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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
|
Problem 6.7. The Amur and Bengal tigers started running in a circle at 12:00, each at their own constant speed. By 14:00, the Amur tiger had run 6 more laps than the Bengal tiger. Then the Amur tiger increased its speed by 10 km/h, and by 15:00, it had run a total of 17 more laps than the Bengal tiger. How many meters is the length of the circle?
|
Answer: 1250.
Solution. In the first 2 hours, the Amur tiger ran 6 more laps, i.e., in 1 hour, it ran 3 more laps. If it had not increased its speed, in the first 3 hours, it would have run 9 more laps. However, the increase in speed resulted in an additional $17-9=8$ laps in the third hour. Since it increased its speed by 10 km/h, the additional distance would be 10 km/h $\cdot$ 1 h $=10$ km, and the length of one lap would then be $\frac{10}{8}$ km, or 1250 meters.
|
1250
|
Algebra
|
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
|
7.2 There were 25 sparrows sitting on two bushes. After 5 sparrows flew from the first bush to the second, and 7 sparrows flew away from the second bush, it turned out that there were twice as many sparrows left on the first bush as on the second. How many sparrows were there on each bush initially?
|
7.2 On the first bush, there were 17 sparrows, and on the second, there were 8.
After 7 sparrows flew away, 18 remained. At this point, there were twice as many sparrows on the first bush as on the second. This means there were 12 sparrows on the first bush and 6 on the second. If we return 7 sparrows to the second bush and then move five sparrows from the second bush to the first, we will return to the initial situation. As a result, we get 12 + 5 = 17 sparrows on the first bush, and 6 + 7 - 5 = 8 sparrows on the second.
|
17
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. What angle do the hour and minute hands form at twenty minutes past one? (Don't forget to justify your answer).
|
7.4 $110^{\circ}$.
At 12:00, the angle between the hour and minute hands is $0^{\circ}$. The minute hand completes a full circle of $360^{\circ}$ in 60 minutes, which is $6^{\circ}$ per minute, and in 20 minutes it will cover $120^{\circ}$. The hour hand moves 12 times slower than the minute hand. Therefore, in 20 minutes, it will cover an angle of $10^{\circ}$. Thus, in 20 minutes, the angle between the hour and minute hands will be $110^{\circ}$.
|
110
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. Find all three-digit numbers $\mathrm{N}$ such that the sum of the digits of the number $\mathrm{N}$ is 11 times smaller than the number $\mathrm{N}$ itself (do not forget to justify your answer).
|
7.5. The number 198 is unique.
From the condition, we get the relation $11 \cdot(a+b+c)=100 a+10 b+c$, or $10 c+b=89 a$. In this relation, the left side is a number less than 100. If $a$ is greater than 1, then the right side will be a number greater than 100. Therefore, $a=1$, $c=8$, $b=9$. All-Russian School Olympiad in Mathematics
|
198
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.7. A green chameleon always tells the truth, while a brown chameleon lies and immediately turns green after lying. In a company of 2019 chameleons (green and brown), each in turn answered the question of how many of them are currently green. The answers were the numbers $1,2,3, \ldots, 2019$ (in some order, not necessarily in the order listed above). What is the maximum number of green chameleons that could have been there initially? (R. Zhenedarov, O. Dmitriev)
|
Answer: 1010.
Solution. Consider two chameleons who spoke in a row. One of them was brown at the moment of speaking; indeed, if both were green, the number of green chameleons would not have changed after the first one spoke, and the second one would have named the same number as the first. We can divide all the chameleons into 1009 pairs who spoke in a row, and one more chameleon; since in each pair there was a brown one, the initial number of green chameleons was no more than $2019-1009=1010$.
It remains to show that there could have been 1010 green chameleons. Let's number the chameleons in the order of their statements. Suppose all the odd-numbered chameleons are green, and all the even-numbered ones are brown. Then the first one will say 1010, the second one will say 1 and turn green. Then the third one will say 1011, and the fourth one will say 2 and turn green, and so on. As a result, the odd-numbered chameleons will say all the numbers from 1010 to 2019, and the even-numbered ones will say all the numbers from 1 to 1009.
Remark. From the first paragraph of the solution, it can be deduced that in any optimal example, the chameleons with even numbers must be exactly brown.
Comment. Only proving that the number of green chameleons was no more than $1010-4$ points.
Only providing an example showing that there could have been exactly 1010 green chameleons initially - 3 points.
|
1010
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.8. In an acute-angled triangle $A B C$, the bisector $B L$ is drawn. The circumcircle of triangle $A B L$ intersects side $B C$ at point $D$. It turns out that point $S$, symmetric to point $C$ with respect to the line $D L$, lies on side $A B$ and does not coincide with its endpoints. What values can $\angle A B C$ take?
(B. Obukhov, zhori)
|
Answer: $60^{\circ}$.
First solution. From the symmetry, triangles $C L D$ and $S L D$ are equal, so $D S = D C$, $\angle C D L = \angle S D L$, and $\angle D L C = \angle D L S$. Since the quadrilateral $A L D B$ is inscribed in a circle, we have $\angle B A L = \angle L D C$ (see Fig. 1). The chords $A L$ and $D L$ of this circle subtend equal angles, so $A L = D L$.
We mark off a segment $A S^{\prime} = D S = D C$ on the ray $A B$. Then triangles $S^{\prime} L A$ and $S L D$ are equal by two sides ($A S^{\prime} = D S$, $A L = D L$) and the angle between them. Therefore, $L S^{\prime} = L S$ and $\angle A S^{\prime} L = \angle D S L$. If points $S^{\prime}$ and $S$ do not coincide, then in the isosceles triangle $L S S^{\prime}$, the angles at the base $S S^{\prime}$ are equal, so $\angle A S^{\prime} L = \angle B S L$. Thus, $\angle D S L = \angle B S L$, which means $D$ lies on $A B$; this is impossible.
Therefore, $S^{\prime} = S$, and $\angle A L S = \angle D L S = \angle D L C$. The sum of these three angles is $180^{\circ}$, so they are each $60^{\circ}$. Hence, $\angle A B C = 180^{\circ} - \angle A L D = 60^{\circ}$.
Fig. 2
Second solution. Let $\Omega$ be the circumcircle of triangle $B C S$ (see Fig. 2). In this triangle, the line $B L$ is the angle bisector of angle $B$, and the line $L D$ is the perpendicular bisector of side $C S$. These lines are distinct; both pass through the midpoint of the arc $C S$ of circle $\Omega$ that does not contain $B$. Therefore, their intersection point $L$ is this midpoint, i.e., $L \in \Omega$.
Since the quadrilateral $A L D B$ is inscribed, we have $\angle A B C = \angle D L C = \beta$. From symmetry, $\angle D L C = \angle D L S = \beta$. Now from circle $\Omega$, we get $180^{\circ} = \angle C B S + \angle S L C = \beta + \angle D L S + \angle D L C = 3 \beta$, from which $\beta = 60^{\circ}$.
Remark. There are other solutions. For example, one can notice that point $L$ is the excenter of triangle $B D S$ (since $B L$ is the angle bisector of the internal angle of this triangle, and $D L$ is the angle bisector of its external angle). This leads to the calculation $\angle A B C = \angle D L C = \angle D L S = 90^{\circ} - \angle A B C / 2$, from which the answer follows.
One can also consider the point $T$, symmetric to point $C$ with respect to $L B$, as well as the projections $S^{\prime}$ and $T^{\prime}$ of point $C$ onto $L D$ and $L B$ respectively. From the inscribed quadrilaterals $A B D L$ and $L T^{\prime} S^{\prime} C$, we have $\angle A B C = \angle D L C = \angle S^{\prime} T^{\prime} C = \angle S T C = \angle B T C$. But triangle $B T C$ is isosceles, so $\angle B T C = 90^{\circ} - \angle A B C / 2$.
Comment. Without additional considerations, it is claimed that triangles $A L S$ and $D L S$ are equal (by two sides and the angle $\angle L A S = \angle L D S$ not between them) - no more than 3 points for the problem.
|
60
|
Geometry
|
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
|
5. Variant 1 Petya wrote 9 different positive integers on the board. It turned out that the arithmetic mean of these numbers is 16. What is the greatest value that the largest number on the board can take
|
Answer: 108.
Solution: The sum of the given numbers is $9 \cdot 16=144$. Since all the numbers are distinct, the sum of the 8 smallest of them is no less than $1+2+\cdots+8=36$. Therefore, the largest number cannot be greater than $144-36=108$. This is possible: $(1+2+\cdots+8+108): 9=16$.
