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class | __index_level_0__ int64 0 742k |
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5.87*. On the sides $BC$, $CA$, and $AB$ of triangle $ABC$, points $A_1$, $B_1$, and $C_1$ are taken, such that the lines $AA_1$, $BB_1$, and $CC_1$ intersect at a single point $P$. Prove that the lines $AA_2$, $BB_2$, and $CC_2$, which are symmetric to these lines with respect to the corresponding angle bisectors, als... | 5.87. We can consider that points \(A_{2}, B_{2}\) and \(C_{2}\) lie on the sides of triangle \(A B C\). According to problem 5.86,
\[
\frac{A C_{2}}{C_{2} B} \cdot \frac{B A_{2}}{A_{2} C} \cdot \frac{C B_{2}}{B_{2} A} = \frac{\sin A C C_{2}}{\sin C_{2} C B} \cdot \frac{\sin B A A_{2}}{\sin A_{2} A C} \cdot \frac{\sin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,013 |
5.88*. Opposite sides of a convex hexagon are pairwise parallel. Prove that the lines connecting the midpoints of opposite sides intersect at one point. | 5.88. Let the diagonals $A D$ and $B E$ of the given hexagon $A B C D E F$ intersect at point $P$; $K$ and $L$ are the midpoints of sides $A B$ and $E D$. Since $A B D E$ is a trapezoid, the segment $K L$ passes through point $P$ (Problem 19.2). By the Law of Sines, $\sin A P K: \sin A K P = A K: A P$ and $\sin B P K: ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,014 |
5.89*. From a certain point $P$, perpendiculars $P A_{1}$ and $P A_{2}$ are dropped to the side $B C$ of triangle $A B C$ and to the altitude $A A_{3}$. Similarly, points $B_{1}, B_{2}$ and $C_{1}, C_{2}$ are defined. Prove that the lines $A_{1} A_{2}, B_{1} B_{2}$ and $C_{1} C_{2}$ intersect at one point or are parall... | 5.89. Consider a homothety with center $P$ and coefficient 2. Since $P A_{1} A_{3} A_{2}$ is a rectangle, under this homothety the line $A_{1} A_{2}$ is transformed into the line $l_{a}$ passing through the point $A_{3}$, and the lines $l_{a}$ and $A_{3} P$ are symmetric with respect to the line $A_{3} A$. The line $A_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,015 |
5.90*. Tangents are drawn through points $A$ and $D$ lying on a circle, intersecting at point $S$. Points $B$ and $C$ are taken on the arc $A D$. Lines $A C$ and $B D$ intersect at point $P$, and lines $A B$ and $C D$ intersect at point $Q$. Prove that the line $P Q$ passes through point $S$. | 5.90. According to problems 5.86 and 5.77, b)
$$
\frac{\sin A S P}{\sin P S D} \cdot \frac{\sin D A P}{\sin P A S} \cdot \frac{\sin S D P}{\sin P D A}=1=\frac{\sin A S Q}{\sin Q S D} \cdot \frac{\sin D A Q}{\sin Q A S} \cdot \frac{\sin S D Q}{\sin Q D A}
$$
But $\angle D A P=\angle S D Q, \angle S D P=\angle D A Q, \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,016 |
5.91*. The incircle of triangle $ABC$ touches its sides at points $A_{1}, B_{1}$, and $C_{1}$. Inside triangle $ABC$, a point $X$ is taken. The line $AX$ intersects the arc $B_{1} C_{1}$ of the incircle at point $A_{2}$; points $B_{2}$ and $C_{2}$ are defined similarly. Prove that the lines $A_{1} A_{2}, B_{1} B_{2}$, ... | 5.91. The second equality from problem 2.59, a) means that
Therefore,
$$
\frac{\sin A_{2} A_{1} C_{1}}{\sin A_{2} A_{1} B_{1}} \cdot \frac{\sin B_{2} B_{1} A_{1}}{\sin B_{2} B_{1} C_{1}} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,017 |
5.92*. Inside triangle $ABC$, a point $X$ is taken. The line $AX$ intersects the circumcircle at point $A_{1}$. In the segment cut off by side $BC$, a circle is inscribed, touching the arc $BC$ at point $A_{1}$, and the side $BC$ at point $A_{2}$. Points $B_{2}$ and $C_{2}$ are defined similarly. Prove that the lines $... | 5.92. According to problem 3.42, a) the segment $A_{1} A_{2}$ is the bisector of triangle $A_{1} B C$. Therefore,
