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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258. A line $p$ intersects the sides of triangle $ABC$ at points $A_{1}, B_{1}, C_{1}$, respectively, and the cevians drawn through a fixed point $P$ intersect the corresponding sides of the triangle at points $P_{1}, P_{2}, P_{3}$. Prove that
$$
\frac{C P_{2}}{P_{2} A}: \frac{C B_{1}}{B_{1} A}+\frac{C P_{1}}{P_{1} B}... | 258. Design a triangle on another plane so that the secant $p$ becomes an improper line. In the new plane, it is necessary to establish the validity of the equality: $\frac{C^{\prime} P_{2}^{\prime}}{P_{2}{ }^{\prime} A^{\prime}}+$ $+\frac{C^{\prime} P_{1}{ }^{\prime}}{P_{1} \mathbf{1}^{\prime} B^{\prime}}=\frac{C^{\pr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,834 |
259. A line intersects sides $C A$ and $C B$ and median $C M_{3}$ of triangle $A B C$ at points $B_{1}, A_{1}, M_{0}$, respectively. Prove that the segments formed satisfy the equation:
$$
\frac{1}{2}\left(\frac{A B_{1}}{B_{1} C}+\frac{B A_{1}}{A_{1} C}\right)=\frac{M_{3} M_{0}}{M_{0} C}
$$
(assume the segments are o... | 259. Use the result of problem № 258, assuming that point $P$ coincides with the centroid of the triangle.
Fig. 28 | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,835 |
260. 261) The opposite sides of quadrilateral $ABCD$ intersect at points $E$ and $F$ respectively. Lines drawn through the vertices of the quadrilateral parallel to a given direction intersect line $EF$ at points $A_{1}, B_{1}, C_{1}, D_{1}$ respectively. Prove that
$$
\frac{1}{A A_{1}}+\frac{1}{C C_{1}}=\frac{1}{B B_... | 260. Consider triangle $B E C$, intersected by the transversal $F D A$, and apply Menelaus' theorem, taking into account that $\frac{E D}{D C}=\frac{d_{D}}{d_{C}-d_{D}}, \frac{C F}{F B}=-\frac{d_{C}}{d_{B}}$ and $\frac{B A}{A E}=\frac{d_{B}-d_{A}}{d_{A}}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,836 |
262. The corresponding vertices of two parallelograms are connected by segments that are divided in equal ratios, measured from the vertices of one parallelogram. Prove that the points of division are also vertices of a parallelogram and that the centers of the three parallelograms lie on the same straight line.
## § ... | 262. Consider the quadrilaterals $A B B_{1} A_{1}$ and $C D D_{1} C_{1}$, whose vertices coincide with the vertices of parallelograms $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$ (Fig. 29). Let points $A_{0}, B_{0}, C_{0}, D_{0}$ divide the segments $A A_{1}, B B_{1}, C C_{1}, D D_{1}$ in the same ratio. Then the segments $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,838 |
263. Prove that for any right triangle, the inequality holds: $a+b<c+h$. | 263. Notice that $b-h_{c}=k(c-a)$, where $k<1$. Use the similarity of triangles $A C H_{3}$ and $A B C$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 43,839 |
264. On the extensions of sides $CA$ and $CB$ of triangle $ABC$, segments $AA_1$ and $BB_1$ are laid off, equal respectively to segments $AL_3$ and $BL_3$, where $L_3$ is the foot of the angle bisector of angle $C$. Prove that point $L_3$ is the incenter of triangle $A_1 B_1 C$. | 264. Use the property of the angle bisector of a triangle and establish the similarity of triangles $A B C$ and $A_{1} B_{1} C$, from which it follows that $A B \| A_{1} B_{1}$. Then show that $A_{1} L$ is the bisector of angle $C A_{1} B_{1}$.
$1 / 48$ Order $\mathrm{N}_{8} 54$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,840 |
266. Prove that the product of the distances from a vertex of a triangle to its incenter and the corresponding excenter is equal to the product of the sides of the triangle that meet at this vertex. | 266. Establish that triangles $O_{3} A C$ and $O^{\prime} B C$ are similar $\left(\angle A C O_{3}=\angle B C O^{\prime}, \angle C B O^{\prime}=\frac{\beta}{2}\right.$ and $\angle A O_{3} C=180^{\circ}-\left(90^{\circ}+\frac{a}{2}+\right.$ $\left.\left.+\frac{\gamma}{2}\right)=90^{\circ}-\frac{\alpha+\gamma}{2}\right)$... | AC\cdotBC=CO_{3}\cdotCO^{\} | Geometry | proof | Yes | Yes | olympiads | false | 43,842 |
268. A tangent is drawn to a circle at the end $M$ of the diameter $M N$, and a segment $A B$ is laid off on this tangent. The lines $A N$ and $B N$ intersect the circle again at points $A_{1}$ and $B_{1}$. Prove that triangles $N A B$ and $N A_{1} B_{1}$ are similar. | 268. Draw the tangent $N K$ at point $N$ and establish the equality of angles $N A_{1} B_{1}$ and $K N B_{1}$, $K N B_{1}$ and $M B N$, using the properties of angles associated with a circle and formed by parallel lines. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,844 |
269. Inside or outside the given angle through its vertex, two rays are drawn, forming equal angles with the sides of the angle. Prove that the product of the distances from two points, taken respectively on these rays, to one side of the angle or its extension is equal to the product of the distances from these same p... | 269. Drop perpendiculars from two taken points to the sides of the angle and consider two pairs of similar triangles formed. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,845 |
270. On the bisector of an angle, a point is given, through which an arbitrary secant is drawn, cutting off segments $a$ and $b$ on the sides of the angle. Prove that the sum $\frac{1}{a}+\frac{1}{b}$ does not depend on the direction of the secant. Formulate and prove the converse theorem. | 270. Draw a line through the given point parallel to one of the sides of the angle, and consider the similar triangles formed. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,846 |
273. Prove that if two sides of a triangle and the height dropped to the third side satisfy the relation:
$$
\frac{1}{h_{3}^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}
$$
then either angle $C$ is a right angle, or the absolute value of the difference between angles $A$ and $C$ is $90^{\circ}$. | 273. The validity of the given relationship for a right triangle is established on the basis that the height dropped to its hypotenuse divides the triangle into two others similar to the given one, and based on the Pythagorean theorem. If the smaller leg $A C$ of the right triangle $A B C$ is reflected relative to the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,849 |
274. On the leg $A C$ of the right triangle $A B C$, a segment $A N$ is laid off, equal to the radius of the inscribed circle. Prove that the line determined by point $N$ and the midpoint $M_{1}$ of the leg $B C$, passes through the incenter of the triangle. | 274. Draw a line through points $N$ and $O$ and show that it will meet the other leg at its midpoint. Indeed, triangles $N O K$ and $M O L$ are similar ( $K$ and $L$ are the points of tangency of the inscribed circle with the legs $A C$ and $B C$ ), and therefore, $\frac{N K}{r}=\frac{r}{M L}$, or $\frac{b-2 r}{r}=\fra... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,850 |
275. In a right triangle and in the two triangles formed by drawing the altitude from the vertex of the right angle, circles are inscribed with radii $r, r_{1}, r_{2}$ respectively. Prove that $r^{2}=r_{1}^{2}+r_{2}^{2}$. | 275. The given triangle and the two obtained are similar, and therefore, $\frac{r_{1}}{a}=\frac{r_{2}}{b}=\frac{r}{c}$. Hence, due to the equality $a^{2}+b^{2}=c^{2}$, we obtain the desired dependence: $r_{1}{ }^{2}+r_{2}{ }^{2}=r^{2}$. | r_{1}^{2}+r_{2}^{2}=r^{2} | Geometry | proof | Yes | Yes | olympiads | false | 43,851 |
276. On the hypotenuse $A B$ (or its extension) of a right triangle $A B C$, a point $P$ is taken such that $A P=m, B P=n, C P=k$. Prove that $a^{2} m^{2}+b^{2} n^{2}=c^{2} k^{2}$. | 276. Draw $P K \perp B C$ and $P L \perp A C$ and consider two pairs of similar triangles $A P L$ and $A B C, B P K$ and $A B C$, from where $P L=\frac{a m}{c}$
and $P K=\frac{b n}{c}$. But $P L^{2}+P K^{2}=K L^{2}$ and $K L=C P$, so $\left(\frac{a m}{c}\right)^{2}+$ $+\left(\frac{b n}{c}\right)^{2}=k^{2}$ or $a^{2} m^... | ^{2}^{2}+b^{2}n^{2}=^{2}k^{2} | Geometry | proof | Yes | Yes | olympiads | false | 43,852 |
277. Prove that the length of the bisector of a right triangle, drawn from the vertex of the right angle, is determined by the formula: $l_{3}=\frac{a b \sqrt{2}}{a+b}$. How will this formula change for the external bisector? | 277. First solution. Drop a perpendicular from the base of the bisector to one of the legs of the triangle and consider the resulting isosceles triangle with a lateral side of $\frac{a b}{a+b}$.