Variant 2 Petya wrote 9 different positive integers on the board. It turned out that the arithmetic mean of these numbers is 17. What is the largest value that the largest number on the board can take?
Answer: 117.
Variant 3 Petya wrote 9 different positive integers on the board. It turned out that the arithmetic mean of these numbers is 18. What is the largest value that the largest number on the board can take?
Answer: 126.
Variant 4 Petya wrote 9 different positive integers on the board. It turned out that the arithmetic mean of these numbers is 19. What is the largest value that the largest number on the board can take?
Answer: 135.
Variant 5 Petya wrote 9 different positive integers on the board. It turned out that the arithmetic mean of these numbers is 20. What is the largest value that the largest number on the board can take?
Answer: 144.
|
108
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 7. Option 1.
Unit cubes were used to assemble a large parallelepiped with sides greater than 3. Two cubes will be called adjacent if they touch by faces. Thus, one cube can have up to 6 neighbors. It is known that the number of cubes that have exactly 6 neighbors is 429. Find the number of cubes that have exactly 4 neighbors.
|
Answer: 108.
Solution: Let $a, b$ and $c$ be the lengths of the sides of the large parallelepiped. Then, the number of cubes with exactly 6 neighbors is: $(a-2)(b-2)(c-2)$. Since each of the factors $a-2, b-2$, and $c-2$ is greater than 1 and their product equals the product of the three prime numbers 3, 11, and 13, these numbers must be 3, 11, and 13 in some order. The number of cubes with exactly 4 neighbors is $4(a-2+b-2+c-2)$ (since each such cube is adjacent to exactly one edge). Thus, we get the answer: $4(3+11+13)=108$.
|
108
|
Combinatorics
|
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 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zone - ring or central field - a certain number of points is awarded.)
|
Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38): 2=34
$$
|
34
|
Logic and Puzzles
|
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 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
|
Answer: 29.
Solution. The side of the largest square (with vertex $A$) is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the length of segment $C D$, which is 5. And its side is greater than the side of the smallest square by the length of segment $E F$, which is 13. In total, the side of the largest square is greater than the side of the smallest square by the length of segment $G H$, which is $11+5+13=29$.
|
29
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
|
Answer: 52.
Solution. Inside the square, we will write the length of its side. Let the sides of the two squares be $a$ and $b$, and we will sequentially calculate the lengths of the sides of the squares.
The sum of the lengths of the sides of the two squares adjacent to the left side of the rectangle is equal to the sum of the lengths of the sides of the two squares adjacent to the right side of the rectangle. We get the equation
$$
\begin{aligned}
(2 a+b)+(3 a+b) & =(12 a-2 b)+(8 a-b) \\
5 a+2 b & =20 a-3 b \\
b & =3 a
\end{aligned}
$$
Thus, to minimize the perimeter of the rectangle, we need to choose $a=1$, $b=3$. It is easy to check that with these values, the rectangle will have dimensions $11 \times 15$, and its perimeter will be 52.
|
52
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.1. A square was cut into four equal rectangles, and from them, a large letter P was formed, as shown in the figure, with a perimeter of 56.
What is the perimeter of the original square
|
Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
From this, we get the equation
$$
\begin{gathered}
28 x=56 \\
x=2
\end{gathered}
$$
The perimeter of the square is $16 x=32$.
|
32
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.2. The numbers from 1 to 9 were placed in the cells of a $3 \times 3$ table such that the sum of the numbers on one diagonal is 7, and on the other - 21. What is the sum of the numbers in the five shaded cells?
|
Answer: 25.
Solution. Note that 7 can be represented uniquely as the sum of numbers from 1 to 9 - this is $1+2+4=7$.
Let's look at the other diagonal with a sum of 21. The largest possible value of the sum in it is $9+8+4=21$ (since the number in the central cell is no more than 4). Therefore, it must contain the numbers $9,8,4$.
Thus, the number 4 is in the central cell, and the numbers 1, 2, 8, and 9 are at the corners. Now it is not difficult to find the sum of the numbers in the shaded cells: $3+4+5+6+7=25$.
|
25
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
|
Answer: 17.
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute-angled ($\angle ABD = \angle CBD < 90^\circ$, $\left.\angle BAD = \angle ADB = \frac{180^\circ - \angle ABD}{2} < 90^\circ\right)$, point $H$ lies on the segment $AB$.
Notice that the right triangles $BDH$ and $BDC$ are equal by the common hypotenuse $BD$ and the acute angle at vertex $B$. Therefore, $BH = BC$ and $DH = CD$.
Now, notice that the right triangles $ADH$ and $EDC$ are also equal by the hypotenuse $AD = ED$ and the leg $DH = CD$. Therefore, $EC = AH$.
Thus, $BD = BA = BH + AH = BC + EC = (7 + 5) + 5 = 17$.
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
|
Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =(A+B)-(B+C+D)+(D+E+F)-(F+G)= \\
& =12^{2}-9^{2}+7^{2}-3^{2}=103
\end{aligned}
$$
|
103
|
Geometry
|
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
|
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
|
Answer: 16.
Solution. Divide the $5 \times 5$ square without the central cell into four $2 \times 3$ rectangles, and each of these into two $1 \times 3$ rectangles.
We get 8 rectangles $1 \times 3$, the sum of the numbers in each of which is 23. Since the sum of all the numbers is 200, we find the number in the central cell as $200 - 23 \cdot 8 = 16$.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
|
Answer: 35.
Solution. Let's start with the assumption that the quadrilateral KMDС is inscribed (several proofs of this fact will be proposed below).
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\angle M D C}{2}=45^{\circ}$. The angle $A M K$ as an external angle for triangle $K D M$ is equal to the sum of angles $M K D$ and $K D A$, so the required angle КDA is $80^{\circ}-45^{\circ}=35^{\circ}$.
Fig. 8: to the solution of problem 10.8
Let's provide the first possible proof of the inscribed nature of quadrilateral KMDC. Consider triangle $M K C$ and its circumscribed circle. Note that point $D$ lies on the bisector of angle $M K C$, and is equidistant from vertices $M$ and $C$ (Fig. 8). However, the bisector of an angle of a non-isosceles triangle and the perpendicular bisector of its opposite side, as is known, intersect at the midpoint of the "smaller" arc of the circumscribed circle of the triangle. In other words, $D$ is the midpoint of the arc $M C$ of the circumscribed circle of triangle $M K C$, not containing point $K$. It should also be noted that $M K \neq K C$ (otherwise triangles $K M D$ and $K C D$ would be equal, but $\angle K M D>90^{\circ}>\angle K C D$ ).
Let's provide the second possible proof of the inscribed nature of quadrilateral KMDC. It will use the fourth criterion for the equality of triangles: if two sides and an angle not between them are equal in two triangles, then these triangles are either equal or the sum of the other two angles not between them is $180^{\circ}$. The fourth criterion is satisfied for triangles $M D K$ and $C D K (M D=D C, D K$ - common, $\angle M K D=\angle C K D)$. However, angles $K M D$ and $K C D$ are not equal (again, the first is obtuse, and the second is acute), so their sum is $180^{\circ}$, which are the opposite angles of quadrilateral KMDC. Therefore, it is inscribed.
## 11th grade
|
35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3. Find the length of the segment $O D_{1}$.
|
Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ to this face, then point $X$ is the center of $\omega$ (point $O$ is equidistant from all points of $\omega$, so the projection of $O$ onto the plane of $\omega$ is also equidistant from them). Let $Y$ be an arbitrary point on $\omega$ (Fig. 10). Triangle $OXY$ is a right triangle; by the problem's condition, $X Y=3$ and $O Y=10$. By the Pythagorean theorem, we get $O X^{2}=10^{2}-3^{2}=91$.
Similarly, we find the squares of the distances from point $O$ to the planes $A_{1} B_{1} C_{1} D_{1}$ and $A D D_{1} A_{1}$, which are both equal to $10^{2}-1^{2}=99$.
By the spatial Pythagorean theorem, the square of the length of segment $O D_{1}$ is equal to the sum of the squares of the distances from point $O$ to the three faces containing point $D_{1}$. Therefore, $O D_{1}^{2}=$ $91+99+99=289$, from which $O D_{1}=17$.
