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7. From 16 m of fabric, 4 men's and 2 children's coats were sewn. How many meters of fabric are needed to sew one men's and one children's coat, if from 18 m of the same fabric, 2 men's and 6 children's coats can be sewn | 7. Let's make a brief record of the problem's condition:
$$
\begin{aligned}
& 4 \text { m, } 2 \text { d }-16 \text { m; } \\
& 2 \text { m, } 6 \text { d }-18 \text { m. }
\end{aligned}
$$
To ensure that the brief record contains the same quantities, we will double the second order:
$$
\begin{aligned}
& 4 \text { m... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,775 |
1. The teacher cut a square sheet of paper with a side of 5 cm into two rectangles. The perimeter of one of these rectangles is $16 \mathrm{cm}$. What is the perimeter of the other? | 1. The first method of solving.
1) $16: 2=8$ cm - the semi-perimeter of the first rectangle;
2) $8-5=3$ cm - the width of the first rectangle;
3) $5-3=2$ cm - the width of the second rectangle;
4) $(2+5) \times 2=14 \text{cm} -$ the perimeter of the second rectangle.
The second method of solving. Before solving by the... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,776 |
2. The sum of three numbers is 1281. From the first number, 329 was subtracted, and to the second number, 401 was added. What needs to be done with the third number so that the sum does not change? Will the solution to the problem change if the word "sum" is replaced with "difference"? | 2. First method of solving. Find out by how much the sum has increased ( $401-329=72$ ). To keep the sum the same, it is necessary to subtract 72 from the third number.
Second method of solving. Let $a, b, c$ be the first, second, and third numbers. According to the condition, their sum is 1281
$$
a+b+c=1281
$$
From... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,777 |
3. Two trains leave from two cities at the same time. The first one travels at 40 km/h, while the second one travels at 48 km/h. How far apart will these trains be from each other after 8 hours, if they are moving in the same direction and the distance between the cities is 892 km? | 3. The trains are moving in the same direction, so they can move in the direction of $A B$ or $B A$. Let's consider each of these cases.
The trains are moving in the direction of $A B$.
1) $40 \times 8=320$ km - the first train traveled;
2) $48 \times 8=384$ km - the second train traveled;
3) $384-320=64$ km - by thi... | 956 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,778 |
4. 15 bags of potatoes and 12 bags of flour weigh 1 ton 710 kg, with the weight of one bag of potatoes being 30 kg less than the weight of one bag of flour. How much does one bag of potatoes and one bag of flour weigh? | 4. The first method of solving.
1) $30 \times 12=360$ kg - this is how much more 12 bags of flour weigh;
2) $1710-360=1350$ kg - this is the weight of potatoes and flour if they were of the same weight;
3) $15+12=27$ bags - the total number of bags if they were all the same;
4) $1350: 27=50$ kg - the weight of one bag ... | one\bag\of\flour\weighs\80\,\\one\bag\of\potatoes\weighs\50\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,779 |
5. It is known that the sum of two three-digit numbers, in the notation of which there are no identical digits, is a three-digit number, where the number of hundreds does not exceed 7 and is greater than 5, the number of tens is odd, and the number of units is 1. In one of the addends, the number of tens is 9, and in t... | 5. According to the condition of the problem, the first addend is of the form $A 9 B$, and the second is $C K 0$. The number of units in the sum is 1, so this number has the form: $M X 1$, i.e., we get the example
$$
A 9 B+C K 0=M X 1 .
$$
From this, we can conclude that $B=1$
$$
A 91+C K 0=M X 1 .
$$
The number of... | 240+391=631,340+291=631,260+391=651,360+291=651,280+491=771,480+291=771,260+491=751,460+29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,780 |
7. A rectangular sheet of iron was divided into two parts such that the first part was four times larger than the second. What is the area of the entire sheet if the first part is $2208 \mathrm{~cm}^{2}$ larger than the second? | 7. 8) $4-1=3$ hours - corresponds to $2208 \mathrm{~cm}^{2}$;
2) $2208: 3=736 \mathrm{~cm}^{2}$ - corresponds to one part;
3) $4+1=5$ hours - total equal parts in the rectangle;
4) $736 \times 5=3680 \mathrm{~cm}^{2}$ - area of the rectangle.
Answer: the area of the rectangle is $3680 \mathrm{~cm}^{2}$.
## EIGHTEENTH | 3680\mathrm{~}^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,782 |
1. In A.S. Pushkin's fairy tale "The Priest and His Workman Balda," it is said: "Balda sat on the mare and rode a verst so that the dust rose in a column." Determine how many tens of meters are contained in the length of the path that Balda rode on the mare. $(1$ verst $=500$ sazhen, 1 sazhen $=3$ arshins, 1 arshin $=$... | 1. 2) $3 \times 500=1500$ arshins - in one verst;
2) $71 \times 1500=106500$ cm - in one verst;
3) $106500: 100=1065$ m - in one verst;
4) $1065: 10=106$ tens (remainder 5) — the number of tens of meters.
Answer: in one verst there are 106 tens of meters. | 106 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,783 |
2. Mark four points on a piece of paper. How many lines can be drawn through them
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。
Note: The last sentence is a repetition of the instruction and should not be included in the translated text. Here is the correct translation:
2. Mark four points on a piece of paper. How many lines ... | 2. To begin with, it is necessary to recall that one and only one straight line passes through two points. After this, we should consider three cases:
1) four points lie on one straight line - one straight line;
2) three points lie on one straight line, and one does not belong to this line - 4 straight lines
3) four po... | Number Theory | proof | Yes | Yes | olympiads | false | 42,784 | |
3. At the mathematics olympiad in two rounds, it is necessary to solve 14 problems. For each correctly solved problem, 7 points are given, and for each incorrectly solved problem, 12 points are deducted. How many problems did the student solve correctly if he scored 60 points? | 3. Let's determine the number of problems solved by the student, given that he received 7 points for each problem: $60: 7=8$ (remainder 4). Since 12 points were deducted for each incorrectly solved problem, the number of correctly solved problems was more than 8. Further, by trial and error, we can establish that there... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,785 |
4. Three friends went to study to become a cook, a carpenter, and a painter. When they met their classmates, they said about their place of study: Victor - cook; Andrey - not a cook; Petr - not a carpenter. It is known that either Petr or Andrey made a mistake. Determine the profession each of them went to study. | 4. We know three statements, and it is said that either Andrey or Petr made a mistake. Let's consider each of these cases.
