problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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397. A tangent is drawn to a circle at the end $M$ of its diameter $MN$. On this tangent, pairs of segments $MA$ and $MB$ are laid out on opposite sides of point $M$, the product of which is constant. Lines passing through point $N$ and points $A$ and $B$ intersect the circle again at points $C$ and $D$. Prove that the... | 397. Establish the similarity of triangles $A M N$ and $M C N$, $B M N$ and $M D N$, and using the formula for the area of a triangle, show that $\frac{M N}{A M}=\frac{C M}{C N}$ and $\frac{M N}{B M}=\frac{M D}{N D}$ (Fig. 49), from which
$$
\frac{M N^{2}}{A M \cdot B M}=\frac{C M \cdot M D}{C N \cdot N D}=\frac{(C M ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,973 |
400. The common internal tangents of two circles determine four points of tangency, which are the vertices of a quadrilateral circumscribed about a circle. Prove that the distance between the centers of the circles is expressed in terms of their radii by the formula:
$$
d^{2}=(R+r)^{2}+4 R r
$$ | 400. Let the points of tangency be denoted as $A, B, C$ and $D$, the radii of the circles $O$ and $O_{1}$ as $R$ and $r$, the distance $O O_{1}$ between their centers as $d$, the segments of the internal tangents $A C$ and $B D$ as $t$, the chords $A B$ and $C D$ of the circles as $2 x$ and $2 y$, and the segments $A D... | ^{2}=(R+r)^{2}+4Rr | Geometry | proof | Yes | Yes | olympiads | false | 43,976 |
401. Through a point $A$ lying outside a circle, two tangents $A T_{1}$ and $A T_{2}$ are drawn to it. From an arbitrary point $M$ on the circle, three rays $M M_{1}, M T_{1}, M T_{2}$ are drawn, forming with the tangent at point $M$ to the circle the angles $\varepsilon, \varphi, \psi$ (where $M_{1}$ is the second poi... | 401. Drop perpendiculars $h_{1}$ and $h_{2}$ from point $M$ to the tangents $A T_{1}$ and $A T_{2}$ (Fig. 51) and, establishing that $\alpha=\varepsilon-2 \varphi$ and $\beta=$ $=2 \psi-\varepsilon-180^{\circ}$, express $h_{1}$ and $h_{2}$ from the resulting triangles:
$h_{1}=A M \cdot \sin \alpha=A M \cdot \sin (\var... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,977 |
403. Given a segment $AB$ equal to $a$. Two circles of radius $a$ are drawn with centers at points $A$ and $B$, intersecting at point $C$. A circle is inscribed in the curvilinear triangle $ABC$, touching the segment $AB$ at point $M$, and the arc $AC$ at point $N$. Prove that the radius of the circle inscribed in the ... | 403. Establish that points $N, O$, and $B$ (Fig. 52) lie on the same straight line ($O$ is the center of the circle of radius $R$ inscribed in the curvilinear triangle $A B C$), and show that the power of point $B$ with respect to the circle $O$ is $\frac{a^{2}}{4}$ or $a(a-2 R)$ and $R=\frac{3}{8} a$. Point $M$ is the... | \frac{27-12\sqrt{2}}{98} | Geometry | proof | Yes | Yes | olympiads | false | 43,979 |
404. Given a triangle $ABC$ and a point $M$ not lying on its sides or their extensions. Tangents are drawn from point $M$ to the circles passing through the triplets of points $(A, B, M)$, $(B, C, M)$, and $(C, A, M)$, intersecting the sides $AB$, $BC$, and $CA$ at points $P$, $Q$, and $R$ respectively. Prove that thes... | 404. Notice that the tangent to the circle at vertex $C$ of the inscribed triangle $A B C$ intersects the opposite side $A B$ at point $C_{1}$, for which $A C_{1}: C_{1} B=A C^{2}: B C^{2}$. Apply the converse of Menelaus' theorem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,980 |
405. Two circles of radii $r_{1}$ and $r_{2}$ touch a circle $O$ of radius $r$ externally (internally) at points, the distance between which is $m$. Prove that the length $t_{12}$ of the common external tangent of the first two circles is given by the formula:
$$
t_{12}=\frac{m}{r} \sqrt{\left(r \pm r_{1}\right)\left(... | 405. Using the Pythagorean theorem, establish the relationship between $t_{12}$ and $O_{1} O_{2}$, and then from triangles $O O_{1} O_{2}$ and $O K_{1} K_{2}$ (where $K_{1}$ and $K_{2}$ are the points of tangency), determine the square of the side $O_{1} O_{2}$ and $m^{2}$, which will allow us to prove the validity of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,981 |
407. On a plane, there is an oriented line $t$ and an oriented circle $O$. Through a point $M$ on the line $t$, two tangents $t_{1}$ and $t_{2}$ are drawn to the circle, directed in accordance with the orientation of the circle. Prove that the product
$$
\operatorname{tg} \frac{\alpha}{2} \cdot \operatorname{tg}-\frac... | 407. Express the ratio of the radius $r$ of a circle to the distance $d$ from its center to a line $t$ in terms of the tangents of the half-angles $\alpha$ and $\beta$, and, using the properties of proportions, establish that $\operatorname{tg} \frac{\alpha}{2} \cdot \operatorname{tg} \frac{\beta}{2}=\frac{r-d}{r+d}$. | \frac{r-}{r+} | Geometry | proof | Yes | Yes | olympiads | false | 43,983 |
408. The vertices of triangles having a common base are located on one straight line. Prove that the geometric locus of the centroids of these triangles is a straight line. | 408. Taking the midpoint of the base as the center of homothety and transforming the given line with a coefficient of $\frac{1}{3}$, we obtain the line which is the desired geometric locus of points. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,984 |
409. A variable secant cuts off segments on the sides of a given angle, measured from the vertex, the sum of which is constant. Prove that the geometric locus of the centroids of the resulting triangles is a segment. | 409. Let a secant cut off segments on the sides of angle $S$ whose sum is equal to $2 a$. Establish that the geometric locus of the midpoints of such secant segments is the segment $M N$, which cuts off segments $S M=S N=a$ on the sides of the angle (see problem No. 12). The desired geometric locus of points is also a ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,985 |
411. A variable $n$-sided polygon deforms in such a way that all its sides remain parallel to themselves, and $n-1$ vertices of it move along $n-1$ lines respectively. Prove that the last vertex in this case describes a straight line. | 411. Find two points of the geometric locus and prove, using the properties of proportional segments, that any third point of the sought geometric locus lies on the line connecting the first two. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,987 |
413. A family of parallelograms is inscribed in a quadrilateral, the sides of which are parallel to the diagonals of the quadrilateral. Show that the geometric locus of the centers of these parallelograms is an open segment. | 413. Ensure that if a parallelogram $M N K L$ is inscribed in the quadrilateral $A B C D$ in the specified manner, then $A M: M B = C N: N B = C K: K D$. Then consider the two lines $A B$ and $C D$ and the similar triplets of points $A, M, B$ and $C, K, D$, from which it follows that the midpoints of segments $A C, M K... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,989 |
415. Given a line $t$ and a point $A$ on it. Prove that the geometric locus of points $M$, for which the ratio $\frac{M A^{2}}{d}$ is a constant ( $d$ - the distance from point $M$ to line $t$ ), is a pair of circles. | 415. Construct a perpendicular to the line $t$ at point $A$ and mark a segment $A N$ on it, equal to $\frac{A M^{2}}{d}$ (Fig. 53). It follows that $\frac{A M}{d}=\frac{A N}{A M}$, and due to the equality of the angles enclosing the proportional sides, triangles $A M M_{1}$ and $A M N$ will be similar, which means that... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,991 |
417. Prove that the geometric locus of points located inside a given angle $\alpha\left(\alpha>60^{\circ}, \alpha \neq 90^{\circ}\right)$, for each of which the sum of the distances to the sides of the angle is equal to the distance to its vertex, is two rays. How will the theorem change if
$$
\alpha=60^{\circ}, \alph... | 417. Draw a line that cuts off equal segments $O A$ and $O B$ on the sides of the given angle $O$ (Fig. 54). If $P$ is a point of the desired geometric locus on the segment $A B$, then $P M+P N=h$, where $M$ and $N$ are the feet of the perpendiculars dropped from $P$ to the sides of the angle or their extensions, and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,993 |
