problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
107. Prove that for any triangle $ABC$ and any point $M$ inside it
a) $\min \left(h_{a}, h_{b}, h_{c}\right) \leqslant d_{a}+d_{b}+d_{c} \leqslant \max \left(h_{a}, h_{b}, h_{c}\right)$, where $\min \left(h_{a}, h_{b}, h_{c}\right)$ denotes the smallest and $\max \left(h_{a}, h_{b}, h_{c}\right)$ the largest of the he... | 107. From the consideration of triangles $MBC$, $M$ and $C$, and $MAB$ (see Fig. 26 on p. 48), we easily obtain
$$
S_{\triangle ABC}=\frac{1}{2} a d_{a}+\frac{1}{2} b d_{b}+\frac{1}{2} c d_{c}, \text{ i.e., } a d_{a}+b d_{b}+c d_{c}=2 S \text{. }
$$
## From this, it follows:
a) If $a \geqslant b \geqslant c$, then
... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 23,866 |
109. Prove that for any triangle $ABC$ and for any point $M$ inside it
$$
d_{a}^{2}+d_{b}^{2}+d_{c}^{2} \geqslant 12 \frac{r^{4}}{R^{2}}
$$
In which case $d_{a}^{2}+d_{b}^{2}+d_{c}^{2}=12 \frac{r^{4}}{R^{2}}$?
If point $M$ coincides with the center of the inscribed circle of the triangle, then the sum $d_{a}^{2}+d_{... | 109. From the result of problem 81 b), it follows that the sum $d_{a}^{3}+d_{b}^{2}+d_{c}^{2}$ reaches its minimum value for a point $M$ inside the triangle $ABC$ for which $d_{a}: d_{b}: d_{c}=a: b: c$, and that this value is $\frac{4 S^{2}}{a^{2}+b^{2}+c^{2}}$. Since, by the results of problems 97 and 95 b), $S \geqs... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 23,868 |
110. Prove that for any triangle $ABC$ and any point $M$ inside it
$$
a R_{a}+b R_{b}+c R_{c} \geqslant 2\left(a d_{a}+b d_{b}+c d_{c}\right)
$$
In which case does this inequality become an equality? | 110. Since, obviously (compare with the beginning of the solution to problem 107),
$$
a d_{a}+b d_{b}+c d_{c}=2 S=a h_{a v}
$$
then
$$
a\left(h_{a}-a_{a}^{\prime}\right)=b d_{b}+c d_{c}
$$
## Similarly, we have
$$
b\left(h_{b}-d_{b}\right)=a d_{a}+c d_{c} \quad \text { and } \quad c\left(h_{c}-d_{c}\right)=a d_{a}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 23,869 |
112*. a) Prove that for any triangle $ABC$ and for any point $M$ inside it,
$$
R_{a} R_{b} R_{c} \geqslant 8 d_{a}^{\prime} d_{b}^{\prime} d_{c}^{\prime}
$$
In what case does this inequality become an equality?
b) Prove that for any triangle $ABC$ and for any point $M$ inside it,
$$
R_{a}+R_{b}+R_{c}>d_{a}^{\prime}... | 112. a) Let
$$
d_{a}^{\prime}=M A_{1}, d_{b}^{\prime}=M B_{1}, \quad d_{c}^{\prime}=M C_{1}
$$
where $A_{1}, B_{1}$, and $C_{1}$ are the points of intersection of the lines $A M, B M$, and $C M$ with the opposite sides of the triangle (see Fig. 26 on p. 48 or Fig. 260 on p. 284). It is clear that the altitudes $A P$ ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 23,871 |
114. a) Prove the following theorem of Pappus ${ }^{1}$: let parallelograms $A B B_{1} A_{1}$ and $A C C_{1} A_{2}$ be constructed on the sides $A B$ and $A C$ of triangle $A B C$, with both these parallelograms lying outside the triangle $A B C$ (Fig. $28, \alpha$) or both lying on the same side of the line $A B$, res... | 114. a) Let the line $A D$ intersect the sides $B C$ and $B_{2} C_{2}$ of the parallelogram $B C C_{2} B_{2}$ at points $K$ and $L$; furthermore, let $E$ and $F$ be the points of intersection of the lines $B B_{2}$ and $A_{1} B_{1}$, $C C_{2}$ and $A_{2} C_{1}$. (See the corresponding figure 28, a, figure 268, where, h... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,873 |
115*. Prove that for any triangle $ABC$ and any point $M$ inside it
$$
R_{a} R_{b} R_{c} \geqslant\left(d_{a}+d_{b}\right)\left(d_{b}+d_{c}\right)\left(d_{c}+d_{a}\right)
$$
In which case does this inequality turn into an equality? | 115. Applying the inequality
$$
a R_{a} \geqslant b d_{b}+c d_{c}
$$
(p. 284), valid for any point $M$ in angle $BAC$, to the point $M'$, symmetric to $M$ with respect to the bisector of angle $A$, we get (compare with the third solution of problem 111b))
$$
a R_{a} \geqslant b d_{c}+c d_{b} .
$$
Adding the last in... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 23,874 |
118. Prove that for each tetrahedron $ABCD$ and every point $M$ inside it:
a) $S_{A} R_{A} + S_{B} R_{B} + S_{C} R_{C} + S_{D} R_{D} \geqslant$
$\geqslant 3\left(S_{A} d_{A} + S_{B} d_{B} + S_{C} d_{C} + S_{D} d_{D}\right)$,
where $S_{A}, S_{B}, S_{C}$, and $S_{D}$ are the areas of the faces $BCD, ACD, ABD$, and $ABC... | 118. a) Since, obviously,
$$
\begin{aligned}
S_{A} h_{A} & =3 V=3 V_{M B C D}+3 V_{M A C D}+3 V_{M A B D}+3 V_{M A B C}= \\
& =S_{A} d_{A}+S_{B} d_{B}+S_{C} d_{C}+S_{D} d_{D}
\end{aligned}
$$
where \( h_{A}=A P \) is the height dropped from vertex \( A \) of the tetrahedron to the face \( B C D \), and (Fig. 272)
$$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 23,877 |
120*. Prove that for any tetrahedron $ABCD$ and any point $M$ inside it
a) $\quad\left(R_{A}+R_{B}+R_{C}+R_{D}\right)\left(\frac{1}{d_{A}}+\frac{1}{d_{B}}+\frac{1}{d_{C}}+\frac{1}{d_{D}}\right) \geqslant 48$;
b) $\left(R_{A}^{2}+R_{B}^{2}+R_{C}^{2}+R_{D}^{2}\right)\left(\frac{1}{d_{A}^{2}}+\frac{1}{d_{B}^{2}}+\frac{1... | 120. It will be more convenient for us to change the order of solving tasks a) and b).
b) We will use the fact that for any $x, y, z, t$ and $x_{1}, y_{1}, z_{1}, t_{1}$, the inequality ${ }^{1}$ holds:
$$
\left(x^{2}+y^{2}+z^{2}+t^{2}\right)\left(x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+t_{1}^{2}\right) \geqslant\left(x x_{1}+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 23,879 |
1. In space, there are 4 points not lying in the same plane. How many planes can be drawn equidistant from these points? | 1. Since the four points do not lie in the same plane, the plane equidistant from these points cannot be located on the same side of all of them. Therefore, there are only two possible cases: 1) three points lie on one side of the considered plane, and the fourth point lies on the other side, and 2) two points lie on e... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,880 |
3. How many spheres exist that touch all the faces of the given triangular pyramid $T$?
| 3. The task is to determine how many points (centers of the sought spheres) are equidistant from the four faces of the pyramid.
