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2.80. On the sides of triangle $A B C$, triangles $A B C^{\prime}, A B^{\prime} C$ and $A^{\prime} B C$ are constructed externally, and the sum of the angles at vertices $A^{\prime}, B^{\prime}$ and $C^{\prime}$ is a multiple of $180^{\circ}$. Prove that the circumcircles of the constructed triangles intersect at one p... | 2.80. Suppose first that the circumcircles of triangles $A^{\prime} B C$ and $A B^{\prime} C$ do not touch and $P$ is their common point, different from $C$. Then $\angle(P A, P B)=$ $=\angle(P A, P C)+\angle(P C, P B)=\angle\left(B^{\prime} A, B^{\prime} C\right)+\angle\left(A^{\prime} C, A^{\prime} B\right)=\angle\le... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,788 |
2.81. a) On the sides $BC$, $CA$, and $AB$ of triangle $ABC$ (or on their extensions), points $A_1$, $B_1$, and $C_1$ are taken, distinct from the vertices of the triangle. Prove that the circumcircles of triangles $AB_1C_1$, $A_1BC_1$, and $A_1B_1C$ intersect at one point.
b) Points $A_1$, $B_1$, and $C_1$ move along... | 2.81. a) Applying the statement of problem 2.80 to triangles $A B_{1} C_{1}, A_{1} B C_{1}$, and $A_{1} B_{1} C$, constructed on the sides of triangle $A_{1} B_{1} C_{1}$, we obtain the required result.
b) Let $P$ be the point of intersection of the specified circles. We will prove that the measure of the angle $\angle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,789 |
2.82. Inside triangle $A B C$, a point $X$ is taken. Lines $A X, B X$, and $C X$ intersect the sides of the triangle at points $A_{1}, B_{1}$, and $C_{1}$. Prove that if the circumcircles of triangles $A B_{1} C_{1}, A_{1} B C_{1}$, and $A_{1} B_{1} C$ intersect at point $X$, then $X$ is the orthocenter of triangle $A ... | 2.82. The circumcircle of triangle $A B_{1} C_{1}$ passes through point $X$, so $\angle B X C=180^{\circ}-\angle A$. This means that point $X$ lies on the circle that is symmetric to the circumcircle of triangle $A B C$ with respect to side $B C$. It is clear that the three circles, symmetric to the circumcircle of the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,790 |
2.83*. On the sides $B C, C A$ and $A B$ of triangle $A B C$, points $A_{1}, B_{1}$ and $C_{1}$ are taken. Prove that if triangles $A_{1} B_{1} C_{1}$ and $A B C$ are similar and oppositely oriented, then the circumcircles of triangles $A B_{1} C_{1}, A_{1} B C_{1}$ and $A_{1} B_{1} C$ pass through the circumcenter of ... | 2.83. As follows from problem 2.81, b), it is sufficient to prove this for one such triangle $A_{1} B_{1} C_{1}$, for example, for the triangle with vertices at the midpoints of the sides of triangle $A B C$. Let $H$ be the point of intersection of the altitudes of triangle $A_{1} B_{1} C_{1}$, i.e., the center of the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,791 |
2.84*. Points $A^{\prime}, B^{\prime}$, and $C^{\prime}$ are symmetric to some point $P$ with respect to the sides $B C, C A$, and $A B$ of triangle $A B C$.
a) Prove that the circumcircles of triangles $A B^{\prime} C^{\prime}, A^{\prime} B C^{\prime}$, $A^{\prime} B^{\prime} C$, and $A B C$ have a common point.
b) ... | 2.84. a) Let $X$ be the intersection point of the circumcircles of triangles $A B C$ and $A B^{\prime} C^{\prime}$. Then $\angle\left(X B^{\prime}, X C\right)=\angle\left(X B^{\prime}, X A\right)+\angle(X A, X C)=$ $=\angle\left(C^{\prime} B^{\prime}, C^{\prime} A\right)+\angle(B A, B C)$. Since $A C^{\prime}=A P=A B^{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,792 |
2.85. Four lines form four triangles.
a) Prove that the circumcircles of these triangles have a common point (Miquel point).
b) Prove that the centers of the circumcircles of these triangles lie on one circle passing through the Miquel point. | 2.85. a) From the condition of the problem, it follows that no three lines intersect at one point. Let the lines $AB, AC, BC$ intersect the fourth line at points $D, E, F$ respectively (Fig. 2.4). Denote by $P$ the point of intersection of the circumcircles of triangles $ABC$ and $CEF$, different from point $C$. We nee... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,793 |
2.86*. A line intersects the sides $A B, B C$, and $C A$ of a triangle (or their extensions) at points $C_{1}, B_{1}$, and $A_{1} ; O, O_{a}, O_{b}$, and $O_{c}$ are the centers of the circumcircles of triangles $A B C, A B_{1} C_{1}, A_{1} B C_{1}$, and $A_{1} B_{1} C$; $H, H_{a}, H_{b}$, and $H_{c}$ are the orthocent... | 2.86. a) Let $P$ be the Miquel point for the lines $AB, BC, CA$ and $A_1B_1$. The angles between the rays $PA, PB, PC$ and the tangents to the circles $S_a, S_b, S_c$ are respectively $\angle(PB_1, B_1A) = \angle(PC_1, C_1A)$, $\angle(PC_1, C_1B) = \angle(PA_1, A_1B)$, $\angle(PA_1, A_1C) = \angle(PB_1, B_1C)$. Since $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,794 |
2.88*. Points $A, B, C$ and $D$ lie on a circle with center $O$. Lines $A B$ and $C D$ intersect at point $E$, and the circumcircles of triangles $A E C$ and $B E D$ intersect at points $E$ and $P$. Prove that:
a) points $A, D, P$ and $O$ lie on the same circle;
b) $\angle E P O=90^{\circ}$ | 2.88. a) Since $\angle(A P, P D)=\angle(A P, P E)+\angle(P E, P D)=\angle(A C, C D)+$ $+\angle(A B, B D)=\angle(A O, O D)$, points $A, P, D$ and $O$ lie on the same circle.
6) It is clear that $\angle(E P, P O)=\angle(E P, P A)+\angle(P A, P O)=\angle(D C, C A)+$ $+\angle(D A, D O)=90^{\circ}$, since the arcs on which... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,796 |
2.90. In triangle $ABC$, the altitude $AH$ is drawn; $O$ is the center of the circumscribed circle. Prove that $\angle OAH=|\angle B-\angle C|$. | 2.90. Let point $A^{\prime}$ be symmetric to point $A$ with respect to the perpendicular bisector of segment $B C$. Then $\angle O A H=\angle A O A^{\prime} / 2=\angle A B A^{\prime}=|\angle B-\angle C|$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,798 |
2.91. Let $H$ be the orthocenter of triangle $ABC$, and $AA'$ be the diameter of its circumcircle. Prove that the segment $A'H$ bisects the side $BC$. | 2.91. Since $A A^{\prime}$ is a diameter, $A^{\prime} C \perp A C$, therefore $B H \| A^{\prime} C$. Similarly, $C H \| A^{\prime} B$. Hence, $B A^{\prime} C H$ is a parallelogram. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,799 |
2.92. Through the vertices $A$ and $B$ of triangle $ABC$, two parallel lines are drawn, and the lines $m$ and $n$ are symmetric to them with respect to the bisectors of the corresponding angles. Prove that the point of intersection of lines $m$ and $n$ lies on the circumcircle of triangle $ABC$. | 2.92. Let $l$ be a line parallel to the two original lines; $D$ be the point of intersection of lines $m$ and $n$. Then $\angle(A D, D B)=\angle(m, A B)+\angle(A B, n)=\angle(A C, l)+$ $+\angle(l, C B)=\angle(A C, C B)$, which means that point $D$ lies on the circumcircle of triangle $A B C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,800 |
2.93. a) From point $A$, lines are drawn tangent to circle $S$ at points $B$ and $C$. Prove that the center of the inscribed circle of triangle $A B C$ and the center of its excircle, touching side $B C$, lie on circle $S$.