$$
\frac{B A_{2}}{C A_{2}}=\frac{B A_{1}}{C A_{1}}=\frac{\sin B A A_{1}}{\sin C A A_{1}}
$$
From the fact that the lines $A A_{1}, B B_{1}$ and $C C_{1}$ intersect at one point, it follows that
$$
\frac{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,018 |
5.93*. a) On the sides $BC$, $CA$, and $AB$ of an isosceles triangle $ABC$ with base $AB$, points $A_1$, $B_1$, and $C_1$ are taken such that the lines $AA_1$, $BB_1$, and $CC_1$ intersect at one point. Prove that
$$
\frac{AC_1}{C_1B} = \frac{\sin ABB_1 \sin CAA_1}{\sin BAA_1 \sin CBB_1}
$$
b) Inside an isosceles tri... | 5.93. a) By Ceva's theorem $\frac{A C_{1}}{C_{1} B}=\frac{C A_{1}}{A_{1} B} \cdot \frac{A B_{1}}{B_{1} C}$, and by the law of sines
$$
\begin{array}{ll}
C A_{1}=\frac{C A \sin C A A_{1}}{\sin A A_{1} B}, & A_{1} B=\frac{A B \sin B A A_{1}}{\sin A A_{1} B} \\
A B_{1}=\frac{A B \sin A B B_{1}}{\sin A B_{1} B}, & B_{1} C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,019 |
5.94*. In triangle $A B C$, the angle bisectors $A A_{1}, B B_{1}$, and $C C_{1}$ are drawn. The angle bisectors $A A_{1}$ and $C C_{1}$ intersect the segments $C_{1} B_{1}$ and $B_{1} A_{1}$ at points $M$ and $N$. Prove that $\angle M B B_{1}=\angle N B B_{1}$.
See also problems $10.56,14.7,14.40$.
## §9. Simson Lin... | 5.94. Let segments $B M$ and $B N$ intersect side $A C$ at points $P$ and $Q$. Then
$$
\frac{\sin P B B_{1}}{\sin P B A}=\frac{\sin P B B_{1}}{\sin B P B_{1}} \cdot \frac{\sin A P B}{\sin P B A}=\frac{P B_{1}}{B B_{1}} \cdot \frac{A B}{P A}
$$
If $O$ is the point of intersection of the angle bisectors of triangle $A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,020 |
5.96*. Points $A, B$ and $C$ lie on the same line, and point $P$ is outside this line. Prove that the centers of the circumscribed circles of triangles $ABP, BCP, ACP$ and point $P$ lie on the same circle. | 5.96. Let $A_{1}, B_{1}$ and $C_{1}$ be the midpoints of segments $P A, P B$ and $P C$; $O_{a}, O_{b}$ and $O_{c}$ be the centers of the circumcircles of triangles $B C P, A C P$ and $A B P$. Points $A_{1}, B_{1}$ and $C_{1}$ are the feet of the perpendiculars dropped from point $P$ to the sides of triangle $O_{a} O_{b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,021 |
5.97*. In triangle $A B C$, the bisector $A D$ is drawn, and perpendiculars $D B^{\prime}$ and $D C^{\prime}$ are dropped from point $D$ to the lines $A C$ and $A B$; point $M$ lies on the line $B^{\prime} C^{\prime}$, such that $D M \perp B C$. Prove that point $M$ lies on the median $A A_{1}$. | 5.97. Let the extension of the bisector $A D$ intersect the circumcircle of triangle $A B C$ at point $P$. Drop perpendiculars $P A_{1}, P B_{1}$ and $P C_{1}$ from point $P$ to the lines $B C, C A$ and $A B$; it is clear that $A_{1}$ is the midpoint of segment $B C$. Under the homothety with center $A$, which maps $P$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,022 |
5.98*. a) From point $P$ on the circumcircle of triangle $A B C$, lines $P A_{1}, P B_{1}$, and $P C_{1}$ are drawn at a given (directed) angle $\alpha$ to the lines $B C, C A$, and $A B$ respectively (points $A_{1}, B_{1}$, and $C_{1}$ lie on the lines $B C, C A$, and $A B$). Prove that points $A_{1}, B_{1}$, and $C_{... | 5.98. a) The solution to problem 5.95 passes without changes in this case as well.