418. Given an equilateral triangle $ABC$. Prove that the geometric locus of points $M$, for each of which the equality
$$
M C^{2}=M A^{2}+M B^{2}
$$
holds, is a circle. | 418. Let point $M$, lying inside triangle $A B C$, belong to the desired locus of points. By rotating $A M$ clockwise around point $A$ by $60^{\circ}$ to position $A D$, we obtain an equilateral triangle $A M D$ and two congruent triangles $A B M$ and $A C D$ (since $A B = A C$, $A M = A D$, and $\angle B A M = \angle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,994 |
419. Prove that the geometric locus of points, each of which has equal sums of distances to one pair of opposite vertices of a rectangle and to the other pair of its vertices, is a pair of perpendicular lines. | 419. Draw the diagonals of rectangle $A B C D$ and consider triangles $A M C$ and $B M D$, which have a common median $M O$ (where $M$ is the point of the sought geometric locus, and $O$ is the point of intersection of diagonals $A C$ and $B D$). Establish the equality $A M^{2}+M C^{2}=$ $=M B^{2}+M D^{2}$ and consider... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,995 |
421. Segments $A C$ and $B C$, not lying on the same line, are divided respectively by points $Q$ and $P$ - one externally and the other internally - such that $C A: A Q = C P: P B$, and points $A$ and $Q$ lie on opposite sides of point $C$. Prove that the geometric locus of the points of intersection of lines $A P$ an... | 421. Apply Menelaus' theorem to triangle $B C Q$ and the transversal $A P$ to establish that the segment $B Q$ is bisected by the point of the sought geometric locus. It follows from this that the line parallel to line $A C$ and bisecting segment $C B$ is the sought geometric locus. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,997 |
422. On a straight line, a segment $A B$ is fixed. Through a given point $M$ on this line, a straight line is drawn, on which points $P$ and $Q$ are chosen such that the ratio $\frac{M P}{M Q}$ is constant. Prove that the geometric locus of the points of intersection of the lines $A P$ and $B Q$ is a straight line. | 422. Use Menelaus' theorem to establish that the ratio $A K: K P$ is constant. ($K$ is the point of intersection of $A P$ and $B Q$). Therefore, the required geometric locus of points will be a line parallel to the line $M P$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,998 |
423. A square $M N K L$ is inscribed in triangle $A B C$ such that its vertices $L$ and $K$ lie on side $A B$ (or its extension), and vertices $M$ and $N$ lie on sides $A C$ and $B C$, respectively. Prove that the geometric locus of vertices $M$ and $N$ of the squares, given that vertex $C$ of the triangle moves along ... | 423. Determine that sides $A C$ and $B C$ (Fig. 55) are divided by points $M$ and $N$ in the ratio $A B: C H_{3}\left(C H_{3}\right.$ - the height of triangle $A B C)$. Therefore, $A M_{i}: M_{i} C_{i}=B N_{i}: N_{i} C_{i}=A B: C_{i} H_{3}^{(i)}$. Then, using the converse of Menelaus' theorem, verify that points $M, M_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,999 |
425. Lines $m$ and $n$ are parallel to two sides of a triangle, and the segments of these lines, enclosed respectively between the other two sides of the triangle, are equal. Prove that the geometric locus of the points of intersection of lines $m$ and $n$ is a segment of a line. | 425. Let $m=K K_{1}, n=L L_{1}$ (points $K$ and $L$ belong to side $A B, A C \neq B C$). If $C_{1}$ is the foot of the angle bisector of angle $C$, then calculate $K C_{1}$ and $L C_{1}$ and prove that $C_{1} K: C_{1} L=a: b=$ $=C_{1} B: C_{1} A$. Complete the given triangle to a parallelogram and establish the homothe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,001 |
427. Inside a circle, there is a point $M$, from which pairs of mutually perpendicular rays are drawn, determining two points by their intersection with the circle. Prove that the geometric locus of the feet of the perpendiculars dropped from point $M$ to the lines connecting these pairs of points is a circle. | 427. Show that the feet of the perpendiculars dropped from point $M$ to the chords are at a constant distance from the midpoint of segment $O M$, equal to $\frac{1}{2} \sqrt{2 R^{2}-\text { OM }^{2}}$. The desired geometric locus is a circle. | \frac{1}{2}\sqrt{2R^{2}-\text{OM}^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 44,003 |
428. On a line, two pairs of points $A, B$, and $C, D$ are given (all four points are distinct). Prove that the geometric locus of points $P$, for which the circles passing through the triplets of points $A, B, P$ and $C, D, P$ touch at point $P$, is a circle. | 428. To establish that the tangent to the circles at point $P$ intersects the line $A B$ at a constant point $M$, using the fact that $M P^{2}=M A \cdot M B=M C \cdot M D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,004 |
429. Through point $A$, which is one of the intersection points of two circles, a variable line is drawn, intersecting the circles again at points $M$ and $N$. Segment $M N$ is divided by point $P$ in the ratio $k$. Prove that the geometric locus of points $P$ is a circle. | 429. In the middle of the segment $A P$ erect a perpendicular, meeting the line of centers $O_{1} O_{2}$ at point $Q$. Verify that $O_{1} Q: Q O_{2} = k$, and thus establish the unique position of point $Q$. From this, it follows that $Q P = Q A = \text{const}$. The required geometric locus is a circle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,005 |
430. Prove that the geometric locus of points, the ratio of whose distances to the diagonals of a trapezoid is inversely proportional to the ratio of their lengths, is a pair of lines, one of which is parallel to the bases of the trapezoid, and the other passes through the midpoints of these bases. | 430. Through the point $O$ of intersection of the diagonals $A C$ and $B D$ of a trapezoid, draw a line parallel to the base $A B$, forming angles $\alpha$ and $\beta$ with the diagonals. The distances $d_{1}$ and $d_{2}$ from any point on this line to the diagonals are proportional to the sines of the angles $\alpha$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,006 |
431. Prove that the geometric locus of points, each of which has a fixed ratio of powers with respect to two given circles, is a circle or, in particular, a straight line. | 431. If this constant is equal to $k(k \neq 1)$, then divide the line of centers of the circles in the ratio - $k$ and take the obtained point as the center of the required circle. Then apply Stewart's theorem (see theorem 26, p. 11). For $k=1$, we have a straight line. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,007 |
432. Given two points. Prove that all lines, relative to which the given points lie on the same side and $m d_{1}+n d_{2}$ is a constant (where $m$ and $n$ are positive numbers, $d_{1}$ and $d_{2}$ are the distances of the given points to the line), touch the same circle. | 432. Use the properties of proportional segments and show that all lines are tangent to a circle of radius $\frac{m d_{1}+n d_{2}}{n+m}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,008 |
433. From a point $M$ inside triangle $ABC$, perpendiculars $MK$, $ML$, and $MN$ are dropped to its sides. Prove that the set of points $M$, for which the segments $MK$, $ML$, and $MN$ can form a triangle, is located inside the triangle whose vertices coincide with the feet of the angle bisectors of triangle $ABC$.
##... | 433. Use the results of problems № 283 and 347 to establish that the sought set consists only of points located inside the triangle whose vertices are the feet of the bisectors of the given triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,009 |
434. 435) Given an angle and its bisector. An arbitrary circle passing through the vertex of the angle cuts off segments $a, b, l$ on the sides of the angle and its bisector, respectively. Prove that the ratio
$$
\frac{a+b}{l}
$$
does not depend on the position of the circle.