b) Prove that the circle passing through vertices $B$ and $C$ of any triangle $A B C$ and the c... | 2.93. a) Let $O$ be the midpoint of the arc of circle $S$ lying inside triangle $ABC$. Then $\angle CBO = \angle BCO$, and by the property of the angle between a tangent and a chord, $\angle BCO = \angle ABO$. Therefore, $BO$ is the bisector of angle $ABC$, i.e., $O$ is the center of the inscribed circle of triangle $A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,801 |
2.94*. On the sides $AC$ and $BC$ of triangle $ABC$, squares $ACA_1A_2$ and $BCB_1B_2$ are constructed externally. Prove that the lines $A_1B$, $A_2B_2$, and $AB_1$ intersect at one point. | 2.94. If angle $C$ is a right angle, the solution to the problem is obvious: $C$ is the intersection point of the lines $A_{1} B, A_{2} B_{2}, A B_{1}$. If, however, $\angle C \neq 90^{\circ}$, then the circumcircles of the squares $A C A_{1} A_{2}$ and $B C B_{1} B_{2}$ have another common point besides $C$ - point $C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,802 |
2.95*. Circles $S_{1}$ and $S_{2}$ intersect at points $A$ and $B$, and the tangents to $S_{1}$ at these points are radii of $S_{2}$. A point $C$ is taken on the inner arc of $S_{1}$ and connected to points $A$ and $B$ by lines. Prove that the second points of intersection of these lines with $S_{2}$ are the endpoints ... | 2.95. Let $P$ and $O$ be the centers of circles $S_{1}$ and $S_{2}$; $\alpha=\angle A P C, \beta=\angle B P C$; lines $A C$ and $B C$ intersect $S_{2}$ at points $K$ and $L$. Since $\angle O A P=\angle O B P=90^{\circ}$, then $\angle A O B=180^{\circ}-\alpha-\beta$. Further, $\angle L O B=180^{\circ}-2 \angle L B O=2 \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,803 |
2.96*. From the center $O$ of a circle, a perpendicular $OA$ is dropped onto a line $l$. Points $B$ and $C$ are taken on the line $l$ such that $AB = AC$. Through points $B$ and $C$, two secants are drawn, the first intersecting the circle at points $P$ and $Q$, and the second at points $M$ and $N$. The lines $PM$ and ... | 2.96. Consider points $M^{\prime}, P^{\prime}, Q^{\prime}$ and $R^{\prime}$, symmetric to points $M, P, Q$ and $R$ with respect to the line $O A$. Since point $C$ is symmetric to point $B$ with respect to $O A$, the line $P^{\prime} Q^{\prime}$ passes through point $C$. The following equalities are easily verified: $\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,804 |
3.1. The lines $P A$ and $P B$ are tangent to a circle with center $O$ ($A$ and $B$ are the points of tangency). A third tangent to the circle is drawn, intersecting the segments $P A$ and $P B$ at points $X$ and $Y$. Prove that the measure of angle $X O Y$ does not depend on the choice of the third tangent. | 3.1. Let the line $X Y$ touch the given circle at point $Z$. The corresponding sides of triangles $X O A$ and $X O Z$ are equal, so $\angle X O A = \angle X O Z$. Similarly, $\angle Z O Y = \angle B O Y$. Therefore, $\angle X O Y = \angle X O Z + \angle Z O Y = (\angle A O Z + \angle Z O B) / 2 = \angle A O B / 2$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,805 |
3.2. The incircle of triangle $ABC$ touches side $BC$ at point $K$, and the excircle - at point $L$. Prove that $CK = BL = (a + b - c) / 2$, where $a, b, c$ are the lengths of the sides of the triangle. | 3.2. Let $M$ and $N$ be the points of tangency of the inscribed circle with sides $AB$ and $BC$. Then $BK + AN = BM + AM = AB$, so $CK + CN = a + b - c$.
Let $P$ and $Q$ be the points of tangency of the excircle with the extensions of sides $AB$ and $BC$. Then $AP = AB + BP = AB + BL$ and $AQ = AC + CQ = AC + CL$. The... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,806 |
3.3. On the base $AB$ of isosceles triangle $ABC$, a point $E$ is taken, and circles are inscribed in triangles $ACE$ and $ECB$, touching segment $CE$ at points $M$ and $N$. Find the length of segment $MN$, if the lengths of segments $AE$ and $BE$ are known. | 3.3. According to problem $3.2 C M=(A C+C E-A E) / 2$ and $C N=(B C+C E-B E) / 2$. Considering that $A C=B C$, we get $M N=|C M-C N|=|A E-B E| / 2$. | \frac{|AE-BE|}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,807 |
3.4. Quadrilateral $ABCD$ has the property that there exists a circle inscribed in angle $BAD$ and tangent to the extensions of sides $BC$ and $CD$. Prove that $AB + BC = AD + DC$. | 3.4. Let the lines $A B, B C, C D$ and $D A$ touch the circle at points $P, Q, R$ and $S$. Then $C Q=C R=x$, so $B P=B C+C Q=B C+x$ and $D S=D C+C R=$ $=D C+x$. Therefore, $A P=A B+B P=A B+B C+x$ and $A S=A D+D S=$ $=A D+D C+x$. Considering that $A P=A S$, we obtain the required result. | AB+BC=AD+DC | Geometry | proof | Yes | Yes | olympiads | false | 26,808 |
3.5. The common internal tangent to circles with radii $R$ and $r$ intersects their common external tangents at points $A$ and $B$ and touches one of the circles at point $C$. Prove that $A C \cdot C B = R r$.
3.6* ${ }^{*}$ To two circles of different radii, common external tangents $A B$ and $C D$ are drawn. Prove t... | 3.5. Let the line $A B$ be tangent to circles with centers $O_{1}$ and $O_{2}$ at points $C$ and $D$. Since $\angle O_{1} A O_{2}=90^{\circ}$, the right triangles $A O_{1} C$ and $O_{2} A D$ are similar. Therefore, $O_{1} C: A C=A D: D O_{2}$. Moreover, $A D=C B$ (see problem 3.2). Consequently, $A C \cdot C B=R r$.
!... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,809 |
3.9. Through a point $P$ lying on the common chord $A B$ of two intersecting circles, chords $K M$ of the first circle and $L N$ of the second circle are drawn. Prove that the quadrilateral $K L M N$ is cyclic. | 3.9. Let $P$ be the point of intersection of the diagonals of a convex quadrilateral $A B C D$. The quadrilateral $A B C D$ is cyclic if and only if $\triangle A P B \sim \triangle D P C$, i.e., $P A \cdot P C = P B \cdot P D$. Since the quadrilaterals $A L B N$ and $A M B K$ are cyclic, then $P L \cdot P N = P A \cdot... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,812 |
3.10. Two circles intersect at points $A$ and $B$; $M N$ is a common tangent to them. Prove that the line $A B$ bisects the segment $M N$. | 3.10. Let $O$ be the point of intersection of the line $A B$ and the segment $M N$. Then $O M^{2}=$ $=O A \cdot O B=O N^{2}$, i.e., $O M=O N$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,813 |
3.11. The line $O A$ is tangent to the circle at point $A$, and the chord $B C$ is parallel to $O A$. The lines $O B$ and $O C$ intersect the circle again at points $K$ and $L$. Prove that the line $K L$ bisects the segment $O A$. | 3.11. Let for definiteness the rays $O A$ and $B C$ be collinear; $M$ - the point of intersection of the lines $K L$ and $O A$. Then $\angle L O M=\angle L C B=\angle O K M$, hence $\triangle K O M \sim \triangle O L M$. Therefore, $O M: K M=L M: O M$, i.e., $O M^{2}=$ $=K M \cdot L M$. Moreover, $M A^{2}=M K \cdot M L... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,814 |
3.12. In parallelogram $A B C D$, diagonal $A C$ is greater than diagonal $B D$; $M$ is a point on diagonal $A C$ such that quadrilateral $B C D M$ is cyclic. Prove that line $B D$ is a common tangent to the circumcircles of triangles $A B M$ and $A D M$. | 3.12. Let $O$ be the point of intersection of the diagonals $A C$ and $B D$. Then $M O \cdot O C = B O \cdot O D$. Since $O C = O A$ and $B O = O D$, we have $M O \cdot O A = B O^2$ and $M O \cdot O A = D O^2$. These equalities mean that $O B$ is tangent to the circumcircle of triangle $A B M$ and $O D$ is tangent to t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,815 |
3.13. Given a circle $S$ and points $A$ and $B$ outside it. For each line $l$ passing through point $A$ and intersecting circle $S$ at points $M$ and $N$, consider the circumcircle of triangle $B M N$. Prove that all these circumcircles have a common point different from point $B$. | 3.13. Let $C$ be the intersection point of the line $A B$ with the circumcircle of triangle $B M N$, distinct from point $B$; $A P$ be the tangent from $A$ to the circle $S$. Then $A B \cdot A C = A M \cdot A N = A P^2$, and thus $A C = A P^2 / A B$, meaning that point $C$ is the same for all lines $l$.