1 case. Suppose Andrey made a mistake. Then Andrey is the cook. And according to the first condition, Viktor is also a cook. Two people cannot be a cook at the same time. In this case, there is no... | Petr-carpenter,Viktor-cook,Andrey-painter | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,786 |
5. Find four consecutive even numbers whose sum is 3924. | 5. Since each even number differs from the previous even number by two, the second number differs from the first by two, the third by four, and the fourth by six. Together, they exceed the first number by $12(2+4+6=12)$.
Let's determine their sum if all were equal to the first number: $3924-12$ = 3912. Since in this c... | 978,980,982,984 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,787 |
6. One side of a rectangle is $5 \mathrm{~cm}$, and the other is the smallest two-digit number that is divisible by 3. For the second rectangle, one side is equal to the shorter side of the first. The area of one rectangle is $25 \mathrm{~cm}^{2}$ greater than the area of the other. Determine the unknown side of the se... | 6. First, let's determine the unknown side of the rectangle. For this, we will find the smallest two-digit number that is divisible by 3, which is 12. According to the problem, one side of the second rectangle is $5 \mathrm{~cm}$. Let's determine the area of the first rectangle $(12 \times 5=60)$. Since the problem doe... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,788 |
7. To number the pages in a mathematics textbook, 390 digits were required. How many pages are in the mathematics textbook? | 7. For writing the first nine pages, 9 digits are needed $(1,2,3,4,5,6,7,8,9)$.
For numbering pages from 10 to 99, 90 two-digit numbers are needed, which use $90 \times 2=180$ digits.
In the next step, we determine how many digits were used to write three-digit numbers $(390-180-9=201)$. Now we find out how many page... | 166 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,789 |
2. The pages of materials for conducting a math club session are numbered. The last page is numbered 128. How many digits were needed to number all the pages? | 2. For numbering the first nine pages, 9 numbers are needed.
For numbering pages from 10 to 99, 90 two-digit numbers are needed, which use $90 \times 2=180$ digits.
Let's find out how many three-digit numbers were used for numbering the pages. There will be $29(128-99=29)$. Determine how many digits were used to writ... | 276 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,791 |
3. If the thought number is multiplied by 6, and then 382 is added to the product, the result will be the largest three-digit number written with two identical even numbers and one odd number. Find the thought number. | 3. Let's determine the largest three-digit number written with two identical even digits and one odd digit — this is 988. Subtract 382 from the obtained number, we get 606. This is the sixfold of the thought number. Dividing 606 by 6, we determine the thought number -101.
Answer: 101. | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,792 |
4. The perimeter of a rectangle is 126 cm, and the difference between its sides is $37 \mathrm{~cm}$. Find the area of the rectangle. | 4. 5) $126: 2=63$ cm - the sum of the length and width of the rectangle;
2) $63+37=100$ cm - the doubled length of the rectangle;
3) $100: 2=50$ cm - the length of the rectangle;
4) $63-50=13$ cm - the width of the rectangle;
5) $50 \times 13=650 \mathrm{~cm}^{2}$ - the area of the rectangle.
Answer: the area of the r... | 650\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,793 |
1. A sheet of cardboard measuring 48 cm in length and 36 cm in width needs to be cut into cards measuring 16 cm in length and 12 cm in width. How can this be done to obtain the maximum number of cards? | 1. In the first step, it is necessary to determine how many cards will fit if the length of each is $16 (48: 16=3)$. Then determine how many cards will fit in the width (36:12 = 3).
Finally, find the number of cards $(3 \times 3=9)$. Then conduct similar reasoning for the case where the cards are laid out by width ( $... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,797 |
2. The sum of three numbers is 121,526. One addend is the largest five-digit number, all digits of which are even; the second is the smallest four-digit number, all digits of which are odd. What is the third number? | 2. We will sequentially consider all the conditions of the problem. Given that the first addend is a five-digit number, it follows that five digits will be used for its notation (XXXXX). Since it must be the largest and all its numbers are even, this will be the number 88888. The second addend is a four-digit number (X... | 31527 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,798 |
3. In each of the four packages, there are balls: white, red, yellow, and green. Each package has a tag, but none of them correspond to reality. Determine the color of the balls that are in the packages. | 3. From the first condition, stating that the tag on each package does not correspond to reality, we can conclude that the yellow package does not contain yellow balls, the white one does not contain white balls, the red one does not contain red balls, and the green one does not contain green balls.
Assume that the wh... | white\in\the\red,\yellow\in\the\white,\green\in\the\yellow,\red\in\the\green | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,799 |
4. (Ancient Indian problem 3-4th century AD). Of the four sacrificial donors, the second gave twice as many coins as the first, the third three times as many as the second, the fourth four times as many as the third, and all together gave 132 coins. How many did the first give? | 4. The first method of solving. By analyzing the condition of the problem, one can conclude that the number of coins given by the first person will not be more than 10. We will then refine this number through trial and error. Let's try the number 5.
$$
5+5 \times 2+5 \times 2 \times 3+5 \times 2 \times 3 \times 4=5+10... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,800 |
5. Nikolai was walking to a math olympiad, covering 100 meters per minute. After 36 minutes, he stepped into School No. 15. Then he remembered that he had left the pen at home, with which he had won the school olympiad, and ran back home at a speed of 3 meters per second. How much time did he spend on the return trip? | 5. 6) $100 \times 36=3600$ m - the distance from home to school;
2) $3 \times 60=180$ m - the number of meters he ran per minute on the way back;
3) $3600: 180=20$ min - the time he spent on the return trip.
Alternatively, the second step could have determined the number of seconds spent on the return trip $(3600: 3=1... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,801 |
7. Cut a square with a side of 13 cm into five rectangles so that all ten numbers expressing the sides of the rectangles are different natural numbers. | 7. In the first step, represent the number 13 as the sum of two addends:
$$
1+12=2+11=3+10=4+9=5+8=6+7
$$
Using the obtained decomposition, we will sequentially break down the first three sides, and select the fourth side. As a result, we get the following solution:
 $65 \times 32=2080 \mathrm{~m}^{2}$ - area of the hall;
2) $250 \times 2080=520000 \text{ g}=520$ kg - amount of paint required to paint the hall;
3) $25 \times 23=575$ kg - total amount of paint available;
4) $575>520-$ the paint will be enough.
