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15. How many triangles exist, the vertices of which can only be the vertices and the intersection point of the diagonals of a given rectangle that is not a square, and all these triangles have a common vertex at a specified vertex of the rectangle (this vertex is fixed)? How many of these triangles are right-angled? Ho... | 15. Consider Figure 20. With the vertex at point $A$, there will be 5 triangles, 3 of which are right triangles. If instead of the intersection point of the diagonals we have another internal point of the rectangle, and the fixed vertex remains the same, i.e., point $A$, then the following cases can arise:
a) This poi... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,674 |
16. Assemble a square using four $1 \times 1$ square tiles, eight $2 \times 2$ square tiles, twelve $3 \times 3$ square tiles, and sixteen $4 \times 4$ square tiles. | 16. The area of all the tiles, and thus of the desired square, is $1 \cdot 4 + 4 \cdot 8 + 9 \cdot 12 + 16 \cdot 16 = 400$ (sq. units). Therefore, we need to form a square with a side length of 20 units. The available tiles can be arranged as shown in Fig. 24.
$. Since $6.25>6$, there will be enough paper. The sheet should be cut as shown in Figure 27.
 | 23. $1 \mathrm{m}^{3}=1000000 \mathrm{~cm}^{3}$; one match has a volume of $5 \cdot 0.2 \cdot 0.2=0.2\left(\mathrm{~cm}^{3}\right)$; $1000000: 0.2=5000000$. Therefore, 5 million matches. | 5000000 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,682 |
24. What is the mutual arrangement of points $A, B, C$ on a plane, if it is known that for any point $M$ on this plane, the distance $A M$ is less than at least one of the distances $B M$ and $C M$? | 24. First, points $A, B$, and $C$ must lie on the same straight line, because otherwise there exists a point inside the triangle $ABC$ that is equidistant from its vertices $A, B$, and $C$ (the center of the circle passing through points $A, B$, and $C$). Second, point $B$ cannot lie between $A$ and $C$, nor can point ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,683 |
25. A scout is in a house with four windows arranged in a rectangular shape. He needs to signal to the sea at night by lighting a window or several windows. How many different signals can he send? | 25. Let's schematically represent the windows and measure them:
a) Lighting all four windows gives one signal;
b) Lighting one of the windows is perceived as one signal, as in the dark, th... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,684 |
26. How many sides can a figure have that is the common part of a triangle and a convex quadrilateral? | 26. As shown in Figure 28, the common part of a triangle and a quadrilateral can be a segment, a triangle, a quadrilateral, a pentagon, a hexagon, and a heptagon. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,685 |
27. Can 1989 sticks of the same length be laid out, without overlapping, to form a closed broken line, if the sticks are placed only in horizontal and vertical positions? | 27. Since the broken line must be closed, starting to lay out the sticks from a certain point, we must return to this point. This means that as many sticks are laid out in one direction as in the opposite direction. For example, if $a$ sticks are laid out downward, then $a$ will be laid out upward, and if $b$ sticks ar... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,686 |
30. What is the maximum number of intersection points that a closed broken line consisting of 7 segments can have? (The common ends of the segments of the broken line are not counted.)
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 30. Take a broken line consisting of two segments. By adding a third segment, it can only intersect one of the two original segments, resulting in one point (1) (Fig. 30). Taking a fourth segment, it can intersect a maximum of two of the three existing segments, resulting in two more points $(2,3)$. A fifth segment can... | 14 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 39,689 |
31. A rectangular children's playground must meet the following requirements: the lengths of the sides must be expressed in whole numbers of meters, and the area must numerically equal the perimeter. What can be the length and width of such a playground? | 31. If the width is $a$ m, and the length is $b$ m, then according to the condition we have $2 a+2 b=a b$, from which $2=b \cdot \frac{a-2}{a}$. The equality is possible when $a>2$ and under the conditions: 1) $a=b=4$; 2) $a=3 ; b=6$. Therefore, the possible sizes of the plots are: 1) $4 \times 4\left(\mathrm{M}^{2}\ri... | 4\times4(\mathrm{M}^{2});6\times3(\mathrm{M}^{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,690 |
32. A tank, having the shape of a rectangular parallelepiped, has the property that if it is placed on any of its side walls, the area of the resulting base does not decrease compared to the given one. What is the maximum possible volume of this tank if its height is 1.5 m? | 32. If the dimensions of the base are $a$ m and $b$ m, then the areas of the lateral faces are $1.5 a \mathrm{M}^{2}$ and $1.5 b \mathrm{M}^{2}$. According to the condition $1.5 a \geqslant a b$ and $1.5 b \geqslant a b$, from which $b \leqslant 1.5$ and $a \leqslant 1.5$. Therefore, the possible maximum volume of the ... | 3.375 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,691 |
34. In what time will the end of the minute hand of a clock describe an arc equal to the length of this hand? | 34. In 1 hour, the tip of the minute hand will describe an arc of length $2 \pi R$, where $R$ is the length of the hand. Then the tip of the minute hand will describe an arc of length $R$ in
$$
\frac{60 \text { min }}{2 \pi R} \cdot R=\frac{30 \text { min }}{\pi} \approx 9.6 \text { min. }
$$ | \frac{30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,693 |
37. On the sides of a square, semicircles are constructed with the sides as diameters (Fig. 11). Determine the area of the shaded figure (four petals), if the side of the square is $a$. | 37. The area of a semicircle is $\frac{\pi a^{2}}{8}$, and the sum of the areas of four such semicircles is $\frac{\pi a^{2}}{2}$. The area of the square is $a^{2}$. Therefore, the desired area is:
$$
\frac{\pi a^{2}}{2}-a^{2} \approx a^{2}\left(\frac{11}{7}-1\right)=\frac{4}{7} a^{2} . \quad\left(\frac{\pi}{2} \appro... | \frac{4}{7}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,696 |
39. Show that in any convex dodecagon, there are two diagonals such that the angle between them is no more than $3^{\circ}$. | 39. If there are two parallel diagonals, the angle between them is $0^{\circ}$, i.e., less than $3^{\circ}$. If no two diagonals are parallel, then by taking a point on the plane and drawing lines through it parallel to each of the diagonals, we will obtain all the angles between the diagonals. There will be a total of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,698 |
40. The lengths of the edges of a rectangular parallelepiped are expressed in natural numbers, and its volume is a prime number greater than 2. Show that the sum of the lengths of the edges emanating from one vertex is an odd number. | 40. Since the volume of the parallelepiped is a prime number and greater than 2, it is odd, i.e., of the form $(2 n+1)$. Let the lengths of the edges be $a, b$ and $c$, then $a \cdot b \cdot c=2 n+1$. Since $a b c$ is a prime number, then $a=b=1$ and $c=2 n+1$, but in this case $(2 n+1)+1+1=2(n+1)+1$ is an odd number. ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,699 |
5. Recalling the autumn cross,
Two squirrels argue for two hours:
- The hare won the race,
And the fox was second!