Second solution. Establish that the hypotenuse is divided by the bisector of the right angle into segments equal to $\frac{b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,853 |
278. Prove that the distance from vertex $C$ of triangle $ABC$ to point $D$, which is symmetric to the center of the circumscribed circle relative to side $AB$, is determined by the formula:
$$
C D^{2}=R^{2}+a^{2}+b^{2}-c^{2}
$$ | 278. The median $C M_{3}$ of triangle $O C D$ is equal to $\frac{C O^{2}+C D^{2}}{2}-\left(\frac{O D}{2}\right)^{2}$. But $C M_{3}$ is also a median of triangle $A B C$, so it is equal to $\frac{a^{2}+b^{2}}{2}-\left(\frac{c}{2}\right)^{2}$. Considering that $\left(\frac{O D}{2}\right)^{2}=R^{2}-\frac{c^{2}}{4}$, we ha... | CD^{2}=R^{2}+^{2}+b^{2}-^{2} | Geometry | proof | Yes | Yes | olympiads | false | 43,854 |
279. Prove that if the angles of triangle $ABC$ satisfy the equation
$$
\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=\frac{9}{4}
$$
then the triangle is equilateral.
46 | 279. From the condition of the problem, based on the sine theorem, it follows that in triangle $ABC$, $a^{2}+b^{2}+c^{2}=9 R^{2}$. However, the distance between the center of the circumscribed circle around the triangle and the orthocenter is $\sqrt{9 R^{2}-a^{2}-b^{2}-c^{2}}$ (see formula 12), therefore, points $O$ an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,855 |
280. The difference between angles $B$ and $A$ of triangle $ABC$ is $90^{\circ}$. Prove that the diameter of the circumcircle of this triangle is determined by the formula: $2 R=\frac{b^{2}-a^{2}}{c}$. | 280. Construct a perpendicular from vertex $B$ to side $A B$, which will intersect side $A C$ at point $D$, and consider the two similar triangles $A B C$ and $B C D$, showing that $B D=\frac{a c}{b}$ and $D C=\frac{a^{2}}{b}$. Then, from the right triangle $A B D$, establish the relationship between the sides of the t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,856 |
281. A secant line passing through vertex $C$ of triangle $ABC$ divides it into two triangles such that the radii of the inscribed circles in these triangles are equal. Prove that the segment of the secant line contained within the triangle is equal to
$$
\sqrt{p(p-c)}
$$ | 281. If a secant meets side $A B$ at point $C_{1}$, then, setting $C_{1} A=c_{1}, C_{1} B=c_{2}, C C_{1}=x$, we find $(a+x) c_{1}=(b+x) c_{2}$ and $c_{2}=\frac{c(a+x)}{2 x+a+b}$. Further, use the proportion $\frac{x}{p-c}=\frac{h}{h-2 r}$, where $h$ is the height, and $r$ is the radius of the inscribed circle of triang... | \sqrt{p(p-)} | Geometry | proof | Yes | Yes | olympiads | false | 43,857 |
282. On the side $AB$ of triangle $ABC$ and on its extension, points $M$ and $N$ are given respectively such that the proportion $\frac{AM}{MB} = \frac{AN}{NB}$ holds. Prove that if angle $MCN$ is a right angle, then $CM$ is the angle bisector of angle $C$ of the given triangle. | 282. If point $S$ is the midpoint of segment $M N$, then establish that $S M^{2}=S A \cdot S B$. Further, prove that $S C$ is a tangent to the circumcircle of triangle $A B C$. Consider two equal angles $S M C$ and $M C S$, one of which is measured by the semisum of two arcs, and the other by half of the arc of the cir... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,858 |
283. On the segment $L_{1} L_{2}$, connecting the bases of the two angle bisectors of triangle $A B C$, an arbitrary point is taken. Prove that the distance from this point to side $A B$ is equal to the sum of the distances from this same point to the other two sides of the triangle. | 283. If we denote by $x, y$ and $z$ the distances from an arbitrary point $P$ to the sides of a triangle and consider the similar triangles formed, then $n y + m z = m n$ (1), where $m$ and $n$ are the distances from points $L_{1}$ and $L_{2}$ to the sides $A C$ and $B C$ of triangle $A B C$. Then, draw a line through ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,859 |
284. Prove that for an acute triangle \(ABC\) the following equality holds:
\[
A H \cdot A H_{1} + B H \cdot B H_{2} + C H \cdot C H_{3} = \frac{1}{2} \left( A B^{2} + B C^{2} + C A^{2} \right)
\]
where \(H, H_{1}, H_{2}, H_{3}\) are the orthocenter and the feet of the altitudes of the triangle. How does the formula ... | 284. Using the cosine theorem and the similarity of triangles, establish that $A H \cdot A H_{1}=\frac{b^{2}+c^{2}-a^{2}}{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,860 |
285. Prove that the incenter of triangle $ABC$ divides the bisector of angle $C$ in the ratio $\frac{a+b}{c}$, measured from the vertex. | 285. First solution. Use Van Aubel's theorem (theorem 5).
