problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
40. Find the geometric locus of points $M$ on a plane, the difference of the squares of the distances from which to two given points $A$ and $B$ on this plane is constant and equal to $a$. For which values of $a$ does the problem have solutions? | Solution. Introduce a rectangular coordinate system as shown in Figure 18, where point $O$ is the midpoint of segment $A B$. Let $M(x ; y)$ be the point of the desired set. By the condition $A M^{2}-B M^{2}=a$, or (considering $A B=2 c$)
$$
(x+c)^{2}+y^{2}-\left((x-c)^{2}+y^{2}\right)=C, \quad \text { or } \quad x=\fr... | \frac{}{4c} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,293 |
42. Prove that if the section of a rectangular parallelepiped by a plane is a triangle, then this triangle is acute. | Solution. Introduce a rectangular coordinate system as shown in Figure 20. Then the vertices of triangle $ABC$ have the following coordinates:
$$
A(x ; 0 ; 0), \quad B(0 ; y ; 0) \quad \text { and } \quad C(0 ; 0 ; z)
$$
Express the sides of triangle $ABC$ through the coordinates of its vertices:
$$
AB=\sqrt{x^{2}+y... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,295 |
43. Does there exist a triangular pyramid with a right-angled triangle as its base, and plane angles at the vertex being right angles? | Solution. Suppose the pyramid in question exists: $S A B C-$ is a pyramid, $\triangle A B C$ is the base $\left(B C=a, \quad A C=b, \quad \angle A C B=90^{\circ}\right.$, $\left.\angle A S B=\angle B S C=\angle C S A=90^{\circ}\right)$.
We introduce a rectangular coordinate system as shown in Figure 21. Then the verti... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,296 |
50. Solve the equation (10th grade)
$$
a \sin x + b \cos x = c
$$
where $a, b$, and $c$ are constants, and $a$ and $b$ are not both zero. | Solution. Method 1. Express $\cos x$ in terms of $\sin x$ (or vice versa), using the identity $\sin ^{2} x+\cos ^{2} x=1$, and square both sides of the equation (it is important to remember that squaring can introduce extraneous roots, so verification is necessary here).
Example. Solve the equation
$$
3 \cos x+4 \sin... | \operatorname{arctg}\frac{4}{3}+\pin,n\in\boldsymbol{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,297 |
52. Solve the equation $2^{x}=-x^{2}+2 x+8$. | Solution. The functions $f(x)=2^{x}$ and $g(x)=-x^{2}+2 x+8$ are defined on the set of all real numbers, where $f(x)=2^{x}$ is an increasing function and has no zeros, while $g(x)=-x^{2}+2 x+8$ (a parabola) is an increasing function on the interval $(-\infty ; 1)$, and a decreasing function on the interval $(1 ;+\infty... | x_{1}\approx-1.9,x_{2}\approx2.7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,298 |
53. Solve the equation $\cos 2x = \lg x$. | Solution. Since $D(\cos )=R$ and $D(\mathrm{lg})=R_{+}$, the roots of the equation, if they exist, must be positive.
The roots of the equation cannot be greater than 10, because for $x>10$ $\lg x>1$, while $\cos 2 x$ cannot be greater than one. Therefore, the roots of the given equation, if they exist, are between 0 a... | x_{1}\approx0.8,x_{2}\approx2.4,x_{3}\approx3.6,x_{4}\approx5.7,x_{5}\approx6.7,x_{6}\approx8.9,x_{7}\approx9.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,299 |
62. Let's find the intervals of monotonicity and the points of extremum of the function $y=|| x^{2}-1|-2|$. | Solution. We will start from the graph of the function $y=x^{2}$. In the system $x_{1} O_{1} y$, we construct the graph of $y=x_{1}^{2}$. We translate the axis $O_{1} x_{1}$ by the vector $\bar{r}_{1}(0 ; 1)$, obtaining the new system $x_{2} O_{2} y$ (Fig. 34). In this system, we will have the graph of the function $y=... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,300 | |
67. Let's find the approximate roots of the equation $x^{3}-4 x+2=0$ with an accuracy of 0.01. | Solution. We will construct the graph of the function $f(x)=x^{3}-4 x+2$. We have:
1) $D(f)=\mathbb{R}$
2) $f^{\prime}(x)=3 x^{2}-4$
3) critical points:
$$
-\frac{2 \sqrt{3}}{3} \approx-1.1 ; \quad \frac{2 \sqrt{3}}{3} \approx 1.1
$$
We will create a table:
| $x$ | $-\infty<x<-1.1$ | $-1.1$ | $-1.1<x<1.1$ | 1.1 | $... | -2.22 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,301 |
69. Without performing division, determine whether:
a) the number 5641713 is divisible by 29;
b) 1379235 is divisible by 11. | Solution. a) $r_{0}=1, r_{1}=10, r_{2}=13, r_{3}=14, r_{4}=24, r_{5}=8$, $r_{8}=22$. Since the number $3 \cdot 1+1 \cdot 10+7 \cdot 13+1 \cdot 14+4 \cdot 24+6 \cdot 8+$ $+5 \cdot 22=372$ is not divisible by 29, the number 5641713 is not divisible by 29.
b) $r_{0}=1, r_{1}=10, r_{2}=1, r_{3}=10, r_{4}=1, r_{5}=10, r_{6}... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,303 |
72. Let's prove that the number $88 ...8$, consisting of 1974 digits, is divisible by 13. | Solution. Here $n+1=1974$ and $a_{0}=a_{1}=a_{2}=\ldots=a_{1973}=8$.
Since (see (4))
$$
\begin{gathered}
8+(3 \cdot 8+4 \cdot 8+8)-(3 \cdot 8+4 \cdot 8+8)+\ldots+ \\
+(3 \cdot 8+4 \cdot 8+0)-(3 \cdot 8+4 \cdot 8+8)=8+0+\underbrace{3+0+\ldots+0}_{328 \text { times }}+ \\
+(3 \cdot 8+4 \cdot 8+0)-(3 \cdot 8+4 \cdot 8+8... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,304 |
74. Let's solve the inequality
$$
\sqrt{x+3}<\sqrt{x-1}+\sqrt{x-2}
$$ | Solution. a) $f(x)=\sqrt{x+3}-\sqrt{x-1}-\sqrt{x-2}$. Let's find the domain of definition:
$$
\left(\{ \begin{array} { l }
{ x + 3 \geqslant 0 , } \\
{ x - 1 \geqslant 0 , } \\
{ x - 2 \geqslant 0 }
\end{array} ) \Leftrightarrow \left(\left\{\begin{array}{l}
x \geqslant-3, \\
x \geqslant 1, \\
x \geqslant 2
\end{arra... | (\frac{2}{3}\sqrt{21};+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 35,306 |
75. Let's solve the inequality
$$
\sqrt{2 x+1} \geqslant \frac{x+1}{\sqrt{1-x}}
$$ | Solution. a) \( f(x) = \sqrt{2 x + 1} - \frac{x + 1}{\sqrt{1 - x}} \).
