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8.2. The part of the graph of a linear function located in the second coordinate quadrant, together with the coordinate axes, forms a triangle. By what factor will its area change if the slope of the function is doubled and the y-intercept is halved?
|
Answer: It will decrease by eight times.
Solution. Let the original linear function be defined by the equation $y=k x+b$. From the condition of the problem, it follows that $k>0$ and $b>0$ (see Fig. 8.2). The points of intersection of its graph with the axes are: $A(0 ; b)$ and $B\left(-\frac{b}{k} ; 0\right)$. Since the triangle $A O B$ cut off by it is a right triangle, its area is $\frac{1}{2} O A \cdot O B=\frac{1}{2} b \cdot \frac{b}{k}=\frac{b^{2}}{2 k}$.
After the specified change in coefficients, the function becomes $y=2 k x+0.5 b$. Its graph intersects the coordinate axes at points $\mathrm{C}\left(0 ; \frac{b}{2}\right)$ and $D\left(-\frac{b}{4 k} ; 0\right)$. The area of the right triangle COD is $\frac{1}{2} O C \cdot O D=\frac{1}{2} \cdot \frac{b}{2} \cdot \frac{b}{4 k}=$ $\frac{b^{2}}{16 k}$. Therefore, the area will decrease by $\frac{b^{2}}{2 k}: \frac{b^{2}}{16 k}=8$ times.
Grading Criteria. «+» A complete and well-reasoned solution is provided «士» A generally correct reasoning is provided, containing minor gaps or inaccuracies that did not affect the answer
«Ғ» Correct reasoning, but computational errors were made
«Ғ» Only the correct answer is provided or the answer was obtained by considering a specific example
«-» The problem is not solved or is solved incorrectly
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. On an $8 \times 8$ chessboard, $k$ rooks and $k$ knights are placed such that no figure attacks any other. What is the largest $k$ for which this is possible?
|
Answer: 5.
Solution: From the condition, it follows that in one row (column) with a rook, no other figure can stand.
Suppose 6 rooks were placed on the board. Then they stand in 6 rows and 6 columns. Therefore, only 4 unpicked cells will remain (located at the intersection of two empty rows and two empty columns). It is impossible to place 6 knights in these cells. Therefore, $k$ is no more than 5.
In Fig. 3, it is shown how to place 5 rooks and 5 knights on the board so that they do not attack each other.
Fig. 3
Remark. There are other examples of placement.
Comment. Only proved that $k$ is no more than $5-4$ points.
Only provided an example of placing 5 knights and 5 rooks - 3 points.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The numbers $\sqrt{2}$ and $\sqrt{5}$ are written on the board. You can add to the board the sum, difference, or product of any two different numbers already written on the board. Prove that you can write the number 1 on the board.
|
Solution: For example, we get $\sqrt{5}-\sqrt{2}$, then $\sqrt{5}+\sqrt{2}$ and $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=5-2=3$, then $\sqrt{2} \cdot \sqrt{5}=\sqrt{10}$, then $\sqrt{10}-3$ and $\sqrt{10}+3$ and finally $(\sqrt{10}-3)(\sqrt{10}+3)=10-9=1$.
Criteria. The goal is achieved if the same numbers are used in some operations (this is prohibited by the condition) - 3 points.
|
1
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
3. From 80 identical Lego parts, several figures were assembled, with the number of parts used in all figures being different. For the manufacture of the three smallest figures, 14 parts were used, and in the three largest, 43 were used in total. How many figures were assembled? How many parts are in the largest figure?
|
Answer: 8 figurines, 16 parts.
Solution. Let the number of parts in the figurines be denoted by $a_{1}43$, so $a_{n-2} \leqslant 13$.
Remove the three largest and three smallest figurines. In the remaining figurines, there will be $80-14-$ $43=23$ parts, and each will have between 7 and 12 parts. One figurine is clearly insufficient, and three would be too many $(7+8+9=24)$. Therefore, 23 parts form 2 figurines. This is possible, and in only one way: $23=11+12$. We have $43=13+14+16$ - the only decomposition with $a_{6} \geqslant 13$.
Criteria. Only the answer - 0 points. Only correct estimates for $a_{3}$ and $a_{n-2}-3$ points. Only a justified answer for the number of figurines - 5 points. Complete solution - 7 points.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The diagonals of the circumscribed trapezoid $A B C D$ with bases $A D$ and $B C$ intersect at point O. The radii of the inscribed circles of triangles $A O D, A O B, B O C$ are 6, 2, and $3 / 2$ respectively. Find the radius of the inscribed circle of triangle $C O D$.
|
Answer: 3
Solution. We will prove a more general statement, that $\frac{1}{r_{1}}+\frac{1}{r_{3}}=\frac{1}{r_{2}}+\frac{1}{r_{4}}$, where $r_{1}, r_{2}, r_{3}$ and $r_{4}$ are the radii of the inscribed circles of triangles $A O D, A O B, B O C$ and $C O D$ respectively.
Let $A B=a, B C=b, C D=c, A D=d, O A=x, O D=y, S_{\triangle A O D}=S$. Let $\frac{b}{d}=k$.
Triangle $C O B$ is similar to triangle $A O D$ with a similarity coefficient $k$, so $S_{\triangle C O B}=k^{2} S$, $O B=k y, O C=k x, S_{\triangle A O B}=S_{\triangle C O D}=k S$. Then
$$
r_{1}=\frac{2 S}{x+y+d}, \quad r_{3}=\frac{2 k^{2} S}{k x+k y+b}, \quad r_{2}=\frac{2 k S}{k y+x+a}, \quad r_{4}=\frac{2 k S}{k x+y+c}
$$
Thus,
$$
2\left(\frac{1}{r_{1}}+\frac{1}{r_{3}}\right)=\frac{x+y+d}{S}+\frac{k x+k y+b}{k^{2} S}=\frac{x+y}{S}+\frac{x+y}{k S}+\frac{d k+\frac{b}{k}}{k S}=\frac{x+y}{S}+\frac{x+y}{k S}+\frac{b+d}{k S}
$$
Here we used that $d k=b$ and $\frac{b}{k}=d$ from the similarity of triangles $A O D$ and $B O C$. Further,
$$
2\left(\frac{1}{r_{2}}+\frac{1}{r_{4}}\right)=\frac{k x+y+c}{k S}+\frac{x+k y+a}{k S}=\frac{x+y}{S}+\frac{x+y}{k S}+\frac{a+c}{k S}=\frac{x+y}{S}+\frac{x+y}{k S}+\frac{b+d}{k S}
$$
where $a+c=b+d$, since the trapezoid is circumscribed. Thus, we have proved that $\frac{1}{r_{1}}+\frac{1}{r_{3}}=$ $\frac{1}{r_{2}}+\frac{1}{r_{4}}$.
By the condition, $\frac{1}{6}+\frac{1}{1.5}=\frac{1}{2}+\frac{1}{r_{4}}$, from which $r_{4}=3$.
Criteria. The solution contains a statement that is not common knowledge and which the reader cannot establish on their own in 5 minutes - no more than 3 points.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
|
Answer: 6.
Solution. Let $a, b, c$ be the natural numbers in the three lower circles, from left to right. According to the condition, $14 \cdot 4 \cdot a = 14 \cdot 6 \cdot c$, i.e., $2a = 3c$. From this, it follows that $3c$ is even, and therefore $c$ is even. Thus, $c = 2k$ for some natural number $k$, and from the equation $2a = 3c$ it follows that $a = 3k$.
It must also hold that $14 \cdot 4 \cdot 3k = 3k \cdot b \cdot 2k$, which means $b \cdot k = 28$. Note that by choosing the number $k$, which is a natural divisor of 28, the natural numbers $a, b, c$ are uniquely determined. The number 28 has exactly 6 natural divisors: $1, 2, 4, 7, 14, 28$. Therefore, there are also 6 ways to place the numbers in the circles.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.8. Given an isosceles triangle $ABC (AB = BC)$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
|
Answer: 4.
Solution. Mark point $K$ on ray $B C$ such that $B E=B K$. Then $A E=C K$ as well.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common side) and the angle between them ($\angle C A E=\angle A C K$ - adjacent to the equal base angles of the isosceles triangle). Therefore, $A K=C E=13$ and $\angle A K C=\angle A E C=60^{\circ}$.
In triangle $A D K$, the angles at vertices $D$ and $K$ are both $60^{\circ}$, so it is an equilateral triangle, and $D K=A K=A D=13$. Therefore, $A E=C K=D K-D C=13-9=4$.
## 8th grade
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-1. A beginner gardener planted daisies, buttercups, and marguerites on their plot. When they sprouted, it turned out that there were 5 times more daisies than non-daisies, and 5 times fewer buttercups than non-buttercups. What fraction of the sprouted plants are marguerites?
|
Answer. Zero. They did not germinate.
Solution. Daisies make up $5 / 6$ of all the flowers, and dandelions make up $1/6$. Therefore, their total number equals the total number of flowers.
Criteria. Only the answer - 0 points. Complete solution - 7 points.
|
0
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-3. Vika has been recording her grades since the beginning of the year. At the beginning of the second quarter, she received a five, after which the proportion of fives increased by 0.15. After another grade, the proportion of fives increased by another 0.1. How many more fives does she need to get to increase their proportion by another 0.2?
|
Answer: 4.
Solution: Let's say Vika had $n$ grades in the first quarter, of which $k$ were fives. Then, after the first five in the second quarter, the proportion of fives increased by $\frac{k+1}{n+1}-\frac{k}{n}=0.15$. Similarly, after the second five, the increase was $\frac{k+2}{n+2}-\frac{k+1}{n+1}=0.1$. Simplifying each equation, we get the system
$$
\left\{\begin{array}{c}
n-k=0.15 n(n+1) \\
n-k=0.1(n+1)(n+2)
\end{array}\right.
$$
In particular, $0.15 n(n+1)=0.1(n+1)(n+2)$, which means $1.5 n=n+2, n=4$. Substituting this value into the first equation, we find that $k=4-0.15 \cdot 4 \cdot 5=1$. Therefore, after receiving two fives in the second quarter, the proportion of fives became $3 / 6=0.5$. Vika wants the proportion of fives to be $0.5+0.2=0.7$ after receiving another $m$ fives, which means $\frac{3+m}{6+m}=0.7$. Solving this equation, we get $m=4$.
Criteria: Answer without justification - 0 points. System of equations formulated - 2 points. Complete solution - 7 points.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A new series of "Kinder Surprises" - chocolate eggs, each containing a toy car - was delivered to the store. The seller told Pete that there are only five different types of cars in the new series, and it is impossible to determine which car is inside by the appearance of the egg. What is the minimum number of "Kinder Surprises" Pete should buy to guarantee having three cars of the same type, regardless of which type?
|
Solution. If Petya buys 10 "Kinder Surprises," in the least favorable situation for him, he will get two cars of each type. If he buys 11 "Kinder Surprises," he will get three cars of one type. Let's prove this. Suppose Petya bought 11 "Kinder Surprises" but did not get three cars of one type. This means the number of cars is no more than 10 (two cars of one type, a total of five types). Contradiction.
Answer. Petya should buy no fewer than 11 "Kinder Surprises."
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Two cyclists are training on a circular stadium. In the first two hours, Ivanov lapped Petrov by 3 laps. Then Ivanov increased his speed by 10 km/h, and as a result, after 3 hours from the start, he lapped Petrov by 7 laps. Find the length of the lap.
|
4. Answer: 4 km. Solution: let I be Ivanov's speed (initial), Π be Petrov's speed, K be the length of the circle. Then 2I - 2Π = 3K and 2I - 2(Ι - Π) = 3K and 3(Ι - Π) = 3K - 10, express Ι - Π from both equations and equate, we get 14K - 20 = 9K, from which K = 4.
Criteria: correct solution - 7 points; the system is set up, but there is no further progress - 2 points; in all other cases - 0 points.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.2. How many rectangular trapezoids $A B C D$ exist, where angles $A$ and $B$ are right angles, $A D=2$, $C D=B C$, the sides have integer lengths, and the perimeter is less than 100?
|
Answer: 5.
Solution. Let $C D=B C=a, A B=b$. Drop a perpendicular from point $D$ to $B C$, and apply the Pythagorean theorem: $(a-2)^{2}+b^{2}=a^{2}$. From this, $b^{2}=4(a-1), a=\frac{b^{2}}{4}+1$. Suppose $A D=2$ is the smaller base of the trapezoid. The perimeter $P=2 a+b+2a$, so $b>2$. Considering that $a=\frac{b^{2}}{4}+1$, $b$ must be even, yielding the permissible values of $b: 4,6,8,10,12$.
Suppose $A D=2$ is not the smaller base of the trapezoid. Then $a=1$ or $a=2$. In the first case, there is a contradiction with the relation $a=\frac{b^{2}}{4}+1$. In the second case, the quadrilateral $A B C D$ becomes a square, which is not considered a trapezoid.