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Find all positive solutions of the equation
$$
x^{101} + 100^{99} = x^{99} + 100^{101}
$$
Do not forget to justify your answer.
|
Solution. The equation is reduced to the form
$$
x^{99}\left(x^{2}-1\right)=100^{99}\left(100^{2}-1\right)
$$
Obviously, the solution is $x=100$. For $0<x<1$, the left side is less than the right side. For $x>1$, the left side is an increasing function, so the required value is taken no more than once. Therefore, there are no other roots. Answer. 100.
|
100
|
Algebra
|
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
|
3. In the box, there are balls of seven colors. One tenth of the balls are red, one eighth are orange, and one third are yellow. There are 9 more green balls than red ones, and 10 more blue balls than orange ones. There are 8 blue balls in the box. The remaining balls are purple. What is the smallest possible number of purple balls?
|
Answer: 25 balls.
Solution. Let the total number of balls be $x$, and the number of violet balls be $y$. Then
$$
\frac{x}{10}+\frac{x}{8}+\frac{x}{3}+\frac{x}{10}+9+\frac{x}{8}+10+8+y=x
$$
from which $\frac{47 x}{60}+27+y=x$, that is, $y=\frac{13 x}{60}-27$, and the smallest value of $y$ is achieved at the smallest value of $x$. Since $y$ is an integer, $x$ must be a multiple of 60. According to the condition, $x$ must also be a multiple of 8, so $x$ is divisible by 120. For $x=120, y=-1$, which is impossible. For $x=240, y=25$, and this is the smallest possible number of violet balls.
Comment. The correct solution and the correct answer are given - 7 points. The correct answer is found by trial and error, and the minimality is not proven - 3 points. If the correct answer is not obtained, but the idea of divisibility is considered in the solution, which is implemented incorrectly - 1 point. For arithmetic errors with correct reasoning, deduct 1-3 points.
|
25
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Vanya and Petya decided to mow the lawn for a football field. Vanya could do it alone in 5 hours, and Petya in 6 hours. They started at 11 o'clock, and stopped mowing simultaneously when their parents called them, but Petya took a one-hour lunch break, and Vanya's lunch break lasted two hours. One tenth of the lawn remained unmowed. At what time did the parents call the boys?
|
Answer: At 15 o'clock.
Solution: Let $t$ be the time from 11 o'clock to the end, then Petya worked $(t-1)$ hours, and Vanya worked $(t-2)$ hours. In one hour, Petya completed $\frac{1}{6}$ of the work, and Vanya completed $\frac{1}{5}$ of the work. We get the equation: $\frac{1}{6} \cdot(t-1)+\frac{1}{5} \cdot (t-2)=\frac{9}{10}$. From this, $t=4$, the time of completion is $11+4=15$.
Comment: The correct solution and the correct answer are given - 7 points. The equation is correctly set up, but due to an arithmetic error, an incorrect answer is obtained - 5 points. The problem is not solved, but there are elements of correct reasoning - 2 points. Only the correct answer and a check that it satisfies the problem's condition are provided - 3 points. Only the answer is provided - 0 points.
|
15
|
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
|
3. A company produces a lemon drink by diluting lemon juice with water. Initially, the company produced a drink containing $15\%$ lemon juice. After some time, the CEO ordered to reduce the lemon juice content to $10\%$. By what percentage will the amount of lemon drink produced increase with the same volumes of lemon supplies?
|
Answer: By $50 \%$.
Solution. Method 1. The content of lemon juice in the drink after the CEO's directive decreased by one and a half times. This means that from the same lemons, one and a half times more lemon drink can be made. In other words, the amount of lemon drink produced will increase by one and a half times or by $50 \%$.
$\underline{2}$ Method. Let $x$ be the amount of drink produced before the CEO's directive. Then the amount of lemon juice in this drink is $-0.15 \cdot x$. Let $y$ be the amount of drink produced after the CEO's directive. Then the amount of lemon juice in this drink is $-0.1 \cdot y$. Since it is assumed that the amount of lemon juice has not changed, we get the equation $0.15 \cdot x = 0.1 \cdot y$. Multiplying both sides of this equation by 10, we get: $y = 1.5 \cdot x$; or: $y = x + 0.5 \cdot x$. Therefore, the amount of drink produced has increased by $50 \%$.
## Grading Criteria.
- Full solution - 7 points.
- The equation is correctly set up according to the problem statement, but further reasoning is missing or incorrect - 2 points.
- The correct answer is obtained using a specific amount (for example, 100 liters) of the produced drink or lemon juice - 1 point.
- Only the correct answer is provided or the correct answer is accompanied by incorrect reasoning - 0 points.
|
50
|
Algebra
|
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
|
5. On a standard graph paper, an angle is drawn (see figure). Find its measure without using measuring instruments. Justify your answer.
|
Answer: $45^{\circ}$.
Solution. Connect the two "extreme" points with a segment (as shown in the figure). The resulting triangle is isosceles because two of its sides, $A B$ and $B C$, are diagonals of three-cell rectangles. Diagonal $A B$ divides the angle of the rectangle at vertex $B$ into two angles that complement each other to a right angle. Triangles $A D B$ and $C E B$ are equal by two legs, so their corresponding angles are equal. Therefore, angle $C B E$ complements angle $A B E$ to a right angle. Thus, triangle $A B C$ is
isosceles and right-angled. Its angles $A$ and $C$ at the base $A C$ are equal by the property of isosceles triangles and have a measure of $45^{\circ}$ by the theorem on the sum of the angles of a triangle.
Comment. If we connect point $B$ to the midpoint of $A C$, we also get an isosceles right-angled triangle. The reasoning is similar.
## Grading Criteria.
- A correct and justified solution - 7 points.
- Generally correct reasoning with minor errors that do not fundamentally affect the solution, and a correct answer - $\mathbf{5}$ points.
- Additional constructions and markings on the diagram that clearly show the solution process, a correct answer, but no reasoning provided - 3 points.
- A correct answer without justification or with incorrect justification - $\mathbf{0}$ points.
|
45
|
Geometry
|
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
|
10.3. Numbers $1,2,3,4,5,6,7$ and 8 are placed at the vertices of a cube, such that the sum of any three numbers belonging to any face is not less than 10. Find the minimum possible sum of four numbers belonging to one face.
|
Answer: 16. Evaluation. Consider an arbitrary face. If the largest number written at the vertex of this face is no more than 5, then the sum of the remaining numbers is no more than $4+3+2=9$. Therefore, the largest number written at the vertex of any face is at least 6, and the minimum possible sum of four numbers belonging to one face is at least 16. An example is shown in the figure, where the minimum sum is on the "near" face.
Comment. Answer without an example - 1 point, example - 2 points, evaluation without an example - 3 points. Answer, example, and evaluation - 7 points.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Given a two-digit number, where the first digit is greater than the second. Its digits were rearranged, and a new two-digit number was obtained. The difference between the first and the second number turned out to be a perfect square. How many possible two-digit numbers satisfy this condition?
|
1. Answer: 13 numbers.
Let the first number be $10a + b$. Then the second number is $10b + a$, where $a > b > 0$. The difference is $9(a - b)$, which implies that the difference $a - b$ is a perfect square. The difference can be 1 or 4. For the first case, we have 8 values for the larger number $(21, 32, \ldots, 98)$, when the difference is $3^2$, and for the second case, there are 5 values $(51, 62, 73, 84, 95)$, when the difference is $6^2$. In total, 13 numbers.
|
13
|
Number Theory
|
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
|
4. On a plane, there are 10 different points. We considered the midpoints of all segments connecting all pairs of points. What is the smallest number of midpoints that could have arisen?
|
4. Answer: 17 midpoints.
Consider the case when the points lie on a single line. Introduce coordinates on it, and let the points be $a_{1}<\cdots<a_{10}$. Consider the midpoints of segments involving $a_{1}$. There are nine of them, and they form an increasing sequence $\frac{a_{1}+a_{2}}{2}<\frac{a_{1}+a_{3}}{2}<\cdots<\frac{a_{1}+a_{10}}{2}$. Similarly, consider the midpoints of segments involving $a_{10}$. There are also nine of them, and they form a sequence $\frac{a_{1}+a_{10}}{2}<\frac{a_{2}+a_{10}}{2}<\cdots<\frac{a_{9}+a_{10}}{2}$. The largest number of the first sequence coincides with the smallest number of the second, and together we have 17 pairwise distinct midpoints. Thus, there are at least 17 midpoints in total.
If we take points with even coordinates $2,4,6, \ldots, 20$, the midpoints will be all points with integer coordinates from 3 to 19 inclusive, and there are exactly 17 of them.