Remark. The ca... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,816 |
3.14. Given a circle $S$, points $A$ and $B$ on it, and a point $C$ on the chord $AB$. For each circle $S'$ that is tangent to the chord $AB$ at point $C$ and intersects the circle $S$ at points $P$ and $Q$, consider the point $M$ of intersection of the lines $AB$ and $PQ$. Prove that the position of point $M$ does not... | 3.14. It is clear that $M C^{2}=M P \cdot M Q=M A \cdot M B$, and the point $M$ lies on the ray $A B$ if $A C>B C$, and on the ray $B A$ if $A C<B C$. Let's assume for definiteness that the point $M$ lies on the ray $A B$. Then $(M B+B C)^{2}=(M B+B A) \cdot M B$. Therefore, $M B=B C^{2} /(A B-2 B C)$, which means the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,817 |
3.15. Two circles touch at point $A$. A common (external) tangent is drawn to them, touching the circles at points $C$ and $D$. Prove that $\angle C A D=90^{\circ}$. | 3.15. Let $M$ be the point of intersection of the line $C D$ and the tangent to the circles at point $A$. Then $M C=M A=M D$. Therefore, point $A$ lies on the circle with diameter $C D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,818 |
3.16. Two circles $S_{1}$ and $S_{2}$ with centers $O_{1}$ and $O_{2}$ touch at point $A$. A line through point $A$ intersects $S_{1}$ at point $A_{1}$ and $S_{2}$ at point $A_{2}$. Prove that $O_{1} A_{1} \| O_{2} A_{2}$. | 3.16. Points $O_{1}, A$ and $O_{2}$ lie on the same line, so $\angle A_{2} A O_{2}=\angle A_{1} A O_{1}$. Triangles $A O_{2} A_{2}$ and $A O_{1} A_{1}$ are isosceles, so $\angle A_{2} A O_{2}=\angle A A_{2} O_{2}$ and $\angle A_{1} A O_{1}=\angle A A_{1} O_{1}$. Therefore, $\angle A A_{2} O_{2}=\angle A A_{1} O_{1}$, i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,819 |
3.17. Three circles $S_{1}, S_{2}$, and $S_{3}$ touch each other pairwise at three different points. Prove that the lines connecting the point of tangency of circles $S_{1}$ and $S_{2}$ with the other two points of tangency intersect circle $S_{3}$ at points that are the endpoints of its diameter. | 3.17. Let $O_{1}, O_{2}$ and $O_{3}$ be the centers of circles $S_{1}, S_{2}$ and $S_{3}; A, B, C$ be the points of tangency of circles $S_{2}$ and $S_{3}, S_{3}$ and $S_{1}, S_{1}$ and $S_{2}; A_{1}$ and $B_{1}$ be the points of intersection of lines $C A$ and $C B$ with circle $S_{3}$. According to the previous probl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,820 |
3.18. Two touching circles with centers $O_{1}$ and $O_{2}$ touch internally a circle of radius $R$ with center $O$. Find the perimeter of triangle $O O_{1} O_{2}$. | 3.18. Let $A_{1}, A_{2}$ and $B$ be the points of tangency of the circles with centers $O$ and $O_{1}, O$ and $O_{2}, O_{1}$ and $O_{2}$. Then $O_{1} O_{2}=O_{1} B+B O_{2}=O_{1} A_{1}+O_{2} A_{2}$. Therefore, $O O_{1}+O O_{2}+$
$+O_{1} O_{2}=\left(O O_{1}+O_{1} A_{1}\right)+\left(O O_{2}+O_{2} A_{2}\right)=O A_{1}+O A_... | 2R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,821 |
3.19. Circles $S_{1}$ and $S_{2}$ touch circle $S$ internally at points $A$ and $B$, and one of the intersection points of circles $S_{1}$ and $S_{2}$ lies on the segment $A B$. Prove that the sum of the radii of circles $S_{1}$ and $S_{2}$ is equal to the radius of circle $S$. | 3.19. Let $O, O_{1}$ and $O_{2}$ be the centers of circles $S, S_{1}$ and $S_{2}$; $C$ be the common point of circles $S_{1}$ and $S_{2}$, lying on the segment $A B$. Triangles $A O B, A O_{1} C$ and $C O_{2} B$ are isosceles, so $O O_{1} C O_{2}$ is a parallelogram and $O O_{1}=O_{2} C=O_{2} B$, hence $A O=A O_{1}+O_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,822 |
3.20. The radii of circles $S_{1}$ and $S_{2}$, touching at point $A$, are $R$ and $r (R>r)$. Find the length of the tangent drawn to circle $S_{2}$ from point $B$ on circle $S_{1}$, given that $A B=a$. (Consider the cases of internal and external tangency.) | 3.20. Let $O_{1}$ and $O_{2}$ be the centers of circles $S_{1}$ and $S_{2} ; X$ - the second intersection point of line $A B$ with circle $S_{2}$. The square of the desired tangent length is $B A \cdot B X$. Since $A B: B X=O_{1} A: O_{1} O_{2}$, then $A B \cdot B X=A B^{2} \cdot O_{1} O_{2} / R=$ $=a^{2}(R \pm r) / R$... | ^2\frac{R\r}{R} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,823 |
3.22. Given four circles $S_{1}, S_{2}, S_{3}$ and $S_{4}$, such that circles $S_{i}$ and $S_{i+1}$ touch externally for $i=1,2,3,4\left(S_{5}=S_{1}\right)$. Prove that the points of tangency form a cyclic quadrilateral. | 3.22. Let $O_{i}$ be the center of the circle $S_{i}, A_{i}$ be the point of tangency of the circles $S_{i}$ and $S_{i+1}$. The quadrilateral $O_{1} O_{2} O_{3} O_{4}$ is convex; let $\alpha_{1}, \alpha_{2}, \alpha_{3}$ and $\alpha_{4}$ be the measures of its angles. It is easy to verify that $\angle A_{i-1} A_{i} A_{i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,825 |
3.23. a) Three circles with centers $A, B, C$, touching each other and a line $l$, are arranged as shown in Fig. 3.2. Let $a, b$, and $c$ be the radii of the circles with centers $A, B, C$. Prove that $1 / \sqrt{c} = 1 / \sqrt{a} + 1 / \sqrt{b}$.
b) Four circles touch each other externally (at six different points). Le... | 3.23. a) Let $A_{1}, B_{1}$ and $C_{1}$ be the projections of points $A, B$ and $C$ onto line $l$; $C_{2}$ be the projection of point $C$ onto line $A A_{1}$. By the Pythagorean theorem, $C C_{2}^{2}=A C^{2}-A C_{2}^{2}$, i.e., $A_{1} C_{1}^{2}=(a+c)^{2}-(a-c)^{2}=4 a c$. Similarly, $B_{1} C_{1}^{2}=4 b c$ and $A_{1} B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,826 |
3.24. Three circles of radius $R$ pass through the point $H; A, B$ and $C$ are the points of their pairwise intersections, different from $H$. Prove that:
a) $H$ is the orthocenter of triangle $A B C$;
b) the radius of the circumcircle of triangle $A B C$ is also equal to $R$. | 3.24. Let \( A_{1}, B_{1} \) and \( C_{1} \) be the centers of the given circles (Fig. 3.12). Then \( A_{1} B C_{1} H \) is a rhombus, which means \( B A_{1} \| H C_{1} \). Similarly, \( B_{1} A \| H C_{1} \), so \( B_{1} A \| B A_{1} \) and \( B_{1} A B A_{1} \) is a parallelogram.