Comment. A correct and justified solution - 7 points. Proven that $b \leq 12$ - 4 points. Odd values of $b$ not eliminated - 3 points deducted. The value $b=2$ not eliminated - 1 point deducted. The correct answer obtained by selecting numbers that satisfy the condition, but not shown that other answers are impossible - 3 points, if there is an incomplete justification - 4-5 points. Some answers found by selection - 1-2 points. The problem is not solved or solved incorrectly - 0 points.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. What is the maximum number of figures consisting of 4 1x1 squares, as shown in the diagram, that can be cut out from a $6 \times 6$ table, if cutting can only be done along the grid lines?
#
|
# Solution:
Example:
The diagram shows that 8 figures can be cut out.
Evaluation:
We will color the even-numbered columns in gray and the odd-numbered columns in white. For each figure cut out, there will be 2 more cells of one color than the other. Therefore, to cut out the entire table, we need the number of figures with three gray cells to be equal to the number of figures with three white cells, which means the number must be even. 36:4=9, which is an odd number, so it is impossible to cut out the entire figure.
Answer: 8 figures.
| Criteria | Points |
| :--- | :---: |
| Complete solution of the problem. | 7 |
| Only the evaluation, without an example. | 5 |
| Correct answer with an example. | 2 |
| Only the correct answer, without evaluation and example. | 0 |
| Incorrect solution. | 0 |
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. Solve the equation $2 \sin ^{2} x+1=\cos (\sqrt{2} x)$.
|
Answer: $x=0$. Solution. The left side of the equation $\geq 1$, and the right side $\leq 1$. Therefore, the equation is equivalent to the system: $\sin x=0, \cos \sqrt{2} x=1$. We have: $x=\pi n, \sqrt{2} x=2 \pi k$ ( $n, k-$ integers). From this, $n=k \cdot \sqrt{2}$. Since $\sqrt{2}$ is an irrational number, the last equality is possible only when $n=k=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. At a round table, 12 people are sitting. Some of them are knights, who always tell the truth, and the rest are liars, who always lie. Each person declared their left neighbor to be a liar. Can we definitely state how many knights and how many liars are at the table?
|
1. Answer: Yes, it is possible.
If a knight is sitting in some place, then he told the truth, and to his left should sit a liar. Conversely, if a liar is sitting in some place, then to his left sits the one who was incorrectly called a liar, that is, a knight. This means that knights and liars alternate around the table, meaning there are 6 of each.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.2. Solve the equation:
$1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$.
|
# Solution.
$1+\frac{1}{x+1}=\frac{x+2}{x+1}$, therefore the given equation is equivalent to the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+1}\right)=x$ under the condition that $\mathrm{x} \neq-2$. Proceeding similarly, we get $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of this equation are the numbers 2 and -2, so the only root of the original equation is the number 2.
## Grading Criteria
- Only the correct answer is provided - 1 point.
- The correct solution process, but the extraneous root is not discarded - 3 points.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. For various positive numbers $a$ and $b$, it is known that
$$
a^{3}-b^{3}=3\left(2 a^{2} b-3 a b^{2}+b^{3}\right)
$$
By how many times is the larger number greater than the smaller one?
|
Solution. Let's consider and transform the difference:
$$
\begin{aligned}
& 0=a^{3}-b^{3}-3\left(2 a^{2} b-3 a b^{2}+b^{3}\right)= \\
& (a-b)\left(a^{2}+a b+b^{2}\right)-3\left(2 a b(a-b)-b^{2}(a-b)\right)= \\
& (a-b)\left(a^{2}+a b+b^{2}-6 a b+3 b^{2}\right)= \\
& (a-b)\left(a^{2}-5 a b+4 b^{2}\right)=(a-b)(a-4 b)(a-b)
\end{aligned}
$$
By the condition \(a \neq b\), we get \(a = 4b\), which means the larger number is 4 times greater.
Answer: 4.
Recommendations for checking. Only the correct answer - 0 points. Answer with verification - 2 points. One factor \(a-b\) is highlighted - no less than 2 points, two factors \(a-b\) are highlighted - no less than 4 points.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. On November 15, a tournament of the game of dodgeball was held. In each game, two teams competed. A win was worth 15 points, a draw 11, and a loss earned no points. Each team played against each other once. At the end of the tournament, the total number of points scored was 1151. How many teams were there?
|
6. Answer: 12 teams. Solution. Let there be $\mathrm{N}$ teams. Then the number of games was $\mathrm{N}(\mathrm{N}-1) / 2$. For each game, a total of 15 or 22 points are earned. Therefore, the number of games was no less than $53(1151 / 22)$ and no more than $76(1151 / 15)$.
Note that if there were no more than 10 teams, then the number of games would not exceed 45. And if there were no fewer than 13 teams, then the number of games would be no less than 78. Therefore, the number of teams was 11 (55 games) or 12 (66 games). In each game, teams score exactly 15 points in total! And 7 additional points if there was a draw. Thus, the total number of draws is (1151-55*15)/7 (a non-integer - this could not have been the case) or (1151-66*15)/7=23. Therefore, the only option is that there were 12 teams.
Grading criteria. Full solution - 7 points. Determined that there were 11 or 12 teams, and one of the cases was not eliminated - 3 points. Justified that the number of games was no less than 53 and no more than 76 without further progress - 2 points. Correct answer with a constructed example - 2 points. Only the correct answer without explanation - 0 points.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. How many positive numbers are there among the first 100 terms of the sequence: $\sin 1^{\circ}, \sin 10^{\circ}, \sin 100^{\circ}, \sin 1000^{\circ}, \ldots ?$
|
2. Note that all members of the sequence, starting from $\sin 1000^{\circ}$, are equal to each other, since the difference between the numbers $10^{k+1}$ and $10^{k}$ for natural $k>2$ is a multiple of 360. Indeed, $10^{k+1}-10^{k}=10^{k}(10-1)=9 \cdot 10^{k}=$ $9 \cdot 4 \cdot 10 \cdot 25 \cdot 10^{k-3} \vdots 360$. By direct verification, we can see that $\sin 1^{\circ}>0$, $\sin 10^{\circ}>0, \sin 100^{\circ}>0, \sin 1000^{\circ}=\sin \left(360^{\circ} \cdot 3-80^{\circ}\right)=\sin \left(-80^{\circ}\right)<0$. Therefore, the rest of the members of the sequence will also be negative. Thus, among the first 100 members of the sequence, there will be exactly three positive members.
Answer: 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. We will call a number greater than 25 semi-prime if it is the sum of some two different prime numbers. What is the maximum number of consecutive semi-prime numbers that can be semi-prime?
|
5. Note that an odd semiprime number can only be the sum of two and an odd prime number.
Let's show that three consecutive odd numbers $2n+1, 2n+3, 2n+5$, greater than 25, cannot all be semiprimes simultaneously. Assume the opposite. Then we get that the numbers $2n-1, 2n+1, 2n+3$ are prime, and all of them are greater than 3. But one of these three numbers is divisible by 3. This leads to a contradiction. Therefore, three consecutive odd numbers $2n+1, 2n+3, 2n+5$, greater than 25, cannot all be semiprimes simultaneously.
Note that among any six consecutive natural numbers, there are three consecutive odd numbers, so there cannot be more than five consecutive semiprime numbers. Five consecutive numbers can be semiprimes; for example, $30=17+13, 31=29+2, 32=19+13, 33=31+2, 34=23+11$. There are other examples as well.
Answer: 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In the office, each computer was connected by wires to exactly 5 other computers. After some computers were infected by a virus, all wires from the infected computers were disconnected (a total of 26 wires were disconnected). Now, each of the uninfected computers is connected by wires to only 3 others. How many computers were infected by the virus?
|
# 5. Answer: 8.
Let $\mathrm{m}$ be the number of infected computers, and $\mathrm{n}$ be the number of uninfected computers. Then, before the infection, there were $5(\mathrm{~m}+\mathrm{n}) / 2$ cables, and after the disconnection, there were $3 \mathrm{n} / 2$ cables (from which it follows that $\mathrm{n}$ is even). The difference between these numbers is 26, leading to the equation $5 \mathrm{~m}+2 \mathrm{n}=52$. This equation has two solutions in natural numbers where $\mathrm{n}$ is even (this can be proven by enumeration): $\mathrm{m}=4, \mathrm{n}=16$ and $\mathrm{m}=8, \mathrm{n}=6$. The first solution does not work: even if all infected computers were connected only to healthy ones, a maximum of $4 \cdot 5=20$ cables would have to be disconnected, not 26.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. On the table, there are candies of three types: caramels, toffees, and lollipops. It is known that there are 8 fewer caramels than all the other candies, and there are 14 fewer toffees than all the other candies. How many lollipops are on the table? Be sure to explain your answer.
|
# Solution.
Method 1. Since there are 8 fewer caramels than other candies, there are 4 fewer caramels than half of the candies. Since there are 14 fewer toffees than all other candies, there are 7 fewer toffees than half of the candies. Thus, if we remove all caramels and toffees, 4 + 7 = 11 candies will remain. Since the remaining candies are exactly the lollipops, there are 11 lollipops.
## Method 2.
Let $л, \kappa, u$ be the number of lollipops, caramels, and toffees, respectively.
According to the problem,
$\pi + \kappa = 8 + u$,
$\pi + u = 14 + \kappa$.
Adding these equations, we get: $2 \pi + \kappa + u = 22 + \kappa + u$
Therefore, $2 \pi = 22$, from which $\pi = 11$.
## Grading Criteria.
Complete correct solution - 7 points.
One arithmetic error in solving the system, but the solution method is correct - 4-5 points.
Correct answer and verification that it fits - 3 points.
The problem is solved with a specific example (e.g., "let there be 30 candies in total, then...") - 2 points.
Correct answer without justification - 2 points.
Other cases - 0 points.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Given an angle of $13^{0}$. How to obtain an angle of $11^{0}$?
|
3. One possible option: lay off the angle of $13^{0}$, 13 times, then the difference between the straight angle and the obtained angle will give the required angle $\left(180^{\circ}-13 \cdot 13^{0}=11^{\circ}\right)$
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3. CONDITION
Given a right triangle $A B C$ with legs $A C=3$ and $B C=4$. Construct triangle $A_{1} B_{1} C_{1}$ by sequentially moving point $A$ a certain distance parallel to segment $B C$ (point $A_{1}$ ), then point $B-$ parallel to segment $A_{1} C$ (point $B_{1}$ ), and finally point $C$ - parallel to segment $A_{1} B_{1}$ (point $C_{1}$ ). What is the length of segment $B_{1} C_{1}$, if it turns out that angle $A_{1} B_{1} C_{1}$ is a right angle and $A_{1} B_{1}=1$?
|
Solution. When a vertex of a triangle is moved parallel to its base, the area of the triangle does not change. Therefore, we sequentially obtain the equality of the areas of triangles $A B C, A_{1} B C, A_{1} B_{1} C$, and finally, $A_{1} B_{1} C_{1}$. Thus, $B_{1} C_{1}=2 S / A_{1} B_{1}=12$.
Answer: 12.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 5. CONDITION
Vladislav Vladimirovich, taking less than 100 rubles, went for a walk. Entering any cafe and having at that moment $m$ rubles $n$ kopecks, he spent $n$ rubles $m$ kopecks ( $m$ and $n$ - natural numbers). What is the maximum number of cafes Vladislav Vladimirovich could visit?
|
Solution. Method one. Let Vladislav Vladimirovich have $a$ rubles $b$ kopecks upon entering the first cafe. It is clear that $b \leqslant a$. Then upon exiting, he will have $a-b-1$ rubles and $b-a+100$ kopecks. Let $a-b=t \leqslant 99$. Thus, Vladislav Vladimirovich now has $t-1$ rubles and $100-t$ kopecks. The condition for the possibility of visiting the second cafe is $t-1 \geqslant 100-t$, or (since $t$ is an integer) $t \geqslant 51$. After visiting the second cafe, Vladislav Vladimirovich has $2 t-102$ rubles and $201-2 t$ kopecks. To be able to visit the third cafe, it is necessary and sufficient that $t \geqslant 76$. Similarly, to visit the fourth cafe, it is necessary and sufficient that $t \geqslant 89$, to visit the fifth $-t \geqslant 95$, to visit the sixth $-t \geqslant 98$, and to visit the seventh $-t$ must be greater than 99. The latter is impossible, while the previous inequality is feasible. Therefore, the answer is 6.
