_{p}^{d}\right|=E_{p}^{s}$. For linear demand functions, using the definition of elasticity, we get: $1.5 \cdot \frac{b P_{e}}{Q_{e}}=\frac{6 P_{e}}{Q_{e}}$. From this, we find that $b=4$.
If a per-unit tax $t=30$ is int... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,548 |
# Problem 1. 15 points
Find the minimum value of the expression
$$
\sqrt{x^{2}-2 \sqrt{3} \cdot|x|+4}+\sqrt{x^{2}+2 \sqrt{3} \cdot|x|+12}
$$
as well as the values of $x$ at which it is achieved.
# | # Solution:
Let $t=\frac{|x|}{2} \geq 0$. Then we need to find the minimum value of the function
$$
f(t)=2\left(\sqrt{t^{2}-\sqrt{3} \cdot t+4}+\sqrt{t^{2}+\sqrt{3} \cdot t+12}\right)
$$
for $t \geq 0$. Transform the function $f(t)$ to the form:
$$
f(t)=2\left(\sqrt{\left(t-\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{4}... | 2\sqrt{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,549 |
# Task 3. 20 points
Retail purchases at wholesale prices are called group buying when people cooperate to buy goods from suppliers without a markup.
The practice of group buying became popular in Russia in the mid-2000s and is still used today. Buyers unite on specialized websites or social networks. In each purchase... | # Solution:
(a) Let's list several factors that can explain the profitability of such purchases
1) Joint purchases allow for significant savings on the acquisition of goods, as they are essentially made at wholesale prices, and overhead costs associated with delivering the goods to the buyer and paying for the servic... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,551 |
# 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 colle... | # 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... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,552 |
# Problem 2.
Find all positive $x, y, z$ for which the expression
$$
\frac{x^{2} y z}{324}+\frac{144 y}{x z}+\frac{9}{4 x y^{2}}
$$
takes the smallest value and among all such $x, y, z$ the expression
$$
\frac{z}{16 y}+\frac{x}{9}
$$
is also minimal. | # Solution:
According to the Cauchy inequality
$$
\frac{x^{2} y z}{324}+\frac{144 y}{x z}+\frac{9}{4 x y^{2}} \geq 3 \sqrt[3]{\frac{x^{2} y z}{324} \cdot \frac{144 y}{x z} \cdot \frac{9}{4 x y^{2}}}=3
$$
Equality is possible only when
$$
\frac{x^{2} y z}{324}=\frac{144 y}{x z}=\frac{9}{4 x y^{2}}
$$
From this,
$$... | 9,\frac{1}{2},16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,554 |
# Problem 3.
Point $M$ lies on the parabola $y=2 x^{2}-3 x+4$, and point $F$ lies on the line $y=3 x-4$.
Find the minimum value of $M F$. | # Solution
Let's draw a tangent to the parabola parallel to the line $\boldsymbol{y}=3 \boldsymbol{x}-4$. We will find the abscissa of the point of tangency:
$$
f^{\prime}(x)=4 x-3=3 \Rightarrow x=1.5
$$
The point $A(1.5 ; 4)$ is the point of tangency of the parabola with the line parallel to $y=3 x-4$. We will find... | \frac{2}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,555 |
# Task 5.
In a Thriving State with a Gini index of 0.1, there are only two equally numerous population groups - the poor and the rich. Within each group, residents do not differ in their incomes. After economic reforms in Thriving, a middle class (residents with average income) emerged in the country, the number of wh... | # Solution:
Let $X$ be the share of the poor in the state's population, and $Y$ be the share of the poor's income in the total income of the state's residents. Then the Gini coefficient can be calculated using the formula:
$$
G_{0}=\frac{\frac{1}{2}-\left[\frac{1}{2} X Y+\frac{1}{2}(Y+1)(1-X)\right]}{\frac{1}{2}}=X-Y... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,557 |
# Problem 1. Maximum 16 points
Solve the equation for $a \neq 0 ; a \neq \pm 1$
$\frac{\left(x-a^{2}\right)\left(x-a^{3}\right)}{a(a+1)(a-1)^{2}}-\frac{(x-a)\left(x-a^{3}\right)}{a(a-1)^{2}}+\frac{(x-a)\left(x-a^{2}\right)}{(a+1)(a-1)^{2}}=x^{3}$ | # Solution
The left side of the equation is a polynomial $P(x)$ of degree no higher than two.
We compute $P(a)=a ; P\left(a^{2}\right)=a^{2} ; P\left(a^{3}\right)=a^{3}$.
Notice that for $a \neq 0 ; a \neq \pm 1$, the numbers $a ; a^{2} ; a^{3}$ are distinct.