Answer: the paint will be enough. | thepaintwillbeenough | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,804 |
2. Two friends walked out of School No. 15 after the Olympiad and went in different directions, pondering a difficult problem. Misha walked 8 km per hour, while Vasya walked half the distance. After 45 minutes, they came up with a solution to the problem and started running towards each other to share the solution. Eac... | 2. The first method of solving.
1) $8: 2=4$ km $/ h$ - Vasya's speed;
2) $8+4=12 \text{km} / h=12000: 60=200 \text{m} /$ min - the speed of moving away from each other;
3) $200 \times 45=9000 \text{m}=9$ km - the distance the boys have traveled;
4) $8+4=12$ km $/ h$ - Misha's new speed;
5) $4+2=6$ km/h - Vasya's new sp... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,805 |
3. Replace the letters with digits so that the equality is correct. Different letters correspond to different digits.
$$
\begin{array}{ccccc}
& \text { A } & \text { B } & \text { A } & \text { C } \\
& + & \text { B } & \text { A } & \text { C } \\
\text { K } & \text { C } & \text { K } & \text { D } & \text { C }
... | 3. Since we are adding a four-digit number to a three-digit number and getting a five-digit number, the first digit in the sum is one $(K=1)$.
$$
\begin{array}{rrrr}
& \text { B } & \text { C } \\
+ & B & A & C \\
1 & C & 1 & \text { C }
\end{array}
$$
When adding the numbers in the units place, the result should be... | A=9,B=5,C=0,D=8,K=1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,806 |
4. Two boats departed from piers $A$ and $C$. The speed of the boat traveling from $A$ is 7 km per hour, and from $C$ is 3 km per hour. Determine after how much time the boats will meet, if the distance between the piers is 20 km, and the speed of the river current from $A$ to $C$ is 2 km per hour. | 4. Since the problem does not specify the direction in which the boats are moving, we need to consider various scenarios: they are moving towards each other; they are moving in the same direction downstream; they are moving in the same direction upstream; they are moving in opposite directions. Given that the boats mus... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,807 |
5. (An old entertaining problem). Divide 46 rubles into 8 parts so that each part is 50 kopecks (50 cents) more than the previous one. | 5. First method of solution. Let the first part be $x$, then the second is $x+50$, the third is $-x+50 \times 2$, the fourth is $-x+50 \times 3$, the fifth is $x+50 \times 4$, the sixth is $-x+50 \times 5$, the seventh is $-x+50 \times 6$, and the eighth is $x+50 \times 7$. We will calculate the sum of all parts.
$$
\... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,808 |
6. A wire 104 centimeters long was bent into a rectangular frame. The length and width of this frame are whole numbers of centimeters. In how many ways can the frame be obtained? | 6. First, let's determine what the sum of the length and width is, that is, we will calculate the semi-perimeter. For this, we divide 104 by 2 and get 52. Let's represent the number 52 as the sum of two addends:
$$
52=1+51=2+50=3+49=\ldots=25+27=26+26 \text {. }
$$
We get 26 ways.
Answer: 26 ways. | 26 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,809 |
7. Before you lies the face of a clock. Divide it into three parts with two straight lines so that the sum of the numbers in each part is the same.
 | 7. Let's find the sum of all numbers depicted on the clock face. For this, we will calculate the sum of the first twelve natural numbers:
$$
1+2+3+\ldots+11+12=78
$$
Determine the sum of the numbers in one part, \( 78: 3=26 \). Now we will select four numbers such that their sum is 26. We get the following options:
... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,810 |
1. It is required to decipher what numbers are encrypted, if the same digits are replaced by the same letters:
$$
\text { KIS }+ \text { KSI = ISK. }
$$ | 1.
$$
\text { KIS+ KSI = ISK. }
$$
From the fact that two three-digit numbers are added and the result is again a three-digit number, we can conclude that I $\neq 0$ and $K<5$, meaning K can take the values $1,2,3$ or 4. We get the following options:
$$
\begin{aligned}
& 1 I S+1 S I=I S 1 ; \\
& 2 I S+2 S I=I S 2 ;... | 495+459=954 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,811 |
2. From points $A$ and $B$, which are 300 km apart, two cars set off simultaneously. The speed of the car that left from $A$ is 40 km/h. Determine the speed of the second car, given that after two hours the distance between them was 100 km. | 2. The problem statement does not specify the direction in which the cars are moving, so all possible scenarios should be considered: moving in opposite directions, towards each other, both in the direction of $A B$, and both in the direction of $B A$.
Assume the cars are moving in opposite directions. This scenario, ... | 140 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,812 |
3. The great commander, Marshal of the Soviet Union, Georgy Konstantinovich Zhukov, was born in the village of Strelkovka, Kaluga Governorate. He lived for 78 years. In the 20th century, he lived 70 years longer than in the 19th century. In what year was G.K. Zhukov born? | 3. Let's make a brief record of the problem statement
1) $78-70=8$ years - doubled the time of life in the 19th century;
2) $8: 2=4$ years - G.K. Zhukov lived in the 19th century;
3) $1900... | 1896 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,813 |
4. If a schoolboy bought 11 pens, he would have 8 rubles left, but for 15 pens, he would be short of 12 rubles 24 kopecks. How much money did the schoolboy have? | 4. 5) $15-11=4$ pens - bought more the second time;
2) $8+12=20$ rubles - amount to four pens;
3) $20: 4=5$ rubles - cost of one pen;
4) $5 \times 11+8=63$ rubles - the schoolboy had.
Answer: the schoolboy had 63 rubles. | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,814 |
5. For several identical pictures, 104 rubles were paid. How much does one picture cost if more than 10 but less than 60 were bought and the price of one picture is a natural number. | 5. First method of solving. Let's factorize the number 104:
$$
104=2 \times 2 \times 2 \times 13
$$
Therefore, the cost of the pictures could be 2, 4, 8, 13, 26, 52, 104 rubles. We will exclude from these numbers those that do not satisfy the conditions that more than 10 but less than 60 were bought.