Fig. 40
=\sqrt{1-x}+\sqrt{2 x-3}-5$. Let's find the domain:
$$
\left(\{ \begin{array} { l }
{ 1 - x \geqslant 0 , } \\
{ 2 x - 3 \geqslant 0 }
\end{array} ) \Leftrightarrow \left(\left\{\begin{array}{l}
x \leqslant 1, \\
x \geqslant 1.5
\end{array}\right) \Leftrightarrow(x \in \varnothing... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 35,308 |
78. Arrange the polynomial $f(x)=x^{4}+2 x^{3}-3 x^{2}-4 x+1$ in powers of $x+1$. | Solution. We assume:
$$
\begin{gathered}
x^{4}+2 x^{3}-3 x^{2}-4 x+1=a_{4}(x+1)^{4}+a_{3}(x+1)^{3}+a_{2}(x+1)^{2}+a_{1}(x+1)+a_{0}, \\
\text { from which } x^{4}+2 x^{3}-3 x^{2}-4 x+1=a_{4} x^{4}+\left(4 a_{4}+a_{3}\right) x^{3}+ \\
+\left(6 a_{4}+3 a_{3}+a_{2}\right) x^{2}+\left(4 a_{1}+3 a_{3}+2 a_{2}+a_{1}\right) x... | f(x)=(x+1)^{4}-2(x+1)^{3}-3(x+1)^{2}+4(x+1)+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,310 |
79. Without performing the operations, represent the product as a polynomial in standard form
$$
(x-1)(x+3)(x+5)
$$ | Solution. The product is a polynomial of the third degree, the coefficient of the leading term (it is not difficult to verify) is 1, and the free term is -15, then we write:
$$
(x-1)(x+3)(x+5)=x^{3}+a x^{2}+b x-15
$$
where $a$ and $b$ are unknown coefficients. To calculate them, let $x=1$ and $x=-3$, then we get:
$$... | (x-1)(x+3)(x+5)=x^{3}+7x^{2}+7x-15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,311 |
80. Given the polynomial $x^{4}+3 x^{3}-15 x^{2}-19 x+30$. Factorize it, knowing that all its roots are integers. | Solution. We will look for the factorization in the form
$$
x^{4}+3 x^{3}-15 x^{2}-19 x+30=(x-a)(x-b)(x-c)(x-d)
$$
assuming that the numbers $a, b, c$, and $d$ are its roots. Expanding the brackets on the right-hand side, we have:
$$
x^{4}+3 x^{3}-15 x^{2}-19 x+30=x^{4}-(a+b+c+d) x^{3}+
$$
$$
+(a b+a c+a d+b c+b d+... | x^{4}+3x^{3}-15x^{2}-19x+30=(x-1)(x-3)(x+2)(x+5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,312 |
81. The difference $\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}$ is an integer. Find this number. | Solution. Since $40 \sqrt{2}-57<0$, then $|40 \sqrt{2}-57|=$ $=57-40 \sqrt{2}$. Then
$$
\begin{aligned}
A & =\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}= \\
& =\sqrt{57-40 \sqrt{2}}-\sqrt{57+40 \sqrt{2}}
\end{aligned}
$$
Let $57-40 \sqrt{2}=(a+b \sqrt{2})^{2}$, where $a$ and $b$ are unknown coefficients. Then
$$
5... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,313 |
82. Let's get rid of the irrationality in the denominator:
$$
\frac{2}{1-\sqrt[3]{3}-\sqrt[3]{9}}
$$ | Solution. Let $\frac{2}{1-\sqrt[3]{3}-\sqrt[3]{9}}=a+b \sqrt[3]{3}+c \sqrt[3]{9}$, then
$$
(1-\sqrt[3]{3}-\sqrt[3]{9})(a+b \sqrt[3]{3}+c \sqrt[3]{9})=2
$$
or
$$
(a-3 b-3 c)+(b-a-3 c) \sqrt[3]{3}+(c-a-b) \sqrt[3]{9}=2
$$
From this, $\left\{\begin{array}{l}a-3 b-3 c=2, \\ b-a-3 c=0, \\ c-a-b=0\end{array}\right.$
Sol... | \frac{1}{5}\cdot(1-2\sqrt[3]{3}-\sqrt[3]{9}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,314 |
89. Simplify the expression:
a) $\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}$
b) $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ | Solution. a) Let $\sqrt{57-40 \sqrt{2}}-\sqrt{57+40 \sqrt{\overline{2}}}=x ; x<0$, since $\sqrt{57-40 \sqrt{2}}<\sqrt{57+40 \sqrt{2}}$.
Square both sides of the equation:
$$
x^{2}=57-40 \sqrt{2}-2 \sqrt{57^{2}-(40 \sqrt{2})^{2}}+57+40 \sqrt{2}
$$
from which
$$
x^{2}=114-2 \sqrt{49}, x^{2}=100, x=-10
$$
(Solution b... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,316 |
90. Prove that $\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}=3$. | Solution. Let the value of the expression on the left side of the equality be denoted by $x$. Reasoning as in the solution of Example 89, b, we obtain the equation $x^{3}-3 x-18=0$, from which $x=3$.
## Solving Equations | 3 | Algebra | proof | Yes | Yes | olympiads | false | 35,317 |
91. Let's solve the equation
$$
(3 x-1) \sqrt{x-4}=17 \sqrt{2}
$$ | Solution. Method 1. The domain of expression (3) is the interval $[4 ;+\infty)$. Squaring both sides of the equation, we obtain the equivalent equation
$$
9 x^{3}-42 x^{2}+25 x-582=0
$$
We find the critical points of the function $f(x)=9 x^{3}-42 x^{2}+25 x-582$:
$$
f^{\prime}(x)=27 x^{2}-84 x+25=0
$$
from which $x... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,318 |
93. a) Solve the equation
$$
(4 x-7) \sqrt{x-3}=25 \sqrt{5}
$$
b) Is the equality true?
$$
\sqrt[20]{3} \cdot \sqrt[4]{6+3 \sqrt[5]{5}+6 \sqrt[3]{25}}=\sqrt[5]{16+9 \sqrt[3]{5}+5 \sqrt[3]{25}} ?
$$ | Solution. b) Let's make a "rough" estimate:
1) $\sqrt[20]{3} \cdot \sqrt[4]{6+3 \sqrt[5]{5}+6 \sqrt[3]{25}}=\sqrt[4]{\sqrt[5]{3}} \times$
$\times \sqrt[4]{6+3 \sqrt[5]{5}+6 \sqrt[3]{25}}=\sqrt[4]{6 \sqrt[5]{3}+3 \sqrt[5]{3} \cdot \sqrt[5]{5}+6 \sqrt[5]{3} \cdot \sqrt[3]{25}} \approx$ $\approx \sqrt[4]{6 \cdot 1.2+3 \... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,319 |
94. Let's calculate the area of the figure bounded by the lines
$$
y=x^{2}-5 x+4 \text{ and } y=2 x-2 .
$$ | Solution. For calculating the area of a curvilinear trapezoid, two formulas are known:
$$
\begin{aligned}
S & =\int_{a}^{b} f(x) d x \\
S & =-\int_{a}^{b} f(x) d x
\end{aligned}
$$
Formula (1) is used when the function $f(x)$ is non-negative, and formula (2) is used when the function is negative.