2) Given an angle and its external bisec... | 434. Use Ptolemy's theorem and prove that $\frac{a+b}{l}$ is equal to the ratio of the base to the lateral side of an isosceles triangle, where the angle at the vertex is constant. Similarly, the second part of the problem is solved. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,010 |
436. Prove that the height $h$ of a right triangle, dropped to the hypotenuse, and the radius $r$ of the inscribed circle are related by the inequality:
$$
h \leqslant r(1+\sqrt{2})
$$ | 436. Drop a perpendicular from the center of the inscribed circle to the height and compare the leg $h-r$ and the hypotenuse $r \sqrt{2}$ of the right triangle. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 44,012 |
440. On the sides $C A$ and $C B$ of triangle $A B C$, squares are constructed outside the triangle. The sides of the squares opposite to the sides of the triangle are extended until they intersect. Prove that the line passing through the obtained point $S$ and the common vertex $C$ of the squares divides the third sid... | 440. Notice that $b: S C=\sin \varphi, \quad a: S C=\sin \psi, \quad$ where $\varphi=$ $=\angle A C C_{1}, \psi=\angle B C C_{1}\left(C_{1}\right.$-the intersection point of lines $A B$ and $\left.C S\right)$. But $A C_{1}: C_{1} B=\left(A C C_{1}\right):\left(B C C_{1}\right)=b \sin \varphi: a \sin \psi$. Therefore, $... | AC_{1}:C_{1}B=b^{2}:^{2} | Geometry | proof | Yes | Yes | olympiads | false | 44,016 |
441. Through the vertex $A$ of an equilateral triangle $A B C$, a secant is drawn, intersecting the side $B C$ at point $M$, and the line through vertex $C$ parallel to side $A B$ at point $N$. Prove that
$$
\frac{1}{m^{2}}-\frac{1}{m n}+\frac{1}{n^{2}}=\frac{1}{a^{2}}
$$
where $m=A M, n=A N, a$ is the side of the tr... | 441. Rotate the secant around point $A$ by $60^{\circ}$ so that triangle $A N K$ is similar to triangle $A B M$. Then $m: a=N K: n$, or $N K=\frac{m n}{a}$, and from triangle $A N K-$ $N K^{2}=m^{2}+n^{2}-m n$. Therefore, $m^{2}-m n+n^{2}=\frac{m^{2} n^{2}}{a^{2}}$. | \frac{1}{^{2}}-\frac{1}{n}+\frac{1}{n^{2}}=\frac{1}{^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 44,017 |
442. In triangle $A B C$, a triangle $A_{1} B_{1} C_{1}$ is inscribed such that the lines $A A_{1}, B B_{1}, C C_{1}$ intersect at one point. Prove that if $C C_{1}$ is the bisector of angle $A_{1} C_{1} B_{1}$, then $C C_{1}$ is the altitude of triangle $A B C$. | 442. Let $B_{1} C_{1}$ intersect $A A_{1}$ at point $P$, and let the lines $A A_{1}$, $B B_{1}$, $C C_{1}$ pass through point $S$ (Fig. 56). Prove that $A P: P S=A A_{1}: S A_{1}$, using Ceva's theorem and Menelaus' theorem applied to triangle $A B S$, point $C$, and the transversal $B_{1} C_{1}$. Considering that $C_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,018 |
443. The sides of triangle $ABC$ are divided in the same ratio $2:(1+\sqrt{5})$ by points $A_{1}, B_{1}, C_{1}$ in the order of traversing its perimeter. Prove that the medians of triangle $A_{1} B_{1} C_{1}$ pass through the vertices of the given triangle. | 443. Solve the problem for an equilateral triangle, for which connect point $C$ with the midpoint $C_{2}$ of segment $A_{1} B_{1}$. Setting $C A_{1}=k x, C B_{1}=x$, where $k=\frac{2}{1+\sqrt{5}}$, calculate the cosine of angle $B_{1} C_{2} C$ from triangle $B_{1} C C_{2}$ and verify that $C C_{2} \| B_{1} C_{1}$. Sinc... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,019 |
444. On the medians of an equilateral triangle \(ABC\), points \(A_1, B_1, C_1\) are taken respectively. Prove that: 1) these points can be chosen such that the line \(C_1 A_1\) passes through point \(A\); the line \(A_1 B_1\) - through point \(B\), the line \(B_1 C_1\) - through point \(C\); 2) the triangle \(A_1 B_1 ... | 444. Let $M_{1} B_{1}=l, B_{1} M_{3}=1-l, M_{3} A_{1}=s . \quad A_{1} M_{2}=$ $=1-s, M_{2} C_{1}=k, C_{1} M_{1}=1-k$, assuming that $A B=B C=C A-$ $=2$. Establish that $l(1+s)=1, s(1+k)=1, k(1+l)=1$. From this, discover that $l=s=k=\frac{1}{2} \cdot(\sqrt{5}-1)$. Therefore, triangle $A_{1} B_{1} C_{1}$ is also equilate... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,020 |
446. Through the center $O$ of an equilateral triangle $A B C$ with side $a$, a line $A_{1} B_{1}$ is drawn, cutting off segments $x$ and $y\left(x=A B_{1}, y=B A_{1}\right)$ from sides $A C$ and $B C$. Prove that these segments satisfy the equation:
$$
3 x y-2 a(x+y)+a^{2}=0
$$ | 446. Express the area of the quadrilateral $A B A_{1} B_{1}$ through $x$, $y$, and $a$ in two ways: $\left(A B A_{1} B_{1}\right)=\left(B_{1} A O\right)+(O A B)+\left(B A_{1} O\right)=$ $=\frac{(a+x+y) a}{12} \frac{\sqrt{3}}{}$ and $\left(A B A_{1} B_{1}\right)=\left(A B A_{1}\right)+\left(A A_{1} B_{1}\right)=\frac{y}... | 3xy-2(x+y)+^{2}=0 | Geometry | proof | Yes | Yes | olympiads | false | 44,022 |
449. The heights $A H_{1}$ and $B H_{2}$ of triangle $A B C$ are extended beyond the vertices, and on their extensions, segments $A A_{2}$ and $B B_{2}$ are laid off, equal to $B C$ and $A C$ respectively. Prove that the segments $A_{2} C$ and $B_{2} C$ are perpendicular and equal. | 449. Establish the equality of triangles $A C A_{2}$ and $B C C_{2}$ by two sides and the included angle. Prove that $\angle A_{2} C A + \angle B_{2} C B = \left|90^{\circ} - \angle C\right|$. Consider the cases when angle $C$ is acute and obtuse. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,025 |
450. In the midpoints $M_{3}, M_{1}, M_{2}$ of the sides $A B, B C, C A$ of triangle $A B C$, perpendiculars are erected, and outside the triangle, segments $M_{3} C_{1}, M_{1} A_{1}, M_{2} B_{1}$ are laid off, respectively equal to half the sides $A B, B C, C A$. Prove that the segments $A A_{1}, B B_{1}, C C_{1}$ are... | 450. The length of the segment $C C_{1}$ can be calculated using the cosine theorem from triangle $C B C_{1}$, and the length of the segment $A_{1} B_{1}$ - from triangle $A_{1} B_{1} C: C C_{1}^{2}=\frac{a^{2}+b^{2}}{2}+2(A B C)$. Verify that $B_{1} C_{1}^{2}+C A_{1}^{2}=$ $=C_{1} A_{1}^{2}+C B_{1}^{2}$. Use the resul... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,026 |
451. Prove that if the altitudes and the radius of the inscribed circle of a triangle satisfy the equation
$$
h_{1}+h_{2}+h_{3}=9 r
$$
then the triangle is equilateral. | 451. Use the equality $h_{1}: r=2 p: a$, as well as the fact that $\frac{a}{b}+\frac{b}{a} \geqslant 2$ and that equality is achieved only when $a=b$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,027 |
453. To prove that for triangle $ABC$ to be acute-angled, it is necessary and sufficient that
$$
\cos ^{2} A+\cos ^{2} B+\cos ^{2} C < 1
$$ | 453. Use the identity $\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+$ $+2 \cos A \cdot \cos B \cdot \cos C=1$, valid for any triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,029 |
454. From the vertex $C$ of triangle $A B C$, a median and an angle bisector are drawn, forming an angle $\alpha$ between them. Prove that
$$
\operatorname{tg}^{2} \frac{C}{2}=\operatorname{tg} \alpha \cdot \operatorname{tg} \beta
$$
where $\beta$ is the acute angle formed by the angle bisector and side $A B$. | 454. Preliminarily prove the validity of the equalities:
$$
\frac{\sin \left(\frac{C}{2}+\alpha\right)}{\sin \left(\frac{C}{2}-\alpha\right)}=\frac{\sin \left(\beta+\frac{C}{2}\right)}{\sin \left(\beta-\frac{C}{2}\right)}
$$
138
using the Law of Sines and the property of the internal angle bisector of a triangle. Con... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,030 |
456. Around an acute-angled triangle $A B C$ with angle $C$ equal to $45^{\circ}$, a circle $O$ is circumscribed. Another circle is circumscribed on side $A B$ as its diameter, intersecting sides $A C$ and $B C$ at points $M$ and $N$ respectively. Prove that segment $M N$ is perpendicular to segment $O C$ and that $M N... | 456. Establish that the center $O$ lies on the second circle and $\checkmark A M O=\smile O N B=90^{\circ}$, and also show that $\cup M O N=90^{\circ}$ (Fig. 58). Therefore, $\quad M N=A O=\frac{A B}{\sqrt{2}}, \cup A M=$ $=\cup O N, \cup M O=\cup N B$, from which $M O \| A N$ and $O N \| M B$. Since $A N \perp C B$ an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,032 |
457. A triangle is inscribed in a circle. The tangents to the circle at the vertices of the triangle form another triangle, the perimeter of which is twice the perimeter of the given triangle. Prove that this condition is necessary and sufficient for the given triangle to be equilateral. | 457. Based on known inequalities (Fig. 59):
Fig. 58 $\sin A + \sin B + \sin C \leqslant \frac{3 \sqrt{3}}{2}, \quad \cos A + \cos B + \cos C \leqslant -^{3}$.