Now consider the general case of the points' arrangement on the plane. Draw lines through all pairs of points, and consider a line that is not perpendicular to any of these lines. Then, in the projection onto this line, all points will be distinct. The midpoint under projection becomes a midpoint, and for points on a line, we have at least 17 distinct midpoints. Therefore, there were at least 17 midpoints in total.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. There is a $10 \times 10$ grid for playing "Battleship". In this case, the game follows unusual rules, and only ships of size $1 \times 2$ are allowed, which can be placed both horizontally and vertically. What is the maximum number of these ships that can be placed on such a field, if the ships must not extend beyond the boundaries of the square, and they cannot touch each other even at a point?
For a complete solution to each problem, 7 points are awarded
The maximum total score is 35
|
5. Answer: 13.
One of the examples of placing 13 ships is shown in the figure:
| | | | | | | | | | |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | |
| | | | | | | | | | |
| | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | |
| | | | | | | | | | |
| | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | | $\mathrm{x}$ | $\mathrm{x}$ | |
| | | | | | | | | | |
| | $\mathrm{x}$ | | $\mathrm{x}$ | | $\mathrm{x}$ | | $\mathrm{x}$ | | |
| $\mathrm{x}$ | | $\mathrm{x}$ | | $\mathrm{x}$ | | $\mathrm{x}$ | | | |
| | | | | | | | | | |
We will prove that it is impossible to place more ships. For each rectangle $1 \times 2$, consider two pairs of parallel lines, located outside its contour at a distance of $1 / 2$. They define a rectangle $2 \times 3$ with an area of 6. Also, for each side of the square $10 \times 10$, consider lines at a distance of $1 / 2$ from the contour of the square, located inside it.
We will say that the framing has been done. From the condition of the problem, it follows that the framings of the rectangles and the square cannot overlap each other. Therefore, all the framings of the ships are contained within the inner square $9 \times 9$. If there were $k$ ships, then their total area together with the framings is no more than the area of the inner square: $6 k \leq 81$. From this, it is clear that $k \leq 13$.
|
13
|
Combinatorics
|
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
|
11.5. On each of ten cards, a real number is written. For each non-empty set of these cards, the sum of all numbers written on the cards in this set was found. It is known that not all of the obtained sums are integers. What is the maximum possible number of integer sums that could have resulted?
|
Answer: 511.
Solution: It is clear that at least one of the cards has a non-integer number, otherwise the sum of the numbers in any set would be an integer. We select one of these cards and call it $a$, and the number on it $-x$. For each set $C$ that does not contain $a$ (including the empty set), we associate the set $C \cup\{a\}$. It is clear that different sets will be mapped to different sets, because if $C \cup\{a\}$ and $D \cup\{a\}$ coincide, then $C$ and $D$ coincide. Therefore, the constructed different pairs do not have common elements.
Notice that any set $C$ will end up in one of these pairs. If it does not contain $\{a\}$, it will fall into a pair by construction; if it does contain $\{a\}$, it is paired with the set from $C$ without $\{a\}$.
Thus, all sets are divided into pairs of the form ( $C, C \cup\{a\}$ ). We discard the pair ( $\varnothing, a$ ) (since neither of these sets gives an integer sum) and look at the remaining ones. The difference in the sums of the numbers on the cards in the sets of a pair differs by the number $x$, so in each pair there is no more than one integer. Since there are $\frac{2^{10}-2}{2}=2^{9}-1=511$ pairs, there are no more than 511 sets with an integer sum.
If our set contains 9 cards with 0 and one with the number $\frac{1}{2}$, then the sum of the numbers on the cards of $2^{9}-1=511$ sets will be 0, which is an integer.
Comment: Only the answer - 1 point.
The answer is given and an example is provided where 511 integer sums are obtained - 2 points.
The solution includes the idea of dividing sets into such pairs that no more than one set can be taken from each - 1 point.
It is proven that there will be no more than 511 integer sums - 5 points.
Remark: 511 integer sums can also be obtained in the case where no integer is written on any of the cards. For example, if $\frac{1}{2}$ is written on each of them, we will get exactly 511 integer sums.
|
511
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. In the store, Sasha bought two pens, three notebooks, and one pencil and paid 33 rubles, Dima bought one pen, one notebook, and two pencils and paid 20 rubles. How much did Tanya pay for four pens, five notebooks, and five pencils?
|
Answer: 73 rubles.
Let $a, b, c$ be the price of a pen, a notebook, and a pencil, respectively. Then, according to the condition, we have $2a + 3b + c = 33$ and $a + b + 2c = 20$. Multiplying the second relation by 2 and adding it to the first equation, we get the answer.
|
73
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zone - ring or central area - a certain number of points is awarded.)
|
Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38): 2=34
$$
|
34
|
Logic and Puzzles
|
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 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?
|
Answer: 29.
Solution. The side of the largest square (with vertex $A$) is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the length of segment $C D$, which is 5. And its side is greater than the side of the smallest square by the length of segment $E F$, which is 13. In total, the side of the largest square is greater than the side of the smallest square by the length of segment $G H$, which is $11+5+13=29$.
|
29
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
|
Answer: 52.
Solution. Inside the square, we will write the length of its side. Let the sides of the two squares be $a$ and $b$, and we will sequentially calculate the lengths of the sides of the squares.
The sum of the lengths of the sides of the two squares adjacent to the left side of the rectangle is equal to the sum of the lengths of the sides of the two squares adjacent to the right side of the rectangle. We obtain the equation
$$
\begin{aligned}
(2 a+b)+(3 a+b) & =(12 a-2 b)+(8 a-b) \\
5 a+2 b & =20 a-3 b \\
b & =3 a
\end{aligned}
$$
Thus, to minimize the perimeter of the rectangle, we need to choose $a=1$, $b=3$. It is easy to verify that with these values, the rectangle will have dimensions $11 \times 15$, and its perimeter will be 52.
|
52
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.1. A square was cut into four equal rectangles, and from them, a large letter P was formed, as shown in the figure, with a perimeter of 56.
What is the perimeter of the original square
|
Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
From this, we get the equation
$$
\begin{gathered}
28 x=56 \\
x=2
\end{gathered}
$$
The perimeter of the square is $16 x=32$.
|
32
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.2. The numbers from 1 to 9 were placed in the cells of a $3 \times 3$ table such that the sum of the numbers on one diagonal is 7, and on the other - 21. What is the sum of the numbers in the five shaded cells?
|
Answer: 25.
Solution. Note that 7 can be represented uniquely as the sum of numbers from 1 to 9 - this is $1+2+4=7$.
Let's look at the other diagonal with a sum of 21. The largest possible value of the sum in it is $9+8+4=21$ (since the number in the central cell is no more than 4). Therefore, it must contain the numbers $9,8,4$.
Thus, the number 4 is in the central cell, and the numbers 1, 2, 8, and 9 are at the corners. Now it is not difficult to find the sum of the numbers in the shaded cells: $3+4+5+6+7=25$.
|
25
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
|
Answer: 17.
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute-angled ($\angle ABD = \angle CBD < 90^\circ$, $\left.\angle BAD = \angle ADB = \frac{180^\circ - \angle ABD}{2} < 90^\circ\right)$, point $H$ lies on the segment $AB$.
Notice that the right triangles $BDH$ and $BDC$ are equal by the common hypotenuse $BD$ and the acute angle at vertex $B$. Therefore, $BH = BC$ and $DH = CD$.
Now, notice that the right triangles $ADH$ and $EDC$ are also equal by the hypotenuse $AD = ED$ and the leg $DH = CD$. Therefore, $EC = AH$.
Thus, $BD = BA = BH + AH = BC + EC = (7 + 5) + 5 = 17$.
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
|
Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =(A+B)-(B+C+D)+(D+E+F)-(F+G)= \\
& =12^{2}-9^{2}+7^{2}-3^{2}=103
\end{aligned}
$$
|
103
|
Geometry
|
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 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers located inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
|
Answer: 16.
Solution. Divide the $5 \times 5$ square without the central cell into four $2 \times 3$ rectangles, and each of them into two $1 \times 3$ rectangles.
We get 8 rectangles $1 \times 3$, the sum of the numbers in each of which is 23. Since the sum of all the numbers is 200, we find the number in the central cell as $200 - 23 \cdot 8 = 16$.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
|
Answer: 35.
Solution. Let's start with the assumption that the quadrilateral KMDС is inscribed (several proofs of this fact will be proposed below).
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\angle M D C}{2}=45^{\circ}$. The angle $A M K$ as an external angle for triangle $K D M$ is equal to the sum of angles $M K D$ and $K D A$, so the required angle КDA is $80^{\circ}-45^{\circ}=35^{\circ}$.
