problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
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From the four inequalities $2x > 70$, $x < 25$, $x > 5$, and $x > 5$, two are true and two are false. Find the value of $x$, given that it is an integer. | Given that $x$ is an integer, the inequalities can be rewritten as: $x>35, x>6$ and $x>5$. If it is not true that $x>5, x>6$ and $x>35$ - one is true and two are false.
Since the third inequality implies the first two, and the second implies the first, there is only one option: the first inequality is true, and the ot... | 6 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2+ $[$ Prime numbers and their properties $]$
Four kids were discussing the answer to a problem.
Kolya said: "The number is 9."
Roman: "It's a prime number."
Katya: "It's an even number."
And Natasha said that the number is divisible by 15.
One boy and one girl answered correctly, while the other two were wrong. ... | If Kolya answered correctly, then both girls were wrong, since the number 9 is odd and does not divide by 15. This means that Roman gave the correct answer. But a prime number does not divide by 15, and the only even prime number is 2.
## Answer
2. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$2+$
12 candidates for mayor were talking about themselves. After some time, one said: "They lied to me once." Another said: "Now it's twice." "Now it's three times" - said the third, and so on up to the 12th, who said: "Now they have lied 12 times." At this point, the host interrupted the discussion. It turned out th... | ## Solution
Assume that the first candidate lied. This means that before his statement, the number of false statements was not equal to 1. After his statement, this number increased by 1, so it became not equal to 2. Therefore, the second candidate also lied. Continuing this reasoning, we get that all candidates lied,... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
In a physics class, the teacher set up the following experiment. He placed 16 weights with masses of $1, 2, 3, \ldots, 16$ grams on a balance scale so that one of the pans was heavier. Fifteen students took turns leaving the classroom and each took one weight, and after each student left, the balance changed its positi... | Since at any given moment the masses on the scales differed by at least 1 gram, in order for the opposite scale to outweigh, it is necessary to remove a weight of no less than two grams. Therefore, no student could have taken a 1-gram weight when leaving the class.
## Answer
A 1-gram weight remained on the scales. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
In a row from left to right, there are 31 wallets, each containing 100 coins. From one of the wallets, some coins were moved: one coin to each of the wallets to the right of it. With one question, you can find out the total number of coins in any set of wallets. What is the minimum number of questions n... | It is enough to get an answer to the question "How many coins are there in total in the wallets with odd numbers?" Indeed, if the answer to it is " $1600+n$ " ( $n>0$ ), then the coins were moved from the wallet with an even number, to the right of which there were exactly $n$ wallets with odd numbers - that is, from t... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Fundamental Theorem of Arithmetic. Prime Factorization]
In the country of Anchuria, where President Miraflores is in power, the time for new presidential elections is approaching. The country has exactly 20 million voters, of which only one percent support Miraflores (the regular army of Anchuria). Miraflores, of co... | We will divide the voters into groups of 5 people. In 66666 such groups, we can place 3 military personnel. As a result, we get 66666 military electors out of 4 million.
We will divide these 4 million electors into groups of 4 people, placing 3 military personnel in 22222 of them. As a result, we get 22222 military el... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
In a line, all integers from 1 to 100 are written in an unknown order. With one question about any 50 numbers, you can find out the order of these 50 numbers relative to each other. What is the minimum number of questions needed to definitely determine the order of all 100 numbers? | To find the desired order $a_{1}, a_{2}, \ldots, a_{100}$ of numbers in a row, it is necessary that each pair $\left(a_{i}, a_{i+1}\right), i=1,2, \ldots, 99$, appears in at least one of the sets about which questions are asked; otherwise, for two sequences $a_{1}, \ldots, a_{i}, a_{i+1}, \ldots, a_{100}$ and $a_{1}, \... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7 wolves eat 7 sheep in 7 days. How many days will it take for 9 wolves to eat 9 sheep? | The number of wolves has increased by the same factor as the number of sheep, so the time of consumption will not change.
## Answer
In 7 days. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
I.v.
In the multiplication example written on the board, the hooligan Petya changed two digits. It became 4$\cdot$5$\cdot$4$\cdot$5$\cdot$4 = 2247. Restore the original example.
# | In the original example, at least one of the multipliers is even.
## Solution
In the resulting example, three multipliers are even, which means that at least one was even in the original example. Therefore, the product was an even number, meaning the last digit of the product was changed. Thus, no more than one digit... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Kalinin }}$ D.:
In the room, there are 20 chairs of two colors: blue and red. On each chair sits either a knight or a liar. Knights always tell the truth, and liars always lie. Each of the seated individuals claimed that they were sitting on a blue chair. Then they somehow rearranged themselves, af... | Initially, all knights sit on blue chairs, and all liars on red ones. Therefore, the number of knights who moved to red chairs is equal to the number of liars who moved to blue chairs. Both groups claimed they were sitting on red chairs. In total, 10 people said they were sitting on red chairs. Therefore, the number of... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The kindergarten received cards for reading lessons: some have "MA" written on them, and others have "NYA". Each child took three cards and started forming words from them. It turned out that 20 children can form the word "MAMA", 30 children can form the word "NYANYA", and 40 children can form the word "MANYA". How man... | Notice that each child has three cards, and there are two different inscriptions on them.
## Solution
Since each child has three cards, and there are only two inscriptions in total, it is necessary that two inscriptions must match, meaning each can form either the word MAMA (20 such children) or the word NYANYA (30 s... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a certain kingdom, there were 32 knights. Some of them were vassals of others (a vassal could have only one suzerain, and a suzerain was always richer than his vassal). A knight who had at least four vassals bore the title of baron. What is the maximum number of barons that could be under these conditions?
(The kin... | Evaluation. 8 barons should have 32 vassals, and the richest knight cannot be anyone's vassal. Example. Let 24 knights be vassals of six barons, and all these barons be vassals of the richest Baron. In total, 7 barons.
## Answer
7 barons. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Yashchenko I.V.