a) Since \( A_{1} B_{1} \perp C H \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,827 |
3.26*. Three circles of the same radius pass through point $P$; $A, B$ and $Q$ are the points of their pairwise intersections. A fourth circle of the same radius passes through point $Q$ and intersects the other two at points $C$ and $D$. The triangles $A B Q$ and $C D P$ are acute, and the quadrilateral $A B C D$ is c... | 3.26. Since $\smile A P+\smile B P+\smile P Q=180^{\circ}$ (see problem 3.25), then $\smile A B=$ $=180^{\circ}-\smile P Q$. Similarly, $\smile C D=180^{\circ}-\smile P Q$, i.e., $\smile A B=\smile C D$, hence $A B=C D$. Moreover, $P Q \perp A B$ and $P Q \perp C D$ (see problem 3.24), therefore $A B \| C D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,829 |
3.27. Tangents $A B$ and $A C$ are drawn from point $A$ to a circle with center $O$. Prove that if segment $A O$ is seen from point $M$ at an angle of $90^{\circ}$, then segments $O B$ and $O C$ are seen from it at equal angles. | 3.27. Points $M, B$ and $C$ lie on a circle with diameter $A O$. Moreover, the chords $O B$ and $O C$ of this circle are equal. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,830 |
3.28. Tangents $A B$ and $A C$ are drawn from point $A$ to a circle with center $O$. A line $K L$ is drawn through point $X$ on segment $B C$, perpendicular to $X O$ (points $K$ and $L$ lie on lines $A B$ and $A C$). Prove that $K X=X L$. | 3.28. Points $B$ and $X$ lie on the circle with diameter $K O$, so $\angle X K O = \angle X B O$. Similarly, $\angle X L O = \angle X C O$. Since $\angle X B O = \angle X C O$, triangle $K O L$ is isosceles, with $O X$ being its height. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,831 |
3.29. On the extension of the chord $K L$ of a circle with center $O$, a point $A$ is taken, and tangents $A P$ and $A Q$ are drawn from it; $M$ is the midpoint of the segment $P Q$. Prove that $\angle M K O=\angle M L O$. | 3.29. It is sufficient to check that $A K \cdot A L = A M \cdot A O$. Indeed, then the points $K, L, M$ and $O$ lie on the same circle, and therefore $\angle M K O = \angle M L O$. Since $\triangle A O P \sim \triangle A P M$, then $A M \cdot A O = A P^{2}$; it is also clear that $A K \cdot A L = A P^{2}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,832 |
3.30*. From point $A$, tangents $A B$ and $A C$ to a circle and a secant intersecting the circle at points $D$ and $E$ are drawn; $M$ is the midpoint of segment $B C$. Prove that $B M^{2}=D M \cdot M E$ and angle $D M E$ is twice the angle $D B E$ or angle $D C E$; furthermore, $\angle B E M=\angle D E C$. | 3.30. Let $O$ be the center of the circle; points $D'$ and $E'$ are symmetric to points $D$ and $E$ with respect to the line $AO$. According to problem 28.7, the lines $ED'$ and $E'D$ intersect at point $M$. Therefore, $\angle BDM = \angle EBM$ and $\angle BEM = \angle DBM$, which means $\triangle BDM \sim \triangle EB... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,833 |
3.31*. Quadrilateral $ABCD$ is inscribed in a circle, and the tangents at points $B$ and $D$ intersect at point $K$, which lies on the line $AC$.
a) Prove that $AB \cdot CD = BC \cdot AD$.
b) A line parallel to $KB$ intersects the lines $BA$, $BD$, and $BC$ at points $P$, $Q$, and $R$. Prove that $PQ = QR$. | 3.31. a) Since $\triangle K A B \sim \triangle K B C$, then $A B: B C=K B: K C$. Similarly, $A D: D C=K D: K C$. Considering that $K B=K D$, we obtain the required result.
b) The problem reduces to the previous one, since
$$
\frac{P Q}{B Q}=\frac{\sin P B Q}{\sin B P Q}=\frac{\sin A B D}{\sin K B A}=\frac{\sin A B D}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,834 |
3.32. Given a circle $S$ and a line $l$ that do not have common points. From a point $P$, moving along the line $l$, tangents $P A$ and $P B$ are drawn to the circle $S$. Prove that all chords $A B$ have a common point.
If a point $P$ lies outside the circle $S$, and $P A$ and $P B$ are tangents to the circle, then th... | 3.32. Drop a perpendicular $O M$ from the center $O$ of the circle $S$ to the line $l$. We will prove that the point $X$, where $A B$ and $O M$ intersect, remains fixed. The points $A, B$, and $M$ lie on a circle with diameter $P O$. Therefore, $\angle A M O = \angle A B O = \angle B A O$, which means $\triangle A M O ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,835 |
3.33*. Circles $S_{1}$ and $S_{2}$ intersect at points $A$ and $B$, and the center $O$ of circle $S_{1}$ lies on $S_{2}$. A line passing through point $O$ intersects segment $A B$ at point $P$, and circle $S_{2}$ at point $C$. Prove that point $P$ lies on the polar of point $C$ with respect to circle $S_{1}$.
## §6. A... | 3.33. Since $\angle O B P = \angle O A B = \angle O C B$, then $\triangle O B P \sim \triangle O C B$, and therefore, $O B^{2} = O P \cdot O C$. Draw a tangent $C D$ from point $C$ to the circle $S_{1}$. Then $O D^{2} = O B^{2} = O P \cdot O C$. Consequently, $\triangle O D C \sim \triangle O P D$ and $\angle O P D = \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,836 |
3.34. Points $C$ and $D$ lie on a circle with diameter $A B$. Lines $A C$ and $B D$, $A D$ and $B C$ intersect at points $P$ and $Q$. Prove that $A B \perp P Q$. | 3.34. The lines $B C$ and $A D$ are altitudes of triangle $A P B$, so the line $P Q$, passing through the point $Q$ of their intersection, is perpendicular to the line $A B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,837 |
3.35*. Lines $P C$ and $P D$ are tangent to a circle with diameter $A B$ (points $C$ and $D$ are the points of tangency). Prove that the line connecting $P$ with the intersection point of lines $A C$ and $B D$ is perpendicular to $A B$. | 3.35. Let the points of intersection of the lines $A C$ and $B D$, and $B C$ and $A D$ be denoted by $K$ and $K_{1}$, respectively. According to the previous problem, $K K_{1} \perp A B$, so it is sufficient to prove that the point of intersection of the tangents at points $C$ and $D$ lies on the line $K K_{1}$.
We wi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,838 |
3.36*. Given the diameter $AB$ of a circle and a point $C$ not lying on the line $AB$. Using only a ruler (without a compass), drop a perpendicular from point $C$ to $AB$, if: a) point $C$ does not lie on the circle; b) point $C$ lies on the circle.