- No, - insists the other squirrel, -
Stop these jokes of yours,
The hare was second, of course,
First was, I remember, - it was!
104
- I, - solemnly spoke the owl, -
Will not meddle in your quar... | 5. If the hare was first or second, then both statements of one of the squirrels would be false. Therefore, it is true that the fox was second, and the moose was first. | Themoosewasfirst,thefoxwas | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,706 |
7. Three indefatigable travelers had to cross a river in a boat that could carry no more than 100 kg, from one bank to the opposite bank. Andrey knew the result of his recent weighing, 54 kg, and that of his friend Oleg, 46 kg. As for Uncle Misha, he weighed 98 kg. How should they act in the most rational way to cross ... | 7. First, Andrey and Oleg cross the river in the boat, then one of them, for example, Andrey, returns back with the boat, in which Uncle Misha then crosses. After this, Oleg returns back with the boat and crosses together with Andrey to Uncle Misha. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,708 |
8. The car is traveling at a speed of $60 \mathrm{km} /$ h. By how much does the speed need to be increased to gain one minute on each kilometer? | 8. If a car is traveling at a speed of $60 \mathrm{km} /$ h, it takes 1 min to cover 1 km. Under such conditions, it is impossible to gain a minute on each kilometer with any increase in speed. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,709 |
9. Vremyankin and Puteykin simultaneously set out from Utrenneye to Vechernoye. The first of them walked half of the time spent on the journey at a speed of $5 \mathrm{km} / \mathrm{h}$, and then at a speed of $4 \mathrm{km} /$ h. The second, however, walked the first half of the distance at a speed of 4 km/h, and then... | 9. Vremyankin walked half of the time at a speed of 5 km/h, therefore, walking at this speed, he covered more than half of the distance, while Puteykin, walking at this speed, only covered half of the distance, which is why Vremyankin arrived at the final destination earlier than Puteykin. | Vremyankin | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,710 |
10. Working in a collective farm orchard harvesting fruits, schoolchildren collected 22 crates, some of which contained apples, others pears, and still others plums. Can it be asserted that there are at least 8 crates, the contents of which are one of the specified types of fruit? | 10. Let's consider the most "unfavorable" case. Suppose there are 7 boxes of apples, pears, and plums each. This makes a total of 21 boxes. If the 22nd box is with apples, then there will be 8 boxes of apples; if this box is with pears, then there will be 8 boxes of pears; if this box is with plums, then there will be ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,711 |
11. In the school, there are 33 classes and 1150 students. Will there be a class in this school with at least 35 students? | 11. If we assume that each class has fewer than 35 students, then the total number of students would not exceed $34 \cdot 33=1122$. In reality, however, there are 1150 students in the school. Therefore, the assumption is incorrect: there is a class with no fewer than 35 students. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,712 |
12. When Svetlana stated that she had 25 copper coins, her older sister Tatyana said that among these coins, there were no fewer than 7 of the same denomination. Is Tatyana right? | 12. Copper coins come in denominations of 1 k., 2 k., 3 k., 5 k., i.e., four types in total. If you take 6 coins of each type, you will have 24 coins in total. Taking one more (the twenty-fifth) coin, you will have 7 coins of one denomination. In other cases, there will be more than 7 coins of one denomination. Therefo... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,713 |
13. In a dark storeroom, there are boots of the same size: 12 pairs of black and as many brown. What is the smallest number of boots that need to be taken so that among them there is at least one pair (left and right boot) of the same color, if in the dark it is impossible to distinguish not only the color of the boot ... | 13. Let's consider the most "unfavorable" case. If you take 24 shoes, they might all be for one foot. By taking one more shoe, it will definitely match one of the previously taken ones. It is possible that you initially took 12 shoes for different feet but of different colors, but in this case, the result will be the s... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,714 |
14. In 7 boxes, there are a total of 100 parts. In any two of these boxes, there is a different number of parts. Is it necessarily true that there will be 3 boxes among them containing no fewer than 50 parts? | 14. Let us have $100=18+17+15+14+13+12+11$. In this case, $18+17+15=50$. If any of the last four addends is replaced by a smaller number, it is necessary to increase the sum of the first three addends, as a result, it will become greater than 50. If, however, at least one of the last four addends is replaced by a large... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,715 |
15. In each of 376 boxes, there are no more than 125 parts. Show that in at least 4 boxes, there is the same number of parts. | 15. Let's consider the most "unfavorable" case: in one box there is one part, in another - two, in the third - three, and so on. There will be 125 such boxes with different contents. But 125 is contained in the number 376 three times with a remainder of $1(376=125 \cdot 3+1)$. Therefore, in this case, there will be 3 b... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,716 |
16. In the box, there are 120 colored pencils: 35 red, 23 green, 14 yellow, 26 blue, 11 brown, and 11 black. What is the smallest number of pencils that need to be taken from the box in the dark (without seeing the pencils) to ensure that there are at least 18 pencils of one color among them? | 16. Let's consider the most "unfavorable" case: among the taken pencils, there could be 14 yellow, 11 brown, 11 black, 17 red, 17 green, and 17 blue. Thus, if we take 87 pencils $14+11+$ $+11+17+17+17=87$, it might happen that among them there will not be 18 pencils of the same color. By taking one more pencil, we will... | 88 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,717 |
17. In a photo lab, there are 130 photographs, of which 35 are from the fourth grade, 30 from the fifth, 25 from the sixth, 20 from the seventh, and the rest are from grades 1-III. The photographs are mixed and cannot be distinguished by touch. What is the smallest number of photographs that need to be taken, without s... | 17. Let's consider the most "unfavorable" case: we take all the photographs of classes I-III $(130-(35+30+25+20)=20)$, then it may turn out that 14 photographs were taken from each of the IV-VII classes, making a total of $14 \cdot 4+20=76$ photographs. Now, if we take one more, then one of the IV-VII classes will have... | 77 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,718 |
18. The traffic inspector noticed that out of 20 cars that passed on the road to the airport, 14 were "Ladas," 15 were dark-colored, 17 were driven by men, and in 18 cars, there were passengers besides the driver. For what minimum number of cars could all 4 of these characteristics be true? | 18. Out of the total number of cars, 6 did not have the first feature (not "Zhiguli"), 5 did not have the second, 3 did not have the third, and 2 did not have the fourth. Therefore, the maximum number of cars that do not have at least one feature is 16 $(6+5+3+2=16)$. The remaining $4(20-16=4)$ have all four features. | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,719 |
20. Schoolchildren helped the collective farm workers to harvest vegetables. The fourth grade collected 18 tons of cabbage and as much carrots as the sixth grade collected onions. The fifth grade worked only on cucumber harvesting and collected a whole number of tons of cucumbers, which is less than the amount of cucum... | 20. The sixth grade collected 9 centners of cucumbers $(18: 2=9)$ and 7 centners of onions, which means the fourth grade collected 7 centners of carrots. The fifth grade collected fewer cucumbers than 9 centners but more than 7 centners, so 8 centners. Thus, the students collected in total $18+7+8+9+7=49$ (centners). | 49 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,721 |
21. In a family, there are 4 children aged 5, 8, 13, and 15, and their names are Tanya, Yura, Svetlana, and Lena. How old is each of them, if one girl attends a kindergarten, Tanya is older than Yura, and the sum of Tanya's and Svetlana's ages is divisible by 3? | 21. The girl who attends the kindergarten is the one who is the youngest, i.e., 5 years old. Tanya cannot be 5 years old, as she is older than Yura. She cannot be 15 years old either, because 15, when added to any of the given numbers, does not divide by 3. She also cannot be 8 years old, since 5 years old is not for Y... | Tanya:13,Yura:8,Svetlana:5,Lena:15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,722 |
22. The chain has broken into five parts,
But I hope you will
Join them as quickly as you can,
Doing the minimum work.
