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56. At the vertices of a regular $n$-gon, numbers are placed: $n-1$ zeros and one one. It is allowed to increase by 1 all the numbers at the vertices of any regular $k$-gon inscribed in the given polygon. Is it possible to make all the numbers equal using such operations? | 93.56. Let $O$ be the center of an $n$-gon, and $a_{i}$ be the number standing at the $i$-th vertex. Then we set
$$
\vec{S}=\sum_{i=1}^{n} a_{i} \overrightarrow{O A_{i}}
$$
It is not hard to see that under any operation described in the condition, the vector $\vec{S}$ does not change. Initially, $\vec{S}=\overrightar... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,483 |
57*. In two urns, there are $2 p+1$ balls. Every second, half of the balls from the urn where an even number of balls lie are moved to the other urn. Let $k<2 p+1$ be some natural number and it is known that the numbers $p$ and $2 p+1$ are prime. Prove that sooner or later, there will be exactly $k$ balls in one of the... | 93.57. Let's move to arithmetic modulo $2 p+1$. Let $x$ be the number of balls in the first urn, and $y$ in the second. Then every second the pair $(x, y)$ changes to the pair $(x / 2, y / 2)$ (do not forget that we are only considering remainders modulo $2 p+1$; it is clear that $y=-x$). Let $m$ be the smallest natura... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,484 |
59. Prove that there exists a function $f(x)$ defined on $[0 ;+\infty]$ such that $f(f(\ldots(x) \ldots))=1+x+2 \sqrt{x}$ (the function $f$ is applied 45 times). | 93.59. For example, $f(x)=\left(\frac{1}{45}+\sqrt{x}\right)^{2}$. | f(x)=(\frac{1}{45}+\sqrt{x})^{2} | Algebra | proof | Yes | Yes | olympiads | false | 37,485 |
62. Prove that for any positive numbers $a_{k}, b_{k}$ $(k=1,2, \ldots, n)$ the inequality
$$
\sum_{k=1}^{n} \frac{a_{k} b_{k}}{a_{k}+b_{k}} \leqslant \frac{A B}{A+B}
$$
holds, where $A=\sum a_{k}$ and $B=\sum b_{k}$. | 93.62. The proof is conducted by induction on n. The base case $(n=2)$
$$
\frac{a_{1} b_{1}}{a_{1}+b_{1}}+\frac{a_{2} b_{2}}{a_{2}+b_{2}} \leqslant \frac{\left(a_{1}+a_{2}\right)\left(b_{1}+b_{2}\right)}{a_{1}+a_{2}+b_{1}+b_{2}}
$$
is equivalent to the inequality
$$
\begin{gathered}
{\left[a_{1} b_{1}\left(a_{2}+b_{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,487 |
64 *. A fly crawls along the contour of each face of a convex polyhedron (thus, there are as many flies as there are faces), and all of them move, circling their face clockwise. It is known that their speeds at any moment are no less than 1 mm/h. Prove that sooner or later, some two flies will collide. | 93.64. Let's mark the center of the $i$-th face - point $A_{i}$ ( $i=1,2, \ldots, n$ ) and connect the centers of adjacent faces with segments.
Denote the resulting graph as $G$. Fix an arbitrary moment in time $t$ and mark some edges of the graph $G$ with arrows as follows: if the $i$-th fly is at this moment on the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,489 |
65 *. On the edge of a $1993 \times 1993$ board, two cells $A$ and $B$ are marked, separated by an odd number of cells. Prove that the number of ways to cover the board with $1 \times 2$ dominoes without cell $A$ is equal to the number of ways to cover the board with dominoes without cell $B$. | 93.65. Let's color the board in a chessboard pattern with black and white, and assume that the corners of the board are black. Then we can consider that both fields $A$ and $B$ are black, because if they were white, the number of ways to color them would be zero in both cases. Consider an arbitrary covering of the enti... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,490 |
Problem 4. Simplify the expression
$$
M=\frac{2}{\sqrt{4-3 \sqrt[4]{5}+2 \sqrt[4]{25}-\sqrt[4]{125}}}
$$ | Solution. Using a calculator, we find the approximate value of the expression $M$. To simplify the calculations, we transform the expression
$$
K=4-3 \sqrt[4]{5}+2 \sqrt[4]{25}-\sqrt[4]{125}
$$
as follows:
$$
\begin{gathered}
K=4-\sqrt[4]{405}+\sqrt[4]{400}-\sqrt[4]{125}=4+\sqrt[4]{400}- \\
-(\sqrt[4]{405}+\sqrt[4]{... | \sqrt[4]{5}+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,494 |
Problem 5. The fifth power of a natural number $n$ consists of the digits $1,2,3,3,7,9$. Find this number. | Solution. Let's form several six-digit numbers (including the smallest and largest) from the given digits and extract their fifth roots using a microcalculator: $\sqrt[5]{123379} \approx 10.42914$; $\sqrt[5]{173923} \approx 11.17047$; $\sqrt[5]{213379} \approx 11.63672$; $\sqrt[5]{337912} \approx 12.75735$; $\sqrt[5]{7... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,495 |
Problem 9. Find the integer part of the expression
$a=\sqrt{1981+\sqrt{1981+\sqrt{1981+\ldots+\sqrt{1981+\sqrt{1981}}}},}$ if the number 1981 appears in it $n$ times ( $n \geqslant 2$ ). | Solution. Using a calculator, we obtain:
$$
\begin{aligned}
& a_{1}=\sqrt{1981} \approx 44.508426 \\
& a_{2}=\sqrt{1981+\sqrt{1981}} \approx 45.005648 \\
& a_{3}=\sqrt{1981+\sqrt{1981+\sqrt{1981}}} \approx 45.011173 \\
& a_{4}=\sqrt{1981+\sqrt{1981+\sqrt{1981+\sqrt{1981}}}} \approx \\
& \approx 45.011234 \\
& a_{5} \a... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,498 |
Task 10.
Find all natural numbers $n$ such that the sum $S(n)$ of the digits in the decimal representation of the number $2^{n}$ is 5. | Solution. To find the solution, we will construct a table of values for the expression $2^{n}$:
| $n$ | $2^{n}$ | $S(n)$ | $n$ | $2^{n}$ | $S(n)$ |
| :---: | ---: | ---: | :---: | ---: | :---: |
| 1 | 2 | 2 | 16 | 65536 | 25 |
| 2 | 4 | 4 | 17 | 131072 | 14 |
| 3 | 8 | 8 | 18 | 262144 | 19 |
| 4 | 16 | 7 | 19 | 524288... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,499 |
Problem 11. It is known that the last digits of the squares of natural numbers can be the digits $0,1,4$, 5, 6, and 9. Is it true that any group of digits can precede the last digit in them, i.e., that for any set of $n$ digits $a_{1}, a_{2}, \ldots, a_{n}$, one can find a natural number whose square ends with the grou... | Solution. To obtain additional information about the properties of squares of natural numbers, we will compile the following table:
| $1^{2}=1$, | $33^{2}=1089$, | $65^{2}=4225$, | $97^{2}=9409$, |
| :---: | :---: | :---: | :---: |
| $2^{2}=4$, | $34^{2}=1156$, | $66^{2}=4356$, | $98^{2}=9604$, |
| $3^{2}=9$, | $35^{2... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,500 |
Problem 1. Prove that the number $17^{n}-11^{n}$ is divisible by 6 for any natural value of $n$. | Solution. By formula (1)
$17^{n}-11^{n}=(17-11) \cdot\left(17^{n-1}+17^{n-2} \cdot 11+\ldots+\right.$ $\left.+11^{n-1}\right)$.
The statement of the problem is proved. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,501 |
3адач а 2. Prove that $2 \cdot 7^{n}+1$ is divisible by 3 for any natural $n$.