Second solution. Use the property of the angle bisector of a triangle twice. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,861 |
287. Prove that the necessary and sufficient condition for the point of tangency of side $AB$ with the inscribed circle in triangle $ABC$ to bisect the segment bounded by the bases of the altitude and the median is the equality: $c=\frac{a+b}{2}$. | 287. Establish that $M H=\frac{b^{2}-a^{2}}{2 c}(b>a), A H=\frac{b^{2}+c^{2}-a^{2}}{2 c}$, $A T=p-a$. | \frac{+b}{2} | Geometry | proof | Yes | Yes | olympiads | false | 43,863 |
289. Through each vertex of a triangle, a line is drawn that divides the perimeter of the triangle in half. Prove that these lines intersect at one point. | 289. Notice that the segments on the sides of the triangle are respectively equal to $p-a, p-b, p-c$, and apply the converse of Ceva's theorem (see theorem 4). | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,865 |
295. On the median $C M_{3}$ of triangle $A B C$, a point $P$ is given, through which lines parallel to sides $C A$ and $C B$ are drawn. Prove that if the segments of these lines, enclosed within the triangle, are equal to each other, then the triangle is isosceles. | 295. Since segments $M N$ and $K L$, enclosed within the triangle, are parallel to sides $C B$ and $C A$, quadrilateral $C K P M$ is a parallelogram, the diagonal $M K$ of which is bisected by the median $C M_{3}$. Therefore, segments $M K$ and $A B$ are parallel, and quadrilateral $A M K B$ is an isosceles trapezoid $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,871 |
296. Through the midpoints of the segments of the medians of a triangle, enclosed between the vertices and the centroid of the triangle, lines are drawn, respectively parallel to the sides of the triangle and forming a second triangle. Prove that the sum of the squares of the distances from any point on the plane to th... | 296. Establish that the second triangle can be obtained from the given one by rotating the latter around its centroid by $180^{\circ}$, and use Leibniz's theorem (see theorem 38). | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,872 |
297. The continuations of the medians of a triangle intersect the circumscribed circle at points $A_{1}, B_{1}, C_{1}$. Prove that
$$
\frac{A G}{G A_{1}}+\frac{B G}{G B_{1}}+\frac{C G}{G C_{1}}=3
$$
where $G$ is the centroid of triangle $A B C$. | 297. Using the concept of the power of a point, establish that
$$
\frac{A G}{G A_{1}}+\frac{B G}{G B_{1}}+\frac{C G}{G C_{1}}=\frac{A G^{2}+B G^{2}+C G^{2}}{R^{2}-d^{2}}
$$
But based on Leibniz's theorem, $A G^{2}+B G^{2}+C G^{2}=3 R^{2}-3 d^{2}$. 116 | 3 | Geometry | proof | Yes | Yes | olympiads | false | 43,873 |
298. On the sides of triangle $A B C$, outside it, similar isosceles triangles $A B C_{1}, B C A_{1}, C A B_{1}$ (angles $A_{1}$, $B_{1}$, $C_{1}$ are equal) are constructed. Prove that the lines $A A_{1}, B B_{1}, C C_{1}$ intersect at one point. | 298. Establish that the line $C C_{1}$ (fig. 30) intersects the side $A B$ at a point $C_{2}$ such that $A C_{2}: C_{2} B = (A C C_{1}):(B C C_{1})$, and the side $B C$ is divided by the line $A A_{1}$ at point $A_{2}$ so that $B A_{2}: A_{2} C = (B A A_{1}):(C A A_{1})$, finally, the point $B_{2}$ of intersection of t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,874 |
301. Prove that the product of the distances from the incenter of a triangle to its vertex and to the corresponding excenter is equal to the product of the diameters of the inscribed and circumscribed circles of this triangle. | 301. To use the fact that the midpoint of the segment $O^{\prime} O_{3}$ belongs to
Fig. 30 of the circumcircle, the concept of the power of a point for the incenter $O^{\prime}$, and Eule... | 4Rr | Geometry | proof | Yes | Yes | olympiads | false | 43,877 |
302. Prove that the distance $d$ from vertex $C$ of triangle $ABC$ to the incenter can be expressed by the formula:
$$
d^{2}=a b-4 R r
$$ | 302. Using the results of problems № 266 and 301, show that $d\left(d+O^{\prime} O_{3}\right)=A C \cdot C B$ and $d \cdot O^{\prime} O_{3}=4 R r$, and therefore, $d^{2}+4 R r=A C \cdot C B$. | ^{2}+4Rr=AC\cdotCB | Geometry | proof | Yes | Yes | olympiads | false | 43,878 |
303. Prove that if: 1) $\frac{1}{h_{3}}=\frac{1}{a}+\frac{1}{b}$, then $\angle C \leqslant 120^{\circ}$, 2) $\frac{1}{m_{3}}=$ $=\frac{1}{a}+\frac{1}{b}$, then $\angle C \geqslant 120^{\circ}$. | 303. Draw through the base $L_{3}$ of the bisector $C L_{3}$ a line parallel to $A C$ and meeting $B C$ at point $N$ (Fig. 31). It is easy to verify that $\frac{1}{C N}=\frac{1}{a}+\frac{1}{b}$ and $C L_{3}=2 C N \cos \frac{C}{2}$.
1) $h_{3}=C N \leqslant C L_{3}$; hence $\cos \frac{C}{2} \geqslant \frac{1}{2}, C \leqs... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,879 |
305. A circle is drawn through two vertices of a triangle and its incenter. Prove that the segment of the tangent, drawn from the third vertex to this circle, is the mean proportional between the sides of the triangle meeting at this vertex. | 305. Determine that the center of the circle passing through vertices $A$ and $B$ of triangle $ABC$ and the incenter $O'$ (Fig. 32) lies at point $M$ where segment $O'O_3$ intersects the circumcircle. Therefore, the power of vertex $C$ with respect to the circle is $\mathrm{CO}' \cdot \mathrm{CO}_3$ or the square of se... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,881 |
306. A circle is circumscribed around triangle $ABC$. The tangent to the circle at vertex $C$ intersects the opposite side at point $D$. Prove that point $D$ divides side $AB$ externally in the ratio equal to the ratio of the squares of the adjacent sides of the triangle. | 306. Consider similar triangles $D B C$ and $A C D$ and use the concept of the power of a point. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,882 |
307. Prove that if between the angles $A$ and $B$ of triangle $ABC$ the following relationships hold: 1) $A=2B$, 2) $A=3B$, then between the sides there exist the respective relationships:
1) $a^{2}=b(b+c)$
2) $c^{2}=\frac{1}{b}(a-b)\left(a^{2}-b^{2}\right)$. | 307. 1) Draw the bisector $A L_{1}$ of angle $A$ and describe a circle around triangle $A B L_{1}$. Establish that since $\angle L_{1} B A = \angle L_{1} A C$, $A C$ is a tangent and the power of point $C$ will be equal to: $C A^{2} = C L_{1} \cdot C B$, or $b^{2} = C L_{1} \cdot a$. Considering that $C L_{1} = \frac{a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,883 |
309. A circle is circumscribed around triangle $ABC$. Prove that the distance from the center $O$ of this circle to the altitude $\mathrm{CH}_{3}$ is
$$
\frac{\left|b^{2}-a^{2}\right|}{2 c}
$$ | 309. To determine the distance $O M$ from the center $O$ to the height $\mathrm{CH}_{3}$,
Fig. 31
$$ | 313. Apply the Law of Cosines to triangle $\mathrm{COH}$, where $C O=R, C H=2 R \operatorname{ctg} C, \angle O C H=|\angle A-\angle B|$, assuming without loss of generality that $\angle C<90^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,889 |
314. Prove that the distance between the bases of the perpendiculars dropped from the base of the altitude to the other two sides of the triangle does not depend on the choice of the altitude. | 314. Determine that the distance between the bases of the perpendiculars is expressed through the height and the sine of the angle of the triangle by the formula $d=h_{3} \sin C$, and note that $h_{1} \sin A=h_{2} \sin B=$ $=h_{3} \sin C$. | h_{3}\sinC | Geometry | proof | Yes | Yes | olympiads | false | 43,890 |
315. The median $C M_{3}$ of triangle $A B C$ forms an acute angle $\varepsilon$ with side $A B$. Prove that
$$
\operatorname{ctg} \varepsilon=\frac{1}{2}|\operatorname{ctg} A-\operatorname{ctg} B|
$$ | 315. Notice that $\operatorname{ctg} \varepsilon=\frac{a^{2}-b^{2}}{2 c}: h_{3}(a>b)$. From this, $\operatorname{ctg} \varepsilon=$ $=\frac{b^{2}-a^{2}}{4 S}$. But $\sin A=\frac{2 S}{b c}, \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$ and $\operatorname{ctg} A=\frac{b^{2}+c^{2}-a^{2}}{4 S}$.