To cut and solder a link-
It takes two minutes...
Remember, the task is given
For a very thoughtful look.
Note. Each link has three rings. | 22. Cut 3 rings of one link and join the remaining 4 links with the three resulting rings (by welding). This will take 6 minutes. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,723 |
23. Approaching the river, the travelers asked the boat owner to ferry them to the opposite bank. Since the boatman was distrustful of foreign money, the travelers offered him a gold chain they had, consisting of 6 links. The boatman agreed, but on the condition that he would transport all the travelers one by one, as ... | 23. The travelers cut the third link, dividing the chain into 3 parts. For the first trip, they paid with one link (the cut one); for the second, they gave the boatman 2 links, receiving one link as "change"; for the third, they gave 3 links, receiving 2 links as "change"; for the fourth, they paid with the cut link; f... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,724 |
24. The residents of city A only tell the truth, the residents of city B only lie, and the residents of city C alternate between truth and lies (i.e., out of two statements made by them, one is true and the other is false). The fire station duty officer was informed over the phone: "We have a fire, come quickly!" "Wher... | 24. If the call was from $\mathrm{A}$, the answer to the question: "Where?" would be: "In city A". It could not have been from B either, because out of the two statements: "There is a fire here" and "In city B", one cannot be true and the other false. Therefore, the call must have been from city B. But since they alway... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,725 |
25. In one book, the following 100 statements were written:
1) “In this book, there is exactly one false statement.”
2) “In this book, there are exactly two false statements ...”
3) “In this book, there are exactly one hundred false statements.”
Which of these statements is true? | 25. If we assume that the first statement is true, this would mean that there is exactly one false statement in the book, and all the others are true, i.e., the second, third, and so on are true, which contradicts the first statement, since, for example, the second statement claims that there are exactly two false stat... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,726 |
26. There are 28 liters of gasoline in a barrel. There are two buckets with a capacity of 7 liters each, into which 6 liters of gasoline need to be poured. In addition, there is a dipper with a capacity of 4 liters. How can the gasoline be distributed? | 26. 27) Fill one bucket from the barrel, then use a dipper to take out 4 liters from the bucket and pour it back into the barrel. As a result, 3 liters will remain in the bucket.
2) Pour 3 liters from this bucket into the dipper, then fill the bucket from the barrel. Use this bucket to top up the dipper, i.e., pour 1 l... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,727 |
28. A problem solved by Poisson (1781-1840) in his youth. Someone has 12 pints (a unit of volume) of honey and wants to pour out half of this amount, but he does not have a container with a capacity of 6 pints. He has 2 containers: one with a capacity of 8 pints, and the other with a capacity of 5 pints. How can he pou... | 28. Let's represent it with such a diagram:
| Vessel | | |
| :---: | :---: | :---: |
| 12 pints | 8 pints | 5 pints |
| 12 | - | - |
| 4 | 8 | - |
| 4 | 3 | 5 |
| 9 | - | - |
| 1 | 8 | 3 |
| 1 | 6 | 3 |
| 6 | 6 | - | | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,728 |
29. How can you use a balance scale and a 200g weight to divide 9 kg of sugar into two bags - 2 kg and 7 kg, if you are allowed to weigh no more than three times? | 29. Place a weight on one pan and distribute the available sugar across the pans so that the scales are in balance. Then, on one pan there will be 4 kg 600 g of sugar, and on the other (where the weight is) - 4 kg 400 g of sugar. The second time, distribute 4 kg 600 g of sugar across the pans. On one of them (without t... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,729 |
30. Three Fridays of a certain month fell on even dates. What day of the week was the 18th of that month? | 30. This month has 5 Fridays, as Fridays in a month alternate between even and odd dates. The first Friday could only have been on the 2nd (otherwise, 3 Fridays cannot fall on even dates), which means the 18th is a Sunday. | Sunday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,730 |
31. In one month, three Wednesdays fell on even dates. What date will the second Sunday of the month be? | 31. If in one month three Wednesdays fall on even dates, this is possible under the condition that the month is not February and starts on a Tuesday. Then Wednesday will be on the 2nd, 16th, and 30th, and the second Sunday will be on the 13th. | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,731 |
32. In which cases does a month have the maximum number of Saturdays? What is this number? | 32. The maximum number of Saturdays in one month is 5, since $7 \cdot 4=28 ; 28<30$ and $28<31$. In February, there can be 5 Saturdays if it is a leap year and the month starts on a Saturday. April, June, September, and November will have 5 Saturdays if they start on a Saturday or a Friday. The remaining months can hav... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,732 |
33. The students of our class spent the first Sunday of the month in the Carpathians, and on the first Sunday after the first Saturday of this month, they went on an excursion to Kyiv. In the following month, on the first Sunday, they made an excursion to the forest, and on the first Sunday after the first Saturday, th... | 33. According to the problem, the first Sunday of the month and the first Sunday after the first Saturday do not coincide. This is possible only if the first Sunday of the month falls on the 1st, and the first Sunday after the first Saturday falls on the 8th. This means that both months started on a Sunday. Since the i... | February1,February8,March1,March8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,733 |
37. Football teams were holding a round-robin tournament. One of the fans of this game, returning from vacation, found the following situation: the total number of points scored by all teams was 44; one team, which scored the fewest points, received one point; the two teams occupying the first and second places scored ... | 37. Since $1+2+3+4+5+6+7+8+8=44$, there are 9 teams participating in the tournament. In total, they should score 72 points ( $8 \cdot 9=72$ ). Therefore, they still need to score 28 points $(72-44=28)$, which can be achieved in 14 games $(28: 2=14)$. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,737 |
38. In a tournament where each pair of teams played each other twice, 4 teams participated. For each win, two points were awarded, for a draw - one, and for a loss - 0. The team that finished in last place scored 5 points. How many points did the team that finished in first place score? | 38. There were 12 games in total $(3 \cdot 4=12)$. In each match, -2 points were awarded, and the total points were -24. Since $5+6+7+8=26>24$, this case is impossible.
