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 |
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182. Let $a+b \sqrt{5}$ be a solution to equation (1). Prove that
a) if $0 < a + b \sqrt{5} < 1$, then $a < 0$ and $b < 0$.
b) if $a + b \sqrt{5} > 1$, then $a > 0$ and $b > 0$. | 182. a) Suppose that $a0$.
b) As just shown, $a \geqslant 0$. Suppose that $b \leqslant 0$. Then $-b \geqslant 0 \geqslant b$ and $a-b \sqrt{5} \geqslant a+b \sqrt{5}>1$. It follows that $1=a^{2}-5 b^{2}=(a+b \sqrt{5})(a-b \sqrt{5})>$ $>1 \cdot 1=1$, leading to a contradiction. | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,759 |
183. Show that there is no integer solution to equation (1) satisfying the inequality
$$
1<a+b \sqrt{5}<9+4 \sqrt{5}
$$ | 183. Let such a solution exist. From problem 182, it follows that $a \geqslant 0, b>0$. Therefore, due to problem 181, $a<9, b<4$. Thus, for $b$ the only possible values are: $b=1,2,3$. By simple verification, we find that none of the numbers $1+5 b^{2}$, where $b=1,2,3$, is a perfect square; therefore, the number $a+\... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,760 |
184. Prove that all integer solutions $p+q \sqrt{5}$ of equation (1), for which $p \geqslant 0$ and $q \geqslant 0$, are obtained by the formula
$$
p+q \sqrt{5}=(9+4 \sqrt{5})^{n}
$$
where the exponent $n$ takes all possible non-negative integer values. | 184. For $n=0 \quad(9+4 \sqrt{5})^{0}=1=1+0 \sqrt{5}$, for $n=1$ $(9+4 \sqrt{5})^{1}=9+4 \sqrt{5}$, and $1+0 \sqrt{5}$ and $9+4 \sqrt{5}$ are solutions of equation (1). For $n \geqslant 2$ the formula $(9+4 \sqrt{5})^{n}$ gives the product of solutions of equation (1), i.e., again a solution $p+q \sqrt{5}$, where $p \g... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,761 |
187. What is the probability that a 6 will not come up in any of the 6 throws of a die? | 187. The probability that six points will not fall when a die is rolled once is $\frac{5}{6}$. The probability of not rolling a six in six rolls is given by the formula (6) on page 126
$$
\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6}=\left(\frac{5}{6}\right)^{6}=... | \frac{15625}{46656}\approx0.84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 24,764 |
188. Let's denote by $p$ the probability of hitting the target with a single shot. Calculate the probability that the target will be hit, i.e., at least one hit will occur, in $n$ shots.
At the beginning of the section, it was noted that unlikely events can be considered practically impossible, and there are different... | 188. Let the desired probability be denoted by $x$. Then the probability of no hits in $n$ shots is $1-x$, and for one shot it is 1 - p (property 3). As in the previous problem, by formula (6)
$$
1-x=(1-p)^{n}
$$
from which
$$
x=1-(1-p)^{n}
$$ | 1-(1-p)^{n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 24,765 |
189. Two players take turns tossing a coin. The one who gets heads first wins. What is the probability that the game will never end? What is the probability that the starting player will win? What is the probability that the second player will win? | 189. Let's denote:
the probability of event $A$: "the starting player wins" - by $x$, the probability of event $B$: "the second player wins" - by $y$, the probability of event $C$: "the game never ends" - by $z$. Events $A, B, C$ are pairwise mutually exclusive and form a complete system, therefore
$$
x+y+z=\mathbf{P... | \frac{2}{3},\frac{1}{3},0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 24,766 |
191. The following experiment is conducted. Two urns that look identical are placed in a room. The left one contains $a$ balls, and the right one contains $b$ balls. People enter the room one by one, and each person moves a ball from the left urn to the right or from the right urn to the left. It is assumed that the pr... | 191. We will represent the number of balls in the left urn using a chip placed on a number line. At the initial
1) A. P. Kiselev, Algebra, Part 2, 23rd edition, Uchpedgiz, M., 1946, p. 78.
18 Law. Zzya. E. Dynkin and V. Uspenskii
moment, the chip is located at point $a$ (Fig. 156). After each unit of time, it moves to... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 24,768 |
195. Prove that:
for all values of $k$, starting from 2,
$$
\sqrt{3 \cdot\left(\frac{1}{2}\right)^{2} \cdot \frac{3}{4} \cdot \frac{1}{2 k}} \leqslant w_{2 k}<\sqrt{3 \cdot\left(\frac{1}{2}\right)^{2} \cdot \frac{1}{2 k}}
$$
for all values of $k$, starting from 3,
$$
\sqrt{5 \cdot\left(\frac{1}{2} \cdot \frac{3}{4}... | 195. To prove, let's write down three products one under the other (cf. formulas (5), (6), and (7) on p. 139):
$$
\begin{aligned}
& \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdots \frac{2 a-3}{2 a-2} \cdot \frac{2 a-3}{2 a-2} \cdot \frac{2 a-1}{2 a} \cdot \frac{2 a-1}{2 a} \cdot \frac{2 a}{2 a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 24,772 |
197. Prove that all numbers in the $n$-th row of the probability triangle do not exceed $\frac{1}{\sqrt{n}}$.
Estimate of an arbitrary element of the triangle. After achieving such a complete success in the approximate calculation of the central elements of the probability triangle, it is natural to try to find equall... | 197. Let $n=2 k$. All members of the $n$-th row are less than the central member, and the central member satisfies the inequality
$$
w_{n}=w_{2 k}<\frac{1}{\sqrt{2 k}}=\frac{1}{\sqrt{n}}
$$
Now let $n=2 k - 1$. In the row with an odd number, there is no central member; the largest in magnitude are the equal members $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,774 |
202. Indicate the number of steps sufficient to assert with an error probability not exceeding 0.001 that the reduced speed of the chip is less than 0.01.
Let us recall now that each movement of the chip is conditioned by the result of tossing a coin. If, in $n$ tosses of the coin, heads appear $l$ times and tails app... | 202. With a probability of error less than 0.001, it can be asserted that the given speed of the chip after $n$ steps will be less than $\frac{2}{\sqrt[3]{0.001} \sqrt{n}}$.
Let's take such an $n$ so that
$$
\frac{2}{\sqrt[3]{0.001} \sqrt{n}} \leqslant 0.01
$$
from which $n \geqslant 4 \cdot 10^{6}$.
Thus, with a p... | 4\cdot10^{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 24,779 |
208. On a line, points $0, \pm 1, \pm 2, \pm 3, \ldots$ are marked.
Fig. 80.
(Fig. 80). A particle, upon reaching point $n$, moves to point $n+1$ with probability $\frac{1}{2}$ and to poi... | 208. Since all points are completely equal, $x$ is the probability that a particle that has left point $n-1$ will ever visit $n$. Similarly, the probability that a particle that has left point $n+1$ will visit $n$ is $y$.
- Suppose the particle starts from point 0 and consider the events:
$A$: "the particle will ever... | 1,1,1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 24,785 |
209. Prove that the particle from problem 208 will visit every point with probability 1.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
---
209. Prove that the particle from problem 208 will visit every point with probabil... | 209. We need to show that with probability 1, the particle will visit point $n$ (for definiteness, let's assume that $n>0$). For $n=1$, this is proven in the previous problem. Suppose the statement is proven for point $n$; let's prove it for point $n+1$.