The geometric locus of points equidistant from the faces of a given dihedral angle is a plane passing through the edge of the dihedral angle and bisecting this angle — the bisector plane of t... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,882 |
4. In how many different ways can the faces of a cube be painted using six given colors (each face must be painted entirely with one color), if only those colorings are considered different that cannot be made to coincide by rotating the cube? | 4. Suppose the faces of a cube are painted green, blue, red, yellow, white, and black. Let's place the cube so that the green face is at the bottom. In this case, the top face can be painted one of the five remaining colors. It is clear that no two colorings, in which the top (i.e., opposite the green) face is painted ... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,883 |
5. In how many different ways can 30 workers be divided into 3 teams of 10 people each? | 5. 10 workers, forming the first brigade, can be chosen from thirty in $C_{30}^{10}=\frac{30 \cdot 29 \cdot \ldots \cdot 21}{10!}$ ways. After this, from the remaining 20 workers, 10 workers for the second brigade can be chosen in $C_{20}^{10}=\frac{20 \cdot 19 \cdots 11}{10!}$ ways. By combining each way of forming th... | 2775498395670 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,884 |
7. The commission consists of 11 people. The materials that the commission works on are stored in a safe. How many locks should the safe have, and how many keys should each member of the commission be provided with, so that access to the safe is possible when a majority of the commission members gather, but not possibl... | 7. According to the problem statement, for any five members of the committee, there should be a lock for which the key is absent from all these committee members (but is present with each of the six absent members, since the presence of any of the nine absent members already makes the meeting of the committee possible)... | 252 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,886 |
8. The numbers from 1 to 1000 are written in a circle. Starting from the first, every 15th number is crossed out (i.e., the numbers $1, 16, 31$, etc.), and during subsequent rounds, already crossed-out numbers are also taken into account. The crossing out continues until it turns out that all the numbers to be crossed ... | 8. After the first cycle, all numbers that give a remainder of 1 when divided by 15 will be crossed out; the last such number will be 991. The first number to be crossed out during the second cycle will be $991+15-1000=6$; subsequently, during the second cycle, all numbers that give a remainder of 6 when divided by 15 ... | 800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,887 |
9. All numbers from 1 to 10000000000 are written in a row. Which numbers will be more - those in which the digit 1 appears, or those in which 1 does not appear? | 9. First solution. Let's calculate the number of numbers in our sequence that do not contain the digit 1. Add the number 0 (zero) at the very beginning of this sequence and omit the last number 10000000000; we will get a sequence of $10^{10}$ numbers, containing one more number, in the notation of which the digit 1 doe... | 6513215600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,888 |
10. All integers from 1 to 222222222 are written in sequence. How many times does the digit 0 appear in the recording of these numbers?
## 2. Factorization of numbers into products and their decomposition into sums
In solving some of the subsequent problems, the following notations prove useful.
The symbol $[x]$ (re... | 10. Obviously, among the first 222222222 integers, there will be 22222222 numbers ending in zero (numbers 10, $20,30, \ldots, 222222220$). Further, among them, there will be 2222222 numbers ending in the digits 00 (numbers $100,200,300, \ldots, 222222200$); 2222222 numbers ending in the digits 01 (numbers 101, 201, 301... | 175308642 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,889 |
11. a) How many integers less than 1000 are not divisible by 5 or 7?
b) How many of these numbers are not divisible by 3, 5, or 7? | 11. a) There are a total of 999 numbers less than 1000 (numbers $1,2,3, \ldots, 999$). Now, let's strike out those that are divisible by 5; there are $\left[\frac{999}{5}\right]=199$ such numbers (numbers $5,10,15,20, \ldots, 995=199.5$). Next, let's strike out all numbers divisible by 7; there are $\left[\frac{999}{7}... | 457 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,890 |
12*. How many integers are there that are less than the number 56700000 and coprime with it? | 12. The problem is very close to the previous one. The factorization of the number 56700000 is $56700000=2^{5} \cdot 3^{4} \cdot 5^{5} \cdot 7$; thus, the problem reduces to determining how many numbers less than 56700000 are not divisible by 2, 3, 5, or 7.
Let's list all numbers from 1 to 56700000. We will strike out... | 12960000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,891 |
13. How many positive integers $x$, less than 10000, are there for which the difference $2^{x}-x^{2}$ is not divisible by 7? | 13. $2^{0}=1$ gives a remainder of $1$ when divided by 7, $2^{1}=2$ gives a remainder of $2$, $2^{2}=4$ gives a remainder of $4$, $2^{3}$ gives a remainder of $1$ again, $2^{4}$ gives a remainder of $2$ again, $2^{5}$ gives a remainder of $4$ again, $2^{6}$ gives a remainder of $1$ for the third time, and so on. Thus, ... | 7142 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,892 |
14. How many different pairs of integers $x, y$, lying between 1 and 1000, are there such that $x^{2}+y^{2}$ is divisible by 49? | 14. If $x^{2}+y^{2}$ is divisible by 49, then $x^{2}+y^{2}$ is also divisible by 7. But $x^{2}$, when divided by 7, can only give remainders of $0, 1, 4$, or 2 (see the solution to problem 13). The remainder from dividing $x^{2}+y^{2}$ by 7 is equal to the sum of the remainders from dividing the numbers $x^{2}$ and $y^... | 10153 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,893 |
15. In how many ways can a million be factored into three factors? Factorizations that differ only in the order of the factors are considered the same. | 15. Since a million is equal to $2^{6} \cdot 5^{6}$, any of its divisors has the form $2^{\alpha} \cdot 5^{\beta}$, and the factorization of a million into 3 factors has the form
$$
1000000=\left(2^{\alpha_{1}} \cdot 5^{\beta_{2}}\right)\left(2^{\alpha_{2}} \cdot 5^{\beta_{2}}\right)\left(2^{\alpha_{3}} \cdot 5^{\beta... | 139 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,894 |
16*. How many different divisors does the number 86400000 have (including 1 and the number 86400000 itself)? Find the sum of all these divisors. | 16. The prime factorization of the number 86400000 is
$$
86400000=2^{10} \cdot 3^{3} \cdot 5^{5}
$$
Therefore, all its divisors have the form $2^{x} \cdot 3^{3} \cdot 5^r$, where $\alpha$ is a non-negative integer not exceeding 10, $\beta$ is a non-negative integer not exceeding 3, and $\gamma$ is a non-negative inte... | 264divisors,sum319823280 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,895 |
17. How many different pairs of integers $A, B$ exist for which the least common multiple is 59400000? | 17. The prime factorization of the number 59400000 is
$$
59400000=2^{6} \cdot 3^{3} \cdot 5^{5} \cdot 11
$$
Therefore, if the least common multiple of two numbers $A$ and $B$ is 59400000, then it must be
$$
A=2^{x_{1}} \cdot 3^{\beta_{1}} \cdot 5 \gamma_{1} \cdot 11^{\delta_{1}}, \quad B=2^{\alpha_{2}} \cdot 3^{\bet... | 1502 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,896 |
18. Find the coefficients of $x^{17}$ and $x^{18}$ after expanding the brackets and combining like terms in the expression
$$
\left(1+x^{5}+x^{7}\right)^{20}
$$ | 18. Let's repeat 20 times the trinomial $1+x^{3}+x^{7}$ and multiply it according to the usual rules. Each term of the resulting sum will represent a product of 20 factors, each equal to either 1, or $x^{5}$, or $x^{7}$. It is essential that in these products of twenty factors, any of the three expressions can stand in... | 3420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,897 |
19. In how many ways can 20 kopecks be exchanged for coins worth 5 kopecks, 2 kopecks, and 1 kopeck? | 19. Let's list all the ways to change 20 kopecks. In such a change, we can use either four 5-kopeck coins, or three such coins, or two, or one, or none.