3.37*. Let $O_{a}, O_{b}$, and $O_{c}$ be the centers of the circumci... | 3.36. a) The line $AC$ intersects the circle at points $A$ and $A_{1}$, and the line $BC$ intersects the circle at points $B$ and $B_{1}$. If $A = A_{1}$ (or $B = B_{1}$), then the line $AC$ (or $BC$) is the desired perpendicular. If this is not the case, then $AB_{1}$ and $BA_{1}$ are the altitudes of triangle $ABC$, ... | notfound | Geometry | proof | Yes | Yes | olympiads | false | 26,839 |
3.39*. In a circle, two perpendicular diameters are drawn, i.e., four radii, and then four circles are constructed, the diameters of which are these radii. Prove that the total area of the pairwise common parts of these circles is equal to the area of the part of the original circle lying outside the constructed four c... | 3.39. It is sufficient to prove this for each of the four parts into which the diameters divide the original circle (Fig. 3.13). Consider a segment in the circle cut off by a chord subtending a central angle of $90^{\circ}$; let $S$ and $s$ be the areas of such segments for the original and the four constructed circles... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,841 |
3.40*. On three segments $O A, O B$, and $O C$ of equal length (point $B$ lies inside the angle $A O C$), circles are constructed with these segments as diameters. Prove that the area of the curvilinear triangle bounded by the arcs of these circles and not containing point $O$ is half the area of (the usual) triangle $... | 3.40. Let the points of intersection of the circles constructed on segments $O B$ and $O C$, $O A$ and $O C$, $O A$ and $O B$ be denoted as $A_{1}, B_{1}, C_{1}$ respectively (Fig. 3.14). $\angle O A_{1} B = \angle O A_{1} C = 90^{\circ}$, therefore points $B, A_{1}$, and $C$ lie on the same line, and since the circles... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,842 |
3.41*. On the sides of an arbitrary acute-angled triangle $ABC$, circles are constructed with these sides as diameters. This results in three "external" curvilinear triangles and one "internal" one (Fig. 3.7). Prove that if the sum of the areas of the "external" triangles minus the area of the "internal" triangle is ta... | 3.41. The considered circles pass through the feet of the altitudes of the triangle, which means that their points of intersection lie on the sides of the triangle. Let $x, y, z$ and $u$ be the areas of the considered curvilinear triangles; $a, b, c, d, e$ and $f$ be the areas of the segments cut off from the circles b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,843 |
3.42. Chord $AB$ divides circle $S$ into two arcs. Circle $S_{1}$ touches chord $AB$ at point $M$ and one of the arcs at point $N$. Prove that:
a) line $MN$ passes through the midpoint $P$ of the second arc;
b) the length of the tangent $PQ$ to circle $S_{1}$ is equal to $PA$. | 3.42. a) Let $O$ and $O_{1}$ be the centers of circles $S$ and $S_{1}$. Triangles $M O_{1} N$ and $P O N$ are isosceles, and $\angle M O_{1} N = \angle P O N$. Therefore, points $P, M$, and $N$ lie on the same line.
 $B E = B D$. Therefore, $\angle D A E + \angle A D E = \angle D E B = \angle B D E = \angle B D C + \angle C D E$. And since $\angle D A B = \angle B D C$, then $\angle A D E = \angle C D E$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,845 |
3.44*. Two circles inscribed in the segment $AB$ of a given circle intersect at points $M$ and $N$. Prove that the line $MN$ passes through the midpoint $C$ of the arc $AB$ complementary to the given segment. | 3.44. Let $O_{1}$ and $O_{2}$ be the centers of the inscribed circles, and $C P$ and $C Q$ be the tangents to them. Then $C O_{1}^{2}=C P^{2}+P O_{1}^{2}=C P^{2}+O_{1} M^{2}$ and, since $C Q=C A=C P$ (problem 3.42, b)), $C O_{2}^{2}=C Q^{2}+Q O_{2}^{2}=C P^{2}+O_{2} M^{2}$. Therefore, $C O_{1}^{2}-C O_{2}^{2}=$ $=M O_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,846 |
3.45*. On the diameter $A B$ of the circle $S$, a point $K$ is taken, and a perpendicular is erected from it, intersecting $S$ at point $L$. Circles $S_{A}$ and $S_{B}$ are tangent to the circle $S$, the segment $L K$, and the diameter $A B$, with $S_{A}$ touching the segment $A K$ at point $A_{1}$, and $S_{B}$ touchin... | 3.45. Let $\angle L A B=\alpha$ and $\angle L B A=\beta\left(\alpha+\beta=90^{\circ}\right)$. According to problem 3.42, b) $A B_{1}=A L$, so $\angle A B_{1} L=90^{\circ}-\alpha / 2$. Similarly, $\angle B A_{1} L=90^{\circ}-\beta / 2$. Therefore, $\angle A_{1} L B_{1}=(\alpha+\beta) / 2=45^{\circ}$. | 45 | Geometry | proof | Yes | Yes | olympiads | false | 26,847 |
3.46*. A circle touching the sides $A C$ and $B C$ of triangle $A B C$ at points $M$ and $N$, respectively, also touches the circumcircle of the triangle (internally). Prove that the midpoint of segment $M N$ coincides with the center of the inscribed circle of triangle $A B C$. | 3.46. Let $A_{1}$ and $B_{1}$ be the midpoints of the arcs $B C$ and $A C$; $O$ be the center of the inscribed circle. Then $A_{1} B_{1} \perp C O$ (see problem 2.19, a) and $M N \perp C O$, so $M N \| A_{1} B_{1}$. We will move points $M^{\prime}$ and $N^{\prime}$ along the rays $C A$ and $C B$ such that $M^{\prime} N... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,848 |
3.47*. Triangles $A B C_{1}$ and $A B C_{2}$ are inscribed in the circle $S$, and the chords $A C_{2}$ and $B C_{1}$ intersect. The circle $S_{1}$ is tangent to the chord $A C_{2}$ at point $M_{2}$, to the chord $B C_{1}$ at point $N_{1}$, and to the circle $S$. Prove that the centers of the inscribed circles of triang... | 3.47. The solution to this problem generalizes the solution to the previous problem. It is sufficient to prove that the center $O_{1}$ of the inscribed circle of triangle $ABC_{1}$ lies on the segment $M_{2}N_{1}$. Let $A_{1}$ and $A_{2}$ be the midpoints of the arcs $BC_{1}$ and $BC_{2}$, $B_{1}$ and $B_{2}$ be the mi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,849 |
3.48. Two circles have radii $R_{1}$ and $R_{2}$, and the distance between their centers is $d$. Prove that these circles are orthogonal if and only if $d^{2}=R_{1}^{2}+R_{2}^{2}$. | 3.48. Let the circles with centers $O_{1}$ and $O_{2}$ pass through the point $A$. The radii $O_{1} A$ and $O_{2} A$ are perpendicular to the tangents to the circles at point $A$, so the circles are orthogonal if and only if $\angle O_{1} A O_{2}=90^{\circ}$, i.e., $O_{1} O_{2}^{2}=O_{1} A^{2}+O_{2} A^{2}$. | ^{2}=R_{1}^{2}+R_{2}^{2} | Geometry | proof | Yes | Yes | olympiads | false | 26,850 |
3.49. Three circles touch each other externally at points $A, B$ and $C$. Prove that the circumcircle of triangle $A B C$ is perpendicular to all three circles. | 3.49. Let $A_{1}, B_{1}$ and $C_{1}$ be the centers of the given circles, and let points $A, B$ and $C$ lie on segments $B_{1} C_{1}, C_{1} A_{1}$ and $A_{1} B_{1}$ respectively. Since $A_{1} B = A_{1} C, B_{1} A = B_{1} C$ and $C_{1} A = C_{1} B$, points $A, B$ and $C$ are the points of tangency of the incircle of tri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,851 |
3.50. Two circles with centers $O_{1}$ and $O_{2}$ intersect at points $A$ and $B$. A line through point $A$ intersects the first circle at point $M_{1}$ and the second circle at point $M_{2}$. Prove that $\angle B O_{1} M_{1}=\angle B O_{2} M_{2}$.
## §10. Radical Axis | 3.50. It is easy to verify that the angle of rotation from vector $\overrightarrow{O_{i} B}$ to vector $\overrightarrow{O_{i} M_{i}}$ (counterclockwise) is equal to $2 \angle\left(A B, A M_{i}\right)$. It is also clear that $\angle\left(A B, A M_{1}\right)=$ $=\angle\left(A B, A M_{2}\right)$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,852 |
3.51. On a plane, a circle \( S \) and a point \( P \) are given. A line passing through point \( P \) intersects the circle at points \( A \) and \( B \). Prove that the product \( PA \cdot PB \) does not depend on the choice of the line.