These numbers a... | 2,4,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,815 |
6. Given a square with a side length of 8. Each side of the square is divided into two equal segments by a point. Find the area of the quadrilateral whose vertices are the constructed points. | 6. The area of the inner quadrilateral can be obtained by subtracting the areas of four triangles from the area of the larger square. Each pair of such triangles forms a square with a side length of $4 \mathrm{~cm}$, and the area of this square is $16 \mathrm{~cm}^{2}$. Thus, the area of the desired quadrilateral is $3... | 32\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,816 |
7. Alyosha claims that he was 9 years old the day before yesterday, and he will turn 12 next year. Is this possible? | 7. Since Alyosha will turn 12 next year, he must be 11 this year. Since he was 9 years old the day before yesterday, such a situation is only possible at the end of one year and the beginning of another. Let's assume he was saying this on January 1, then on December 30, he was still 9 years old. Therefore, on December ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,817 |
2. A string 60 dm long was cut into three pieces. The length of one of them is 1 cm more than another and 1 cm less than the third. Find the length of each piece. | 2. First method of solving. Let's make a drawing according to the problem statement.
1) $600-3=597$ cm - the length of the string if all parts were as small as the smallest one
2) $597: 3... | 199 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,819 |
3. A sports team from one of the schools in Kaluga participated in ten shooting competitions, with each new competition they scored 2 points more than in the previous one. In the last competition, they scored 10 times more points than in the first. How many points did they score in the first and last competitions? | 3. Let the number of points scored in the first competition be denoted by $a$, then in each subsequent competition, the team scored two points more than in the previous one:
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a$ | $a+2$ | $a+2... | 220 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,820 |
6. From a class of 20 people, a team of three students is formed to participate in mathematics, Russian language, and informatics olympiads. In this class, all students excel academically. How many ways are there to form the team of olympiad participants, if each member of the team participates in one olympiad? | 6. We can select a participant for the mathematics olympiad in twenty ways, for the Russian language olympiad in nineteen ways, and for informatics in eighteen ways. In total, this results in 6840 ways $(20 \times 19 \times 18=6840)$.
Answer: 6840 ways to form the team. | 6840 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,823 |
1. From School No. 15 and School No. 24, which are 153 meters apart, two friends Ivan and Petr started running towards each other at the same time. Ivan's speed is 46 meters per minute. After one and a half minutes of running, they had 24 meters left to run before meeting. What was Petr's speed? | 1. 2) $46: 2 \times 3=69$ m - Ivan ran in one and a half minutes;
2) $153-(69+24)=60$ m - the distance that Pete ran;
3) $60: 3 \times 2=40$ m/min - Pete's speed.
Answer: Pete's speed is $40 \mathrm{m} /$ min. | 40\mathrm{}/\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,825 |
2. How many three-digit numbers less than 500 do not contain 1? | 2. Let's determine which digit can represent the hundreds place of $-2.3$ or 4. Consider numbers of the form $2 x x$. There will be a total of one hundred such numbers. The digit 1 can be in the tens or units place. If 1 is in the tens place ($21 x$), we get the numbers: $210, 211, 212, 213, 214, 215, 216, 217, 218, 21... | 243 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,826 |
3. Without changing the order of the digits, place operation signs between them so that the result of these operations is 1
$$
12345678=1 .
$$ | 3. First, it is necessary to recall how one can obtain a unit - by multiplying a unit by a unit, dividing a unit by a unit, or finding the difference between two consecutive numbers. Based on this, the search for a solution should be conducted. Here is one of the solutions: $12: 3: 4 \times 56: 7: 8=1$.
Answer: $12: 3... | 12:3:4\times56:7:8=1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,827 |
5. Three girlfriends went to the Olympiad. One of them was in a red dress, another in a yellow one, and the third in a blue one. Their notebooks were of the same colors. It is known that only Svetlana had the same color for her dress and notebook. Neither Tanya's dress nor her notebook were red, and Irina had a yellow ... | 5. Since Tanya did not have the red notebook, and Irina had the yellow one (also not the red one), the red notebook could only belong to Svetlana. Given that Svetlana has the red notebook and Irina has the yellow one, Tanya must have the blue notebook. Svetlana was in a red dress because the color of her dress and note... | Svetlana | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,829 |
1. The perimeter of a rectangle is 40 cm, and its area does not exceed $40 \mathrm{~cm}^{2}$. The length and width of the rectangle are expressed as natural numbers. By how much will the area of the rectangle increase if its perimeter is increased by 4 cm? | 1. Let's determine what the semi-perimeter of the rectangle is equal to $(40: 2=20)$. Since the sum of the length and width is 20, the sides can be 19 and 1, 18 and 2, 17 and 3, etc. Only the first two cases satisfy the second condition of the problem: 19 and $1(19 \times 1=19)$, 18 and 2 $(18 \times 2=36)$.
If the pe... | 2,4,21,36,38\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,832 |
2. Two friends collected 300 stamps. One collected 5 per week, the other - 3. The second boy collected for 20 weeks longer than the first. How many days did each boy collect stamps? | 2. 3) $20 \times 3=60 \text{m}$ - the second boy collected in 20 weeks;
2) $300-60=240 \text{m}$ - the boys collected during the time the first boy was collecting;
3) $5+3=8$ m - they collected per day;
4) $240: 8=30$ days - the first boy collected;
5) $30+20=50$ days - the second boy collected.
Answer: the first boy ... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,833 |
4. Every 24 hours, the clock runs ahead by 3 minutes. They were set accurately. After what shortest time will the clock hands show the exact time again? | 4. To make the clock show the exact time, it is necessary for it to run ahead by 12 hours. Let's find out how many minutes are in 12 hours $\operatorname{cax}(12$ hours $=60$ min. $\times 12=720$ min.). Determine the number of days $-720: 3=240$.