. Then, in the right triangle $\mathrm{MCH}_{3}$, the acute angles of which are equal to $\angle A$ and $90^{\circ}-\angle A$, a line is drawn from the vertex $M$ of the right angle at an a... | \operatorname{tg}A\cdot\operatorname{tg}B | Geometry | proof | Yes | Yes | olympiads | false | 43,894 |
319. Through the vertex $C$ of triangle $A B C$, two lines are drawn inside it, forming equal angles with sides $C A$ and $C B$ and intersecting the third side $A B$ at points $M$ and $N$ respectively. Prove that
$$
\frac{C A^{2}}{C B^{2}}=\frac{A M \cdot A N}{B M \cdot B N}
$$ | 319. Express the distances from points $M$ and $N$ to the sides $A C$ and $B C$ of triangle $A B C$ in terms of $A M, B M, A N, B N$, and then use the result of problem № 269 and the Law of Sines. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,895 |
320. The internal and external bisectors of angle $C$ of triangle $ABC$ are equal. Prove that there is a relationship between the sides of this triangle: $\left(b^{2}-a^{2}\right)^{2}=c^{2}\left(a^{2}+b^{2}\right)$. Verify the validity of the converse theorem. | 320. Establish that the bisector $C L_{3}$ of angle $C$ divides the base $A B$ into parts equal to $\frac{b c}{a+b}$ and $\frac{a c}{a+b}$, and then use the Law of Sines for $\triangle A D C$ and perform a series of identical transformations: $\frac{b c}{(a+b) \sin \frac{c}{2}}=\frac{b}{\sin \varphi}(\varphi=\angle A D... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,896 |
322. A line parallel to side $AB$ is drawn through the incenter of triangle $ABC$. Prove that the segment of this line contained within the triangle is equal to
$$
\frac{c(a+b)}{a+b+c}
$$ | 322. The sought segment, parallel to side $A B$, is divided by the incenter into two others, which are respectively equal to $\frac{b c}{2 p}$ and $\frac{a c}{2 p}$, and the sum of them is $\frac{(a+b) c}{a+b+c}$. | \frac{(+b)}{+b+} | Geometry | proof | Yes | Yes | olympiads | false | 43,898 |
323. Prove that for any triangle $ABC$ the following dependence is valid:
$$
O^{\prime} A \cdot O^{\prime} B \cdot O^{\prime} C=4 R r^{2}
$$
where $O^{\prime}$ is the incenter, and $r$ and $R$ are the radii of the inscribed and circumscribed circles of this triangle. | 323. Expressing $O A, O B$ and $O C$ in terms of the radius $r$ of the inscribed circle and the sines of the half-angles of the triangle, and the latter in terms of the sides of the triangle $A B C$, establish that
$$
O A \cdot O B \cdot O C=\frac{r^{3} a b c}{(p-a)(p-b)(p-c)}=\frac{r^{3} a b c p}{S^{2}}=\frac{r^{3} a... | 4Rr^{2} | Geometry | proof | Yes | Yes | olympiads | false | 43,899 |
324. In the plane of an equilateral triangle, a random line is given. Prove that the distances $m, n, p$ from the vertices of the triangle to the line and the height of the triangle are related by the equation:
$$
(m-n)^{2}+(n-p)^{2}+(p-m)^{2}=2 h^{2}
$$ | 324. Without loss of generality, a line can be drawn through one of the vertices of the triangle (Fig. 35). If this line forms an angle $\varphi$ with one side, then with the other side it forms an angle $60^{\circ}+\varphi$, and with the third side $60^{\circ}-\varphi$. After this, it remains to prove that $a^{2}\left... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,900 |
325. In the plane of an equilateral triangle with side $a$, a point is given whose distances from the vertices of the triangle are $m$, $n$, and $p$. Prove that the specified segments satisfy the equation:
$$
a^{4}+m^{4}+n^{4}+p^{4}=a^{2} m^{2}+a^{2} n^{2}+a^{2} p^{2}+m^{2} n^{2}+n^{2} p^{2}+m^{2} p^{2}
$$ | 325. Denote the angles $A B M$ and $M B C$ as $\alpha$ and $\beta$ and establish that $\cos (\alpha \pm \beta) = \frac{1}{2}$, or $\cos \alpha \cdot \cos \beta \pm \sin \alpha \cdot \sin \beta = \frac{1}{2}$. By squaring the last equation and performing transformations, establish that $4 \cos ^{2} \alpha + 4 \cos ^{2} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,901 |
326. In triangle $A B C$, a triangle $A_{1} B_{1} C_{1}$ is inscribed such that points $A_{1}, B_{1}, C_{1}$ divide the sides $B C, C A$, and $A B$ respectively in equal ratios when traversing the contour of the triangle in a certain direction. Prove that both triangles can be equilateral only simultaneously. | 326. If triangle $A B C$ is equilateral, it is easy to prove that triangle $A_{1} B_{1} C_{1}$ with vertices at the points of division of sides $A B, B C$ and $C A$ is also equilateral. To prove the converse, we divide the sides of the equilateral triangle $A_{1} B_{1} C_{1}$ in the same ratio $\lambda\left(A C_{1}: C_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,902 |
328. Prove that if the angles $A, B, C$ of triangle $ABC$ satisfy the equation:
$$
3 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \cos \frac{C}{2} + \sin \frac{3A}{2} \cdot \sin \frac{3B}{2} \cdot \cos \frac{3C}{2} = 0
$$
then the sides of the triangle are related by the dependence:
$$
a^{3} + b^{3} = c^{3}
$$ | 328. Given that $\sin \frac{A}{2} \cdot \sin \frac{B}{2}=\frac{1}{2}\left(\cos \frac{A-B}{2}-\sin \frac{C}{2}\right)$, transform the left side of the given equation into the expression: $\quad 3 \sin A+3 \sin B-3 \sin C-\sin 3 A-\sin 3 B+$ $+\sin 3 C, \quad$ from which $2(\sin A+\sin B-\sin C)+(\sin A-\sin 3 A)+$ $+(\s... | ^{3}+b^{3}=^{3} | Geometry | proof | Yes | Yes | olympiads | false | 43,904 |
331. Prove that the lines connecting the vertices of a triangle with the projections of the incenter of the triangle onto its corresponding perpendicular bisectors intersect at one point.