From this, it follows that $P(x)=x$.
It remains to solv... | x_{1}=0;x_{2,3}=\1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,559 |
# 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... | # 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 trian... | 307 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,560 |
# Task 3. Maximum 16 points
The results of trading in the shares of companies "a", "b", "c", "d", "e" were predicted by two financial analysis gurus. Trying to predict the results of the day's trading, one of the analysts calculated that the share prices at the end of the day would be ranked in descending order as "a"... | # Solution:
Consider the second guru's prediction.
Obviously, if a correctly indicated pair includes one correctly indicated element, then the other element of the pair is also correctly indicated.
The sequence "d", "a", "e", "s", "b" (let's write the result of the second guru in descending order) contains four pair... | edacb | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,561 |
1. A solution and justified answer are provided, in which one of the letters is in the correct position 3 points.
The text above has been translated into English, preserving the original text's line breaks and format. | Answer: in descending order dabce - 3 points. | dabce | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,562 |
# 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 ch... | # 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 $n$ i... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,568 |
# Task 6. Maximum 16 points
Many universal online stores, which may have several million items in their range, deliver orders to their customers for free. At the same time, large orders and orders with only two items can be split and delivered to the customer in two batches, on different days. Explain why, with free d... | # Solution
Large online stores typically have many warehouses for storing goods, located in different areas of the city, regions of the country, and many delivery points. The delivery of goods to customers from an online store can be influenced by a multitude of different factors. Let's provide some examples.
1) Diff... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,569 |
2. Between Moscow and Saint Petersburg, there are seven settlements: Yaroslavl, Tver, Klin, Vyshny Volochek, Shestihino, Zavidovo, and Bologoye. It is known that Moscow is connected by direct railway communication with seven points, Saint Petersburg with five, Tver with four, Yaroslavl with two, Bologoye with two, Shes... | Solution: Let's build a graphical interpretation of the problem. The sum of the degrees of the vertices of the graph must be even. Let the degree of the vertex BB be x, and Klin be y. Then the degrees of the vertices of the railway graph are $7,5,4,2,2,2,1, x, y$. The sum of all vertices $7+5+4+2+2+2+1+x+y=23+x+y$. Sin... | 2,3,4,or\5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,571 |
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 t... | 90894 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,572 |
4. Find all values of the parameter $\alpha$, for each of which the system
$$
\left\{\begin{array}{c}
\sqrt{-5 y+4+4 \alpha}-\sqrt{\alpha-x}=0 \\
y-\sqrt{x}=0
\end{array}\right.
$$
has at least one solution with respect to the variables $y$ and $x$. | Solution. Taking into account the domain of definition (ODZ) for this problem, the original system transforms into:
$$
\left\{\begin{array}{c}
-5 y+4+4 \alpha \geq 0 \\
\alpha-x \geq 0 \\
y \geq 0 \\
x \geq 0 \\
-5 y+4+4 \alpha=\alpha-x \\
y=\sqrt{x}
\end{array}\right.
$$
Let's explicitly express the constraints on t... | Nosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,573 |
5. In recent years, with the emergence of non-state pension funds in our country, a comprehensive program for increasing the future pension of citizens has been established. In particular, a number of such funds offer their clients the opportunity to create an individual pension plan. According to this plan, the client... | # Solution:
Before retirement, over 20 years, the client will deposit $7000 * 12 = 84000$ rubles into their account each year.
After the first year of savings, the amount on their pension account will be $84000 + 84000 * 0.09 = 91560$ rubles.
After 2 years: $91560 + 84000 + (91560 + 84000) * 0.09 = 191360$ rubles, a... | 26023.45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,574 |
6. In her fourth year at Hogwarts, Hermione was outraged by the infringement of house-elf rights and founded the Association for the Restoration of Elven Independence. Of course, even the brightest and noblest idea requires funding for promotion, so Hermione decided to finance her campaign by producing merchandise, sta... | # Solution:
Since the prices must be the same for the faculties, Hermione Monopolist must take into account the aggregate demand for badges, which is given by:
$$
Q(P)=q_{w s}+q_{s}=\left\{\begin{array}{l}
26-2 P, P \geq 13 \\
36-3 P, P<13
\end{array}\right.
$$
Then the marginal revenue of Hermione's monopoly is giv... | 7.682 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,575 |
7. Suppose you live in a small town and are very familiar with all its residents. Since the townspeople, quite rightly, consider you to be economically savvy, some of them have come up with the idea of you opening a recruitment agency where you alone would represent all employable residents in the labor market. For thi... | # Solution:
(a) It is more advantageous for you to choose the scheme where you have a leadership position: you make the decision first, anticipating how the company will then act. Let's show this.