b) Let \( A_{1} \) and \( B_{1} \) be the feet of the perpendiculars dropped from point \( P \) to the lines \( B C \) and \( C A \), and let points \( A_{2} \) and \( B_{2} \) on the lines \( B C \) and \( A C \) be such that \( \angle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,023 |
5.99*. a) From point $P$ on the circumcircle of triangle $A B C$, perpendiculars $P A_{1}$ and $P B_{1}$ are dropped to the lines $B C$ and $A C$. Prove that $P A \cdot P A_{1}=2 R d$, where $R$ is the radius of the circumcircle, and $d$ is the distance from point $P$ to the line $A_{1} B_{1}$.
b) Let $\alpha$ be the ... | 5.99. a) Let the angle between the lines $P C$ and $A C$ be $\varphi$. Then $P A=2 R \sin \varphi$. Since points $A_{1}$ and $B_{1}$ lie on the circle with diameter $P C$, the angle between the lines $P A_{1}$ and $A_{1} B_{1}$ is also $\varphi$. Therefore, $P A_{1}=d / \sin \varphi$, and thus $P A \cdot P A_{1}=2 R d$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,024 |
5.100*. Let $A_{1}$ and $B_{1}$ be the projections of point $P$ on the circumcircle of triangle $A B C$ onto the lines $B C$ and $A C$. Prove that the length of the segment $A_{1} B_{1}$ is equal to the length of the projection of segment $A B$ onto the line $A_{1} B_{1}$. | 5.100. Points $A_{1}$ and $B_{1}$ lie on the circle with diameter $P C$, so $A_{1} B_{1}=$ $=P C \sin A_{1} C B_{1}=P C \sin C$. Let the angle between the lines $A B$ and $A_{1} B_{1}$ be $\gamma$
and $C_{1}$ - the projection of point $P$ onto the line $A_{1} B_{1}$. The lines $A_{1} B_{1}$ and $B_{1} C_{1}$ coincide,... | PC\sinC | Geometry | proof | Yes | Yes | olympiads | false | 27,025 |
5.101*. On a circle, points $P$ and $C$ are fixed; points $A$ and $B$ move along the circle such that the angle $A C B$ remains constant. Prove that the Simson lines of point $P$ with respect to triangles $A B C$ are tangent to a fixed circle. | 5.101. . Let $A_{1}$ and $B_{1}$ be the feet of the perpendiculars dropped from point $P$ to the lines $B C$ and $A C$. The points $A_{1}$ and $B_{1}$ lie on a circle with diameter $P C$. Since $\sin A_{1} C B_{1}=\sin A C B$, the chords $A_{1} B_{1}$ of this circle have a fixed length. Therefore, the lines $A_{1} B_{1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,026 |
5.102*. Point $P$ moves along the circumcircle of triangle $A B C$. Prove that in this case the Simson line of point $P$ relative to triangle $A B C$ rotates by an angle equal to half the angular measure of the arc traversed by point $P$. | 5.102. Let $A_{1}$ and $B_{1}$ be the feet of the perpendiculars dropped from point $P$ to the lines $B C$ and $C A$. Then $\angle\left(A_{1} B_{1}, P B_{1}\right)=\angle\left(A_{1} C, P C\right)=\smile B P / 2$. It is also clear that for all points $P$ the lines $P B_{1}$ have the same direction. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,027 |
5.103*. Prove that the Simson lines of two diametrically opposite points of the circumcircle of triangle $ABC$ are perpendicular, and their point of intersection lies on the nine-point circle (see problem 5.117$)$.