Consider the event $A_{n+1}$: "the particle will eventually visi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,786 |
17. Prove that if $a$ and $b$ are two sides of a triangle, $\alpha$ is the angle between them, and $l$ is the length of the angle bisector, then $l=\frac{2 a b \cos \frac{\alpha}{2}}{a+b}$. | 17. The bisector divides the triangle into two, the areas of which are respectively $\frac{a l}{2} \sin \frac{\alpha}{2}, \frac{b l}{2} \sin \frac{\alpha}{2}$, and the area of the entire triangle is $\frac{a b}{2} \sin \alpha$; therefore, $\frac{a l}{2} \sin \frac{\alpha}{2}+\frac{b l}{2} \sin \frac{\alpha}{2}=\frac{a ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 24,791 |
39. In trapezoid $A B C D$ with sides $|A B|=a$, $|B C|=b$, the bisector of angle $A$ is drawn. Determine whether it intersects: base $B C$ or lateral side $C D$. | 39. If $a>b$, then the bisector intersects the lateral side $C D$; if $a<b$, then the base $B C$. | If\b,\then\the\bisector\intersects\the\lateral\side\CD;\if\b,\then\the\base\BC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,795 |
40. Find the length of a segment of a line parallel to the bases of a trapezoid and passing through the point of intersection of the diagonals, if the bases of the trapezoid are equal to $a$ and $b$. | 40. $\frac{2 a b}{a+b}$. 41. $\arccos \frac{1-k}{1+k}$. 42. $\frac{a+b}{4} \sqrt{3 b^{2}+2 a b-a^{2}}$. 43. $a^{2}$. 44. $\frac{1}{2} \sqrt{\frac{S}{2}} \cdot$ 45. $\left(\sqrt{S_{1}}+V \overline{S_{2}}\right)^{2} \cdot 46.90^{\circ}+\frac{\alpha}{2} \cdot 47 . \frac{|a-b|}{a+b} \times$ $\times \sqrt{a^{2}+b^{2}} .48 .... | \frac{2}{+b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,796 |
64. To a circle of radius $R$ from an external point $M$, tangents $M A$ and $M B$ are drawn, forming an angle $\alpha$. Determine the area of the figure bounded by the tangents and the smaller arc of the circle. | 64. $R^{2}\left[\operatorname{ctg} \frac{\alpha}{2}-\frac{1}{2}(\pi-\alpha)\right]$. 65. $\frac{a}{4} \sqrt{10}$. 66. $\frac{a\left(4 \sin ^{2} \alpha+1\right)}{8 \sin \alpha}$. 67. $2 r^{2}(2 \sqrt{\overline{3}}+3)$. 68. $\frac{a^{2}+4 r^{2}}{4 r}$, 69. $\frac{3 a}{2(5+\sqrt{13})}$, 70. $\frac{a \sqrt[V]{10}}{4}$. 71.... | R^{2}[\operatorname{ctg}\frac{\alpha}{2}-\frac{1}{2}(\pi-\alpha)] | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,798 |
84. The common chord of two intersecting circles is seen from their centers at angles of $90^{\circ}$ and $60^{\circ}$. Find the radii of the circles if the distance between their centers is $a$. | 84. There are two possible cases: both centers are located on opposite sides of the common chord, or on the same side. Accordingly, there are two pairs of answers: $a(\sqrt{3}+1), a \frac{\sqrt{2}}{2}(\sqrt{3}+1)$ and $a(\sqrt{3}-1), a \frac{\sqrt{2}}{2}(\sqrt{3}-1)$. | (\sqrt{3}+1),\frac{\sqrt{2}}{2}(\sqrt{3}+1)(\sqrt{3}-1),\frac{\sqrt{2}}{2}(\sqrt{3}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,799 |
129. In a parallelogram, there are two circles of radius 1, each touching the other and three sides of the parallelogram. It is also known that one of the segments of a side of the parallelogram from a vertex to the point of tangency is $\sqrt{3}$. Find the area of the parallelogram. | 129. $\frac{4}{3}(2 \sqrt{\overline{3}}+3)$. 130. $\frac{2 R^{2} \sin ^{3} \alpha \sin \beta}{\sin (\alpha+\beta)}$. 131. $\frac{3 \sqrt{\overline{3}}(\sqrt{\overline{1}}-1)}{32 \pi}$. 132. $1,1, \quad 133 . \frac{a^{2}}{16 R}$. 134. $90^{\circ}$. 135. $30^{\circ}$. 136. $\frac{a \sqrt{ } \overline{7}}{4}$. 137. $\frac... | \frac{4}{3}(2\sqrt{3}+3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,807 |
1. In triangle $ABC$, the median $AD$ is drawn. $\widehat{D A C} + \widehat{A B C} = 90^{\circ}$. Find $\widehat{B A C}$, given that $|A B| = |A C|$. | 1. Take a point $A_{1}$ on the line $B A$ such that $\left|A_{1} B\right|=\left|A_{1} C\right|$. The points $A_{1}, A, D$ and $C$ lie on the same circle $\left(\widehat{D A_{1} C}=90^{\circ}-\widehat{A B C}=\widehat{D A C}\right)$. Therefore, $\widehat{A_{1} A C}=\widehat{A_{1} D C}=90^{\circ}$, which means $\widehat{B... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,810 |
2. Three circles with radii 1, 2, and 3 touch each other externally. Find the radius of the circle passing through the points of tangency of these circles. | 2. Note that the circle passing through the points of tangency is inscribed in a triangle with vertices at the centers of the circles. By equating the expressions for the area of the triangle obtained using Heron's formula and as the product of the semiperimeter and the radius of the inscribed circle, we find \( r=1 \)... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,811 |
4. In an equilateral triangle $A B C$ the side is equal to $a$. On side $B C$ lies point $D$, and on $A B$ point $E$ such that $|B D|=\frac{1}{3} a,|A E|=|D E|$. Find the length of $C E$. | 4. If $|E D|=|A E|=x$, then $|B E|=a-x$. Writing the cosine theorem for $\triangle B D E$:
$$
x^{2}=\frac{a^{2}}{9}+(a-x)^{2}-\frac{2}{3} a(a-x) \cdot \frac{1}{2}
$$
we find $x=\frac{7}{15} a$.
Answer: $: C E \left\lvert\,=\frac{13}{15} a\right.$. | \frac{13}{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,813 |
5. In a right triangle $A B C$, from the vertex of the right angle $C$, the bisector $C L,|C L|=a$, and the median $C M,|C M|=b$ are drawn. Find the area of triangle $A B C$. | 5. Denoting the legs by $x$ and $y$, we obtain the system of equations
$$
\left\{\begin{aligned}
a & =\frac{x y \sqrt{2}}{x+y} \\
4 b^{2} & =x^{2}+y^{2}
\end{aligned}\right.
$$
from which we will find $x y$.
Answer: $\frac{a^{2}+a \sqrt{a^{2}+8 b^{2}}}{4}$. | \frac{^{2}+\sqrt{^{2}+8b^{2}}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,814 |
6. A circle is inscribed in a trapezoid. Find the area of the trapezoid if the lengths of $a$ one of the bases and segments $b$ and $d$, into which one of the lateral sides is divided by the point of tangency (segment $b$ is adjacent to the given base) are known. | 6. Let (Fig. 1) $A B C D$ be a trapezoid, $|D C|=a, 0$ be the center of the circle. $A D$ is a lateral side, which is divided by the point of tangency $K$ into segments $|A K|=d,|K D|=b$. In $\triangle A O D$, the angle $A O D$ is a right angle, and the height $O K$ is the radius of the circle, so $r=$ $=|O K|=\sqrt{\o... | \frac{^{2}+(-b)}{-b}\sqrt{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,815 |
7. In a trapezoid, the diagonals are equal to 3 and 5, and the segment connecting the midpoints of the bases is equal to 2. Find the area of the trapezoid. | 7. Let (Fig. 2) $ABCD$ be the given trapezoid ($BCAD$). Draw a line through $C$ parallel to $BD$, and denote by $K$ the point of intersection of this line with the line $AD$. The area of $\triangle ACK$ is equal to the area of the trapezoid, the sides $AC$ and $CK$ are equal to the diagonals of the trapezoid, and the m... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,816 |
8. In triangle $A B C$, the lengths of the sides are known: $|A B|=12,|B C|=13,|C A|=15$. On side $A C$, a point $M$ is taken such that the radii of the circles inscribed in triangles $A B M$ and $B C M$ are equal. Find the ratio $|A M|:|M C|$. | 8. Let $|A M|:|M C|=k$. The condition of equality of the radii of the circles inscribed in triangles $A B M$ and $B C M$ means that their areas are in the same ratio as their perimeters. From this, since the ratio of the areas is $k$, we get ; $B M \left\lvert\,=\frac{13 k-12}{1-k}\right.$. From this equality, in parti... | \frac{22}{23} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,817 |