In Mexico, ecologists have succeeded in passing a law according to which each car must not be driven at least one day a week (the owner reports to the police the car's number and the "day off" for the car). In a certain family, all adults wish to drive daily (each for their own business!). How many car... | b) If no more than one car "rests" each day, then there are no more than 7 cars in total.
## Solution
a) Five cars are not enough, because on the day when one of the cars is "resting," someone will have no car to ride in. Six cars, obviously, are enough.
b) If no more than one car "rests" each day, then the total nu... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Around a round table, 30 people are sitting - knights and liars (knights always tell the truth, while liars always lie). It is known that each of them has exactly one friend at the same table, and a knight's friend is a liar, while a liar's friend is a knight (friendship is always mutual). When asked "Is your friend si... | All those sitting at the table are paired as friends, which means there are an equal number of knights and liars. Consider any pair of friends. If they are sitting next to each other, the knight will answer "Yes" to the given question, and the liar will answer "No." If they are not sitting next to each other, their ans... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Mathematical logic (other) ] [ Divisibility of numbers. General properties ]
On a meadow, ladybugs gathered. If a ladybug has six spots on its back, it always tells the truth, and if it has four spots, it always lies, and there were no other ladybugs on the meadow. The first ladybug said: "Each of us has the same nu... | If the first ladybug tells the truth, then the second and third should also tell the truth, as they should have the same number of spots on their backs as the first. But the second and third ladybugs contradict each other, so at least one of them is lying. Therefore, the first ladybug is also lying.
Suppose each of th... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Raskin M.A.
On an island, there live chameleons of five colors. When one chameleon bites another, the color of the bitten chameleon changes according to some rule, and the new color depends only on the color of the biter and the color of the bitten. It is known that $\$ 2023 \$$ red chameleons can agree on a sequence ... | Let's start with an example of rules under which at least 5 red chameleons would be necessary for the described recoloring. Let's number the colors so that red is the first color and blue is the last. Then, let the rules be as follows: a chameleon of color $k 1$ can reduce the number of its color by a bite from a chame... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic operations. Numerical identities ] Joke problems
Calculate the product
$$
\left(100-1^{2}\right)\left(100-2^{2}\right)\left(100-3^{2}\right) \ldots\left(100-25^{2}\right)
$$
# | Pay attention to the factors hidden behind the ellipsis.
## Solution
The product is 0, since among the factors there will be (100 - $10^{2}$).
## Answer
0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ [equations in integers $]$ ]
In the room, there are three-legged stools and four-legged chairs. When people sat on all these seats, there were 39 legs in the room.
How many stools are in the room? | According to the condition, there are five- and six-legged creatures in the room, with a total of 39 legs. The number of legs of the five-legged creatures ends in 0 or 5. But it cannot end in 0: then the number of legs of the six-legged creatures would end in 9. In such a case, the number of five-legged creatures can b... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
|Rabbits sawed several logs. They made 10 cuts and got 16 chunks. How many logs did they saw
# | Recall problem 89914 - the problem about how the rabbits made 10 cuts to split one log.
## Solution
From each log, you get one more chunk than the number of cuts. Since there are 6 more chunks, it means there were 6 logs. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Text problems ] [Algorithm theory]
Pouring milk. From an 8-liter bucket filled with milk, you need to measure out 4 liters using empty 3-liter and 5-liter buckets. | Let's record the solution in a table.
| Eight-liter bucket | Five-liter bucket | Three-liter bucket |
| :---: | :---: | :---: |
| 8 | | |
| 3 | 5 | |
| | 5 | |
| 3 | 2 | 3 |
| :---: | :---: | :---: |
| 6 | 2 | |
| 6 | | 2 |
| 1 | 5 | 2 |
| 4 | 4 | 3 | | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Three mad painters started painting the floor each in their own color. One managed to paint 75% of the floor red, another 70% green, and the third 65% blue. What part of the floor is definitely painted with all three colors? | Evaluation. 25% of the floor is not painted red, 30% of the floor is not painted green, and 35% of the floor is not painted blue.
25+30+35=90. From this, it follows that at least 10% of the floor is painted with all three colors.
An example where exactly 10% is painted with all three colors is clear from the evaluati... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Andjans A.
In a table with $m$ rows and $n$ columns, a horizontal move is defined as a permutation of the elements in the table such that each element remains in the same row it was in before the permutation; similarly, a vertical move is defined ("row" in the previous definition is replaced with "column"). Determine ... | The case $m=1$ or $n=1$ is trivial.
Let $m \neq 1$ and $n \neq 1$. We will show that two moves are insufficient. Suppose in the initial table, the top-left
square is \begin{tabular}{|l|l|}
\... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Zkov G.
A bank serves a million customers, the list of whom is known to Ostap Bender. Each has a six-digit PIN code, and different customers have different codes. In one move, Ostap Bender can choose any customer he has not yet chosen and peek at the digits of the code at any $N$ positions (he can choose different pos... | It is not difficult to do when $N=3$. Since any combination of the first three digits occurs exactly 1000 times, by looking at these digits for everyone except Koreiko, Bender will know them for Koreiko as well. Then it is sufficient to look at the last three digits of Koreiko's code.
We will prove that when $N<3$, it... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In a box, there are cards numbered with natural numbers from 1 to 2006. On the card numbered 2006 lies the card numbered 2005, and so on down to 1. In one move, it is allowed to take one top card (from any box) and place it either at the bottom of an empty box, or on a card with a number one greater. How many empty box... | The answer follows from the general fact: let the number of cards be $n$, where $2^{\mathrm{k}-1} \leq n < 2^{\mathrm{k}}$. If we assume that all cards can be placed in $k$ boxes, then card $a$ still needs to be moved so that all cards end up in one box. For this, at some point, it will be necessary to occupy the "top"... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Let's call a natural number "remarkable" if it is the smallest among all natural numbers with the same digit sum as it has.
How many three-digit remarkable numbers exist?
# | "Remarkable" numbers with the sum of digits from 1 to 18 are either single-digit or two-digit. Therefore, remarkable three-digit numbers have a sum of digits from 19 to 27. Each such sum corresponds to one remarkable number. Consequently, there are nine three-digit "remarkable" numbers.