There is only one way to change 20 kopecks using four 5-kopeck coins (since four 5-kopeck coins already make 20 kopecks).
If we use three 5-kopeck coins, we need to ... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,898 |
20. In how many ways can $n$ kopecks be made using coins of 1 kopeck and 2 kopecks? | 20. When composing $n$ kopecks from coins worth 1 kopeck and 2 kopecks, we cannot use more than $\frac{n}{2}$ coins worth 2 kopecks. If we take fewer than $\frac{n}{2}$ or exactly $\frac{n}{2}$ such coins, then the way of composing is already uniquely determined (if we take $q$ coins worth 2 kopecks, where $q \leqslant... | [\frac{n}{2}]+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,899 |
21*. In how many ways can $n$ kopecks be made from coins:
a) of denomination 1 kopeck, 2 kopecks, and 3 kopecks?
b) of denomination 1 kopeck, 2 kopecks, and 5 kopecks? | 21. a) When composing \( n \) kopecks from coins worth 1 kopeck, 2 kopecks, and 3 kopecks, we can either not use any 3 kopeck coins, use one such coin, use two, three, and so on up to \(\left[\frac{n}{3}\right]\) coins. In the first case, we need to compose \( n \) kopecks from coins worth 1 kopeck and 2 kopecks (this ... | (\frac{(n+4)^{2}}{20}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,900 |
22**. In how many ways can you make up a ruble using coins of 1, 2, 5, 10, 20, and 50 kopecks? | 22. Since from coins of 10, 20, and 50 kopecks, only a whole number of tens of kopecks can be formed, then from 1, 2, and 5 kopeck coins, a whole number of tens of kopecks must also be formed. Therefore, the following cases are possible:
1) The ruble is composed of coins worth 10, 20, and 50 kopecks,
2) Coins worth 10,... | 4562 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,901 |
23. In how many ways can the number $n$ be represented as the sum of two positive integers, if representations differing only in the order of the addends are considered the same?
Represent the above text in English, please keep the original text's line breaks and format, and output the translation result directly. | 23. If the number $n$ is represented as the sum of two addends
$$
n=x+y
$$
then one of these addends will not be greater than $\frac{n}{2}$. This addend can take the values $1,2,3, \ldots,\left[\frac{n}{2}\right] ;$ all these cases are distinct, since the second addend in this case will be no less than $\frac{n}{2}$.... | [\frac{n}{2}] | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,902 |
24. How many integer solutions does the inequality
$$
|x|+|y|<100 ?
$$
have? Here, for $x \neq y$, the solutions $x, y$ and $y, x$ should be considered different. | 24. We need to find the number of pairs of integers \(x, y\) such that \(|x| + |y|\) is 0, 1, 2, 3, ..., or 99. Let's count the number of pairs \(x, y\) for which \(|x| + |y| = k\). Here, \(|x|\) can take \(k + 1\) different values, namely \(0, 1, 2, \ldots, k-1, k\); in this case, \(|y|\) will be \(k, k-1, k-2, \ldots... | 19801 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 23,903 |
25. In how many ways can the number $n$ be represented as the sum of three positive integer addends, if representations differing in the order of the addends are considered different? | 25. The problem reduces to finding the number of positive integer solutions to the equation $x+y+z=n$.
First, note that the equation $x+y=l$ (where $l$ is a positive integer) has $l-1$ positive integer solutions. Indeed, in this case, $x$ can be equal to $1,2, \ldots, l-1$ (it cannot be equal to $l$, because then $y$ ... | \frac{(n-1)(n-2)}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,904 |
26***. In how many ways can the number $n$ be represented as the sum of three positive integer addends, if representations differing only in the order of the addends are considered the same? | 26. Arrange the terms of this sum in ascending order. It is clear that the first (smallest) term can vary from 1 to $\left[\frac{n}{3}\right]$, i.e., there are $\left[\frac{n}{3}\right]$ possibilities for it.
If the first term is $x$, then the second cannot be less than $x$ and greater than $\left[\frac{n-x}{2}\right]... | (\frac{n^{2}}{12}) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,905 |
27*. How many positive integer solutions does the equation
$$
x+y+z=n
$$
have that satisfy the inequalities
$$
x \leqslant y+z, \quad y \leqslant x+z, \quad z \leqslant x+y ?
$$ | 27. Since $z=n-x-y$, the solution will be fully determined if we know $x$ and $y$. Substituting $n-x-y$ for $z$ in our inequalities:
1) The inequality $z \leqslant x+y$ gives
$$
n-x-y \leqslant x+y
$$
from which
$$
n \leqslant 2(x+y) ; \quad x+y \geqslant \frac{n}{2}
$$
i.e.
$$
y \geqslant \frac{n}{2}-x
$$
2) The... | \frac{n^2-1}{8} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,906 |
29*. a) How many different positive integer solutions does the equation
$$
x_{1}+x_{2}+x_{3}+\ldots+x_{m}=n
$$
have?
b) How many different non-negative integer solutions does the equation
$$
x_{1}+x_{2}+x_{3}+\ldots+x_{m}=n
$$
have?
Note. A particular case of problem 29a) (corresponding to $m=3$) is problem 25.
... | 29. The first solution to problems a) and b). First of all, note that the number of positive (respectively non-negative) integer solutions to the equation
$$
x_{1}+x_{2}+\ldots+x_{m}=n
$$
is equal to the difference between the number of positive (non-negative) integer solutions to the inequality
$$
x_{1}+x_{2}+\ldot... | C_{n+-1}^{-} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,908 |
32. a) What is the maximum number of rooks that can be placed on a chessboard of $n^{2}$ squares so that no two rooks threaten each other? In how many different ways can this be done?
b) What is the minimum number of rooks that can be placed on a chessboard of $n^{2}$ squares so that these rooks threaten all the squar... | 32. a) A chessboard consisting of $n^{2}$ squares (see Fig. 39, where the case $n=8$ is shown) contains $n$ rows and $n$ columns. To ensure that no two rooks placed on this board threaten each other, it is necessary that no two rooks stand on the same row or the same column. It is clear, therefore, that the total numbe... | 33514312 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,911 |
37. a) What is the maximum number of kings that can be placed on a chessboard of 64 squares so that no two kings threaten each other?
b) The same question for a chessboard of $n^{2}$ squares. | 37. a) Let's divide the chessboard into 16 squares, each consisting of four cells, as shown in Fig. 45, a. When placing kings so that no two of them threaten each other, in each of these 16 squares, there can be no more than one king. Therefore, it is impossible to place more than 16 kings so that they do not threaten ... | [\frac{n+1}{2}]^2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,915 |
38. a) What is the smallest number of kings that can be placed on a chessboard of 64 squares so that these kings threaten all the squares free of figures?
b) The same question for a chessboard of $n^{2}$ squares. | 38. a) Let's divide the board into nine parts as indicated in Fig. $46, \alpha$. In each of these parts, there is a cell (marked with a circle in Fig. 46, $a$) that can only be threatened by kings standing on cells of the same part. Therefore, for all fields of the board to be under threat, there must be at least one k... | [\frac{n+2}{3}]^{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,916 |
39. a) What is the largest number of queens that can be placed on a chessboard of 64 squares so that no two queens threaten each other?