9. A circle of radius 1 is inscribed in triangle $A B C$, where $\cos \widehat{A B C}=0.8$. This circle touches the midline of triangle $A B C$, parallel to side $A C$. Find the length of side $A C$. | 9. It follows from the condition that the height to side $AC$ is equal to two diameters of the inscribed circle, i.e., 4. If $M, N$ and $K$ are the points of tangency with $AB, BC$ and $CA$, then
$$
|BM|=|BN|=r \operatorname{ctg} \frac{\widehat{ABC}}{2}=1 \cdot \sqrt{\frac{1+0.8}{1-0.8}}=\sqrt{9}=3
$$
If $|MA|=|AK|=x... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,818 |
10. Given an equilateral triangle $ABC$ with area $S$. Three lines are drawn parallel to its sides at an equal distance from them, intersecting inside the triangle and forming a triangle $A_{1} B_{1} C_{1}$ with area $Q$. Find the distance between the parallel sides of triangles $ABC$ and $A_{1} B_{1} C_{1}$. | 10. If $Q \geqslant \frac{1}{4} S$, then the desired distance will be $\frac{\sqrt[1]{3}}{3} \times$ $\times(\sqrt{S}-\sqrt{Q})$. If $Q<\frac{1}{4} S$, then there are two possible answers: $\frac{\sqrt[4]{3}}{3}(\sqrt{S} \pm \sqrt{Q})$ | \frac{\sqrt[4]{3}}{3}(\sqrt{S}\\sqrt{Q}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,819 |
11. Sides $A B$ and $C D$ of quadrilateral $A B C D$ are perpendicular and are diameters of two equal tangent circles with radius $r$. Find the area of quadrilateral $A B C D$, if $\frac{|B C|}{|A D|}=k$. | 11. Let $M$ be the point of intersection of the lines $A B$ and $C D$, and let $O_{1}$ and $O_{2}$ be the centers of the given circles. Suppose $|B M| = |x|$, $|M C| = |y|$, and choose the signs of $x$ and $y$ such that the following relations hold:
$$
\begin{aligned}
|B C|^{2}= & x^{2}+y^{2}, \quad|A D|^{2}=(x+2 r)^{... | 3r^{2}|\frac{1-k^{2}}{1+k^{2}}| | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,820 |
12. Two tangent circles are inscribed in an angle of magnitude $\alpha$. Determine the ratio of the radius of the smaller circle to the radius of a third circle that is tangent to the first two and one of the sides of the angle. | 12. Let (Fig. 3) \(O_{1}, O_{2}\) and \(O\) be the centers of the circles, \(M_{1}, M_{2}, M\) be the points of tangency with the side of the angle, \(r_{1}, r_{2}\) and \(r\) be the radii of the circles \((r_{1}<r_{2})\). By drawing a line through \(O_{1}\) parallel to \(M_{1} M_{2}\) until it intersects \(O_{2} M_{2}... | (\sqrt{\frac{1-\sin\frac{\alpha}{2}}{1+\sin\frac{\alpha}{2}}}+1)^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,821 |
13. In triangle $A B C$, a circle is constructed on the midline $D E$, parallel to $A B$, as its diameter, intersecting sides $A C$ and $B C$ at points $M$ and $N$. Find $M N$, if $|B C|=a,|A C|=b,|A B|=c$. | 13. If $\hat{C}=\alpha$, then $|C N|=\frac{b}{2} \cos \alpha,|C M|=\frac{a}{2} \cos \alpha\left(\alpha<90^{\circ}\right)$; therefore, $\triangle C M N$ is similar to $\triangle C A B$ with a similarity coefficient of $\frac{\cos \alpha}{2}$, so $|M N|=|A B| \frac{\cos \alpha}{2}=\frac{c}{2} \frac{a^{2}+b^{2}-c^{2}}{2 a... | \frac{(^{2}+b^{2}-^{2})}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,822 |
14. In triangle $ABC$, the difference between the internal angles $\hat{A}-\hat{B}=\varphi$ is given. It is known that the height dropped from vertex $C$ to side $AB$ is equal to the difference $|BC|-|AC|$. Find the angles of triangle $ABC$. | 14. If the height $|C D|=h$, then $|B C|=\frac{h}{\sin \hat{B}},|A C|=\frac{h}{\sin \hat{A}}$. By the condition $\frac{h}{\sin \hat{B}}-\frac{h}{\sin \hat{A}}=h$, hence $\sin \hat{A}-\sin \hat{B}=\sin \hat{A} \sin \hat{B}$, $2 \sin \frac{\hat{A}-\hat{B}}{2} \cos \frac{\hat{A}+\hat{B}}{2}=\frac{1}{2}[\cos (\hat{A}-\hat{... | \sin\frac{\hat{C}}{2}=1-\sin\frac{\varphi}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,823 |
15. Find the area of the rhombus $ABCD$, if the radii of the circles circumscribed around triangles $ABC$ and $ABD$ are $R$ and $r$. | 15. If the acute angle of a rhombus is $\alpha$, then the diagonals of the rhombus will be $2 r \sin \alpha$ and $2 R \sin \alpha$. On the other hand, the ratio of the diagonals is the tangent of half this angle, i.e., $\operatorname{tg} \frac{\alpha}{2}=\frac{r}{R}$, $\sin \alpha=\frac{2 R r}{R^{2}+r^{2}}$.
Answer: $... | \frac{8R^{3}r^{3}}{(R^{2}+r^{2})^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,824 |
16. Given an angle of magnitude $\alpha$ with vertex at $A$ and a point $B$ at distances $a$ and $b$ from the sides of the angle. Find the length of $A B$. | 16. If $M$ and $N$ are the bases of the perpendiculars dropped from $B$ to the sides of the angle, then the quadrilateral $A M B N$ is cyclic, and $A B$ is the diameter of the circle circumscribed around $\triangle M B N$, so $\widehat{M B N}=\pi-\alpha$ if $B$ is inside the given angle or its vertical, and $\widehat{M... | |AB|=\frac{\sqrt{^{2}+b^{2}+2\cos\alpha}}{\sin\alpha}, | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,825 |
17. Given the lengths $h_{a}$ and $h_{b}$ of the altitudes of triangle $ABC$ dropped from vertices $A$ and $B$, and the length $l$ of the angle bisector of angle $C$. Find the angle $\hat{C}$. | 17. Let $|A C|=b, \mid B C_{1}=a$. We have $b=\frac{h_{a}}{\sin \hat{C}}, a=$ $=\frac{h_{b}}{\sin \hat{C}}, l=\frac{2 a b \cos \frac{\hat{C}}{2}}{a+b}$ (see problem 17, section I). Therefore, $\boldsymbol{l}=\frac{2 h_{a} h_{b} \cos \frac{\hat{C}}{2}}{\sin \hat{C}\left(h_{a}+h_{b}\right)}$, from which $\sin \frac{\hat{... | \sin\frac{\hat{C}}{2}=\frac{h_{}h_{b}}{(h_{}+h_{b})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,826 |
18. A circle is circumscribed around a right-angled triangle. Another circle of the same radius touches the legs of this triangle, with one of the points of tangency being the vertex of the triangle. Find the ratio of the area of the triangle to the area of the common part of the two given circles. | 18. Let (Fig. 4) \(O_{1}\) be the center of the second circle. From the condition, its radius is equal to one of the legs; let the leg \(|C A|=R\); since \(R\) is the radius
Fig. 4. of the ... | \frac{3\sqrt{3}}{5\pi-3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,827 |
19. Circles with radii $R$ and $r$ touch each other internally. Find the side of a regular triangle, one vertex of which coincides with the point of tangency, and the other two lie on different given circles. | 19. Let the side of the triangle be $x$ and the sides emanating from the common point of the circles form angles $\alpha$ and $\beta$ with the line passing through the centers, where $\alpha \pm \beta=60^{\circ}$; then $\cos \alpha=\frac{x}{2 R}, \cos \beta=\frac{x}{2 r}$ (or vice versa). Finding $\sin \alpha$ and $\si... | \frac{Rr\sqrt{3}}{\sqrt{R^2+r^2-Rr}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,828 |
20. Two circles with radii $R$ and $r(R>r)$ are externally tangent at point $A$. A straight line through point $B$, taken on the larger circle, is tangent to the smaller circle at point $C$. Find $|B C|$, if $|A B|=a$. | 20. Draw the line $B A$ and denote by $D$ the second point of intersection with the smaller circle. Consider the arcs $A \bar{B}$ and $A \bar{D}$ (smaller than a semicircle). Since the common tangent to the circles at point $A$ forms equal angles with $A B$ and $A D$, the central angles corresponding to these arcs are ... | \sqrt{\frac{R+r}{R}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,829 |
22. The diagonals of quadrilateral $A B C D$ intersect at point $M$, the angle between them is $\alpha$. Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the centers of the circumcircles of triangles $A B M, B C M, C D M$, $D A M$ respectively. Determine the ratio of the areas of quadrilateral $A B C D$ and $\mathrm{O}_{1} \mathrm{... | 22. Note that $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ is a parallelogram with angles $\alpha$ and $\pi-\alpha\left(O_{1} O_{4} \perp A C\right.$ and $O_{2} O_{3} \perp A C$, hence, $O_{1} O_{4} O_{2} O_{3}$ and so on). If $K$ is the midpoint of $A M$ and $L$ is the midpoint of $M C$, then $\left|O... | 2\sin^{2}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,831 |