| 5.103. Let $P_{1}$ and $P_{2}$ be diametrically opposite points on the circumcircle of triangle $ABC$; $A_{i}$ and $B_{i}$ be the feet of the perpendiculars dropped from point $P_{i}$ to the lines $BC$ and $AC$; $M$ and $N$ be the midpoints of sides $AC$ and $BC$; $X$ be the intersection point of the lines $A_{1} B_{1}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,028 |
5.104*. Points $A, B, C, P$ and $Q$ lie on a circle with center $O$, and the angles between the vector $\overrightarrow{O P}$ and the vectors $\overrightarrow{O A}, \overrightarrow{O B}, \overrightarrow{O C}$ and $\overrightarrow{O Q}$ are $\alpha, \beta, \gamma$ and $(\alpha+\beta+\gamma) / 2$. Prove that the Simson l... | 5.104. If point $R$ of the given circle is such that $\angle(\overrightarrow{O P}, \overrightarrow{O R})=(\beta + \gamma) / 2$, then $O R \perp B C$. It remains to check that $\angle(O R, O Q)=\angle\left(P A_{1}, A_{1} B_{1}\right)$. But $\angle(O R, O Q)=\alpha / 2$, and $\angle\left(P A_{1}, A_{1} B_{1}\right)=\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,029 |
5.105*. Points $A, B, C$ and $P$ lie on a circle with center $O$. The sides of triangle $A_{1} B_{1} C_{1}$ are parallel to the lines $P A, P B, P C\left(P A \| B_{1} C_{1}\right.$ and so on). Through the vertices of triangle $A_{1} B_{1} C_{1}$, lines are drawn parallel to the sides of triangle $A B C$.
a) Prove that... | 5.105. Without loss of generality, we can assume that the circumcircles of triangles $ABC$ and $A_{1}B_{1}C_{1}$ coincide. Let's define the angles $\alpha, \beta$, and $\gamma$ as in the condition of problem 5.104. We will show that the points with angular coordinates $(\alpha+\beta+\gamma) / 2, (-\alpha+\beta+\gamma) ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,030 |
5.106*. Chord $P Q$ of the circumcircle of triangle $A B C$ is perpendicular to side $B C$. Prove that the Simson line of point $P$ with respect to triangle $A B C$ is parallel to line $A Q$. | 5.106. Let the lines $A C$ and $P Q$ intersect at point $M$. In triangle $M P C$, draw the altitudes $P B_{1}$ and $C A_{1}$. Then $A_{1} B_{1}$ is the Simson line of point $P$ with respect to triangle $A B C$. Moreover, according to problem $1.53$, $\angle\left(M B_{1}, B_{1} A_{1}\right) = \angle(C P, P M)$. It is al... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,031 |
5.107*. The heights of triangle $ABC$ intersect at point $H$; $P$ is a point on its circumcircle. Prove that the Simson line of point $P$ with respect to triangle $ABC$ bisects the segment $PH$. | 5.107. Let's draw the chord $P Q$, perpendicular to $B C$. Let points $H^{\prime}$ and $P^{\prime}$ be symmetric to points $H$ and $P$ with respect to the line $B C$; point $H^{\prime}$ lies on the circumcircle of triangle $A B C$ (problem 5.10). First, we will prove that $A Q \| P^{\prime} H$. Indeed, $\angle\left(A H... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,032 |
5.108*. Quadrilateral $A B C D$ is inscribed in a circle; $l_{a}$ is the Simson line of point $A$ with respect to triangle $B C D$, and lines $l_{b}, l_{c}$, and $l_{d}$ are defined similarly. Prove that these lines intersect at one point. | 5.108. Let $H_{a}, H_{b}, H_{c}$ and $H_{d}$ be the orthocenters of triangles $B C D, C D A, D A B$ and $A B C$. Lines $l_{a}, l_{b}, l_{c}$ and $l_{d}$ pass through the midpoints of segments $A H_{a}, B H_{b}, C H_{c}$ and $D H_{d}$ (see problem 5.107). The midpoints of these segments coincide with a point $H$ such th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,033 |
5.109*. a) Prove that the projections of point $P$ on the Simson lines of triangles $BCD$, $CDA$, $DAB$, and $BAC$ lie on one line (the Simson line of the inscribed quadrilateral).