1) Let's predict how the company will respond after you name the wage rate, and take this into account when we choose the... | 0.24 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 7,576 |
# Variant № 1
## Task 1
Santa Claus is preparing gifts for children for New Year. In his gift cabinet, there are five shelves, and on each shelf, there are n different items. Santa Claus takes any item from the first shelf, then from the second, third, fourth, and fifth. This results in one set of items, which define... | # Solution:
The number of options by simple counting is n^5-n. Let's factorize. $\mathrm{N}^{\wedge} 5-\mathrm{N}=$ (n^2-1)(n^2+1)n. This is the product of three consecutive numbers $n, n+1, n-1$ and the number $\mathrm{n}^{\wedge} 2+1$. Therefore, divisibility by 2 and 3 is obvious. To prove divisibility by 5, consid... | 386.05 | Combinatorics | proof | Yes | Yes | olympiads | false | 7,577 |
# Variant № 2
Task 1
A construction company has built a cottage village consisting of three identical streets. The houses in the village, according to the project, are identical and are offered for sale at the same price. The business plan of the company, which includes expenses for all stages of construction, contai... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,578 | |
6. In her fourth year at Hogwarts, Hermione was outraged by the infringement of house-elf rights and founded the Association for the Restoration of Elven Independence. Of course, even the brightest and noblest idea requires funding for promotion, so Hermione decided to finance her campaign by producing merchandise, sta... | # Solution:
Since the prices must be the same for the faculties, Hermione Monopolist must take into account the aggregate demand for badges, which is given by:
$$
Q(P)=q_{w s}+q_{s}=\left\{\begin{array}{l}
26-2 P, P \geq 13 \\
36-3 P, P<13
\end{array}\right.
$$
Then the marginal revenue of Hermione's monopoly is giv... | 7.682 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,580 |
# Task 3. Maximum 20 points
The wizard has revealed the secret of wisdom. For those who would like to know exactly where the secret is hidden, he left a clue in his magic book: $5 \cdot$ BANK $=6 \cdot$ GARDEN
Each letter in this clue represents a certain digit. Find these digits and substitute them into the GPS coor... | # Solution
The hint is presented as an equation. Let's write it down and solve it in integers.
$$
5 * \mathrm{~S} * 1000+5 * \mathrm{~A} * 100+5 * \mathrm{H} * 10+5 * \mathrm{~K}=6 * \mathrm{C} * 100+6 * \mathrm{~A} * 10+6 * D
$$
Using the rules of multiplication and divisibility, we can say that D is either 5 or 0,... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,582 |
# Task 1. 15 points
The distances between the transport nodes, where cargo is received and transferred, are respectively: 2 km; $\sqrt{7}$ km, and 3 km. A warehouse, located within the transport zone (a triangle with vertices at the points of cargo reception and transfer), is connected to the transport nodes by straig... | # Solution
The solution to the problem reduces to finding the minimum distance from some point inside a triangle with sides 2 km, $\sqrt{7}$ km, and 3 km, and then multiplying this distance by 2 (the truck travels to and from each node from the warehouse without repeating nodes).
Let in triangle $\mathrm{ABC}$, the b... | 2\sqrt{19} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,583 |
# Task 2. 15 points
Solve the equation
$$
\sqrt{\frac{x^{3}+5}{1+\sqrt{5}}}=x
$$
# | # Solution
The equation is equivalent to the system
$$
\sqrt{\frac{x^{3}+5}{1+\sqrt{5}}}=x \Leftrightarrow\left\{\begin{array} { c }
{ \frac { x ^ { 3 } + 5 } { 1 + \sqrt { 5 } } = x ^ { 2 } } \\
{ x \geq 0 , }
\end{array} \Leftrightarrow \left\{\begin{array} { c }
{ x ^ { 3 } + 5 = x ^ { 2 } ( 1 + \sqrt { 5 } ) , ... | \frac{1+\sqrt{1+4\sqrt{5}}}{2},\sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,584 |
# Task 3. 15 points
Solve the inequality
$f(g(x))>g(f(x))$, if $f(x)=\left\{\begin{array}{cc}x, & x \geq 0 \\ 0, & x<0\end{array} ; g(x)=\left\{\begin{array}{cc}2-x, & x<2 \\ 0, & x \geq 2\end{array}\right.\right.$.
# | # Solution
$f(g(x)) > g(f(x))$ for $x < 0$.
Answer: $x < 0$.
## Criteria:
Correct solution and correct answer provided - 15 points.