Since $2 \cdot 7^{n}+1=2\left(7^{n}-1\right)+3$ and the divisibility of $7^{n}-1$ by 3 follows from formula (1), the statement of the problem is proved.
3 ддача 3. Prove that $3^{2 n+1}+2^{n+2}$ is divisible by 7 for any natural value of $... | Solution. Obviously, $3^{2 n+1}+2^{n+2}=9^{n} \cdot 3+2^{n} \cdot 4=3\left(9^{n}-2^{n}\right)+3 \cdot 2^{n}+4 \cdot 2^{n}=3\left(9^{n}-2^{n}\right)+7 \cdot 2^{n}$.
The number $9^{n}-2^{n}$ is divisible by 7 due to formula (1). The number $7 \cdot 2^{n}$ is also divisible by 7. The problem is solved.
Mathematical indu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,502 |
Problem 4. Prove that the number $8^{n}+6$ is divisible by 7 for any integer $n \geqslant 1$. | Solution. For $n=1$, the statement of the problem is true. Assume that it is true for $n=k(k \geqslant 1)$, i.e.
$$
8^{k}+6=7 m
$$
where $m$ is a natural number.
Now let's check that the statement of the problem is also true for $n=k+1$, i.e.,
$$
8^{k+1}+6=7 t
$$
where $t$ is a natural number.
From equation (1), ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,503 |
Problem 5. Prove that for any natural $n$ the expression $3^{2 n+2}+2^{6 n+1}$ is divisible by 11. | Solution. For $n=1$, the statement of the problem is true. Assume that the statement is true for $n=k(k>1)$, i.e.,
$$
3^{2 k+2}+2^{6 k+1}=11 m
$$
where $m$ is a natural number.
We will prove that the statement of the problem is also true for $n=k+1$, i.e.,
$$
3^{2(k+1)+2}+2^{6(k+1)+1}=11 p
$$
where $p$ is a natura... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,504 |
Problem 6. Can the numbers $n^{2}+3 n+39$ and $n^{2}+n+$ +37 ( $n$ - a natural number) both be divisible by 49? | The first solution. If for some value of $n$ the expressions $n^{2}+3 n+39$ and $n^{2}+n+37$ are divisible by 49, then their difference should also be divisible by 49:
$$
\left(n^{2}+3 n+39\right)-\left(n^{2}+n+37\right)=2(n+1)
$$
The expression $2(n+1)$ is divisible by 49 if $n+1=49 k$ (where $k$ is a natural number... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,505 |
Problem 7. Consider all possible seven-digit numbers $K$, written with the digits $1,2,3,4,5,6,7$ in any order. Do there exist two such numbers $M$ and $N$ among these numbers such that $M$ is divisible by $N$? | Solution. First, the sum of the digits of the number $K$ is 28. Therefore, $M$ and $N$ do not divide by 3 and 6.
Secondly, 7654321 is the largest of the numbers $K$, and 1234567 is the smallest. Since 7654321 : 1234567 ≈ 6.3, it is clear that when dividing $M$ by $N$, the result can only be 2, 4, or 5.
If the number ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,506 |
3 Task 1. For which natural $n(n \geqslant 2)$ is the equality
$$
-\sqrt[2]{17 \sqrt{5}+38}+\sqrt[2]{17 \sqrt{5}-38}=\sqrt{20} ?
$$ | Solution. Using a calculator, we find:
$$
\begin{gathered}
17 \sqrt{5}+38 \approx 38.013154+38 \approx 76.013154>1 \\
17 \sqrt{5}-38 \approx 0.0131541$, and the function $\varphi(n)=\sqrt[n]{17 \sqrt{5}-38}$ is increasing, and $\varphi(n)<1$. To discover some properties of the function $\psi(n)=f(n)+\varphi(n)$, we wi... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,508 |
T a s k 2. Find the natural roots of the equation
$17(x y z t+x y+x t+z t+1)-54(y z t+y+t)=0$. | Solution.
We solve this equation with respect to $x$:
$$
17 x=54-17(z t+1):(y z t+y+t) .
$$
From here,
$$
54-17 x=\frac{17}{y+\frac{t}{z t+1}}
$$
The right-hand side of equation (1) is an integer and positive, since $y, z, t, x$ are natural numbers. Therefore, $x \leqslant 3$. When $x=1$, the left-hand side of equ... | 3,1,=2,5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,509 |
Problem 4. Given the equation
$$
x^{3}-3 x y^{2}+y^{3}=n
$$
Prove that if a natural number $n$ is such that the given equation has an integer solution, then it has at least three integer solutions. | Solution. Let's try to discover some properties of the integer solutions of equation (1).
1) Let, for example, $\left(x_{1} ; y_{1}\right)=(0 ; 1)$. This ordered pair of numbers is a solution of the equation
$$
x^{3}-3 x y^{2}+y^{3}=1
$$
Using a calculator, it is easy to notice that the integer solutions of equation... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,511 |
Problem 5. Find at least one solution to the equation
$$
a^{3}+b^{4}=c^{5}
$$
in natural numbers. | Solution. What can be taken as a working hypothesis, following which it will be possible to outline ways to find a solution? Let's make a table:
| $n$ | $n^{3}$ | $n^{4}$ | $n^{5}$ |
| ---: | ---: | ---: | ---: |
| | | | |
| | 1 | 1 | 1 |
| 2 | 8 | 16 | 32 |
| 4 | 27 | 81 | 243 |
| 5 | 64 | 256 | 1024 |
| 6 | 216... | =256,b=64,=32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,512 |
Problem 6. For which integers $m$ and $n$ does the equality
$$
(5+3 \sqrt{2})^{m}=(3+5 \sqrt{2})^{n} ?
$$
hold? | Solution. Obviously, the pair $(0 ; 0)$ is a solution to equation (1).
To find other solutions, we use a calculator.
$$
5+3 \sqrt{2} \approx 9.242639 ; 3+5 \sqrt{2}=10.071065
$$
Thus, $3+5 \sqrt{2}>5+3 \sqrt{2}$. Therefore, $n$ and $m$ are either natural numbers or negative. Moreover, if the pair $(n ; m)$ is a solu... | (0;0) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,513 |
Problem 1. Prove that the remainder of dividing the polynomial $M(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}$ by the binomial $x-a$ is equal to the value of the polynomial $M(x)$ at $x=a$, i.e., it is equal to $M(a)$. | Proof. Let $M(x)=(x-a) P(x)+$ $+K(x)$, where $P(x)$ and $K(x)$ are polynomials. If $x-a=0$, i.e., $x=a$, then $M(a)=0 \cdot P(a)+K(a)$, i.e., $K(a)=M(a)$.
This property of polynomials is known as the Remainder Theorem (named after the French mathematician Étienne Bézout (1730-1783), who first formulated and proved thi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,514 |
Problem 3. Factorize
$$
M(x ; y ; z)=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y) .
$$ | Solution. Let's assign some values to the variables $x$ and $y$. Let $x=y=1$. Then $M(1 ; 1 ; z)=$ $=(1-z)+(z-1)=0$. Let $x=y=2$. Then $M(2,2, z)=8(2-z)+8(z-2)=0$. This suggests that $M(x, y, z)$ is divisible by $(x-y)$ or $(y-x)$. Thus, we come to the hypothesis that $M(x, y, z)$ is divisible by $(x-z)$ and $(y-z)$.