Method two. To visit a cafe, it is necessary that the number of rubles in Vladislav Vladimirovich's wallet is not less than the number of kopecks. Let the difference between the number of rubles and kopecks be $p$ $(0 \leqslant p$ (in the absence of an example for 6 cafes) | 4 points |
| an example for 6 cafes is provided, but its optimality is not proven | 3 points |
| any examples for fewer than 6 cafe visits or proof that 8 or more cafes cannot be visited | 0 points |
| answer without justification | 0 points |
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Option 1.
In the Ivanov family, both the mother and the father, and their three children, were born on April 1st. When the first child was born, the parents' combined age was 45 years. The third child in the family was born a year ago, when the sum of the ages of all family members was 70 years. How old is the middle child now, if the sum of the ages of the children is 14 years?
|
Answer: 5.
Solution.
If the first child is older than the second child by $x$ years, and the middle child is older than the third child by $y$ years, then $70-45=3(x+y)+y$, because the age of each parent and the eldest child increased by $(x+y)$ years by the time the third child was born, and the age of the second child increased by $y$ years. Similarly, $(x+y+1)+(y+1)+1=14$. We have the system of equations $3x+4y=25, x+2y=11$, from which we find $x=3, y=4$.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. The children went to the forest to pick mushrooms. If Anya gives half of her mushrooms to Vitya, all the children will have the same number of mushrooms, and if instead Anya gives all her mushrooms to Sasha, Sasha will have as many mushrooms as all the others combined. How many children went to pick mushrooms
|
Answer: 6 children.
Solution: Let Anya give half of her mushrooms to Vitya. Now all the children have the same number of mushrooms (this means that Vitya did not have any mushrooms of his own). For Sanya to now get all of Anya's mushrooms, he needs to take the mushrooms from Vitya and Anya. Then he will have the mushrooms of three children: Vitya, Anya, and his own. The same number will be with the others, which means that besides Vitya, Anya, and Sanya, three more children went to the forest.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one corner, the second carpet $6 \times 6$ - in the opposite corner, and the third carpet $5 \times 7$ - in one of the remaining corners (all dimensions are in meters).
Find the area of the part of the hall covered by carpets in three layers (give the answer in square meters).
|
Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 units from the bottom side.
The first carpet intersects this rectangle horizontally between the 5th and 8th meters from the left side of the square room, and vertically between the 4th and 6th meters from the top side. In the end, we get a rectangle $2 \times 3$, the area of which is 6.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
|
Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
Notice that triangle $A B D$ is equal to triangle $E B D$ by three sides: $B D$ is a common side, $A D=D E, A B=B E$ from the equality of triangles $A B C$ and $E B D$. Then $\angle D A B=$ $\angle B E D=\angle B A C$ and $\angle A B D=\angle D B E=\angle A B E=\frac{1}{3} \cdot 360^{\circ}=120^{\circ}$.
Since $A B=B E$, triangle $A B E$ is isosceles with an angle of $120^{\circ}$, so $\angle B A E=\frac{1}{2}\left(180^{\circ}-120^{\circ}\right)=30^{\circ}$. Therefore,
$$
\angle B A C=\angle D A B=\angle D A E-\angle B A E=37^{\circ}-30^{\circ}=7^{\circ}
$$
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
|
Answer: -6.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the condition, it is clear that $x_{1} < 0$ and $x_{2} > 0$. Since $x_{1}$ and $x_{2}$ are the roots of the quadratic trinomial $f(x)$, by Vieta's theorem, we have $x_{1} \cdot x_{2} = 12b$, from which we get $b = \frac{x_{1} \cdot x_{2}}{12} < 0$.
Let $H$ be the point with coordinates $(3 ; 0)$ (Fig. 11). Clearly, in the isosceles triangle $A T C$, the segment $T H$ is the height, and therefore it is also the median. Thus, $3 - x_{1} = A H = H C = x_{2} - 3$, from which we get $x_{1} = 6 - x_{2}$.
Let $M$ be the point with coordinates $(0, 3)$. Since $T H = T M = 3$ and $T A = T B$, the right triangles $A T H$ and $B T M$ are equal by the leg and hypotenuse. Therefore, $3 - x_{1} = H A = M B = 3 - b$, that is, $x_{1} = b = \frac{x_{1} \cdot x_{2}}{12}$ (by Vieta's theorem), from which we find $x_{2} = 12$. Finally, $x_{1} = 6 - x_{2} = 6 - 12 = -6$ and $b = x_{1} = -6$.
Another solution. As in the previous solution, let the abscissas of points $A$ and $C$ be $x_{1}$ and $x_{2}$, respectively; we will also use the fact that point $B$ has coordinates $(0 ; b)$. Immediately, we understand that $O A = |x_{1}| = -x_{1}$, $O C = |x_{2}| = x_{2}$, and $O B = |b| = -b$.
Let's find the second intersection of the circle with the y-axis, let this be point $D$ with coordinates $(0 ; d)$ (Fig. 12). The chords $A C$ and $B D$ of the circle intersect at the origin $O$; from the properties of the circle, we know that $O A \cdot O C = O B \cdot O D$. We get $-x_{1} \cdot x_{2} = -b \cdot d$, from which, replacing $x_{1} \cdot x_{2}$ with $12b$ by Vieta's theorem, we get $d = 12$.
Fig. 12: to the solution of problem 10.7
It remains to note that triangle $BTD$ is isosceles, and the midpoint of its base, point $M$, has coordinates $(0 ; 3)$. Reflecting point $D(0 ; 12)$ relative to it, we get $B(0 ; -6)$.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Andrey, Boris, Vasily, Gennady, and Dmitry played table tennis in pairs such that every two of them played with every other pair exactly once. There were no draws in the tennis matches. It is known that Andrey lost exactly 12 times, and Boris lost exactly 6 times. How many times did Gennady win?
Om vem: Gennady won 8 times.
|
Solution. The first pair can be formed in $5 \times 4: 2=10$ ways, the second pair can be formed in $3 \times 2: 2=3$ ways. In total, we get $10 \times 3: 2=15$ games. Andrei played in 4 pairs, and they played with 3 pairs. Therefore, Andrei played $4 \times 3=12$ times. According to the problem, he lost 12 times, which means he lost all his games. Together with him, Boris lost 3 times in a pair. Since Boris won 6 times against Andrei when playing with Vasily (2 times), with Gennady, and with Dmitry, the rest of the games he lost, that is, 3 times with Andrei and once each with Vasily (against Gennady and Dmitry), with Gennady, and with Dmitry. Therefore, Gennady lost 3 times with Andrei and once with Boris, a total of 4 times. Thus, he won 8 times.
Criteria. If the solution is incorrect - 0 points.
If the reasoning is correct but there is a computational error - 3 points. If the solution is correct - 7 points.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4. Masha wrote a three-digit number on the board, and Vera wrote the same number next to it, but she swapped the last two digits. After that, Polina added the obtained numbers and got a four-digit sum, the first three digits of which are 195. What is the last digit of this sum? (The answer needs to be justified.)
Answer: 4.
|
Solution. Let Masha write the number $100 x+10 y+z$. Then Vera wrote the number $100 x+10 z+y$, and the sum of these numbers is $200 x+11 y+11 z$. For $x \leqslant 8$ this expression does not exceed 1798, and therefore cannot start with 195. Thus, $x=9$. Then $11(y+z)$ is a three-digit number starting with 15. Among three-digit numbers starting with 15, only 154 is divisible by 11, so the last digit of the sum is 4.
Note. Masha could have written the numbers 959, 968, 977, 986, or 995.
## Criteria
1 p. The correct answer is provided.
2 p. It is determined that Masha's number starts with the digit 9, but there is no further progress.
2 p. An example of a three-digit number that satisfies the condition of the problem is provided.
3 p. It is determined that Masha's number starts with the digit 9, and an example of a three-digit number that satisfies the condition of the problem is provided.
4 p. The correct answer and justification are provided.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.5.1. Points $A, B, C, D, E, F$ on the diagram satisfy the following conditions:
- points $A, C, F$ lie on the same line;
- $A B=D E=F C$
- $\angle A B C=\angle D E C=\angle F C E$
- $\angle B A C=\angle E D C=\angle C F E$
- $A F=21, C E=13$.
Find the length of segment $B D$.
|
# Answer: 5.
Solution. Note that triangles $A B C, D E C$ and $F C E$ are equal by the second criterion of triangle congruence.
Since $\angle D E C=\angle F C E$, lines $D E$ and $A F$ are parallel. Therefore, $\angle A C B=\angle C D E=$ $\angle C A B$, so all three triangles are isosceles, and their lateral sides are equal to 13.
It remains to calculate the answer:
$$
B D=B C-C D=13-A C=13-(A F-C F)=13-(21-13)=5 .
$$
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.6.1. In a row, there are 10 boxes. These boxes contain balls of two colors: red and blue. In some boxes, all the balls may be of the same color; there are no empty boxes. It is known that in each subsequent box (from left to right), there are no fewer balls than in the previous one. It is also known that there are no two boxes with the same set of red and blue balls. How many blue and how many red balls are in the rightmost box, if the total number of red balls in all the boxes is 11 and the total number of blue balls is 13?
|
Answer: 1 red ball, 3 blue balls.
Solution. Among all the boxes, there can be:
- a maximum of 2 boxes with one ball (one with a red ball, the other with a blue ball),
- a maximum of 3 boxes with two balls (one with two red balls, another with one red and one blue, and the third with two blue balls),
- a maximum of 4 boxes with three balls (one with three red balls, another with two red and one blue, the third with one red and two blue, and the fourth with three blue balls).
Thus, in the first nine boxes, there will be at least $2 \cdot 1 + 3 \cdot 2 + 4 \cdot 3 = 20$ balls. Then, in the rightmost box, there will be no more than $11 + 13 - 20 = 4$ balls. However, from the above reasoning, it follows that if a box contains 1, 2, or 3 balls, it must be one of the first nine boxes. Therefore, the rightmost box must contain exactly 4 balls, and the first nine boxes must contain exactly 20 balls.
It is not difficult to understand that the first nine boxes contain 10 red and 10 blue balls. Then, in the rightmost box, there will be $11 - 10 = 1$ red and $13 - 10 = 3$ blue balls. $\square$
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The distance between points A and B is 90 km. At 9:00, a bus left point A for point B at a speed of 60 km/h. Starting from 9:00, every 15 minutes, buses leave point B towards point A at a speed of 80 km/h. The bus that left point A, after traveling 45 km, reduces its speed to 20 km/h due to a breakdown and continues at this speed. A passenger in this bus is considering whether to continue the slow journey to point B or return to point A in the first passing bus. How much time does the passenger have to make a decision?
|
2. A bus leaving point A will travel 45 km in $\frac{45}{60}$ hours. During this time, a bus leaving point B at 9:00 will travel $\frac{45}{60} \cdot 80=60$ km and will be closer to point A than the bus leaving point A. A bus leaving point B at 9:15 will travel $\left(\frac{45}{60}-\frac{15}{60}\right) \cdot 80=40$ km and will be 5 km away from the bus leaving point A. Therefore, the time it will take for them to meet is $\frac{5}{20+80}=\frac{1}{20}$ hours. The passenger has 3 minutes to make a decision.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In the country, there are 20 cities. An airline wants to organize two-way flights between them so that from any city, it is possible to reach any other city with no more than $\mathrm{k}$ transfers. At the same time, the number of air routes should not exceed four. What is the smallest $\mathrm{k}$ for which this is possible?
|
5. $\mathrm{k}=2$. At least two transfers will be required. From an arbitrary city A, one can reach no more than four cities without a transfer, and with one transfer - no more than $4 \times 3=12$ cities. That is, if using no more than one transfer, one can fly to no more than 16 other cities, but 19 are required.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.1. Semyon was solving the quadratic equation $4 x^{2}+b x+c=0$ and found that its two roots are the numbers $\operatorname{tg} \alpha$ and $3 \operatorname{ctg} \alpha$ for some $\alpha$. Find $c$.
|
Answer: 12
Solution. By Vieta's theorem, $c / 4$ equals the product of the roots. Considering that $\operatorname{tg} \alpha \cdot \operatorname{ctg} \alpha=1$, we get $c / 4=3$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.5. On a line, 5 points $P, Q, R, S, T$ are marked, exactly in that order. It is known that the sum of the distances from $P$ to the other 4 points is 67, and the sum of the distances from $Q$ to the other 4 points is 34. Find the length of the segment $P Q$.
|
Answer: 11.