In our example, we ... | 20\frac{5}{6} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,320 |
96. Let's calculate the area of the figure bounded by the lines
$$
y=x^{3}, \quad y=-1, \quad y=-3 \text{ and } x=0.
$$ | Solution. Method 1. Perform a rotation around point $O$ by an angle of $-90^{\circ}$; in this case, the given figure (Fig. 45) will transform into an equal curvilinear trapezoid bounded by the lines
$$
y=(-x)^{\frac{1}{3}}, y=0, x=-3 \text { and } x=-1
$$
By formula (1), we have:
$$
S=\int_{-3}^{-1}(-x)^{\frac{1}{3}... | \frac{3}{4}(3\sqrt[3]{3}-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,322 |
97. Let's calculate the area of the figure bounded by the lines
$$
y=1, y=0, y=\log _{2} x \text { and } x=0
$$ | Solution. Apply symmetry with respect to the line $y=x$ (Fig. 48); the image of the given figure will be a curvilinear trapezoid bounded by the lines $y=2^{x}, y=0, x=0$ and $x=1$, hence, by formula (1)
$$
S=\int_{0}^{1} 2^{x} d x=\frac{1}{\ln 2}
$$ | \frac{1}{\ln2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,323 |
99. Let's find the length $L$ of the arc of the curve: $x^{2}+y^{2}=R^{2}$. | Solution. This curve is a circle of radius $R$ centered at the origin (Fig. 53).
First, we find the length $l$ of the semicircle located in the first and second quadrants. Noting that its equation will be $y = \sqrt{R^2 - x^2}$, by formula (3) we have:
$$
\begin{gathered}
l = \int_{-R}^{R} \sqrt{1 + \left(\left(\sqrt... | 2\piR | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,324 |
103. Find the first, second, and third differences for the sequence with the general term $a_{n}=n^{2}+1$. | Solution. Let's determine the differences of the given sequence:
In this example, we had a sequence of the 2nd order. As can be seen from the solution, the first differences form an arithme... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,325 |
104. Define the dependence of variable $y$ on variable $x$ by a formula based on the table:
| $x$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $y$ | -1 | -3 | 1 | 17 | 51 | 109 | 197 | 321 | 487 | | | Solution. Let's determine the differences of the given sequence:
$$
\begin{gathered}
-1, \quad-3, \quad 1, \quad 17, \quad 51, \quad 109,197,421,487, \quad \cdots \\
-2, \quad 4,16, \quad 34, \quad 58, \quad 88,124, \quad 166, \quad \cdots \\
6, \quad 12, \quad 18, \quad 24, \quad 30, \quad 36, \quad 42, \ldots
\end{g... | x^{3}-3x^{2}+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,326 |
106. Find the sum: $1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}$. | Solution. Let's denote the sum by $S_{n}$:
$$
S_{n}=1^{3}+3^{3}+5^{3}+\ldots(2 n-1)^{3}
$$
We need to find an expression for the general term $S_{n}$ that would allow us to compute the sum $S_{n}$ for any $n$ without raising all numbers from 1 to $2 n-1$ to the third power.
Consider the first differences of the sequ... | n^{2}(2n^{2}-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,327 |
108. What is the maximum number of parts that $n$ lines can divide a plane? | Solution. After conducting an analysis, we establish that the number of parts will be maximal if no two of these lines are parallel and no three intersect at the same point.
Let the desired number of parts be denoted by $y_{n}$. We will construct a table:
| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | $\ldots$ |
| :---: | :... | y_{n}=\frac{1}{2}(n^{2}+n+2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,328 |
1. Exercise on the multiplication table. We build the following sequence of digits.
Let the first digit be 2, the next 3,
$$
2 \cdot 3=6,
$$
the third digit in the sequence will be 6;
$$
3 \cdot 6=18
$$
the fourth digit will be 1, and the fifth 8;
$$
6 \cdot 1=6, \quad 1 \cdot 8=8
$$
the sixth digit is 6, follow... | 1. First, note that in this sequence, each odd digit can only appear between two even digits.
Indeed, suppose that two consecutive members $c, d$ of the considered sequence are odd. Then there can be two cases:
either the number $\overline{c \bar{d}}$ is the product of two members $a, b$ of this sequence, preceding t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,329 |
2. An interesting property of numbers. Let's take an arbitrary natural number in the decimal system (for example, 2583) and calculate the sum of the squares of its digits ( $2^{2}+5^{2}+8^{2}+3^{2}=102$ ). We will do the same with the resulting number $\left(1^{2}+0^{2}+2^{2}=5\right)$ and continue this process.
$$
\l... | 2. Let an arbitrary natural $n$-digit number be denoted by
$$
L=10^{n-1} a_{n}+10^{n-2} a_{n-1}+\ldots+10^{2} a_{3}+10 a_{2}+a_{1}
$$
and the sum of the squares of its digits by
$$
L_{1}=a_{n}^{2}+a_{n-1}^{2}+\ldots+a_{3}^{2}+a_{2}^{2}+a_{1}^{2}
$$
We have:
$$
\begin{aligned}
& L-L_{1}=\left(10^{n-1}-a_{n}\right) ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,330 |
3. Divisibility by 11. Prove that for any natural number $k$ the number
$$
5^{5 k+1}+4^{5 k+2}+3^{5 k}
$$
is divisible by 11. | 3. When dividing the powers $5^{\alpha}, 4^{\beta}, 3^{\gamma}$ by 11 (where $\alpha, \beta, \gamma$ denote non-negative integers not exceeding 5), we obtain the following remainders:
| number | remainder | number | remainder | number | remainder |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 5 | | | | | |... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,331 |
4. Divisibility of numbers. The number
$$
3^{105}+4^{105}
$$
is divisible by $13, 49, 181$ and 379, but not divisible by 5 and 11.