## Answer
9 numbers. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the pond, 30 pikes were released, which gradually eat each other. A pike is considered full if it has eaten no less than three pikes (full or hungry). What is the maximum number of pikes that can become full? | The number of pikes eaten is not less than three times the number of satiated pikes.
## Solution
Let $s$ be the number of satiated pikes. Then they together have eaten no fewer than $3s$ pikes. Since each pike can only be eaten once, and at least one pike remains at the end, $3s < 30$. Therefore, $s \leq 9$.
We will... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Each of the thirty sixth-graders has one pen, one pencil, and one ruler. After their participation in the Olympiad, it turned out that 26 students lost a pen, 23 - a ruler, and 21 - a pencil. Find the smallest possible number of sixth-graders who lost all three items. | From the condition, it follows that four sixth-graders have a pen, seven have a ruler, and nine have a pencil. Thus, at least one item can be owned by no more than $4+7+9=20$ people. Therefore, no fewer than $30-20=10$ people have lost all three items.
All three items will be lost by exactly 10 people if each of the o... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Lobanov M. S.
Around a round rotating table, on which there are 8 white and 7 black cups, 15 gnomes are sitting. They have put on 8 white and 7 black caps. Each gnome takes a cup whose color matches the color of their cap and places it in front of them, after which the table is rotated randomly. What is the maximum nu... | Let's consider an arbitrary arrangement of cups and write down their colors in a row. Below this row, we will also write down all its different cyclic shifts — a total of 14. We will count how many color matches there will be in the same position in the original arrangement and in the arrangements obtained by shifts. F... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
## In the city, there are 57 bus routes. It is known that:
1) from any stop, you can get to any other stop without transferring;
2) for each pair of routes, there is one, and only one, stop where you can transfer from one of these routes to the other
3) each route has no fewer than three stops.
How many stops does ea... | Let there be $n$ stops on some route $a$. Take a stop $B$ that route $a$ does not pass through. From $B$, there is a route to each of the $n$ stops on route $a$, and there is exactly one such route, since two different routes cannot have two common stops. Each route passing through $B$ intersects route $a$. Therefore, ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kalinin D.A.
Find the maximum number of colors in which the edges of a cube can be painted (each edge in one color) so that for each pair of colors there are two adjacent edges painted in these colors. Adjacent are considered edges that have a common vertex.
# | There are several ways to color the edges of a cube in six colors while satisfying the condition of the problem. Here is one of them:
We will show that more than six colors are not possible... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Graph Traversal ]
There is a group of islands connected by bridges in such a way that from any island, you can reach any other island. A tourist visited all the islands, crossing each bridge exactly once. He visited the Triplet Island three times. How many bridges lead to the Triplet Island if the tourist
a) did no... | a) If the Triple tourist entered the island 3 times and exited 3 times, that means he used 6 bridges.
b) In this case, the tourist entered the island twice, but exited three times.
## Answer
a) 6 bridges; b) 5 bridges; c) 4 bridges. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Pascal's Triangle and Newton's Binomial ]
How many times greater is the sum of the numbers in the 101st row of Pascal's Triangle compared to the sum of the numbers in the 100th row?
# | Let in the 100th row stand the numbers $c_{0}, c_{1}, \ldots, c_{100}$. Then in the 101st row stand the numbers $c_{0}, c_{0}+c_{1}, c_{1}+c_{2}, \ldots, c_{99}+$ $c_{100}, c_{100}$
their sum is $2 c_{0}+2 c_{1}+\ldots+2 c_{100}$.
## Otvet
Twice. | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Graph Theory (Miscellaneous)]
Several Top Secret Objects are connected by an underground railway in such a way that each Object is directly connected to no more than three others, and from each Object, it is possible to reach any other by underground travel with no more than one transfer. What is the maximum number o... | Evaluation. From this Object, you can reach three Objects in one "move", and with a transfer - to another $2 \cdot 3=6$ Objects. Therefore, there are no more than 10 Objects.
An example with 10 Objects is shown in the figure.
 })] \\ [\text { Extreme Principle (other) })]\end{array}\right.$
In how many ways can the numbers from 1 to 100 be permuted so that adjacent numbers differ by no more than 1? | Where can the number 1 be placed?
## Solution
Next to the number 1, only the number 2 can stand, so 1 must be at the edge. Suppose 1 is at the beginning. Then the next number is 2, the next is 3 (no other numbers can be next to 2), the next is 4, and so on. We get the arrangement $1, 2, \ldots, 99, 100$.
If 1 is at ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Announce $A$.
$N$ friends simultaneously learned $N$ pieces of news, with each person learning one piece of news. They started calling each other and exchanging news.
Each call lasts 1 hour. Any number of news items can be shared in one call.
What is the minimum number of hours required for everyone to learn all the... | a) A piece of news known to one of the friends will be known to no more than two (including the first) after 1 hour, no more than four after the second hour, ..., and no more than 32 after the 5th hour. Therefore, it will take no less than 6 hours.
We will show that 6 hours are sufficient. The conversations can procee... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Combinations and Permutations ]
[ $\underline{\text { Directed Graphs }}$ ]
In the discussion, 15 deputies participated. Each of them, in their speech, criticized exactly $k$ of the remaining 14 deputies.
For what smallest $k$ can we assert that there will be two deputies who criticized each other? | Consider a directed graph where the vertices correspond to deputies, and an edge leading from $A$ to $B$ means that deputy $A$ has criticized deputy $B$.
## Solution
If each deputy has criticized 8 others, then the number of edges in the graph is $15 \cdot 8 = 120$, which is greater than the number of pairs $C_{15}^{... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Sorting in ascending (descending) order ] [ Classical combinatorics (other). $\quad]$
a) A traveler stopped at an inn, and the owner agreed to accept rings from a golden chain the traveler wore on his wrist as payment for his stay. However, he set a condition that the payment should be daily: each day the owner shou... | a) It is enough to cut two rings so that pieces of three and six rings are separated. On the third day, the traveler gives the piece of three rings and receives two rings as change, and on the sixth day, the piece of six rings and receives five rings as change.
b) Arrange the resulting pieces of the chain (not countin... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A kindergarten received cards for reading lessons: some have "MA" written on them, and others have "NYA".