Therefore, $5+6+6+7=24$. This means the winning team scored 7 points. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,738 |
39. In a chess tournament, each of the 8 participants plays one game with each other. All participants scored a different number of points (an integer), and the second-place finisher scored as many points as all the chess players who finished from fifth to eighth place combined. How did the players who finished in thir... | 39. The participant who took second place, out of a total of 28 points scored by all participants together, scored no more than six points, as otherwise, they would have taken either first place or shared the first and second places, which contradicts the condition. The only possible scenario is: $28=7+6+5+4+3+2+1+0$. ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,739 |
40. In a football tournament, the team "Torpedo", which took first place, scored 5 points, with a ratio of goals scored to goals conceded of 7:0. In second place is the team "Azovets", which scored 2 goals and conceded 3. They scored 4 points. In third place is "Energiya", with 2 points, 2 goals scored, and 6 goals con... | 40. "Builder" has one draw and the rest are losses. "Torpedo" can have 2 wins and one draw. There can be no more draws, as "Azovets" and "Energia" have an even number of points. The draw between "Torpedo" and "Builder" can only be a zero-zero, i.e., 0:0, since "Torpedo" has 7:0. Then it is easy to fill in the table for... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,740 |
41. In the table of a football tournament of four teams, only some records have been preserved: Restore the table:
| Team | «Start» | «Vimpel» | «Avangard» | «Zvezda» | Points | Overall score | Place |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| «Start» | $\mathrm{X}$ | $1: 1$ | | | | $: 3$ ... | 41. "Vympel" has one draw and lost the rest, as it received one point. "Avangard" conceded only one goal, so the second numbers in the columns with "Start" and "Vympel" are zeros, and the first numbers are obviously ones; then in the column with "Zvezda" it is also 1, as the total number of goals scored by "Avangard" i... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,741 |
42. Four football teams "Start", "Comet", "Rocket", and "Banner" each played one game with each other. After the rain, only some records remained on the table. But the referee remembered that the other matches ended with scores of $2: 0, 1: 1, 2: 2, 3: 1$, and $5: 3$. Help the referee restore the table. (In the "Goals"... | 42. "Start" won all matches, scoring 6 points. "Comet" has only 2 goals. One match she won 1:0 (against "Vimpel"), so she drew another match 1:1, and lost the third with a "clean" score. She could only lose to "Start" 0:2, so the draw was with "Rocket". "Rocket" lost to "Start" and, obviously, with a score of 3:5, as s... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,742 |
43. There are 4 watermelons of different weights. How, using a balance scale without weights, can you arrange them in ascending order of weight in no more than five weighings | 43. Let the masses of the watermelons be $a_{1}, a_{2}, a_{3}, a_{4}$. With one weighing, compare, for example, $a_{1}$ and $a_{2}$, and with the second weighing, compare $a_{3}$ and $a_{4}$. Suppose $a_{1}>a_{2}$ and $a_{3}>a_{4}$. With the third weighing, compare $a_{1}$ and $a_{3}$. If $a_{1}>a_{3}$, then $a_{1}>a_{... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,743 |
44. Out of four identical-looking parts, one differs in mass from the other three, but it is unknown whether its mass is greater or lesser. How can you identify this part with two weighings on a balance scale without weights? | 44. Let the masses of these parts be denoted as $a_{1}, a_{2}, a_{3}, a_{4}$. Take any two parts. Let, for example, the masses of these parts be $a_{1}$ and $a_{2}$. Compare them. In this case, there can be two scenarios: $a_{1} \neq a_{2}$ or $a_{1}=a_{2}$.
1) Let $a_{1} \neq a_{2}$. If $a_{1}>a_{2}$, then by placing,... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,744 |
45. Can a knight's move in chess get from the bottom left square of the chessboard to the top right, visiting each square exactly once? | 45. If the knight is on a black square, after making one move, it will land on a white square; the third square it lands on will be black, i.e., the same as the first; the fourth will be white, and so on. This means that the squares with even numbers are of a different color than the first. The knight, having visited a... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,745 |
47. The Badger invited guests: Bear, Lynx, and Squirrel, And they gifted the Badger A candlestick and a plate.
When he invited Lynx, Squirrel, Mouse, and Wolf, He received as a gift A candlestick and a needle. They were invited again Wolf, Mouse, and Sheep, And he received in gift A needle and a ring.
He invited again... | 47. Let's schematically represent the condition of the problem as follows:
By comparing schemes 1) and 2), we get that the bear gave the plate, and the sable or the squirrel gave the candle... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,747 |
48. On the way back from the Palace of Sports, the following dialogue took place between Nina and her coach:
- Do you have a brother?
- I have three brothers,- answered Nina.
- How old are they?
- The product of the numbers representing the age of each of them is 36, and the sum is the number of that bus standing at t... | 48. The total number of triplets of numbers, the product of which is 36, will be: 1) 1,4 , 9 (then $1+4+9=14$ ); 2) $1,6,6$ (sum 13); 3) $1,3,12$ (sum 16); 4) $1,2,18$ (sum 21); 5) 2, 2, 9 (sum 13); 6) 3, 3, 4 (sum 10); 7) 2, 3, 6 (sum 11); 8) 1, 1, 36 (sum 38).
If the coach had seen a bus number different from 13, he... | 2,2,9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,748 |
Problem 3. The bisector of angle $A$ of triangle $A B C$ intersects the circumscribed circle at point W. Prove that $b+c \leq 2 A W$.