23. In a parallelogram with area $S$, the bisectors of its internal angles are drawn. The area of the quadrilateral formed by their intersection is $Q$. Find the ratio of the lengths of the sides of the parallelogram. | 23. Show that the bisectors of a parallelogram intersect to form a rectangle, the diagonals of which are parallel to the sides of the parallelogram and equal to the difference of the sides of the parallelogram. Consequently, if $a$ and $b$ are the sides of the parallelogram, $\alpha$ is the angle between them, then $S=... | \frac{S+Q+\sqrt{Q^{2}+2QS}}{S} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,832 |
24. In triangle $ABC$, a point $M$ is taken on side $AC$, and a point $N$ is taken on side $BC$. Segments $AN$ and $BM$ intersect at point $O$. Find the area of triangle $CMN$, if the areas of triangles $OMA$, $OAB$, and $OBM$ are $S_{1}$, $S_{2}$, and $S_{3}$, respectively. | 24. If $x$ is the area of triangle $O M N$, and $y$ is the area of triangle $C M N$, then
$\frac{|O N|}{|O A|}=\frac{x}{S_{1}}=\frac{S_{3}}{S_{2}}, \quad x=\frac{S_{1} S_{3}}{S_{2}}, \quad \frac{|A M|}{|M C|}=\frac{S_{1}+x}{y}=\frac{S_{1}+S_{2}}{S_{3}+x+y}$, from which $y=\frac{S_{1} S_{3}\left(S_{1}+S_{2}\right)\left... | \frac{S_{1}S_{3}(S_{1}+S_{2})(S_{3}+S_{2})}{S_{2}(S_{2}^{2}-S_{1}S_{3})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,833 |
25. The point of intersection of the medians of a right triangle lies on the circle inscribed in this triangle. Find the acute angles of the triangle. | 25. Let in triangle $ABC$ angle $C$ be right, $M$ be the point of intersection of the medians, $O$ be the center of the inscribed circle, $r$ be its radius, and $\widehat{CBA} = \alpha$; then
$$
\begin{aligned}
& |AB| = r \left( \operatorname{ctg} \frac{\alpha}{2} + \operatorname{ctg} \left( \frac{\pi}{4} - \frac{\alp... | \frac{\pi}{4}\\arccos\frac{4\sqrt{6}-3\sqrt{2}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,834 |
26. The circle inscribed in triangle $A B C$ divides the median $B M$ into three equal parts. Find the ratio of the sides $|B C|:|C A|:|A B|$. | 26. Let the segments of the median have length $a$. Denote by $x$ the smaller of the segments into which the side corresponding to the median is divided by the point of tangency. Now all sides can be expressed in terms of $a$ and $x$. The sides enclosing the median: $a \sqrt{2}+x$, $3 a \sqrt{2}+x$, the third side: $2 ... | 10:5:13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,835 |
27. In triangle $A B C$, a perpendicular line passing through the midpoint of side $A B$ intersects line $A C$ at point $M$, and a perpendicular line passing through the midpoint of $A C$ intersects line $A B$ at point $N$. It is known that $|M N|=|B C|$ and line $M N$ is perpendicular to line $B C$. Determine the angl... | 27. Let $|BC|=a, \hat{C}>\hat{B}, D$ and $E$ be the midpoints of $AB$ and $AC$. The quadrilateral $EMDN$ is cyclic (since $\widehat{MEN}=\widehat{MDN}=\left.=90^{\circ}\right), |MN|=a, |ED|=\frac{a}{2}, MN$ is the diameter of the circle circumscribed around $MEND$. Therefore, $\widehat{DME}=30^{\circ}, \widehat{CAB}=$ ... | \hat{A}=60,\hat{B}=15,\hat{C}=105 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,836 |
28. The area of trapezoid $ABCD$ is $S$, the ratio of the bases $|AD|:|BC|=3$; on the line intersecting
the extension of base $AD$ beyond point $D$, segment $EF$ is located such that $AE\|DF, BE\|CF$ and $|AE|:|DF|=$ $=|CF|:|BE|=2$. Determine the area of triangle EFD. | 28. Let $K$ and $M$ be the points of intersection of line $E F$ with $A D$ and $B C$. Suppose $M$ lies on the extension of $B C$ beyond point $B$. If $|A D|=3 a,|B C|=a$, then from the similarity of the corresponding triangles, it follows that $|D K|=|A D|=3 a,|M B|=|B C|=a$ (Fig. 5, a). Additionally, $|M E|=|E F|=|F K... | \frac{1}{12}or\frac{9}{20} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,837 |
29. Side $B C$ of triangle $A B C$ is equal to $a$, the radius of the inscribed circle is $r$. Find the area of the triangle if the inscribed circle touches the circle constructed on $B C$ as its diameter. | 29. Let $O$ be the center of the inscribed circle, $M$ the midpoint of $BC$, and $K, L, M$ the points of tangency of the inscribed circle with the sides $AC$, $AB$, and $BC$ of the triangle, respectively. Denote $|AK|=|AL|=x$, $|CK|=$ $=|CN|=y$, $|BL|=|BN|=z$, and $y+z=a$. By the condition, $|OM|=$ $=\frac{a}{2}-r$. Th... | \frac{^2r}{-r} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,838 |
30. Given an equilateral triangle $A B C$ with side $a, B D$ is its height. On $B D$, a second equilateral triangle $B D C_{1}$ is constructed, and on the height $B D_{1}$ of this triangle, a third equilateral triangle $B D_{1} C_{2}$ is constructed. Find the radius of the circumcircle of triangle $C C_{1} C_{2}$. Prov... | 30. We will prove that if $C_{1}$ and $C_{2}$ (Fig. 6) are on the other side of $B C$ from vertex $A$, then the center of the circle circumscribed around $\triangle C C_{1} C_{2}$ is at point $O$ on side $A B$, with $|B O|=\frac{1}{4}|A B|$. By drawing the altitude $C M$ from vertex $C$, we obtain that $B C_{1} C M$ is... | \frac{}{4}\sqrt{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,839 |
31. The sides of a parallelogram are equal to $a$ and $b(a \neq b)$. Through the vertices of the obtuse angles of this parallelogram, lines perpendicular to the sides are drawn. These lines intersect to form a parallelogram similar to the original one. Find the cosine of the acute angle of the given parallelogram. | 31. Consider two cases: 1 - when the bases of the perpendiculars lie on the sides of the parallelogram, and 2 - when one of the perpendiculars does not intersect the side to which it is dropped. In the 1 -st case, we arrive at a contradiction, and in the 2 -nd case, we get that $\cos \alpha=\frac{2 a b}{a^{2}+b^{2}}$. | \cos\alpha=\frac{2}{^{2}+b^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,840 |
32. In triangle $KLM$, the bisectors $KN$ and $LP$ intersect at point $Q$. Segment $PN$ has a length of 1, and vertex $M$ lies on the circle passing through points $N, P, Q$. Find the sides and angles of triangle $PNQ$. | 32. Expressing angle $P Q N$ in terms of the angles of the triangle and considering that $\widehat{P M N}+\widehat{P Q N}=180^{\circ}$, we find $\widehat{P M N}=60^{\circ}$, hence $\widehat{N P Q}=\widehat{Q M N}=$ $=30^{\circ}, \widehat{P N Q}=\widehat{P M Q}=30^{\circ}$, i.e., triangle $P Q N$ is isosceles with angle... | |PQ|=|QN|=\frac{1}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,841 |
33. On the diagonal $AC$ of the convex quadrilateral $ABCD$, there is the center of a circle of radius $r$, which is tangent to the sides $AB, AD$, and $BC$. On the diagonal $BD$, there is the center of a circle of the same radius $r$, which is tangent to the sides $BC, CD$, and $AD$. Find the area of the quadrilateral... | 33. From the condition, it follows that $A B C D$ is a trapezoid ($B C \| A D$), $A C$ is the bisector of angle $B A D$; therefore, $|A B|=|B C|$; similarly, $|B C|=|C D|$. Let $|A B|=|B C|=|C D|=a,|A D|=b$. The distance between the midpoints of the diagonals is $2 r$, hence, $\frac{b-a}{2}=2 r$. Draw the height $B M$ ... | 4r^{2}(\sqrt{2}+1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,842 |
34. The radius of the circumcircle of an acute-angled triangle $A B C$ is 1. It is known that the center of the circle passing through the vertices $A, C$ and the orthocenter of triangle $A B C$ lies on this circumcircle. Find the length of side $A C$. | 34. Let the angles $A, B$ and $C$ be denoted by $\alpha, \beta$ and $\gamma$. Let $H$ be the orthocenter, and $O$ be the center of the circle passing through $A$, $H$ and $C$. Then