Correctly found functions $f(g(x))$ and $g(f(x))$, bu... | x<0 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,585 |
# Task 4. 20 points
At the beginning of 2020, Alexander, due to fear of uncertainty, bought several kilograms of buckwheat at a price of 70 rub/kg. By the beginning of 2022, Alexander had 1 kg of buckwheat left, and its price had risen to 100 rub/kg. It is known that at the beginning of 2020, Alexander could open annu... | # Solution:
(a) We know that 1 kg of buckwheat remained, which Alexander could have bought at the beginning of 2022. Let's calculate whether it would have been profitable for him. We will consider the option where he sequentially adds money to annual deposits, and the option where he immediately puts them into a two-y... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,586 |
# 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 c... | # 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 t... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,587 |
# Task 6. 15 points
The practice of buying goods at retail but at wholesale prices is called group buying when people cooperate to purchase goods from suppliers without a markup.
The practice of group buying became popular in Russia in the mid-2000s and is still used today. Buyers come together on specialized website... | # Solution:
Here are several factors that can explain the profitability of such purchases.
1) Joint purchases allow for significant savings on the acquisition of goods, as they are essentially made at wholesale prices, and the overhead costs associated with delivering the goods to the buyer and paying for the service... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,588 |
# Task 1. 15 points
The distances between the piers on the island, where the loading and unloading of fish catches take place, are 4 km, $\sqrt{13}$ km, and 3 km, respectively. A refrigeration unit, located within the transport zone (a triangle with vertices at the loading-unloading points), is connected to the piers ... | # Solution
The solution to the problem reduces to finding the minimum distance from some point inside a triangle with sides 4 km, $\sqrt{13}$ km, and 3 km, and then multiplying this distance by 2 (the refrigerated truck travels to and from the refrigerator to each pier for loading and unloading without repeating ports... | 2\sqrt{37} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,589 |
# Task 2. 15 points
Solve the equation
$$
\sqrt{\frac{x^{3}+7}{1+\sqrt{7}}}=x
$$
# | # Solution:
The equation is equivalent to the system
$$
\begin{aligned}
& \qquad \sqrt{\frac{x^{3}+7}{1+\sqrt{7}}}=x \Leftrightarrow\left\{\begin{array} { c }
{ \frac { x ^ { 3 } + 7 } { 1 + \sqrt { 7 } } = x ^ { 2 } , } \\
{ x \geq 0 , }
\end{array} \Leftrightarrow \left\{\begin{array} { c }
{ x ^ { 3 } + 7 = x ^ ... | \frac{1+\sqrt{1+4\sqrt{7}}}{2},\sqrt{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,590 |
# Task 4. 20 points
At the beginning of 2015, Vladimir bought several kilograms of buckwheat at a price of 70 rub/kg. By the beginning of 2017, Vladimir had 1 kg of buckwheat left, and its price had risen to 85 rub/kg. It is known that at the beginning of 2015, Vladimir could open annual deposits at an interest rate o... | # Solution
(a) We know that 1 kg of buckwheat remained, which Vladimir could have bought at the beginning of 2017. Let's calculate whether it would have been profitable for him. We will consider the option where he sequentially adds money to annual deposits, and the option where he immediately puts them into a two-yea... | 92.57 | Other | math-word-problem | Yes | Yes | olympiads | false | 7,592 |
# 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 c... | # 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 t... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,593 |
# Task 6. 15 points
The practice of buying goods at retail but at wholesale prices is called group buying when people cooperate to purchase goods from suppliers without a markup.
The practice of group buying became popular in Russia in the mid-2000s and is still used today. Buyers come together on specialized website... | # Solution:
Let's consider several factors that can explain the profitability of such purchases.
1) Joint purchases allow for significant savings on the acquisition of goods, as they are essentially made at wholesale prices, and the overhead costs associated with delivering the goods to the buyer and paying for the s... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,594 |
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 \Leftrightar... | Answer: 1. 6. 2.6 3.74. 7.
## Checking Criteria: | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,595 |
# Assignment 4. 20 points
## Variant 1
In country Alpha, only two goods, X and Y, are produced from a single production factor - factor Z, of which there are exactly 100 units in Alpha. The amount of good X produced from factor Z can be described by the function $X=\frac{\sqrt{Z_{X}}}{2}$, and one unit of good Y is p... | # Solution:
(a) Since the country can sell and buy goods on the world market, to achieve the consumption of the largest quantity of good sets, it needs to achieve the highest income from trade. For this, the country needs to use its resources as efficiently as possible. First, let's find the equation of the production... | \alpha>2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,598 |
# Assignment 5. 20 points
It can be observed that white electrical sockets and switches are usually $20-30 \%$ cheaper than sockets and switches of other colors of the same quality from the same manufacturers. How might economists explain this price difference in these goods? Provide two different explanations. | # Solution:
Here are possible explanations.