Each child took three cards and started forming words from them. It turned out that 20 children could form the word "MAMA", 30 children could form the word "NYANYA", and 40 children could form the word "MANYA". Ho... | Notice that each child has three cards, and there are two different inscriptions on them.
## Solution
Since each child has three cards, and there are only two inscriptions in total, it is necessary that two inscriptions must match, meaning each child can form either the word MAMA (20 such children) or the word NYANYA... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tompsongo A.K.
In a certain kingdom, there were 32 knights. Some of them were vassals of others (a vassal could have only one suzerain, and the suzerain was always richer than his vassal). A knight who had at least four vassals bore the title of baron. What is the maximum number of barons that could be under these con... | Evaluation. 8 barons should have 32 vassals, and the richest knight cannot be anyone's vassal. Example. Let 24 knights be vassals of six barons, and all these barons be vassals of the richest Baron. In total, 7 barons.
## Answer
7 barons. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
|
| | $[$ Pigeonhole Principle (other) $)]$ | |
In a company of 10 people, 14 pairwise arguments have occurred. Prove that it is still possible to form a group of three friends. | The total number of ways to choose a company of three people is $C_{10}^{3}=120$. Each quarrel destroys no more than eight such companies, so the number of destroyed companies is no more than $8 \cdot 14=112$. Therefore, at least 8 friendly companies remain. | 8 | Combinatorics | proof | Yes | Yes | olympiads | false |
$\begin{aligned} & \text { [Systems of points and segments. Examples and counterexamples] } \\ & {[\quad \underline{\text { Classical combinatorics (other) }}]}\end{aligned}$
On a plane, 10 equal segments were drawn, and all their points of intersection were marked. It turned out that each point of intersection divide... | On each segment, there are no more than two points. On the other hand, each intersection point belongs to at least two segments. Therefore, there are no more than $10 \cdot 2: 10=10$ points. An example with 10 points is shown in the figure.
 all the others? | A total of $6 \cdot 5=30$ auditions are needed. No more than 9 auditions can take place in one concert. Therefore, there must be at least four concerts.
Example: Musicians with numbers (4,5,6), (2,3,6), (1,3,5), and (1,2,4) perform in the concerts.
## Answer
In 4 concerts. | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Folkoro
In a certain state, the airline system is arranged in such a way that each city is connected by air routes
to no more than three other cities, and from any city, you can reach any other city with no more than one
transfer. What is the maximum number of cities that can be in this state? | Evaluation. From a fixed city $A$, one can directly reach no more than three cities, and with one transfer - no more than $3 \cdot 2=6$ additional cities. Thus, the total number of cities can be no more than ten. For an example of a network of 10 cities, see the figure below.
.]
In a dark room, on a shelf, there are four pairs of socks of two different sizes and two different colors lying in a random order. What is the smallest number of socks that need to be moved from the shelf to a suitcase, without leaving the room, to ensure that the suitcase... | First, note that seven socks are sufficient. Indeed, only one sock is not taken, which means all pairs of socks except one are taken. Therefore, in the suitcase, there are two pairs of different colors and sizes, specifically the pair that differs from the one not taken only by color, and the pair that differs from the... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the number of zeros with which the number $11^{100}-1$ ends.
# | Use the binomial expansion of the expression $(10+1)^{100}$.
## Solution
$S=(10+1)^{100}-1=\ldots+(100 \cdot 99 \cdot 98: 6) \cdot 10^{3}+(100 \cdot 99: 2) \cdot 10^{2}+100 \cdot 10+1=\ldots+(33 \cdot 49) \cdot 10^{5}+495000+1000=A$ +496000, where all terms in place of the ellipsis are divisible by $10^{4}$, meaning ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Planar graphs. Euler's formula ]
In the country of Lake, there are seven lakes connected by ten non-intersecting channels, such that one can sail from any lake to any other. How many islands are there in this country?
# | Consider a planar graph, where the vertices correspond to lakes, and the edges - to channels. Let it divide the plane into $F$ pieces. By Euler's formula (problem $\underline{30759}$) $7-10+F=2$, so $F=5$. One of these pieces is the mainland, and the rest are islands.
## Answer
4 islands. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Frankin B.R.
In a tournament with 20 athletes, 10 referees were involved. Each athlete played against each other once, and each match was officiated by exactly one referee. After each game, both participants took a photo with the referee. A year after the tournament, a stack of all these photos was found. It turned ou... | Let's call a person suspicious if we cannot determine whether they are an athlete or an referee. Note that each referee was only photographed with athletes, that is, no more than 20 people. On the other hand, each athlete was photographed with all other athletes and at least one referee, that is, no fewer than 20 peopl... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A.K.
In the plane, there is an open, non-self-intersecting broken line with 31 segments (adjacent segments do not lie on the same straight line). Through each segment, a line containing this segment was drawn. As a result, 31 lines were obtained, some of which may have coincided. What is the smallest number of differe... | Evaluation. Except for the ends, the broken line has 30 vertices, and each is the intersection of two lines. If there are no more than eight lines, then there are no more than $7 \cdot 8: 2=28$ intersection points - a contradiction.
Example - in the figure.
. With one sword strike, one can cut all the necks coming out of some head $A$ of the hydra. But in this case, from head $A$, one neck will instantly grow to all heads with which $A$ was not connected. Heracles... | Let's move on to a graph where heads are vertices, necks are edges, and a strike on the necks coming out of head $A$ is called inverting vertex $A$.