Answer: 240 days. | 240 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,835 |
5. For encrypting texts, the famous Swiss mathematician Leonhard Euler (1707-1783) used Latin squares. The figure below shows a Latin square containing four rows and four columns.
| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 2 | | | |
| 3 | | | |
| 4 | | | |
Fill in the remaining cells in the table so t... | 5. Let's provide one of the solutions
| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 2 | 1 | 4 | 3 |
| 3 | 4 | 1 | 2 |
| 4 | 3 | 2 | 1 | | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,836 |
6. In this century, the 200th anniversary of the birth of the famous Russian mathematician, a native of the Kaluga province P.L. Chebyshev, will be celebrated. In the number that records his year of birth, the sum of the numbers in the hundreds and thousands place is three times the sum of the numbers in the tens and u... | 6. Since in the 21st century the 200th anniversary of P.L. Chebyshev's birth will be celebrated, he was born in the 19th century - 18th. The sum of the numbers in the thousands and hundreds place is $9(1+8=9)$. The sum of the numbers in the tens and units place is three times less than 9. Therefore, it is $3(a+b=3)$. T... | 1821 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,837 |
7. How many numbers less than 2011 are divisible by 117 and 2? | 7. When solving this problem, one can reason as follows. First, list the numbers less than 2011 and divisible by 117 (2), then among them, select those that are divisible by 2 (117), and determine how many there are.
Since the number is divisible by both 2 and 117, it will also be divisible by their product $2 \times ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,838 |
1. A rectangular section of road with a length of 800 m and a width of 50 m was paved with asphalt. For every 100 m$^{2}$ of road, 3 tons 800 kg of asphalt were used. How many tons of asphalt were used? | 1. 2) $800 \times 50=40000 \mathrm{~m}^{2}$ - the area of the road;
2) $40000: 100=400$ times - the road contains $100 \mathrm{m}^{2}$ segments;
3) $3800 \times 400=1520000 \mathrm{~kg}=1520 \mathrm{~T}$ - the amount of asphalt used.
Answer: 1520 tons of asphalt were used. | 1520 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,839 |
2. What digits stand in the natural number sequence at the thirteenth and one hundred twentieth positions?
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。 | 2. Let's write down the first fifteen numbers:
$$
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
$$
To fill the first nine positions, 9 digits are required, and then two digits are needed to write each two-digit number. Starting from the tenth position, two-digit numbers are written. For each of them, two places are allocated. ... | 6 | Number Theory | proof | Yes | Yes | olympiads | false | 42,840 |
3. Two cars are driving towards each other on the Kaluga - Moscow highway. The speed of the first car is 60 km/h, and the speed of the second car is $1 / 5$ more than that of the first. Before reaching a bridge 2 km long, which can only be crossed by one car at a time, the first car has 120 km left to travel, and the s... | 3. 4) $60: 5=12$ km/h - the second car is faster by this speed;
2) $60+12=72$ km/h - the speed of the second car;
3) $120: 60=2$ h - the time it takes for the first car to reach the bridge
4) $180: 72=2$ (remainder 36) h - the time it takes for the second car to reach the bridge
5) $60: 2=30$ km - the distance the firs... | theywillnotinterfere | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,841 |
4. A two-digit number was increased by 3, and it turned out that the sum is divisible by 3. When 7 was added to this same two-digit number, the resulting sum was divisible by 7. If 4 is subtracted from this two-digit number, the resulting difference is divisible by four. Find this two-digit number. | 4. Since the sum of a two-digit number and three is divisible by three and the second addend is also divisible by three, the unknown two-digit number is divisible by three. Similarly, we can establish that the two-digit number is also divisible by seven and four. Thus, it is divisible by $84(3 \times 4 \times 7=84)$. A... | 84 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,842 |
5. At $17-00$ the speed of the racing car was 30 km/h. Every 5 minutes thereafter, the speed increased by 6 km/h. Determine the distance traveled by the car from $17-00$ to $20-00$ of the same day. | 5. Let's write down the sequence of numbers expressing the speed at which the car was traveling on the highway: $30,36,42,48,54 \ldots 240$. To determine the distance, we need to calculate the sum:
$$
\begin{aligned}
30: 12 & +36: 12+42: 12+\ldots .+234: 12+240: 12= \\
& =(30+36+42+. .+234+240): 12
\end{aligned}
$$
I... | 405 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,843 |
6. Point $K$ is the midpoint of side of rectangle $A B C D$. The area of the shaded part is $76 \mathrm{~cm}^{2}$. What is the area of the entire rectangle?
178 | 6. The remaining part is equal to the three shaded ones. In total, we have four equal parts.
1) $76 \times 4=304 \mathrm{~cm}^{2}$ - the area of the entire rectangle.
Answer: the area of the entire rectangle is $304 \mathrm{~cm}^{2}$. | 304\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,844 |
1. Pete, Vasya, Kolya, and Tolya counted their catch after fishing. Tolya caught more than Kolya. Pete and Vasya together caught as much fish as Tolya and Kolya. Pete and Kolya together caught less than Vasya and Kolya. Who among the fishermen could have been the first in the amount of fish caught? | 1. Let's introduce the notation. Let the letters P, V, K, T represent the amount of fish caught by Petya, Vasya, Kolya, and Tolya, respectively. From the problem statement, it is known that Tolya caught more fish than Kolya ($K < T$). Since Petya and Kolya together caught less fish than Vasya and Kolya (P + K < V + K),... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,846 | |
7. Without rearranging the digits on the left side of the equation, insert two "+" signs between them so that the equation $8789924=1010$ becomes true. | 7. To narrow down the search options, it should be based on the fact that the sum of the units of each number should give a ten. Example solution: $87+899+24=1010$.
Answer: $87+899+24=1010$.
## TWENTY-THIRD | 87+899+24=1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,852 |
1. Three classmates, Vitya, Borya, and Ivan, attend sports sections: football, hockey, and tennis. Each of them attends only one section. Ivan and the tennis player live in neighboring houses. Borya is older than the football player, and the hockey player is the same age as one of the boys. Which sports sections does e... | 1. This problem involves three classmates (Vitya, Boris, Ivan) and three sports sections (football, hockey, tennis). Let's create a table:
| | V | B | I |
| :--- | :--- | :--- | :--- |
| F | | | |
| X | | | |
| T | | | |
Based on the conditions of the problem, let's fill in the table. Ivan and the tennis pl... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,853 |
2. A cube with a side of 8 dm is cut into cubes with sides of 4 cm. How many cubes will be obtained? | 2. Let's find out how many 4 cm cubes fit into one side of the larger cube $(80: 4=20)$. Then, in the first layer of cubes, there will be 400 $(20 \times 20=400)$. We have 20 such layers. In total, there will be 8000 cubes $(400 \times 20=8000)$.