1) Consumers who would like to purchase colored switches and outlets generally value them higher than white switches and outlets of the same quality, meaning they are willing to pay more for them. However, companies that could produce (or sell) items of different colors, wh... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,599 |
3. In all other cases - 0 points
*Note: the numerical assessment of the free area (solution) is not the only possible one, for example, the "gap" can be more than 10 m.
## Task 2 (12 points)
Crocodile Gena and Old Lady Shapoklyak entered into a futures contract, according to which Gena agreed to invest in the 1st pr... | # Solution:
1) It is clear that the expression defines the equation of a certain line $l$ in the plane $x O p$. Consider the expression
$$
\begin{gathered}
p^{2}-12 p+x^{2}-14 x+69=0 \Leftrightarrow(p-6)^{2}+(x-7)^{2}-36-49+69=0 \\
\Leftrightarrow(p-6)^{2}+(x-7)^{2}=4^{2}
\end{gathered}
$$
It defines in the plane $x... | 1)2.6,2)16840 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,600 |
3. In all other cases - 0 points
## *Important: the numerical assessment of the free area (solution) is not the only possible one, for example, the "gap" can be more than 10 m.
## Assignment 2 (12 points)
Crocodile Gena and Old Lady Shapoklyak entered into a futures contract, according to which Gena agreed to invest... | # Solution:
1) It is clear that the expression $4 x_{1}-3 p_{1}-44=0 \Leftrightarrow x=\frac{3}{4} p+11$ defines the equation of a certain line $l$ in the plane $x O p$. Consider the expression
$$
\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}+... | 13080 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,603 |
# Problem 2. Maximum 14 points
Solve the equation:
$$
\log _{4}\left(4^{\sqrt{2} \sin x}+4^{\sqrt{2} \cos x}\right)+\log _{\left(\operatorname{tg}^{4} x+1\right)^{2}} \sqrt{2}=\log _{16} \frac{\operatorname{ctg}^{4} x}{\operatorname{ctg}^{4} x+1}
$$ | Solution
Let's carry out a series of simple transformations
$$
\begin{gathered}
\log _{4}\left(4^{\sqrt{2} \sin x}+4^{\sqrt{2} \cos x}\right)+\frac{1}{2} \cdot \frac{1}{2} \cdot \log _{\operatorname{tg}^{4} x+1} 2-\frac{1}{4} \cdot \log _{2} \frac{\operatorname{ctg}^{4} x}{\operatorname{ctg}^{4} x+1}=0 \\
\log _{4}\l... | \frac{5\pi}{4}+2\pik,k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,607 |
# 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$ ... | # 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 trian... | 307 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,608 |
# Problem 4. Maximum 14 points
A high-tech Japanese company has presented a unique robot capable of producing construction blocks, which can be sold for 90 monetary units each. Due to a shortage of specialized chips, it is impossible to replicate this robot or even repair the existing one if it becomes inoperative. If... | # Solution
(a) Let's write down the company's objective function. Let $Q_{d}$ be the number of blocks the robot produces in a day. $Q_{d}=L \cdot \frac{1}{\sqrt{L}}=\sqrt{L} \rightarrow \max$. Notice that the function $Q_{d}(L)$ is monotonically increasing, which means $Q_{d}$ reaches its maximum value when $L=L_{\max... | 24,16,L^{*}=9,TR_{\max}=1650,30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,609 |
# 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. H... | # 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$ i... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,610 |
# Problem 6. Maximum 14 points
Often in movie theaters and entertainment centers, popcorn is sold in containers of different sizes. For example, customers are offered three options: small, medium, and large, which hold 50, 70, and 130 grams of popcorn at prices of 200, 400, and 500 rubles, respectively.
(a) Explain h... | # Solution:
(a) The decoy effect in the given example manifests in encouraging the customer to buy the largest container of popcorn. The "decoy" in this case can be considered the medium-sized container of popcorn. If this option were not available, some customers would choose the smallest container for 200 rubles, wh... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,611 |
# Task 7. Maximum 10 points
One of the companies that sells office equipment and computers has launched a promotion under which every buyer of a laptop or desktop computer receives wireless headphones with a charging case as a gift. Ivan Ivanovich decided to buy a laptop, which costs 130,000 rubles. When forming the o... | # Solution:
(a) Companies that deliver goods to end customers typically have many warehouses for storing goods, located in different areas of the city, regions of the country, and many delivery points. The delivery of goods to customers from even one online store on different days can be influenced by a multitude of d... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,612 |
# Task 4. Maximum 20 points
A client of a brokerage firm deposited 12,000 rubles into a brokerage account at an exchange rate of 60 rubles per dollar, instructing the broker to invest the amount in bonds of foreign banks with a guaranteed yield of $12\%$ per year in dollars.