10 | Proof. In the right triangle $A T W$ (Fig. 4) the leg $A T$ is not greater than the hypotenuse $AW$. But since $A T=\frac{b+c}{2}$ (Problem 2), the inequality $b+c \leq 2 A W$ is proved. | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,750 |
Problem 4. The bisector of angle $A$ of triangle $ABC$ intersects the circumscribed circle at point W. Prove the validity of the formula: $A W=\frac{b+c}{2 \cos \frac{A}{2}}$. | Proof. Since $A W$ (Fig. 4) is the bisector of angle $A$, from the right triangle $A T W$ we find $A W$:
$$
A W=\frac{A T}{\cos \frac{A}{2}}=\frac{b+c}{2 \cos \frac{A}{2}}
$$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,751 |
Problem 7. Construct triangle $ABC$ from the following elements: a, $A, b+c$.
This ancient problem, according to al-Biruni [3], belongs to Menelaus (1st century AD), who formulated it as follows: inscribe a two-segment broken line of a given length in a given arc. | Al-Biruni's Solution. On the segment \( BC = a \) and angle \( A \), we construct a segment. We mark point \( D \) - the midpoint of the chord \( CB \) (Fig. 7). On \( DC \), as a diameter, we describe a circle and from point \( C \) with the compass set to \( CK = \frac{b+c}{2} \) (by Archimedes' theorem), we make a m... | b+ | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,753 |
Problem 8. In and $C$ are fixed points on a circle, along which point $A$ moves arbitrarily. Point $K$ is the midpoint of the broken line $CAB$. Find the geometric locus of points $K$. | Solution. Let $K_{i}$ be the point of the desired locus. Let $D$ and $W$ be the midpoints of the upper and lower arcs $BC$ (Fig. 8). On $DB$ and $DC$, and on $BW$ and $CW$, as diameters, describe circles. By Archimedes' theorem, all points $K_{i}$ will be
12
located on the arcs $l_{1}$ and $l_{2}$, situated in the upp... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,754 |
Problem 9. Triangle $ABC$ is inscribed in a circle. $D$ is the midpoint of the arc $BC$ not containing $A$ (Fig. 9). $DK$ is perpendicular to $AC$. $M_{1}$ is the midpoint of side $BC$. Prove that $KM_{1}$ is parallel to $l_{a}$ ($l_{a}$ - the angle bisector of $\angle A$). | Proof. Draw a line $t$ from point $B$ parallel to $l_{a}$ until it intersects the extension of $A C$ at point $E$. It is not difficult to show that $A E=A B=c$. Then $C E=b+c$. But by Archimedes' theorem, $C K=\frac{b+c}{2}$. Therefore, $K M_{1}$ is the midline in triangle $C B E$, that is,
; after wh... | S_{ABC}=\frac{4}{3}S_{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,758 |
Problem 14. In triangle $ABC$, the lengths of its heights are known: $h_{a} ; h_{b} ; h_{c}$. Find the area of this triangle. | Solution. From the formula for the area of a triangle
$$
S=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c}
$$
we have:
$$
\begin{gathered}
a=\frac{2 S}{h_{a}} ; b=\frac{2 S}{h_{b}} ; c=\frac{2 S}{h_{c}} \\
p=\frac{a+b+c}{2}=S\left(\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}\right)
\end{gathered}
$$
... | \frac{1}{\sqrt{(\frac{1}{h_{}}+\frac{1}{h_{b}}+\frac{1}{h_{}})(\frac{1}{h_{}}+\frac{1}{h_{b}}-\frac{1}{h_{}})(\frac{1}{h_{}}+\frac{1}{h_{}}-\frac{1}{h_{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,759 |
Task 15. What does Heron's formula for a triangle look like, where the sides are expressed as irrational numbers? | Solution. According to Heron's formula
$$
16 S^{2}=(a+b+c)(b+c-a)(a+c-b)(b-c+a)
$$
or
$$
\begin{gathered}
16 S^{2}=\left((a+c)^{2}-b^{2}\right) \cdot\left(b^{2}-(c-a)^{2}\right)= \\
\left(a^{2}+2 a c+c^{2}-b^{2}\right) \cdot\left(b^{2}-c^{2}+2 a c-a^{2}\right)
\end{gathered}
$$
or
$$
16 S^{2}=(2 a c)^{2}-\left(a^{... | \frac{1}{4}\sqrt{4^{2}^{2}-(^{2}+^{2}-b^{2})^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,760 |
Problem 16. Prove the inequality for the elements of an arbitrary triangle:
$$
\frac{1}{(p-a)^{2}}+\frac{1}{(p-b)^{2}}+\frac{1}{(p-c)^{2}} \geq \frac{1}{r^{2}}
$$ | Proof. According to the inequality
$$
x^{2}+y^{2}+z^{2} \geq x y+y z+x z
$$
we have:
18
$$
\begin{gathered}
\frac{1}{(p-a)^{2}}+\frac{1}{(p-b)^{2}}+\frac{1}{(p-c)^{2}} \geq \\
\geq \frac{1}{(p-a)(p-b)}+\frac{1}{(p-b)(p-c)}+\frac{1}{(p-a)(p-c)}= \\
=\frac{p-c+p-a+p-b}{(p-a)(p-b)(p-c)}= \\
=\frac{3 p-2 p}{(p-a)(p-b)(... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 39,761 |
Problem 17. Prove the inequality
$$
(x+y+z)(x+y-z)(x+z-y)(z+y-x) \leq 4 x^{2} y^{2}
$$
where $x, y, z$ are the sides of some triangle. | Proof. Since $x, y, z$ are the sides of some triangle, it is necessary to prove the following inequality:
$2 p(2 p-2 x)(2 p-2 y)(2 p-z) \leq 4 x^{2} y^{2}$, or $16 S^{2} \leq 4 x^{2} y^{2}$, or $S \leq \frac{1}{2} x y$.