Solution. From the condition of the problem, it is known that
$$
P Q+P R+P S+P T=67 \quad \text { and } \quad Q P+Q R+Q S+Q T=34
$$
Let's find the difference of these quantities:
$$
\begin{aligned}
33 & =67-34=(P Q+P R+P S+P T)-(Q P+Q R+Q S+Q T)= \\
& =(P Q-Q P)+(P R-Q R)+(P S-Q S)+(P T-Q T)=0+P Q+P Q+P Q=3 \cdot P Q
\end{aligned}
$$
from which $P Q=11$.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Uncle bought a New Year's gift for each of his nephews, consisting of a candy, an orange, a pastry, a chocolate bar, and a book. If he had bought only candies with the same amount of money, he would have bought 224. He could have bought 112 oranges, 56 pastries, 32 chocolate bars, and 16 books with the same amount of money. How many nephews does Uncle have? Justify your answer.
|
2. Answer. 8. Solution. Let's express the prices of all items in terms of the price of a candy. An orange costs as much as two candies, a pastry - as 4 candies, a chocolate bar - as 224:32 = 7 candies, a book - as 14 candies. The total price of the gift is equal to the price of $1+2+4+7+14=28$ candies, and the number of nephews the uncle has is $224: 28=8$.
Grading criteria. Correct solution - 7 points. In all other cases - 0 points.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Two brothers sold a flock of sheep, receiving as many rubles for each sheep as there were sheep in the flock. Wishing to divide the proceeds equally, they began to take 10 rubles from the total sum in turns, starting with the older brother. After the older brother took 10 rubles again, the younger brother was left with less than 10 rubles. To ensure an equal division, the older brother gave the younger brother his knife. For how many rubles was the knife valued?
|
6. Answer: 2 rubles. Solution. Let there be $n$ sheep in the flock. Then the brothers earned $n^{2}$ rubles. From the condition, it follows that the number of tens in the number $n^{2}$ is odd. Represent the number $n$ as $10 k+m$, where $k-$ is the number of tens, and $m-$ is the number of units in it. Then $n^{2}=100 k^{2}+20 k m+m^{2}$. Thus, the oddness of the number of tens in the number $n^{2}$ is equivalent to the oddness of the number of tens in the number $m^{2}$. By checking the squares of single-digit numbers, we ensure that the number of tens is odd only for $4^{2}$ and $6^{2}$. In both of these cases, the number $n^{2}$ ends in 6, meaning the younger brother received 4 rubles less than the older brother. To ensure an equal division in this situation, the older brother should give the younger brother 2 rubles.
Grading criteria. Correct solution - 7 points. The fact about the oddness of the number of tens in $m^{2}$ is established - 2 points. In other cases - 0 points.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 2. Option 1.
Tourists Vitya and Pasha are walking from city A to city B at equal speeds, while tourists Katya and Masha are walking from city B to city A at equal speeds. Vitya met Masha at 12:00, Pasha met Masha at 15:00, and Vitya met Katya at 14:00. How many hours after noon did Pasha meet Katya?
|
Answer: 5.
Solution: The distance between Masha and Katya and their speeds do not change, and the speeds of Vitya and Pasha are equal. Vitya met Katya 2 hours after Masha, so Pasha will also meet Katya 2 hours after Masha, i.e., at 5:00 PM - 5 hours after noon.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 4. Variant 1
If from the discriminant of the quadratic polynomial $f(x)=a x^{2}+2 b x+c$ we subtract the discriminant of the quadratic polynomial $g(x)=$ $(a+1) x^{2}+2(b+2) x+c+4$, the result is 24. Find $f(-2)$.
|
Answer: 6.
Solution. We have: $D_{1}-D_{2}=4\left(b^{2}-a c-(b+2)^{2}+(a+1)(c+4)\right)=4(-4 b+4 a+c)=4 f(-2)$.
## Variant 2
If the discriminant of the quadratic trinomial $f(x)=a x^{2}+2 b x+c$ is subtracted from the discriminant of the quadratic trinomial $g(x)=$ $(a+1) x^{2}+2(b+2) x+c+4$, the result is 28. Find $f(-2)$.
Answer: 7.
## Variant 3
If the discriminant of the quadratic trinomial $f(x)=a x^{2}+2 b x+c$ is subtracted from the discriminant of the quadratic trinomial $g(x)=$ $(a+1) x^{2}+2(b+3) x+c+9$, the result is 16. Find $f(-3)$.
Answer: 4.
## Variant 4
If the discriminant of the quadratic trinomial $f(x)=a x^{2}+2 b x+c$ is subtracted from the discriminant of the quadratic trinomial $g(x)=$ $(a+1) x^{2}+2(b-3) x+c+9$, the result is 20. Find $f(3)$.
Answer: 5.
## Variant 5
If the discriminant of the quadratic trinomial $f(x)=a x^{2}+2 b x+c$ is subtracted from the discriminant of the quadratic trinomial $g(x)=$ $(a+1) x^{2}+2(b+4) x+c+16$, the result is 8. Find $f(-4)$.
Answer: 2.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Variant 1. Point $I$ is the center of the circle inscribed in triangle $A B C$ with sides $B C=6$, $C A=8, A B=10$. Line $B I$ intersects side $A C$ at point $K$. Let $K H$ be the perpendicular dropped from point $K$ to side $A B$. Find the distance from point $I$ to line $K H$.
|
Answer: 2.
Solution.
By the converse of the Pythagorean theorem, angle $C$ is a right angle. Then, triangles $B K C$ and $B K H$ are congruent by the hypotenuse and an acute angle. Therefore, these triangles are symmetric with respect to the line $B K$. Thus, the distance from $I$ to $K H$ is equal to the distance from $I$ to $C K$, i.e., the radius of the inscribed circle in triangle $A B C$. For a right triangle, this radius is given by $\frac{A C + C B - B A}{2} = 2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's number, if it is known that after he ran away, 3 people remained in the line? (After each command, one or several players ran away, after which the line closed, and there were no empty spaces between the remaining players.)
|
Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
|
Answer: 3.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals are bisected by their intersection point $L$, it is a parallelogram (in particular, $AC = DX$). Therefore, $DX \parallel AC$. Since $AC \parallel ED$ by the condition, the points $X, D, E$ lie on the same line.
Since $AC \parallel EX$, then $\angle EAX = \angle CAX = \angle AXE$, i.e., triangle $AEX$ is isosceles, $EA = EX$. Then
$$
ED = EX - XD = EA - AC = 15 - 12 = 3
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1 The sum of 100 numbers is 1000. The largest of these numbers was doubled, and some other number was decreased by 10. After these actions, the sum of all the numbers did not change. Find the smallest of the original numbers.
|
9.1 The sum of 100 numbers is 1000. The largest of these numbers was doubled, and some other number was decreased by 10. After these actions, the sum of all the numbers did not change. Find the smallest of the original numbers.
Otvet: 10.
$\boldsymbol{P e s h e n i e : ~ P u s t ь ~} M$ - the largest number, $t$ - some other number. According to the condition $2 M+t-10=M+t$, from which $M=10$. All other numbers are not greater than 10, and if at least one of them is less than 10, then the sum of all 100 numbers will be less than $100 \cdot 10=1000$. Therefore, all the numbers out of these 100 are equal to 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3 Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be?
|
9.3 Two parabolas with different vertices are the graphs of quadratic trinomials with leading coefficients $p$ and $q$. It is known that the vertex of each parabola lies on the other parabola. What can $p+q$ be equal to?
Answer: 0.
Solution: Let $\left(x_{1}, y_{1}\right)$ be the coordinates of the vertex of one parabola, $\left(x_{2}, y_{2}\right)$ - the other. Then the equations of the parabolas can be represented as
$$
y=p\left(x-x_{1}\right)^{2}+y_{1} \quad \text { and } y=q\left(x-x_{2}\right)^{2}+y_{2} \text {. }
$$
The point ( $x_{2}, y_{2}$ ) lies on the first parabola:
$$
y_{2}=p\left(x_{2}-x_{1}\right)^{2}+y_{1},
$$
and the point ( $x_{1}, y_{1}$ ) lies on the second parabola:
$$
y_{1}=q\left(x_{1}-x_{2}\right)^{2}+y_{2} \text {. }
$$
Adding the obtained equations, we find
$$
y_{2}+y_{1}=p\left(x_{2}-x_{1}\right)^{2}+y_{1}+q\left(x_{1}-x_{2}\right)^{2}+y_{2}
$$
From which (since $\left.\left(x_{2}-x_{1}\right)^{2}=\left(x_{1}-x_{2}\right)^{2}\right)$
$$
(p+q)\left(x_{1}-x_{2}\right)^{2}=0
$$
Since $x_{1} \neq x_{2}$ (if $x_{1}=x_{2}$, then from (*) $y_{2}=y_{1}$ and the vertices of the parabolas coincide), then $p+q=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Masha and Alina are playing on a $5 \times 5$ board. Masha can place one chip in some cells. After that, Alina covers all these cells with L-shaped pieces consisting of three cells (non-overlapping and not extending beyond the boundaries of the square, L-shaped pieces can only be placed along the grid lines). If Alina manages to cover all the cells with chips, she wins; otherwise, Masha wins. What is the minimum number of chips Masha needs to place so that Alina cannot win?
|
Answer: 9 chips.
Solution. Example. Masha can place chips in the cells indicated in the figure (a). Then, Alina will need nine corners, as one corner cannot cover more than one cell with a chip. However, nine corners without overlapping cannot be placed on the board, since $27>25$.
Estimate. If Masha places fewer than nine chips, then at least one of the cells indicated in the figure (a) will not have a chip. Then, Alina will be able to cover all cells of the board except for that one. For example, if there is no chip in cell 1,
a)
b) or 2, or 3 in the figure (b). Then, Alina can place 6 corners and completely cover part of the board, meaning all other chips will be covered. In the area shaded gray, two corners can be placed so that only one of the cells 1, 2, or 3 - the one without a chip - remains free. If the chip is not placed next to another side, the figure needs to be rotated.
Comment. A complete and justified solution is provided - 7 points. The correct answer and an example of the table are provided, but the estimate of the minimum value is missing or incorrect - 3 points. Only the answer is provided - 0 points.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. All natural numbers, the sum of the digits in the representation of which is divisible by 5, are listed in ascending order: $5,14,19,23,28,32, \ldots$ What is the smallest positive difference between consecutive numbers in this sequence? Provide an example and explain why it cannot be smaller.
---
The smallest positive difference between consecutive numbers in this sequence is 1. An example is the pair of numbers 14 and 15. The sum of the digits of 14 is $1 + 4 = 5$, which is divisible by 5. The sum of the digits of 15 is $1 + 5 = 6$, which is not divisible by 5. However, the next number in the sequence is 19, and the sum of its digits is $1 + 9 = 10$, which is divisible by 5. Therefore, the smallest difference is 1, as seen between 14 and 15.
To explain why it cannot be smaller: The smallest possible difference between two numbers is 1. If the difference were 0, the numbers would be the same, which is not possible for consecutive numbers in the sequence. If the difference were greater than 1, there would be a gap between the numbers, which means there would be a number in between that could potentially have a sum of digits divisible by 5, contradicting the requirement that the sequence lists all such numbers in ascending order. Therefore, the smallest possible difference is 1.
|
Answer. The smallest difference is 1, for example, between the numbers 49999 and 50000.
Solution. The difference cannot be less than 1, as we are talking about the difference between different natural numbers.
Comment. How to guess the solution.
It is clear that if two adjacent numbers differ only in the units place, then the difference between them is 5 (for example, 523 and 528). Therefore, the numbers must differ in other places as well. We can try to take the larger number as a round number, then the numbers will differ in at least two places. For example, take 50, the previous number is 46, and the difference is 4. If we take 500, the previous number is 497, and the difference is 3. It remains to choose such a number of zeros so that the difference is 1.
## Grading criteria.
- An example of the required numbers with a difference of 1 is provided - 7 points.
- Reasoning is provided that allows constructing consecutive numbers in the sequence with the minimum difference, and numbers with a difference of 2 are constructed, but numbers with a difference of 1 are not constructed - 4 points.
- Examples are provided showing that the difference can be less than 4, but without justification of minimality - 2 points.
- Other cases - $\mathbf{0}$ points.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task № 2
The captain's assistant, observing the loading of the ship, smoked one pipe after another from the start of the loading. When $2 / 3$ of the number of loaded containers became equal to $4/9$ of the number of unloaded containers, and the ship's bells struck noon, the old seafarer began to light another pipe. When he finished smoking it, the ratio of the number of loaded containers to the number of unloaded containers became the inverse of the ratio that existed before he started smoking that pipe. How many pipes did the second assistant smoke during the loading (assuming that the loading speed, as well as the smoking speed, remained constant throughout the process.)
|
Answer: The assistant smoked 5 pipes.
## Solution.
Let $x$ be the part of containers that were loaded by noon, and $y$ be the remaining part of containers. Then from the conditions we get: $\left\{\begin{array}{l}\frac{2}{3} x=\frac{4}{9} y \\ x+y=1\end{array} \Rightarrow x=\frac{2}{5}, \quad y=\frac{3}{5}\right.$.