If there is a vertex $X$ of degree no more than 10, it is sufficient to invert its neighbors, and it will separate. If there is a vertex connected to all vertices except ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A cube with a side of 10 is divided into 1000 smaller cubes with an edge of 1. In each small cube, a number is written, and the sum of the numbers in each column of 10 cubes (in any of the three directions) is 0. In one of the cubes (denoted as $A$), the number 1 is written. Three layers, parallel to the faces of the c... | Through the given cube $A$ pass one horizontal layer $G$ and two vertical layers. The sum of all numbers in 81 vertical columns not included in the last two layers is 0. From this sum, we need to subtract the sum $S$ of the numbers lying in the cubes at the intersection of these columns with $G$ (there are 81 such cube... | -1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In Anchuria, a unified state exam is taking place. The probability of guessing the correct answer to each question on the exam is 0.25. In 2011, to obtain a certificate, one needed to answer three questions correctly out of 20. In 2012, the School Administration of Anchuria decided that three questions were too few. No... | If a graduate guesses the answers, the Unified State Exam (EGE) can be considered a Bernoulli scheme with a success probability of $p=0.25$ and a failure probability of $q=0.75$. In 2011, to pass the exam, one needed to answer at least three questions correctly. It is more convenient to find the probability of the oppo... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Coloring Vertex Degree
20 phones are connected by wires in such a way that each wire connects two phones, each pair of phones is connected by no more than one wire, and no more than two wires extend from each phone. It is necessary to color the wires (each wire entirely in one color) so that the wires extending from e... | We will represent phones as points and the wires connecting them as segments connecting these points. Since no more than two segments come out of each point, the resulting graph breaks down into open paths and simple cycles. Note that an open path can be painted with two colors, and a simple cycle with three. For this,... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[The product of the lengths of the segments of chords and the lengths of the segments of secants] [Properties and characteristics of an isosceles triangle.] [Thales' theorem and the theorem of proportional segments]
On the extension of side $A D$ of rhombus $A B C D$ beyond point $D$, point $K$ is taken. Lines $A C$ a... | Let $O$ be the center of the circle. Use the similarity of triangles $K Q O$ and $K B A$.
## Solution
Let $O$ be the center of the circle, $F$ be the second intersection point of the circle with the line $A$. Since triangle $A O Q$ is isosceles $(O A=O Q=6)$, then
$\angle A Q O=\angle O A Q=\angle B A Q$. Therefore,... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In triangle $A B C$, the bisector $C D$ of the right angle $A C B$ is drawn; $D M$ and $D N$ are the altitudes of triangles $A D C$ and $B D C$, respectively.
Find $A C$, given that $A M=4, B N=9$.
# | Note that $M C=M D=D N$. From the similarity of triangles $A M D$ and $D N B$, it follows that $A M: M D=D N: N B$, that is, $M C^{2}=A M \cdot N B=36$. Therefore,
$A C=A M+M C=4+6=10$.
## Answer
10. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## [ Quadrilaterals (extreme properties).]
For what value of the height does a rectangular trapezoid with an acute angle of $30^{\circ}$ and a perimeter of 6 have the maximum area
# | Express the area of the given trapezoid through its height and apply the Cauchy inequality.
## Solution
Let $h$ be the height of the trapezoid, $3 x$ be the sum of the bases. Then the larger lateral side of the trapezoid is $2 h$, and the perimeter is $3 x + 3 h = 6$. The area of the trapezoid is $\frac{3 x h}{2} = \... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
What is the greatest possible number of rays in space emanating from a single point and forming obtuse angles with each other
# | Answer: 4. We will consider vectors instead of rays. We can assume that the first vector has coordinates (1, $0,0)$. Then the other vectors have coordinates ( $x_{i}, y_{i}, z_{i}$ ), where $x_{i}0$, and thus, $y_{\mathrm{i}}2$ the inequality $x_{\mathrm{i}} x_{\mathrm{j}}+y_{\mathrm{i}} y_{\mathrm{j}}+z_{\mathrm{i}} z... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
From a point located outside a circle, two mutually perpendicular tangents are drawn to the circle. The radius of the circle is 10. Find the length of each tangent.
# | ## Solution
The quadrilateral formed by the given tangents and radii drawn to the points of tangency is a square.
## Answer
10. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left.\frac{\text { [Inscribed and Circumscribed Circles ]}}{\text { Rhombuses Properties and Characteristics }}\right]$
The side of the rhombus is 8 cm, and the acute angle is $30^{\circ}$. Find the radius of the inscribed circle. | The diameter of the inscribed circle is equal to the height of the rhombus.
## Solution
The diameter of the inscribed circle is equal to the height of the rhombus, and the height, dropped from a vertex to the opposite side, is the leg of a right triangle opposite the angle of $30^{\circ}$.
Therefore, the height of t... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A circle is inscribed in a right angle. A chord connecting the points of tangency is equal to 2. Find the distance from the center of the circle to this chord.
# | The diagonals of a square are equal.
## Solution
The quadrilateral formed by the tangents and radii drawn to the points of tangency is a square.
The desired segment is half the diagonal of this square, i.e., 1.
. The six resulting triangles will be isosceles.
## Answer
All... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Vassily H.S.
Consider all possible closed broken lines with six segments, all vertices of which lie on a circle.
a) Draw such a broken line that has the maximum possible number of self-intersections.
b) Prove that a broken line of this kind cannot have a greater number of self-intersections. | a) The figure shows a six-segment broken line with seven self-intersection points.
b) We will prove that such a broken line cannot have more than seven self-intersections. Let the total numb... | 7 | Geometry | proof | Yes | Yes | olympiads | false |
9
A point lying inside a circumscribed $n$-gon is connected by segments to all vertices and points of tangency. The triangles formed by this are alternately painted red and blue. Prove that the product of the areas of the red triangles is equal to the product of the areas of the blue triangles. | Let $h_{1}, \ldots, h_{n}$ be the distances from a given point to the corresponding sides, and $a_{1}, \ldots, a_{n}$ be the distances from the vertices of the polygon to the points of tangency. Then the product of the areas of both the red and blue triangles is $a_{1} \ldots a_{n} h_{1} \ldots h_{n} / 2^{n}$.