But $\frac{1}{2}(\angle B + ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,033 |
458. Prove that in any triangle, through any of its vertices, a secant can be drawn that divides the triangle into two triangles such that the circles inscribed in them touch each other. Find the ratio in which the secant divides the side of the triangle. | 458. Notice that the base of the secant divides the segment, the ends of which coincide with the points of tangency of the inscribed circles, in half. The required secant passes through the point of tangency of the circle inscribed in the given triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,034 |
459. The circle inscribed in triangle $ABC$ touches its sides $AB, BC, CA$ at points $M, N, K$ respectively. The midpoints $M_{1}, N_{1}, K_{1}$ of the radii $O^{\prime} M, O^{\prime} N, O^{\prime} K$ are connected respectively to the vertices $C, A, B$. Prove that the lines $A N_{1}, B K_{1}, C M_{1}$ intersect at one... | 459. Establish that $C M_{1}$ intersects $A B$ at point $C_{1}$ (Fig. 60), for which
$$
\frac{A C_{1}}{C_{1} B}=\frac{2 p(p-a)+b c}{2 p(p-b)+a c}
$$
Similarly,
$$
\frac{B A_{1}}{A_{1} C}=\frac{2 p(p-b)+a c}{2 p(p-c)+a b}, \quad \frac{C B_{1}}{B_{1} A}=\frac{2 p(p-c)+a b}{2 p(p-a)+b c}
$$
Use Ceva's theorem (Theorem... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,035 |
461. Prove that a line drawn through vertex $C$ of triangle $ABC$ parallel to the line passing through its incenter and the midpoint of side $AB$ bisects the perimeter of the triangle. | 461. Connect the center $O$ with the point of tangency $T_{3}$ of the inscribed circle and drop the altitude $\mathrm{CH}_{3}$. Consider similar triangles $\mathrm{CH}_{3} P_{3}$ and $O T_{3} M_{3}\left(O M_{3} \| C P_{3}\right)$. Prove that $A P_{3}=p-b$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,037 |
462. Prove that if $d_{1}, d_{2}, d_{3}$ are the distances from the incenter of a triangle to its vertices, then
$$
d_{1}^{2}: d_{2}^{2}: d_{3}^{2}=\frac{p-a}{a}: \frac{p-b}{b}: \frac{p-c}{c} .
$$
Determine the ratio:
$$
d_{\mathbf{1}}^{2}: \frac{p-a}{a}
$$ | 462. Setting $x^{2}=\frac{a \cdot d_{1}^{2}}{p-a}$, note that $p-a=r \operatorname{ctg} \frac{A}{2}$, $d_{1}=\frac{r}{\sin \frac{A}{2}}$. Finally, $x^{2}=4 R r$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,038 |
463. The sides of a triangle are intersected by a line at points $A_{1}$, $B_{1}, C_{1}$. On the segments $A A_{1}, B B_{1}, C C_{1}$, as diameters, circles are constructed. Prove that these three circles belong to the same pencil. | 463. Notice that the centers of three circles are located on one straight line (Gauss line), which serves as the axis of symmetry for these circles. Let two circles pass through point $M$. According to problem № 448, the segments $A A_{1}, B B_{1}, C C_{1}$ are seen from point $M$ at right angles. Therefore, the third ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,039 |
464. On the side of a triangle, a point is taken from which perpendiculars are dropped to the other two sides. Prove that the distance between the bases of these perpendiculars will be the smallest if the point on the side coincides with the base of the triangle's altitude. Express the length of this smallest segment i... | 464. The distance $d$ between the bases of the perpendiculars is determined by the formula: $d=2 R_{1} \cdot \sin C ; d$ will be the smallest when $2 R_{1}=h_{3}$. | 2R_{1}\cdot\sinC | Geometry | proof | Yes | Yes | olympiads | false | 44,040 |
465. Prove that if a line is drawn through vertex $C$ of triangle $ABC$ perpendicular to the bisector of this angle, then the product of the distances from vertices $A$ and $B$ to this line will be greater than the product of the distances from these same vertices to any other line passing through the same vertex. | 465. The product of the distances is $a b \cos \varphi \cdot \cos \psi(\varphi+\psi=C$ or $\left.180^{\circ}-C\right)$. The maximum of the product $\cos \varphi \cdot \cos \psi$ is achieved when $\varphi=\psi=$ $=\frac{C}{2}$. The desired line is the external bisector of the triangle at vertex $C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,041 |
466. Through the vertex $B$ of the parallelogram $A B C D$, a secant is drawn, intersecting the extensions of sides $D A$ and $D C$ at points $P$ and $Q$ respectively. Prove that the sum of segments $P A$ and $C Q$ will be the smallest if each of these segments is equal to $\sqrt{a b}$ (where $a$ and $b$ are the length... | 466. If $P A=x, C Q=y$, then $x y=a b$. Setting $x+y=s$, we find that $s=\frac{x^{2}+a b}{x}$, or $s=\left(\sqrt{x}-\sqrt{\frac{a b}{x}}\right)^{2}+2 \sqrt{a b}$;
$$
s_{\min }=2 \sqrt{a b} \text { when } x=y=\sqrt{a b}
$$ | 2\sqrt{} | Geometry | proof | Yes | Yes | olympiads | false | 44,042 |
467. Through the vertex $B$ of the trapezoid $A B C D$, a secant is drawn, intersecting the extensions of sides $D A$ and $D C$ of the trapezoid at points $P$ and $Q$ respectively. Prove that if $P A = \sqrt{cd} - (a - c)$ ($a$ - the larger base of the trapezoid, $c$ - the smaller base, vertex $A$ belongs to the larger... | 467. If $P A=x, C Q=y$, then $(a+x): c=(d+y): d$, from which $a y+x y=c d+c y$. Letting $x+y=s$, we find:
$$
\begin{gathered}
s=\frac{x^{2}+x(a-c)+c d}{x+a-c}=x+\frac{c d}{x+a-c} \\
s=\left(\sqrt{x+a-c}-\sqrt{\frac{c d}{x+a-c}}\right)^{2}+2 \sqrt{c d}-(a-c)
\end{gathered}
$$
From this, $s_{\min }=2 \sqrt{c d}-(a-c)$ ... | 2\sqrt{}-(-) | Geometry | proof | Yes | Yes | olympiads | false | 44,043 |
468. Prove that a line passing through the midpoint of side $AB$ of triangle $ABC$ and its incenter divides the segment connecting vertex $C$ with the point of tangency of the inscribed circle with side $AB$ in half. | 468. Four points $A, C_{1}, B$ and $C$ are vertices of a degenerate circumscribed quadrilateral; therefore, the midpoint $M_{3}$ of side $A B$ and the midpoint $N$ of segment $C C_{1}$ are midpoints of the diagonals and lie on the same line with the center of the inscribed circle (see theorem 21). | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,044 |
472. Prove that the tangents drawn to the circle at the vertices of an inscribed scalene triangle intersect its opposite sides at points lying on the same straight line. | 472. Tangents drawn to the circle at the vertices of triangle $ABC$ meet the opposite sides at points $A_{1}$, $B_{1}$, and $C_{1}$. Establish that $A C_{1}: C_{1} B = -A C^{2}: C B^{2}$, $B A_{1}: A_{1} C = -B A^{2}: A C^{2}$, and $C B_{1}: B_{1} A = -C B^{2}: B A^{2}$, and use the converse of Menelaus' theorem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,048 |
473. Prove that if the circles inscribed in the triangles into which a given triangle is divided by its median are equal, then the given triangle is isosceles. | 473. First solution. Use the formula $r=$ $=\frac{S}{p}$, where $r$ is the radius of the inscribed circle, $S$ is the area, and $p$ is the semiperimeter of the triangle. The areas of the considered triangles and the radii of the inscribed circles are equal according to the condition. Therefore, the perimeters of these ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,049 |
474. Using the formula
$$
e^{2} f^{2}=a^{2} c^{2}+b^{2} d^{2}-2 a b c d \cos (A+C)
$$