$$
\begin{aligned}
& \widehat{H O C}=2 \widehat{H A C}=2\left(90^{\circ}-\gamma\right) \\
& \widehat{H O A}=2 \widehat{H C A}=2\left(90^{\... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,843 |
35. In triangle $ABC$, points $M, N$, and $P$ are taken: $M$ and $N$ are on sides $AC$ and $BC$, and $P$ is on segment $MN$, such that
$$
\frac{|AM|}{|MC|}=\frac{|CN|}{|NB|}=\frac{|MP|}{|PN|}
$$
Find the area of triangle $ABC$, if the areas of triangles $AMP$ and $BNP$ are $T$ and $Q$. | 35. Denoting the ratio $\frac{|A M|}{|M C|}=\lambda$, we will have $S_{M C P}=\frac{T}{\lambda}$, $S_{C P N}=\lambda Q, \frac{S_{M C P}}{S_{C P N}}=\lambda$, i.e., $\frac{T}{Q}=\lambda^{3}$,
$$
\begin{aligned}
S_{A B C} & =\frac{|A C|}{|M C|} \cdot \frac{|B C|}{|C N|} S_{C M N}=\frac{(\lambda+1)^{2}}{\lambda}\left(\fr... | (T^{1/3}+Q^{1/3})^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,844 |
37. In an isosceles triangle \(ABC (|AB| = |BC|)\), a perpendicular to \(AE\) is drawn through the endpoint \(E\) of the bisector \(AE\) until it intersects the extension of side \(AC\) at point \(F\) (with \(C\) between \(A\) and \(F\)). It is known that \(|AC| = 2m, |FC| = m/4\). Find the area of triangle \(ABC\). | 37. If $\widehat{B A C}=\widehat{B C A}=2 \alpha$, then by the Law of Sines we find
$$
|A E|=\frac{2 m \sin 2 \alpha}{\sin 3 \alpha},|A F|=\frac{|A E|}{\cos \alpha}=\frac{2 m \sin 2 \alpha}{\sin 3 \alpha \cos \alpha}
$$
Thus, $\frac{9}{4} m=\frac{2 m \sin 2 \alpha}{\sin 3 \alpha \cos \alpha}$, from which $\cos 2 \alp... | \frac{5^{2}\sqrt{11}}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,846 |
39. From point $K$, located outside a circle with center $O$, two tangents $K M$ and $K N$ are drawn to this circle ($M$ and $N$ are the points of tangency). A point $C(|M C|<|C N|)$ is taken on the chord $M N$. A line perpendicular to segment $O C$ is drawn through point $C$, intersecting segment $N K$ at point $B$. I... | 39. Let $A$ be the point of intersection of the lines $B C$ and $K M$. The quadrilateral $O N B C$ is inscribed $\left(\widehat{(O C B}=\widehat{O N B}=90^{\circ}\right)$, therefore, $\widehat{O B C}=\widehat{O N C}=\frac{\alpha}{2}$. Similarly, the quadrilateral $C M A O$ is inscribed and $\widehat{C A O}=\widehat{C M... | \frac{\sqrt{R^{2}+b^{2}-2Rb\cos\frac{\alpha}{2}}}{\sin\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,847 |
40. Pentagon $A B C D E$ is inscribed in a circle. Points $M, Q, N$ and $P$ are the feet of the perpendiculars dropped from vertex $E$ to sides $A B, B C, C D$ (or their extensions) and diagonal $A D$, respectively. It is known that $|E P|=d$, and the ratio of the area of triangle $M Q E$ to the area of triangle $P N E... | 40. Points $E, M, B$ and $Q$ (Fig. 7) lie on the same circle
Fig. 7. with diameter $B E$, and points $E, P$, $D$ and $N$ lie on the circle with diameter $E D$.
Thus, $\widehat{E M Q}=\wide... | \sqrt{k} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,848 |
41. Given a rectangular trapezoid. It is known that a certain line, parallel to the bases, divides it into two trapezoids, each of which can have a circle inscribed in it. Determine the bases of the original trapezoid if its lateral sides are equal to $c$ and $d(d>c)$. | 41. Extend the non-parallel sides of the trapezoid until they intersect, we will get three similar triangles, and the similarity coefficient between the middle and the larger triangle and between the smaller and the middle one is the same. Let this coefficient be $\lambda$, the larger base - $x$, the radius of the larg... | \frac{-\sqrt{^{2}-^{2}}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,849 |
42. On the lateral sides $K L$ and $M N$ of an isosceles trapezoid $K L M N$, points $P$ and $Q$ are chosen respectively such that the segment $P Q$ is parallel to the bases of the trapezoid. It is known that a circle can be inscribed in each of the trapezoids $K P Q N$ and $P L M Q$, and the radii of these circles are... | 42. Drop perpendiculars from the centers of the circles to one of the lateral sides and draw a line through the center of the smaller circle parallel to this side. This will form a right triangle with the hypotenuse \( R + r \), one leg \( R - r \), and an acute angle at this leg \(\alpha\), which is equal to the acute... | 2R\sqrt{\frac{R}{r}},2r\sqrt{\frac{r}{R}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,850 |
43. In triangle $ABC$, all sides of which are different, the bisector of angle $A$ intersects side $BC$ at point $D$. It is known that $|AB|-|BD|=a, |AC|+$ $+|CD|=b$. Find $|AD|$. | 43. Let's take a point $K$ on side $A B$ such that $|B K| \neq |B D|$, and a point $E$ on the extension of $A C$ such that $|C E|=|C D|$. We claim that $\triangle A D K$ is similar to $\triangle A D E$. If $\hat{A}, \hat{B}$, and $\hat{C}$ are the measures of the angles of $\triangle A B C$, then
$$
\begin{aligned}
& ... | |AD|=\sqrt{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,851 |
44. Using the result of the previous problem, prove that the square of the length of the bisector of a triangle is equal to the product of the lengths of the sides enclosing it, minus the product of the segments of the third side, into which it is divided by the bisector. | 44. Using the notations from the previous problem
$$
\begin{aligned}
|A D|^{2} & =(|A C|+|C D|)(|A B|-|B D|)= \\
& =|A C| \cdot|A B|-|C D| \cdot|B D|+(|A B| \cdot|C D|-|A C| \cdot|B D|) .
\end{aligned}
$$
But the term in parentheses is zero, since (see problem 9, section I) $\frac{|A B|}{|A C|}=\frac{|B D|}{|C D|}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 24,852 |
45. Given a circle with diameter $A B$. A second circle with center at $A$ intersects the first circle at points $C$ and $D$ and the diameter at point $E$. On the arc $C E$, not containing point $D$, a point $M$ is taken, different from points $C$ and $E$. The ray $B M$ intersects the first circle at point $N$. It is k... | 45. Extend $B N$ and $C N$ to intersect the second circle again at points $K$ and $L$ respectively. $|M N|=|N K|$, since $\widehat{A N B}=90^{\circ}$ and $M K$ is a chord of the circle with center at $A$. $\widehat{L N K}=\widehat{B N C}=\widehat{B N D}$ (since the corresponding arcs are equal). Therefore, $|L N|=|N D|... | \sqrt{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,853 |
46. In triangle $ABC$, angle $\hat{B}$ is $\pi / 4$, and angle $\hat{C}$ is $\pi / 6$. Circles are constructed on medians $BM$ and $CN$ as diameters, intersecting at points $P$ and $Q$. Chord $PQ$ intersects side $BC$ at point $D$. Find the ratio $|BC|:|DC|$. | 46. Note (Fig. 8) that \( P Q \perp C B \). Let \( T \) be the intersection point of \( M N \) and \( P Q \), and \( L \) and \( K \) be the feet of the perpendiculars dropped from \( C \) and \( B \) to the line \( M N \) ( \( L \) and \( K \) lie on the circles constructed with \( C N \) and \( B M \) as diameters). ... | \frac{|BD|}{|DC|}=\frac{1}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,854 |
47. Let $A B$ be the diameter of a circle, $O$ its center, $|A B|=2 R, C$ a point on the circle, $M$ a point on $A C$. From $M$, a perpendicular $M N$ is dropped onto $A B$ and a perpendicular to $A C$ is erected, intersecting the circle at point $L$ (segment $C L$ intersects $A B$). Find the distance between the midpo... | 47. Let (Fig. 9) $\breve{B C}=2 \alpha, \breve{B L}=2 \beta$. Then
$$
|A C|=2 R \cos \alpha, \quad|C L|=2 R \sin (\alpha+\beta)
$$
$|C M|=|C L| \cos \left(90^{\circ}-\beta\right)=2 R \sin (\alpha+\beta) \sin \beta$,
$$
\begin{aligned}
A M \mid & =|A C \cdot|-|C M|=2 R[\cos \alpha-\sin (\alpha+\beta) \sin \beta]= \\
... | \sqrt{\frac{R^{2}}{4}+R} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,855 |