... | M(x,y,z)=(x+y+z)(x-y)(y-z)(z-x) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,516 |
Problem 4. Factorize $M(x, y, z)=$ $=x^{3}+y^{3}+z^{3}-3 x y z$. | Solution. If $x=y=1$, then
$$
\begin{gathered}
M(1,1, z)=2+z^{3}-3 z=\left(z^{3}-z\right)+(2-2 z)= \\
=(z-1)^{2}(z+2)
\end{gathered}
$$
If $x=1, z=2$, then $M(1, y, 2)=y^{3}-6 y+9$. It is easy to notice that the number -3 is a root of the equation $y^{3}-6 y+9=0$. Therefore, by the theorem of Bezout, we get:
$$
y^{3... | M(x,y,z)=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,517 |
Problem 6. Factorize
$M(a, b, c, d)=\left((a-c)^{2}+(b-d)^{2}\right)\left(a^{2}+b^{2}\right)-(a d-b c)^{2}$. | Solution. Let $a=b=1$. Then $M(1,1, c, d)=$ $=c^{2}+d^{2}+4+2 c d-4 c-4 d=(c+d-2)^{2}$.
Let $a=1, b=2$. Then $M(1,2, c, d)=c^{2}+4 d^{2}+$ $+4 c d-10 c-20 d+25=(c+2 d-5)^{2}$. But $5=2^{2}+1^{2}$. Therefore,
$$
M(1,2, c, d)=\left(c+2 d-1^{2}-2^{2}\right)^{2}
$$
The considered particular cases allow us to make the as... | M(,b,,)=(+-^{2}-b^{2})^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,519 |
Problem 1. In triangle $ABC$, the measures of angles $B$ and $C$ are both $40^{\circ}$. Prove that if segment $BD$ is the angle bisector of angle $B$, then $BD + DA = BC$. | First solution. The angle $B D C$ is equal to $120^{\circ}$ (Fig. 1). Therefore, $B C>B D$ and on side $B C$ we can lay off a segment $B E$, equal to $B D$. Now let's prove that $E C=A D$. We have $\angle B D E=\angle B E D=80^{\circ}$. Therefore, $\angle E D C=$ $=120^{\circ}-80^{\circ}=40^{\circ}=\angle D C E$ and $E... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,520 |
Problem 3. In a right triangle $ACB$ (where $\angle C$ is the right angle), the bisector $AD=\frac{C \sqrt{3}}{3}$, and the hypotenuse $AB=c$. Find the legs $AC$ and $CB$ of this triangle. | First solution. Let: $AC = x, BC = y$, $CD = z$. We obtain the following system of equations:
$$
\left\{\begin{aligned}
x^{2} + z^{2} & = c^{2} \\
x^{2} + z^{2} & = \frac{c^{2}}{3} \\
\frac{y - z}{z} & = \frac{c}{x}
\end{aligned}\right.
$$
From equation (3) we find:
$$
z = \frac{yx}{c + x}
$$
From equations (2) and... | AC=0.5,\BC=0.5\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,522 |
Problem 5. Prove that the bases of the altitudes $A A_{1}$, $B B_{1}$, $C C_{1}$ of an acute-angled triangle $A B C$ are the vertices of a triangle $A_{1} B_{1} C_{1}$, for which the bisectors of the internal angles lie on the altitudes of triangle $A B C$. | Solution. Let point $M$ be the intersection of the altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ of an acute triangle $A B C$. Angles $A C_{1} M$ and $A B_{1} M$ are right angles. Therefore, points $C_{1}$ and $B_{1}$ lie on a circle with diameter $A M$. Angles $C_{1} A M$ and $C_{1} B_{1} B$ are equal because they are i... | \frac{ac}{b} | Geometry | proof | Yes | Yes | olympiads | false | 37,523 |
Problem 7. On the diagonals $A C$ and $C E$ of a regular hexagon $A B C D E F$, points $M$ and $K$ are taken such that $A M: A C = C K: C E = n$ (Fig. 6). Points $B, M$, and $K$ lie on the same straight line. Find $n$. | Solution. As point $K$ moves along the diagonal $EC$ (from $E$ to $C$), the ratio $CK: CE$ decreases from 1 to 0, while the ratio $AM: AC$ increases from 0.5 ( $AM_1$: $AC=0.5$ ) to 1. Therefore, there exists a unique point $K$ on diagonal $CE$ such that
$$
AM: AC = CK: CE
$$
By trial, we hypothesize that equality (1... | \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,524 |
Problem 8. A regular triangle $ABC$ is inscribed in a circle (Fig. 7). A point $X$ is marked on the arc $AmC$ of this circle. Prove that $AX + XC = BX$. | Solution. On the extension of the chord $A X$ (beyond point $X$), construct the segment $X Y = X C$.
Now the problem reduces to proving the equality of segments $A Y$ and $B X$. The first is a side of triangle $A C Y$, and the second is a side of triangle $B X C$.
Instrumental measurements suggest that these triangle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,525 |
Problem 9. In trapezoid $A B C D$, segments $B C$ and $A D$ are the bases. $B C=9.5 ; A D=20 ; A B=5$; $C D=8.5$. Find its area. | Solution. To solve the problem, it is necessary to find the length of the height of this trapezoid. Place the trapezoid $ABCD$ relative to a rectangular coordinate system as shown in Figure 8. Then $A(0; 0)$, $D(20; 0)$, $B(x; y)$, $C(x+9.5; y)$. According to the problem's conditions,
$$
\begin{gathered}
AB^2 = (x-0)^... | 59 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,526 |
Problem 10. Given a circle with center $O$ and diameter $A B=4$ (Fig. 9). Point $C$ is the midpoint of the radius $O B$. Construct points $M$ and $P$ on the circle, symmetric with respect to the line $A B$, such that the segment $C P$ is perpendicular to the segment $A M$. | Solution. Place the circle relative to the coordinate axes as shown in Figure 9. Then $A(-2 ; 0), B(2 ; 0), C(1 ; 0)$.
Let $M(x ; y)$. Then $P(x ;-y)$. According to the problem, the lines $A M$ and $C P$ are perpendicular. Therefore,
$$
\overline{A M} \cdot \overline{C P}=0
$$
Find the coordinates of the vectors: $\... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,527 | |
Problem 11. Find the area of triangle $ABC$, if $A(3, -1)$, $B(1, -3)$, $C(-6, 6)$. | Solution. Let's complete triangle $A B C$ to form a rectangle $C M K P$ (Fig. 10). Obviously,
$$
\begin{gathered}
S_{A B C}=S_{C P K M}-S_{C M B}-S_{B A K}-S_{A C P}= \\
=9 \cdot 9-0.5 \cdot 9 \cdot 4-0.5 \cdot 2 \cdot 2-0.5 \cdot 9 \cdot 7=29.5
\end{gathered}
$$ | 29.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,528 |
Problem 13. In triangle $A B C$, side $A C=$ $=5 \text{ cm}, B C-A B=2 \text{ cm}, \angle A: \angle C=2$. Find the lengths of sides $A B$ and $B C$. | Solution. Let $\angle A C B=\varphi$, then $\angle A=$ $=2 \varphi, \angle B=180^{\circ}-3 \varphi$. By the Law of Sines:
$$
\begin{gathered}
\frac{\sin 3 \varphi}{5}=\frac{\sin 2 \varphi}{B C}=\frac{\sin \varphi}{A B} \\
A B=\frac{5 \sin \varphi}{\sin 3 \varphi}, B C=\frac{5 \sin 2 \varphi}{\sin 3 \varphi}
\end{gathe... | AB=4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,530 |
Task 14. Find the ratio of the legs $CB$ and $CA$ of the right triangle $ACB$, if it is known that one half of the hypotenuse (from the midpoint of the hypotenuse to the vertex) is seen from the center $X$ of the inscribed circle at a right angle. | Solution. The angle $A X B$ is equal to $135^{\circ}$ (Fig. 13). Therefore, point $X$ is the intersection of the circumcircles of triangle $A X B$ and the right triangle $A X O(\overline{A O}=\overline{O B})$. After this, the construction of a triangle similar to triangle $A C B$ is clear.