But since $S=\frac{1}{2} x y \sin \angle(x ; y)$, and $\sin \angle(x ; y) \leq 1$, the inequality ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 39,762 |
Problem 18. Among all triangles of a given perimeter, find the one that has the maximum area. | Solution. According to Heron's formula
$$
S^{2}=p \cdot(p-a) \cdot(p-b) \cdot(p-c)
$$
and by the condition of the problem $p=\frac{a+b+c}{2}=$ const. Note that:
$$
(p-a)+(p-b)+(p-c)=3 p-2 p=p=\text { const. }
$$
It is known that the product of three positive factors is maximized when they are equal, given a constan... | =b= | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,763 |
Problem 19. Prove the inequality for triangle $ABC$ with area $S$ and perimeter $2p$:
$$
S \leq \frac{p^{2}}{3 \sqrt{3}}
$$ | Proof. By the Cauchy inequality for three non-negative terms, we have:
$$
\frac{(p-a)+(p-b)+(p-c)}{3} \geq \sqrt[3]{(p-a)(p-b)(p-c)}
$$
or
$$
\frac{p}{3} \geq \sqrt[3]{(p-a)(p-b)(p-c)}
$$
After cubing, we get: $\frac{p^{3}}{27} \geq(p-a)(p-b)(p-c)$, or $\frac{p^{4}}{27} \geq p(p-a)(p-b)(p-c)$, i.e., $S^{2} \leq \fr... | S\leq\frac{p^{2}}{3\sqrt{3}} | Inequalities | proof | Yes | Yes | olympiads | false | 39,764 |
Problem 20. Prove the formula
$$
a b + b c + a c = r^{2} + p^{2} + 4 R r
$$
where $a, b, c$ are the sides, and $R$ and $r$ are the radii of the circumcircle and incircle of a certain triangle. | Proof. By Heron's formula $S^{2}=p \cdot(p-a) \cdot(p-b) \cdot(p-c)$, from which
$$
\begin{gathered}
p r^{2}=(p-a)(p-b)(p-c)=p^{3}-p^{2}(a+b+c)+p(a b+b c+a c)-a b c= \\
=-p^{3}+p(a b+b c+a c)-4 p r R
\end{gathered}
$$
After dividing by $p$ we get: $r^{2}=-p^{2}+a b+b c+a c-4 r R$, which is equivalent to the proof of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,765 |
Problem 21. In a triangle with a perimeter of $2 \sqrt{3}$, the product of its three angle bisectors is 1, and the radius of the inscribed circle is $\frac{1}{3}$. Find the angles of the triangle. | Solution. We will use the formula for the angle bisector of a triangle:
20
Example 3. Application of Heron's formula
$$
I_{a}=\frac{2 \sqrt{b c p(p-a)}}{b+c} \text { (prove it!). }
$$
Similarly,
$$
I_{b}=\frac{2 \sqrt{a c p(p-b)}}{a+c} ; I_{c}=\frac{2 \sqrt{a b p(p-c)}}{a+b}
$$
Then, taking into account the condi... | 60;60;60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,766 |
Problem 22. Find the minimum value of the expression $(x+y) \cdot(x+z),$ if $x, y, z-$ are positive numbers and $x \cdot y \cdot z \cdot(x+y+z)=1$. | Solution. Let $a=y+z ; b=x+z ; c=x+y$. Consider a triangle with sides $a, b, c$ (it is not difficult to verify that such a triangle exists for any $x, y, z$). Then:
$$
\begin{gathered}
(x+y) \cdot(x+z)=b \cdot c \\
x y z(x+y+z)=(p-a)(p-b)(p-c) p=S^{2}=1
\end{gathered}
$$
But $b \cdot c=\frac{2 S}{\sin A}=\frac{2}{\si... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,767 |
Problem 23. For angles $A, B$, and $C$ of an arbitrary triangle, prove the formula:
$$
\operatorname{ctg} \frac{A}{2}+\operatorname{ctg} \frac{B}{2}+\operatorname{ctg} \frac{C}{2}=\operatorname{ctg} \frac{A}{2} \cdot \operatorname{ctg} \frac{B}{2} \cdot \operatorname{ctg} \frac{C}{2}
$$
## Drawings in the sand | Proof. It is not difficult to show that $\operatorname{ctg} \frac{A}{2}=\frac{p-a}{r}$. Similarly, $\operatorname{ctg} \frac{B}{2}=\frac{p-b}{r}, \operatorname{ctg} \frac{C}{2}=\frac{p-c}{r}$. Substitute these values into the formula that needs to be proved. We have:
$$
\frac{p-a}{r}+\frac{p-b}{r}+\frac{p-c}{r}=\frac{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,768 |
Problem 30. In a right triangle $ABC$, angle $C$ is a right angle. The line $K_{1} K_{3}$ intersects line $BC$ at point $N$, and $K_{2} K_{3}$ intersects line $AC$ at point $T$. Prove that the area of triangle $TCN$ is half the area of triangle $ABC$.
| Solution. Let $A K_{3}=A K_{2}=m_{\text {; }}$
$B K_{3}=B K_{1}=n$ (Fig. 15).
$\Delta A J K_{2}=\Delta T K_{2} C$ (prove it!).
Then $T C=m$. Similarly, from the equality of $\triangle B J K_{1}$ and $\Delta N K_{1} C$, it follows that $C N=n$. Then:
$$
S_{T C N}=\frac{1}{2} m n
$$
Since $S_{A B C}=m \cdot n$, then... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,771 |
Problem 31. In quadrilateral BECN, $\quad BE=CE$ (Fig. 16)
\[
\angle BHC=\beta, \angle BEC=\varphi
\]
(outside the quadrilateral).
Given that $\varphi=2\beta$. Prove that points B, C, and H lie on a circle with center E.
. Suppose that point $H$ does not belong to this circle. Then point $D$ belongs to it and $\varphi=2 \angle B D C$ (the central angle is twice the corresponding inscribed angle). But by the condition, $\varphi=2 \beta$. It is n... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,772 |
Problem 32. Chords $AB$ and $CD$ of a circle with center $O$ intersect at point $K$ and are divided into segments $a, b, c, d$ (Fig. 21). Prove that $ab = cd$. | Proof. Let
$A K=a ; B K=b ; C K=c ; D K=d$.
Connect the ends of the chords and point $K$ with the center of the circle. Then, by formula I for $\triangle A O B$ and $\triangle C O D$, respectively, we get:
. }
$$
From which
$$
A K \cdot K B=R^{2}-d^{2}=\text { const } .
$$
 | R^2-^2 | Geometry | proof | Yes | Yes | olympiads | false | 39,774 |
Problem 35. From the vertex $A$ of an isosceles triangle $ABC$ ($AB = AC$), a random line is drawn, intersecting $BC$ at point $D$ and the circumcircle of $\triangle ABC$ at point $E$ (Fig. 24). Prove that $AE \cdot AD = \text{const}$. Find $AE \cdot AD$ if $AB = AC = b$.
$$
AE \cdot AD = (AD + DE) \cdot AD =
$$
. The... | ^2=2bm | Geometry | proof | Yes | Yes | olympiads | false | 39,775 |
Problem 37. The base of an isosceles triangle relates to its lateral side as 4:3 (Fig. 28). In what ratio does the height, drawn to the lateral side, divide it? | Solution. According to formula II:
$$
\begin{aligned}
& 4 x=\sqrt{2 \cdot 3 x \cdot m} \text {, or } 16 x^{2}=6 x m, \\
& \text { hence } m=\frac{8}{3} x . \text { Then } \\
& \quad n=3 x-\frac{8}{3} x=\frac{1}{3} x, \text { therefore, } n: m=1: 8 .