6) ${ }^{* * *}$ The same question for a chessboard of $n^{2}$ squares. | 39. a) On each vertical of a chessboard, there can be no more than one queen, so it is impossible to place more than eight queens on a 64-square chessboard such that no two of them threaten each other. Eight queens, none of which threaten each other, can be placed (one such arrangement is shown, for example, in Fig. 47... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,917 |
41. a) Each vertex at the base of the triangle is connected by straight lines to \( n \) points located on the opposite side. Into how many parts do these lines divide the triangle?
b) Each of the three vertices of the triangle is connected by straight lines to \( n \) points located on the opposite side of the triang... | 41. a) First, let's draw $n$ lines connecting vertex $B$ with $n$ points located on side $AC$ - these lines will divide the triangle into $n+1$ parts. Now, we will draw lines connecting vertex $C$ with points located on side $AB$ (Fig. 57). Each of these lines (the number of which is $n$) will intersect the previously ... | 3n^2+3n+1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,919 |
42*. What is the greatest number of parts into which:
a) $n$ lines can divide a plane?
b) $n$ circles can divide a plane?
43**. What is the greatest number of parts into which:
a) $n$ planes can divide space?
b) $n$ spheres can divide space? | 42. a) It is clear that $n$ lines will divide the plane into the maximum number of parts if all these lines intersect (i.e., no two of them are parallel) and no three of them intersect at the same point. Therefore, we only need to determine how many parts $n$ pairwise non-parallel lines, no three of which intersect at ... | \frac{n^2+n+2}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,920 |
44*. In how many points do the diagonals of a convex n-gon intersect if no three of them intersect at the same point? | 44. First solution. Consider the diagonal $A_{1} A_{k}$ of the $n$-gon $A_{1} A_{2} A_{3} \ldots A_{n}$. On one side and the other of this diagonal, there are respectively $k-2$ vertices of the $n$-gon (vertices $A_{2}, A_{3}, A_{4}, \ldots, A_{k-1}$) and $n-k$ vertices (vertices $A_{k+1}, A_{k+2}, \ldots, A_{n}$). The... | \frac{n(n-1)(n-2)(n-3)}{24} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,921 |
46. a) How many different rectangles, in terms of size or position, consisting of whole cells, can be drawn on a chessboard of 64 cells?
b) The same question for a chessboard of $n^{2}$ cells. | 46. a) First, let's determine how many rectangles can be found on a chessboard, composed of whole numbers of
Fig. 61. cells and having a given width \( k \) and a given height \( l \). Eac... | [\frac{n(n+1)}{2}]^2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,923 |
47. a) How many different squares, in terms of size or position, consisting of whole cells, can be drawn on a chessboard of 64 cells?
b) The same question for a chessboard of $n^{2}$ cells. | 47. a) This problem is close to the previous one. The number of all possible squares consisting of $k^{2}$ cells that can be chosen on a chessboard of 64 cells is $(9-k)^{2}$ (see the solution to problem 46 a)). From this, it follows that the number of all squares is $8^{2}+7^{2}+6^{2}+\ldots+1^{2}=$
$$
=64+49+36+25+1... | 204 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,924 |
48*. No three diagonals of a certain convex $n$-gon intersect at one point. How many triangles, different in size or position, can be drawn so that all their sides lie on the sides or diagonals of this $n$-gon? | 48. It is evident that the vertices of the sought triangles are either the vertices or the intersection points of the diagonals of the original $n$-gon. Let us consider the four possible cases separately:
$1^{\circ}$. All three vertices of the triangle are vertices of the $n$-gon.
$2^{\circ}$. Two vertices of the tri... | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,925 | |
49**. (Cayley's Problem). How many convex $k$-gons are there, all vertices of which coincide with the vertices of a given convex $n$-gon, and all sides are its diagonals? | 49. For there to exist at least one such $k$-gon, $n$ must be no less than $2k$ (since each pair of vertices of the $k$-gon must be separated by at least one vertex of the $n$-gon).
Let the vertices of the $n$-gon be denoted by $A_{1}, A_{2}, \ldots, A_{n-1}, A_{n}$, and let us count how many $k$-gons, satisfying the ... | \frac{n(n-k-1)!}{k!(n-2k)!} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,926 |
53. a) A circle is divided into $p$ equal sectors, where $p$ is a prime number. In how many different ways can these $p$ sectors be painted using $n$ colors, if it is allowed to paint several (and even all) sectors with the same color, and two colorings are considered different only if they cannot be made to coincide b... | 53. a) Each of the $p$ sectors can be painted with $n$ colors. Therefore, two sectors can be painted in $n^{2}$ different ways (each of the $n$ ways to paint one sector can be combined with each of the $n$ ways to paint the second sector), three sectors can be painted in $n^{3}$ different ways (each of the $n^{2}$ ways... | \frac{n^{p}-n}{p}+n | Combinatorics | proof | Yes | Yes | olympiads | false | 23,930 |
55. Apply the binomial theorem of Newton to find the value of the following sums:
a) $C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\ldots+C_{n}^{n}$
b) $C_{n}^{0}-C_{n}^{1}+C_{n}^{2}-\ldots+(-1)^{n} C_{n}^{n}$
c) $C_{n}^{0}+\frac{1}{2} C_{n}^{1}+\frac{1}{3} C_{n}^{2}+\ldots+\frac{1}{n+1} C_{n}^{n}$
d) $C_{n}^{1}+2 C_{n}^{2}+3 C_{... | 55. a) $C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\ldots+C_{n}^{n}=(1+1)^{n}=2^{n}$.
b) $C_{n}^{0}-C_{n}^{1}+C_{n}^{2}-\ldots+(-1)^{n} C_{n}^{n}=(1-1)^{n}=0$.
c) Using the fact that
$$
\frac{n+1}{k+1} C_{n}^{k}=\frac{(n+1) n(n-1) \ldots(n-k+1)}{1 \cdot 2 \cdot 3 \ldots k(k+1)}=C_{n}^{k+1}
$$
Therefore,
$$
\begin{aligned}
& (n... | 2^n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,932 |
56. Apply the binomial theorem of Newton to find the values of the following sums (the ellipsis at the end of these sums indicates that the series continues until the terms lose their meaning, i.e., until the upper index becomes greater than the lower one):
a) $C_{n}^{0}+C_{n}^{2}+C_{n}^{4}+C_{n}^{6}+\ldots$
b)
$c_{n}... | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,933 | ||
57. Factorial Binomial Theorem. Let us introduce the notation:
$$
a(a-h)(a-2 h) \ldots[a-(n-1) h]=a^{n \mid h}
$$
so that, in particular, $a^{n 10}=a^{n}, a^{1 / h}=a$. Prove that in these notations the following equality holds:
$$
(a+b)^{n h}=a^{n / h}+C_{n}^{1} a^{n-1 \mid h} b+C_{n}^{2} a^{n-2 / h} b^{\eta / h}+\... | 57. The theorem is proved by mathematical induction in a way completely analogous to the usual proof of the binomial formula of Newton. It is easy to verify that for $n$, equal to 1 and 2, the factorial binomial theorem is true ${ }^{1}$ ). Now suppose that this theorem is true for the exponent $n$, i.e., that
 $C_{n}^{0} C_{m}^{k}+C_{n}^{1} C_{m}^{k-1}+C_{n}^{2} C_{m}^{k-2}+\ldots+C_{n}^{k} C_{m}^{0}$;
b) $C_{m}^{0} C_{n}^{k}-C_{m+1}^{1} C_{n}^{k-1}+C_{m+2}^{2} C_{n}^{k-2}-\cdots+(-1)^{k} C_{m+k}^{k} C_{n}^{0}$.