b) Prove that analogously, by induction, the Simson line of an inscribed $n$-gon can be defined as the line containing the projections of ... | 5.109. a) Let $B_{1}, C_{1}$, and $D_{1}$ be the projections of point $P$ onto the lines $A B, A C$, and $A D$. The points $B_{1}, C_{1}$, and $D_{1}$ lie on a circle with diameter $A P$. The lines $B_{1} C_{1}, C_{1} D_{1}$, and $D_{1} B_{1}$ are the Simson lines of point $P$ with respect to the triangles $A B C, A C ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,034 |
5.110. Let $A_{1} B_{1} C_{1}$ be the pedal triangle of point $P$ with respect to triangle $A B C$. Prove that $B_{1} C_{1}=B C \cdot A P / 2 R$, where $R$ is the radius of the circumcircle of triangle $A B C$. | 5.110. Points $B_{1}$ and $C_{1}$ lie on the circle with diameter $A P$. Therefore, $B_{1} C_{1}=$ $=A P \sin B_{1} A C_{1}=A P(B C / 2 R)$ | B_{1}C_{1}=AP\cdot\frac{BC}{2R} | Geometry | proof | Yes | Yes | olympiads | false | 27,035 |
5.112*. Inside an acute-angled triangle $A B C$, a point $P$ is given. Dropping perpendiculars $P A_{1}, P B_{1}$ and $P C_{1}$ to the sides, we obtain $\triangle A_{1} B_{1} C_{1}$. Performing the same operation for it, we get $\triangle A_{2} B_{2} C_{2}$, and then $\triangle A_{3} B_{3} C_{3}$. Prove that $\triangle... | 5.112. It is clear that $\angle C_{1} A P=\angle C_{1} B_{1} P=\angle A_{2} B_{1} P=\angle A_{2} C_{2} P=\angle B_{3} C_{2} P=$ $=\angle B_{3} A_{3} P$ (the first, third, and fifth equalities follow from the cyclic nature of the corresponding quadrilaterals; the other equalities are obvious). Similarly, $\angle B_{1} A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,037 |
5.113*. Triangle $A B C$ is inscribed in a circle of radius $R$ with center $O$. Prove that the area of the pedal triangle of point $P$ with respect to triangle $A B C$ is $\frac{1}{4}\left|1-\frac{d^{2}}{R^{2}}\right| S_{A B C}$, where $d=P O$. | 5.113. Let $A_{1}, B_{1}$ and $C_{1}$ be the feet of the perpendiculars dropped from point $P$ to the lines $B C, C A$ and $A B$; $A_{2}, B_{2}$ and $C_{2}$ be the points of intersection of the lines $P A, P B$ and $P C$ with the circumcircle of triangle $A B C$. Let $S, S_{1}$ and $S_{2}$ be the areas of triangles $A ... | S_{1}=\frac{1}{4}|1-\frac{^{2}}{R^{2}}|S | Geometry | proof | Yes | Yes | olympiads | false | 27,038 |
5.114*. From point $P$, perpendiculars $P A_{1}, P B_{1}$, and $P C_{1}$ are dropped to the sides of triangle $A B C$. Line $l_{a}$ connects the midpoints of segments $P A$ and $B_{1} C_{1}$. Lines $l_{b}$ and $l_{c}$ are defined similarly. Prove that these lines intersect at one point. | 5.114. Points $B_{1}$ and $C_{1}$ lie on the circle with diameter $P A$, so the midpoint of segment $P A$ is the center of the circumcircle of triangle $A B_{1} C_{1}$. Therefore, $l_{a}$ is the perpendicular bisector of segment $B_{1} C_{1}$. Hence, the lines $l_{a}, l_{b}$, and $l_{c}$ pass through the center of the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,039 |
5.115*. a) Points $P_{1}$ and $P_{2}$ are isogonally conjugate with respect to triangle $ABC$. Prove that their pedal triangles have a common circumcircle, and that its center is the midpoint of segment $P_{1} P_{2}$.