We construct an arbitrary se... | 4:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,531 |
Problem 15. The bisectors $A M$ and $B H$ of triangle $A B C$ intersect at point $O$. It is known that $A O: M O = \sqrt{3}$ and $H O: B O = \sqrt{3} - 1$. Find the angles of triangle $A B C$. | Solution. Consider triangle $B O A$ as the image of an isosceles triangle $B O A$ with angle $B O A$ equal to $120^{\circ}$.
Construct such a triangle $B O A$. On the extension of segment $B O$ (beyond point $O$), lay off segment $H O = (\sqrt{3}-1) B O$. On the extension of segment $A O$ (beyond point $O$), lay off s... | \angleA=60,\angleB=90,\angleC=30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,532 |
Problem 17. The side of a regular triangle $ABC$ is 4. Point $D$ is the midpoint of side $BC$. A line passing through $B$ intersects side $AC$ at point $M$. Perpendiculars from points $D$ and $A$ to line $BM$ are $DH$ and $AK$. Calculate the length of segment $AM$, if
$$
AK^4 - DH^4 = 15
$$ | Instruction. Equation (1) has a unique solution. Let $\angle A B M=x$. Then $D H=2 \sin \left(60^{\circ}-x\right)$, $A K=4 \sin x$. Equation (1) takes the form:
$$
256 \sin ^{4} x-16 \sin ^{4}\left(60^{\circ}-x\right)=15
$$
Solving equation (2) on a microcalculator, we get $x=30^{\circ}$, and therefore, $A M=2$. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,533 |
Problem 18. In triangle $A B C$, $A C=B C$, $\angle A C B=30^{\circ}$. Point $D$ lies on side $B C$ and $B D=3, D C=6$. Find a point $P$ on side $A C$ of this triangle such that angle $D P C$ is 1.5 times the angle $A P B$. | Solution. \(AB = 2AC \sin 15^{\circ} = 18 \sin 15^{\circ}\). Let \(\angle APB = x\). Then \(\angle DPC = 1.5x\). Applying the Law of Sines to the corresponding triangles, we get:
\[
AP: \sin \left(75^{\circ} + x\right) = 18 \sin 15^{\circ}: \sin x ; AP =
\]
\[
\begin{aligned}
& = 18 \sin 15^{\circ} \sin \left(75^{\ci... | 3\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,534 |
Problem 19. In triangle $ABC$, the difference between the internal angles $A-B=60^{\circ}$. It is known that the height $CH$ from $C$ to side $AB$ is equal to $CH=BC-AC$. Find the angles of the triangle. | Solution. Point $H$ can coincide with point $A$, lie inside segment $A B$, or lie on ray $B A$ (outside segment $B A$).
1) If point $H$ coincides with point $A$, then $A=90^{\circ}$, $B=30^{\circ}, C=60^{\circ}$.
2) If $A>90^{\circ}$, then as angle $A$ increases, angle $B$ also increases. In this case, $B C$ decreases... | A=90,B=30,C=60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,535 |
3 Task 1. Solve the system of equations
$$
\left\{\begin{array}{l}
|y-x|=2(x+y) \\
|x|+|y|=4
\end{array}\right.
$$ | Solution. The left side of equation (1) is non-negative. Therefore, $x+y \geqslant 0$, i.e., $y \geqslant -x$.
If $|y-x|=0$, i.e., $y=x$, then $x=0$.
If $y>x$, then equation (1) takes the form $y-$ $-x=2(x+y)$. From this, $y=-3x$.
If $y<x$, then from equation (1) we get $x-y=$ $=2(x+y)$ and $y=-x: 3$.
Thus, the gra... | (-1,3),(3,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,536 |
3 Task 2. Solve the system of equations
$$
\left\{\begin{array}{l}
|x|+x=|y|+y \\
|x-1|=x+|y|
\end{array}\right.
$$ | Solution. First, let's investigate equation (1):
If $x \geqslant 0$ and $y \geqslant 0$, then $2 x=2 y$, i.e., in this case, the graph of equation (1) is the bisector OA of the first coordinate angle (Fig. 2).
If $x<0$ and $y \geqslant 0$, then $0=2 y$, i.e., in this case, the graph of equation (1) will be the ray $O... | (\frac{1}{3};\frac{1}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,537 |
Problem 3. Solve the inequality
$$
a^{5}-1.3 a^{2}-2.8 a-3.1 \geqslant 0
$$ | Solution. We investigate the extremum and monotonicity of the continuous function
$$
f(a)=a^{5}-1.3 a^{2}-2.8 a-3.1
$$
We find:
$$
f^{\prime}(a)=5 a^{4}-2.6 a-2.8
$$
We determine the roots of the equation
$$
5 a^{4}-2.6 a-2.8=0
$$
as follows: we transform it to the form $5 a^{4}=$
$=2.6 a+2.8$. We sketch the grap... | 1.616 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,538 |
Problem 4. Solve the equation
$$
3 x^{4}-2 x^{3}+4 x^{2}-4 x+12=0 .
$$ | Solution. Transform equation (1) to the form:
$$
3 x^{4}-2 x^{3}=-4 x^{2}+4 x-12
$$
We will plot the graphs of the functions
$$
f(x)=3 x^{4}-2 x^{3}, \varphi(x)=-4 x^{2}+4 x-12 .
$$
For this, we find:
$$
\begin{gathered}
f^{\prime}(x)=12 x^{3}-6 x^{2}=6 x^{2}(2 x-1) \\
\varphi^{\prime}(x)=4(-2 x+1)
\end{gathered}
... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,539 |
Task 2. Name the set of solutions to the inequality
$$
|5-x|<|x-2|+|7-2 x| \text {. }
$$ | Solution. Transform the given inequality to the form:
$$
|x-2|+|7-2 x|-|5-x|>0
$$
Construct the graph of the function
$$
f(x)=|x-2|+|7-2 x|-|5-x| \text { (Fig. 1). }
$$
To simplify the construction, let's recall the following:
1) The expressions $|x-2|,|7-2 x|,|5-x|$ are equal to zero at the points $2 ; 3.5 ; 5$, ... | (-\infty;2),(3.5;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,540 |
Problem 4. Solve the inequality
$$
\frac{7}{|x-1|-3} \geqslant|x+2|
$$ | Solution. We construct the graphs of the functions $f(x)=|x+2|$, $\varphi(x)=|x-1|-3$, $\psi(x)=\frac{7}{\varphi(x)}$ (Fig. 3). The graphs of these functions can be easily constructed without using derivatives, as they are monotonic on the corresponding intervals.