This value, taken with a plus sign for a point \( P \) outside the circle and w... | 3.51. Draw another line through point $P$ intersecting the circle at points $A_{1}$ and $B_{1}$. Then $\triangle P A A_{1} \sim \triangle P B_{1} B$, so $P A: P A_{1}=P B_{1}: P B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,853 |
3.52. Prove that for a point $P$ lying outside the circle $S$, its power with respect to $S$ is equal to the square of the length of the tangent drawn from this point. | 3.52. Draw a tangent $PC$ through point $P$. $\triangle PAC \sim \triangle PCB$, therefore $PA: PC = PC: PB$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,854 |
3.53. Prove that the power of a point $P$ with respect to a circle $S$ is equal to $d^{2}-R^{2}$, where $R$ is the radius of $S$, and $d$ is the distance from the point $P$ to the center of $S$. | 3.53. Let the line passing through point $P$ and the center of the circle intersect the circle at points $A$ and $B$. Then $P A = d + R$ and $P B = |d - R|$. Therefore, $P A \cdot P B = \left|d^2 - R^2\right|$. It is also clear that the value $d^2 - R^2$ and the power of point $P$ with respect to the circle $S$ have th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,855 |
3.54*. On a plane, two non-concentric circles $S_{1}$ and $S_{2}$ are given. Prove that the geometric locus of points for which the power with respect to $S_{1}$ is equal to the power with respect to $S_{2}$ is a straight line.
This line is called the radical axis of the circles $S_{1}$ and $S_{2}$. | 3.54. Let $R_{1}$ and $R_{2}$ be the radii of the circles. Consider a coordinate system in which the centers of the circles have coordinates $(-a, 0)$ and $(a, 0)$. According to problem 3.53, the powers of a point with coordinates $(x, y)$ relative to the given circles are $(x+a)^{2}+y^{2}-R_{1}^{2}$ and $(x-a)^{2}+y^{... | (R_{1}^{2}-R_{2}^{2})/4 | Geometry | proof | Yes | Yes | olympiads | false | 26,856 |
3.55*. Prove that the radical axis of two intersecting circles passes through their points of intersection.
3.56 . On a plane, there are three circles, the centers of which do not lie on the same line. Draw the radical axes for each pair of these circles. Prove that all three radical axes intersect at one point.
This... | 3.55. The power of the intersection points of circles relative to each of them is zero, so it lies on the radical axis. If there are two intersection points, then they uniquely define the radical axis. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,857 |
3.57*. On a plane, there are three pairwise intersecting circles. Through the intersection points of any two of them, a line is drawn. Prove that these three lines either intersect at one point or are parallel. | 3.57. According to problem 3.55, the lines containing the chords are radical axes. According to problem 3.56, the radical axes intersect at one point if the centers of the circles do not lie on the same line. Otherwise, they are perpendicular to this line. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,858 |
3.58*. Given two non-concentric circles $S_{1}$ and $S_{2}$. Prove that the set of centers of circles that intersect both these circles at right angles is their radical axis, from which (if the given circles intersect) their common chord is removed. | 3.58. Let $O_{1}$ and $O_{2}$ be the centers of the given circles, and $r_{1}$ and $r_{2}$ their radii. A circle $S$ of radius $r$ with center $O$ is orthogonal to the circle $S_{i}$ if and only if $r^{2}=O O_{i}^{2}-r_{i}^{2}$, i.e., the square of the radius of the circle $S$ is equal to the power of the point $O$ wit... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,859 |
3.60*. On the sides $BC$ and $AC$ of triangle $ABC$, points $A_1$ and $B_1$ are taken; $l$ is a line passing through the common points of the circles with diameters $AA_1$ and $BB_1$. Prove that:
a) the line $l$ passes through the point $H$ of intersection of the altitudes of triangle $ABC$
b) the line $l$ passes thr... | 3.60. a) Let $S_{A}$ and $S_{B}$ be circles with diameters $A A_{1}$ and $B B_{1}$; $S$ be a circle with diameter $A B$. The common chords of the circles $S$ and $S_{A}$, $S$ and $S_{B}$ are the heights $A H_{a}$ and $B H_{b}$, so they (or their extensions) intersect at point $H$. According to problem 3.57, the common ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,861 |
3.61*. The extensions of sides $A B$ and $C D$ of quadrilateral $A B C D$ intersect at point $F$, and the extensions of sides $B C$ and $A D$ intersect at point $E$. Prove that the circles with diameters $A C, B D$, and $E F$ have a common radical axis, and that the orthocenters of triangles $A B E, C D E, A D F$, and ... | 3.61. In triangle $CDE$, let's draw the altitudes $CC_1$ and $DD_1$; let $H$ be their point of intersection. The circles with diameters $AC$ and $BD$ pass through points $C_1$ and $D_1$ respectively, so the power of point $H$ relative to each of these circles is equal to its power relative to the circle with diameter $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,862 |
3.62*. Three circles intersect pairwise at points $A_{1}$ and $A_{2}, B_{1}$ and $B_{2}, C_{1}$ and $C_{2}$. Prove that $A_{1} B_{2} \cdot B_{1} C_{2} \cdot C_{1} A_{2}=A_{2} B_{1} \cdot B_{2} C_{1} \cdot C_{2} A_{1}$. | 3.62. The lines $A_{1} A_{2}, B_{1} B_{2}$ and $C_{1} C_{2}$ intersect at some point $O$ (see problem 3.57). Since $\triangle A_{1} O B_{2} \sim \triangle B_{1} O A_{2}$, we have $A_{1} B_{2}: A_{2} B_{1}=O A_{1}: O B_{1}$. Similarly, $B_{1} C_{2}: B_{2} C_{1}=O B_{1}: O C_{1}$ and $C_{1} A_{2}: C_{2} A_{1}=O C_{1}: O ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,863 |
3.63*. On the side $B C$ of triangle $A B C$, a point $A^{\prime}$ is taken. The perpendicular bisector of segment $A^{\prime} B$ intersects side $A B$ at point $M$, and the perpendicular bisector of segment $A^{\prime} C$ intersects side $A C$ at point $N$. Prove that the point symmetric to point $A^{\prime}$ with res... | 3.63. Let $B^{\prime}$ and $C^{\prime}$ be the points of intersection of the lines $A^{\prime} M$ and $A^{\prime} N$ with the line through point $A$ parallel to $B C$ (Fig. 3.17). Since triangles $A^{\prime} B M$ and $A^{\prime} N C$ are isosceles, $\triangle A B C = \triangle A^{\prime} B^{\prime} C^{\prime}$. Since $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,864 |
3.64*. Solve problem 1.67 using the properties of the radical axis.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.64. Let $AC$ and $BD$ be tangents; $E$ and $K$ be the points of intersection of lines $AC$ and $BD$, $AB$ and $CD$; $O_{1}$ and $O_{2}$ be the centers of the circles (Fig. 3.18). Since $AB \perp O_{1}E$, $O_{1}E \perp O_{2}E$, and $O_{2}E \perp CD$, then $AB \perp CD$, and therefore, $K$ is the point of intersection ... | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,865 | |
3.67*. Prove that the diagonals $A D, B E$ and $C F$ of a circumscribed hexagon $A B C D E F$ intersect at one point (Brianchon).
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.67. Let the convex hexagon $A B C D E F$ touch a circle at points $R, Q, T, S, P, U$ (point $R$ lies on $A B$, $Q$ on $B C$, and so on).
Choose an arbitrary number $a>0$ and construct points $Q^{\prime}$ and $P^{\prime}$ on lines $B C$ and $E F$ such that $Q Q^{\prime} = P P^{\prime} = a$, and vectors $\overrightarr... | proof | Combinatorics | MCQ | Yes | Yes | olympiads | false | 26,867 |
3.68*. Given four circles $S_{1}, S_{2}, S_{3}$, and $S_{4}$, such that circles $S_{i}$ and $S_{i+1}$ touch externally for $i=1,2,3,4\left(S_{5}=S_{1}\right)$. Prove that the radical axis of circles $S_{1}$ and $S_{3}$ passes through the point of intersection of the common external tangents to $S_{2}$ and $S_{4}$. | 3.68. Let $A_{i}$ be the point of tangency of circles $S_{i}$ and $S_{i+1}, X$ be the intersection point of lines $A_{1} A_{4}$ and $A_{2} A_{3}$. Then $X$ is the intersection point of the common external tangents to circles $S_{2}$ and $S_{4}$ (see problem 5.66). Since quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is cyclic... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,868 |
3.69*. a) Circles $S_{1}$ and $S_{2}$ intersect at points $A$ and $B$. The power of point $P$ with respect to circle $S_{1}$ relative to circle $S_{2}$ is $p$, the distance from point $P$ to the line $A B$ is $h$, and the distance between the centers of the circles is $d$. Prove that $|p|=2 d h$.