After smoking one pipe, the ratio became reversed, meaning the part of loaded containers became equal to $3 / 5$, and thus, during this time, they loaded $1 / 5$ of the containers. Therefore, the captain's assistant smoked 5 pipes during the loading.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 4
Find the area of the figure defined by the inequality
$$
|x|+|y|+|x-y| \leq \mathbf{2}
$$
|
Answer: The area of the figure is 3 sq.units.
## Solution.
It is easy to see that if the point $(\boldsymbol{x}, \boldsymbol{y})$ satisfies the original inequality $|\boldsymbol{x}|+|\boldsymbol{y}|+|\boldsymbol{x} \boldsymbol{y}| \leq 2$, then the point $(-x,-\boldsymbol{y})$ will also satisfy this inequality, since $|-\boldsymbol{x}|+|-y|+|-\boldsymbol{x}+\boldsymbol{y}| \leq \mathbf{2}$ and $|\boldsymbol{x}|+|\boldsymbol{y}|+|-(x-y)| \leq \mathbf{2}$, or $|\boldsymbol{x}|+|\boldsymbol{y}|+|\boldsymbol{x}-\boldsymbol{y}| \leq \mathbf{2}$. Geometrically, this will mean the symmetry of the figure with respect to the origin.
Due to this, we will construct the figure in the 1st and 2nd quadrants of the coordinate plane and perform a central symmetry transformation.
We will open the modulus for points in the 1st quadrant. In this case, there are two possibilities: $\left\{\begin{array}{c}x \geq 0 \\ y \geq 0 \\ x-y \geq 0 \\ x+y+x-y \leq 2\end{array}\right.$ and $\left\{\begin{array}{c}x \geq 0 \\ y \geq 0 \\ x-y \leq 0 \\ x+y-x+y \leq 2\end{array}\right.$. In the first case, we get $x \leq 1$ and $y \leq x$.
Therefore, we have a triangle lying below the line $y=x$.
In the second case, $y \leq 1$ and $y \geq x$. Therefore, we have a triangle lying above the line $\mathrm{y}=\mathrm{x}$. Combining the figures, we get a square with sides lying on the coordinate axes and equal to 1. For points in the second quadrant, there is only one situation:
That is, $x-y \leq 1, y \geq x-1$. We get a triangle formed by the coordinate axes and the line $y=x-1$. As a result, we get the figure: The area of this figure is 3 (sq.units) as the area of two squares with side 1 and two right isosceles triangles with legs 1.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (7 points) In a bag, there are 15 balls (see the figure). Color each ball in one of three colors: blue, green, or red - so that two of the statements are true, and one is false:
- there is one more blue ball than red balls;
- there are an equal number of red and green balls;
- there are 5 more blue balls than green balls.
## Write in detail how you reasoned.
|
# Answer.
7 blue balls, 6 red balls, 2 green balls.
Solution. We will prove that the second statement cannot be true. Indeed, if the first and second statements are true, then if we remove one blue ball, the number of balls of each color should be equal.
But \(15-1=14\) balls cannot be evenly divided into 3 colors. Now, let's assume the second and third statements are true. Then if we remove 5 blue balls, the number of balls of each color should again be equal.
But \(15-5=10\) balls cannot be evenly divided into 3 colors.
Thus, only the first and third statements can be true.
We can reason in different ways from here.
First method. If we add 1 red ball to the bag, the number of blue and red balls will be equal, and if we add 5 more green balls, the number of balls of each color will be the same, specifically, there will be \((15+1+5):3=7\) balls of each color.
Now we can count how many balls of each color were in the bag: 7 blue balls, \(7-1=6\) red balls, and \(7-5=2\) green balls.
Second method. From the true statements 1 and 3, it follows that there are 4 fewer green balls than red balls. If we remove 5 blue balls and 4 red balls from the bag, the number of balls of each color will be the same, specifically, there will be \((15-5-4):3=2\) balls of each color. Thus, there were 2 green, 6 red, and 7 blue balls in the bag.
It is also possible to solve the problem using an equation.
## Grading Criteria.
- Any correct and complete solution (the correct statements are chosen, the number of balls of each color is calculated, and explanations are provided) 7 points.
- The correct statements are indicated but not justified, and the number of balls of each color is correctly found based on this - 4 points.
- The correct statements are justified, but the number of balls of each color is not found or is found incorrectly - 3 points.
- An incomplete enumeration of color options is performed, and the correct answer is found - 1 point.
- Only the answer is provided - 0 points.
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (7 points) Four girls are singing songs, accompanying each other. Each time, one of them plays the piano while the other three sing. In the evening, they counted that Anya sang 8 songs, Tanya - 6 songs, Olya - 3 songs, and Katya - 7 songs. How many times did Tanya accompany? Justify your answer.
|
Answer. Twice.
Solution. If we add up the specified number of songs sung, each song will be counted 3 times (from the perspective of each of the three singing girls). Thus, we can find out how many songs were sung in total: $(8+6+3+7): 3=8$. It is known that Tanya sang 6 out of 8 songs, so she accompanied $8-6=2$ times.
## Grading Criteria.
- Any correct solution - 7 points.
- The total number of songs sung is correctly found, but the answer to the problem is not obtained or is incorrect - 3 points.
- The correct answer is obtained by incomplete enumeration - 1 point.
- Only the answer is provided - 0 points.
Maximum score for all completed tasks - 35.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Between the digits of the number 987654321, signs + should be placed so that the sum is 99. In how many ways can this be achieved?
|
3. Answer: in two ways.
It is clear that there cannot be three-digit addends, and there must be at least one two-digit addend, since the sum of all digits is 45. Let there be one two-digit addend, and we group the digits \(a+1\) and \(a\). Then the two-digit addend is \(10(a+1) + a = 11a + 10\), and the sum of the remaining digits is \(45 - (a+1) - a = 44 - 2a\). In total, we get \(11a + 10 + 44 - 2a = 99\), which means \(a = 5\). This gives the solution \(9 + 8 + 7 + 65 + 4 + 3 + 2 + 1 = 99\).
Now let there be two two-digit addends. We will assume that in one case we group the digits \(b+1\) and \(b\), and in the other \(a+1\) and \(a\), where \(b > a+1\). The two-digit numbers are \(11b + 10\) and \(11a + 10\) respectively, and the sum of the single-digit addends is \(45 - (2b + 1) - (2a + 1) = 43 - 2(a + b)\). Together we have \(11(a + b) + 20 + 43 - 2(a + b) = 99\), from which \(a + b = 4\). Only one option fits, \(a = 1\), \(b = 3\), which gives the second solution \(9 + 8 + 7 + 6 + 5 + 43 + 21 = 99\).
There are no more solutions, as the sum of three two-digit addends will be no less than \(65 + 43 + 21\), which exceeds 99.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.3. Let $n$ - be a natural number. What digit stands immediately after the decimal point in the decimal representation of the number $\sqrt{n^{2}+n}$?
|
Answer: 4.
Solution. We will prove that $(n+0.4)^{2} < 0.8$. The last inequality is true for any natural $n$.
2) $n^{2}+n < (n+0.5)^{2} \Leftrightarrow n < n+0.25$, which is obvious.
Evaluation Criteria.
“+” A complete and justified solution is provided
“Ғ” The correct answer is provided, but only one of the two required inequalities is proven
“-” The correct answer is obtained by considering specific examples
“-” Only the answer is provided
“-” The problem is not solved or is solved incorrectly
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. The tsar has eight sons, and they are all fools. Every night the tsar sends three of them to guard the golden apples from the firebird. The tsareviches cannot catch the firebird, blame each other for it, and therefore no two of them agree to go on guard duty a second time. For how many nights can this continue at most?
|
Answer: 8 nights.
Solution. Evaluation. Consider any of the sons. In each guard, he is with two brothers, so after his three appearances, there will be one brother with whom he has not been on guard and will not be able to go, as there will be no third for them. This situation is "symmetric," meaning if son A has not been on guard with son B after his three appearances, then son B has also not been on guard with son A after his three appearances. Therefore, there are at least four pairs of sons who have not and will not be in the same guard. The total number of possible pairs is $\frac{8 \cdot 7}{2}=28$, so there can be no more than $28-4=24$ pairs in the guards. Since three pairs of sons are involved each night, the number of nights can be no more than $24: 3=8$.
Example. Let the sons be denoted as: 1, 2, 3, 4, A, B, C, D. Then eight triplets satisfying the problem's condition are: A12, B23, C34, D41, AB4, BC1, CD2, DA3.
An example can also be provided in the form of a diagram (see Fig. 10.5). Eight sons are represented by points, and eight triplets satisfying the problem's condition are represented by lines.
Grading criteria.
“+” A complete and justified solution is provided.
“士” A generally correct reasoning is provided, containing minor gaps or inaccuracies, and the correct answer is obtained.
“干” The correct answer and evaluation are provided, but the example is missing or incorrect.
“Ғ” The correct answer and example are provided, but the evaluation is missing or incorrect.
“-” Only the answer is provided.
“-” The problem is not solved or is solved incorrectly.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. Arkady, Boris, and Vasili decided to run the same distance, consisting of several laps. Arkady ran each lap 2 minutes faster than Boris, and Boris ran 3 minutes faster than Vasili, and all of them ran at a constant speed. When Arkady finished the distance, Boris had one lap left to run, and Vasili had two laps left. How many laps did the distance consist of? Provide all possible answers and explain why there are no others.
|
Answer: 6.
Solution: Let the distance be $n$ laps, and Arkady takes $t$ minutes per lap. Then in $n t$ minutes, Arkady runs the entire distance. In this time, Boris runs one lap less, and Vasily runs two laps less. Therefore, $n t=(n-1)(t+2)$ and $n t=(n-2)(t+5)$. From this, it follows that $2 n=t+2$ and $5 n=2 t+10$, so $n=5 n-4 n=(2 t+10)-2 \cdot(t+2)=6$. Note that in this case, $t=10$, and all conditions are satisfied.
Comment: Only the answer - 2 points.
Remark: From the text of the problem, it follows that the problem has at least one solution. Therefore, if it is proven that there can be no other solutions besides $n=6$, it is not necessary to check this solution.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's number, if it is known that after he ran away, 3 people remained in the line? (After each command, one or several players ran away, after which the line closed, and there were no empty spaces between the remaining players.)
|
Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
|
Answer: 3.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals are bisected by their intersection point $L$, it is a parallelogram (in particular, $AC = DX$). Therefore, $DX \parallel AC$. Since $AC \parallel ED$ by the problem's condition, the points $X, D, E$ lie on the same line.
Since $AC \parallel EX$, then $\angle EAX = \angle CAX = \angle AXE$, i.e., triangle $AEX$ is isosceles, $EA = EX$. Then
$$
ED = EX - XD = EA - AC = 15 - 12 = 3
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. What is the total number of emeralds in the boxes?
#
|
# Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. Then the emeralds are in the two remaining boxes, and there are a total of $5+7=12$.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.1. In the picture, nine small squares are drawn, with arrows on eight of them. The numbers 1 and 9 are already placed. Replace the letters in the remaining squares with numbers from 2 to 8 so that the arrows from the square with the number 1 point in the direction of the square with the number 2 (the number 2 can be in square $A$ or $B$), the arrows from the square with the number 2 point in the direction of the square with the number 3, and so on, the arrows from the square with the number 8 point in the direction of the square with the number 9.
Construct the correspondence.
- In square $A$
- In square $B$
- In square $C$
- In square $D$
- In square $E$
- In square $F$
- In square $G$
- stands the number 2.
- stands the number 3.
- stands the number 4.
- stands the number 5.
- stands the number 6.
- stands the number 7.
- stands the number 8.
|
Answer: In square $A$ there is the number 6, in $B-2$, in $C-4$, in $D-5$, in $E-3$, in $F-8$, in $G-7$.
Solution. Let's order all the squares by the numbers in them. This "increasing chain" contains all nine squares.
Notice that in this chain, immediately before $C$ can only be $E$ (only the arrows from $E$ point to $C$). Similarly, immediately before $E$ can only be $B$ (not $C$, because $C$ is after $E$). Immediately before $B$ can only be 1. Therefore, the numbers in squares $B, E, C$ are $2,3,4$ respectively. Then 5 is definitely in $D$, 6 is definitely in $A$, 7 is definitely in $G$, and 8 is definitely in $F$.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.4. Given a square $A B C D$. Point $L$ is on side $C D$ and point $K$ is on the extension of side $D A$ beyond point $A$ such that $\angle K B L=90^{\circ}$. Find the length of segment $L D$, if $K D=19$ and $C L=6$.
|
Answer: 7.
Solution. Since $ABCD$ is a square, then $AB=BC=CD=AD$.