(a) Determine the amount in rubles that th... | # Solution and criteria for checking:
(a) The brokerage account received 12000 / $60=200$ dollars (1 point)
The stocks generated an income of $200 * 0.12=24$ dollars over the year (1 point)
At the end of the year, the account had $200+24=224$ dollars (1 point)
The broker's commission amounted to $24 * 0.25=6$ dolla... | 16742.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,613 |
# 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 l... | # 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 t... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,616 |
# 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 c... | # 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 t... | 8640 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,617 |
# Problem 2.
Find all positive $x, y$ for which the expression
$$
\frac{x y}{2}+\frac{18}{x y}
$$
takes the smallest value, and among all such $x, y$ the expression
$$
\frac{y}{2}+\frac{x}{3}
$$
is also minimal. | # Solution:
According to the Cauchy inequality,
$$
\frac{x y}{2}+\frac{18}{x y} \geq 2 \sqrt{\frac{x y}{2} \cdot \frac{18}{x y}}=6
$$
Equality is possible only when
$$
\frac{x y}{2}=\frac{18}{x y}
$$
From this,
$$
x y=6 \Rightarrow y=\frac{6}{x}
$$
Substituting the obtained value into the expression
$$
\frac{y}... | 3,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,618 |
12. $\quad$ Maximum 15 points. Let $[x]$ denote the integer part of the number $x$ (i.e., the greatest integer not exceeding $x$). Solve the system of equations:
$$
\left\{\begin{array}{c}
{[x+y-3]=2-x} \\
{[x+1]+[y-7]+x=y}
\end{array}\right.
$$ | # Solution:
Since for any $x \in \mathbb{R}, n \in \mathbb{Z}$ the equality holds:
$$
[x+n]=[x]+n
$$
the system is equivalent to the system:
$$
\left\{\begin{array}{c}
{[x+y]-5=-x} \\
{[x]+[y]+x-6=y}
\end{array}\right.
$$
From the first equation, it follows that $x \in \mathbb{Z}$ (it is sufficient to note that ea... | (3,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,622 |
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^{... | # 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 \... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,623 |
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, ... | # 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 al... | 127 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,624 |
4. Maximum 15 points. The company "Superheroes, Inc" urgently called Superman and Flash to perform some construction work to build protective structures against a hurricane. It is known that Flash's productivity is twice that of Superman. The payment for each superhero depends only on the time they spend on the work, a... | # Solution:
9/10 of the work was completed together by the superheroes, with 3/10 done by Superman and 6/10 by Flash, since his productivity is twice as high. Since 1/10 of the entire work was completed by Superman in 1 minute, 3/10 of the work will take him 3 minutes. At the same time, Flash will also work for 3 minu... | Superman:22.5,Flash:30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,625 |
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 ... | # 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 manu... | 328710 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,626 |
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 asses... | # 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... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,627 |
Task 7.
## Maximum 15 points
In the modern world, every consumer often has to make decisions about replacing old equipment with more energy-efficient models. 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 ene... | # Solution and Grading Scheme:
a) Expenses for 10 months when installing an energy-saving bulb independently: 120 rubles + 12 (W) * 100 (hours) / 1000 * 5 (rubles/kWh) * 10 (months) = 180 rubles.
Expenses for 10 months when turning to an energy service company:
$(12 + (60 - 12) * 0.75)($ W) * 100 (hours) $/ 1000 * 5... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,631 |
3. The remaining $60 \%$ of voters, all else being equal, are willing to support Ratibor, but unfortunately, they would not refuse to sell their vote to Nikifor for a certain payment. It is known that if Nikifor offers 1 monetary unit per vote, he will gain only one additional supporter; if he offers 2 monetary units, ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,632 | |
1. Draw $K M$ until it intersects with $C B$ at point $N$. Let $A M=2 y, M D=3 y, A K=x$, $K B=3 x$. Then $B L=4 y, L C=y$. Triangles $A K M$ and $N K B$ are similar, hence $N B=6 y$, since $A K: K B=1: 3$. Therefore, $N L=6 y+4 y=10 y$. Triangles $N Q L$ and $A Q M$ are similar, so $A Q: Q L=A M: N L=2 y: 10 y=1: 5$. ... | Answer: 1. 1:5, 1:2. 2. 3:13, 1:3. 3. 2:7, 1:8. 4. 3:10, $1: 12$
## Grading Criteria:
All solved correctly - 20 points.