$. We need to find the greatest number of candies, so Maria will have 33 candies. Sasha will have 68 candies $(35... | 136 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,855 |
4. Students from one of the schools are tasked with making 455 Christmas tree ornaments for kindergartens in the city. All students should make the same number of ornaments. More than 10, but fewer than 70 people participated in making the ornaments. How many students made the ornaments, and how many did each of them m... | 4. Let's represent the number 455 as the product of several numbers. Since the number ends in 5, it is divisible by $5 (455: 5=91)$. The number 91 is also divisible by $7 (91: 7=13)$. Thus, the number 455 can be represented as the product of the following numbers: $455=5 \times 7 \times 13$.
Now let's represent the nu... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,856 |
5. I have thought of a natural number from 1 to 8. You want to guess it by asking questions, to which I can only answer "yes" or "no." What three questions would you ask to guess the number?
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。 | 5. All numbers from 1 to 8 can be divided into two groups: $1,2,3,4$ and $5,6,7,8$. Therefore, the first question will determine which group of numbers the chosen number belongs to. For this, we can ask questions like: "Is this number greater than or equal to 5? Is this number greater than 4? Is this number less than 5... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,857 |
7. How many different products that are divisible by 10 (order does not matter) can be formed from the numbers $2,3,5,7,9$? Numbers in the product do not repeat! | 7. For a number to be divisible by 10, it must be divisible by 2 and 5.
Thus, the product must necessarily include 2 and 5. We get the first product $-2 \times 5$.
Now, let's add one more factor to this product:
$$
2 \times 5 \times 3, 2 \times 5 \times 7, 2 \times 5 \times 9
$$
Add one more factor to the obtained ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,859 |
1. From 2 kg 600 grams of copper, three items were made: a spoon, a candlestick, and a teapot. Each item is three times heavier than the previous one. What is the weight of the teapot? | 1. Let's make a brief record of the problem statement.
1) $1+3+9=13$ (parts) - corresponds to 2 kg 600 grams;
2) $2600: 13=200$ (grams) - weight of the spoon;
3) $200 \times 3=600$ (grams... | 1800 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,860 |
2. An old problem. A landlord, having calculated that a cow is four times more expensive than a dog, and a horse is four times more expensive than a cow, took 200 rubles with him to the city and spent all the money to buy a dog, two cows, and a horse. How much does each of the purchased animals cost? | 2. Let's write the conditions of the problem as equalities:
$$
K=4 C, L=4 K
$$
Substituting $K$ with $4C$ in the second equality, we get:
$$
\text { L }=16 \mathrm{C} .
$$
Since a dog, two cows, and a horse cost 200 rubles, we can write the following equation:
$$
C+2 K+L=200
$$
Using the previously established re... | the\dog\costs\8\rubles,\the\cow\32\rubles,\\the\horse\128\rubles | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,861 |
3. From the numbers 1 to 15 inclusive, form such pairs of numbers where the difference between the first and second numbers is divisible by 7. | 3. Among the numbers from 1 to 15, there are only two that are divisible by 7 - 7 and 14. Now let's form pairs of numbers, the difference of which is 7 or 14.
$$
\begin{array}{lll}
8-1=7 ; & 9-2=7 ; & 10-3=7 \\
11-4=7 ; & 12-5=7 ; & 13-6=7 \\
14-7=7 ; & 15-8=7 ; & 15-1=14
\end{array}
$$
In total, we have nine pairs.
... | 81,92,103,114,125,136,147,158,151 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,862 |
4. In the fourth grades of School No. 15, 100 students are studying. 37 of them have enrolled in the football section, 40 - in the swimming section, 15 people have enrolled in both sections. How many students have not enrolled in any section? | 4. 5) $40-15=25$ (students) - enrolled only in the swimming section;
2) $37-15=22$ (students) - enrolled only in the football section;
3) $25+22+15=62$ (students) - enrolled in the football and tennis sections;
4) $100-62=38$ (students) - did not enroll in any section.
Answer: 38 students did not enroll in any section... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,863 |
5. Find the number of three-digit numbers ending in zero that give a remainder of 1 when divided by 3, given that the tens or hundreds digit is divisible by 4. | 5. According to the condition, three-digit numbers end with zero. Therefore, their general form is: ас 0. Since the number of tens or hundreds is divisible by 4, we get the following two groups of numbers:
$$
\mathrm{a} 00, \mathrm{a} 40, \mathrm{a} 80 \text { and } 4 \mathrm{c} 0,8 \mathrm{c} 0 .
$$
Using the last c... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,864 |
7. Find the pattern that the numerical sequence follows: $1,2,3,4,5,8,7,16,9 \ldots$, and write down the next five numbers. | 7. Let's break down the given numerical sequence into two:
$$
1,3,5,7,9, \ldots \text { and } 2,4,8,16 \ldots
$$
Each term in the first sequence is obtained by adding two to the previous term, while each term in the second sequence is obtained by multiplying the previous term by two. Now, knowing the pattern, we can ... | 32,11,64,13,128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,866 |
1. On a line, three points $A, B$, and $C$ are chosen, such that $A B=3, B C=5$. What can $A C$ be? (There are different possibilities.)