What is... | 3 | Geometry | proof | Yes | Yes | olympiads | false |
[The triangle formed by the bases of two altitudes and a vertex]
The side of the triangle is $\sqrt{2}$, and the angles adjacent to it are $75^{\circ}$ and $60^{\circ}$.
Find the segment connecting the bases of the altitudes drawn from the vertices of these angles. | Let $B M$ and $C N$ be the altitudes of triangle $A B C$, $B C=\sqrt{2}$, $\angle B=75^{\circ}$, $\angle C=60^{\circ}$. Then $\angle A=45^{\circ}$.
Triangle $A M N$ is similar to triangle $A B C$ with a similarity coefficient of $\cos 45^{\circ}$. Therefore, $M N=$ $B C \cdot \cos 45^{\circ}=1$.
## Answer
1.
Probl... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Algebraic problems on the triangle inequality ]
In a triangle, two sides are equal to 3.14 and 0.67. Find the third side, given that its length is an integer.
# | Apply the triangle inequality.
## Solution
Let the integer $n-$ be the length of the third side. Then
$$
3.14-0.67<n<3.14+0.67
$$
From this, we find that $n=3$.
## Answer
3.
## Problem | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
[ Homothetic Circles]
Inside an angle, there are three circles $\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}$, each of which touches the two sides of the angle, and circle $S_{2}$ touches circles $S_{1}$ and $S_{3}$ externally. It is known that the radius of circle $S_{1}$ is 1, and the radius of circle $S_{3}$ is... | Homothety with the center at the vertex of the angle, translating circle $S_{1}$ to circle $S_{2}$, also translates circle $S_{2}$ to circle $S_{3}$.
## Solution
Consider the homothety $\mathrm{H}$ with the center at the vertex of the angle, translating circle $\mathrm{S}_{1}$ to circle $\mathrm{S}_{2}$. Circle $\mat... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ inscribed quadrilaterals]
Three consecutive sides of an inscribed quadrilateral are in the ratio 1:2:3. Find its sides if it is known that the perimeter is 24 m.
# | Since the sums of the opposite sides of a circumscribed quadrilateral are equal to each other, two parts fall on the fourth side. Therefore, the sides are $\frac{1}{8}, \frac{1}{4}, \frac{3}{8}$ and $\frac{1}{4}$ of the perimeter, i.e., 3, 6, 9, 6.
## Answer
3 cm, 6 cm, 9 cm, 6 cm. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Dirichlet's Principle (finite number of points, lines, etc.).] [ Central angle. Arc length and circumference length ]
On a circle of radius 1, a point $O$ is marked and a notch is made to the right with a compass of radius $l$. From the resulting point $O_{1}$, another notch is made in the same direction with the s... | We will prove by induction on $n$ that the number of different arcs after $n$ cuts does not exceed 3. For $n=2$, this is obvious. Let $A_{k}$ denote the cut with number $k$. Suppose $n$ cuts have been made and the point $A_{n}$ falls on the ARC $A_{k} A_{l}$. Then the point $A_{n-1} 1$ falls on the arc $A_{k}-1 A_{l-} ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.] [ Volume of a parallelepiped ]
A cube with a side of 1 m was sawn into cubes with a side of 1 cm and laid in a row (in a straight line). What length did the row turn out to be
# | We get $100 \times 100 \times 100=1000000$ (cm) or 10000 m = 10 km. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9}
On a ruler, three marks are made: 0, 2, and 5. How can you measure a segment equal to 6 using it?
# | $5-2=3$, so you can lay down a segment of length 3. Two such segments give a segment equal to 6.
Send a comment | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Sharygin I.F.
A rectangle is composed of six squares (see the right figure). Find the side of the largest square if the side of the smallest one is 1.
 | The side of the largest square is equal to the sum of the sides of two squares: the next one in the clockwise direction and the smallest one.
## Solution
Note that the side of the largest square is equal to the sum of the sides of two squares: the next one in the clockwise direction and the smallest one. Denoting the... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Prokopenko D:
Around triangle $A B C$, a circle $\Omega$ is described. Let $L$ and $W$ be the points of intersection of the angle bisector of angle $A$ with side $B C$ and circle $\Omega$, respectively. Point $O$ is the center of the circumcircle of triangle $A C L$. Restore triangle $A B C$ if the circle $\Omega$ and... | Let $O^{\prime}$ be the center of $\Omega$. Then the lines $O^{\prime} O$ and $O^{\prime} W$ are perpendicular to the sides $A C$ and $B C$, so the directions of these sides are known. Moreover,
$\angle C O L=2 \angle C A L=2 \angle L C W$ and, therefore, $\angle O C W=90^{\circ}$ (see the figure). Consequently, $C$ i... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 |
In triangle $A B C$, it is known that $\angle B A C=75^{\circ}, A B=1, A C=\sqrt{6}$. A point $M$ is chosen on side $B C$, such that $\angle$ ВАМ $=30^{\circ}$. Line $A M$ intersects the circumcircle of triangle $A B C$ at point $N$, different from $A$. Find $A N$. | Denote $A N$ as $x$, express $B N$ and $C N$ through the circumradius of triangle $A B C$, and apply the Law of Cosines to triangles $A B N$ and $A C N$ (or use Ptolemy's theorem).
## Solution
## First Method.
Let $A N=x, R$ be the circumradius of the triangle. Then
$$
B N=2 R \sin 30^{\circ}=R, C N=2 R \sin 45^{\c... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Trapezoids Auxiliary similar triangles $]$
Point $M$ is located on the lateral side $A B$ of trapezoid $A B C D$, such that $A M: B M=2: 1$. A line passing through point $M$ parallel to the bases $A D$ and $B C$ intersects the lateral side $C D$ at point $N$. Find $M N$, if $A D=18, B C=6$. | Let's draw the diagonal $B D$. Suppose it intersects the segment $M N$ at point $P$. According to the theorem of proportional segments,
$B P: P D = C N: N D = B M: A M = 1: 2$. Triangle $B M P$ is similar to triangle $B A D$ with a similarity ratio of $B M / B A = 1 / 3$, and triangle $D N P$ is similar to triangle $D... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[
Thales' Theorem and the Theorem of Proportional Segments
] Difficulty [Ratio of Areas of Triangles with a Common Base or Common Height]
In triangle $ABC$, with an area of 6, a point $K$ is taken on side $AB$, dividing this side in the ratio $AK: BK = 2: 3$, and a point $L$ is taken on side $AC$, dividing $AC$ in t... | Draw a line through point $K$ parallel to $BL$, until it intersects side $AC$ at point $M$. By Thales' theorem, $LM = 3/5 AL = CL$. Therefore, $CQ = KQ$.