Fig. 8: to the solution of problem 10.8
Let's provide the first possible proof of the inscribed nature of quadrilateral KMDC. Consider triangle $M K C$ and its circumscribed circle. Note that point $D$ lies on the bisector of angle $M K C$, and is equidistant from vertices $M$ and $C$ (Fig. 8). However, the bisector of an angle of a non-isosceles triangle and the perpendicular bisector of its opposite side, as is known, intersect at the midpoint of the "smaller" arc of the circumscribed circle of the triangle. In other words, $D$ is the midpoint of the arc $M C$ of the circumscribed circle of triangle $M K C$, not containing point $K$. It should also be noted that $M K \neq K C$ (otherwise triangles $K M D$ and $K C D$ would be equal, but $\angle K M D>90^{\circ}>\angle K C D$ ).
Let's provide the second possible proof of the inscribed nature of quadrilateral KMDC. It will use the fourth criterion for the equality of triangles: if two sides and an angle not between them are equal in two triangles, then these triangles are either equal or the sum of the other two angles not between them is $180^{\circ}$. The fourth criterion is satisfied for triangles $M D K$ and $C D K (M D=D C, D K$ - common, $\angle M K D=\angle C K D)$. However, angles $K M D$ and $K C D$ are not equal (again, the first is obtuse, and the second is acute), so their sum is $180^{\circ}$, which are the opposite angles of quadrilateral KMDC. Therefore, it is inscribed.
## 11th grade
|
35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3. Find the length of the segment $O D_{1}$.
|
Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ to this face, then point $X$ is the center of $\omega$ (point $O$ is equidistant from all points of $\omega$, so the projection of $O$ onto the plane of $\omega$ is also equidistant from them). Let $Y$ be an arbitrary point on $\omega$ (Fig. 10). Triangle $OXY$ is a right triangle; by the problem's condition, $X Y=3$ and $O Y=10$. By the Pythagorean theorem, we get $O X^{2}=10^{2}-3^{2}=91$.
Similarly, we find the squares of the distances from point $O$ to the planes $A_{1} B_{1} C_{1} D_{1}$ and $A D D_{1} A_{1}$, which are both equal to $10^{2}-1^{2}=99$.
By the spatial Pythagorean theorem, the square of the length of segment $O D_{1}$ is equal to the sum of the squares of the distances from point $O$ to the three faces containing point $D_{1}$. Therefore, $O D_{1}^{2}=$ $91+99+99=289$, from which $O D_{1}=17$.
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
|
Answer: 52.
Solution. All sides of a square are equal, and its perimeter is 24, so each side is $24: 4=6$. The perimeter of the rectangle is 16, and its two largest sides are each 6, so the two smallest sides are each $(16-6 \cdot 2): 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimeter is $2 \cdot(8+18)=$ 52.
|
52
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How much does the triangular weight weigh?
|
Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ grams.
|
60
|
Algebra
|
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 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
|
Answer: 22.
Solution. Note that there are a total of 91 coins, so after all moves, each box should have exactly 13 coins. At least 7 coins need to be moved from the box with 20 coins. Now consider the boxes adjacent to the box with 20 coins. Initially, they have a total of 25 coins, and at least 7 more coins will be transferred from the box with 20 coins. In the end, these two boxes should have a total of 26 coins, so at least \(25 + 7 - 26 = 6\) coins need to be moved from these boxes. Next, consider the boxes with 17 and 18 coins, from which at least 4 and 5 coins need to be moved, respectively. In total, there should be at least \(7 + 6 + 4 + 5 = 22\) moves.
Let's provide an example of how to equalize all the boxes in exactly 22 moves. From the box with 20 coins, we move 3 coins to the box with 10 (which then becomes 13), and 4 coins to the box with 15 (which then becomes 19). From the box that now has 19 coins, we move 6 coins to the box with 5 coins (which then becomes 11). From the box with 17 coins, we move 2 coins to the box with 11, and 2 coins to the box with 6 (which then becomes 8). Finally, we move 5 coins from the box with 18 coins to the box with 8 coins. In the end, each box has exactly 13 coins. It is easy to verify that exactly 22 moves were made.
|
22
|
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
|
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued faces, the same number of dots is marked. All dots were erased from all faces except nine, as shown in the diagram. What is the total number of dots that were initially marked on the surface of the figure?
|
Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, the dots were not erased, while on the other, they were erased. It is not difficult to understand that originally there were the same number of dots on these faces.
Let's calculate the total number of dots that were originally on all 9 pairs of faces. We get
$$
2 \cdot(1+1+6+2+2+5+3+3+4)=54
$$
There are 6 faces $1 \times 1$ left, about which we still know nothing. However, it can be noticed that they are divided into pairs according to the following principle: in one pair, there will be faces from one $1 \times 1 \times 1$ cube. In each such pair, the sum of the numbers was 7. Then we get that the number of dots on the surface of the figure originally was
$$
54+3 \cdot 7=75
$$
|
75
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
|
Answer: 83.
Solution. Mark a point $K$ on the ray $AB$ such that $AK = AC$. Then the triangle $KAC$ is equilateral; in particular, $\angle AKC = 60^{\circ}$ and $KC = AC$. At the same time, $BK = AK - AB = AC - AB = AD$. This means that triangles $BKC$ and $DAC$ are equal by two sides and the angle $60^{\circ}$ between them (Fig. 2).
It remains to note that the angle $\angle ABC$ - the exterior angle of triangle $BKC$ - is equal to the exterior angle at vertex $D$ of triangle $DAC$, which is calculated as the sum of two interior angles: $60^{\circ} + 23^{\circ} = 83^{\circ}$.
|
83
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.5. On the base $AC$ of isosceles triangle $ABC (AB = BC)$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
|
Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angle at vertex \(B\) is \(30^\circ\), so \(HM = \frac{1}{2} BM = 1.5\). Then \(AC = 2AH = 2(AM + MH) = 2 \cdot (7 + 1.5) = 17\).
Fig. 4: to the solution of problem 9.5
Another solution. Mark a point \(K\) on \(MC\) such that \(\angle BKM = 60^\circ\) (Fig. 4; such a point lies exactly on the segment \(MC\) because \(\angle BCM = \angle BAM = \angle BMC - \angle ABM < 60^\circ\)). Note that in triangle \(BMK\), two angles are \(60^\circ\), so it is equilateral and \(BK = MK = BM = 3\). Also note that triangles \(ABM\) and \(CBK\) are equal because \(BC = AB\), \(\angle A = \angle C\), \(\angle AMB = \angle CKB = 120^\circ\) (therefore, \(\angle ABM = \angle CBK\)). Then \(CK = AM = 7\) and \(AC = AM + MK + KC = 7 + 3 + 7 = 17\).
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
|
Answer: 18.
Solution. Let the lines $B M$ and $A D$ intersect at point $K$ (Fig. 5). Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 5: to the solution of problem 9.8
In the isosceles triangle $A D H$, draw the height and median $D S$. Then, in triangle $A H K$, the segment $D S$ passes through the midpoint of side $A H$ and is parallel to $H K$. Therefore, $D S$ is the midline of this triangle, and $A D = D K = 18$.
Remark. If in a right triangle a point on the hypotenuse is equidistant from two vertices of the triangle, then it is equidistant from all three vertices of the triangle.
## 10th Grade
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.3. On the side $AD$ of rectangle $ABCD$, a point $E$ is marked. On the segment $EC$, there is a point $M$ such that $AB = BM, AE = EM$. Find the length of side $BC$, given that $ED = 16, CD = 12$.
|
Answer: 20.
Solution. Note that triangles $A B E$ and $M B E$ are equal to each other by three sides. Then $\angle B M E=\angle B A E=90^{\circ}$.
Fig. 6: to the solution of problem 10.3
From the parallelism of $A D$ and $B C$, it follows that $\angle B C M=\angle C E D$ (Fig. 6). Therefore, right triangles $B C M$ and $C E D$ are equal by an acute angle and the leg $B M=A B=C D$. Using the Pythagorean theorem for triangle $C D E$, we conclude
$$
B C=C E=\sqrt{C D^{2}+E D^{2}}=\sqrt{12^{2}+16^{2}}=20
$$
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure?
|
Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D=$ $180^{\circ}-\angle C A D-\angle A D C=180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ}$.