\end{aligned}
$$
}{2}$, or
$$
h=\sqrt{\frac{b(2 b-m)}{2}}
$$
$S_{A B C}=\frac{1}{2} a h=\frac{1}{2} \sqrt{2 b m} \cdot \sqrt{\frac{b(2 b-m)}{2}}$, or:
$$
4 S_{A B ... | \frac{b}{2}\sqrt{\frac{2b}{2b-}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,777 |
Problem 39. The height drawn to the base of an isosceles triangle is equal to 5 cm. The height drawn to the lateral side is 6 cm. Find the base. | Solution. Let $A D=5$ cm and $C K=6 \mathrm{~cm}$
(Fig. 30). By formula II $a^{2}=2 b m$, but
$$
\begin{aligned}
& b=\sqrt{25+\frac{a^{2}}{4}}(\text { from } \triangle A D C) ; \\
& \left.m... | 7.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,778 |
Problem 40. Prove geometrically that:
a) $\operatorname{ctg} 30^{\circ}+\operatorname{ctg} 75^{\circ}=2$;
b) $\operatorname{tg} 15^{\circ}=4 \sin ^{2} 15^{\circ}$. | Solution. Consider $\triangle ABC (AB = AC)$ with an angle of $30^{\circ}$ at the vertex (Fig. 31). Draw $CK$ - the height to the lateral side. Clearly, $CK = \frac{b}{2}$ as the leg opposite the $30^{\circ}$ angle.
a) By formula II $a^{2} = 2bm$, hence
$$
m = \frac{a^{2}}{2b}
$$
. Prove that the radii $\mathrm{CO}_{2}$ and $\mathrm{BO}_{1}$ are parallel. | Solution. It is not hard to see that this problem is the inverse of Archimedes' lemma on parallel diameters. We propose to prove it independently.
How does the lemma on parallel diameters help in solving problems? Let's consider some examples.
34 | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,781 |
Problem 44. Construct a circle that is tangent to a given circle and a given line at a given point $M$ on the line. | Construction instruction. Draw the diameter $AB$ of the given circle, perpendicular to the given line. By connecting point $A$ and point $M$, we find point $E$ (Fig. 33), which, according to the lemma, is the point of tangency of the desired and given circles. Then, by drawing a perpendicular from point $M$ and extendi... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,782 |
Problem 45. Through the given point $K$, draw a circle that is tangent to the given circle at the given point $T$ on it. | Construction instruction. Extend $K T$ until it intersects the given circle at point $E$ (Fig.35). Draw the diameter $E F$ of the given circle. Through point $K$, draw a line parallel to $E F$ and extend $F T$ until it intersects this line at point $N$. Then
 | Construction instruction. Let $O_{1}$ and $\mathrm{O}_{2}$ be the centers of the given circles (Fig. 36). We will indicate the sequence of actions for constructing the desired circle. Draw a line containing the diameter $O_{1} T_{\text{, obviously, it also contains the diameter of the desired circle. }}$ Parallel to $O... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,784 |
Problem 47. Two circles touch externally at point $A$. Through point $A$, a secant is drawn, intersecting the first circle at point $B$, and the second circle at point $C$. Prove that the tangents to the circles at points $B$ and $C$ are parallel (Fig. 37). | Solution. By the converse of Archimedes' Lemma, the radii of the circles drawn to points $B$ and $C$ are parallel, hence the corresponding tangents are also parallel,
 | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,785 |
Problem 48. Three circles touch each other pairwise at three different points. Prove that the lines connecting the point of tangency of two circles with the other two points of tangency intersect the third circle at points that are the endpoints of its diameter. | Solution. According to the lemma about parallel diameters, $M_{3}$ is parallel to $O_{1} C$ (Fig. 38) and $O_{3} K$ is parallel to $\mathrm{CO}_{2}$. Points $O_{1}, C_{1}$ and $O_{2}$ lie on the same line, so points $M, O_{3}$, and $K$ also lie on the same line. Therefore, points $M$ and $K$ are the endpoints of the di... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,786 |
Problem 49. Two circles touch externally at point $A$.
A secant is drawn through point $A$, intersecting the first circle at point $B$ and the second circle at point $C$. Prove that the rat... | Solution. Let $AB$ be a chord of the circle with center $\mathrm{O}_{2}$, and $\mathrm{AC}$ a chord of the circle with center $O_{1}$. By Archimedes' lemma, $O_{1} C$ is parallel to $O_{2} B$. It is then not difficult to show that triangles $O_{1} C A$ and $A B O_{2}$ are similar. Therefore, the ratio of $A C$ to $A B$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,787 |
Problem 50. Two circles with radii $R$ and $r$ touch externally at point $A$. A secant is drawn through point $A$, intersecting the first circle at point $B$ and the second circle at point $C$. A tangent $B D$ is drawn from point $B$ to the second circle. Find the length of $B D$, if $B A=a$. | Solution. From the previous problem, it is known that $A C=\frac{a \cdot r}{R}$, where $R$ and $r$ are the radii of the first and second circles, respectively. Then $B D^{2}=(a+x) a=\left(a+\frac{a \cdot r}{R}\right) a=a^{2} \frac{R+r}{R}$.
Answer. $B D=a \sqrt{\frac{R+r}{R}}$. | BD=\sqrt{\frac{R+r}{R}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,788 |
Problem 51. Two circles, with radii $2r$ and $r$, touch internally at point K. A third circle of radius $r$ touches the smaller circle at point $M$ and intersects the larger circle again at point $T$, as shown in Fig. 39. Prove that points $K, M$, and $T$ lie on the same straight line. | Solution. Since $O_{2} O_{3}=O_{1} T=2 r$ and $T O_{3}=O_{1} O_{2}=r_{1} \quad$ then $\quad T O_{1} O_{2} O_{3} \quad-$ is a parallelogram. Therefore, $T O_{3}$ is parallel to $O_{1} K$. From this, by Archimedes' lemma, it follows that points $K, M$ and $T$ lie on the same line.
Another clear application of the lemma ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,789 |
Problem 52. Two circles touch each other internally at point $A$. Draw a chord $BC$ of the larger circle that touches the smaller circle at point $D$. Prove that $AD$ is the bisector of $\angle BAC$ (Fig. 40).