Additional construction made, similarity of triangles proven - from 5 to 10 points. In all other cases - 0 points.
# | 1:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,634 |
# 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... | # 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 ... | 326 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,637 |
# 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 s... | # 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... | 104 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,638 |
# 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... | # 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 ... | 326 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,639 |
# 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 s... | # 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... | 104 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,640 |
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 cy... | 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 condit... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,641 |
2. A natural number, not ending in zero, had one of its digits erased. As a result, the number decreased by 6 times. Find all numbers for which this is possible. | Answer: 108 or $12a$ when $a=1,2,3,4$.
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 $m=0$ and $n=a=0$. Then the equation will take the form $m=10^{k-1}(2a+8n)$. By the condition, the numbe... | 108or12awhen=1,2,3,4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,642 |
3. Given real numbers $x_{1}, \ldots, x_{n}$. Find the maximum value of the expression
$$
A=\left(\sin x_{1}+\ldots+\sin x_{n}\right) \cdot\left(\cos x_{1}+\ldots+\cos x_{n}\right)
$$ | Answer: $\frac{n^{2}}{2}$.
Solution. Note that for any $a_{1}, \ldots, a_{n}$
$$
\left(\sum_{k=1}^{n} a_{k}\right)^{2} \leqslant n \sum_{k=1}^{n} a_{k}^{2}
$$
From this, by the Cauchy inequality,
$$
A \leqslant \frac{1}{2}\left(\left(\sum_{k=1}^{n} \sin x_{k}\right)^{2}+\left(\sum_{k=1}^{n} \cos x_{k}\right)^{2}\ri... | \frac{n^{2}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,643 |
4. On the segment $A B$ of length 10, a circle $\omega$ is constructed with $A B$ as its diameter. A tangent to $\omega$ is drawn through point $A$, and a point $K$ is chosen on this tangent. A line through point $K$, different from $A K$, is tangent to the circle $\omega$ at point $C$. The height $C H$ of triangle $A ... | Answer: 8.
Solution. Let $O$ be the center of $\omega$. Note that
$$
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 triang... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,644 |
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 defi... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,645 |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 1, 4, and 4, and the angles at their vertices are $-4 \operatorname{arctg} \frac{1}{3}, 4 \operatorname{arctg} \frac{9}{11}$, and $4 \operatorname{arctg} \frac{9}{11}$ respectively (the angle at the vertex ... | # Answer: $\frac{5}{3}$.
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 ... | \frac{5}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,646 |
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 neck... | 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 prob... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,647 |
2. For a natural number ending not in zero, one of its digits was replaced by zero (if it was the leading digit, it was simply erased). As a result, the number decreased by 6 times. Find all numbers for which this is possible. | Answer: $12 a$ when $a=1,2,3,4$.
Solution. Represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, $k, m, n-$ are non-negative integers, and $m0$, otherwise $m=0$ and $a=0$. Then the equation transforms to $m=2 a \cdot 10^{k-1}$. Due to the condition, the number $m$ does not en... | 12awhen=1,2,3,4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,648 |
3. Given numbers $x_{1}, \ldots, x_{n}$ from the interval $\left[0, \frac{\pi}{2}\right]$. Find the maximum value of the expression
$$
A=\left(\sqrt{\sin x_{1}}+\ldots+\sqrt{\sin x_{n}}\right) \cdot\left(\sqrt{\cos x_{1}}+\ldots+\sqrt{\cos x_{n}}\right) .
$$ | Answer: $\frac{n^{2}}{\sqrt{2}}$.
Solution. Note that for any $a_{1}, \ldots, a_{n}$
$$
\left(\sum_{k=1}^{n} a_{k}\right)^{4} \leqslant\left(n \sum_{k=1}^{n} a_{k}^{2}\right)^{2} \leqslant n^{3} \sum_{k=1}^{n} a_{k}^{4}
$$
From this, by the Cauchy-Schwarz inequality,
$$
A^{2} \leqslant \frac{1}{2}\left(\left(\sum_{... | \frac{n^{2}}{\sqrt{2}} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,649 |
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 $Q$. Fi... | 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 tr... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,650 |
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,... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,651 |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 1, 12, and 12, and the vertex angles are $-4 \operatorname{arctg} \frac{1}{3}, 4 \operatorname{arctg} \frac{2}{3}$, and $4 \operatorname{arctg} \frac{2}{3}$ respectively (the vertex angle of a cone is the a... | Answer: $\frac{40}{21}$.