Fig. 7: to the solution of problem 10.6
Draw segment $C M$ (Fig. 7). Since in a right triangle, the median to the hypotenuse is equal to half of it, we have $C M=D M=A M=B M$. Triangle $M C D$ is isosceles with an angle of $60^{\circ}$ at the base, so it is equilateral, and $\angle C M D=60^{\circ}$. Then $\angle B M C=180^{\circ}-\angle A M B-\angle C M D=180^{\circ}-70^{\circ}-60^{\circ}=50^{\circ}$. Since triangle $B M C$ is isosceles with vertex $M$, we have $\angle C B M=$ $\frac{1}{2}\left(180^{\circ}-\angle B M C\right)=\frac{1}{2}\left(180^{\circ}-50^{\circ}\right)=65^{\circ}$. Finally,
$$
\angle B C A=180^{\circ}-\angle C B M-\angle B O C=180^{\circ}-65^{\circ}-80^{\circ}=35^{\circ} \text {. }
$$
Remark. There are other solutions that use the fact that $A B C D$ is a cyclic quadrilateral with the center of the circumscribed circle $M$.
|
35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
|
Answer: 58.
Solution. As is known, in a circle, inscribed angles subtended by equal chords are either equal or supplementary to $180^{\circ}$. Since $B C=C D$ and $\angle B A D<180^{\circ}$, we get that $\angle B A C=\angle D A C$.
Fig. 9: to the solution of problem 11.5
Since $\angle B D A=\angle B C A=64^{\circ}$, we get that $\angle B D O=\angle A D O=32^{\circ}$. Therefore, point $O$ lies on the two angle bisectors of triangle $A B D$, i.e., it is the point of intersection of the angle bisectors (Fig. 9). Then
$$
\angle B O C=\angle B A O+\angle A B O=\frac{\angle B A D+\angle A B D}{2}=\frac{180^{\circ}-\angle B D A}{2}=\frac{180^{\circ}-64^{\circ}}{2}=58^{\circ}
$$
Remark. The solution could also have been completed differently: since $O$ is the point of intersection of the angle bisectors of triangle $A B D$, by the trident lemma $C D=C B=C O$, from which $\angle C O B$ is easily found.
|
58
|
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
|
2. Agronomist Bilbo noticed that if the length of his rectangular field were 20 meters longer, the perimeter of the field would be twice as large. However, if the width of the field were twice as large, the perimeter of the field would be 18 meters larger. What is the area of the field?
#
|
# Answer: $99 \mu^{2}$.
## Solution:
## Method № 1.
If the width of the field has doubled, then the perimeter has simply increased by twice the width of the field, but according to the problem, this change is 18 meters - which means the width of the field is 9 meters. When the length of the field increases by 20 meters, the perimeter of the field increases by 40 meters, but (according to the problem) this change is equal to the perimeter of the field (since the perimeter has doubled). Therefore, the perimeter of the field is 40 meters. Then the length and width together make 20 meters, and the width is 9 meters, so the length is 11 meters, and the area is $9 \cdot 11 = 99\left(\mu^{2}\right)$.
## Method № 2.
Essentially the same, but with symbolic notation. Let the length of the field be $x$, and the width of the field be $y$. Then the perimeter is $2x + 2y$. When the length of the field increases by 20 meters, we get a rectangle with length $x + 20$ and width $y$, its perimeter is $2(x + 20) + 2y$, which, according to the problem, equals $2(2x + 2y)$.
We get the equation: $2(x + 20) + 2y = 2(2x + 2y)$, from which $2x + 2y = 40$, then $x + y = 20(m)$.
When the width of the original field is doubled, we get a rectangular field with length $x$, width $2y$, and perimeter $2x + 4y$, which, according to the problem, equals $2x + 2y + 18$.
From the equation $2x + 4y = 2x + 2y + 18$ we get that $2y = 18, y = 9$ (meters).
Then $x = (x + y) - y = 20 - 9 = 11$ (meters), and the area is $9 \cdot 11 = 99\left(\mu^{2}\right)$.
## Criteria:
A correct answer without justification and without an example of suitable length and width - give 1 point. A correct answer without justification, but with an indication of the length and width at which it is obtained, give 2 points.
|
99
|
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
|
1. A palindrome is a natural number that reads the same from left to right as it does from right to left. Does there exist a five-digit palindrome that is equal to the sum of two four-digit palindromes?
|
Answer: For example, $6006+5005=11011$.
Other examples are possible: 7777+4444=12221.
Criteria. Answer without an example: 0 points.
|
11011
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Which three-digit numbers are more numerous: those in which all digits have the same parity, or those in which adjacent digits have different parity
|
Answer: equally
Instructions. Method 1. We will consider the numbers of interest by tens. In each ten, the hundreds and units digits are the same (they are of the same parity). There are exactly 10 tens: five of each parity, and thus, from this ten, five numbers will be added to each type. Therefore, the numbers of each type are the same.
Method 2. Let's count how many three-digit numbers have digits of the same parity. Even: the hundreds digit can be chosen in 4 ways (the number cannot start with zero), and the tens and units digits can be chosen in 5 ways each. In total, $4 \cdot 5 \cdot 5 = 100$ ways. Odd: the hundreds digit can be chosen in 5 ways, as well as the tens and units digits. In total, $5 \cdot 5 \cdot 5 = 125$ ways. Therefore, the number of numbers with digits of the same parity is $100 + 125 = 225$.
Let's count how many three-digit numbers have adjacent digits of different parity. EOE: the hundreds digit can be chosen in 4 ways (the number cannot start with zero), and the tens and units digits can be chosen in 5 ways each. In total, $4 \cdot 5 \cdot 5 = 100$ ways. OEO: the hundreds digit can be chosen in 5 ways, as well as the tens and units digits. In total, $5 \cdot 5 \cdot 5 = 125$ ways. Therefore, the number of numbers with adjacent digits of different parity is $100 + 125 = 225$.
There are 225 of each, and thus, they are equally numerous.
Criteria. Answer without justification: 0 points.
|
225
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. On a line, seven points A, B, C, D, E, F, G are marked in the given order. It turned out that $A G=23 \mathrm{~cm}, B F=17 \mathrm{~cm}$ and $\mathrm{CE}=9 \mathrm{~cm}$. Find the sum of the lengths of all segments with endpoints at these points.
|
Answer: 224 cm.
Instructions. There are five pairs of segments that sum up to AG and AG itself. There are three pairs of segments that sum up to BF and BF itself. There is one pair of segments that sum up to CE and CE itself. Therefore, the total length of all segments is $6 \cdot 23 + 4 \cdot 17 + 2 \cdot 9 = 138 + 68 + 18 = 224($ cm $)$.
Criteria. Correct answer without justification: 2 points.
For a computational error with correct justification, 3 points are deducted.
|
224
|
Geometry
|
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
|
10.2. Given a parallelogram $A B C D$, where $A B<A C<B C$. Points $E$ and $F$ are chosen on the circle $\omega$ circumscribed around triangle $A B C$ such that the tangents to $\omega$ at these points pass through $D$; moreover, segments $A D$ and $C E$ intersect. It turns out that $\angle A B F=\angle D C E$. Find the angle $A B C$.
(A. Yakubov, S. Berlov)
|
Answer: $60^{\circ}$.
Solution. Since $D$ lies outside $\omega$, angle $A B C$ is acute. Let $A^{\prime}$ be the second intersection point of $D C$ and $\omega$. Since $B C>A C$, we have $\angle D C A=\angle C A B>\angle C B A=\angle D A^{\prime} A$; thus, $A^{\prime}$ lies on the extension of segment $D C$ beyond point $C$. Note that $E_{E C A^{\prime}}=$ $=2\left(180^{\circ}-\angle E C A^{\prime}\right)=2 \angle E C D=2 \angle A B F=\overline{A C F}$ (see Fig. 2).
Fig. 2
Let $\ell$ be the bisector of angle $E D F$. Since $D E$ and $D F$ are tangents to $\omega$, line $\ell$ passes through the center $O$ of circle $\omega$. Perform a reflection about $\ell$; under this reflection, $\omega$ maps to itself. Since $\overline{E C A^{\prime}}=\overline{A C F}$, points $A$ and $A^{\prime}$ map to each other under this reflection. Therefore, $\angle D A A^{\prime}=\angle D A^{\prime} A$. On the other hand, since point $A^{\prime}$ lies on $\omega$, we have $\angle A A^{\prime} C=\angle A B C=\angle A D A^{\prime}$. Thus, all three angles of triangle $D A A^{\prime}$ are equal, from which it follows that $\angle A B C=\angle A D A^{\prime}=60^{\circ}$.
|
60
|
Geometry
|
math-word-problem
|
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
|
1. Option 1. To a confectionery factory for cake packaging, 5 rolls of ribbon, each 60 m long, were delivered. How many cuts need to be made to get pieces of ribbon, each 1 m 50 cm long?
|
Answer: 195.