b) The powers of poin... | 3.69. a) Consider a coordinate system with the origin $O$ at the midpoint of the segment connecting the centers of the circles, and the $O x$ axis directed along this segment. Let the point $P$ have coordinates $(x, y); R$ and $r$ be the radii of the circles $S_{1}$ and $S_{2}; a=d / 2$. Then $(x+a)^{2}+y^{2}=R^{2}$ an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,869 |
4.1. Prove that the medians divide a triangle into six triangles of equal area.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 4.1. Triangles adjacent to one side have equal bases and a common height, so they are equal in area. Let $M$ be the point of intersection of the medians of triangle $ABC$. The line $BM$ cuts each of the triangles $ABC$ and $AMC$ into two equal-area triangles, so $S_{ABM}=S_{BCM}$. Similarly, $S_{BCM}=S_{CAM}$. | Logic and Puzzles | other | Yes | Yes | olympiads | false | 26,870 | |
4.3. Inside the triangle $A B C$, find a point $O$ such that the areas of triangles $B O L, C O M$, and $A O N$ are equal (points $L, M$, and $N$ lie on the sides $A B, B C$, and $C A$, respectively, and $O L \parallel B C, O M \parallel A C$, and $O N \parallel A B$; see Fig. 4.1).
$ and $S_{A C E}=S_{A E O}+S_{C E O}+S_{... | S_{ABCDEF}=2S_{ACE} | Geometry | proof | Yes | Yes | olympiads | false | 26,875 |
4.7*. Inside a convex quadrilateral $A B C D$, there exists a point $O$ such that the areas of triangles $O A B, O B C, O C D$, and $O D A$ are equal. Prove that one of the diagonals of the quadrilateral bisects the other.
## §2. Calculation of areas | 4.7. Let $E$ and $F$ be the midpoints of the diagonals $A C$ and $B D$. Since $S_{A O B}=S_{A O D}$, point $O$ lies on the line $A F$. Similarly, point $O$ lies on the line $C F$. Suppose the intersection point of the diagonals is not the midpoint of either of them. Then the lines $A F$ and $C F$ have only one common p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,876 |
4.8. The height of a trapezoid, whose diagonals are perpendicular to each other, is 4. Find the area of the trapezoid, given that the length of one of its diagonals is 5. | 4.8. Let the diagonal $AC$ of trapezoid $ABCD$ with base $AD$ be equal to 5. Extend triangle $ACB$ to parallelogram $ACBE$. The area of trapezoid $ABCD$ is equal to the area of right triangle $DBE$. Let $BH$ be the height of triangle $DBE$. Then $EH^{2}=BE^{2}-BH^{2}=5^{2}-4^{2}=3^{2}$ and $ED=BE^{2} / EH=25 / 3$. Ther... | \frac{50}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,877 |
4.9. Each diagonal of the convex pentagon $A B C D E$ cuts off a triangle of unit area from it. Calculate the area of the pentagon $A B C D E$. | 4.9. Since $S_{A B E}=S_{A B C}$, then $E C \| A B$. The other diagonals are also parallel to the corresponding sides. Let $P$ be the intersection point of $B D$ and $E C$. If $S_{B P C}=x$, then $S_{A B C D E}=S_{A B E}+S_{E P B}+S_{E D C}+S_{B P C}=3+x$ (since $S_{E P B}=S_{A B E}=1$, as $A B P E$ is a parallelogram)... | \frac{\sqrt{5}+5}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,878 |
4.10. In rectangle $A B C D$, two different rectangles are inscribed, sharing a common vertex $K$ on side $A B$. Prove that the sum of their areas is equal to the area of rectangle $A B C D$. | 4.10. The centers of all three rectangles coincide (see problem 1.7), so the two smaller rectangles have a common diagonal $K L$. Let $M$ and $N$ be the vertices of these rectangles lying on side $B C$. Points $M$ and $N$ lie on the circle with diameter $K L$. Let $O$ be the center of this circle. $O_{1}$ is the projec... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,879 |
4.12*. In the triangle $T_{a}=\triangle A_{1} A_{2} A_{3}$, the triangle $T_{b}=$ $=\triangle B_{1} B_{2} B_{3}$ is inscribed, and in the triangle $T_{b}$, the triangle $T_{c}=\triangle C_{1} C_{2} C_{3}$ is inscribed, with the sides of triangles $T_{a}$ and $T_{c}$ being parallel. Express the area of triangle $T_{b}$ ... | 4.12. Let the areas of triangles $T_{a}, T_{b}$, and $T_{c}$ be $a, b$, and $c$. Triangles $T_{a}$ and $T_{c}$ are homothetic, so the lines connecting their corresponding vertices intersect at one point $O$. The coefficient $k$ of similarity of these triangles is $\sqrt{a / c}$. Clearly, $S_{A_{1} B_{3} O}: S_{C_{1} B_... | \sqrt{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,881 |
4.13. On the sides of triangle $A B C$, points $A_{1}, B_{1}$, and $C_{1}$ are taken, dividing its sides in the ratios $B A_{1}: A_{1} C=p, C B_{1}: B_{1} A=q$ and $A C_{1}: C_{1} B=$ $=r$. The intersection points of segments $A A_{1}, B B_{1}$, and $C C_{1}$ are located as shown in Fig. 4.3. Find the ratio of the area... | 4.13. Using the result of problem 1.3, it is easy to verify that
$$
\frac{B Q}{B B_{1}}=\frac{p+p q}{1+p+p q}, \quad \frac{B_{1} R}{B B_{1}}=\frac{q r}{1+q+q r}, \quad \frac{C R}{C C_{1}}=\frac{q+q r}{1+q+q r}, \quad \frac{C P}{C C_{1}}=\frac{p r}{1+r+p r}
$$
It is also clear that
$$
\frac{S_{P Q R}}{S_{R B_{1} C}}=... | \frac{(1-p)^{2}}{(1+p+pq)(1+q+)(1+r+pr)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,882 |
4.14. The diagonals of quadrilateral $ABCD$ intersect at point $O$. Prove that $S_{AOB}=S_{COD}$ if and only if $BC \| AD$. | 4.14. If $S_{A O B}=S_{C O D}$, then $A O \cdot B O=C O \cdot D O$. Therefore, $\triangle A O D \sim \triangle C O B$ and $A D \| B C$. These arguments are reversible. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,883 |
4.15. a) The diagonals of a convex quadrilateral $ABCD$ intersect at point $P$. The areas of triangles $ABP$, $BCP$, $CDP$ are known. Find the area of triangle $ADP$.
b) A convex quadrilateral is divided by its diagonals into four triangles, the areas of which are expressed as integers. Prove that the product of these... | 4.15. a) Since $S_{A D P}: S_{A B P}=D P: B P=S_{C D P}: S_{B C P}$, then $S_{A D P}=$ $=S_{A B P} \cdot S_{C D P} / S_{B C P}$.