relating the sides, diagonals, and the sum of opposite angles of a quadrilateral, derive Stewart's Theorem (see Theorem No. 26). | 474. To obtain the dependence
$$
B D^{2} \cdot A C=A B^{2} \cdot D C+B C^{2} \cdot A D-A C \cdot A D \cdot D C
$$
consider a degenerate quadrilateral $A B C D$ with angle $D$ equal to
$180^{\circ}$. Separately verify the case when point $D$ lies on the extension of $A C$. | BD^{2}\cdotAC=AB^{2}\cdotDC+BC^{2}\cdotAD-AC\cdotAD\cdotDC | Geometry | proof | Yes | Yes | olympiads | false | 44,050 |
475. Given a quadrilateral $A B C D$, where angle $D$ is equal to $30^{\circ}$ and $A B=B C=C A$. Prove that a right triangle can be constructed on the segments $D A, D B, D C$. | 475. Given that $\angle B+\angle D=90^{\circ}$, use the fact that $A B^{2} \cdot C D^{2}+B C^{2} \cdot A D^{2}=A C^{2} \cdot B D^{2}$, and the converse of the Pythagorean theorem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,051 |
476. Using Ptolemy's theorem, prove Stewart's theorem, and vice versa (see theorems No. 17 and No. 26). | 476. Let it be required to determine the distance from point $P$ on side $A B$ of triangle $A B C$ to vertex $C$. Extend segment $P C$ to intersect the circumcircle of this triangle at point $D$, and consider the inscribed quadrilateral $A C B D$ and two pairs of similar triangles formed.
. Prove that: 1) the square of the sum of the distances
from this point to the vertices of the triangle is equal to $\frac{1}{2}\left(a^{2}+b^{2}+c^{2}+4(A B C)\righ... | 478. 1) On the side $A B$ outside the triangle $A B C$, construct an equilateral triangle $A B P$ and prove that $C P = A M + B M + C M$. Then use the law of cosines applied to triangle $A P C$ 2) Also use it for triangles $A M B, B M C, C M A$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,054 |
479. The tangent $t$ to the inscribed circle of triangle $ABC$ intersects its sides $AC$ and $AB$ at points $B_1$ and $C_1$, respectively. Prove that the segments $x$ and $y \left(AC_1 = x, AB_1 = y\right)$ satisfy the equation:
$$
p x y - bc(x + y) + bc(p - a) = 0
$$
Prove that if the segments $x$ and $y$ cut off by... | 479. From the condition of the problem, it follows that $B_{1} C_{1}+a=(b-y)+$ $+(c-x)$, or $B_{1} C_{1}-a=(x-c)-(b-y)$, or $B_{1} C_{1}-$ $-a=(y-b)-(c-x)$. In all three cases, we have:
$$
\begin{aligned}
B_{1} C_{1}^{2}= & x^{2}+y^{2}-\frac{x y}{b c} \cdot\left(b^{2}+c^{2}-a^{2}\right)=(x+y)^{2}- \\
& -4(p-a)(x+y)+4(... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,055 |
480. From a point $A$ lying outside a circle, two tangents $A T_{1}$ and $A T_{2}$ and a secant $B D$ ( $B$ and $D$ - points of intersection with the circle) are drawn. Prove that the points $A, B, C$ and $D$ form a harmonic set, that is,
$$
\frac{A B}{B C}=\frac{A D}{D \bar{C}}
$$
where point $C$ is the point of int... | 480. From the similarity of triangles $A B T_{1}$ and $A T_{1} D$, $A B T_{2}$ and $A D T_{2}$, establish that
$$
\frac{A B}{A D}=\frac{B T_{1} \cdot B T_{2}}{D T_{1} \cdot D T_{2}}
$$
Then show that the ratio $\frac{B C}{D C}$ is equal to the ratio of the areas of triangles $B T_{1} T_{2}$ and $D T_{1} T_{2}$, i.e.,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,056 |
481. At the ends of the diameter $M N$ of the circle $O$, tangents $M K$ and $N L$ are drawn. On the tangent $M K$, equal segments $A B$ and $B C$ are sequentially laid out, and from points $A, B$, and $C$, tangents are drawn to the circle, intersecting the tangent $N L$ at points $A_{1}, B_{1}, C_{1}$, respectively. P... | 481. Connect $A, A_{1}$ and $T$ ( $T$ - the point of tangency of $A A_{1}$ with the circle) with the center $O$. $A O$ and $A_{1} O$, as the bisectors of adjacent angles, are perpendicular, and $OT$ is the altitude of the right triangle $A O A_{1}$, hence $O T^{2}=A T \cdot T A_{1}$, or $r^{2}=M A \cdot N A_{1}$. Simil... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,057 |
482. From point $S$ to circle $O$, tangents $S T_{1}$ and $S T_{2}$ are drawn, which are intersected by two other tangents at points $M, M_{1}$ and $N, N_{1}$ respectively. Prove that
$$
\frac{T_{1} N}{N M}: \frac{T S}{S M}=\frac{S N_{1}}{N_{1} M_{1}}: \frac{S T_{2}}{T_{2} M_{1}}
$$ | 482. Let $S M=x, S M_{1}=y, S N=c, S N_{1}=b, N N_{1}=a$ (Fig. 62). From the condition that the line $M M_{1}$ is tangent to the circle inscribed in the triangle $S N N_{1}$, it follows that $a+M M_{1}=M N+M_{1} N_{1}$. From this equality, it follows that $p x y-b c(x+y)+b c(p-a)=0$. Verify that the relation between th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,058 |
483. On a circle, two sets of points $A, B, C, D$ and $A_{1}$, $B_{1}, C_{1}, D_{1}$ are given, arranged such that the lines $A A_{1}, B B_{1}, C C_{1}$, and $D D_{1}$ intersect at one point. Prove that if one set of points is harmonic, then the other set of points is also harmonic. | 483. If one set of four points is harmonic, then the product of the opposite sides of the corresponding quadrilateral are equal. According to problem № 387, a quadrilateral with vertices at another set of four points has the same property, and therefore the second set of four points is also harmonic. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,059 |
484. On the plane, there is a closed convex line with the property that the segments of the tangents drawn from each point located in the external region relative to the line are equal to each other. Prove that this line is a circle. | 484. Describe a triangle $ABC$ around a curve, then $CA_{1}=CB_{1}$, $AB_{1}=AC_{1}$, $BC_{1}=BA_{1}\left(A_{1}, B_{1}, C_{1}\right.$ - points of tangency) and
$$
AB_{1}^{2}+BC_{1}^{2}+CA_{1}^{2}=AC_{1}^{2}+BA_{1}^{2}+CB_{1}^{2}
$$
From this it follows that the perpendiculars erected at points $A_{1}$, $B_{1}$, $C_{1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,060 |
485. The circle $O_{3}$ is tangent to the circles $O_{1}$ and $O_{2}$. Prove that the line passing through the points of tangency passes through one of the centers of similarity of the circles $O_{1}$ and $O_{2}$. Consider separately the case when the radii of the circles $O_{1}$ and $O_{2}$ are equal. | 485. Since the points of tangency are the centers of similarity of the circles, the line passing through two centers of similarity will necessarily pass through the third center of similarity (see Theorem 20). | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,061 |
487. Two circles of radii $R$ and $r$ touch each other internally (externally) ($R > r$). Prove that the radius $r_{0}$ of the circle that touches the given two circles and their line of centers is given by the formula:
$$
r_{0}=\frac{4 R r(R \mp r)}{(R \pm r)^{2}}
$$
where the upper signs correspond to internal tang... | 487. Connect the centers of three circles in pairs and consider the resulting triangle with sides $R-r$, $r+r_{0}$, $R-r_{0}$, and with height $r_{0}$. Use the Pythagorean theorem. | r_{0}=\frac{4Rr(R\r)}{(R\r)^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 44,063 |
489. From a point $M$ lying outside a circle, a tangent $M T$ and a secant intersecting the circle at points $P$ and $Q$ are drawn. Through the endpoints $C$ and $D$ of the chord $C D$, parallel to the tangent $M T$, lines $C P$ and $C Q$ are drawn, intersecting the latter at points $A$ and $B$. Prove that $T M^{2}=M A... | 489. Show that triangles $M A P$ and $M B Q$ are similar and $M A \cdot M B=$ $=M P \cdot M Q$, and since $M T^{2}=M P \times$ $\times M Q$, then $M A \cdot M B=M T^{2}$. | MA\cdotMB=MT^2 | Geometry | proof | Yes | Yes | olympiads | false | 44,065 |