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle ABL$ to $90^{\circ}$. Therefore, right triangles $ABK$ and $CBL$ are congruent by the acute angle and the leg $AB = BC$ (Fig. 1). Consequently, $AK = CL = 6$. Then
$$
LD = CD - CL = AD - CL = (KD - AK) - CL = KD - 2 \cdot CL = 19 - 2 \cdot 6 = 7
$$
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Russia, Germany, and France decided to build the Nord Stream 2 gas pipeline, 1200 km long, agreeing to finance this project equally. In the end, Russia built 650 kilometers of the pipeline, Germany built 550 kilometers of the pipeline, and France contributed its share in money. Germany received 1.2 billion euros from France. How much should Russia receive from France?
|
Solution. Each country was supposed to build 400 kilometers of the gas pipeline. Thus, Russia built 250 km of the pipeline for France, and Germany built 150 km of the pipeline. Therefore, the money from France should be distributed between Russia and Germany in a ratio of 5:3. Hence, Russia will receive 1.2 billion $\times \frac{5}{3}=2$ billion euros from France.
Answer: 2 billion euros.
## Recommendations for checking
It is derived that Russia will receive 1.2 billion $\times \frac{5}{3}$, but it is calculated incorrectly - 6 points.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The numbers $1,2,3,4,5,6,7,8,9$ are written into the cells of a $3 \times 3$ table. After that, all possible sums of numbers standing in adjacent (by side) cells are written down in a notebook. What is the smallest number of different numbers that could have been written in the notebook?
|
# Solution:
Consider the number in the central cell of the table. Next to it are 4 different neighbors - they give 4 different sums with the central number, so there are already at least 4 different sums written down. An example where there are exactly 4 of them exists (one is shown on the right, the sums are $8,9,10,11$).
## Criteria:
There is an example with 4 different sums (to be checked!) - 2 points.
There is an estimate - that is, an explanation of why it is impossible to have fewer than 4 different sums - 3 points. Both a correct estimate and a correct example - 7 points.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. From a paper square $8 \times 8$, p seven-cell corners were cut out. It turned out that it was impossible to cut out any more such corners. For what smallest $n$ is this possible? A seven-cell corner is obtained by cutting out a $3 \times 3$ square (in cells) from a $4 \times 4$ square.
|
Answer: $\mathrm{n}=3$.
Instructions. Example. The figures show two examples of placing three corners so that no more can be cut out.
Cells belonging to the same corner are marked with the same number. Other examples are possible.
Estimate. It remains to show that if two corners have been cut out, then another one can be cut out. If there is a row and a column with no cells of the corners, then taking the cell at their intersection as the "vertex" of the corner and directing the sides of the corner towards the more distant sides, we get a corner that can be cut out. If such a row and column do not exist, then let the columns be covered (otherwise, rotate the board by $90^{\circ}$). In this case, there are two rectangles $1 \times 4$ covering the columns. Cutting the board along the horizontal axis of symmetry, we get a corner in each part. In one of the parts (in any), there will be a rectangle adjacent to the short edge, with no cells of the corner, and at least four cells away from the covering rectangle. One of its edge cells serves as the vertex of the corner that can be cut out.
Criteria. Example without estimate: 3 points.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. From a three-digit number, the sum of its digits was subtracted. The same operation was performed on the resulting number, and so on, 100 times. Prove that the result will be zero. (6 points)
|
Solution. Since $\overline{a b c}-(a+b+c)=9 \cdot(11 a+b)$, the first difference is divisible by 9. The sum of its digits is divisible by 9, which means the second, and similarly, all other differences will be divisible by 9.
The sum of the digits of a three-digit number divisible by 9 can be 9, 18, or 27. Therefore, after 100 operations, the number will either become 0 or decrease by at least 900. Thus, any number less than 900 will become zero.
Suppose the number is not less than 900. Then, after the first move, a number divisible by 9 will be obtained, ranging from $900-9=891$ to $999-27=972$. There are 9 such numbers. By checking, one can verify that they will also turn into 0 after 99 operations.
|
0
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
10.5. From the digits $1,2,3,4,5,6,7,8,9$, nine (not necessarily distinct) nine-digit numbers are formed; each digit is used exactly once in each number. What is the maximum number of zeros that the sum of these nine numbers can end with? (N. Agakhanov)
|
Answer: Up to 8 zeros.
Solution: We will show that the sum cannot end with 9 zeros. Each of the numbers formed is divisible by 9, since the sum of its digits is divisible by 9. Therefore, their sum is also divisible by 9. The smallest natural number divisible by 9 and ending with nine zeros is $9 \cdot 10^{9}$, so the sum of our numbers is not less than $9 \cdot 10^{9}$. This means that one of them is not less than $10^{9}$, which is impossible.
It remains to show how to form numbers whose sum ends with eight zeros. For example, we can take eight numbers equal to 987654321, and one number 198765432. Their sum is $81 \cdot 10^{8}$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. Let $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1$. Find all values that the number $x+y$ can take, and prove that no other values are possible.
|
Solution: Consider the function $f(x)=x+\sqrt{x^{2}+1}$. Since for any real $x$ we have $\sqrt{x^{2}+1}>\sqrt{x^{2}}=|x| \geqslant -x$, this function is positive everywhere. Let $a>0$. Solving the equation $f(x)=a$, we find that the value $a$ is taken by the function at the unique point $x=\frac{a^{2}-1}{2 a}$. Our equation has the form $f(x) f(y)=1$. Therefore, if $f(x)=a$, then $f(y)=1 / a$. Then $x=\frac{a^{2}-1}{2 a}$, $y=\frac{\frac{1}{a^{2}}-1}{2 \cdot \frac{1}{a}}=\frac{1-a^{2}}{2 a}$. Hence, $x+y=0$.
Answer: 0.
Note: The problem has other solutions, including those involving the trigonometric substitution $x=\operatorname{tg} \alpha, y=\operatorname{tg} \beta$.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct justified answer | 7 points |
| In the correct solution approach, errors are made in transformations, possibly leading to an incorrect answer | 4 points |
| The idea of considering the function $f(x)=x+\sqrt{x^{2}+1}$, not brought to the solution of the problem | 2 points |
| Correct answer supported by correct examples (in any number), but not proven in the general case | 1 point |
| Correct answer without justification or incorrect answer | 0 points |
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
8.3. The numbers $a, b, c$ satisfy the relation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$. Find $(a+b)(b+c)(a+c)$.
|
Answer: 0. Solution: Move $\frac{1}{a}$ to the right side, we get $\frac{b+c}{b c}=\frac{-(b+c)}{a(a+b+c)}$. If $b+c \neq 0$, then we will have (multiplying by the denominator)
$$
a^{2}+a b+a c+b c=0 \Leftrightarrow a(a+b)+c(a+b)=0 \Leftrightarrow(a+b)(a+c)=0
$$
Thus, in any case $(a+b)(b+c)(a+c)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one corner, the second carpet $6 \times 6$ - in the opposite corner, and the third carpet $5 \times 7$ - in one of the remaining corners (all dimensions are in meters).
Find the area of the part of the hall covered by carpets in three layers (give the answer in square meters).
|
Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 units from the bottom side.
The first carpet intersects this rectangle horizontally between the 5th and 8th meters from the left side of the square room, and vertically between the 4th and 6th meters from the top side. In the end, this results in a rectangle $2 \times 3$, the area of which is 6.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
|
Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
Notice that triangle $A B D$ is equal to triangle $E B D$ by three sides: $B D$ is a common side, $A D=D E, A B=B E$ from the equality of triangles $A B C$ and $E B D$. Then $\angle D A B=$ $\angle B E D=\angle B A C$ and $\angle A B D=\angle D B E=\angle A B E=\frac{1}{3} \cdot 360^{\circ}=120^{\circ}$.
Since $A B=B E$, triangle $A B E$ is isosceles with an angle of $120^{\circ}$, so $\angle B A E=\frac{1}{2}\left(180^{\circ}-120^{\circ}\right)=30^{\circ}$. Therefore,
$$
\angle B A C=\angle D A B=\angle D A E-\angle B A E=37^{\circ}-30^{\circ}=7^{\circ}
$$
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
|
Answer: -6.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the condition, it is clear that $x_{1} < 0$ and $x_{2} > 0$. Since $x_{1}$ and $x_{2}$ are the roots of the quadratic trinomial $f(x)$, by Vieta's theorem, we have $x_{1} \cdot x_{2} = 12b$, from which we get $b = \frac{x_{1} \cdot x_{2}}{12} < 0$.
Let $H$ be the point with coordinates $(3 ; 0)$ (Fig. 11). Clearly, in the isosceles triangle $A T C$, the segment $T H$ is the height, and therefore it is also the median. Thus, $3 - x_{1} = A H = H C = x_{2} - 3$, from which we get $x_{1} = 6 - x_{2}$.
Let $M$ be the point with coordinates $(0, 3)$. Since $T H = T M = 3$ and $T A = T B$, the right triangles $A T H$ and $B T M$ are equal by the leg and hypotenuse. Therefore, $3 - x_{1} = H A = M B = 3 - b$, that is, $x_{1} = b = \frac{x_{1} \cdot x_{2}}{12}$ (by Vieta's theorem), from which we find $x_{2} = 12$. Finally, $x_{1} = 6 - x_{2} = 6 - 12 = -6$ and $b = x_{1} = -6$.
Another solution. As in the previous solution, let the abscissas of points $A$ and $C$ be $x_{1}$ and $x_{2}$, respectively; we will also use the fact that point $B$ has coordinates $(0 ; b)$. Immediately, we understand that $O A = |x_{1}| = -x_{1}$, $O C = |x_{2}| = x_{2}$, and $O B = |b| = -b$.
Let's find the second intersection of the circle with the y-axis, let this be point $D$ with coordinates $(0 ; d)$ (Fig. 12). The chords $A C$ and $B D$ of the circle intersect at the origin $O$; from the properties of the circle, we know that $O A \cdot O C = O B \cdot O D$. We get $-x_{1} \cdot x_{2} = -b \cdot d$, from which, replacing $x_{1} \cdot x_{2}$ with $12b$ by Vieta's theorem, we get $d = 12$.
Fig. 12: to the solution of problem 10.7
It remains to note that triangle $BTD$ is isosceles, and the midpoint of its base, point $M$, has coordinates $(0 ; 3)$. Reflecting point $D(0 ; 12)$ relative to it, we get $B(0 ; -6)$.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.4 The pirates, led by John Silver on Treasure Island, found Billy Bones' chest, which contained 40 coins worth 1 ducat each and 40 coins worth 5 ducats each. John Silver has not yet decided how to divide this money among all the pirates (he does not want to take anything for himself). For what maximum number of pirates can any of his decisions on dividing the coins be implemented (each pirate must receive a whole number of ducats; it is possible that some pirates will not receive any money at all)? Justify your answer.
|
Solution. We will show that if there are no more than 11 pirates, Silver can divide the coins as he wishes. Indeed, let the $i$-th pirate need to receive $S_{i}$ coins. $S_{i}=5 x_{i}+a_{i}$, for some integers $x_{i}$ and $a_{i}$ (where $a_{i}$ is the remainder of the division of $S_{i}$ by 5). Note that the sum of all $S_{i}$ is 240 ducats, so the sum of all $a_{i}$ is divisible by 5. Each of the numbers $a_{i}$ is no more than 4, so the sum of $a_{i}$ is no more than 44, and being divisible by 5, it is no more than 40. We will give out all $a_{i}$ in one-ducat coins - there will be enough. Now it remains to give each pirate a sum that is a multiple of 5 ducats. This is easily done since the number of one-ducat coins left will be a multiple of 5.
If there are 12 or more pirates, the captain will not be able to pay, for example, such sums: 10 pirates 4 ducats each, one 3 ducats, and one the rest, 197 ducats. Indeed, the first 11 pirates can only be paid with one-ducat coins, and for this, no less than 43 are needed.
Answer: 11 pirates.
| THERE IS IN THE SOLUTION | SCORE |
| :--- | :---: |
| Correct and justified answer | 7 points |
| Correct justification that for any number of pirates less than 12 and any decision by Silver, the distribution is possible plus an example of distribution for 12 people, which is not possible (if its impossibility is not proven) | 4 points |
| Correct justification that for any number of pirates less than 12 and any decision by Silver, the distribution is possible | 3 points |
| Correct example of when it will not be possible to distribute the coins for 12 pirates (with proof of impossibility) | 2 points |
| Correct example of when it will not be possible to distribute the coins for 12 pirates (without proof of impossibility) | 1 point |
| Correct answer without justification or with incorrect justification OR examples of the impossibility of distributing money for a number of pirates greater than 12 | 0 points |
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. Uncle Chernomor assigns 9 or 10 of his thirty-three bogatyrs (knights) to duty each evening. What is the smallest number of days after which it can happen that all bogatyrs have been on duty the same number of times?
|
# 10.5. 7.
Let \( m \) and \( n \) be the number of days when 9 and 10 heroes were on duty, respectively. Let \( k \) be the number of days each hero was on duty. Then \( 9m + 10n = 33k \). Note that from this, \( n \) must be divisible by 3.