## § 13. Metric Relations in a Quadrilateral | 331. Determine that the line passing through vertex $C$ and the projection of the incenter onto the perpendicular bisector of side $A B$ divides it into segments of lengths $p-b$ and $p-a$. Use Ceva's theorem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,907 |
332. Prove that if the bisectors of two opposite angles of a quadrilateral intersect on one diagonal, then the bisectors of the other two angles intersect on the other diagonal. | 332. If the bisectors of angles $A$ and $C$ intersect at point $M$ on diagonal $B D$, then $\frac{A B}{A D}=\frac{B M}{M D}=\frac{B C}{C D}$, or $A B \cdot C D=A D \cdot B C$. From this, it follows that $\frac{A B}{B C}=\frac{A D}{D C}$, i.e., the bisectors of angles $B$ and $D$ will divide the diagonal in the same rat... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,908 |
334. The sums of the squares of the opposite sides of a quadrilateral are equal. Prove that the diagonals of the quadrilateral are perpendicular. | 334. Assume the opposite and drop perpendiculars $B P$ and $D Q$ to the diagonal $A C$. Then $A P^{2}-P^{2} C=A B^{2}-B C^{2}$ and $A Q^{2}-$ $-Q C^{2}=A D^{2}-D C^{2}$, but by the condition $A B^{2}-B C^{2}=A D^{2}-D C^{2}$. Therefore, $A P^{2}-P C^{2}=A Q^{2}-Q C^{2}$, which is possible under one condition: $P \equiv... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,910 |
336. Each of the two opposite sides of a quadrilateral is divided into parts proportional to the other two adjacent sides of the quadrilateral. Prove that the latter two sides are equally inclined to the line connecting the points of division. | 336. Points $M$ and $N$ divide sides $A B$ and $C D$ such that $\frac{A M}{M B} = \frac{D N}{N C} = \frac{A D}{B C}$ (Fig. 36). We will subject sides $A D$ and $B C$ to a parallel translation in the direction of side $C D$ by vectors $\overrightarrow{D N}$ and $\overrightarrow{C N}$ so that points $C$ and $D$ coincide ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,912 |
337. Prove that the sum of the squares of the diagonals of a trapezoid is equal to the sum of the squares of its lateral sides, increased by twice the product of the bases. | 337. Apply the cosine theorem twice, considering the angles α and $180^{\circ}-\alpha$; similarly, apply the cosine theorem twice, considering the angles β and $180^{\circ}-\beta$. The angles are opposite the diagonals. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,913 |
342. The diagonals of a quadrilateral are perpendicular. From the point of intersection of the diagonals, 52 perpendiculars are dropped to the sides of the quadrilateral, which are extended to intersect the opposite sides at points $M, N, K, L$. Prove that $M N K L$ is a rectangle, the sides of which are parallel to th... | 342. Let $S$ be the point of intersection of the diagonals $A C$ and $B D$ (Fig. 37), $S P \perp D C, S Q \perp A D$ and the lines $S P$ and $S Q$ intersect $A B$ and $B C$ respectively at points $M$ and $N$. Draw the altitudes $A U$ and $C V$ of triangle $A C D$, intersecting at point $H$ on $D S$. Notice that $A M: M... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,918 |
345. Prove that the diagonals $e$ and $f$ and the pairs of opposite sides $a, c$ and $b, d$ of a quadrilateral satisfy the inequality:
$$
e^{2}+f^{2} \leqslant b^{2}+d^{2}+2 a c
$$
Determine under what conditions equality holds. | 345. Since the square of the median line is calculated by the formula $\mu_{2}{ }^{2}=\frac{1}{4}\left(a^{2}+c^{2}-b^{2}-d^{2}+e^{2}+f^{2}\right) \quad$ and $\mu_{2}{ }^{2} \leqslant \frac{(a+c)^{2}}{4}$, then $a^{2}+c^{2}-$ $-b^{2}-d^{2}+e^{2}+f^{2} \leqslant a^{2}+2 a c+c^{2}$, from which $e^{2}+f^{2} \leqslant b^{2}... | e^{2}+f^{2}\leqslantb^{2}+^{2}+2 | Inequalities | proof | Yes | Yes | olympiads | false | 43,921 |
348. The products of the cosines of opposite angles of a quadrilateral are equal. Prove that the quadrilateral is a trapezoid or a parallelogram. | 348. Transform the equality $\cos A \cdot \cos C = \cos B \cdot \cos (A + B + C)$ into the form $\sin (A + B) = \sin (B + C)$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,924 |
349. In the plane, there is a quadrilateral $A B C D$. For three points $M$, located in its plane but not lying on the same line, the equality holds: $M A^{2}+M C^{2}=M B^{2}+M D^{2}$. Prove that the given quadrilateral is a rectangle. | 349. Connect point $M$ with the midpoints of the diagonals of the quadrilateral and use the geometric locus of points (see page 13, problem 48). Establish the coincidence of the midpoints of the diagonals and the equality of the latter. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,925 |
350. The lateral sides of a quadrilateral are rotated about their midpoints by $90^{\circ}$. Prove that if the given quadrilateral is a trapezoid, then the four resulting points are the vertices of a quadrilateral with equal diagonals, and conversely. | 350. Continue the sides
Fig. 38 sides $A D$ and $B C$ (Fig. 38) until they intersect at point $S$ and connect it to the ends $M$ and $N, P$ and $Q$ of the segments obtained by rotating side... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,926 |
351. Prove that the diagonals of a trapezoid are expressed through its sides by the formulas:
$$
e^{2}=a c+\frac{a d^{2}-c b^{2}}{a-c}, \quad f^{2}=a c+\frac{a b^{2}-c d^{2}}{a-c}
$$
where $a$ and $c$ - are the lengths of the bases of the trapezoid. | 351. Translate side $A D$ parallel to itself onto segment $D C$ and from triangle $A C B$ by Stewart's theorem (theorem 26) determine: $a d^{2}=e^{2}(a-c)+b^{2} c-a c(a-c)$, or $e^{2}=a c+\frac{a d^{2}-c b^{2}}{a-c}$. Similarly determine $f^{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,927 |
353. The sides $AB$ and $D\mathcal{C}$ of the quadrilateral $ABCD$ are divided into parts proportional to the other two adjacent sides of the quadrilateral. Prove that the square of the distance between the points of division is equal to
$$
\frac{bd}{(b+d)^{2}}\left(e^{2}+f^{2}+2bd-a^{2}-c^{2}\right) .
$$
Prove that ... | 353. Points $M$ and $N$ divide the sides $AB$ and $CD$ of quadrilateral $ABCD$ such that $-\frac{AM}{MB}=\frac{DN}{NC}=\frac{AD}{BC}$. Using Stewart's theorem for triangles $ANB$, $DBC$, and $DAC$, as well as the result from problem № 336 and performing a series of transformations, show that $MN^{2}=\frac{AD \cdot BC}{... | \frac{2bd\cos\frac{\psi}{2}}{b+} | Geometry | proof | Yes | Yes | olympiads | false | 43,929 |
354. The bisector of the angle formed by the non-parallel sides of a trapezoid, extended to their point of intersection, is constructed. Prove that the segment \( t \) of this bisector, enclosed between the bases of the trapezoid, is expressed through its sides by the formula:
\[
t^{2}=\frac{4 b d}{(b+d)^{2}}(p-a)(p-c... | 354. The lateral sides $AD$ and $BC$ of trapezoid $ABCD$ intersect at point $S$, forming an angle whose bisector meets the bases $CD$ and $AB$ at points $M$ and $N$. Establish that $\frac{DM}{MC}=\frac{AN}{NB}=\frac{AD}{BC}$, and use the results from problems № 337 and 353. Consequently,
\[
\begin{aligned}
t^{2}= & \f... | ^{2}=\frac{4}{(b+)^{2}}(p-)(p-) | Geometry | proof | Yes | Yes | olympiads | false | 43,930 |
356. Prove that for any inscribed quadrilateral, the absolute value of the difference of the diagonals is not greater than the absolute value of the difference of a pair of opposite sides. | 356. Using the result of problem № 345 and Ptolemy's theorem, we will have:
$$
\left\{\begin{array}{l}
e^{2}+f^{2} \leqslant b^{2}+d^{2}+2 a c \\
2 e f=2 b d+2 a c
\end{array}\right.