490. Through the ends of the diameter $A B$ of a certain circle, lines are drawn intersecting at point $M$. Line $A M$ intersects the circle at point $D$, and line $B M$ at point $C$. Prove that
$$
A B^{2}=A M \cdot A D \pm B M \cdot B C
$$ | 490. Draw lines $A C$ and $B D$, intersecting at point $N$, and line $M N$, meeting $A B$ at point $K$. Then consider the quadrilaterals $A D N K$, $D M C N$, and $C B K N$, around which circles can be circumscribed, and show that $B K \cdot A B = B N \cdot B D = B M \cdot B C$ and $A M \cdot A D = A C \cdot A N = A B ... | AB^{2}=AM\cdotAD\BM\cdotBC | Geometry | proof | Yes | Yes | olympiads | false | 44,066 |
491. An isosceles trapezoid is circumscribed around a circle. A tangent intersects one pair of opposite sides of the trapezoid at points \( P \) and \( R \), and the other pair at points \( Q \) and \( S \). Prove that
\[
\frac{P Q}{Q R}=\frac{R S}{S R}
\] | 491. Let the lateral sides $AD$ and $BC$ of the trapezoid (Fig. 64) intersect at point $M$. Suppose $AB=c, BM=AM=b, PA=x$, $AQ=y, DQ=z, DC=u$. The segments $x$ and $y$ satisfy the equation: $\quad pxy - bc(x+y) + bc(p-b) = 0, \quad$ where $\quad p$ is the semiperimeter of triangle $ABM$, and point $P$ lies on side $AB$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,067 |
492. A circle is inscribed in a trapezoid. A line passing through the point of tangency on one base of the trapezoid and the point of intersection of its lateral sides intersects the circle at a point where a tangent is constructed. Prove that the latter bisects the second base of the trapezoid. | 492. Using the transformation of homothety, establish that the line passing through the intersection point $F$ of the non-parallel sides $B C$ and $A D$ of trapezoid $A B C D$ and the point of tangency $T$ on the base $C D$, intersects the base $A B$ at point $K$, which is symmetric to the point of tangency $L$ of the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,068 |
493. Prove that the bisectors of the angles formed by the opposite sides of a cyclic quadrilateral intersect at a point lying on the line passing through the midpoints of the diagonals of the quadrilateral. | 493. Use the solution of problem No. 366. Determine that the angle bisectors of angles \(E\) and \(F\) (Fig. 65), formed by the opposite sides of the given quadrilateral \(ABCD\), intersect these sides at points that are the vertices of a parallelogram, the sides of which are parallel to the diagonals of the given quad... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,069 |
495. A quadrilateral $ABCD$ is inscribed in a circle, and the tangents at opposite vertices intersect at points $M$ and $N$. Prove that: 1) the opposite sides of the quadrilateral intersect on the line $MN$ and 2) the resulting intersection point divides the segment $MN$ in the ratio $MA: ND$. | 495. Establish that if tangents (MA and ND) are drawn from two points (M and N) to a circle, then the line (AD) determined by the points of tangency (A and D) will divide the segment (MN) connecting the given points in the ratio of the segments of the tangents (MA: ND). Depending on the arrangement of the tangents, the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,071 |
496. A quadrilateral with mutually perpendicular diagonals is inscribed in a circle $O$ of radius $R$, and the point of intersection $M$ of the diagonals is at a distance $d_{0}$ from the center $O$. Prove that the radii of the circles inscribed and circumscribed around the quadrilateral, whose vertices are the bases o... | 496. If $a_{1}, b_{1}, c_{1}, d_{1}$ are the sides of the second quadrilateral (Fig. 66), then $a_{1}=\frac{u d_{1}}{2 R}, b_{1}=\frac{u_{1} d_{2}}{2 R}, \quad c_{1}=\frac{v d_{1}}{2 R}, d_{1}=\frac{v_{1} d_{2}}{2 R}, \quad$ where $d_{1}, d_{2}$ are the diagonals, and $u, u_{1}, v, v_{1}$ are the segments of these diag... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,072 |
500. In the plane of quadrilateral $ABCD$, a point $M$ is taken. A circle is drawn through points $A, B$, and $M$, and the tangent to this circle at point $M$ intersects side $AB$ or its extension at point $P$. The tangent at point $M$ to the circle drawn through points $B, C$, and $M$ meets side $BC$ or its extension ... | 500. Let's construct circles with centers at points $P, Q, R, S$ and radii $P M, Q M, R M, S M$ (Fig. 67). According to the condition, the centers of these circles lie on the same line. Therefore, the circles belong to the same pencil. Points $A$ and $B$ are inverse with respect to the first circle, points $B$ and $C$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,076 |
501. Using the result of problem № 127 (the converse of Archimedes' theorem), prove the identity
$$
\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cdot \cos \frac{\alpha - \beta}{2}
$$
under the condition that the sum of the angles $\alpha$ and $\beta$ does not exceed $180^{\circ}$. | 501. The perpendicular dropped from the midpoint $M$ of the arc $A C B$ to the chord $A C$ divides it in half at point $P$. If $\angle B A C=\alpha$, $\angle A B C=\beta, \angle M A B=\gamma, \angle C A M=\delta$, then
$$
\gamma=\frac{\alpha+\beta}{2}, \quad \delta=\frac{\beta-\alpha}{2}, \quad(\beta \geqslant \alpha)... | \sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cdot\cos\frac{\alpha-\beta}{2} | Algebra | proof | Yes | Yes | olympiads | false | 44,077 |
502. Prove that triangle $ABC$ is equilateral if
$$
R(\tan A + \tan B + \tan C) = 2p
$$ | 502. Let's describe a circle around triangle $ABC$ and draw tangents to it at the vertices of the triangle, forming another triangle, the perimeter of which is equal to $2 R(\operatorname{tg} A+\operatorname{tg} B+\operatorname{tg} C)$. Thus, the perimeter of the circumscribed triangle is twice the perimeter of the ins... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,078 |
503. A circle of radius $r$ is inscribed in a triangle. Three circles are constructed, each touching two sides of the triangle and the inscribed circle. Prove that the radii $R_{1}, R_{2}, R_{3}$ of these three circles satisfy the relation:
$$
\sqrt{R_{1} R_{2}}+\sqrt{R_{2} R_{3}}+\sqrt{R_{3} R_{1}}=r
$$ | 503. It is evident that $\left(r+R_{1}\right):\left(r-R_{1}\right)=1: \sin \frac{A}{2}$ (Fig. 68). From this, $R_{1}=\frac{1-\sin \frac{A}{2}}{1+\sin \frac{A}{2}} \cdot r$, or $R_{1}=\operatorname{tg}^{2}\left(45^{\circ}-\frac{A}{4}\right) \cdot r$. Therefore,
$$
\begin{gathered}
\sqrt{R_{1} R_{2}}+\sqrt{R_{2} R_{3}}+... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,079 |
504. Points $A$ and $B$ have equal powers with respect to a circle. A secant line through point $A$ intersects the circle at points $M$ and $N$, and a secant line through point $B$ intersects the circle at points $P$ and $Q$. Prove that the lines $M P$ and $N Q$ intersect the line $A B$ at points $E$ and $F$, which als... | 504. From the equality of angles $A M E$ and $F P B$ (Fig. 69), it follows that $A M: B P = A A_{1}: B B_{1}$, where $A A_{1} \perp Q E, B B_{1} \perp N F$.