Solution: From one roll, 40 pieces of ribbon, each 1 m 50 cm long, can be obtained. For this, 39 cuts are needed. Therefore, a total of $5 \cdot 39=195$ cuts are required.
|
195
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 2. Option 1
Given a rectangular grid. We will call two cells adjacent if they share a side. Let's count the number of cells that have exactly four adjacent cells. It turned out to be 23. How many cells have exactly three adjacent cells?
|
Answer: 48.
Solution. Let $a$ and $b$ be the sides of the rectangle. The total number of cells that have exactly 4 neighbors by side is $(a-2)(b-2)$, and on the other hand, there are 23 such cells. Since 23 is a prime number, the numbers $a-2$ and $b-2$ are equal to the numbers 1 and 23 in some order. The number of cells that have exactly three neighbors is $2(a-2+b-2)$. Therefore: $2(1+23)=48$.
## Variant 2
Given a grid rectangle. We will call two cells neighboring if they share a side. Let's count the number of cells that have exactly four neighboring cells by side. It turned out to be 29. How many cells have three neighboring cells?
Answer: 60.
## Variant 3
Given a grid rectangle. We will call two cells neighboring if they share a side. Let's count the number of cells that have exactly four neighboring cells by side. It turned out to be 31. How many cells have three neighboring cells?
Answer: 64.
## Variant 4
Given a grid rectangle. We will call two cells neighboring if they share a side. Let's count the number of cells that have exactly four neighboring cells by side. It turned out to be 53. How many cells have three neighboring cells?
Answer: 108.
## Variant 5
Given a grid rectangle. We will call two cells neighboring if they share a side. Let's count the number of cells that have exactly four neighboring cells by side. It turned out to be 59. How many cells have three neighboring cells?
Answer: 120.
|
48
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3. Variant 1
Three rectangles A, B, and C are drawn on the sheet (see figure).
Rectangles A and B have the same width, while rectangles B and C have the same length (width - top to bottom, length - left to right). The length of rectangle B is 3 cm longer than the length of rectangle A, and the area of B is 12 cm² larger than the area of A. The width of rectangle C is 3 cm larger than the width of rectangle B, and the area of C is 24 cm² larger than the area of B. Find the area of rectangle A in square centimeters.
|
Answer: 20.
Solution: Let rectangle A have a length of $a$ cm and a width of $b$ cm. If the length is increased by 3 cm, the area increases by $3 b$. Therefore, $3 b=12, b=4$. The area of rectangle B is larger than that of A by $24 \mathrm{~cm}^{2}$, so the length of rectangle B is $24: 3=8$ cm. Therefore, the length of rectangle A is $a=8-3=5$ cm. The area is $a b=20$ cm $^{2}$.
## Variant 2
On a sheet, three rectangles A, B, and C are drawn (see figure).
Rectangles A and B have the same width, and rectangles B and C have the same length (width - top to bottom, length - left to right). The length of rectangle B is 2 cm longer than that of rectangle A, and the area of B is 10 cm ${ }^{2}$ larger than that of A. The width of rectangle C is 2 cm wider than that of rectangle B, and the area of C is 16 cm² larger than that of B. Find the area of rectangle A in square centimeters.
Answer: 30.
## Variant 3
On a sheet, three rectangles A, B, and C are drawn (see figure).
Rectangles A and B have the same width, and rectangles B and C have the same length (width - top to bottom, length - left to right). The length of rectangle B is 3 cm longer than that of rectangle A, and the area of B is 15 cm² larger than that of A. The width of rectangle C is 2 cm wider than that of rectangle B, and the area of C is 26 cm $^{2}$ larger than that of B. Find the area of rectangle A in square centimeters.
Answer: 50.
## Variant 4
On a sheet, three rectangles A, B, and C are drawn (see figure).
Rectangles A and B have the same width, and rectangles B and C have the same length (width - top to bottom, length - left to right). The length of rectangle B is 4 cm longer than that of rectangle A, and the area of B is 28 cm² larger than that of A. The width of rectangle C is 3 cm wider than that of rectangle B, and the area of C is 39 cm $^{2}$ larger than that of B. Find the area of rectangle A in square centimeters.
Answer: 63.
## Variant 5
On a sheet, three rectangles A, B, and C are drawn (see figure).
A
B
C
Rectangles A and B have the same width, and rectangles B and C have the same length (width - top to bottom, length - left to right). The length of rectangle B is 2 cm longer than that of rectangle A, and the area of B is 16 cm $^{2}$ larger than that of A. The width of rectangle C is 4 cm wider than that of rectangle B, and the area of C is 56 cm $^{2}$ larger than that of B. Find the area of rectangle A in square centimeters.
Answer: 96.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 4. Variant 1
The train was moving from point A to point B at a constant speed. Halfway through the journey, a breakdown occurred, and the train stopped for 15 minutes. After that, the driver had to increase the train's speed by 4 times to arrive at point B on schedule. How many minutes does the train travel from point A to point B according to the schedule?
|
Answer: 40.
Solution. Let's represent the train's path as a segment divided into 8 parts. Denote the intervals it travels in equal time with arcs.
Since the train stopped for 15 minutes, its travel time with the breakdown is 15 minutes less than the scheduled time. This time difference corresponds to \(4-1=3\) intervals. Therefore, one interval is \(15 \div 3 = 5\) minutes, and the entire journey takes \(5 \times 8 = 40\) minutes.
## Variant 2
A train was traveling from point A to point B at a constant speed. Halfway through the journey, a breakdown occurred, and the train stopped for 12 minutes. After that, the driver had to increase the train's speed by 5 times to arrive at point B on schedule. How many minutes does the train travel from point A to point B according to the schedule?
Answer: 30.
## Variant 3
A train was traveling from point A to point B at a constant speed. Halfway through the journey, a breakdown occurred, and the train stopped for 14 minutes. After that, the driver had to increase the train's speed by 3 times to arrive at point B on schedule. How many minutes does the train travel from point A to point B according to the schedule?
Answer: 42.
## Variant 4
A train was traveling from point A to point B at a constant speed. Halfway through the journey, a breakdown occurred, and the train stopped for 20 minutes. After that, the driver had to increase the train's speed by 6 times to arrive at point B on schedule. How many minutes does the train travel from point A to point B according to the schedule?
Answer: 48.
## Variant 5
A train was traveling from point A to point B at a constant speed. Halfway through the journey, a breakdown occurred, and the train stopped for 21 minutes. After that, the driver had to increase the train's speed by 4 times to arrive at point B on schedule. How many minutes does the train travel from point A to point B according to the schedule?
Answer: 56.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 5. Option 1.
An apple, three pears, and two bananas together weigh 920 g; two apples, four bananas, and five pears together weigh 1 kg 710 g. How many grams does a pear weigh?
|
Answer: 130
Solution: From the first condition, it follows that two apples, four bananas, and six pears together weigh 1 kg 840 g. Therefore, a pear weighs $1840-1710=130$ g.
|
130
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 7. Option 1
Diligent Masha wrote down in a row all natural numbers from 372 to 506 inclusive. Then she calculated two sums: first, the sum of all odd numbers in this row, and then the sum of all even numbers. After that, she subtracted the smaller sum from the larger one. What result did she get?
|
Answer: 439.
Solution. Let's divide all numbers from 372 to 506, except 506, into pairs such that each even number corresponds to the next odd number (this is possible because the last number is 505). In each pair, the odd number will be 1 greater than the even one. There will be (505 - 371) : 2 = 67 pairs in the range from 372 to 505, which means that among the numbers from 372 to 505 inclusive, the sum of the odd numbers is 67 greater. But there is still the number 506, which is even. Therefore, the sum of the even numbers is 506 - 67 = 439 greater.
## Variant 2
Diligent Masha wrote down in a row all natural numbers from 126 to 432 inclusive. Then she calculated two sums: first, of all the odd numbers in this row, and then - of all the even numbers. After that, she subtracted the smaller sum from the larger one. What result did she get?
Answer: 279.
## Variant 3
Diligent Masha wrote down in a row all natural numbers from 245 to 683 inclusive. Then she calculated two sums: first, of all the odd numbers in this row, and then - of all the even numbers. After that, she subtracted the smaller sum from the larger one. What result did she get?
Answer: 464.
## Variant 4
Diligent Masha wrote down in a row all natural numbers from 579 to 833 inclusive. Then she calculated two sums: first, of all the odd numbers in this row, and then - of all the even numbers. After that, she subtracted the smaller sum from the larger one. What result did she get?
Answer: 706.
## Variant 5
Diligent Masha wrote down in a row all natural numbers from 286 to 978 inclusive. Then she calculated two sums: first, of all the odd numbers in this row, and then - of all the even numbers. After that, she subtracted the smaller sum from the larger one. What result did she get?