The lines $x=-2$ and $x=4$ are vertical asymptotes of ... | [-\sqrt{7}-2;-2),(4;5] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,542 |
Problem 5. Solve the inequality $x-2 \leqslant \sqrt{x}$.
 | Solution. We construct the graphs of the functions $f(x)=x-2$ and $\varphi(x)=\sqrt{x}$ (Fig. 4). It then becomes clear that the equation $x-2=\sqrt{x}$ has only one positive root $x_{0}>2$, and the solution to the given inequality is the interval $\left[0 ; x_{0}\right]$, since the inequality is defined only for non-n... | [0,4] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,543 |
Problem 6. Name the set of solutions of the inequality
$$
\sqrt{7+x}>5-x
$$ | Solution. We construct the graphs of continuous monotonic functions:
$$
f(x)=\sqrt{7+x} \text { and } \varphi(x)=5-x .
$$
The function $f(x)$ is defined on the interval $[-7 ;+\infty)$, is continuous and monotonically increasing. It varies from 0 to $+\infty$. The function $\varphi(x)$ is defined for all real numbers... | (2;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,544 |
Problem 7. Solve the inequality $\sqrt[4]{x-1}+$ $+\sqrt[4]{x+14} \geqslant 3$ | Solution. The left side of the inequality is defined on the interval $[1 ;+\infty)$. The function $f_{1}(x)=\sqrt[4]{x-1}$ is continuous and monotonic (on the specified interval it increases from 0 to $+\infty$). The function $f_{2}(x)=\sqrt[4]{x+14}$ is continuous and monotonic (on the interval $[1 ;+\infty$) it incre... | x\geqslant2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,545 |
Problem 8. Solve the inequality
$$
\frac{2 \sqrt{x+1}}{1-2 \sqrt{3-x}}<1
$$ | Solution. The expression $2 \sqrt{x+1}$ is defined on the half-interval $[-1 ;+\infty)$. The expression $1-2 \sqrt{3-x}$ is defined and non-zero on the intervals ( $-\infty$; $2.75$ ) and ( 2, $75 ; 3]$, so the inequality (1) is defined on the intervals $[-1 ; 2.75$ ) and $(2.75 ; 3]$.
We construct the graphs of the f... | [-1;2.75) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,546 |
Problem 9. Solve the equation
$$
\log _{3} x=1+\frac{-3}{2 x-1}
$$
 | Solved. We construct the graphs of the functions $f(x)=\log _{3} x, \quad \varphi(x)=$ $=1+\frac{-3}{2 x-1}$ (Fig. 5).
On the half-interval $(-\infty ; 0]$, the function $f(x)$ is not defined, so there are no solutions to equation (1) on this interval.
On the interval $(0 ; 0.5)$, the function $\varphi(x)$ is positiv... | nosolution | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,547 |
Problem 4. Solve the inequality
$$
\sqrt[4]{x^{2}+2 x+10}+x^{4} \leqslant 62-32 x
$$ | Solution. Since $x^{2}+2 x+10=(x+1)^{2}+9>0$ for any value of $x$, the given inequality is defined for any real value of $x$. Let's investigate some properties of the continuous function
$$
f(x)=\sqrt[4]{(x+1)^{2}+9}+x^{4}+32 x-62
$$
If $x \geqslant 0$, then the continuous function $f(x)$ is increasing. Therefore, th... | x_{1}=-\sqrt{7}-1,x_{2}=\sqrt{7}-1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,550 |
Problem 5. Solve the inequality
$$
2 \sqrt[5]{2 x+8}+\sqrt[5]{303-5 x} \geqslant 7
$$ | Let's investigate the extremum and monotonicity of the continuous function
$$
f(x)=2 \sqrt[5]{2 x+8}+\sqrt[5]{303-5 x}
$$
We find:
$$
f^{\prime}(x)=\frac{0.8}{\sqrt[5]{(2 x+8)^{4}}}-\frac{1}{\sqrt[5]{(303-5 x)^{4}}}
$$
To find the critical points of this function, we solve the equation \( f^{\prime}(x)=0 \):
$$
\f... | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,551 | |
Task 2. Find all values of $a$ for which the equation
$$
2 \lg (x+3)=\lg (a x)
$$
has a unique solution. | The first solution. Obviously, $x>-3, a x>0$, $x \neq 0$. From equation (1) we get:
$$
(x+3)^{2}=a x, a=\frac{(x+3)^{2}}{x}=\left(\sqrt{x}+\frac{3}{\sqrt{x}}\right)^{2}, \text { if } x>0
$$
Obviously,
$$
\left(\sqrt{x}+\frac{3}{\sqrt{x}}\right)=\left(\sqrt{x}-\frac{3}{\sqrt{x}}\right)^{2}+12
$$
From here it is clea... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,553 |
Task 3. Find those real values of the parameter $a$ for which the inequality
$$
x^{2}-(a+1) x+a+1>0
$$
is true for all values of $x$ satisfying the condition $|x| \leqslant 1$. | First solution. Solve the inequality (1) with respect to the parameter:
$$
\begin{aligned}
& x^{2}-a x-x+a+1>0 \\
& x^{2}-x+1>a x-a \\
& a(x-1)\frac{x^{2}-x+1}{x-1}=\frac{x(x-1)+1}{x-1}=x+\frac{1}{x-1}
\end{aligned}
$$
We construct the graph of the function $a(x)=x+\frac{1}{x-1}$ on $-1 \leqslant x \leqslant 1$.
For... | -1<+\infty | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,554 |
3 Task 4. Solve the equation
$$
\begin{aligned}
& x^{-1}+a x=1 \\
& x^{-1}+a x>1 \\
& x^{-1}+a x<1
\end{aligned}
$$
with respect to $x$ ( $a$ - parameter) and investigate the properties of their solutions. | Solution. Obviously, $x \neq 0$. First, we solve equation (1) and inequalities (2) and (3) with respect to the parameter $a$:
$$
a=\frac{x-1}{x^{2}}
$$
From inequality (2) we obtain:
$$
\begin{aligned}
& a>\frac{x-1}{x^{2}}, \text { if } x>0 \\
& a>0 \\
& a>\frac{x-1}{x^{2}}, \text { if } x<0
\end{aligned}
$$
We in... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,555 |
Problem 1. How many solutions does the equation $\sqrt{x}+x^{3}=5$ have? | Solution. The function $f(x)=\sqrt{x}+x^{3}$ is defined for $x \geqslant 0$, continuous, monotonic, non-negative, and varies from 0 to $+\infty$. Therefore, the given equation has only one solution $x_{1}$. Obviously, $1<x_{1}<2$. | x_{1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,563 |
Problem 2. Does there exist a root $x_{0}$ of the equation $\sin 2 x+5 \sin x+5 \cos x+1=0$, if $0 \leqslant x_{0} \leqslant 0.25 \pi$? | Solution. Does not exist. If $0 \leqslant x \leqslant 0.25 \pi$, then all terms on the left side of this equation are non-negative. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,564 |
Problem 3. Equation
$$
\sin 2 x+5 \sin x+5 \cos x+1=0
$$
was transformed as follows:
$$
\begin{gathered}
2 \sin x \cos x+5 \sin x+5 \cos x+1=0 \\
\cos x(2 \sin x+5)=-1-5 \sin x \\
\cos x=\frac{-1-5 \sin x}{2 \sin x+5} \\
\cos x=\frac{11.5}{2 \sin x+5}-2.5
\end{gathered}
$$
Prove that equations (1), (2), (3) are equ... | Solution. If $0<x<\frac{\pi}{6}$, then the functions $\operatorname{tg} 2 x$ and $\operatorname{tg} 3 x$ are positive and increasing. From this, it is clear that on $\left(0 ; \frac{\pi}{6}\right)$ the left side of equation (1) is negative, while the right side is positive. | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,565 |
Problem 6. How many solutions does the equation
$$
\arcsin 2x + \arcsin x = \frac{\pi}{3} ?