From the equality of angles $A N F$ and $E Q B$, it follows that $A N: B Q = A A_{2}: B B_{2}$, where $A A_{2} \perp P N, B B_{2} \perp M Q$. Therefore, $\frac{A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,080 |
506. A quadrilateral inscribed in a circle is divided by one diagonal into two triangles, into which circles are inscribed with radii $r_{1}$ and $r_{2}$, and by the second diagonal into two other triangles, into which circles are inscribed with radii $r_{3}$ and $r_{4}$. Prove that $r_{1}+r_{2}=r_{3}+r_{4}$. | 506. According to Euler's theorem for each of the triangles, we have: $O O_{1}^{2}=$ $=R^{2}-2 R r_{1}, O O_{2}^{2}=R^{2}-2 R r_{2}, O O_{3}^{2}=$ $:=R^{2}-2 R r_{3}, O O_{4}^{2}=R^{2}-2 R r_{4}$. Since the centers $O_{1}, O_{2}, O_{3}$, and $O_{4}$ are the vertices of a rectangle (see problem № 200) and, moreover, the... | r_{1}+r_{2}=r_{3}+r_{4} | Geometry | proof | Yes | Yes | olympiads | false | 44,082 |
507. A circle is circumscribed around triangle $ABC$. The bisector of angle $C$ of the triangle meets the circle at point $C_{1}$. Prove that if $C C_{1}=R+2 r$, then either angle $C$ is $60^{\circ}$, or
$$
h_{3}=\frac{R}{2}+2 r
$$ | 507. Around $\triangle ABC$, a circumcircle $O$ and an incircle $O'$ are described. Using Euler's theorem and the definition of the power of point $O'$ with respect to circle $O$, establish that $2 R r = C O' \cdot O' C_1$, i.e., considering the condition $C C_1 = C O' + O' C = R + 2 r$, there are two cases: 1) $R = O'... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,083 |
508. In the plane of triangle $ABC$, a point $P$ is given. Prove that a triangle can be constructed from the segments $PA \cdot \sin A, PB \cdot \sin B, PC \cdot \sin C$, which degenerates for points $P$ located on the circumcircle of the given triangle. | 508. According to Theorem 37 (Fig. 70), we have that
$$
P A^{2} \cdot B C^{2} \leqslant P B^{2} \cdot A C^{2} + P C^{2} \cdot A B^{2} + 2 P B \cdot A C \cdot P C \cdot A B
$$
or $P A \cdot B C \leqslant P B \cdot A C + P C \cdot A B$. Replacing the sides of the triangle $ABC$ with the sines of the opposite angles in ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,084 |
509. A quadrilateral is inscribed in circle $O$. A pair of its opposite sides intersect at point $E$, through which a line $t$ is drawn perpendicular to $O E$. Prove that the other two sides, as well as the diagonals, intersect line $t$ at points that are symmetric with respect to point $E$. | 509. If the diagonals $C A$ and $B D$ of quadrilateral $A B C D$ (Fig. 71) intersect line $t$ at points $P$ and $P^{\prime}$, then to prove the symmetry of points $P$ and $P^{\prime}$ with respect to the intersection point $M \equiv E$ of sides $B A$ and $C D$, it is sufficient to establish that their powers with respe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,085 |
510. Each of the exterior angles of a triangle is divided into three equal parts. To each side, two trisectors are adjacent, forming angles with this side, each of which is equal to one third of the exterior angle of the triangle. Prove that the points of intersection of these pairs of trisectors are the vertices of an... | 510. Denote the external angles of triangle $ABC$ as $3 \alpha, 3 \beta, 3 \gamma$ respectively. Then $\alpha + \beta + \gamma = 120^{\circ}$. Establish that the side of the constructed triangle is equal to $8 R \cdot \sin \alpha \cdot \sin \beta \cdot \sin \gamma$, where $R$ is the radius of the circumcircle of the gi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,086 |
511. In the plane of triangle $ABC$, a point $M$ is given. Circles $(A, B, M)$, $(B, C, M)$, $(C, A, M)$ are constructed. Tangents to these circles at point $M$ meet the sides $AB$, $BC$, $CA$ at points $C_1$, $A_1$, $B_1$ respectively. Prove that the obtained points lie on one straight line. Consider the case of degen... | 511. Establish that $A C_{1}: C_{1} B=-\left(M A^{2}: M B^{2}\right)$, and use Menelaus' theorem (theorem 3). | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,087 |
512. On the line $l$ there are three points $A, B, C$ given sequentially $(AB=m, BC=n)$, through which lines are drawn forming an equilateral triangle. Prove that the largest of all equilateral triangles has a side length equal to
$$
\frac{2}{\sqrt{3}} \sqrt{m^{2}+m n+n^{2}}
$$ | 512. Through point $C$ we draw a line $l_{c}$ at an angle $\alpha$ to line $l$, through point $B$ - a line $l_{b}$ at an angle $60^{\circ}+\alpha$, through point $A$ - a line $l_{\alpha}$ at an angle $120^{\circ}+\alpha$, forming an equilateral triangle. The side $x$ of this triangle is equal to $m \cos \alpha-\frac{m+... | \frac{2}{\sqrt{3}}\sqrt{^{2}+n+n^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 44,088 |
513. In the plane of triangle $ABC$, two points $M$ and $N$ are given. Prove that
$$
M N^{2}=p_{1} r_{1}^{2}+p_{2} r_{2}^{2}+p_{3} r_{3}^{2}-\left(a^{2} p_{2} p_{3}+b^{2} p_{3} p_{1}+c^{2} p_{1} p_{2}\right)
$$
where $r_{1}=N A, r_{2}=N B, r_{3}=N C, p_{1}=\frac{M A_{1}}{A A_{1}}, p_{2}=\frac{M B_{1}}{B B_{1}}, p_{3}... | 513. Let $C M$ intersect $A B$ at point $C_{1}$. Using Ceva's theorem, establish that $A C_{1}: C_{1} B=p_{2}: p_{1}$. Using Stewart's theorem, calculate $N C_{1}^{2}$ from triangle $A B N$. Then, based on Van Aubel's theorem (see theorem 5), prove that $C M: M C_{1}=\frac{p_{1}+p_{\mathrm{e}}}{p_{\mathbf{s}}}$. Using ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,089 |
514. In the plane of triangle $ABC$, a point $M$ is given, the distances $d_{1}, d_{2}, d_{3}$ from which to the vertices of the triangle satisfy the equation:
$$
d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = \frac{a^{2} + b^{2} + c^{2}}{3}
$$
Prove that the point $M$ is the centroid of the triangle. | 514. According to the formula obtained from solving problem № 513, we have: $\quad M G^{2}=\frac{1}{3}\left(d_{1}{ }^{2}+d_{2}{ }^{2}+d_{3}{ }^{2}\right)-\frac{1}{9}\left(a^{2}+b^{2}+c^{2}\right) . \quad$ Considering the condition of the problem, we establish that $M G=0$ and $M \equiv G$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,090 |
515. Prove that the distances $d_{1}, d_{2}, d_{3}$ from the incenter of a triangle to its vertices satisfy the equation:
$$
\frac{d_{1}^{2}}{b c}+\frac{d_{2}^{2}}{c a}+\frac{d_{3}^{2}}{a b}=1
$$