 | $\triangle$ If point $B$ is between points $A$ and $C$, then this distance is equal to $3+5=8$. But there is another case when $B$ is outside the segment $A C$. By drawing a picture, we convince ourselves that in this case the distance is equal to $5-3=2 . \triangleleft$ | 8or2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,867 |
2. On a line, four points $A, B, C$, $D$ are chosen, such that $A B=1, B C=2, C D=4$. What can $A D$ be? List all possibilities. | $\triangleright$ First, let's look at what the distance between points $A$ and $C$ might be. As in the previous problem, there are two possibilities (point $B$ is inside $A C$ or outside) - and it turns out to be either 3 or 1. Now we get two problems: in one, $A C=3$ and $C D=4$, and in the other, $-A C=1, C D=4$. Eac... | 1,3,5,7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,868 |
3. On a wooden ruler, three marks are made: 0, 7, and 11 centimeters. How can you measure a segment of (a) 8 cm; (b) 5 cm using it? | $\Delta$ Using divisions by 7 and 11, it is easy to measure 4 centimeters. Doing this twice, we get a segment of 8 centimeters. Measuring 5 centimeters is a bit more complicated: knowing how to measure 8 and 7, we can measure 1 centimeter. Doing this 5 times, we get 5 centimeters. $\triangleleft$
We can do it differen... | 8=2\times45=3\times11-4\times7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,869 |
4. Point $B$ lies on segment $A C$ of length 5. Find the distance between the midpoints of segments $A B$ and $B C$.
---
Note: The translation preserves the original formatting and structure of the text. | $\triangleright$ Point $B$ can be chosen in different ways. For example, let it divide segment $A C$ into parts of 1 cm and $4 \text{~cm}$. The midpoints of these segments ($A B$ and $B C$) are 0.5 cm and 2 cm away from $B$, respectively, totaling $2.5 \text{~cm}$. Trying other positions of point $B$, one can verify th... | 2.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,870 |
5. In a village, two houses $A$ and $B$ stand along a straight road, 50 meters apart from each other. At what point on the road should a well be built so that the sum of the distances from the well to the houses is minimized? | $\triangle$ If we act fairly, we should build the well in the middle between the houses: then the distance to each house will be 25 meters, and in total 50. However, from the point of view of the problem's condition, any point $X$ on the road segment between $A$ and $B$ is equally good, since the segments $A X$ and $X ... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,871 |
6. The same problem for three and four houses standing along the road at intervals of 50 meters: where should the well be built in this case? | $\triangle$ For three houses: if the middle house is not considered, then all points on the segment between the extreme houses
would be equally good. Therefore, the best
. However, from the perspective of the given probl... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,873 |
14. There is a wooden set square with an angle of $40^{\circ}$. How to construct an angle (a) $80^{\circ}$; (b) $160^{\circ}$; (c) $20^{\circ}$ using it? | $\Delta$ Constructing an angle of $80^{\circ}$ is simple - you need to lay off an angle of $40^{\circ}$ twice. This will result in angles of $120^{\circ}$ and $160^{\circ}$ (which we need). Notice now that $20^{\circ}$ is the complement of $160^{\circ}$ to a straight angle, so if we extend one of the sides of the $160^... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,874 |
18. How many degrees does the minute hand turn in a minute? The hour hand? | $\Delta$ In one hour, the minute hand makes a full revolution, and in half an hour, it turns by $180^{\circ}$ (a straight angle). Half an hour is 30 minutes, so in one minute, it turns by $1 / 30$ of the angle in $180^{\circ}$, which is $180 / 30=6^{\circ}$.
The hour hand moves $1 / 12$ of a circle in one hour, meanin... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,877 |
19. What angle do the minute and hour hands form exactly at 3:05? | $\triangleright$ In the five minutes that have passed after 3 o'clock, the minute hand will turn by $30^{\circ}$. However, for the correct answer, we need to consider that the hour hand will also turn. In one hour, it turns by one hour mark on the clock face, which is also $30^{\circ}$. Therefore, in 5 minutes, it will... | 62.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,878 |
21. Prove that the bisectors of two adjacent angles are perpendicular. Prove that the bisectors of two vertical angles lie on the same line. | $\Delta$ Consider two adjacent angles, each of which is divided by a bisector into two equal parts. Thus, the straight angle is divided into four parts - two equal halves of one angle and two equal halves of the other. If we leave only two of these four parts (one from each pair of equal parts), they will form half of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,879 |
22. Five rays emanating from a single point divide the plane into five equal angles. Find the measure of these angles. | $\Delta$ If the sum of these five angles is a full circle, that is, two straight angles, which is $2 \times 180^{\circ}=360^{\circ}$. Each of them is one fifth of $360^{\circ}$, that is, $360 / 5=72^{\circ}$. $\triangleleft$
This problem and its solution require some comments. The reason is that different geometry tex... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,880 |
23. Four rays emanate from a point, dividing the plane into four angles. It turns out that (counting counterclockwise) the first angle is equal to the third, and the second is equal to the fourth. Prove that the rays form two lines (continuing each other). | $\triangleright$ Let's denote the first and third angles (which are equal) as $\alpha$, and the second and fourth angles as $\beta$. Then $\alpha+\beta+\alpha+\beta=360^{\circ}$, which means $2 \alpha+2 \beta=360^{\circ}$, so $2(\alpha+\beta)=360^{\circ}$, and $\alpha+\beta=180^{\circ}$. Adjacent angles that sum up to ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,881 |
45. Prove that in a triangle, any side is less than the sum of the other two sides. (A triangle is a figure formed by three points not lying on the same straight line, connected by segments. These points are called vertices, and the segments are called sides of the triangle.) | $\Delta$ In fact, this problem is simply a restatement of the above statement: if $A, B$ and $C$ are the vertices of a triangle, $AC$ is one of its sides, and $AB$ and $BC$ are the other two sides, then it is precisely this that needs to be proven: $AC < AB + BC$. (Point $B$ does not lie on the segment $AC$, since poin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,882 |
46. The distance $A B$ is 5, and the distance $B C$ is 3. Can the distance $A C$ be 9? Can the distance $A C$ be 1? | $\triangleright$ The answer to both questions is negative. In the first case, this is impossible because $AC$ cannot be greater than $AB + BC = 5 + 3 = 8$. In the second case, it is a bit more complicated, and we need to apply the triangle inequality to the same points in a different order. If $AC = 1$, then it follows... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,883 |
47. On the plane, four points $A, B, C$, and $D$ are given. Prove that $A D \leqslant A B + B C + C D$. | $\checkmark$ This inequality could be called the quadrilateral inequality and read as follows: a side of a quadrilateral is not greater than the sum of the other three sides. (Or: the broken path $A B C D$ is not shorter than the straight path $A D$.) It is a consequence of the triangle inequality, and to see this, it ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,884 |