How to check this? | 4. The expression $a^{n}+b^{n}$ is divisible by $a+b$ if $n$ is an odd number. Therefore, the number
$$
3^{105}+4^{105}=\left(3^{3}\right)^{35}+\left(4^{3}\right)^{35}
$$
is divisible by $3^{3}+4^{3}=7 \cdot 13$. Similarly, from the equalities
$$
\begin{aligned}
& 3^{105}+4^{105}=\left(3^{5}\right)^{21}+\left(4^{5}\... | 3^{105}+4^{105}\equiv2\quad(\bmod11) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,332 |
6. Arrangement of numbers. Find 10 such numbers $x_{1}$, $x_{2}, x_{3}, x_{4}, \ldots, x_{10}$, so that
the number $x_{1}$ is within the interval $[0,1]$,
the numbers $x_{1}$ and $x_{2}$ are each in the first and second halves of this interval,
the numbers $x_{1}, x_{2}, x_{3}$ are each in one of the three equal par... | 6. Many sequences of numbers $x_{1}$, $x_{2}, \ldots, x_{10}$ can satisfy the conditions of the problem. Here are two of them:
$$
\begin{array}{llllllllll}
0.95 & 0.05 & 0.34 & 0.74 & 0.58 & 0.17 & 0.45 & 0.87 & 0.26 & 0.66 \\
0.06 & 0.55 & 0.77 & 0.39 & 0.96 & 0.28 & 0.64 & 0.13 & 0.88 & 0.48
\end{array}
$$
The numb... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,334 |
8. Proportion. The numbers $A, B, C, p, q, r$ are related by the following ratios:
$$
A: B=p, \quad B: C=q, \quad C: A=r
$$
Write the proportion
$$
A: B: C=\square: \square: \square
$$
such that the expressions in the blanks consist of $p, q, r$, and these expressions can be transformed into one another by cyclic p... | 8. We can accept
$$
A: B: C=\sqrt[3]{\frac{\bar{p}}{r}}: \sqrt[3]{\frac{q}{p}}: \sqrt[3]{\frac{r}{q}}
$$
Indeed, due to the easily provable equality
$$
p q r=1
$$
we have:
$$
A: B=\sqrt[3]{\frac{p^{2}}{q r}}=p, \quad B: C=\sqrt[3]{\frac{q^{2}}{r p}}=q, \quad C: A=\sqrt[3]{\frac{r^{2}}{p q}}=r
$$
Using the equalit... | 4\max{x,y,z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,336 |
9. Irrationality of the root. Prove in an elementary way that the positive root of the equation
$$
x^{5}+x=10
$$
is irrational. | 9. For $x>0$ the left side of the equation increases with the increase of $x$, and it is easy to notice that at $x=1.5$ it is less than 10, and at $x=1.6$ - more than 10. Therefore, the root of the equation lies within the interval $(1.5,1.6)$. Let's write it as an irreducible fraction $p / q$. Then the equation will t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,337 |
10. Inequality. Prove the inequality
$$
\begin{gathered}
\frac{A+a+B+b}{A+a+B+b+c+r}+\frac{B+b+C+c}{B+b+C+c+a+r}> \\
>\frac{c+c+A+a}{C+c+A+a+b+r}
\end{gathered}
$$
where all letters represent positive numbers. | 10. First, note that the following lemma holds: if $p, q, x, y$ are positive numbers, then the inequalities
$$
\frac{1}{p}>\frac{1}{q}, \quad x>y
$$
imply the inequality
$$
\frac{x}{x+p}>\frac{y}{y+q}
$$
Indeed, by the condition
$$
\frac{1}{p}>\frac{1}{q}>0 \quad \text { and } \quad x>y>0
$$
therefore,
$$
\frac{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,338 |
11. Numerical sequences. Find a sequence $a_{0}, a_{1}, a_{2}, \ldots$ of positive numbers such that $a_{0}=1$ and $a_{n}-a_{n+1}=a_{n+2}$ for $n=0,1,2, \ldots$ Show that there is only one such sequence.
G L A B A II
POINTS, POLYGONS, CIRCLES, ELLIPSES | 11. The equation $1-z=z^{2}$ has one and only one positive root $z=\frac{\sqrt{5}-1}{2}$. Therefore, this $z$ satisfies the equations
$$
1-z=z^{2}, z-z^{2}=z^{3}, \ldots, z^{n}-z^{n+1}=z^{n+2}, \ldots
$$
Denoting the number $z^{n}$ by $a_{n}(n=0,1,2, \ldots)$ we obtain the required sequence. Since $0<\mathbf{z}<1$, t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,339 |
12. Points on a Plane. Several (or several dozen) points are given on a plane. Each of them is connected by a straight line segment to the nearest point; in this case, there is no doubt which point is the nearest, as it is assumed that all distances are different.
Prove that the resulting figure does not contain a clo... | 12. I. Suppose that the obtained figure contains a closed polygon $A B C D E . . . M N$.
Suppose, in addition, that $A N A B$, we can reason similarly, then the assumption we have made leads to a contradiction. Hence, it follows that our figure cannot contain a closed polygon.
II. Suppose that the formed figure conta... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,340 |
13. Angle Research. Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive numbers. Choose a ray $O X$ on the plane, and cut off $O P_{1}=x_{1}$ on it, then perpendicular to $O P_{1}$ cut off $P_{1} P_{2}=x_{2}$, then perpendicular to $O P_{2}$ cut off $P_{2} P_{3}=x_{3}$, and so on up to $P_{n-1} P_{n}=x_{n}$. In this process,... | 13. To simplify, let's introduce the following notations:
$$
r=\sqrt{x_{1}^{2}+\ldots+x_{k-1}^{2}}, \quad a=x_{k}, \quad b=x_{k+1}
$$
We will prove that if $a<b$, then the order
$$
x_{i}, x_{2}, \ldots, x_{k-1}, b, a, x_{k+2}, \ldots, x_{n}
$$
^{2}\right]
$$
and if $\angle A=120^{\circ}$, then
$$
S=\frac{\sqrt{3}}{12}\left[a^{2}-(b-c)^{2}\right]
$$ | 14. If in triangle $ABC$ with sides $a, b, c$ we have $\angle A=60^{\circ}$, then $\angle B+\angle C=120^{\circ}$, and six such triangles can be arranged in a wreath (Fig. 7), bounded from the outside by a regular hexagon with side $a$, and from the inside by a regular hexagon with side $b-c$. By calculating the areas ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,342 |
16. Division of a Triangle. Divide a triangle into 19 triangles so that the same number of sides meet at each vertex of the resulting figure (as well as at the vertices of the large triangle).
In this problem, the number 19 cannot be replaced by a larger number, but it can be replaced by smaller numbers. Which ones? | 16. To divide a triangle into a certain number of triangles so that the same number of sides meet at each vertex of the resulting figure, we can use regular polyhedra whose faces are triangles. These can be the following polyhedra: the regular tetrahedron, octahedron, and icosahedron, and only these.
If inside a tetra... | 4,8,19 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,344 |
17. Triangles. In this problem, $n$ denotes a natural number. On the plane, there are $3 n$ points, no three of which lie on the same line. Is it possible to form $n$ triangles from these points (taking them as vertices) that do not intersect and do not contain each other?
Similar problems can be posed for quadrilater... | 17. The answer to the question posed in the problem is positive. Since there is only a finite number of lines, each passing through some two of the given $3 n$ points, we can choose a line that does not pass through any of the given $3 n$ points so that all these points lie on one side of it. If we move this line acros... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,345 |
18. Triangular Network (I). As is known, the entire plane can be covered with a network of equilateral triangles.
Can one place a plus or minus sign at each node of this network so that for each triangle forming the network, the following rule applies: if two vertices of the triangle have the same sign, then the third... | 18. The answer to the question posed in the problem is positive: plus and minus signs can be placed at the nodes of the network so that the condition of the problem is satisfied.
We start with the observation that if a solution to the problem exists, then only two cases are possible:
1) in the three vertices of each
... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,346 |
19. Triangular Network (II). Prove that it is impossible to cover the entire plane with a network of triangles such that five triangles meet at each vertex. | 19. Suppose the opposite, i.e., that it is possible to cover the plane with a network of triangles such that at each node \( W \) five triangles meet (Fig. 22). Then the sum of the angles at vertex \( W \) of any four of these
$$
Imagine that this rectangle is cut out of paper and lies on the table, facing us with its longer side. Cut off the largest square that can be cut from... | 20. Let $O A_{0}=a, O B_{0}=$ $=b$ (Fig. 27). According to the problem, $b=a q$, where
$$
q=\frac{1}{2}(\sqrt{5}-1)
$$
The side of the first cut-off square is $a q$, the side of the second is $a q^{2}$, the side of the third is $a q^{3}$, and so on.