Answer: 632.
|
439
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 8. Variant 1
Tanya had a set of identical sticks. She formed a large triangle from them, with each side consisting of 11 sticks, and laid out a pattern inside the triangle such that the triangle was divided into smaller triangles with a side of 1 stick (the figure shows an example of such a pattern for a triangle with a side of 3 sticks). How many sticks did Tanya use in total?
|
Answer: 198.
Solution (1st method). All the sticks can be divided into those that go horizontally, at an angle to the right, and at an angle to the left:
In the first such row, there is 1 stick, in the second - 2 sticks, in the third - 3 sticks, and so on. There are 11 rows in total, so there are $3 \cdot(1+2+\ldots+11)=3 \cdot 66=198$ sticks in total.
Solution (2nd method). Notice that all the sticks can be divided into groups of 3, forming a triangle with a side of 1 stick:
Then, on the first level from the top, there will be one such triangle, on the second level - 2 triangles, and so on. On the bottom level, there will be 11 triangles (because the base of the triangle will consist of 11 sticks). In total, we get: $3 \cdot(1+2+\ldots+11)=3 \cdot 66=198$ sticks.
## Variant 2
Tanya had a set of identical sticks. She formed a large triangle from them, with each side consisting of 17 sticks, and laid out a pattern inside the triangle so that the triangle was divided into small triangles with a side of 1 stick (the figure shows an example of such a pattern for a triangle with a side of 3 sticks). How many sticks did Tanya use in total?
Answer: 459.
## Variant 3
Tanya had a set of identical sticks. She formed a large triangle from them, with each side consisting of 15 sticks, and laid out a pattern inside the triangle so that the triangle was divided into small triangles with a side of 1 stick (the figure shows an example of such a pattern for a triangle with a side of 3 sticks). How many sticks did Tanya use in total?
Answer: 360.
## Variant 4
Tanya had a set of identical sticks. She formed a large triangle from them, with each side consisting of 19 sticks, and laid out a pattern inside the triangle so that the triangle was divided into small triangles with a side of 1 stick (the figure shows an example of such a pattern for a triangle with a side of 3 sticks). How many sticks did Tanya use in total?
Answer: 570.
## Variant 5
Tanya had a set of identical sticks. She formed a large triangle from them, with each side consisting of 14 sticks, and laid out a pattern inside the triangle so that the triangle was divided into small triangles with a side of 1 stick (the figure shows an example of such a pattern for a triangle with a side of 3 sticks). How many sticks did Tanya use in total?
Answer: 315.
|
198
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. Answer. $\angle A=\angle C=72^{\circ}, \angle B=36^{\circ}$.
|
Solution.
Let $O$ be the common center of the given circles. From the condition, it follows that $BO$ and $CO$ are the bisectors of angles $ABC$ and $BCD$ ($O$ is the incenter) and, moreover, $AO = BO = CO$ ($O$ is the circumcenter). Therefore, if $\angle ABO = \alpha$, then $\angle OBC = \alpha, \angle OCB = \angle OBC = \alpha, \angle BCD = 2\alpha$ (CD is the bisector). Hence, $\angle OAC = \angle OCA = 3\alpha$. Thus, the sum of the angles of triangle ABC is $10\alpha$, from which $\alpha = 18^\circ$.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.6. On the board, three natural numbers were written: two ten-digit numbers $a$ and $b$, as well as their sum $a+b$. What is the maximum number of odd digits that could have been written on the board?
(I. Bogdanov, P. Kozhevnikov)
|
Answer: 30.
Solution: Note that the number $a+b$ has no more than 11 digits, so in total, no more than 31 digits are written on the board. At the same time, all three numbers $a, b, a+b$ cannot be odd simultaneously. Therefore, one of their last three digits is even, which means that there are no more than 30 odd digits written.
Let's provide an example showing that there could be exactly 30 odd digits:
$5555555555+5555555555=11111111110$.
Remark. There are many examples with 30 odd digits, for instance, $3999999999+7999999999=11999999998$.
Comment. Only the answer without any correct explanation - 0 points. Only the proof that there are no more than $30-3$ points. Only a correct example with 30 odd digits - 4 points.
|
30
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. From two settlements, which are 60 km apart, two cars set off in the same direction at the same time. The first car, traveling at a speed of 90 km/h, caught up with the second car after three hours. What is the speed of the second car?
|
Solution. The closing speed of the cars is $60: 3=20(\kappa m / h)$, so the speed of the second car is $90-20=70(km / h)$.
Answer: 70 km $/$ h.
Comment. The correct answer was obtained through correct reasoning - 7 points. The closing speed was correctly found, but the wrong answer was given - 3 points.
|
70
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. On a grid sheet of size $100 \times 100$, several non-overlapping cardboard isosceles right triangles with a leg of 1 were placed; each triangle occupies exactly half of one of the cells. It turned out that each unit segment of the grid (including boundary segments) is covered by exactly one leg of a triangle. Find the maximum possible number of cells that do not contain any triangle.
(D. Khramtsov)
|
Answer. $49 \cdot 50=2450$ cells
Solution. Let $n=50$. We will call a triangle upper if it is located above the line containing its horizontal leg, and lower otherwise. Number the horizontal lines of the grid from bottom to top with numbers from 0 to $2n$.
Denote by $u_{k}$ (respectively $d_{k}$) the number of segments on the $k$-th line that are part of upper (respectively lower) triangles; then $u_{k}+d_{k}=2n$ and $u_{0}=d_{2n}=2n$. Moreover, the vertical segments of the grid located between the $k$-th and $(k+1)$-th lines participate in exactly $u_{k}+d_{k+1}$ triangles, so $u_{k}+d_{k+1}=2n+1$. From this, it is not difficult to obtain that $d_{k}=k$ and $u_{k}=2n-k$ for all $k$.
Now consider the cells located between the $k$-th and $(k+1)$-th lines of the grid. At least $u_{k}=2n-k$ of these cells contain an upper triangle, and at least $d_{k+1}=k+1$ of them contain a lower triangle. Therefore, the number of free cells in this row is no more than $2n-\max(u_{k}, d_{k+1})$, which is no more than $k$ for $k<n$ and no more than $(2n-1)-k$ for $k \geq n$. In total, the number of free cells is no more than $2(0+1+\ldots+(n-1))=n(n-1)$.
It remains to provide an example where this estimate is achieved. Figure 7 shows an example for $n=4$. The example for $n=50$ is constructed similarly: a "rectangle" of cells with sides of $n+1$ and $n$ cells, parallel to the diagonals of the board, is highlighted, its cells are painted in a checkerboard pattern (so that the corner cells of the rectangle are black), and two triangles are placed in each black cell (thus, $n(n-1)$ white cells remain free); in the remaining four "corners" of the board, the triangles are placed so that the right angle of the triangle is directed in the same direction as the entire "corner".
|
2450
|
Combinatorics
|
math-word-problem
|
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
|
8.5. Kolya drew 10 segments and marked all their intersection points in red. Counting the red points, he noticed the following property: on each segment, there are three red points. a) Provide an example of the arrangement of 10 segments with this property. b) What is the maximum number of red points for 10 segments with this property?
|
Answer: b) 15. Solution. a) See the example in the figure. As another example, you can take two copies of the right part of the figure from problem 7.5. b) The example given in the figure shows that 15 red points can be obtained. Let's prove that this is the maximum possible number. Number all 10 segments and write down the numbers of the segments to which each red point belongs near each red point. Since a red point belongs to at least two segments, the total number of recorded numbers is at least $2 N$, where $N$ is the number of red points (the numbers
are recorded more than once). But according to the condition, each segment number is recorded at three red points, i.e., a total of exactly 30 numbers are recorded (each of the 10 segment numbers is recorded three times). Therefore, $2 N \leq 30$, i.e., $N \leq 15$. Note. From the proof, it follows that the maximum $N=15$ is achieved only in the case where each red point belongs to exactly two segments (in the second example from part a), where there are points of intersection of three segments, $N=14$).
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and then count how many dots he got. He got 10 dots.
Then the teacher decided to complicate the task and asked to count the number of dots, but for the letter "P" obtained in the same way from a square with a side of 10 cm. How many dots will Zhenya have this time?
|
Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$.
|
31
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4.5. Hooligan Dima laid out a structure in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the diagram.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire structure to burn?
(Fire spreads along the toothpicks at a constant speed. The fire continues to spread from each burned toothpick to all adjacent unburned toothpicks.)
|
Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
|
65
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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