b) According to problem a) $S_{A D P} \cdot S_{C B P}=S_{A B P} \cdot S_{C D P}$. Therefore
$$
S_{A B P} \cdot S_{C B P} \cdot S_{C D P} \cdot S_{A D P}=\left(S_{A D P} \cdot S_{C B P}\righ... | S_{ABP}\cdotS_{CBP}\cdotS_{CDP}\cdotS_{ADP}=(S_{ADP}\cdotS_{CBP})^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,884 |
4.16*. The diagonals of quadrilateral $ABCD$ intersect at point $P$, and $S_{ABP}^{2}+S_{CDP}^{2}=S_{BCP}^{2}+S_{ADP}^{2}$. Prove that $P$ is the midpoint of one of the diagonals. | 4.16. After reducing by $\sin ^{2} \varphi / 4$, where $\varphi-$ is the angle between the diagonals, the given equality of areas can be rewritten as $(A P \cdot B P)^{2}+(C P \cdot D P)^{2}=(B P \cdot C P)^{2}+$ $+(A P \cdot D P)^{2}$, i.e., $\left(A P^{2}-C P^{2}\right)\left(B P^{2}-D P^{2}\right)=0$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,885 |
4.17*. In a convex quadrilateral $A B C D$, there exist three interior points $P_{1}, P_{2}, P_{3}$, not lying on the same line, and possessing the property that the sum of the areas of triangles $A B P_{i}$ and $C D P_{i}$ is equal to the sum of the areas of triangles $B C P_{i}$ and $A D P_{i}$ for $i=1,2,3$. Prove t... | 4.17. Suppose that quadrilateral $A B C D$ is not a parallelogram; for example, lines $A B$ and $C D$ intersect. According to problem 7.2, the set of points $P$ lying inside quadrilateral $A B C D$, for which $S_{A B P}+S_{C D P}=S_{B C P}+S_{A D P}=S_{A B C D} / 2$, is a segment. Therefore, points $P_{1}, P_{2}$ and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,886 |
4.18. Let $K, L, M$ and $N$ be the midpoints of the sides $AB, BC, CD$ and $DA$ of a convex quadrilateral $ABCD$; the segments $KM$ and $LN$ intersect at point $O$. Prove that
$$
S_{AKON} + S_{CLOM} = S_{BKOL} + S_{DNOM}
$$ | 4.18. It is clear that $S_{A K O N}=S_{A K O}+S_{A N O}=\left(S_{A O B}+S_{A O D}\right) / 2$. Similarly, $S_{C L O M}=\left(S_{B C O}+S_{C O D}\right) / 2$. Therefore, $S_{A K O N}+S_{C L O M}=S_{A B C D} / 2$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,887 |
4.19. Points $K, L, M$, and $N$ lie on the sides $A B, B C, C D$, and $D A$ of parallelogram $A B C D$, respectively, and segments $K M$ and $L N$ are parallel to the sides of the parallelogram. These segments intersect at point $O$. Prove that the areas of parallelograms $K B L O$ and $M D N O$ are equal if and only i... | 4.19. If the areas of parallelograms $K B L O$ and $M D N O$ are equal, then $O K \cdot O L = O M \cdot O N$. Considering that $O N = K A$ and $O M = L C$, we get $K O: K A = L C: L O$. Therefore, $\triangle K O A \sim \triangle L C O$, which means point $O$ lies on diagonal $A C$. These arguments are reversible. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,888 |
4.20. On the sides $AB$ and $CD$ of quadrilateral $ABCD$, points $M$ and $N$ are taken such that $AM: MB = CN: ND$. Segments $AN$ and $DM$ intersect at point $K$, and segments $BN$ and $CM$ intersect at point $L$. Prove that $S_{KMLN} = S_{ADK} + S_{BCL}$. | 4.20. Let $h_{1}, h$ and $h_{2}$ be the distances from points $A, M$ and $B$ to the line $C D$. According to problem 1.1, b) $h=p h_{2}+(1-p) h_{1}$, where $p=A M / A B$. Therefore, $S_{D M C}=$ $=h \cdot D C / 2=\left(h_{2} p \cdot D C+h_{1}(1-p) \cdot D C\right) / 2=S_{B C N}+S_{A D N}$. Subtracting $S_{D K N}+S_{C L... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,889 |
4.21. On side $A B$ of quadrilateral $A B C D$, points $A_{1}$ and $B_{1}$ are taken, and on side $C D$ - points $C_{1}$ and $D_{1}$, such that $A A_{1}=B B_{1}=p A B$ and $C C_{1}=$ $=D D_{1}=p C D$, where $p<0.5$. Prove that $S_{A_{1} B_{1} C_{1} D_{1}} / S_{A B C D}=1-2 p$. | 4.21. According to problem $4.20 S_{A B D_{1}}+S_{C D B_{1}}=S_{A B C D}$. Therefore $S_{A_{1} B_{1} C_{1} D_{1}}=$ $=(1-2 p) S_{A B D_{1}}+(1-2 p) S_{C D B_{1}}=(1-2 p) S_{A B C D}$. | S_{A_{1}B_{1}C_{1}D_{1}}=(1-2p)S_{ABCD} | Geometry | proof | Yes | Yes | olympiads | false | 26,890 |
4.22*. Each side of a convex quadrilateral is divided into five equal parts, and the corresponding points of opposite sides are connected (Fig. 4.4). Prove that the area of the middle (shaded) quadrilateral is 25 times smaller than the area of the original.
). Therefore, it is sufficient to prove that the areas of parallelograms $K L M N$ and $K L_{1} M N_{1}$ are equal... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,893 |
4.25*. A square is divided into four parts by two perpendicular lines, the point of intersection of which lies inside it. Prove that if the areas of three of these parts are equal, then the areas of all four parts are equal.
## §5. Various Problems | 4.25. Let the lines $l_{1}$ and $l_{2}$ divide the square into four parts with areas $S_{1}, S_{2}, S_{3}$, and $S_{4}$. For the first line, the areas of the parts into which it divides the square are $S_{1}+S_{2}$ and $S_{3}+S_{4}$, and for the second line, they are $S_{2}+S_{3}$ and $S_{1}+S_{4}$. Since by the condit... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,894 |
4.26. Given a parallelogram $A B C D$ and some point $M$. Prove that $S_{A C M}=\left|S_{A B M} \pm S_{A D M}\right|$. | 4.26. All three considered triangles have a common base $A M$. Let $h_{b}, h_{c}$ and $h_{d}$ be the distances from points $B, C$ and $D$ to the line $A M$. Since $\overrightarrow{A C}=$ $=\overrightarrow{A B}+\overrightarrow{A D}$, then $h_{c}=\left|h_{b} \pm h_{d}\right|$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,895 |
4.27. On the sides $AB$ and $BC$ of triangle $ABC$, parallelograms are constructed externally; $P$ is the point of intersection of the extensions of their sides parallel to $AB$ and $BC$. On side $AC$, a parallelogram is constructed, one of whose sides is equal and parallel to $BP$. Prove that its area is equal to the ... | 4.27. We can consider that $P$ is the common point of the parallelograms constructed on the sides $A B$ and $B C$, i.e., these parallelograms have the form $A B P Q$ and $C B P R$. It is clear that $S_{A C R Q}=S_{A B P Q}+S_{C B P R}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,896 |
4.28*. A point $O$ inside a regular hexagon is connected to the vertices. The six resulting triangles are alternately colored red and blue. Prove that the sum of the areas of the red triangles is equal to the sum of the areas of the blue ones. | 4.28. Let the side of the given hexagon be $a$. The extensions of the red sides of the hexagon form an equilateral triangle with side $3a$, and the sum of the areas of the red triangles is half the product of $a$ and the sum of the distances from point $O$ to the sides of this triangle, so it is equal to $a^{2} \cdot 3... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,897 |
4.29*. The extensions of sides $A D$ and $B C$ of a convex quadrilateral $A B C D$ intersect at point $O ; M$ and $N$ are the midpoints of sides $A B$ and $C D, P$ and $Q$ are the midpoints of diagonals $A C$ and $B D$. Prove that:
a) $S_{P M Q N}=\left|S_{A B D}-S_{A C D}\right| / 2$
b) $S_{O P Q}=S_{A B C D} / 4$. | 4.29. a) The area of parallelogram $P M Q N$ is equal to $B C \cdot A D \sin \alpha / 4$, where $\alpha$ is the angle between lines $A D$ and $B C$. The heights of triangles $A B D$ and $A C D$, dropped from vertices $B$ and $C$, are $O B \sin \alpha$ and $O C \sin \alpha$, respectively. Therefore, $\left|S_{A B D}-S_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,898 |
4.30*. On the sides $A B$ and $C D$ of a convex quadrilateral $A B C D$, points $E$ and $F$ are taken. Let $K, L, M$ and $N$ be the midpoints of segments $D E, B F, C E$ and $A F$. Prove that the quadrilateral $K L M N$ is convex and its area does not depend on the choice of points $E$ and $F$. | 4.30. Segments $K M$ and $L N$ are the midlines of triangles $C E D$ and $A F B$, so they have a common point - the midpoint of segment $E F$. Moreover, $K M = C D / 2, L N = A B / 2$ and the angle between lines $K M$ and $L N$ is equal to the angle $\alpha$ between lines $A B$ and $C D$. Therefore, the area of quadril... | AB\cdotCD\sin\alpha/8 | Geometry | proof | Yes | Yes | olympiads | false | 26,899 |
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