 For more on the issues addressed in this problem, see, for example, I. M. Yaglom, D. O. Shklyarsky, N.... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,350 |
23. Lattice Points. The definition of an integer lattice is given in the condition of problem 22. It is clear that by choosing a suitable radius, one can always ensure that a circle of this radius centered at the point $(\sqrt{2}, \sqrt{3})$ passes through a given lattice point. It is required to prove, however, that o... | 23. If on a circle of radius $r$ with center at the point $(\sqrt{2}, \sqrt{3})$ there are two points $(x, y)$ and $(u, v)$ of our lattice, then
$$
(x-\sqrt{2})^{2}+(y-\sqrt{3})^{2}=(u-\sqrt{2})^{2}+(v-\sqrt{3})^{2} \quad\left(=r^{2}\right)
$$
from which
$$
c \sqrt{2}+d \sqrt{3}=u^{2}+v^{2}-x^{2}-y^{2}=n
$$
where $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,351 |
24. Lattice points enclosed within a circle. In this problem, we are dealing with a lattice located inside a circle $K$, more precisely, with those points of the integer lattice that are enclosed within this circle (but not on its circumference).
Prove that for any non-negative integer $n$ there exists a circle contai... | 24. The definition of an integer lattice is given in the condition of problem 22.
For any $r$, let $f(r)$ denote the number of lattice points located inside (not on the boundary!) the circle of radius $r$ centered at the point $(\sqrt{2}, V \overline{3})$. From the previous problem, it follows that as $r$ increases, t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,352 |
26. Polygon. On the plane, there are $n$ points; no three of them lie on the same line. Is it always possible to find a closed $n$-sided polygon with non-intersecting sides, whose vertices are these points? | 26. Suppose that for $n$ points, no three of which lie on the same line, one can find a closed $n$-gon whose vertices are these points; let $n+1$ points be arranged so that no three of them lie on the same line. Among these points, there is one, for example, point $P$, which can be separated by a line from the other po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,354 |
27. Points and a Circle. On a plane, there are four points through which neither a circle nor a line can be drawn. Can these points be labeled with the letters $A, B, C, D$, such that point $D$ lies inside the circle passing through points $A, B, C$? | 27. Let's draw circles through points 1, 2, 3 and 1, 2, 4. If point 4 lies inside the circle 123 or point 3 lies inside the circle 124, the answer to the question posed in the problem will be positive. Therefore, let's assume that neither of these possibilities occurs. It is easy to notice that in this case, the arc of... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,355 |
28. Geometric problem. Given an ellipse, the length of the major axis is $2a$, and the length of the minor axis is $2b$. Draw a closed curve of the same length as the ellipse, enclosing an area that is greater than the area of the ellipse by $(a-b)^{2}$.
## CHAPTER III
## SPACE, POLYHEDRA, PLANES | 28. The desired curve is obtained by dividing the ellipse into four parts and combining them as shown in Fig. 37.
Here is another solution that does not require "flipping" the parts, i.e., moving out of the plane of the ellipse. By connecting the vertices of the ellipse with chords alternately (Fig. 38), we obtain a r... | (-b)^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,356 |
29. Space Division. Through a fixed point in space, we draw planes to divide the space into as many parts as possible. One plane will divide the space into two parts, two intersecting planes will divide it into four parts, and three planes intersecting at a point and having no other common point will divide the space i... | 29. Instead of the entire space, we will divide the tar, through the center of which we draw planes. On the surface of the sphere (bounded by its sphere), intersecting great circles will arise. We will take one of them as the әquator and project all other circles from the center of the sphere onto a plane tangent to th... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,357 |
30. Two Projections. Imagine a plane $\Pi_{1}$ tangent to the Earth at the North Pole $N$, and a plane $\Pi_{2}$ tangent to the Earth at the South Pole $S$. We can draw one map by projecting each point on the Earth's surface from $N$ onto $\Pi_{2}$, and another map by projecting each point from $S$ onto $\Pi_{1}$; this... | 30. It is easy to verify that the transformation in question in the problem is an inversion *). Indeed, let $\Pi_{1}$ and $\Pi_{2}$ be the tangent planes to the Earth's sphere at points $N$ and $S$ (Fig. 39). Suppose, according to the transformation defined in the problem, point $P_{1}$ on plane $\Pi_{1}$ and point $\d... | NP_{1}\cdotNP_{2}=4r^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,358 |
31. Cube. Holding a model of a cube in your hand so that it can rotate around its longest axis (i.e., around the line connecting opposite vertices), you can wind black yarn around it without any gaps. The yarn will shade only half of the cube (why?). The same can be done with another axis; there are four in total, and ... | 31. The yarn wound around a cube rotating about one of its axes (Fig. 41) will remain only on those edges that do not have common points with the axis of rotation; the yarn will cover half of each face of the cube, i.e., half of the cube's surface.
Now, let's rotate the cube sequentially around each of the four axes, ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,359 |
32. Geodesics. This problem does not require knowledge of mathematics. Let's place a rubber band (so-called "prescription", used in pharmacies for packaging medicines) on a stationary cube in such a way that it holds on the cube and does not cross itself.
The line along which this rubber band will lie is called a geod... | 32. We will prove that through each point on the surface of a cube, there pass four different geodesics, and in total, we have seven families of geodesic lines.
If we assume that the cube is smooth, then a rubber band wrapped around it will be arranged in such a way that the perimeter of the polygon it forms will reac... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,360 |
33. Molecular motion. Inside a cubic box, a material particle moves without the influence of external forces, reflecting off the walls of the box according to the classical law (the angle of incidence equals the angle of reflection, i.e., the normal to the face at the point of reflection is the bisector of the angle fo... | 33. Look at the arrangement of six cubes shown in Fig. 46. We consider that cube $B$ arose from $A$ by reflection from their common face, similarly $C$ from $B$, $D$ from $C$, $E$ from $D$, and $F$ from $E$. Instead of considering the motion of a particle in cube $\boldsymbol{A}$ according to the laws of reflection, we... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,361 |
34. Unfolding a Cube. Models of polyhedra are made from flat nets. In a net, faces are adjacent to each other along edges, and the model is constructed by folding the cardboard net along the edges. A regular tetrahedron has two such different nets. How many does a cube have? | 34. All existing nets (a total of 11) are shown in Fig. 49. The first six solutions give nets in which four faces of the cube are arranged in one strip of the net. No other solutions of this type exist. The next four nets are those in which there are three faces in one
 and the bodie... | 35. Bodies formed by truncating the eight corners of a cube have 14 faces (the number of faces of the cube + the number of vertices of the cube). Eight of them are triangles, and six are octagons (Fig. 50).
 different from a cube, all faces of which are congruent rhombi? | 36. It is not hard to notice that the answer to the question posed in the problem will be positive: a hexahedron that meets the conditions of the problem is a parallelepiped in which the edges are equal and the three dihedral angles at one vertex are equal,
86
Suppose we have a rhombus with an acute angle $\alpha$ an... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,364 |
37. Tetrahedron. There are 6 rods of different lengths, and it is known that in any order, a tetrahedron (triangular pyramid) can be formed from them. How many different tetrahedrons can result from this? | 37. The answer depends on whether we consider two tetrahedra different if one can be obtained from the other by a mirror reflection. We will show that in the first case, there are 60 different tetrahedra; in the second case, the number will obviously be reduced to 30.