Sometimes when derivin... | 58. a) Using the fact that
$$
C_{n}^{i}=\frac{n(n-1) \ldots(n-i+1)}{i!}=\frac{n^{i \mu}}{i!}
$$
we get:
$$
C_{n}^{i} C_{m}^{k-i}=
$$
$$
=\frac{n^{i / 1} m^{k-i \mid 1}}{i!} \frac{\left.m^{k-i}\right)!}{(k-i)!} \frac{k!}{i!(k-i)!} n^{i \mid 1} m^{k-i \mid 1}=
$$
$$
=\frac{1}{k!} C_{k}^{i} n^{i \mid 1} m^{k-i \mid 1... | C_{n--1}^{k} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,935 |
59. a) Use the described geometric scheme to prove the relation
$$
C_{n}^{m}+C_{n-1}^{m-1}+C_{n-2}^{m-2}+\ldots+C_{n-m}^{0}=C_{n+1}^{m}
$$
b) Show that the following generalization of the previous relation holds:
$C_{n}^{m} C_{k}^{0}+C_{n-1}^{m-1} C_{k+1}^{1}+C_{n-2}^{m-2} C_{k+2}^{2}+\ldots+C_{n-m}^{0} C_{k+m}^{m}=... | 59. a) Let's consider the shortest paths leading from the intersection $(0,0)$ to the intersection $(n-m+1, m)$ (Fig. 68). The number
Fig. 68.
of such paths is obviously equal to $C_{n+1}^... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 23,936 |
62. In the city, there are 10,000 bicycles with various numbers from 1 to 10,000. What is the probability that the number of the first bicycle encountered will not contain the digit 8? | 62. Since the first encountered cyclist can with equal probability turn out to be any of the 10000 cyclists in the city, the total number of equally probable outcomes of the experiment here is 10000. It remains to calculate the number of "favorable outcomes," i.e., the number of numbers that do not contain the digit 8.... | 0.6561 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,938 |
63. a) The word "ремонт" is made up of letters from a cut-out alphabet. Then the cards with individual letters are thoroughly shuffled, after which four of them are randomly drawn and laid out in a row in the order of drawing. What is the probability of getting the word "море" in this way?
b) A similar operation is pe... | 63. a) The considered experiment consists in sequentially drawing some four cards from a group of six cards. In this case, the first draw can result in six different outcomes (the first can be any of the six cards), the second in five different outcomes (the second can be any of the five cards remaining after the first... | \frac{1}{30} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,939 |
64 *. In a urn, there are tickets with the digits $0,1,2,3,4,5$, $6,7,8$ and 9. Five tickets are randomly drawn and laid out in a row one after another in the order of drawing. What is the probability that the resulting number will be divisible by 396? | 64. Let the sequentially drawn digits be $x, y, z, t, u$; the number $N$ has the form $\overline{x y z t u}$ (the bar above indicates that $x, y, z, t, u$ are the digits of the number $N$). Analogously to the solution of the previous problem, we conclude that the total number of different outcomes of the experiment, wh... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,940 |
65. Suppose you have forgotten one digit of a phone number you need and dial it at random. What is the probability that you will have to make no more than two calls? | 65. The considered experiment consists in randomly dialing a forgotten digit; if a wrong number is dialed, another digit is randomly dialed. Thus, there are two different types of outcomes here: either two numbers are dialed, or only one (the latter will be the case if the correct number is dialed the first time). We h... | 0.2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,941 |
68. In the USSR Cup football games, after the qualifying matches, the football teams are divided by lot into two equal groups. In each group, the winner of that group is determined separately, and the two winners meet in the cup final. Let the total number of teams that have passed the qualifying matches be 20.
a) Wha... | 68. a) $\frac{10}{19}$.
b) $\frac{C_{16}^{6}}{\frac{1}{2} C_{20}^{10}}=\frac{28}{323} \approx 0.08 ; \quad\left(\frac{28}{323}\right)^{2} \approx 0.0064$.
c) $\frac{3 C_{16}^{8}}{\frac{1}{2} C_{20}^{10}}=\frac{135}{323} \approx 0.42$. | )\frac{10}{19};b)\frac{28}{323}\approx0.08,(\frac{28}{323})^2\approx0.0064;)\frac{135}{323}\approx0.42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,944 |
69. What is more likely: to win 3 games out of 4 or 5 games out of 8 against an equally matched opponent? | 69. The total number of possible outcomes of a sequence of four matches we get by combining a win or loss in the first match with a win or loss in the second, third, and fourth matches. This number is equal to $2^{4}=16$. All these outcomes are equally probable, as the opponents are of equal strength, and for each sepa... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,945 |
70. a) From a box containing $n$ white and $m$ black balls, $k$ balls are drawn at random. What is the probability that exactly $r$ of the drawn balls will be white?
b) Apply the result of part a) to find the value of the sum
$$
C_{n}^{0} C_{m}^{k}+C_{n}^{1} C_{m}^{k-1}+C_{n}^{2} C_{m}^{k-2}+\ldots+C_{n}^{k} C_{m}^{0... | 70. a) Here, we consider an experiment consisting of drawing $k$ balls from a box containing $n+m$ balls. Choosing any $k$ balls from $n+m$ can be done in $C_{n+m}^{k}$ different ways. This is the number of equally probable outcomes of our experiment. Favorable outcomes will be those in which exactly $r$ white balls an... | C_{n+}^{k} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,946 |
71*. a) Banach's problem ${ }^{1}$. A person simultaneously bought two boxes of matches and put them in his pocket. After that, every time he needed to light a match, he randomly took one or the other box. After some time, upon emptying one of the boxes, the person discovered that it was empty. What is the probability ... | 71. a) At the moment when a person discovers that the matchbox taken out of the pocket is empty, the second box can contain 0 matches (i.e., be also empty), 1 match, 2 matches, 3 matches, and so on up to \( n \) matches inclusive. The considered experiment consists of the person repeatedly drawing one match at random f... | 2^{2n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,947 |
72*. Two hunters $A$ and $B$ went duck hunting. Suppose each of them hits a duck they encounter as often as they miss. Hunter $A$ encountered 50 ducks during the hunt, and hunter $B$ encountered 51 ducks. What is the probability that hunter $B$'s catch will exceed hunter $A$'s catch? | 72. Since each of the hunters $A$ and $B$ hits and kills a duck as often as they miss, the probability of killing a duck for each of these hunters with each individual shot is $1 / 2$, i.e., the same as the probability of a coin landing heads. Thus, our problem is equivalent to the following: $A$ tosses a coin 50 times... | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,948 |
73. a) Two hunters simultaneously saw a fox and shot at it simultaneously. Suppose that each of these hunters at such a distance usually hits and kills the fox in one out of three cases. What is the probability that the fox will be killed?
b) Solve the same problem for the case of three hunters, assuming that the accu... | 73. a) The experiment described in the problem involves two hunters shooting at a fox simultaneously. The frequency of hits and misses of one hunter, of course, does not depend on the results of the simultaneous shot of the second: if they shoot at the fox many times simultaneously, the first will hit the target one-th... | \frac{5}{9},\frac{19}{27},1-(\frac{2}{3})^n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,949 |
74. A hunter shoots at a fox running away from him for the first time from a distance of $100 \mu$; let the probability of hitting the fox from this distance be ${ }^{1} / 2$ (i.e., from a distance of $100 m$ the hunter hits the running fox as often as he misses). In the event of a miss, the hunter reloads his gun and ... | 74. First, let's calculate the probability of hitting the target on the second and third shots. The second shot is taken from a distance of $150 \mu$, and the third shot from a distance of $200 \mu$. Since the probability of hitting the target is inversely proportional to the square of the distance, and the probability... | \frac{95}{144} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,950 |
75**. The Problem of the Four Liars. It is known that each of the four people $A, B, C$, and $D$ tells the truth only one time out of three. If $A$ states that $B$ denies that $C$ claims that $D$ lied, then what is the probability that $D$ actually told the truth?