$$
From this, $(e-f)^{2} \leqslant(b-d)^{2}$, or $|e-f| \leqslant|b-d|$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,932 |
357. Prove that if the sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of its diagonals, then the quadrilateral is a parallelogram. | 357. Use Euler's theorem (Theorem 39), and prove that the distance between the midpoints of the diagonals of a quadrilateral is zero. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,933 |
358. A circle of radius $r$ is circumscribed around an isosceles trapezoid with bases $2a$ and $2c$. Prove that $r^{2}=ac$. | 358. Connect the center $O$ of the inscribed circle with vertices $A$ and $D$ of trapezoid $A B C D$ and with $K$ - the point of tangency of the lateral side with this circle. From the right triangle $A O D$, it follows that $r^{2}=O K^{2}=A K \cdot K D=a c$. | r^{2}=ac | Geometry | proof | Yes | Yes | olympiads | false | 43,934 |
359. On the line passing through the points of intersection $E$ and $F$ of the opposite sides of a cyclic quadrilateral, a point $P$ is taken arbitrarily, from which perpendiculars are dropped to the sides of the quadrilateral. Prove that the lines connecting the feet of the perpendiculars dropped to opposite sides are... | 359. $P A', P B', P C', P D'$ - perpendiculars dropped from point $P$ to the sides $D A, A B, B C, C D$. Considering right triangles and inscribed quadrilaterals $E A' C' P$ and $P B' D' F$, establish that $\angle D A' C'=\angle E P C'=\psi, \quad \angle C' P B'=\angle A B E=$ $=\angle A D C=\delta, \angle F P B'=\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,935 |
361. A quadrilateral is inscribed in a circle, the products of whose opposite sides are equal. Prove that the distances from the point of intersection of the diagonals of the quadrilateral to its sides are proportional to these sides. | 361. If point $M$ of intersection of diagonals $A C$ and $B D$ (Fig. 39) is at distances $m_{a}, m_{b}, m_{c}$, $\boldsymbol{m}_{d}$ from sides $A B, B C, C D$ and $D A$, then the ratio of the areas of triangles $A M B$ and $B M C$ will be
 at points $P$ and $R$, and the other pair at points $Q$ and $S$. Prove that both pairs of points divide each other harmonically. | 362. From the similarity of triangles $P A Q$ and $Q D R$, it follows that $\frac{P Q}{Q R}=\frac{P A}{D R}$, and from the similarity of triangles $P S B$ and $R S C$, we have that $\frac{P S}{R S}=$ $=\frac{P B}{R C}$ (Fig.40). Then establish the equality of the ratios $\frac{P A}{D R}$ and $\frac{P B}{R C}$, by conne... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,938 |
363. Prove that the squares of the distances from the center of the circle inscribed in a quadrilateral to two of its opposite vertices are in the ratio of the products of the sides of the quadrilateral meeting at the corresponding vertices.
Verify the validity of the theorem for quadrilaterals circumscribed about a c... | 363. Since $\angle A O D + \angle B O C = 2 d$ and $\angle C O E + \angle B O C = 2 d$, then $\angle A O D = \angle C O E$ ( $E$ and $F$ are the points of intersection of lines $B O$ and $D O$ with diagonal $A C$ ). Then from triangle $A O C$ we have:
$$
\frac{A O^{2}}{O C^{2}} = \frac{A F \cdot A E}{C E \cdot C F} = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,939 |
364. On the sides of a convex quadrilateral with perpendicular diagonals, similar isosceles triangles are constructed outside it, the axes of symmetry of which are the perpendicular bisectors of the sides of the given quadrilateral. Prove that the vertices of these isosceles triangles, from which the sides of the given... | 364. Apply the cosine theorem twice for the quadrilateral (see theorem 36). Note that: 1) the midlines of this quadrilateral are equal; 2) the sums of the distances from two adjacent vertices of the quadrilateral, located on one side of the midline, to this midline are equal to each other; 3) the sums of the squares of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,940 |
365. Prove that the bisectors of the angles formed by the extensions of the opposite sides of a cyclic quadrilateral intersect the sides of this quadrilateral at points that are the vertices of a rhombus, the sides of which are parallel to the diagonals of the quadrilateral. | 365. Using the property of angles with the vertex outside the circle and the fact that angles $BMC$ and $ANB$ are bisected by the bisectors intersecting sides $AD$ and $BC$, $CD$ and $AB$ of quadrilateral $ABCD$ at points $K$ and $L$, $P$ and $Q$, respectively, establish the perpendicularity of $KL$ and $PQ$. Consequen... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,941 |
366. The bisectors of the angles formed by the extensions of opposite sides of a convex quadrilateral intersect a certain pair of adjacent sides of the quadrilateral at points $M$ and $N$ respectively. Prove that if the line $M N$ is parallel to a diagonal of the quadrilateral, then a circle can be circumscribed around... | 366. According to the condition $\frac{A M}{M D}=\frac{A N}{N B}$ (Fig. 41); but $\frac{A M}{M D}=\frac{A E}{D E}$, $\frac{A N}{N B}=\frac{A F}{F B}$. Therefore, $\frac{A E}{D E}=\frac{A F}{F B}$, or $\frac{A E}{A F}=\frac{D E}{F B}$. But the sides $A E$ and $A F$ of triangle $A E F$ are inversely proportional to its h... | \angleB=\angleDor\angleB+\angleD=180 | Geometry | proof | Yes | Yes | olympiads | false | 43,942 |
367. Quadrilateral $A B C D$ is inscribed in circle $O$ and circumscribed about circle $O^{\prime}$. Prove that the points of tangency of the inscribed circle divide the opposite sides $A B$ and $D C$ ( $A D$ and $B C$ ) in the same ratios. | 367. Establish that the right triangles $A M O_{1}$ and $O_{1} N C$ are similar (where $M$ and $N$ are the points of tangency of the circle $O_{1}$ with the sides $A B$ and $(D)$, from which it follows that $A M \cdot N C=O_{1} M \cdot O_{1} N=r^{2}$. Similarly, show that $M B \cdot D N=r^{2}$. Therefore, $\frac{A M}{M... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,943 |
368. A quadrilateral is inscribed in a circle and circumscribed about a circle. Prove that the lines connecting the points of tangency of opposite sides are perpendicular.