## Answer
4. Send a comment
## Problem | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10}
Points $A, B, C, D$ are marked on a sheet of paper. A recognition device can perform two types of operations with absolute accuracy: a) measure the distance in centimeters between two given points; b) compare two given numbers. What is the minimum number of operations this device needs to perform to definitely det... | To determine whether $A B C D$ is a rectangle, it is sufficient to check the equalities $A B=C D$, $B C=A D$, and $A C=B D$ - a total of 9 operations (3 operations for each equality: two measurements and one comparison).
A rectangle $ABCD$ will be a square if $A B=B C$ - for this, one more, the 10th, operation of comp... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
On the table, there are 9 apples, forming 10 rows with 3 apples in each (see the figure).
It is known that the weights of nine rows are the same, while the weight of the te... | Before spending money, let's think
Let $l_{1}, l_{2}, l_{3}$ be the weights of the three highlighted (see fig. a) diagonal rows, and $v_{1}, v_{2}, v_{3}$ be the weights of the three vertical rows. Then $l_{1}+l_{2}+l_{3}=v_{1}+v_{2}+v_{3}:$ both are simply the sum of the weights of all nine apples. But at least five ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Emelyanov L.A.
A square is cut into $n$ rectangles of size $a_{i} \times b_{i}, i=1, \ldots, n$.
For what smallest $n$ can all the numbers in the set $\left\{a_{1}, b_{1}, \ldots, a_{n}, b_{n}\right\}$ be distinct? | First, let's show that no rectangle (in particular, a square) can be cut into two, three, or four rectangles with different sides.
Obviously, if a rectangle is cut into two rectangles, they share a common side. Suppose the rectangle is cut into three rectangles. Then one of them contains two vertices of the original r... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8,9
In a convex quadrilateral $A B C D$, diagonal $A C$ is drawn, $A D=7, B C=3, \angle A C D=60^{\circ}$. It is known that points $A, B, C, D$ lie on the same circle, and the perpendicular from point $A$ to side $C D$ bisects angle $\angle B A D$. Find the diagonal $A C$. | Find the sine of angle $A D C$.
## Solution
Let $A M$ be the perpendicular dropped from point $A$ to side $C D$. The ray $A C$ passes between the sides of angle $B A M$, so
$$
\angle B A M=\angle B A C+\angle C A M .
$$
Let $R$ be the radius of the circumscribed circle of quadrilateral $A B C D$. Then
$$
\begin{al... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Rhombi. Properties and characteristics ] [Area of a triangle (using two sides and the angle between them).]
In an acute-angled triangle $A B C$, point $D$ is chosen on side $A B$ such that $\angle D C A=45^{\circ}$. Point $D_{1}$ is symmetric to point $D$ with respect to line $B C$, and point $D_{2}$ is symmetric to... | $\angle B C D=30^{\circ}$, quadrilateral $D B D_{1} C-$ is a rhombus.
## Solution
Let $Q$ and $P$ be the points of intersection of segments $D D_{1}$ and $D_{1} D_{2}$ with lines $B C$ and $A C$. Denote $\angle A C B=\gamma$. Then
$$
\begin{gathered}
\angle D_{1} C P=\angle D_{2} C P=\angle A C B=\gamma \\
\angle Q ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 | |
A circle can be inscribed in quadrilateral $A B C D$. Let $K$ be the intersection point of its diagonals. It is known that $A B>B C>K C, B K=4+\sqrt{2}$, and the perimeter and area of triangle $B K C$ are 14 and 7, respectively.
Find $D C$. | Applying Heron's formula, find the sides of triangle $B K C$. Prove that $A C \perp B D$. Also prove that if the diagonals of a convex quadrilateral are perpendicular to each other, then the sums of the squares of the opposite sides are equal.
## Solution
Let $p=7$ be the semi-perimeter, $S=7$ be the area of triangle... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ $\left.\begin{array}{l}\text { Inscribed quadrilaterals } \\ \text { [Quadrilateral: calculations, metric relations.] }\end{array}\right]$
In the quadrilateral $A B C D$, a circle can be inscribed. Let $K$ be the point of intersection of its diagonals. It is known that $B C>A B>B K, K C=\sqrt{7}-1$, the cosine of an... | Let $\angle K B C=\alpha, B K=x, B C=y$. Let $P=2 \sqrt{7}+4$ be the perimeter of triangle $B K C$. Then
$$
x+y=P-K C=2 \sqrt{7}+4-(\sqrt{7}-1)=5+\sqrt{7}
$$
By the cosine rule,
$$
K C^{2}=B K^{2}+B C^{2}-2 \cdot B K \cdot B C \cdot \cos \alpha, \text { or }(\sqrt{7}-1)^{2}=x^{2}+y^{2}-2 x y \cdot \frac{\sqrt{7}+1}{... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In triangle $K L M$, the ratio of the radii of the circumscribed and inscribed circles is 3. The inscribed circle touches the sides of triangle $K L M$ at points $A, B$, and $C$. Find the ratio of the area of triangle $K L M$ to the area of triangle $A B C$.
# | Let $O$ be the center of the circle inscribed in triangle $KLM$. Then point $O$ lies inside triangle $ABC$ and
$$
S_{\triangle \mathrm{ABC}}=S_{\triangle \mathrm{BOC}}+S_{\triangle \mathrm{AOC}}+S_{\triangle \mathrm{AOB}}.
$$
Next, apply the Law of Sines and formulas for the area of a triangle.