Consider Fig. 55, which shows a tetrahedron with e... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,365 |
39. Octahedron. Is it possible to construct an octahedron, all faces of which are congruent quadrilaterals? Can a decahedron (10-faced polyhedron) and, in general, a $2n$-faced polyhedron (where $n>3$ is any integer) be constructed with the same property? | 39. A $2n$-gon, meeting the conditions of the problem, exists, and it is not difficult to indicate a way to construct it. For this purpose, we
Fig. 61. consider two parallel circular disks... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,367 |
40. Distance on a Surface. We can associate with each pair of points on a closed convex surface the shortest of the arcs connecting them (it is not excluded that such an arc is not unique, for example, for any sphere, there are infinitely many shortest arcs between diametrically opposite points of the sphere). The dist... | 40. As the required tetrahedron, consider a pyramid whose base is an equilateral triangle \(ABC\),
Fig. 62.
and whose lateral faces are isosceles triangles \(ABT\), \(BCT\), and \(CAT\) w... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,368 |
41. The journey of a fly. A fly landed on the top of a regular dodecahedron model (a twelve-faced polyhedron) and decided to walk around it, moving along the edges of the dodecahedron; in doing so, it managed to visit all the vertices without visiting any of them twice and returned to the starting vertex at the end of ... | 41. A polyhedron forms a spatial network, the sides of which are edges, the nodes are vertices, and the cells are faces of the polyhedron.
Fig. 63.
, bounded by the same number of pairwise congruent faces (faces of the same body do n... | 43. Let's take an arbitrary convex polyhedron. One of its faces will be taken as the base of a pyramid with such small dihedral angles at the base that the polyhedron, obtained from the convex polyhedron by attaching this pyramid, remains convex and it is possible to make a recess in the original polyhedron, symmetrica... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,371 |
44. Non-convex polyhedron. Can a non-convex polyhedron be bounded by congruent quadrilaterals?
44a. Models of regular polyhedra. From the six edges of a regular tetrahedron, four edges can be selected to form a closed skew, i.e., non-planar quadrilateral. This quadrilateral can be considered a model of the tetrahedron... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,372 | |
45. A Task from Wonderland. Lewis Carroll, a mathematician and famous children's writer, was the author of amusing absurdities. For example, he recommended using a map with a scale of 1:1, as it is enough to spread it on the ground to know where you are at any moment: one just needs to read the inscription where one is... | 45. The three conical planet models discussed in the problem are obtained from a rectangular grid of meridians and parallels (Fig. 112), by cutting it in one of three ways, as indicated in Fig. 113, and then rolling the remaining part into a cone with the vertex at point $N$.
In Fig. 114, we see the cone from "above,"... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,373 |
46. Three Spheres and a Line. Three spheres have a common point $P$, and it is known that no line passing through point $P$ is tangent to all three spheres simultaneously. Show that these spheres have another common point. | 46. Among the three given spheres, there are no two that touch each other. Indeed, otherwise, the common tangent plane of these two spheres passing through point $P$ would intersect the third sphere along a circle. The tangent to this circle at point $P$ would obviously be tangent to all three spheres, which contradict... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,374 |
49. Holiday Ham. Three neighbors pooled 15 rubles each and bought a ham (without skin, fat, and bones). One of them divided it into three parts, insisting that the parts were equal in weight. Another claimed she only trusted the scales at the corner store; there it turned out that the supposedly equal parts, when recal... | 49. The dispute can be resolved as follows.
The right of first choice of a piece of ham will be given to the third participant in the purchase. She, of course, will choose the piece that on the home scales weighed no less than each of the two remaining pieces, and therefore, the one which, in her opinion, is worth no ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,377 |
50. Flatbread division. Any flatbread, regardless of its shape, can be divided into four equal parts by two mutually perpendicular cuts. In other words, for any flat region with area \( P \), one can find two mutually perpendicular lines such that each of the four quadrants they form contains a part of that region with... | 50. A system of two arbitrary mutually perpendicular lines will be briefly called a cross. Let us mark one ray, i.e., one half-line, on this cross (in Fig. 123 it is indicated by an arrow). A cross, positioned arbitrarily relative to a planar region, can always be parallel-shifted so that the quarters of the region adj... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,378 |
51. Dividing a Triangular Cake. Pavel and Gavel must divide a triangular cake between themselves. Gavel set the condition that he would cut his portion with a straight cut, and Pavel agreed on the condition that he would mark a point $P$ in advance, through which the cut must pass. Since the cake has the same thickness... | 51. First, note that if Pavel chooses point \( P \) at the centroid of the triangle (Fig. 124), then the best division for
Fig. 124.
![](https://cdn.mathpix.com/cropped/2024_05_21_5d77f48... | 5:4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,379 |
53. How old is Sofia Sergeyevna? Our acquaintance Sofia Sergeyevna is not old, as she was born after World War I, but she does not like to answer directly when asked how old she is.
When asked on July 27, 1950, how old she was, she replied: I am only one year old, because I only celebrate my birthday when it falls on ... | 53. Let's denote the days of the week as I, II, ..., VII and construct a table, agreeing that the day of the week that fell on a certain specific date in 1911 is denoted by the number I.
| $\boldsymbol{A}$ | B | c | $D$ | $A$ | $B$ | $c$ | D |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1911 | ... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,381 |
54. How many fish are in the pond? A certain ichthyologist wanted to determine how many fish in the pond were suitable for catching. For this purpose, he cast a net with a pre-selected mesh size and, pulling it out, found 30 fish, marked each of them, and threw them back into the pond. The next day, he cast the same ne... | 54. Let $n$ denote the number of fish in the pond suitable for catching. Then the ratio of the number of marked fish to the total number of fish is $30 / n$.