## § 14. Metric Relations in a Circle | 368. Since $\angle M O_{1} F + \angle B = 180^{\circ}$ and $\angle M O_{1} E = 2 \cdot \angle M N E$ ( $M, E, N$ and $F$ are the points of tangency of the circle $O_{1}$ with the sides $A B, B C$, $C D$ and $D A$ ), then $\angle M N E = \frac{180^{\circ} - \angle B}{2}$. Similarly, $\angle F E N = \frac{180^{\circ} - \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,944 |
370. A circle is circumscribed around triangle $A B C$. A line $H_{1} H_{2}$ is drawn through the feet of the altitudes $A H_{1}$ and $B H_{2}$, intersecting the circle at points $P$ and $Q$, and the extension of the third side at point $S$. Prove that $S P \cdot S Q=S H_{1} \cdot S H_{2}$. | 370. Utilize the properties of a quadrilateral inscribed in a circle and establish the similarity of the following pairs of triangles: $S A P$ and $S E Q, S A B_{1}$ and $S A_{1} B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,946 |
371. A tangent is drawn to a circle at point $M$ of the diameter $M N$. Through the ends of the chord $A B$, parallel to $M N$, the lines $N A$ and $N B$ are drawn, intersecting the tangent at points $P$ and $Q$. Prove that the product $M P \cdot M Q$ is independent of the choice of the parallel chord. | 371. Determine that the triangles are similar and $M P \cdot M Q=M N^{2}$. | MP\cdotMQ=MN^2 | Geometry | proof | Yes | Yes | olympiads | false | 43,947 |
372. At the ends of the diameter $A B$ of circle $O$, tangents are drawn. The tangent line drawn at point $M$ of circle $O$ intersects the tangents and diameter $A B$ at points $K, L$, and $N$ respectively. Prove that $\frac{K M}{M L}=\frac{K N}{N L}$. | 372. Consider similar triangles $A N K$ and $B N L$ and use the equality of tangent segments drawn from a point to a circle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,948 |
373. From point $S$ to circle $O$, two tangents $S T_{1}$ and $S T_{2}$ are drawn. The tangent line drawn at any point $M$ on the circle intersects $S T_{1}, S T_{2}, T_{1} T_{2}$ at points $K, L, N$ respectively. Prove that
$$
\frac{K M}{M L}=\frac{K N}{N L}
$$ | 373. Through point $K$, draw a line parallel to $S T_{2}$ until it intersects line $T_{1} T_{2}$ at point $P$, and establish that $P K = K T_{1} = K M$ and $T_{2} L = L M$. Considering these equalities and the similarity of triangles $N K P$ and $N L T_{2}$, show that $\frac{K N}{N L} = \frac{K P}{L T_{2}} = \frac{K M}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,949 |
374. Through a point $M$ inside a circle, a chord $A B$ is drawn. The distances from point $M$ to the tangents to the circle at points $A$ and $B$ are $m$ and $n$. Prove that the sum $\frac{1}{m}+\frac{1}{n}$ does not depend on the direction of the chord. Prove that if point $M$ lies outside the circle, then the differ... | 374. Draw the diameter $B C$ through point $B$. Triangles $A B C$, $A M N$, and $M B K$ are similar ( $N$ and $K$ are the feet of the perpendiculars dropped from point $M$ to the tangents $A S$ and $B S$ ). Therefore, $\frac{1}{m}=\frac{B C}{A M(M B+A M)}$ and $\frac{1}{n}=\frac{B C}{M B(A M+M B)}$, or $\frac{1}{m}+\fr... | \frac{BC}{AM\cdotMB} | Geometry | proof | Yes | Yes | olympiads | false | 43,950 |
375. Inside the circle $O$, there is a point $M$, through which a chord $A B$ is drawn. Prove that the product
$$
\operatorname{tg} \frac{\angle A O M}{2} \cdot \operatorname{tg} \frac{\angle B O M}{2}
$$
does not depend on the direction of the chord. Express this product in terms of the radius of the circle and the ... | 375. Using the cosine theorem, calculate $A M$ and $B M$ from triangles $A O M$ and $B O M$ and use the fact that $A M \cdot M B = R^2 - O M^2$. Establish that
$$
\operatorname{tg} \frac{\angle A O M}{2} \cdot \operatorname{tg} \frac{\angle B O M}{2} = \frac{R-d}{R+d}
$$ | \frac{R-}{R+} | Geometry | proof | Yes | Yes | olympiads | false | 43,951 |
376. From point $S$, tangents $S T_{1}$ and $S T_{2}$ are drawn to circle $O$. A point $M$ is taken on the chord $T_{1} T_{2}$, and a line is drawn through $M$ perpendicular to $O M$. Prove that the segment of this line contained within the angle $T_{1} S T_{2}$ is bisected by point $M$. | 376. Let a line perpendicular to $O M$ intersect the tangents $T_{1} S$ and $T_{2} S$ at points $K$ and $L$ respectively (Fig. 42).
Fig. 42 Establish that the quadrilaterals $T_{1} K O M$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,952 |
377. From points $A$ and $B$, lying outside a circle, tangents $A T_{1}$ and $B T_{2}$ are drawn to it. Prove that the segment $A B$ is divided by the line $T_{1} T_{2}$ in the ratio $A T_{1}: B T_{2}$. Consider two cases corresponding to the internal and external division $\left(A T_{1} \neq B T_{2}\right)$. | 377. Establish that $T_{1} T_{2}$ is the bisector of angle $B T_{2} C\left(T_{2} C \|\right.$ $\| T_{1} A$ ), and then, using the property of the bisector of a triangle and considering the similar triangles formed, show that $T_{1} T_{2}$ divides the segment $A B$ in the ratio $A T_{1}$ :
$: B T_{2}$ (Fig. 43). If $T_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,953 |
378. In the vertex $A$ of the inscribed triangle $ABC$, a tangent is drawn. A line parallel to side $BC$ intersects the other two sides of the triangle at points $M$ and $N$, the circle at points $P$ and $Q$, and the tangent at point $S$. Prove that $S M \cdot S N = S P \cdot S Q$. | 378. Determine that the circle passing through points $A$, $M, N$, is tangent to the line $A S$ at point $A$. Therefore, the power of point $S$ with respect to these circles is the same, hence $A S^{2}=$ $=S M \cdot S N=S P \cdot S Q$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,954 |
379. In the vertex $A$ of the inscribed triangle $ABC$, a tangent is drawn. A line parallel to side $AB$ intersects the other two sides at points $M$ and $N$, the circle at points $P$ and $Q$, and the tangent at point $S$. Prove that $NS \cdot MN = NP \cdot NQ$. | 379. Determine that points $A, M, C, S$ lie on the same circle, and show that each of the desired products is equal to $A N \cdot N C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,955 |
381. Through the point $A$ of intersection of two circles, the secants $C A D$ and $E A F$ are drawn. Prove that the ratio of the segments $E F$ and $C D$ is equal to the ratio of their distances from the point $B$ - the second point of intersection of these circles. | 381. Establish that triangles $B C D$ and $B E F$ are similar, and use the fact that the corresponding altitudes of similar triangles are proportional to the corresponding sides | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,957 |
383. A quadrilateral is inscribed in a circle, where the products of the opposite sides are equal. Prove that the tangents to the circle at the opposite vertices of the quadrilateral intersect on the diagonals or are parallel to them. | 383. Prove by contradiction. Assume that the tangents to the circle at points $A$ and $C$ meet the diagonal $D B$ at points $M$ and $M_{0}$ (Fig. 44). Then establish that $\frac{d^{2}}{a^{2}}=\frac{D M}{M B}$ and $\frac{c^{2}}{b^{2}}=\frac{D M_{0}}{M_{0} B}$. But since $\frac{d^{2}}{a^{2}}=\frac{c^{2}}{b^{2}}$, the poi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,959 |