$$ | Solution. The function $f(x)=\arcsin 2 x$ is defined on the interval $[-0.5 ; 0.5]$, and the function $\varphi(x)=\arcsin x$ is defined on the interval $[-1 ; 1]$. Therefore, the left side of equation (1) is defined on the interval $[-0.5 ; 0.5]$. The monotonic and continuous function $y=\arcsin 2 x+\arcsin x$ changes ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,566 |
Problem 11. Prove that the equation $\sqrt{x+3}+$ $+\sqrt{3 x-2}=7$ has a unique root $x_{1}$. | Solution. The function $f(x)=\sqrt{x+3}+\sqrt{3 x-2}$ is defined for $x \geqslant \frac{2}{3}$, increasing, continuous, $f\left(\frac{2}{3}\right)=\sqrt{\frac{11}{3}}<7$. It is easy to notice that $x_{1}=6$. | x_{1}=6 | Algebra | proof | Yes | Yes | olympiads | false | 37,567 |
Task 13. Find the minimum value of the function
$$
\psi(x)=\sqrt{15-12 \cos x}+\sqrt{7-4 \sqrt{3} \sin x}
$$
on the interval $[0 ; 0.5 \pi]$. | Solution. Since $|\cos x| \leqslant 1$ and $|\sin x| \leqslant 1$, the function $\psi(x)$ is defined on the entire interval $[0 ; 0.5 \pi]$. To find the properties of the functions $f(x)=\sqrt{15-12 \cos x}$, $\varphi(x)=\sqrt{7-4 \sqrt{3} \sin x}, \quad \psi(x)=f(x)+\varphi(x) \quad$, we create a table of their values... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,568 |
Task 14. Determine the necessary and sufficient conditions that the numbers $a, b, c$ must satisfy so that the equation $x^{3}+a x^{2}+b x+c=$ $=0(1)$. has three real roots forming an arithmetic progression. | Solution. Let $x=p+t$. After this, the given equation (1) is transformed into:
$$
\begin{gathered}
t^{3}+(a+3 p) t^{2}+\left(3 p^{2}+2 a p+b\right) t+ \\
+\left(p^{3}+a p^{2}+b p+c\right)=0
\end{gathered}
$$
If $p=-a: 3$, i.e., $a+3 p=0$, then the equation takes the form
$$
t^{3}+\left(b-\frac{a^{3}}{3}\right) t+\le... | \frac{}{3}-\frac{2^{3}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,569 |
Problem 15. Solve the equation
$$
\frac{5}{\sqrt{1-x^{2}}}=\frac{1}{x}+\frac{85}{12}
$$
104 | The positive function $P(x)=\frac{5}{\sqrt{1-x^{2}}}$ is defined on $(-1 ; 0)$ and $(0 ; 1)$. The positive function $F(x)=\frac{1}{x}+\frac{85}{12}$ is considered on $\left(-1 ;-\frac{12}{85}\right)$ and $(0 ; 1)$. To form a hypothesis about the number of roots of the equation, we construct tables of the functions $P(x... | 0.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,570 |
Problem 16. Solve the equation
$$
\lg \left(x^{2}+9\right)-3 \cdot 2^{x}+5=0
$$ | Solution. The equation does not have negative roots because on $(-\infty ; 0)$ the function $P(x)=\lg \left(x^{2}+\right.$ $+9)+5$ is decreasing, the function $K(x)=3 \cdot 2^{x}$ is increasing, and $P(0)>K(0)$. To hypothesize about the number of roots of the given equation, we construct tables of the functions $P(x)$ ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,571 |
Problem 17. Calculate the smallest root $x_{0}$ of the equation
$$
x^{2}-\sqrt{\lg x+100}=0
$$
(with a relative error of no more than $10^{-390} \%$ ). | Solution. Equation (1) is defined on $[10^{-100}; +\infty)$. Let $P(x) = x^2 - \sqrt{\lg x + 100}$. Obviously, $P(10^{-100}) > 0; P(1) < 0; P(0.1) < 0$.
Replace equation (1) with the equivalent equation $x^4 = \lg x + 100$. Denote: $M(x) = x^4, \quad F(x) = \lg x + 100$. We find $M'(x) = 4x^3, F'(x) = 1/(x \ln 10)$.
... | 10^{-100}<x_0<10^{-100+10^{-396}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,572 |
Problem 18. Find the roots of the equation $F(x)=$ $=2^{2 x}-5 \cdot 3^{x}+29=0$ (to the nearest 0.01). | Solution. The given equation is equivalent to the equations:
$$
4^{x}+29=5 \cdot 3^{x}, \ln \left(4^{x}+29\right)=\ln 5+x \ln 3
$$
Let $P(x)=\ln \left(4^{x}+29\right)-x \ln 3-\ln 5$.
Solving the equation $P^{\prime}(x)=0$, we get $4^{x}=29 \ln 3 / \ln \frac{4}{3}$. Hence, $x \ln 4=\ln \frac{29 \ln 3}{\ln (4 / 3)}$ a... | x_1=2,x_2\approx5.55 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,573 |
Problem 4. Construct the parallel projection of an equilateral triangle $A B C$, with a side length of 4 cm, if the plane of this triangle is parallel to the projection plane. | Solution. The parallel projection of triangle $A B C$ is a triangle equal to the given one. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,577 |
Problem 7. Given a regular quadrilateral pyramid $M A B C D$, all edges of which are equal to each other. Point $P$ is the midpoint of edge $M C$. The diagonals $B D$ and $A C$ of the base intersect at point $H$. Construct the image of this pyramid on a plane parallel to the face $B C M$, taking the line $H P$ as the d... | Solution. Triangle $B C M$ is equilateral and its plane is parallel to the projection plane. Therefore, we first construct the equilateral triangle $B C M$ (Fig. 1). Segment $P H$ is represented by point $P=H^{1}$ (the midpoint of segment $M C$). Line $A M$ is parallel to line $H P$. Therefore, edge $A M$ is represente... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,579 |
Problem 8. Determine the shape of the parallel projection of a regular tetrahedron $D A B C$ onto a plane parallel to the edges $A B$ and $C D$. | Solution. Intersecting lines $A B$ and $C D$ are mutually perpendicular and $A B = C D$. There exists a unique plane $\alpha$ to which the lines $A B$ and $C D$ are parallel. Any line $l$ intersecting the plane $\alpha$ can be taken as the direction of projection.
If the line $l$ is parallel to the edge $B C$ or $A C$... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,580 |
Task 10. Can two parallel non-coincident lines be considered the image of intersecting lines? | Solution. The answer to the question of the problem will be affirmative if the projecting line is parallel to two parallel planes, each of which passes through one of the given skew lines. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,582 |
Problem 15. Prove that the orthogonal projection of a regular tetrahedron $D A C B$ with edge $a$ onto a plane will have the maximum area when this plane is parallel to two skew edges $D C$ and $A B$ of the tetrahedron. | Solution. Edges $D C$ and $A B$ of the given regular tetrahedron are equal and mutually perpendicular. If lines $D C$ and $A B$ are parallel to the projection plane, then the orthogonal projection of the regular tetrahedron is a square $A B C D$, the diagonals of which are equal to $a$. The area of this square is $0.5 ... | 0.5^{2} | Geometry | proof | Yes | Yes | olympiads | false | 37,585 |
Problem 16. Given a parallelogram $A B C D$. Point $M$ is the midpoint of segment $A B$. Point $K$ lies on segment $A D$ and $A K: K D=1: 2$. Find in what ratio point $P$, the intersection of segments $A C$ and $M K$, divides segment $M K$. | Solution. Consider this parallelogram as the image of a square $ABCD$ with a side length of 6. The segment $AP$ is the angle bisector of the right angle of triangle $KAM$. Therefore, $KP: PM = AK: AM = \frac{1}{3} AD: 0.5 \quad AB = 2: 3$. | 2:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,586 |
Task 17. Given a regular hexagonal pyramid $M A B C D E F$. Point $K$ bisects edge $B M$. Find the ratio in which the plane $F E K$ divides edge $A M$ (at point $X$). | Solution. We construct the orthogonal projection of the given pyramid, taking the line $F E$ as the projecting line (Fig. 8). The plane $F E K$ is represented by the line $K E$. Since in the figure the segments $K E$ and $M A$ are medians of the triangle $M B E$, then $M X: X A=2: 1$.