48. From the vertex $A$ of triangle $A B C$, perpendiculars $A M$ and $A N$ are dropped to the bisectors of the external angles at $B$ and $C$ of the triangle. Prove that the length of the segment $M N$ is equal to the semiperimeter of triangle $A B C$. | 48. Let the lines $A M$ and $A N$ intersect the line $B C$ at points $D$ and $E$. It is easy to see that triangles $A B D$ and $A C E$ are isosceles (the bisector is also the altitude), i.e., $|D E|$ equals the perimeter of triangle $A B C$, and $M N$ is the midline in triangle $A D E$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 24,856 |
49. Three circles pass through two given points on the plane each. Let $O_{1}, O_{2}, O_{3}$ be their centers. A line passing through one of the points, common to all three circles, intersects them again at points $A_{1}, A_{2}, A_{3}$. Prove that $\frac{\left|A_{1} A_{2}\right|}{\left|A_{2} A_{3}\right|}=$ $=\frac{\le... | 49. Let one of the intersection points through which the line passes be denoted as $C$. Let $B_{1}, B_{2}, B_{3}$ be the feet of the perpendiculars dropped from $O_{1}, O_{2}, O_{3}$ to the line, and let $K$ and $M$ be the points of intersection of the lines parallel to $A_{1} A_{3}$, passing through $O_{1}$ and $O_{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 24,857 |
50. Given a triangle $ABC$. The tangent to the circumcircle of this triangle at point $B$ intersects the line $AC$ at point $M$. Find the ratio $|AM|:|MC|$, if $|AB|:|BC|=k$. | 50. We have
$$
\begin{aligned}
\frac{|M A|}{|M C|} & =\frac{S_{A B M}}{S_{M B C}}=\frac{\frac{1}{2}|M B| \cdot |R A| \sin \widehat{M B A}}{\frac{1}{2}|M B| \cdot|B C| \sin \widehat{M B C}}= \\
& =\frac{|B A|^{2}}{|B C|^{2}} \frac{|B C|}{\sin \widehat{M B C}} \frac{\sin \widehat{M B A}}{|B A|}
\end{aligned}
$$
But $\s... | k^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,858 |
51. On a straight line, points $A, B, C$, and $D$ are sequentially arranged, such that $|A C|=\alpha|A B|, \quad|A D|=\beta|A B|$. A circle is drawn through $A$ and $B$, and $C M$ and $D N$ are two tangents to this circle ( $M$ and $N$ are points on the circle, lying on opposite sides of the line $A B$ ). In what ratio... | 51. From problem 50, it follows that \(\frac{|A M\rangle^{2}}{|M B|^{2}}=\frac{|A C|}{|B C|}, \frac{|A N|^{2}}{|N B|^{2}}=\frac{|A D|}{|B D|^{2}}\). If \(K\) is the intersection point of \(M N\) and \(A B\), then
\[
\begin{aligned}
\frac{|A K|}{|K B|} & =\frac{S_{A M N}}{S_{B M N}}=\frac{|A M| \cdot|A N| \sin \widehat... | \sqrt{\frac{\alpha\beta}{(\alpha-1)(\beta-1)}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,859 |
52. $A B C D$ is a circumscribed quadrilateral, the lengths of the segments from $A$ to the points of tangency are $a$, the lengths of the segments from $C$ to the points of tangency are $b$. In what ratio does the diagonal $A C$ divide the diagonal $B D$? | 52. Let $K, L, M$ and $N$ be the points of tangency of the sides $A B, B C, C D$ and $D A$ with the circle. Denote by $P$ the intersection point of $A C$ and $K M$. If $\widehat{A K M}=\varphi$, then $\widehat{K M C}=180^{\circ}-\varphi$. Therefore,
$\frac{|A P|}{|P C|}=\frac{S_{A K M}}{S_{K M C}}=\frac{\frac{1}{2}|A ... | \frac{}{b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,860 |
53. Point $K$ lies on the base $AD$ of trapezoid $ABCD$, such that $|AK|=\lambda|AD|$. Find the ratio $|AM|:|AD|$, where $M$ is the point of intersection with $AD$ of the line passing through the points of intersection of the lines $AB$ and $CD$ and the lines $BK$ and $AC$.
Taking $\lambda=1 / n, n=1,2,3, \ldots$, obt... | 53. Let $P$ and $Q$ be the points of intersection of $B K$ and $A C$, $A B$ and $D C$. The line $Q P$ intersects $A D$ at point $M$ and $B C$ at point $N$. Using the similarity of corresponding triangles, we can write the following equalities:
$$
\frac{|A M|}{|M D|}=\frac{|B N|}{|N C|}=\frac{|M K|}{|A M|}=\frac{|A K|-... | \frac{|AM|}{|AD|}=\frac{\lambda}{\lambda+1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,861 |
54. In a right triangle $ABC$ with hypotenuse $|AB|=c$, a circle is constructed with the height of the triangle $CD$ as its diameter. Tangents to this circle, passing through points $A$ and $B$, touch it at points $M$ and $N$ and intersect when extended at point $K$. Find $|MK|$. | 54. Let $|K M|=$, $K N|=x| A D|=$, $y| D B|=$, $z$. Then $|C D|=$ $=\sqrt{y z}, y+z=c$. The radius of the inscribed circle in $\triangle A K B$ is $\frac{1}{2}|C D|=\frac{1}{2} \sqrt{y z}$. Express the area of triangle $A K B$ using Heron's formula and $s=p r$. We get the equation
$$
\sqrt{(x+y+z) x y z}=(x+y+z) \frac... | \frac{}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,862 |
55. On the sides $A B, B C$ and $C A$ of triangle $A B C$, points $C_{1}, A_{1}$ and $B_{1}$ are taken such that $\left|A C_{1}\right|:\left|C_{1} B\right|=$ $=\left|B A_{1}\right|:\left|A_{1} C\right|=\left|C B_{1}\right|:\left|B_{1} A\right|=k$. On the sides $A_{1} B_{1}$, $B_{1} C_{1}$ and $C_{1} A_{1}$, points $A_{... | 55. Draw a line through $A_{2}$ parallel to $A C$. Let $R$ be the point of intersection of this line with $A B$. From the fact that $\frac{|A R|}{|R C_{1}|} = \frac{|B_{1} A_{2}|}{|A_{2} C_{1}|} = \frac{1}{k}$, and $\frac{|A C_{1}|}{|C_{1} B|} = k$, we find that $\frac{|A R|}{|A B|} = \frac{k}{(k+1)^{2}}$. Similarly, b... | \frac{k^{2}-k+1}{(k+1)^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 24,863 |
56. In triangle $ABC$, $R$ and $r$ are the radii of the circumcircle and the incircle, respectively. Let $A_{1}, B_{1}$, $C_{1}$ be the points of intersection of the angle bisectors of triangle $ABC$ with the circumcircle. Find the ratio of the areas of triangles $ABC$ and $A_{1} B_{1} C_{1}$. | 56. Let's use the following formula for the area of a triangle: $S=2 R^{2} \sin \hat{A} \sin \hat{B} \sin \hat{C}$, where $\hat{A}, \hat{B}$, and $\hat{C}$ are the angles of the triangle. Then the area of triangle $A_{1} B_{1} C_{1}$, where $A_{1}$, $B_{1}$, and $C_{1}$ are the points of intersection of the angle bisec... | \frac{2r}{R} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,864 |
57. There are two triangles with respectively parallel sides and areas $S_{1}$ and $S_{2}$, where one of them is inscribed in triangle $A B C$, and the other
is circumscribed around it. Find the area of triangle $A B C$. | 57. Let $O$ be the center of similarity of the inscribed and circumscribed triangles, $M_{1}$ and $M_{2}$ be two corresponding vertices ( $M_{1}$ lies on side $A B$ ), and segment $O A$ intersects the inscribed triangle at point $K$. Then $S_{O M_{1} K}=\lambda S_{1}, S_{O M_{2} A}=\lambda S_{2}, \frac{S_{O M_{1} A}}{S... | S_{ABC}=\sqrt{S_{1}S_{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,865 |
58. Determine the measure of angle $\hat{A}$ of triangle $ABC$, if it is known that the bisector of this angle is perpendicular to the line passing through the point of intersection of the altitudes and the center of the circumscribed circle of this triangle. | 58. Let $O$ be the center of the circumscribed circle, $H$ the intersection point of the altitudes of $\triangle ABC$. Since the line $OH$ is perpendicular to the bisector of angle $A$, it intersects sides $AB$ and $AC$ at points $K$ and $M$ such that $|AK|=|AM|$. Thus, $|AO|=|OB|$ and $\widehat{AOB}=2 \hat{C}$ (assumi... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,866 |
60. Given $\triangle A B C$. On the ray $B A$, take a point $D$ such that $|B D|=|B A|+|A C|$. Let $K$ and $M$ be two points on the rays $B A$ and $B C$ respectively such that the area of $\triangle B D M$ is equal to the area of $\triangle B C K$. Find $\widehat{B K M}$, if $\widehat{B A C}=\alpha$. | 60. The condition $S_{\triangle B D M}=S_{\triangle B C K}$ means that
or
$$
|B D| \cdot |B M| = |B K| \cdot |B C|
$$
$$
(|B A| + |A C|) |B M| = |B K| \cdot |B C| .