The second time, the ichthyologist caught 40 fish, two of which were marked. The ratio of the number of marked fish to the total number of fish caught is $1 / 20... | 600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,382 |
55. Roller Calibration. One of the components of a gasoline engine has the shape of a roller. To measure the thickness of the roller, a steel plate is used, in which 15 holes are drilled in a row with precisely established dimensions. The first hole has a diameter of 10 mm, and each subsequent hole has a diameter 0.04 ... | 55. In any case, four trials are sufficient if the following method is adopted: we will compare the roller with the average hole, i.e., the eighth in order, then, depending on the result, with the fourth or twelfth, and so on. The result of each trial will be an answer "Yes" (if the roller fits into the hole) or "No" (... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,383 |
57. The tape on the tube. 25 m of tape 0.1 mm thick was tightly wound on a cardboard tube - a roll with a diameter of 1 dm was obtained. What is the diameter of the tube? | 57. The cross-sectional area of the roller is $25 \pi \mathrm{cm}^{2}$, the tape covers an area of $25 \mathrm{~cm}^{2}$, so the core has an area of $25(\pi-1)$ cm ${ }^{2}$. Let $d$ be the diameter of the tube (without the tape); then from the equation
$$
\pi \frac{d^{2}}{4}=25(\pi-1)
$$
it follows that
$$
d=10 \sq... | 10\sqrt{\frac{\pi-1}{\pi}}\mathrm{}\approx8.26\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,385 |
58. Clock with identical hands. It is known that, when determining the time without a clock, no one makes an error of more than six hours.
A clockmaker inserted two identical hands into the clock, so that it is impossible to distinguish the small hand from the large one. What will be the greatest error that threatens ... | 58. As is known, the time shown by the clock is fully determined by the coordinate of the hour hand on the clock face (Fig. 127); the minute hand has only an auxiliary significance, namely, it creates something like a vernier on the clock face, which allows the accuracy of determining the coordinate of the hour hand to... | 5\frac{7}{13} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,386 |
59. Giants and dwarfs. During a physical education class in a class where all students were of different heights, the teacher, having lined up the class in a rectangular formation, said: "Now we will see who among you is the tallest dwarf." He found the shortest in each row, and when these "dwarfs" stepped out of the f... | 59. 60) It could happen that the same boy turned out to be the shortest giant and the tallest dwarf. Moreover, a class consisting of $k m$ students of different heights ($k$ and $m$ - natural numbers greater than $1$; $k$ denotes the number of rows, $m$ denotes the number of columns), can be arranged in a rectangular f... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,387 |
60. Students of classes $\boldsymbol{A}$ and $\boldsymbol{B}$. In a school, there are two classes: $A$ and $B$. Students of class $A$ boast that they are taller than the students of class $B$, while students of class $B$ are considered better mathematicians. One day, when a student from class $A$ looked down on a stude... | 60. The questions posed in the task will be denoted by the numbers under which they appear, and we will use the symbol $p \rightarrow q$ (implication) to denote that a "yes" answer to question $p$ implies a "yes" answer to question $q$. Then the following implications are true:
$$
\begin{gathered}
1 \rightarrow 2,1 \r... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,388 |
61. Statistics. A certain statistician decided to investigate how non-smoking compartments are used on railways in different countries. He identified the following possibilities:
a) smokers most often travel in smoking compartments,
$a^{\prime}$ ) not $a$ («not $a$ » denotes the negation of proposition $a$ ),
b) non-... | 61. The symbols in question in the problem are as follows:
| $1^{\circ} a b c d$, | $5^{\circ} a^{\prime} b c d$, | $9^{\circ} a b^{\prime} c d^{\prime}$, | $13^{\circ} a^{\prime} b c^{\prime} d^{\prime}$, |
| :--- | :--- | :--- | :--- |
| $2^{\circ} a b c d^{\prime}$, | $6^{\circ} a b c^{\prime} d^{\prime}$, | $10^{\... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,389 |
62. Blood Groups. As is well known (Landsteiner, Janeway, Moss, and others), there are 4 blood groups in humans: $O, A, B, AB$ (terminology of Dungern and Hirschfeld; this classification makes it possible to determine whether a given person can serve as a donor for a certain other person). Let the symbol $X \rightarrow... | 62. Let's arrange the symbols denoting blood groups in one row and one column: place donors in the column and recipients in the row, and fill the resulting table with the signs "十" and "−" as follows:
| | | Recipients | | | |
| :---: | :---: | :---: | :---: | :---: | :---: |
| | | 0 | $A$ | $B$ | $A B$ |
| | 0... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 35,390 |
66. Tying Boxes. In confectionery shops, to tie boxes of candies, they proceed as follows: the ribbon runs diagonally and forms one closed diagonal (non-planar) octagon; on the lid, two parallel segments of the ribbon are visible, and at the bottom, the ribbon runs in a similar manner. Knowing all the measurements of t... | 66. If we draw the trace of a ribbon on a box in the form of a rectangular parallelepiped, then on the net of the box this trace will form a straight line (see Fig. 131, where two faces of the box are repeated). From the figure, it is easy to notice that:
$1^{\circ}$ the length of the ribbon is $2 \sqrt{(a+c)^{2}+(b+c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,394 |
67. Beam Balance. A beam balance is a thin rod (of constant thickness) made of wood (or metal) from homogeneous material. At one end, a relatively heavy weight is attached, and at the other end, a hook that supports the items to be weighed. The rod is marked with notches to form a scale, from which we can read (for exa... | 67. Let us consider the case when we have a 1 kg weight.
Let \( A C = l \) (Fig. 132) denote the distance between the center of gravity \( C \) of the unloaded balance and the point \( A \) where the weight is suspended. Let \( Q \) be the weight of the balance. The center of gravity of the balance is determined exper... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,395 |
70. Practical task. The factory territory is flat, but gently sloping. We have a leveling instrument consisting of a horizontal telescope rotating around a vertical axis (with angles of rotation read from the scale of the horizontal circle), and a staff, which we sight to read the difference in levels, as well as the d... | 70. The task is solved as follows: we install a theodolite at point $O$ of the given section (Fig. 147) and measure the slope of the section in an arbitrarily chosen direction $O C_{1}$. For this purpose, we install a staff at point $A_{1}$ in the same direction and
.
Connect points $A$ and $B$ with a segment. We have $A M\alpha \text { and } \gamma>\beta
$
Adding these inequalities and the equality $\gamma=\gamma$, we get
$
3 \gamma>\alpha+\beta+\gamma \te... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,399 |
72. Railway Network (I). There are five cities; no three of them lie on the same straight line. These cities need to be connected by a railway network consisting of four straight roads; at that, railway lines can be built one over another on viaducts.
How many such different railway networks exist? | 72. There can exist three types of railway networks, as shown in Fig. 149.
1) In the first case, each city can be a node where four lines converge, so there will be five different networks of this kind.
2) In the second case, a city that will be a node where three lines converge can be connected to three other cities i... | 125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,400 |
74. Test Flight. A new type of aircraft took off from Oslo, heading for the shortest route to airport $X$, located in South America, right on the equator. Witnesses to the departure from Oslo see the aircraft disappear on the horizon at a point lying exactly to the west.
What is the length of the flight path? At what ... | 74. The shortest path on the Earth (sphere) is the arc of a great circle, i.e.
Fig. 155. circle formed by the intersection of the sphere with a plane passing through the center of the spher... | 10,000 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,402 |
75. The Sun and the Moon. The distance of the Sun from the Earth is 387 times greater than the distance of the Moon from the Earth. By how many times does the volume of the Sun exceed the volume of the Moon? | 75. During a total solar eclipse, the sizes of the Sun and the Moon appear to be approximately the same to us. From this, it follows that the diameter of the Sun is 387 times larger than the diameter of the Moon, and the ratio of their volumes is $387^{3} \approx 58 \cdot 10^{6}$. | 58\cdot10^{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,403 |
77. Chessboard. Let a square or rectangular chessboard have an odd number of cells (for example, 49 or 63). Cells that share a side will be called adjacent.