. Then the segment $DE$, parallel to the line $BC_{1}$, will be represented by the point $D=E$ (at point $E$, the lines $B_{1}A$ and $A_{1}C$ intersect). Since the segments $DE$ and $BC_{... | 1:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,588 |
Problem 19. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. In what ratio does the point $E$, which lies on the plane passing through vertex $A$ and the centers $K$ and $H$ of the faces $A_{1} B_{1} C_{1} D_{1}$ and $B_{1} C_{1} C B$, divide the edge $B_{1} C_{1}$? | Solution. We represent the lower base of the given cube as square $A B C D$, and the line $A K$ as point $A=K$ (Fig. 10). On this diagram, the plane $A K H$ is depicted as the line $A H$. The lines $B_{1} C_{1}$ and $H A$ intersect at point $E$. For triangle $B_{1} C A$, the segments $A H$ and $B_{1} C_{1}$ are medians... | B_{1}E:EC_{1}=2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,589 |
Problem 20. The base of the quadrilateral pyramid $M A B C D$ is the parallelogram $A B C D$. Point $K$ bisects edge $D M$. Point $P$ lies on edge $B M$ and $B P: P M=1: 3$. The plane $A P K$ intersects edge $M C$ at point $X$. Find the ratio of segments $M X$ and $X C$. | Solution. For the image of the face $M B C$, we take the right triangle $M B C$ (Fig. 11), and for the direction of projection - the line $A P$. The section $A P X K$ of the pyramid $M A B C D$ is depicted on the drawing as the segment $P K$. To solve the problem, we supplement the rectangular trapezoid $M B C D$ to fo... | MX:XC=3:4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,590 |
Problem 26. In quadrilateral $ABCD$, a trapezoid $MKPE$ is inscribed (Fig. 15), with its parallel sides parallel to diagonal $AC$. Prove that the lines $ME$, $BD$, and $KP$ intersect at one point. | Solution. We take the quadrilateral $A B C D$ (together with its diagonals) as the image of a triangular pyramid. The trapezoid $M K P E$ is a section of this pyramid by the plane $M K P$. This plane is not parallel to the line $B D$, since the non-parallel sides $M E$ and $K P$ of the trapezoid $M K P E$ are not paral... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,596 |
Problem 27. Given a triangle $ABC$ and an arbitrary point $M$ inside it (Fig. 16). Prove that if parallelograms $M B B_{1} C$ and $M A A_{1} B_{1}$ are constructed, then the diagonal $M A_{1}$ of the latter passes through the centroid $O$ of the given triangle. | Solution. Consider the triangle $ABC$ together with the segments $BM, CM, MA$, and $MO$ as the image of a regular triangular pyramid $MABC$ (with base $ABC$ and height $MO$).
Fig. 17
. Clearly, $FD = MP = \frac{1}{3} PB$.
From similar triangles $TFD$ and $TPB$, it follows that $TD: DB = 1: 2$. Now, we transfer point $T$ to Figure 23.
![](https://cdn.mathpix.com... | TD:DB=1:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,604 |
P r o b l e m 1. "Prove that segments $A B$ and $C D$ are parallel if $A(1 ; 3), B(2 ;-1), C(0 ; 4), D(2 ;-4)$. | Solution. We find $A B=(1 ;-4)$ and $C D=(2 ;-8)$. From here $\overline{A B}=0.5 \overline{C D}$.
Problem 2. Find the coordinates of vertex $D$ of parallelogram $A B C D$, if $A(-1 ; 2 ; 3) ; B(-2 ; 4 ; 6)$, $C(6 ;-5 ; 8)$.
Solution. Quadrilateral $A B C D$ is a parallelogram. Therefore, $A D=B C=(8 ;-9 ; 2)$. To fin... | D(7,-7,5) | Geometry | proof | Yes | Yes | olympiads | false | 37,606 |
Problem 3. Does the point $C(7 ; 9 ; 0)$ belong to the line $A B$, if $A(-1 ; 1 ; 2)$ and $B(3 ; 6 ;-1)$? | Solution. If point $C$ lies on the line $A B$, then there exists a number $k$ such that $\overline{A C}=k \overline{A B}$, i.e., $(8 ; 8 ;-2)=k(4 ; 5 ;-3)$, which is impossible. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,607 |
Problem 4. Write the coordinates of point $C$, which divides the segment $A B$ in the ratio $2: 1$, starting from point $A$, if $A(1 ;-1 ; 2)$ and $B(7 ;-4 ;-1)$. | Solution. Obviously, $\overline{A C}=\frac{2}{3} \overline{A B}=\frac{2}{3}(6 ; \quad-3$; $-3)=(4 ;-2 ;-2)$. Therefore, $C(1+4 ;-1-2 ; 2-2)$, i.e., $C(5 ;-3 ; 0)$. | C(5;-3;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,608 |
Task 5. Can the vector $\overline{D_{1} K}$ be decomposed into the vectors $\overline{D_{1} D}$ and $\overline{D_{1} A_{1}}$, if $K(-3 ; 7 ;-7), D_{1}(-3$; $10 ;-5), D(-5 ; 6 ;-1), A_{1}(1 ; 6 ;-7)$? | Solution. If $\overline{D_{1} K}=k \overline{D_{1} D}+n \overline{D_{1} A}$, then $(0 ;-3$; $-2)=k(-2 ;-4 ; 4)+n(4 ;-4 ;-2)$. From this, we obtain the system of equations:
$$
\left\{\begin{aligned}
0 & =-2 k+4 n \\
-3 & =-4 k-4 n \\
-2 & =4 k-2 n
\end{aligned}\right.
$$
This system of equations is inconsistent. There... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,609 |
Problem 6. Find the equation of the plane $ABF$, if $F(-4; 8; -3), A(-1; 2; -3), B(3; 4; 1)$. | Solution. If $M(x, y, z)$ is an arbitrary point on the plane $A B F$, then $\overline{B M}=k B A+n B F$ or $(x-3 ; y-4; z-1)=k(-4 ;-2 ;-4)+n(-7 ; 4 ;-4)$.
From this, we obtain the system of equations:
$$
\left\{\begin{array}{l}
x-3=-4 k-7 n \\
y-4=-2 k+4 n \\
z-1=-4 k-4 n
\end{array}\right.
$$
Multiply both sides of... | 5z-2y-4x+15=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,610 |
Problem 7. Find the distance $B_{1} H$ from point $B_{1}$ to the line $D_{1} B$, if $B_{1}(5 ; 8 ;-3), D_{1}(-3 ; 10 ;-5)$, $B(3 ; 4 ; 1)$. | Solution. $\quad$ Let $\overline{D_{1} H}=n \overline{D_{1} B}=n(6 ; \quad-6 ; 6)$. Further,
$$
\begin{aligned}
\overline{H B_{1}} & =\overline{D_{1} B_{1}}-\overline{D_{1} H}=(8 ;-2 ; 2)-n(6 ;-6 ; 6)= \\
& =(8-6 n ;-2+6 n ; 2-6 n) .
\end{aligned}
$$
According to the problem, $\overline{D_{1} B} \cdot \overline{H B}_... | 2\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,611 |
Problem 8. Form the equation of the plane $\alpha$, passing through the midpoint $M$ of the segment $A D$ and perpendicular to the line $C B$, if $A(-1 ; 2 ;-3), D(-5 ; 6$; $-1), C(-3 ; 10 ;-5), B(3 ; 4 ; 1)$. | Solution. Obviously, $M(-3 ; 4 ;-2)$. Let $P(x ; y ; z)$ be an arbitrary point on the plane $\alpha$. Then $\overline{M P} \cdot \overline{C B}=0$, or $(x+3 ; y-4 ; z+2) \cdot(6 ;-6 ; 6)=0$, or $6(x+3)-6(y-4)+6(z+2)=0$.