48. Prove that in a triangle, any side is less than half the perimeter (sum of the sides). | $\triangleright$ It is convenient to denote the sides of the triangle by letters $a, b$, and $c$. Then the semi-perimeter will be $(a+b+c) / 2$ and we need to prove that the side of the triangle (let's say $a$) is less than, that is, we need to prove that
$$
a<\frac{a+b+c}{2}
$$
It is more convenient to multiply both... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,885 |
49. Prove that in a quadrilateral, any diagonal is less than half the perimeter (the sum of the four sides). | $\triangleright$ Let the diagonal of a quadrilateral with sides $a, b, c, d$ be $x$ (see the figure). It divides it into two triangles. We can write two inequalities for these triangles:
$$
x < a + b ; x < c + d
$$
If we add them, we get $2x < a + b + c + d$. It remains to divide both sides by two. $\triangleleft$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,886 |
51. Four houses $A, B, C$ and $D$ are located at the vertices of a quadrilateral (see figure). Where should a well $X$ be dug so that the sum of the distances from it to the four houses is the smallest? | $\triangle$ If we only care about the residents of houses $A$ and $C$, forgetting about the interests of the residents of $B$ and $D$, then the optimal points will be points $X$ on the segment $A C$ (and from the perspective of the sum $A X+X C$, they are all equally good). Similarly, from the perspective of the sum
![... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,888 |
52. Inside triangle $ABC$, a point $D$ is taken. Prove that $AD + DC \leqslant AB + BC$. | $\triangleright$ To solve this problem, we will make an additional construction, as they say: extend the segment $A D$ beyond point $D$ until it intersects with side $B C$. Let's call the intersection point $E$. Now, we will take $A E+E C$ as an intermediate quantity for comparison. The path $A-B-C$ is longer than the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,889 |
53. Prove that the length of a broken line (a continuous line composed of several segments) is not less than the distance between its ends. (Thus, a straight path is always shorter than an indirect one.) | $\Delta$ Let's straighten our path gradually. Take some segment of the path consisting of two links $P Q$ and $Q R$ and replace the segment $P-Q-R$ with the segment $P-R$. By the triangle inequality, the path will be shortened, and the number of links in the broken line will decrease by 1. After several such reductions... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,890 |
54. A polygon was cut into two parts along a straight line. Prove that the perimeter of each part is less than the perimeter of the original polygon. (In other words, if you cut a piece off a paper polygon with a straight cut, its perimeter will decrease.) | $\Delta$ The perimeter of each of the parts differs from the full perimeter of the original polygon in that the broken path segment is replaced by a straight line. Because of this, the perimeter can only decrease. $\triangleleft$
This problem requires some explanation. Generally speaking, a straight line can divide a ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,891 |
55. One convex polygon lies inside another. Prove that the perimeter of the inner polygon is less than the perimeter of the outer polygon. | $\checkmark$ The inner polygon can be cut out from the outer one by making cuts along the sides. Each such cut removes a piece from the remaining polygon and (as we saw in problem 54) reduces the perimeter, until the inner polygon is finally obtained. $\triangleleft$
## A few more problems | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,892 |
67. How can this property be used to check the equality of figures on two pictures in a book (without cutting the figures out of the book)? | $\triangleright$ Draw one picture on a transparent film (tracing paper) and compare it with another. $\triangleleft$ | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,894 |
76. In a quadrilateral, the point of intersection of the diagonals divides each of them in half. Prove that the opposite sides of this quadrilateral are equal. | $\triangleright$ Let $A B C D$ be the given quadrilateral, and $O$ be the point of intersection of its diagonals. On the diagram, mark pairs of equal segments by condition with single and double hatch marks, respectively. Additionally, mark the vertical (and therefore equal) angles $A O B$ and $C O D$. Now we have all ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,897 |
78. In triangle $ABC$, the median $AM$ (the segment connecting vertex $A$ to the midpoint $M$ of side $BC$) is extended beyond point $M$ by a distance equal to $AM$. Find the distance from the resulting point to points $B$ and $C$, if sides $AB$ and $AC$ are equal to 5 and 4, respectively. | $\triangle$ Let $N$ be the end of the extended median, so that $N$ lies on the extension of $A M$ beyond point $M$ and $A M=M N$. Then, in the quadrilateral $A B N C$, the diagonals are bisected by their point of intersection, so we can refer to an already solved problem and see that the opposite sides of the quadrilat... | NB=4,NC=5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,899 |
79. Prove that in an arbitrary triangle $ABC$ the median $AM$ does not exceed half the sum of the sides $AB$ and $AC$. | $\checkmark$ We need to prove that the doubled median does not exceed the sum of $A B + A C$. In the previous problem, the doubled median $A N$ was constructed, which is a side of the triangle $A B N$, the other two sides of which are $A B$ and $B N = A C$. It remains to refer to the triangle inequality. $\triangleleft... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,900 |
80. Using the first criterion of triangle congruence, prove that in two congruent triangles, the medians drawn to the corresponding sides are equal. | $\triangle$ Let $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ be these triangles (equal sides and angles are marked with the same letters), $A M$ and $A^{\prime} M^{\prime}$ - medians drawn in them (so $M$ is the midpoint of $B C$, and $M^{\prime}$ is the midpoint of $B^{\prime} C^{\prime}$). Let's look at the triangl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,901 |
81. Using the second criterion of triangle congruence, prove that in two congruent triangles, the bisectors that divide the corresponding angles in half are equal | $\Delta$ (In the problem, of course, we are talking about the segments of the bisectors to the opposite side.) Let $A D$ and $A^{\prime} D^{\prime}$ be such segments. Following the solution of the previous problem, consider triangles $A B D$ and $A^{\prime} B^{\prime} D^{\prime}$ - in them, the sides $A B$ and $A^{\pri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,902 |
82. In a quadrilateral, opposite sides are equal in pairs. Prove that the opposite angles in it are equal. (Such quadrilaterals are called parallelograms, and we will return to them later). | $\triangle$ Let $A B C D$ be such a quadrilateral. Draw a diagonal in it, for example, $A C$. It will divide it into two triangles $A B C$ and $A D C$, to which the third criterion of congruence of triangles can be applied - side $A C$ is common to both, and the other two pairs of sides are equal by condition. Therefor... | proof | Geometry | proof | Yes | Yes | olympiads | false | 42,903 |
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