Note. It is naturally assumed in the problem that $C$ ... | \frac{1}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 23,951 | |
76. a) In some rural areas of Rossin, there once existed the following divination. A girl holds six blades of grass in her hand so that the ends of the blades stick out at the top and bottom; a friend ties these blades of grass in pairs at the top and bottom separately. If, as a result, all six blades of grass are tied... | 76. a) Six ends of blades of grass can be paired with each other in $5 \cdot 3 \cdot 1=15$ different ways (the first end can be tied to any of the five remaining; after this, one of the four remaining can be tied to any of the three others; and after this, the two remaining ends can be tied together in only one way). S... | \frac{8}{15} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,952 |
78***. a) Someone wrote $n$ letters and sealed them in envelopes without writing the addresses beforehand. After that, he no longer knew which letter was in which envelope, and therefore wrote $n$ addresses on the envelopes at random. What is the probability that at least one of the recipients will receive the letter i... | 1-\frac{1}{e} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,954 | |
79**. a) A train consists of $n$ cars. Each of $k$ passengers chooses a car at random. What is the probability that each car will have at least one passenger?
- b) Under the conditions of part a), what is the probability that exactly $r$ cars of the train will be occupied?
c) Using the result from part a), find the v... | 79. a) This problem is solved similarly to the first solution of problem 78a). Here, the experiment involves each of the $k$ passengers independently choosing one of the $n$ train cars at random. In this case, one passenger has $n$ different possibilities for choosing a car, two passengers have $n^2$ possibilities, thr... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,955 |
83. What is the probability that a randomly chosen positive integer will be coprime with 6? What is the probability that at least one of two randomly chosen numbers will be coprime with 6? | 83. Among any six consecutive integers, one is divisible by 6, and the others give remainders of 1, 2, 3, 4, and 5 when divided by 6. Thus, among any six consecutive integers, two are coprime with 6 (namely, those that give remainders of 1 and 5 when divided by 6). Therefore, the probability that a randomly chosen numb... | \frac{5}{9} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,957 |
84. a) What is the probability that the square of a randomly chosen integer will end in the digit 1? What is the probability that the cube of a randomly chosen integer will end in the digits 11?
b) What is the probability that the tenth power of a randomly chosen integer ends in the digit 6? What is the probability th... | 84. a) The square of an integer ends in the digit 1 if the number itself ends in 1 or 9. This means that among any 10 consecutive integers, exactly two will have squares that end in 1. Therefore, the desired probability is 0.2.
The cube of an integer, as it is easy to verify, ends in the digit 1 only if the number its... | 0.2,0.01,0.2,0.4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,958 |
85. What is the probability that $C_{n}^{7}$, where $n$ is a randomly chosen integer greater than seven, is divisible by 7? What is the probability that $C_{n}^{7}$ is divisible by 12? | 85. $C_{n}^{7}=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}$; therefore, the probability that $C_{n}^{7}$ is divisible by 7 is equal to the probability that the product $n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$ is divisible by 49. But the latter, obviously, will occur if and only if... | \frac{91}{144} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,959 |
86. What is the probability that $2^{n}$, where $n$ is a randomly chosen positive integer, ends in the digit 2? What is the probability that $2^{n}$ ends in the digits 12? | 86. Writing out all consecutive powers of two
$2,4,8,16,32,64,128,256,512,1024,2048,4096, \ldots$, we can see that the last digits of these numbers repeat periodically with a period of 4. Therefore, out of four consecutive powers of two, exactly one will end in 2; hence, the probability that $2^{n}$ ends in 2 is $\fra... | 0.05 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 23,960 |
90***. a) Chebyshev's problem ${ }^{1}$ ). Prove that the probability that two randomly chosen integers are coprime is $\frac{6}{\pi^{2}}$, where $\pi \approx 3.14$ is the ratio of the circumference of a circle to its diameter.
b) Prove that the probability that four randomly chosen integers have a common divisor is $... | 90. a) In the solution to problem 89a), it was shown that the probability that two randomly chosen numbers are coprime is equal to one divided by the sum of the series
$$
1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots
$$
But the sum of this series is $\frac{\pi^{2}}{6}$ (see problem 143a)). From this, it fo... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 23,962 |
92. The Meeting Problem. Two people agreed to meet at a certain place between 12 o'clock and 1 o'clock. According to the agreement, the one who arrives first waits for the other for 15 minutes, after which he leaves. What is the probability that these people will meet, if each of them chooses the moment of their arriva... | 92. All possible outcomes of the experiment considered in this problem consist in the first person arriving at some moment of time $x$ between 12 o'clock and 1 o'clock and the second person arriving at another moment of time $y$ within the same interval. Thus, these outcomes are determined by a pair of numbers $x, y$, ... | \frac{7}{16} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,964 |
94*. A rod of length $l$ is broken into three parts at two randomly chosen points. What is the probability that the length of none of the resulting pieces will exceed a given value $a$? | 94. This problem is a generalization of the previous one. Indeed, problem 93 can obviously be formulated as follows: what is the probability that none of the three parts of a rod broken at two randomly chosen points will exceed half the total length of the rod? In this problem, the length $\frac{l}{2}$ is replaced by a... | \begin{pmatrix}0,\text{if}0\leqslant\leqslant\frac{}{3},\\(3\frac{}{}-1)^{2},\text{if}\frac{}{3}\leqslant\leqslant\frac{}{2},\\1-3(1-\frac{}{})^ | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,966 |
95. Three points $A, B$ and $C$ are taken at random on a circle. What is the probability that triangle $A B C$ will be acute? | 95. Since all points on the circle are equal to each other, we can consider that the first of the three points (point $A$) is somehow fixed. The positions of points $B$ and $C$ will then be determined by the lengths of the arcs $A B$ and $A C$, measured in a certain direction (for example, counterclockwise). Now, let's... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,967 |
96*. A piece is broken off from each of three identical rods; the break points of all three rods are chosen at random. What is the probability that a triangle can be formed from the three resulting pieces? | 96. The possible outcomes of the experiment are determined by a triplet of numbers $x, y, z$, each chosen randomly between 0 and $l$, where $l$ is the length of the considered rods. Considering these numbers as coordinates of a point in space, we can represent the set of all outcomes as a collection of points within a ... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,968 |
97**. A piece is randomly broken off from each of three identical rods. What is the probability that these three pieces can be used to form an acute triangle? | 97. In this problem, we consider the same experiment as in the previous one. As before, the set of all possible outcomes of the experiment is represented by the set of points of the cube \(O A B C D E F G\) with side length \(l\); the probability of an event is equal to the ratio of the volume of the part of the cube c... | 1-\frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,969 |
98****. A rod is broken into three parts; two break points are chosen at random. What is the probability that the three resulting pieces can form an acute triangle? | 98. This problem considers the same experiment as problem 93. As in the second solution of problem 93, all possible outcomes of this experiment will be characterized by two numbers: $A K=x$ and $K L=z$ (see diagram $80, a$). In this case, the set of all outcomes is represented by the set of points of the triangle $O S ... | 3\ln2-2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,970 |
99***. A rod is broken into two parts at a randomly chosen point; then the larger of the two resulting parts is again broken into two parts at a randomly chosen point. What is the probability that a triangle can be formed from the three resulting pieces? | 99. For simplicity, let's assume that the length of our rod is 2 (this is legitimate, as we can always take half the length
Fig. 87. of the rod as the unit of length). Let \( AB \) be our r... | 2\ln2-1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,971 |
100***. Buffon's Problem ${ }^{1}$. A plane is divided by parallel lines, which are at a distance of $2a$ from each other. A thin needle of length $2a$ is thrown randomly onto the plane. Prove that the probability that the needle will intersect one of the lines is $\frac{2}{\pi} \approx 0.637$ (where $\pi=3.14 \ldots$ ... | 100. The position of the needle after its fall on a plane is determined by the distance $y$ from the center of the needle $A B$ to the nearest of the parallel lines and the angle $x$ between the direction of the needle and the direction of these lines (Fig. 89). Clearly, $0 \leqslant y \leqslant a$,
, but after the closure of any two routes, there would inevitably ... | 101. Possibly. Consider, for example, 10 lines on a plane such that no two of them are parallel and no three intersect at the same point. We will consider these lines as bus routes, and their points of intersection as stops. In this case, from each stop, one can travel to any other: if the stops lie on the same line, t... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23,973 |
103*. The bus network of the city, consisting of several (more than two) routes, is organized in such a way that:
$1^{\circ}$ each route has at least three stops,
$2^{\circ}$ from any stop to any other stop, one can travel without transferring, and
$3^{\circ}$ for each pair of routes, there is one (and only one) sto... | 103. a) Let \( n \) be the number of stops on one of the bus routes in the city. We need to prove that in this case, each route will have exactly \( n \) stops and that exactly 2 routes will pass through each stop.