384. At point $K$ of circle $O$, a tangent is drawn, on which two equal segments $K M$ and $M N$ are sequentially laid out from point $K$. The line passing through point $N$ and point $L$, which is diametrically opposite to point $K$, intersects the circle at point $P$. Prove that the line $M P$ is a tangent to circle ... | 384. To establish that triangle $K N P$ is a right triangle, and consequently, $K M=M P=M N$ and $\angle P K M=\angle K P M$; since angle $P K M$, formed by the tangent and the chord, is equal to the angle between the same chord $K P$ and the line $P M$, the latter is tangent to the circle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,960 |
385. A tangent is drawn to a circle at the end $M$ of the diameter $M N$, and on it, equal segments $A B$ and $B C$ are laid out successively. Lines passing through point $N$ and points $A, B, C$ intersect the circle at points $A_{1}, B_{1}, C_{1}$. Prove that the tangents to the circle at points $A_{1}$ and $C_{1}$ in... | 385. Establish the similarity of triangles $A B N$ and $A_{1} B_{1} N$, and $B C N$ and $B_{1} C_{1} N$ (draw the tangent to the circle at point $N$), from which it follows that $\frac{A B}{A_{1} B_{1}}=\frac{N B}{N A_{1}}$ and $\frac{B C}{B_{1} C_{1}}=\frac{N B}{N C_{1}}$. Multiplying the last equalities, we obtain th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,961 |
386. A quadrilateral is inscribed in a circle. Prove that if the tangents to the circle at two opposite vertices intersect on one diagonal, then the other two tangents, drawn at the remaining vertices, intersect on the other diagonal or are parallel to it. | 386. Consider two pairs of similar triangles and establish that if the tangents intersect on the diagonal, then the products of the opposite sides of the quadrilateral are equal. Then connect the point of intersection of the other two tangents with a vertex of the quadrilateral and use proof by contradiction to show th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,962 |
388. Through the vertex $A$ of the parallelogram $A B C D$, an arbitrary circle is drawn, intersecting the line $A B$ at point $B^{\prime}$, the line $A C$ at point $C^{\prime}$, and the line $A D$ at point $D^{\prime}$. Prove that $A C \cdot A C^{\prime}=A B \cdot A B^{\prime}+A D \cdot A D^{\prime}$. | 388. By Ptolemy's theorem in the quadrilateral $A B^{\prime} C^{\prime} D^{\prime}$ (Fig. 45) $A C^{\prime} \cdot B^{\prime} D^{\prime}=A B^{\prime} \cdot C^{\prime} D^{\prime}+A D^{\prime} \times$ $\times B^{\prime} C^{\prime}$. Establishing the similarity of triangles $A C D$ and $B^{\prime} C^{\prime} D^{\prime}$ an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,964 |
389. Prove the converse of Ptolemy's theorem: if the product of the diagonals of a quadrilateral is equal to the sum of the products of its opposite sides, then a circle can be circumscribed around such a quadrilateral. | 389. Let the product of the diagonals of quadrilateral $A B C D$ be equal to the sum of the products of the opposite sides (Fig. 46). Construct point $M$ such that $\angle B A M = \angle B D C$ and $\angle A B M = \angle C B D$. From the similarity of triangles $A B M$ and $B C D$, establish that $A B : B D = A M : D C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,965 |
390. Through the point $E$, where two opposite sides of a cyclic quadrilateral intersect, a tangent $ET$ to the circle is drawn. Prove that if one diagonal of the quadrilateral is parallel to the tangent, then the other diagonal bisects the segment $E T$, and conversely. | 390. If sides $A D$ and $B C$ of quadrilateral $A B C D$ intersect at point $E$ and $A C \| E T$, then $E M^{2}=M B \cdot M D=M T^{2}$, which follows from the similarity of triangles $E M D$ and $E M B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,966 |
391. Given two intersecting circles and a line passing through their points of intersection. Prove that the square of the tangent segment, drawn from any point on one circle to the other, is proportional to the distance from this point to the line. | 391. Connect an arbitrary point $P$ of circle $O_{1}$ (from which a tangent $P T$ to circle $O_{2}$ is drawn) with the centers $O_{1}$ and $O_{2}$ (Fig. 47). Denoting the projection of segment $P O_{2}$ onto the line of centers by $K O_{2}$, and the projection of radius $O_{2} A$, drawn to the point of intersection of ... | PT^{2}=2O_{2}O_{1}\cdot | Geometry | proof | Yes | Yes | olympiads | false | 43,967 |
392. Two circles touch each other externally at point $A$. From point $B$, taken on one of the circles, a tangent $B T$ is drawn to the other circle. Prove that the ratio $B A: B T$ does not depend on the choice of point $B$ on the first circle. | 392. Using Stewart's theorem, determine $BA$ from triangle $OBO_{1}$ and replace $O_{1}B^{2}$ with $BT^{2}-r_{1}^{2}$. After performing the transformations, show that $\frac{BA}{BT}=\sqrt{\frac{r}{r+r_{1}}}$. | \sqrt{\frac{r}{r+r_{1}}} | Geometry | proof | Yes | Yes | olympiads | false | 43,968 |
393. Two circles of different radii touch at point $K$, through which a chord $K A$ is drawn in one circle, and a perpendicular chord $K B$ in the other. Prove that the line $A B$ passes through the center of homothety of the circles. How will the theorem change if the circles are equal? | 393. Establish the parallelism of radii $O_{1} A$ and $O_{2} B$ using the property of angles associated with the circle. Consider the internal and external tangency of the given circles. If the circles are equal, then $A B$ maintains a constant length $(2 R)$ and a constant direction. | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,969 |
394. Two circles touch each other externally at point $K$. A common external tangent to these circles touches them at points $T_{1}$ and $T_{2}$, respectively. Prove that the distances from the chords $K T_{1}$ and $K T_{2}$ to the corresponding centers are in the ratio of the cubes of these chords. | 394. Since $\frac{T_{1} K^{2}}{T_{1} T_{2}^{2}}=\frac{r}{r+r_{1}}$ and $\frac{T_{2} K^{2}}{T_{1} T_{2}^{2}}=\frac{r_{1}}{r+r_{1}}$ (see the hint to problem No. 392), then $\frac{T_{1} K^{2}}{T_{2} K^{2}}=\frac{r}{r_{1}}$. Then establish the similarity of triangles $O K M$ and $O_{1} K N$ ($M$ and $N$ are the feet of th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 43,970 |
396. Prove that for any chord $AB$ of a given circle, the ratio $AB^2 : d$, where $d$ is the distance from point $A$ to the tangent to the circle at point $B$, is a constant. | 396. Since $\frac{A B}{d}=\frac{1}{\sin \alpha}(\alpha-$ angle between the chord and the tangent) and $A B=2 R \cdot \sin \beta$ ( $\beta$ - angle at which the chord is seen from any point on the circle), then due to the equality $\alpha=\beta$ we have: $\frac{A B^{2}}{d}=2 R$ | \frac{AB^{2}}{}=2R | Geometry | proof | Yes | Yes | olympiads | false | 43,972 |
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