Firstly, the equation \( 9m + 10n = 33 \) has no solutions. Indeed, \( n \) cannot be greater than 3, so \( n \) is either 0 or 3, but \( 33 - 10n \) is not divisible by 9.
Secondly, the equation \( 9m + 10n = 66 \) has a unique solution in non-negative integers \( m = 3, n = 4 \). This is easy to verify by substituting \( n \) with 0, 3, 6. Note that \( m + n = 7 \).
Furthermore, if \( k \geq 3 \), then from the inequalities \( 10(m + n) \geq 9m + 10n = 33k \geq 99 \) it follows that \( m + n > 8 \).
It remains to provide an example of a duty schedule for 7 days. Assign the heroes numbers from 1 to 33. First day: 1-9; second day: 10-18; third day: 19-27; fourth day: 27-33 and 1-3; in the remaining days, 10 heroes in order.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. Let all numbers $x, y, z$ be non-zero. Find all values that the expression
$$
\left(\frac{x}{|y|}-\frac{|x|}{y}\right) \cdot\left(\frac{y}{|z|}-\frac{|y|}{z}\right) \cdot\left(\frac{z}{|x|}-\frac{|z|}{x}\right)
$$
can take.
|
Solution: By the Pigeonhole Principle, among the numbers $x, y$, and $z$, there will be two numbers of the same sign. Then the corresponding bracket will be equal to 0, and the entire product will also be equal to 0.
Answer: Only the number 0.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| correct and justified answer | 7 points |
| some (not all) cases of modulus expansion are correctly analyzed | 3 points |
| an example is provided showing that 0 can be obtained | 1 point |
| correct answer without justification (or with incorrect justification) | 0 points |
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.6. All natural numbers from 1 to 20 were divided into pairs, and the numbers in each pair were added. What is the maximum number of the ten resulting sums that can be divisible by 11? Justify your answer.
|
Solution: The number 11 is the only number in the set that is divisible by 11, so adding it to any other number will disrupt divisibility by 11. Therefore, all 10 sums cannot be divisible by 11. One example where nine sums are divisible by 11 is as follows:
$(1,10),(2,20),(3,19),(4,18),(5,17),(6,16),(7,15),(8,14),(9,13),(11,12)$.
Answer: 9 sums.
Recommendations for checking:
| present in the work | points |
| :--- | :--- |
| Presence of a correct example and proof of its optimality | 7 points |
| There is a proof that 10 sums cannot be obtained (in the absence of an example with 9 sums) | 3 points |
| An example is provided showing that 9 sums can be obtained (in the absence of a proof of its optimality) | 1 point |
| Correct answer without justification (with incorrect justification) | 0 points |
| Examples of a smaller number of sums (in any quantity) | not evaluated |
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Seven friends entered a cafe and ordered 3 small cups of coffee, 8 medium cups, and 10 large cups. The volume of a small cup is half the volume of a medium cup, and the volume of a large cup is three times the volume of a small cup. How should the friends divide the cups of coffee among themselves so that everyone drinks an equal amount of coffee? Pouring coffee from one cup to another is not allowed.
|
Solution: Let's call the amount of coffee in a small cup a "norm". Then, in total, we have $3+8 \times 2+10 \times 3=49$ norms. Since there are seven friends, each should get 7 norms.
We divide as follows: 1 small + 2 large - 3 people; 2 medium + 1 large - 4 people.
Criteria. A correct example - 7 points. No additional reasoning is required. If the result is obtained that each person needs 7 "norms", but the example is not shown (or the example is incorrect) - $\mathbf{1}$ point.
Incorrect example, lack of example - 0 points.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In a football tournament, 12 teams participated. By September, they had played several games, and no two teams had played each other more than once. It is known that the first team played exactly 11 games. Three teams played 9 games each. One team played 5 games. Four teams played 4 games each. Two teams played only one game each. And
## MATHEMATICS
9th GRADE
the information about the twelfth team was lost. How many games did the 12th team play?
|
Answer: 5 games.
Solution. Let the first team K1 play 11 games - i.e., once with everyone. Teams K2 and K3 played 1 game each - these are games with team K1.
There are 9 teams left (K4-K12). Three of these teams (K4, K5, K6) played 9 games each. One of these games was with K1. And 8 with all teams K4-K12 (except themselves).
Teams K7, K8, K9, K10 played 4 games each. It is clear that these games were with teams K1, K4, K5, K6. (And there were no other games for these teams)
Team K11 played 5 games. 4 of them were with K1, K4, K5, K6. And it is certain that she did not play with teams K2, K3, K7, K8, K9, K10. Thus, she played the fifth game with team K12.
In total, team K12 played 5 games (with K1, K4-6, K11) (and did not play with K2, K3, K7-K10).
Criteria. Correct solution - 7 points.
Fully correctly described who played with whom, but there is no answer to the question of the problem (or it is somehow incorrect) - 6 points.
Noted that the team that played 11 games played with everyone (without further progress) - 0 points. If it is also noted that two teams played only with it - **1** point.
A school student can use graphs for their solution without explaining the terms used.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Given a convex quadrilateral ABCD. Point $M$ is the midpoint of side BC, and point $N$ is the midpoint of side CD. Segments AM, AN, and $MN$ divide the quadrilateral into four triangles, the areas of which, written in some order, are consecutive natural numbers. What is the maximum possible area of triangle $\mathrm{ABD}$?
|
Answer: 6.
Solution. Estimation. Let $n, n+1, n+2, n+3$ be the areas of the four triangles. Then the area of quadrilateral $ABCD$ is $4n+6$. $MN$ is the midline of triangle $BCD$, so $S_{BCD} = 4S_{MCN}$, but $S_{MCN} \geq n$, hence $S_{BCD} \geq 4n$. Then $S_{ABD} = S_{ABCD} - S_{BCD} \leq 6$.
Example. If $ABCD$ is an isosceles trapezoid with bases $BC=4$ and $AD=6$ and height 2, then $S_{ABD} = 6$. The areas of triangles $CMN$, $ABM$, $AND$, and $AMN$ are 1, 2, 3, and 4, respectively, i.e., they are consecutive natural numbers.
Criteria. A fully correct solution - 7 points.
A correct solution, but the example is not shown to be suitable (the fact that the areas of the four specified triangles are consecutive natural numbers is not justified) - 5 points.
It is proven that the area does not exceed 6, but there is no example where the area is 6 (or it is incorrect) - 4 points.
An example is obtained where the area of triangle $ABD$ is 6 (with justification and verification of calculations), but there is no explanation of why it cannot be more than 6 - **3** points.
The example is correct, but without justification that it fits the conditions of the problem (that the areas of the required triangles are 4 consecutive numbers) - **1** point.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Three jumps of a two-headed dragon are equal to 5 jumps of a three-headed one. But in the time it takes for the two-headed dragon to make 4 jumps, the three-headed one makes 7 jumps. Which one runs faster? Justify your answer.
#
|
# Answer. Three-headed.
Solution. Consider the time it takes for a two-headed dragon to make 3*4=12 jumps. In this time, a three-headed dragon makes $3 * 7=21$ jumps. Since 12=4*3, 12 jumps of the two-headed dragon are equal to 4*5=20 jumps of the three-headed dragon. Thus, in the same amount of time, the three-headed dragon moves 21 jumps, while the two-headed dragon moves 20 jumps of the three-headed dragon. Therefore, the three-headed dragon runs faster.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Three lines intersect at one point 0. Outside these lines, a point M is taken and perpendiculars are dropped from it to them. The points $\mathrm{H}_{1}, \mathrm{H}_{2}$ and $\mathrm{H}_{3}$ are the bases of these perpendiculars. Find the ratio of the length of the segment OM to the radius of the circle circumscribed around the triangle $\mathrm{H}_{1} \mathrm{H}_{2} \mathrm{H}_{3}$.
Answer: 2.
|
Solution
$2, \prime \prime$
First, consider two intersecting lines and a point $M$ lying outside these lines. $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$ are the feet of the perpendiculars dropped from point $\mathrm{M}$ to these lines. The possible cases are:
1) $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$ lie on opposite sides of the line $0 M$
2) $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$ lie on the same side of the line $0 M$;
3) $\mathrm{H}_{1}$ or $\mathrm{H}_{2}$ coincides with point 0
In the first case, $\angle \mathrm{MH} \mathrm{H}_{1} \mathrm{O} + \angle \mathrm{MH} \mathrm{H}_{2} \mathrm{O} = 90^{\circ} + 90^{\circ} = 180^{\circ}$. The quadrilateral $\mathrm{MH}_{1} \mathrm{OH}_{2}$ is cyclic, i.e., points $\mathrm{M}, \mathrm{H}_{1}, \mathrm{O}$, and $\mathrm{H}_{2}$ lie on the same circle, with the diameter of this circle being segment $OM$. In the second case, the quadrilateral $\mathrm{MH}_{1} \mathrm{H}_{2} \mathrm{O}$ is also cyclic. Points $M, \mathrm{H}_{1}, \mathrm{O}$, and $\mathrm{H}_{2}$ again lie on the same circle with diameter $OM$. In the third case, it is obvious that for points $M, H_{1}, 0$, and $H_{2}$, we get the same result.
Draw a third line and mark point $\mathrm{H}_{3}$. Then, obviously, points $\mathrm{H}_{1}$, $\mathrm{H}_{2}$, and $\mathrm{H}_{3}$ lie on the circle with diameter $OM$, and the desired ratio is 2.
## Comments on Grading
The problem is solved, but in proving that $\mathrm{M}, \mathrm{H}_{1}, \mathrm{O}$, and $\mathrm{H}_{2}$ lie on the same circle, only case 1) or case 2) is considered -5 points.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-2. In square $A B C D$, a segment $C E$ is drawn such that the angles shown in the diagram are $7 \alpha$ and $8 \alpha$. Find the value of angle $\alpha$ in degrees. If necessary, round the answer to 0.01 or write the answer as a common fraction.
|
Answer: $9^{\circ}$.
Solution. In triangle $D F E$, the angles are $7 \alpha, 8 \alpha$ and $45^{\circ}$.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha + 8 \alpha + 45^{\circ} = 180^{\circ}$, from which $\alpha = \frac{1}{15} \cdot 135^{\circ} = 9^{\circ}$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
|
Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then $\angle A C O=\angle O A C=90^{\circ}-67^{\circ}=23^{\circ}$ (here we used the fact that triangle $A C D$ is a right triangle: angle $A C D$, which subtends the diameter, is a right angle).
Thus, $x=32^{\circ}-23^{\circ}=9^{\circ}$.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. On the board, there are 2017 digits. From these, several numbers were formed, the sums of the digits of these numbers were calculated, and then the sum of all the numbers was subtracted by the sum of the sums of their digits. The resulting number was broken down into digits, and the above operation was repeated again. After performing this operation several times, only one digit remained on the board for the first time. What is this digit?
|
Solution. Since the difference between a number and the sum of its digits is divisible by 9, the first operation will result in a number that is a multiple of 9. Moreover, if we take the sum of several numbers and subtract the sum of the digits of these numbers, the result will also be a multiple of 9. Continuing the calculations, we will get numbers that are multiples of 9 but decreasing in absolute value. In the end, the first single-digit number we obtain must be divisible by 9, and that is only 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. The numbers from 1 to 20 are arranged in a circle. We will paint a number blue if it is divisible without a remainder by the number to its left. Otherwise, we will paint it red. What is the maximum number of blue numbers that could be in the circle?
|
# Solution.
Evaluation. It is obvious that numbers cannot be blue if the number to their left is greater than or equal to 11. That is, no more than 10 numbers can be blue.
Example. As an example, both any correct arrangement and a correct algorithm are counted. An example of a correct algorithm.
1) write down the number 20
2) if the last written number $=2 \mathrm{k}$, then write the number $\mathrm{k}$ to its left, otherwise write the largest unused number to its left.
3) while there are unused numbers - repeat step 2.
## Criteria.
7 points for having both the evaluation and a correct example
4 points for only the example
3 points for only the evaluation.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. It is known that $a b c=1$. Calculate the sum
$$
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a}
$$
|
Solution: Note that
$$
\frac{1}{1+a+a b}=\frac{1}{a b c+a+a b}=\frac{1}{a(1+b+b c)}=\frac{a b c}{a(1+b+b c)}=\frac{b c}{1+b+b c}
$$
Similarly, by replacing 1 with the number $a b c$, we have
$$
\frac{1}{1+c+c a}=\frac{a b}{1+a+a b}=\frac{a b^{2} c}{1+b+b c}=\frac{b}{1+b+b c} .
$$
Then
$$
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a}=\frac{b c}{1+b+b c}+\frac{1}{1+b+b c}+\frac{b}{1+b+b c}=1
$$
Answer: 1.