On each cell of the chessboard, we place one pawn, then collect the pawns and place them again on the cells of the chessboard.
Is it possible for each pawn to en... | 77. A cell adjacent to a white cell is black, and vice versa. On a chessboard with an odd number of cells, the number of white cells is not equal to the number of black cells, therefore the required movement of pawns in the problem is impossible. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,405 |
80. Elliptical Billiards. On an elliptical billiard table, ball $A$ lies at the edge, and ball $B$ is on the segment $s$ connecting the foci of the ellipse. It is necessary to hit ball $\boldsymbol{A}$ so that, after bouncing off the edge, it collides with ball $B$; however, it is forbidden for ball $A$ to cross segmen... | 80. Let's use the following property: the tangent to an ellipse at any point forms equal angles with the segments connecting this point to the foci of the ellipse. This means that a billiard ball sent from focus $F_{1}$, after bouncing off the edge, should hit focus $F_{2}$ (on Fig. 161, the path of the billiard ball i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,408 |
81. Sports Problem (I). In a class, there are 25 students. Out of them, 17 can ride a bicycle, 13 can swim, and 8 can ski. No student masters all three sports, but both cyclists, swimmers, and skiers have good or satisfactory grades in mathematics, which is particularly noteworthy, as six students have unsatisfactory g... | 81. There are 25 students in the class. Since 6 students have unsatisfactory grades in mathematics, there are 19 students with grades at least satisfactory; there are no more than 19 athletes in the class. The number of "athlete-students" (if we count one sport practiced by one student as a unit) in the class is \(17+1... | )0,b)2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,409 |
82. Sports Problem (II). Three runners $A, B$ and $C$ systematically competed in a 200-meter race, noting the order in which they reached the finish line after each run. Summarizing the results, they found that in the majority of races $A$ outpaced $B$, and that in the majority of cases $B$ ran faster than $C$, and als... | 82. On the first day, the runners completed the race in the order $A, B, C$. On the second day, the order was $B, C, A$, and on the third day, it was $C, A, B$.
The results of these competitions are such that $A$ overtook $B$ in two out of three possible cases; similarly, in two out of three cases, $B$ overtook $C$ an... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,410 |
83. Theory of sports tournaments. Dr. Silvester Sharadek's chess club has 10 members. Every year, tournaments are held to classify the players. Each plays against each other until the first decisive game (draws are not taken into account). We will say that "A wins against B" if A has won against B in the classification... | 83. It is impossible for all club members to form one class, i.e., for them all to have the same number of wins, because the number of matches (45) is not divisible by 10.
It is also impossible to divide the athletes into nine classes. Indeed, if there were nine classes, one class would have two players, and the other... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,411 |
84. Union of Volleyball Teams. The best volleyball teams form a union that organizes seasonal matches; each team plays once against any other. It may happen that one of the teams wins all the others, but this is not necessarily the case. Therefore, let's agree to call a champion a team that has defeated any other team ... | 84. First, we will prove by mathematical induction that there is always some team that is the champion. Let there be $n$ teams in the sports league. Gather the captains of these teams in one room, and ask one of them (let's call him $K$) to leave the room along with the captains of all the teams directly defeated by te... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,412 |
85. Tournaments. The so-called cup system for determining the winner from 8 athletes consists of dividing the players into pairs by drawing lots. 4 matches determine four winners, who participate in the second round; the third round of the competition is the final. The winner of the final match receives the first prize... | 85. The question should be understood as follows: in a cup system, it can happen that a player who deserves the second prize, because he is stronger than all the other players except the champion, meets the champion before the final, i.e., in the case of eight players, in the first or second round; in such a case, the ... | 0.5714 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,413 |
86. The Cyclist and the Pedestrians. The director of the collective farm sent two messengers on foot: one with a letter to the post office in the city, and the other - a quarter of an hour later - in the opposite direction, to the neighboring village council. But suddenly he realized that he had mixed up the letters an... | 86. In Fig. 162, on the horizontal axis, we plot the road: to the left - to the post office, to the right - to the village council. On the vertical axis, we plot time. The lines $O P_{1}$ and $O O^{\prime} P_{2}$ represent the paths of the messengers. If the cyclist first catches up with the messenger who left earlier,... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,414 |
87. Four dogs. Four dogs $A, B, C$, and $D$ stand at the corners of a square meadow and suddenly start chasing each other as indicated by the arrows in Fig. 2.
Each dog runs straight towards the next: $A$ towards $B$, $B$ towards $C$, $C$ towards $D$, and $D$ towards $A$. The side of the meadow is 100 m, and the speed... | 87. Since each dog runs at a right angle to the direction of the dog chasing it, and the one chasing runs straight toward the fleeing one, the chasing dog approaches the next dog at a speed of $10 \mathrm{m} /$ s and will catch it after 10 seconds. As a result, the path of each dog is $100 \mathrm{~m}$. At every moment... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,415 |
88. The Chase (I). Ship $P$ has spotted ship $Q$, which is sailing in a direction perpendicular to $P Q$, maintaining its course. Ship $P$ is chasing $Q$, always heading directly towards $Q$; the speed of both ships is the same at any moment (but can vary over time). Without calculations, it is clear that $P$ is sailin... | 88. Let $\alpha$ denote the instantaneous angle between the direction $P Q$ and the path of ship $Q$ (Fig. 163), and $v$ - the speed of ships $P$ and $Q$ at that moment. The mutual approach of the ships is influenced by the speed $v$ of ship $P$, directed towards $Q$, and the component $v \cos \alpha$ of the speed of s... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,416 |
89. Pursuit (II). Ship $O_{1}$ notices another ship $O_{2}$, which at the moment of observation is moving perpendicular to the line $O_{1} O_{2}$. Ship $O_{2}$ does not notice the signals from $O_{1}$ and maintains its original course and speed $v_{2}$. Ship $O_{1}$ wants to attract attention, as it needs help, and wit... | 89. Let $P_{1}, P_{2}$ and $O_{1}$, $O_{2}$ denote the positions of both ships at the initial moment (when the second ship is first sighted by the first) and at the moment when their distance is the smallest. If so, the first ship sailed from point $P_{1}$ to $O_{1}$ along the straight line $P_{1} O_{2}$ (Fig. 164). Mo... | \sqrt{1-k^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,417 |
90. Are the conditions of the problem really incomplete? Someone, not paying close attention to the previous problem, recounted it to Dr. Sharadek and asked how to determine the course based on the conditions of the problem. Unfortunately, he forgot which of the ships had the greater speed and which had the lesser, but... | 90. If $k$ has the same value as in the previous problem, then the course of the signaling ship is determined by the angle $\alpha$ (Fig. 164) as follows:
$$
\sin \alpha=k=\frac{v_{1}}{v_{2}}
$$
Fig. 164.
If, on the contrary, $k$ denotes the ratio of the speed of the detected ship to the speed of the signaling ship,... | \sin\alpha=k | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,418 |
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