$$
Draw a line through $M$ parallel to $A C$; let $L$ be the point of intersection of this line with $B A$. We will prove that $|L M| = |K L|$; from th... | \alpha/2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,868 |
61. In trapezoid $A B C D$, the lateral side $A B$ is perpendicular to $A D$ and $B C$, and $|A B|=\sqrt{|A D| \cdot|B C|}$. Let $E$ be the point of intersection of the non-parallel sides of the trapezoid, $O$ be the point of intersection of the diagonals, and $M$ be the midpoint of $A B$. Find $\widehat{E O M}$. | 61. Let $|A D|=a,|B C|=b$. Drop a perpendicular $O K$ from $O$ to $A B$. Now we can find $|B K|=\sqrt{\overline{a b}} \frac{b}{b+a},|B E|=\sqrt{\overline{a b}} \frac{b}{a-b}$, $|M K|=\frac{\sqrt{a b}}{2}-\sqrt{\overline{a b}} \frac{b}{b+a}=\sqrt{\overline{a b}} \frac{a-b}{2(a+b)},|E K|=|B E|+|B K|=$ $=\sqrt{a \bar{b}} ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,869 |
62. On a plane, there are two lines intersecting at point $O$, and two points $A$ and $B$. Denote the bases of the perpendiculars dropped from $A$ to the given lines as $M$ and $N$, and the bases of the perpendiculars dropped from $B$ as $K$ and $L$. Find the angle between the lines $M N$ and $K L$, if $\widehat{A O B}... | 62. Note that points $A, M, N$ and $O$ lie on the same circle (Fig. 11). Therefore, $\widehat{N M O}=\widehat{O A N}=90^{\circ}-\widehat{A O N}$. This means that when $O A$ is rotated around $O$
 the angle bisector of $\angle A$ and extend $B M$ until it intersects with the bisector at point N. Since $|B N|=|N C|$, then $\widehat{B N C}=120^{\circ} ;$ therefore, the angles $\widehat{B N A}, \widehat{C N A}$ are also $120^{\circ}$ each, $\widehat{N C A}=\widehat{N C M}=20^{\circ}, \quad$... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,874 |
67. In triangle $ABC$, given are $\widehat{ABC}=100^{\circ}, \widehat{ACB}=$ $=65^{\circ}$. On $AB$, a point $M$ is taken such that $\widehat{MCB}=55^{\circ}$, and on $AC$ - a point $N$ such that $\widehat{NBC}=80^{\circ}$. Find $\widehat{NMC}$. | 67. Describe a circle around $\triangle M C B$ (Fig. 13) and extend $B N$ until it intersects with the circle
Fig. 12. at point $M_{1} ;\left|C M_{1}\right|=|C M|$, since the angles subtend... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,875 |
68. In triangle $ABC$, it is given that $|AB|=|BC|, \widehat{ABC}=$ $=20^{\circ}$; on $AB$ a point $M$ is taken such that $\widehat{MCA}=60^{\circ}$; on side $CB$ - a point $N$ such that $\widehat{NAC}=50^{\circ}$. Find $\widehat{NMA}$. | 68. Let's take a point $K$ on $BC$ (Fig. 14) such that $\widehat{K A C}=60^{\circ}$, $M K \| A C$. Let $L$ be the intersection point of $A K$ and $M C$; $\triangle A L C$ is equilateral, $\triangle A N C$ is isosceles (calculate the angles), so $\triangle L N C$ is also isosceles, $\widehat{L C N}=20^{\circ}$. Now let'... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,876 |
69. In triangle $ABC$, given are $\widehat{ABC}=70^{\circ}, \widehat{ACB}=$ $=50^{\circ}$. On $AB$, a point $M$ is taken such that $\widehat{MCB}=40^{\circ}$, and on $AC$-point $N$ such that $\widehat{NBC}=50^{\circ}$. Find $\widehat{NMC}$. | 69. Let's take point $K$ (Fig. 15) such that $\widehat{K B C}=\widehat{K C B}=30^{\circ}$, and denote by $L$ the intersection point of the lines $M C$ and $B K$.
Fig. 15. Since $\triangle ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,877 |
70. Let $M$ and $N$ be the points of tangency of the inscribed circle with the sides $B C$ and $B A$ of triangle $A B C$, and let $K$ be the point of intersection of the angle bisector of angle $A$ with the line $M N$. Prove that $\widehat{A K C}=50^{\circ}$. | 70. Let $O$ be the center of the inscribed circle; points $C, O, K$ and $M$ lie on the same circle: $\left(\widehat{C O K}=\frac{\hat{A}}{2}+\frac{\hat{C}}{2}=90^{\circ}-\frac{\hat{B}}{2}=\right.$ $=\widehat{K M B}=180^{\circ}-\widehat{K M C}$; if point $K$ is on the extension of $N M$, then $\widehat{C O K}=\widehat{C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 24,878 |
71. In a convex hexagon $A B C D E F$, where $|A B|=|B C|,|C D|=|D E|,|E F|=|F A|$, the angles $\hat{B}=\alpha, \hat{D}=\beta, \hat{F}=\gamma$ are known. Determine the angles of triangle $B D F$, if $\alpha+\beta+\gamma=2 \pi$. | 71. Let $K$ be the point of intersection of the circle centered at $B$ with radius $|A B|$ and the circle centered at $F$ with radius $|A F|$. Then $\widehat{A K C}=180^{\circ}-\frac{\alpha}{2}, \widehat{A K E}=180^{\circ}-\frac{\gamma}{2}$; considering the condition, we get that $\widehat{C K E}=180^{\circ}-\frac{\bet... | \frac{\alpha}{2},\frac{\beta}{2},\frac{\gamma}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,879 |
72. Let $P$ and $Q$ be two distinct points on the circumcircle of triangle $ABC$ such that $|PA|^2 = |PB| \cdot |PC|$, $|QA|^2 = |QB| \cdot |QC|$ (one of the points is on the arc $\breve{AB}$, the other is on the arc $\widehat{AC}$). Find the difference $\widehat{PAB} - \widehat{QAC}$, if the difference of the angles $... | 72. If $P$ lies on the arc $\overrightarrow{A B}$, and $Q$ lies on the arc $\overrightarrow{A C}$, then for the angles $\widehat{P A B}=\dot{\varphi}, \widehat{Q A C}=\psi$ we obtain two relations:
$$
\begin{aligned}
&\left\{\begin{aligned}
\sin ^{2}(\hat{C}-\varphi) & =\sin \varphi \sin (\hat{B}+\hat{C}-\varphi) \\
\... | \frac{\pi-\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,880 |
73. On a given circle, two fixed points $A$ and $B$ are taken, $\overrightarrow{A B}=\alpha$. A random circle passes through points $A$ and $B$. A line $l$ is also drawn through $A$, intersecting the circles again at points $C$ and $D$ ($C$ is on the given circle). The tangents to the circles at points $C$ and $D$ (whe... | 73. We will prove (Fig. 16) that $\triangle C M N$ is similar to $\triangle C A B$. We have $\widehat{M C N}=\widehat{C B A}$. Since the quadrilateral CBDM is inscribed, then
$\frac{|C M|}{|C B|}=\frac{\sin \widehat{C B M}}{\sin \widehat{C M B}}=\frac{\sin \widehat{C D M}}{\sin \widehat{C D B}}=$ $=\frac{\sin \widehat... | \frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,881 |
74. Prove that if one angle in a triangle is equal to $120^{\circ}$, then the triangle formed by the bases of its angle bisectors is a right triangle. | 74. Let $\widehat{A B C}=120^{\circ}, B D, A E, C M$-angle bisectors of $\triangle A B C$. We will show that $D E$ is the bisector of angle $B D C$, and $D M$ is the bisector of angle $B D A$. For this, it is sufficient to show that
$$
\frac{|B E|}{|E C|}=\frac{|B D|}{|D C|}
$$
(similarly for point $M$). But this fol... | proof | Geometry | proof | Yes | Yes | olympiads | false | 24,882 |
75. In quadrilateral $A B C D$, it is given that $\widehat{D A B}=150^{\circ}$, $\widehat{D A C}+\widehat{A B D}=120^{\circ}, \widehat{D B C}-\widehat{A B \bar{D}}=60^{\circ}$. Find $\widehat{B D C}$. | 75. Let $\widehat{A B D}=\alpha, \widehat{B D C}=\varphi$. By the condition, $\widehat{D A C}=$ $=120^{\circ}-\alpha, \widehat{B A C}=30^{\circ}+\alpha, \widehat{A D B}=30^{\circ}-\alpha, \widehat{D B C}=60^{\circ}+\alpha$. Using the Law of Sines for triangles $A B C, B C D, A C D$, we get
$$
\begin{aligned}
& \frac{|... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,883 |
76. On the side $C B$ of triangle $A B C$, a point $D$ is taken such that $|C D|=\alpha|A C|$. The radius of the circumcircle of $\triangle A B C$ is $R$. Find the distance between the center of the circumcircle of $\triangle A B C$ and the center of the circumcircle of $\triangle A D B$. | 76. Prove that if $O$ and $O_{1}$ are the centers of the circumcircles of $\triangle A B C$ and $\triangle A D B$, then $\triangle A O O_{1}$ is similar to $\triangle A C D$.