Recommendations for checking:
| present in the work | points |
| :--- | :--- |
| Correct justified answer | 7 points |
| Correct answer obtained by considering special cases (in any number) | 1 point |
| Correct answer without justification OR incorrect answer OR algebraic transformations not leading to the answer | 0 points |
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.6. On a certain segment, its endpoints and three internal points were marked. It turned out that all pairwise distances between the five marked points are different and are expressed in whole centimeters. What is the smallest possible length of the segment? Justify your answer.
|
Solution: There are 5 points, so there are 10 pairwise distances. If all of them are expressed as positive whole numbers of centimeters and are distinct, at least one of them is not less than 10. Therefore, the length of the segment is not less than 10. Suppose the length of the segment is exactly 10. Then the pairwise distances are all numbers from 1 to 10. Among them, there is a length of 9. This means that there must be a point that is 1 cm away from one of the segment's ends, let's say from the left end - see the top diagram, marked points are black. Then points that are 1 cm away from the right end and 2 cm away from the left (they are crossed out in the diagram) cannot be marked, otherwise there would be two segments of 1 cm. To achieve a length of 8 cm, a point must be marked 2 cm away from the right end. Now, points that are 1 or 2 cm away from the marked points (white points in the diagram) cannot be marked. Only two points remain, but each of them is the midpoint of a segment between some already marked points, so they cannot be marked either. Therefore, it is impossible to mark 5 points on a segment of length 10 in the required manner, and the length of the segment must be strictly greater than 10 cm.
This length is a whole number of centimeters, as it represents the distance between marked points - the ends of the segment. Therefore, it is not less than 11 cm. We will show the required arrangement of marked points on a segment of length 11. Let the marked points (from left to right) be points \(A, B, C, D\), and \(E\). Let \(AB = 1\), \(BC = 3\), \(CD = 5\), \(DE = 2\) - all distances in centimeters (see the bottom diagram). Then \(AC = 4\), \(AD = 9\), \(AE = 11\), \(BD = 8\), \(BE = 10\), \(CE = 7\) - all 10 distances are different.
Answer: 11 centimeters.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |
| Correct justified answer | 7 points |
| Proven that marking points in the required manner on a segment of length 10 is impossible, but no example on a segment of length 11 cm | 5 points |
| Provided an example of the correct arrangement of points on a segment of length 11, but not justified that a length of 10 is impossible | 2 points |
| Correct answer without justification | 1 point |
| Incorrect examples of point placement OR correct placement of points on segments longer than 11 cm | 0 points |
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 4.1. Condition:
In front of the elevator stand people weighing 50, 51, 55, 57, 58, 59, 60, 63, 75, and 140 kg. The elevator's load capacity is 180 kg. What is the minimum number of trips needed to get everyone up?
|
Answer: 4 (or 7)
## Solution.
In one trip, the elevator can move no more than three people, as the minimum possible weight of four people will be no less than $50+51+55+57=213>180$. Note that no one will be able to go up with the person weighing 140 kg, so a separate trip will be required for his ascent. For the remaining nine people, at least $\frac{9}{3}=3$ trips will be needed, as no more than 3 people can fit in the elevator. The trips will be as follows: $(50,51,75),(55,57,63),(58,59,60)$ and (140).
In this problem, 7 is also accepted as a correct answer if the solution takes into account not only the elevator's ascents but also its descents to pick up people.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 4.2. Condition:
In front of the elevator stand people weighing 150, 60, 70, 71, 72, 100, 101, 102, and 103 kg. The elevator's load capacity is 200 kg. What is the minimum number of trips needed to get everyone up?
|
Answer: 5 (or 9)
## Solution
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $60+70+71=201>200$. Note that no one will be able to go up with the person weighing 150 kg, so a separate trip will be required for his ascent. For the remaining nine people, at least $\frac{8}{2}=4$ trips will be needed, as no more than 2 people can fit in the elevator. Thus, the trips will be: $(60,103),(70,102),(71,101),(72,100)$ and (150).
In this problem, 9 is also considered a correct answer if the solution takes into account not only the elevator's ascents but also its descents to pick up people.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 4.3. Condition:
In front of the elevator stand people weighing 150, 62, 63, 66, 70, 75, 79, 84, 95, 96, and 99 kg. The elevator's load capacity is 190 kg. What is the minimum number of trips needed to get everyone up?
|
# Answer: 6 (or 11)
## Solution.
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $62+63+66=191>190$. Note that no one will be able to go up with the person weighing 150 kg, so a separate trip will be required for his ascent. For the remaining ten people, at least $\frac{10}{2}=5$ trips will be needed, as no more than 2 people can fit in the elevator. Thus, the trips will be: (62, 99), $(63,96),(66,95),(70,84),(75,79)$ and (150).
In this problem, 11 is also considered a correct answer if the solution takes into account not only the elevator's ascents but also its descents to pick up people.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 4.4. Condition:
In front of the elevator stand people weighing $130,60,61,65,68,70,79,81,83,87,90,91$ and 95 kg. The elevator's load capacity is 175 kg. What is the minimum number of trips needed to get everyone up?
|
# Answer: 7 (or 13)
## Solution.
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $60+61+65=186>175$. Note that no one will be able to go up with the person weighing 135 kg, so a separate trip will be required for his ascent. Six trips will be needed for the remaining ten people, as no more than 2 people can fit in the elevator. The trips will be as follows: $(60,95),(61,91),(65,90),(68,87),(70,83),(79,81)$ and $(130)$.
In this problem, 13 is also considered a correct answer if the solution takes into account not only the elevator's ascents but also its descents to pick up people.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 5.1. Condition:
In the warehouse, there are 8 cabinets, each containing 4 boxes, each with 10 mobile phones. The warehouse, each cabinet, and each box are locked. The manager has been tasked with retrieving 52 mobile phones. What is the minimum number of keys the manager should take with them?
|
# Answer: 9
Solution. To retrieve 52 phones, at least 6 boxes need to be opened. To open 6 boxes, no fewer than 2 cabinets need to be opened. Additionally, 1 key to the warehouse is required. In total, $6+2+1=9$ keys need to be taken by the manager.
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 8.2. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 2 meters away from me."
- The rightmost in the row said: "My fellow tribesman in this row stands 2 meters away from me."
It is known that there are two knights and two liars in the row. What distances could the second and third islanders from the left have mentioned? List all possible options.
## Options for matching:
Second islander $\quad 1$ m
Third islander $\quad 2$ m
3 m
$4$ m
#
|
# Answer:
The second islander -1 m
The third islander -1 m.
## Solution.
Let's number the islanders from left to right. Suppose the first one is a knight. Then from his statement, it follows that the third one is also a knight; by the principle of exclusion, the second and fourth must be liars. The fourth said that his fellow islander stands two meters away. This is true, but the fourth is a liar, which leads to a contradiction. Therefore, the first one must be a liar.
Notice that a similar situation is created for the fourth and the first: if we look at the line from right to left, the fourth will be the first in it, and the same reasoning will apply to him. That is, the fourth must also be a liar. It follows that the second and third are knights, as the liars are the first and fourth.
From the above, it follows that the second and third are knights, who will say "one meter."
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 8.3. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 1 meter away from me."
- The second from the left said: "My fellow tribesman in this row stands 2 meters away from me."
It is known that there are two knights and two liars in the row. What distances could the third and fourth islanders from the left have named? List all possible options.
## Options for matching:
Third islander $\quad 1$ m
Fourth islander $\quad 2$ m
$3 \mathrm{M}$
$4 \mathrm{M}$
#
|
# Answer:
The third islander -1 m; 3 m; 4 m.
The fourth islander -2 m.
## Solution.
Let's number the islanders from right to left. Suppose the first one is a knight. Then, from his statement, it follows that the second one is also a knight. However, the second one said that his fellow tribesman is two meters away from him, which is a lie, since the first one, who should be the fellow tribesman of the second, is standing next to him. Therefore, we conclude:
the first one is a liar.
Now, from the first one's statement, it follows that the second one must be a knight, otherwise he would be a liar and the first one would have told the truth. If the second one is a knight, then from his statement, it follows that the fourth one is also a knight. This leaves the third one to be only a liar.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9-1. Postman Pechkin calculated that he walked half the distance (at a speed of 5 km/h) and only a third of the time he was cycling (at a speed of 12 km/h). Did he make a mistake in his calculations?
|
Answer. Mistaken.
Solution. Let's denote the entire distance Pechkin traveled as $2 S$ km. Then, on foot, he covered a distance of $S$ km and spent $S / 5 = 0.2 S$ (hours) on it. According to the problem, this constituted $2 / 3$ of the total time spent, meaning the entire journey took $0.2 S : 2 / 3 = 0.3 S$ (hours), and he cycled for $0.1 S$ hours. Therefore, his speed should be $S / (0.1 S) = 10$ (km/h).
Criteria. Only the answer - 0 points. Complete solution - 7 points.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9-2. The school volleyball team played several matches. After they won another match, the share of victories increased by $1 / 6$. To increase the share of victories by another 1/6, the volleyball players had to win two more consecutive matches. What is the minimum number of victories the team needs to achieve to increase the share of wins by another 1/6?
|
Answer: 6.
Solution: Let the team initially play $n$ matches, of which $k$ were won. Then, after the next win, the share of victories increased by $\frac{k+1}{n+1}-\frac{k}{n}=\frac{1}{6}$. Similarly, after two more wins, the increase was $\frac{k+3}{n+3}-\frac{k+1}{n+1}=\frac{1}{6}$. Simplifying each equation, we get the system
$$
\left\{\begin{array}{c}
n-k=\frac{n(n+1)}{6} \\
2(n-k)=\frac{(n+1)(n+3)}{6}
\end{array}\right.
$$
Dividing the second equation by the first, we get that $\frac{n+3}{n}=2$, from which $n=3$. Substituting this value into the first equation, we find that $k=3-3 \cdot 4 / 6=1$. Thus, at the beginning, the share of victories was $1 / 3$, after the next win - 2/4, and after two more wins, the team had $1+1+2=4$ victories in $3+1+2=6$ games. If the team wins $m$ more games, their share of victories will be $\frac{4+m}{6+m}$, which should match $\frac{4}{6}+\frac{1}{6}=\frac{5}{6}$. Solving the corresponding equation, we get that $m=6$.
Criteria: Only the answer - 0 points. The system of equations is set up - 5 points. Complete solution - 7 points.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. From the digits $1,2,3,4,5,6,7,8,9$, nine (not necessarily distinct) nine-digit numbers are formed; each digit is used exactly once in each number. What is the maximum number of zeros that the sum of these nine numbers can end with?
(N. Agakhanov)
|
Answer: Up to 8 zeros.
Solution: We will show that the sum cannot end with 9 zeros. Each of the numbers formed is divisible by 9, since the sum of its digits is divisible by 9. Therefore, their sum is also divisible by 9. The smallest natural number divisible by 9 and ending with nine zeros is $9 \cdot 10^{9}$, so the sum of our numbers is not less than $9 \cdot 10^{9}$. This means that one of them is not less than $10^{9}$, which is impossible.
It remains to show how to form numbers whose sum ends with eight zeros. For example, we can take eight numbers equal to 987654321, and one number 198765432. Their sum is $81 \cdot 10^{8}$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.2. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 10 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 11 girls in this parallel, and there are more boys than girls? Pizzas can be divided into parts.
|
# Answer: 11
Solution. Let the number of boys be $m$, and the number of pizzas that the girls received be $x$. If each boy had eaten as much as each girl, the boys would have eaten 5 pizzas. Then $m: 11 = 5: x$, from which we get $m x = 55$. The number 55 has only one divisor greater than 11, which is 55. Therefore, $m = 55, x = 1$, and the answer is $10 + 1 = 11$ pizzas (unfortunately, only one pizza was ordered for 11 girls, but this was by their choice).
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5.4. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 10 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 17 girls in this parallel, and there are more boys than girls? Pizzas can be divided into parts.
|
Answer: 11
Solution. Let the number of boys be $m$, and the number of pizzas that the girls got be $x$. If each boy had eaten as much as each girl, the boys would have eaten 5 pizzas. Then $m: 17 = 5: x$, from which $m x = 85$. The number 85 has only one divisor greater than 17, which is 85. Therefore, $m=85, x=1$, the answer is $10+1=11$ pizzas.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved two seats to the right, Galia had moved one seat to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. Which seat did Anya occupy before she got up?
|
# Answer: 2
Solution. Let's see how the seat number of everyone except Anya has changed. Varya's seat number increased by 2, Galia's decreased by 1, and the sum of Diana's and Eli's seat numbers did not change. At the same time, the total sum of the seat numbers did not change, so Anya's seat number must have decreased by 1. But then the seat left for her cannot be seat 5 (since there is no seat 6), so the seat left for her must be seat 1, and initially, she had seat 2.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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