## Solution
Let the ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Chords and secants (etc.). $]$
A chord subtends an arc of $90^{\circ}$ and is equal to 16. Find its distance from the center. | The foot of the perpendicular dropped from the center of the circle to the given chord bisects it.
## Solution
Let $M$ be the foot of the perpendicular dropped from the center $O$ of the circle to the chord $A B$. Then $M$ is the midpoint of $A B$ and the right triangle $O M A$ is isosceles. Therefore, $O M = M A = 8... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The radius of the circle is 13, the chord is 10. Find its distance from the center.
# | A diameter perpendicular to a chord bisects it. Therefore, the square of the desired distance is $13^{2}-5^{2}=$ $12^{2}$.
Answer
12. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\begin{aligned} & \text { [ Congruent triangles. Criteria for congruence ] } \\ & \text { [Central symmetry helps solve the problem].] }\end{aligned}$
Segments $A B$ and $C D$ intersect at point $O$, which is the midpoint of each of them. What is the length of segment $B D$ if segment $A C=10$? | Triangle $B O D$ is equal to triangle $A O C$ by two sides and the included angle. Therefore, $B D=A C=10$.
## Answer
10. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Properties and characteristics of an isosceles triangle. ]
The median $A M$ of triangle $A B C$ is perpendicular to its bisector $B K$. Find $A B$, if $B C=12$. | Let $P$ be the intersection point of $B K$ and $A M$. In triangle $A B M$, the bisector $B P$ is an altitude.
## Solution
Let $P$ be the intersection point of segments $B K$ and $A M$. In triangle $A B M$, the bisector $B P$ is an altitude, so triangle $A B M$ is isosceles. Therefore, $A B = B M = 1 / 2 B C = 6$.
##... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The area of an isosceles trapezoid is 32. The cotangent of the angle between the diagonal and the base is 2. Find the height of the trapezoid.
# | The projection of the diagonal of an isosceles trapezoid onto its base is equal to the midline of the trapezoid.
## Solution
Let $h$ be the height we are looking for. The projection of the diagonal of the trapezoid onto its larger base is $2h$. Therefore, the midline of the trapezoid is also $2h$. According to the pr... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Properties and characteristics of tangents ] [ Pythagorean Theorem (direct and inverse) ]
Find the geometric locus of points from which tangents are drawn to a given circle, equal to a given segment.
# | Consider a right triangle with vertices at the considered point, the center of the given circle, and the point of tangency.
## Solution
Let the tangent line drawn from point $M$ to the given circle be $d$ (i.e., the segment with endpoints at point $M$ and the point of tangency is equal to the given length), and the r... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Tangent Circles [ Equilateral (Equiangular) Triangle ]
In a given circle with a radius of 3, six equal circles are inscribed, each touching the given circle; moreover, each of these six circles touches two adjacent ones. Find the radii of the circles.
# | Consider a triangle with vertices at the centers of the external and two adjacent internal circles.
## Solution
Let $r$ be the radius of one of the six internal circles. The triangle with vertices at the centers of the external and two adjacent internal circles is equilateral, so $3-r=2r$. Therefore, $r=1$.
 base $AD$ of trapezoid $ABCD$, according to the midline theorem, is 12. Through the vertex $C$ of the smaller base $BC$, draw a line parallel to the lateral side $AB$, until it intersects... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$C D$ is the median of triangle $A B C$. The circles inscribed in triangles $A C D$ and $B C D$ touch the segment $C D$ at points $M$ and $N$. Find $M N$, if $A C - B C = 2$. | If the circle inscribed in triangle $P Q R$ touches side $P Q$ at point $S$, then $P S=1 / 2(P Q+P R-R Q)$.
## Solution
Since $A D=D B$, and
$$
C M=1 / 2(A C+C D-A D) \text{ and } C N=1 / 2(B C+C D-B D)
$$
then
$$
\begin{aligned}
M N=|C M-C N| & =|1 / 2(A C+C D-A D)-1 / 2(B C+C D-B D)|= \\
& =1 / 2|A C-B C|=1 / 2 ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Acute angle $A$ of rhombus $A B C D$ is $45^{\circ}$, the projection of side $A B$ on side $A D$ is 12. Find the distance from the center of the rhombus to side $C D$. | Use the theorem of the midline of a triangle.
## Solution
Let $M$ be the projection of vertex $B$ of the rhombus $A B C D$ onto side $A D$. According to the problem, $A M=12$. The acute angle at vertex $A$ of the right triangle $A M B$ is $45^{\circ}$, so $B M=A M$.
Let $P$ and $Q$ be the projections of the center $... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Inside triangle $A B C$, there are points $P$ and $Q$ such that point $P$ is at distances 6, 7, and 12 from the lines $A B, B C, C A$ respectively, and point $Q$ is at distances 10, 9, and 4 from the lines $A B, B C, C A$ respectively. Find the radius of the inscribed circle of triangle $A B C$. | Prove that the midpoint of segment $P Q$ is the center of the inscribed circle of triangle $A B C$.
## Solution
Let $D, E, F$ be the feet of the perpendiculars dropped from points $P, Q, O$ to line $A B$ respectively. By the given condition, $P D=6, Q E=10$. $O F$ is the midline of trapezoid $D P Q E$, so the length ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Construction on a projection drawing ]
The base of a right prism $A B C A 1 B 1 C 1$ is a right triangle $A B C\left(\angle B=90^{\circ}, A B=B C=10\right)$; $A A 1=B B 1=C C 1=12$. Point $M$ is the midpoint of the lateral edge $A A 1$. A plane is drawn through points $M$ and $B 1$, forming an angle of $45^{\circ}$ ... | Let the lines $M E$ and $A 1 C 1$ intersect at point $P$. Then the plane of the section intersects the plane of the base $A 1 B 1 C 1$ along the line $B 1 P$ (Fig.1). Suppose that point $P$ lies on the extension of the edge $A 1 C 1$ beyond point $A 1$. Drop the perpendicular $A 1 K$ from point $A 1$ to the line $B 1 P... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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