Prove the converse theorem. | 515. Use the formula given in problem № 513, keeping in mind that $M \equiv N \equiv O^{\prime}$. To prove the converse theorem, use the same formula and, based on the given relationship, discover that the distance of the given point from the incenter of the triangle is zero. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,091 |
518. Through the centroid of a triangle, an arbitrary secant is drawn. Prove that the ratio of the area of the resulting triangle to the area of the quadrilateral is not greater than $\frac{4}{5}$. | 518. Draw a secant through the centroid of the triangle, parallel to one of its sides, and establish that it divides the triangle into a triangle and a quadrilateral, the ratio of whose areas is 4:5. Then use the result of the previous problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,094 |
519. On the extensions of sides $A B, B C, C A$ of triangle $A B C$, points $C_{1}, A_{1}, B_{1}$ are taken respectively such that $A B = B C_{1}, B C = C A_{1}, C A = A B_{1}$. Prove that the ratio of the areas of triangles $A B C$ and $A_{1} B_{1} C_{1}$ is $1: 7$. | 519. Connect points $A$ and $A_{1}$, $B$ and $B_{1}$, $C$ and $C_{1}$, and consider triangles $A A_{1} B_{1}$, $A A_{1} C$, $C A_{1} C_{1}$, etc., equal in area to triangle $A B C$. | 1:7 | Geometry | proof | Yes | Yes | olympiads | false | 44,095 |
520. Through the vertices of triangle $A B C$, parallel lines are drawn intersecting its opposite sides at points $A_{1}, B_{1}, C_{1}$ respectively. Prove that the ratio of the area of the given triangle to the area of triangle $A_{1} B_{1} C_{1}$ is $1: 2$. | 520. Since $\left(A_{1} B_{1} C_{1}\right)=\left(A_{1} B_{1} C\right)+\left(B_{1} C_{1} C\right)+\left(C_{1} A_{1} C\right)$, and $\left(B_{1} C_{1} C\right)=\left(C_{1} C B\right)$ and $\left(C_{1} A_{1} C\right)=\left(C C_{1} A\right)$, then $\left(A_{1} B_{1} C_{1}\right)=\left(A_{1} B_{1} C\right)+$ $+(A C B)=2(A C... | 1:2 | Geometry | proof | Yes | Yes | olympiads | false | 44,096 |
521. To prove that for a point $N$, lying inside triangle $ABC$, to be the centroid of this triangle, it is necessary and sufficient that triangles $MAB$, $MBC$, $MCA$ are equal in area. | 521. Using the property of the median to divide the area of a triangle in half; show that if point $G$ is the centroid of the triangle, then the areas of triangles $G A B, G B C, G C A$ are equal. Then establish that if the areas of these triangles are equal,
}+\frac{1}{(B M S)}=\text { const }
$$ | 522. Draw a secant, the segment $A_{1} B_{1}$ of which, enclosed within the angle $A S B$, is bisected at point $M$ (Fig. 73). Establish that
where $\left(S A_{1} M\right)=\left(S M B_{1}\... | \frac{1}{(SAM)}+\frac{1}{(SMB)}=\frac{2}{(SA_1M)} | Geometry | proof | Yes | Yes | olympiads | false | 44,098 |
523. In triangle $A B C$, triangle $A_{1} B_{1} C_{1}$ is inscribed and triangle $A_{2} B_{2} C_{2}$ is circumscribed around it, with the corresponding sides of the latter two triangles being parallel. Prove that
$$
(A B C)^{2}=\left(A_{1} B_{1} C_{1}\right) \cdot\left(A_{2} B_{2} C_{2}\right)
$$ | 523. Establish that (Fig. 74) $(A B C):\left(A_{1} B_{1} C_{1}\right)=A_{2} B_{2}: A_{1} B_{1}$, $(A B C):\left(A_{2} B_{2} C_{2}\right)=A_{1} B_{1}: A_{2} B_{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,099 |
524. On the median $C M_{3}$ of triangle $A B C$, an arbitrary point $P$ is given. Lines $A P$ and $B P$ intersect the sides of the triangle at points $A_{1}$ and $B_{1}$, respectively. Prove that the areas of triangles $A A_{1} C$ and $B B_{1} C$ are equal. | 524. Determine that the segment $A_{1} B_{1}$ is bisected by the median $C M_{3}$, and compare the areas of two pairs of triangles: $A P C, B P C$ and $C P A_{1}, C P B_{1}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,100 |
526. From the vertices of triangle $A B C$, parallel inclined lines $A A_{1}, B B_{1}, C C_{1}$ are drawn to a line. The segments of these inclined lines are divided by points $A_{2}, B_{2}, C_{2}$ respectively in the same ratio $k$, measured from their bases. Prove that the ratio of the areas of the oriented triangles... | 526. Each of the triangles $A B C$ and $A_{2} B_{2} C_{2}$ is divided by one of the obliques into two triangles, which have a common side respectively. Prove that the ratio of the common side of the first pair of triangles to the common side of the second pair of triangles is $(k+1): k$, and consider the equality of th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,102 |
527. On the sides $AC$ and $BC$ of triangle $ABC$, points $M$ and $N$ are given such that $\frac{AM}{\overline{MC}}=\frac{CN}{NB}=k$. On the segment $MN$, a point $P$ is taken such that $MP: PN=k$. Prove that
$$
(AM P)^{\frac{1}{3}}+(BN P)^{\frac{1}{3}}=(ABC)^{\frac{1}{3}}
$$ | 527. Let $(A M P)=x,(B N P)=y,(M C P)=u,(C P N)=v$, $(A B C)=s$. Then $u=k v, x=k u, v=k y, S=\frac{(k+1)^{2}}{k}(u+v)$. From the obtained equations, eliminate $u, v, k$ (Fig. 75).
Fig. 74
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,103 |
528. Prove that the ratio of the area of a given quadrilateral to the area of the quadrilateral whose vertices are at the midpoints of the sides of the given one is 2. | 528. The first method. Use the fact that the area of a quadrilateral is expressed by the formula $(A B C D)=\frac{1}{2} e f \sin \varphi$, and the second quadrilateral is a parallelogram with sides $\frac{e}{2}$, $\frac{f}{2}$ and an angle $\varphi$ between them.
The second method. Notice that the midline of a triangl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,104 |
530. Prove that if two opposite vertices of a quadrilateral are moved by equal segments in the same direction, then the areas of the given quadrilateral and the resulting quadrilateral will be equal. | 530. The diagonals of both quadrilaterals are respectively equal and form equal angles in each of them. Use the known formula for the area of a quadrilateral. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,106 |
531. The midlines of a quadrilateral divide it into four quadrilaterals. Prove that the sum of the areas of one pair of non-adjacent quadrilaterals is equal to the sum of the areas of the other pair. | 531. Connect the point of intersection of the midlines with the vertices of the quadrilateral and consider the eight resulting triangles. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,107 |
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