. $O_{1} D$ is perpendicular to $B C$, s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,790 |
Problem 54. In the segment including angle BAC, a circle is inscribed (Fig.41), touching the bisector of angle BAC at point N. Prove that point $N$ coincides with the incenter of triangle BAC. | Solution. Let $W$ be the midpoint of the arc $BC$ not containing $A$. It is known that $BW = CW = JW$, where $J$ is the incenter of triangle $ABC$. According to problem 53, the length of the tangent to any circle inscribed in the segment $BAC$ is $BW$. Therefore, $BW = CW = NW$, but point $N$
. A chord of the larger circle is tangent to the two smaller circles. Point $C$ is the midpoint of this chord. Prove that $X$ is the incenter of triangle $ABC$.
Figure 42. | Solution. Through point $C$, draw the diameter $M N$, which is obviously perpendicular to the given chord. Through point $X$, draw the diameters of the smaller circles $K X$ and $X L$. Then, by the lemma of parallel diameters, points $B, X, M$ lie on the same line and points $A, X, N$ also lie on the same line. Connect... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,793 |
Problem 57. In an inscribed quadrilateral, one of the diagonals is a diameter. Prove that the projections of opposite sides onto the other diagonal are equal. | Solution. In Fig. $46 A B C D$ is a cyclic quadrilateral, $A C$ is a diameter. $B H$ and $D M$ are the projections of sides $A B$ and $C D$ onto diagonal $D B$. According to the lemma about perpendiculars on a secant, segments $B H$ and $D M$ are equal.
. Line $\mathrm{H}_{2} \mathrm{H}_{3}$ is a secant of this circle, then, by the lemma about perpendiculars to the secant, segme... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,796 |
Problem 59. On a straight line, points $A, B$, and $C$ are given, with point $C$ lying between $A$ and $B$. On the segments $AB$, $AC$, and $BC$, as diameters, three circles are constructed (Fig. 48). Through point $C$, a chord $MQ$ of the larger one of them is drawn. Prove that the smaller circles intercept equal segm... | Solution. Let $M Q$, a chord of the larger circle, intersect the smaller circles at points $N$ and $P$. We need to prove that $M N = P Q$. Since $A C$ is a diameter, $\mathrm{AN} \perp \mathrm{MQ}$, and similarly, $B P \perp M Q$. Then, by Archimedes' lemma, the perpendiculars $A N$ and $B P$ from the ends of the diame... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,797 |
Problem 60. The circle inscribed in triangle $ABC$ touches side $BC$ at point $K_{1}$. The excircle of triangle $ABC$ touches $BC$ at point $T_{1}$. Prove that $C T_{1} = B K_{1}$ (Fig. 49). | Solution. Let $J$ and $J_{a}$ be the centers of the inscribed and excircle, respectively.
$$
\angle J C J_{a}=\angle J B J_{a}=90^{\circ}, \quad \text { since } \quad \text { these }
$$
are the angles between the bisectors of adjacent angles. Therefore, a circle can be circumscribed around quadrilateral $C{ }^{2} J_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,798 |
Problem 62. Given a circle with center $O$ and diameter $AB$ (Fig. 51). On the tangent to the circle at point $A$, point $C$ is given. From point $C$, a second tangent $CD$ is drawn. Prove that segment $DB$ is parallel to segment $OC$. | Solution. Since $AB$ is the diameter, the line $DB$ is perpendicular to $DA$, but the line $OC$ is also perpendicular to $DA$ (show this yourself), so the lines $DB$ and $OC$ are parallel.
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,800 |
Problem 63. In an isosceles trapezoid $ABCD$, a circle is inscribed, touching the sides of the trapezoid at points $T_{1}, T_{2}, T_{3}, T_{4}$ (see Fig.53). $K$ is the point of intersection of $CT_{3}$ and $T_{1}T_{2}$. In what ratio does point $K$ divide $T_{1}T_{2}$? | Solution. By Archimedes' Lemma ( $T_{3} T_{4}$ is the diameter, $T_{2} E$ is the perpendicular dropped from the point of tangency $T_{2}$ ), the line $C T_{3}$ bisects $T_{2} E$. Therefore, $T_{1} K$ is to $K T_{2}$ as $3: 1$.
. The angle $A C B$ is a right angle (see problem 83), so if we construct a circle with diameter $A B$, it will touch... | \frac{2Rr}{R+r} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,803 |
Problem 69. Two circles intersect at points $A$ and $B$. The center $O$ of the first circle lies on the second. A secant ODC is drawn through $O$ (see Fig. 62). Prove that point $D$ is the incenter of triangle $A B C$. | Proof. Since $O A=O B$, it is obvious that $CO$ is the bisector of angle
50
$A C B$. According to problem 67, $D B$ divides the arc $A C$ in half, therefore, $D B$ is the bisector of angle $A B C$. Thus, point $D$ is the incenter of triangle $A B C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,804 |
Problem 72. A plane intersects a sphere with center $O$ along a circle. Points $A$ and $B$ are taken on the sphere, on opposite sides of the plane, such that the radius $OA$ is perpendicular to the given plane. A plane is drawn through $AB$. It intersects the circle at points $X$ and $Y$. Prove that the product $B X \c... | Proof. In the section of the sphere by a plane passing through $AB$, a circle is obtained containing points $A, B, X$, and $Y$. Moreover, the values $AB=a$, $AX=AY=b$ do not depend on the choice of the plane, and $\angle ABX = \angle ABY$ (Fig. 65). Draw a circle with center $A$ and radius $b$. It will intersect the la... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,807 |
Problem 75. In a circle of radius $P$, two mutually perpendicular chords $M N$ and $P Q$ (Fig. 78) are drawn. Find the distance between points $M$ and $P$, if $N Q=m$. | Solution. According to Archimedes' Lemma
$$
a^{2}+b^{2}+c^{2}+d^{2}=4 R^{2}
$$
But $\quad a^{2}+c^{2}=M P^{2}$ and $\quad b^{2}+d^{2}=m^{2}$. Then we have: $M P^{2}+m^{2}=4 R^{2}$, from which
. But also
$$
n^{2}+t^{2}=a^{2}+b^{2}+c^{2}+d^{2}=4 R^{2}
$$
$$
\text { Therefore, }
$$
$$
m^{2}+k^{2}=n^{2}+t^{2}=4 R^{2} \text {. }
$$
Let us note in parentheses that, generally speaking, a stronger fact holds, whi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,810 |
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