Let's denote the route with \( n \) stops by \( a \), and the stops on this route by \( A_{1}, A_{2}, \l... | 8 | Combinatorics | proof | Yes | Yes | olympiads | false | 23,975 |
105*. On a plane, there are $n$ pairwise non-parallel lines arranged such that through each intersection point of two of these lines, there passes another one of them. Prove that all $n$ lines intersect at one point. | 105. We will prove the theorem by contradiction. Suppose that not all $n$ lines intersect at one point, i.e.
Fig. 97. that through some point $A$ of intersection of two of our lines, at lea... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,977 |
106**. On a plane, there are $n$ points arranged in such a way that on every line connecting two of these points, there lies at least one more of them. Prove that all $n$ points lie on one line.
Note. It can be proven that the statements of problems 105 and 106 are consequences of each other. See Chapter II of O. A. V... | 106. The solution is close to the solution of problem 105, but it is somewhat more difficult to come up with. Suppose the statement of the problem is false, i.e., that there exist three points $A$, $B$, and $C$ among the given $n$ points that do not lie on the same line. Draw a line $MN$ through point $A$ that does not... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,978 |
114**. On a plane, a network of lines is drawn, at each node of which no more than 10 of them meet (a node is a point where three or more lines meet). The lines are colored with different colors so that no two adjacent lines (lines meeting at the same node) are colored the same color. Prove that such a coloring can alw... | 114. It is not difficult to show that indeed for some networks 14 different colors may not be enough; for this, it is sufficient to consider a network with three nodes, each two of which are connected by five lines (Fig. 113). Any two lines of this network converge at one node, from which it follows that the number of ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 23,983 |
116*. A triangle is divided into smaller triangles, adhering to the conditions specified in problem 115a). Prove that if an even number of triangles meet at each vertex of the division, then all vertices can be numbered with the digits 1, 2, and 3 such that the vertices of each triangle in the division are numbered wit... | 116. Let us prove even more than required by the problem. Namely, we will show that if an arbitrary polygon \( \mathbf{M} \) is divided into triangles in accordance with the conditions of the problem (i.e., such that no two triangles share a part of one of their sides) and at each vertex of the division, an even number... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,984 |
119. a) Prove that any convex polygon of area 1 can be enclosed in a parallelogram of area 2.
b) Prove that a triangle of area 1 cannot be enclosed in a parallelogram of area less than 2. | 119. a) Let \( AB \) be a side of a convex polygon \( M \) with area \( 1 \), and \( C \) be the point of the polygon \( M \) farthest from \( AB \) (or one of such points if \( M \) has a side parallel to \( AB \)). Draw the line \( AC \) (Fig. 122); this line divides the polygon \( M \) into two parts \( M_1 \) and \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,987 |
121*. a) Let $M$ be a convex polygon and $l$ be an arbitrary line. Prove that in $M$ one can inscribe a triangle, one side of which is parallel to $l$ and the area of which is not less than $3/8$ of the area of $M$.
b) Let $M$ be a regular hexagon and $l$ be a line parallel to one of its sides. Prove that in $M$ one c... | 121. a) Let's draw two lines parallel to $l$ and located on opposite sides of the convex polygon $M$, and then move them so that they pass through the vertices $A$ and $B$ of the polygon; in this case, the polygon $M$ will be enclosed within the strip formed by the two lines
=\cos (n \operatorname{arc} \cos x)
$$
represents a polynomial in $x$ of degree $n$ with the leading coefficient $2^{n-1}$. Find all roots of the equation $T_{n}(x)=0$ and all values of $x$ between -1 and +1 for which the poly... | 130. We have (see, for example, problem 138b)): $\cos n \alpha=\cos ^{n} \alpha-C_{n}^{2} \cos ^{n-2} \alpha \sin ^{2} \alpha+C_{n}^{4} \cos ^{n-4} \alpha \sin ^{4} \alpha-\ldots$
Now let
$$
\alpha=\arccos x
$$
In this case
$$
\cos \alpha=x, \quad \sin \alpha=\sqrt{1-x^{2}}
$$
and, consequently,
$$
\begin{aligned... | proof | Algebra | proof | Yes | Yes | olympiads | false | 23,994 |
131. Find the quadratic trinomial
$$
x^{2}+p x+q
$$
the deviation of which from zero on the interval from -1 to +1 has the smallest possible value.
$132^{* *}$. Prove that the deviation from zero of the polynomial
$$
x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n-1} x+a_{n}
$$
of degree $n$ with the leading coeffic... | 131. It is easy to see that the polynomial
$$
P(x)=x^{2}+p x+q=\left(x+\frac{p}{2}\right)^{2}+q-\frac{p^{2}}{4}
$$
takes its minimum value $q-\frac{p^{2}}{4}$ at $x=-\frac{p}{2}$. Let us now consider two cases separately.
$1^{\circ} .\left|-\frac{p}{2}\right| \geqslant 1$ (Fig. 130, a). Suppose our polynomial takes ... | P_{0}(x)=x^{2}-\frac{1}{2} | Algebra | proof | Yes | Yes | olympiads | false | 23,995 |
137. Simplify the expression
$$
\cos \frac{\alpha}{2} \cos \frac{\alpha}{4} \cos \frac{\alpha}{8} \ldots \cos \frac{\alpha}{2^{n}}
$$ | 137. Multiply our expression from the end by $\sin \frac{\alpha}{2^{n}}$. We get:
$\cos \frac{\alpha}{2} \cos \frac{\alpha}{4} \ldots \cos \frac{\alpha}{2^{n-1}}\left(\cos \frac{\alpha}{2^{n}} \sin \frac{\alpha}{2^{n}}\right)=$
$$
\begin{aligned}
& =\frac{1}{2} \cos \frac{\alpha}{2} \cos \frac{\alpha}{4} \ldots\left(... | \frac{1}{2^{n}}\frac{\sin\alpha}{\sin\frac{\alpha}{2^{n}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 23,998 |
138. Prove that
a) $\sin n \alpha=C_{n}^{1} \sin \alpha \cos ^{n-1} \alpha-C_{n}^{3} \sin ^{3} \alpha \cos ^{n-3} \alpha+$
$$
+C_{n}^{5} \sin ^{5} \alpha \cos ^{n-5} \alpha-\ldots
$$
for example,
$\sin 6 \alpha=6 \sin \alpha \cos ^{5} \alpha-20 \sin ^{3} \alpha \cos ^{3} \alpha+6 \sin ^{5} \alpha \cos \alpha$.
b) ... | 138. To derive the formulas for tasks a) and b), it is sufficient to expand the brackets in the left-hand side of de Moivre's formula
$$
(\cos \alpha+i \sin \alpha)^{n}=\cos n \alpha+i \sin n \alpha
$$
(see footnote on p. 210), using the binomial formula of Newton, and then equate the real parts and the coefficients ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 23,999 |
139. Form the equations whose roots are the numbers:
a) $\operatorname{ctg}^{2} \frac{\pi}{2 m+1}, \operatorname{ctg}^{2} \frac{2 \pi}{2 m+1}, \operatorname{ctg}^{2} \frac{3 \pi}{2 m+1}, \ldots, \operatorname{ctg}^{2} \frac{(m-1) \pi}{2 m+1}$, $\operatorname{ctg}^{2} \frac{m \pi}{2 m+1}$
b) $\operatorname{ctg} \frac{... | 139. a) The formula from problem 138a) can be rewritten in the following form:
\[
\sin n \alpha = \sin^n \alpha \left( C_n^1 \operatorname{ctg}^{n-1} \alpha - C_n^3 \operatorname{ctg}^{n-3} \alpha + C_n^5 \operatorname{ctg}^{n-5} \alpha - \ldots \right).