Answer: $\alpha R$. | \alphaR | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,884 |
77. A circle is circumscribed around a right triangle $A B C$ ( $\widehat{C}==90^{\circ}$ ). Let $C D$ be the altitude of the triangle. A circle centered at $D$ passes through the midpoint of the arc $\widetilde{A B}$ and intersects $A B$ at point $M$. Find $|C M|$, if $|A B|=c$. | 77. If $K$ is the midpoint of $\breve{A B}$, and $O$ is the center of the circle, $|A B|=2 R=c$, then
$$
\begin{aligned}
& |C M|^{2}=|C D|^{2}+|D M|^{2}=|C D|^{2}+|D K|^{2}= \\
& =|A D| \cdot|D B|+R^{2}+|D O|^{2}=(|O A|-|D O|)(|O B|+|D O|)+R^{2}+ \\
& \quad+|D O|^{2}=(R-|D O|)(R+|D O|)+R^{2}+|D O|^{2}=2 R^{2}
\end{ali... | \frac{}{2}\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,885 |
78. Find the perimeter of triangle $ABC$, if $|BC|=$ $=a$ and the segment of the line tangent to the inscribed circle and parallel to $BC$, contained within the triangle, is equal to $b$. | 78. Let $KM$ be a segment parallel to $BC$, and $N$ and $L$ be the points of tangency of the inscribed circle with sides $AC$ and $BC$. As is known (see problem 18, section I), $|AN|=|AL|=p-a$, where $p$ is the semiperimeter of $\triangle ABC$. On the other hand, $|AN|=|AL|$ is the semiperimeter of $\triangle AKM$, sim... | \frac{2a^2}{-b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,886 |
79. In a triangle, three lines parallel to its sides and tangent to the inscribed circle have been drawn. They cut off three triangles from the given one. The radii of the circumcircles of these triangles are $R_{1}, R_{2}, R_{3}$. Find the radius of the circumcircle of the given triangle. | 79. Prove that if $a, b, c$ are the lengths of the sides of a triangle, then the perimeters of the cut-off triangles will be $2(p-a), 2(p-b)$, $2(p-c)$. Consequently, if $R$ is the radius of the circumscribed circle, then $R_{1}+R_{2}+R_{3}=\left(\frac{p-a}{p}+\frac{p-b}{p}+\frac{p-c}{p}\right) R=R$.
Answer: $R=R_{1}+... | R_{1}+R_{2}+R_{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,887 |
80. In a circle of radius $R$, two chords $A B$ and $A C$ are drawn. On $A B$ or its extension, a point $M$ is taken, the distance from which to the line $A C$ is equal to $|A C|$. Similarly, on $A C$ or its extension, a point $N$ is taken, the distance from which to the line $A B$ is equal to $|A B|$. Find $|M N|$. | 80. If $\widehat{B A C}=\alpha$, then $|A M|=\frac{|A C|}{\sin \alpha},|A N|=\frac{|A B|}{\sin \alpha}$, i.e.
Fig. 17. $|A M|:|A N|=|A C|:|A B| ; \quad$ thus, $\triangle A M N$ is similar ... | 2R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,888 |
81. Given a circle of radius $R$ with center $O$. Two other circles touch the given one internally and intersect at points $A$ and $B$. Find the sum of the radii of the two latter circles, given that $\widehat{O A B}=90^{\circ}$. | 81. Let (Fig. 17) $O_{1}$ and $O_{2}$ be the centers of intersecting circles. Denote their radii by $x$ and $y$, $|O A|=a$. Since triangles $A O O_{1}$ and $A O O_{2}$, as follows from the condition, are equal in area, expressing their areas using Heron's formula, considering that $\left|O_{1} A\right|=x$, $\left|0 O_{... | x+R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,889 |
82. In a circle of radius $R$, two intersecting perpendicular chords are drawn.
a) Find the sum of the squares of the four segments of these chords, into which they are divided by the point of intersection.
b) Find the sum of the squares of the lengths of the chords, if the distance from the center of the circle to t... | 82. Let $AB$ and $CD$ be given chords, and $M$ be their point of intersection.
a) The arcs $\overline{AC}$ and $\overline{BD}$ together form a semicircle; therefore, $|AC|^{2} + |BD|^{2} = 4R^{2}$; thus,
$|AM|^{2} + |MC|^{2} + |MB|^{2} + |MD|^{2} = |AC|^{2} + |BD|^{2} = 4R^{2}$,
b) $|AB|^{2} + |CD|^{2} = (|AM| + |MB... | 4R^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,890 |
83. Given two concentric circles with radii $r$ and $R (r < R)$. Through a point $P$ on the smaller circle, a line is drawn intersecting the larger circle at points $B$ and $C$. The perpendicular to $BC$ at point $P$ intersects the smaller circle at point $A$. Find $|PA|^2 + |PB|^2 + |PC|^2$. | 83. If $M$ is the second intersection point of $B C$ with the smaller circle, then $|B M|=|P C|(M-$ between $B$ and $P), |B P|=|M P|+|B M|$, $|B P| \cdot|P C|=R^{2}-r^{2},\left.|M P+| P A\right|^{2}=4 r^{2}, \quad|P A|^{2}+|P B|^{2}+$ $+|P C|^{2}=|P A|^{2}+(|P B|-|P C|)^{2}+2|P B| \cdot|P C|=\left(|P A|^{2}+|M P|^{2}\r... | 2(R^2+r^2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,891 |
84. In a semicircle, two intersecting chords are drawn from the ends of the diameter. Prove that the sum of the products of the segment of each chord, adjacent to the diameter, by the entire chord is equal to the square of the diameter. | 84. Let us denote (Fig. 18) the lengths of the chord segments as shown in the figure, the diameter by $2 r$; using the fact that angles subtended by the diameter are right angles, and $x y=u v$, we get
$$
\begin{aligned}
x(x+y)+u(u+v)=x^{2}+x y & +u v+u^{2}= \\
& =(u+v)^{2}+x^{2}-v^{2}=(u+v)^{2}+m^{2}=4 r^{2}
\end{ali... | 4r^{2} | Geometry | proof | Yes | Yes | olympiads | false | 24,892 |
85. Let $a, b, c$ and $d$ be the lengths of the sides of a cyclic quadrilateral (with $a$ and $c$ being opposite sides), and $h_{a}, h_{b}, h_{c}$ and $h_{d}$ be the distances from the center of the circumscribed circle to the respective sides. Prove that if the center of the circle is inside the quadrilateral, then $a... | 85. If $\alpha, \beta, \gamma, \delta$ are the arcs corresponding to the sides $a, b, c$, and $d$, then the equality to be proven corresponds to the trigonometric identity $\sin \frac{\alpha}{2} \cos \frac{\gamma}{2}+\cos \frac{\alpha}{2} \sin \frac{\gamma}{2}=\sin \frac{\beta}{2} \cos \frac{\delta}{2}+\cos \frac{\beta... | proof | Geometry | proof | Yes | Yes | olympiads | false | 24,893 |
86. The opposite sides of a quadrilateral inscribed in a circle intersect at points $P$ and $Q$. Find the length of the segment $|P Q|$, if the tangents to the circle drawn from $P$ and $Q$ are equal to $a$ and $b$. | 86. Let (Fig. 19) $ABCD$ be a cyclic quadrilateral. Describe a circle around $\triangle ADP$. Denote by $M$ the point of intersection of this circle with the line $PQ$. We have $\widehat{DMQ}=\widehat{DAP}=\widehat{BCD}$. Therefore, the quadrilateral $CDMQ$ is cyclic. Since, by the condition, the tangents drawn from $P... | |PQ|=\sqrt{^{2}+b^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,894 |
88. Quadrilateral $ABCD$ is circumscribed around a circle. The point of tangency of the circle with side $AB$ divides this side into segments $a$ and $b$, and the point of tangency of the circle with side $AD$ divides it into segments $a$ and $c$. Within what limits can the radius of the circle vary? | 88. The radius of the inscribed circle is bounded by the magnitudes of the radii of two limiting cases. It cannot be less than
Fig. 20. the radius of the circle inscribed in a triangle with... | \sqrt{\frac{}{+b+}}<r<\sqrt{++} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 24,896 |
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