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 ... | \frac{40}{21} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,652 |
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 ... | 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 $m$ ... | 65 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,653 |
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,
$$
m+10^{k} a+10^{k+1} n=9\left(m+10^{k} n\right) \Lon... | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,654 |
3. The numbers $x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{n}$ satisfy the condition $x_{1}^{2}+\ldots+x_{n}^{2}+y_{1}^{2}+\ldots+y_{n}^{2} \leqslant 2$. Find the maximum value of the expression
$$
A=\left(2\left(x_{1}+\ldots+x_{n}\right)-y_{1}-\ldots-y_{n}\right) \cdot\left(x_{1}+\ldots+x_{n}+2\left(y_{1}+\ldots+y_{n}\r... | Answer: $5 n$.
Solution 1. For $k=1, \ldots, n$ let $t_{k}=3 x_{k}+y_{k}, s_{k}=x_{k}-3 y_{k}$. Then
$$
\begin{aligned}
\left.4 A=\left(4 x_{1}+\ldots+4 x_{n}-2 y_{1}-\ldots-2 y_{n}\right) \cdot\left(2 x_{1}+\ldots+2 x_{n}+4 y_{1}+\ldots+4 y_{n}\right)\right) & = \\
=\left(\sum_{k=1}^{n}\left(3 x_{k}+y_{k}\right)+\su... | 5n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,655 |
4. In triangle $A B C$, a circle $\omega$ with radius $r$ is inscribed, touching side $A B$ at point $X$. On the circle, point $Y$ is marked, diametrically opposite to point $X$. Line $C Y$ intersects side $A B$ at point $Z$. Find the area of triangle $A B C$, given that $C A + A Z = 1$. | Answer: $r$.
Solution. Draw a tangent to $\omega$ through point $Y$, intersecting segments $A C$ and $B C$ at points $P$ and $Q$ respectively. Let $M$ and $N$ be the points where $\omega$ t... | r | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,656 |
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 n... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,657 |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 32, 48, and 48, and the vertex angles are $\frac{\pi}{3}$, $\frac{2 \pi}{3}$, and $\frac{2 \pi}{3}$ respectively (the vertex angle of a cone is the angle between its generators in the axial section). A sphe... | Answer: $13(\sqrt{3}+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, $C$ be its projection on the table, and $R$ be the r... | 13(\sqrt{3}+1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,658 |
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 o... | 112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,659 |
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... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,660 |
3. The numbers $x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{n}$ satisfy the condition $x_{1}^{2}+\ldots+x_{n}^{2}+y_{1}^{2}+\ldots+y_{n}^{2} \leqslant 1$. Find the maximum value of the expression
$$
A=\left(3\left(x_{1}+\ldots+x_{n}\right)-5\left(y_{1}+\ldots+y_{n}\right)\right) \cdot\left(5\left(x_{1}+\ldots+x_{n}\right)... | Answer: $17 n$.
Solution. For $k=1, \ldots, n$ let $t_{k}=4 x_{k}-y_{k}, s_{k}=x_{k}+4 y_{k}$. Then
$$
A=\left(\sum_{k=1}^{n}\left(4 x_{k}-y_{k}\right)-\sum_{k=1}^{n}\left(x_{k}+4 y_{k}\right)\right) \cdot\left(\sum_{k=1}^{n}\left(4 x_{k}-y_{k}\right)+\sum_{k=1}^{n}\left(x_{k}+4 y_{k}\right)\right)=\left(\sum_{k=1}^{... | 17n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,661 |
4. In a right triangle $ABC$, a height $BH$ is dropped onto the hypotenuse $AC$. Points $X$ and $Y$ are the centers of the circles inscribed in triangles $ABH$ and $CBH$, respectively. The line $XY$ intersects the legs $AB$ and $BC$ at points $P$ and $Q$. Find the area of triangle $BPQ$, given that $BH = h$. | Answer: $\frac{h^{2}}{2}$.
Solution 1. The lines $H X$ and $H Y$ are the angle bisectors of the right angles $A H B$ and $B H C$, hence
$$
\angle A H X=\angle B H X=\angle B H Y=\angle C H ... | \frac{^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,662 |
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 $1 \times 1$ squa... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,663 |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are 72, 28, and 28, and the angles at their vertices are $-\frac{\pi}{3}$, $\frac{2 \pi}{3}$, and $\frac{2 \pi}{3}$ respectively (the angle at the vertex of a cone is the angle between its generators in the axi... | Answer: $\frac{\sqrt{3}+1}{2}$.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the cones, $O$ be the center of the sphere, $C$ be its projection on the table, and $R$ be the radius of the sphere. The point $C$ is equidistant from the points of tangency of the bases of the cones, so it lies at the i... | \frac{\sqrt{3}+1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,664 |
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