\]
Let now \( n = 2m + 1 \) be odd. If \( \alpha \) is equal to... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 24,000 |
140. Prove that
a) $\operatorname{ctg}^{2} \frac{\pi}{2 m+1}+\operatorname{ctg}^{2} \frac{2 \pi}{2 m+1}+$
$$
\begin{array}{r}
+\operatorname{ctg}^{2} \frac{3 \pi}{2 m+1}+\ldots+\operatorname{ctg}^{2} \frac{m \pi}{2 m+1}= \\
=\frac{m(2 m-1)}{3}
\end{array}
$$
b) $\operatorname{cosec}^{2} \frac{\pi}{2 m+1}+\operatornam... | 140. a) Since the equation
$$
C_{2 m+1}^{1} x^{m}-C_{2 m+1}^{3} x^{m-1}+C_{2 m+1}^{5} x^{m-2}-\ldots=0
$$
has roots $\operatorname{ctg}^{2} \frac{\pi}{2 m+1}, \quad \operatorname{ctg}^{2} \frac{2 \pi}{2 m+1}, \ldots, \operatorname{ctg}^{2} \frac{m \pi}{2 m+1}$
(see problem 139a)), the polynomial
$$
C_{2 m+1}^{1} x^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,001 |
141. Prove that
$$
\sin \frac{\pi}{2 m} \sin \frac{2 \pi}{2 m} \sin \frac{3 \pi}{2 m} \cdot \cdots \sin \frac{(m-1) \pi}{2 m}=\frac{\sqrt{\bar{m}}}{2^{m-1}}
$$
and
$$
\sin \frac{\pi}{4 m} \sin \frac{3 \pi}{4 m} \sin \frac{5 \pi}{4 m} \ldots \sin \frac{(2 m-1) \pi}{4 m}=\frac{\sqrt{2}}{2^{m}}
$$ | 141. From the solution of problems 139v), g) it follows that
$$
\begin{array}{r}
C_{2 m}^{1}(1-x)^{m-1}-C_{2 m}^{3}(1-x)^{m-2} x+C_{2 m}^{5}(1-x)^{m-3} x^{2}-\ldots= \\
\quad=A\left(x-\sin ^{2} \frac{\pi}{2 m}\right)\left(x-\sin ^{2} \frac{2 \pi}{2 m}\right) \ldots\left(x-\sin ^{2} \frac{(m-1) \pi}{2 m}\right)
\end{ar... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,002 |
143. a) Derive from the identities of problems 140a) and b) the Euler's formula
$$
\frac{\pi^{2}}{6}=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots
$$
b) What is the sum of the infinite series
$$
1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+\ldots ?
$$ | 143. a) Since for angles of the first quadrant $\operatorname{cosec} \alpha > \frac{1}{\alpha} > \operatorname{ctg} \alpha$ (see problem 136a)), it follows from the identities of problems 140a) and b) that
$$
\begin{aligned}
& \frac{m(2 m-1)}{3}< \\
& <\left(\frac{2 m+1}{\pi}\right)^{2}+\left(\frac{2 m+1}{2 \pi}\right... | \frac{\pi^{4}}{90} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 24,004 |
149. Let $S_{1}$ and $S_{2}$ be the areas of curvilinear trapezoids bounded by the hyperbola $y=\frac{1}{x}$, the x-axis, and the lines $x=a_{1}, x=b_{1}$ and $x=a_{2}, x=b_{2}$ respectively (Fig. 22). Prove that if $\frac{b_{1}}{a_{1}}=\frac{b_{2}}{a_{2}}$, then $S_{1}=S_{2}$.
$ and $A_{2} D_{2}\left(O A_{2}=a_{2}, O D_{2}=b_{2}\right)$ on the x-axis each into $n$ equal parts. We will construct a rectangle on each of these parts as indicated in Fig. 142. The area of the $k$-th of the first rectangles is... | proof | Calculus | proof | Yes | Yes | olympiads | false | 24,009 |
150. Prove that for any positive $z_{1}$ and $z_{2}$
$$
F\left(z_{1} z_{2}\right)=F\left(z_{1}\right)+F\left(z_{2}\right)
$$ | 150. Let's consider several possible cases separately.
$1^{\circ} . z_{1}$ and $z_{2}$ are both greater than 1 (Fig. 143, a). Since $\frac{z_{1} z_{2}}{z_{2}}=\frac{z_{1}}{1}$, by the result of problem 149, the areas of the two curvilinear trapezoids shaded in Fig. $143, a$,
$ takes the value 1 at some point located between 2 and 3.
The value of $z$ for which $F(z)=1$ will henceforth always be denoted by the letter e. From problem 151, it follows that $2<e<3$.
The number $e$ plays a significant role in mathematics and often appears in various questions t... | 151. Note that by its very definition, the function $F(z)$ is an increasing function: for $z_{2}>z_{1}$, it is obvious that $F\left(z_{2}\right)>F\left(z_{1}\right)$. Since $F(1)=0$ and $F(z)$ changes continuously as $z$ increases and cannot "skip" any value,
^{n}
$$
where $e$ is the common limit of the sequences in problems 156a) and b).
Problem 157 can also be formulated as follows: prove that
$$
\lim _{n \rightarrow \infty}\left(1+\frac{z}{n}\right)^{n}=\l... | 157. Let $z$ be positive first. Consider the curvilinear trapezoid $A B C D$ bounded by the hyperbola $y=\frac{1}{x}$, the x-axis, and the lines $x=1$ and $x=1+\frac{z}{n}$ (Fig. 152, a). The area of this trapezoid is enclosed between the areas of the rectangles $A B E D$ and $A F C D$, i.e., between $\frac{z}{n} \cdot... | proof | Calculus | proof | Yes | Yes | olympiads | false | 24,016 |
159. Find the value of the following limits:
a) $\lim _{n \rightarrow \infty} n \ln \left(1+\frac{1}{n}\right)$;
b) $\lim _{n \rightarrow \infty} n \log _{a}\left(1+\frac{1}{n}\right)$,
c) $\lim _{n \rightarrow \infty} n(\sqrt[n]{a}-1)$.
$160^{* * *}$. Prove that for any positive integer $n$, the number $n!$ lies wi... | 159. a) As we know, $\ln \left(1+\frac{1}{n}\right)$ is equal to the area of the curvilinear trapezoid $A B C D$ bounded by the hyperbola $y=\frac{1}{x}$, the x-axis, and the lines $x=1$ and $x=1+\frac{1}{n}$ (see Fig. 151 on p. 478). Since the area of this trapezoid is less than the area of the rectangle $A B E D$ and... | 1,\frac{1}{\ln},\ln | Calculus | proof | Yes | Yes | olympiads | false | 24,017 |
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