Thus, the equation of the plane $\alpha$ is: $z-y+x+9=0$. | z-y+x+9=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,612 |
Problem 9. Form the equation of the plane $\alpha$, relative to which the points $D(-5 ; 6 ;-1)$ and $B(3 ; 4 ; 1)$ are symmetric. | Solution. The plane $\alpha$ intersects the segment $DB$ at point $P(-1; 5; 0)$, which is the midpoint of segment $DB$. Let $M(x; y; z)$ be an arbitrary point on the plane $\alpha$. Then $\overline{DB} \cdot \overline{PM}=0$, i.e., $(8; -2; 2) \times (x+1; y-5; z)=0$, or $8(x+1)-2(y-5)+2z=0$.
Thus, the equation of the... | 4x-y+z+9=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,613 |
Problem 10. Find the coordinates of the point $M(x ; y ; z)$, symmetric to the point $B_{1}(5 ; 8 ;-3)$ with respect to the line defined by the points $B(3 ; 4 ; 1)$ and $A_{1}(1 ; 6 ;-7)$ (Fig. 1). | First solution. $\overline{B H}=0.5 \overline{B B}_{1}+0.5 \overline{B M} ; \overline{B H}=$ $=k \overline{B A_{1}}$.
\[
\begin{gathered}
k \overline{B A_{1}}=0.5 \overline{B B_{1}}+0.5 \overline{B M} \\
\overline{B_{1} M} \cdot \overrightarrow{B A_{1}}=0
\end{gathered}
\]
From equations (1) and (2) we get:
$k(-2 ; ... | M(-1;2;-3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,614 |
Problem 11. Find the distance $A_{1} H$ from the point $A_{1}(1 ; 6 ;-7)$ to the plane $A B_{1} K$, defined by the points $A(-1 ; 2 ;-3), B_{1}(5 ; 8 ;-3)$ and $K(-1 ; 8 ;-6)$ (Fig. 2). | The first solution. Let $P(0 ; y ; z)$ be such a point that $\overline{A P} \cdot \overline{A B_{1}}=0$ and $\overline{A P} \cdot \overline{A K}=0$. Since $\overline{A P}=(1 ; y-2$; $z-3), \overline{A B_{1}}=(6 ; 6 ; 0), \overline{A K}=(0 ; 6 ;-3)$, the last two equations take the form:
$$
6+6(y-2)=0 ; 6(y-2)-3(z+3)=0... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,615 |
P r o b l e m 12. The medians $C K$ and $A M$ of triangle $A B C$ intersect at point $P$. Prove that $M P: P A=$ $=K P: P C=1: 2$. | Solution. Let point $P$ belong to segment $A M$ and $A P: P M=2: 1$. Then $\overline{C P}=\frac{1}{3} \overline{C A}+\frac{2}{3} \overline{C M}=$ $=\frac{1}{3}(\overline{C A}+\overline{C B})$. Further, $\overline{C K}=0.5(\overline{C A}+\overline{C B})$. From this, it is clear that $\overline{C P}=\frac{2}{3} \overline... | 2:3 | Geometry | proof | Yes | Yes | olympiads | false | 37,616 |
Problem 15. Given triangle $ABC$, $\overline{BD}=3 \overline{DC}$, $\overline{AE}=1.5 \overline{EC}$, $\overline{BK}=2 \overline{KA}$ (Fig. 7). Prove that the lines $AD$, $BE$, and $CK$ intersect at one point. | Solution. Let $\overline{B A}$ and $\overline{B C}$ be the basis vectors and point $M$ be the intersection of lines $A D$ and $C K$. Clearly,
$$
\overline{B E}=0.4 \overline{B A}+0.6 \overline{B C}=0.2(2 \overline{B A}+3 \overline{B C})
$$
Due to the collinearity of points $K, M, C$ and $A, M, D$, we have:
$$
\overl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,618 |
Problem 16. Quadrilateral $ABCD$ is a trapezoid with bases $AB$ and $CD$. $\overline{AM}=\overline{MD}$, $\overline{BK}=\overline{KC}$. Prove that lines $AK$ and $CM$ are not parallel (Fig. 8). | S o l u t i o n. We choose vectors $\overline{A B}$ and $\overline{A D}$ as the basis vectors. Let $\overline{D C}=k \overline{A B}$. Then $\overline{M C}=0.5 \overline{A D}+k \overline{A B}$; $\overline{A K}=0.5(\overline{A B}+\overline{A C})=0.5(\overline{A B}+\overline{A D}+k \overline{A B})=\overline{M C}+$ $+0.5(1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,619 |
3 a d a c h a 19. A circle with radius 1 and center at point $O$ is circumscribed around an equilateral triangle $A B C$. Prove that the sum of the squares of the distances from any point $M$ on the circle to the vertices of this triangle is equal to 6 (Fig. 9). | Solution. Obviously,
$\overline{C M}^{2}+\overline{M A}^{2}+\overline{M B}^{2}=(\overline{O M}-\overline{O C})^{2}+(\overline{O A}-\overline{O M})^{2}+$ $+(\overline{O B}-\overline{O M})^{2}=\overline{O M}^{2}+\overline{O C}^{2}-2 \overline{O M} \cdot \overline{O C}+\overline{O A}^{2}+$ $+\overline{O M}^{2}-2 \overlin... | 6 | Geometry | proof | Yes | Yes | olympiads | false | 37,622 |
Problem 20. In a convex quadrilateral $A B C D$, sides $A B$ and $C D$ are perpendicular. Prove that
$$
B C^{2}+A D^{2}=A C^{2}+B D^{2}
$$ | Solution. Vectors $\overline{A B}, \overline{A C}, \overline{A D}$ determine the position on the plane of the vertex of the given quadrilateral. We rewrite equality (1) as:
$$
\overline{B C}^{2}-\overline{A C}^{2}=\overline{D B}^{2}-\overline{A D}^{2}
$$
We express the vectors involved in this equality in terms of ve... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,623 |
Problem 22. Prove that the bisectors of two dihedral angles of a trihedral angle and the bisector of the angle adjacent to the third dihedral angle lie in the same plane. | Solution. Let us lay off equal segments $O A, O B, O C$ on the edges of a trihedral angle with vertex $O$ (Fig. 11). Let $\overline{O B}_{1}=-\overline{O B}$. Triangles $B_{1} O C, A O B, A O C$ are isosceles, so their bisectors $O K, O M, O P$ are also medians. Taking the vectors $\overline{O A}, \overline{O B}$ and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,625 |
Problem 23. The base of the quadrilateral pyramid $M A B C D$ is the parallelogram $A B C D, \overline{D K}=$ $=\overline{K M}, \overline{B P}=0.25 \overline{B M}$; point $X$ is the intersection of line $M C$ and plane $A K P$. Find the ratio $M X: X C$. | Solution. We choose the following vectors as the basis vectors
Fig. 11
=0$ and $\overline{A D}(\ove... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,628 |
Problem 26. $A B C D A_{1} B_{1} C_{1} D_{1}$ is a cube (Fig. 13). $A B=$ $=1 ; \overline{D M}=\frac{8}{15} \overline{D B}_{1}$. The plane $\alpha$ passes through point $M$ and is perpendicular to the line $B_{1} D$. Find the ratio in which the plane $\alpha$ divides the edge $A_{1} D_{1}$. | Let the plane $\alpha$ and the line $A_{1} D_{1}$ intersect at point $X$. According to the problem,
$$
\overline{M X} \cdot \overline{D B_{1}}=0
$$
Consider the basis vectors $\overline{D A}, \overline{D C}$, and $\overline{D D}_{1}$.
$\overline{D B}_{1}=\overline{D D}_{1}+\overline{D A}+\overline{D C}$. Let $\overl... | D_{1}X:XA_{1}=3:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,629 |
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