b) Prove that this statement remains true if, instead of perpendiculars from points $P_{1}$ and $P_{2... | 5.115. a) Drop perpendiculars \(P_{1} B_{1}\) and \(P_{2} B_{2}\) from points \(P_{1}\) and \(P_{2}\) to \(A C\), and perpendiculars \(P_{1} C_{1}\) and \(P_{2} C_{2}\) to \(A B\). We will prove that points \(B_{1}, B_{2}, C_{1}\), and \(C_{2}\) lie on the same circle. Indeed, \(\angle P_{1} B_{1} C_{1} = \angle P_{1} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,040 |
5.116*. Let $H$ be the orthocenter of triangle $ABC$, $O$ the circumcenter, and $M$ the centroid. Prove that point $M$ lies on segment $OH$, and that $OM: MH = 1: 2$. (The line containing points $O, M$, and $H$ is called the Euler line). | 5.116. Let $A_{1}, B_{1}$ and $C_{1}$ be the midpoints of sides $B C, C A$ and $A B$. Triangles $A_{1} B_{1} C_{1}$ and $A B C$ are similar, with a similarity coefficient of 2. The altitudes of triangle $A_{1} B_{1} C_{1}$ intersect at point $O$, so $O A_{1}: H A=1: 2$. Let $M^{\prime}$ be the intersection point of seg... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,041 |
5.117*. Prove that the midpoints of the sides of a triangle, the feet of the altitudes, and the midpoints of the segments connecting the orthocenter with the vertices lie on one circle (the nine-point circle), and that the center of this circle is the midpoint of the segment $O H$.
保留源文本的换行和格式，直接输出翻译结果如下：
5.117*. Pro... | 5.117. Let $A_{1}, B_{1}$ and $C_{1}$ be the midpoints of sides $B C, C A$ and $A B$; $A_{2}, B_{2}$ and $C_{2}$ be the feet of the altitudes; $A_{3}, B_{3}$ and $C_{3}$ be the midpoints of the segments connecting the orthocenter with the vertices. Since $A_{2} C_{1}=C_{1} A=A_{1} B_{1}$ and $A_{1} A_{2} \| B_{1} C_{1}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,042 |
5.118*. The altitudes of triangle $ABC$ intersect at point $H$.
a) Prove that triangles $ABC$, $HBC$, $AHC$, and $ABH$ have a common nine-point circle.
b) Prove that the Euler lines of triangles $ABC$, $HBC$, $AHC$, and $ABH$ intersect at one point.
c) Prove that the centers of the circumcircles of triangles $ABC$, $... | 5.118. a) For example, let's prove that triangles $ABC$ and $HBC$ have a common nine-point circle. Indeed, the nine-point circles of both triangles pass through the midpoint of side $BC$ and the midpoints of segments $BH$ and $CH$.
b) The Euler line passes through the center of the nine-point circle, and the nine-poin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,043 |
5.119*. Which sides does the Euler line intersect in an acute and obtuse triangle
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 5.119. Let $A B>B C>C A$. It is easy to verify that for acute and obtuse triangles, the point $H$ of intersection of the altitudes and the center $O$ of the circumscribed circle are located exactly as shown in Fig. 5.5 (i.e., for an acute triangle, the point $O$ lies inside the triangle $B H C_{1}$, and for an obtuse t... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 27,044 | |
5.120*. a) Prove that the circumcircle of triangle $ABC$ is the nine-point circle of the triangle formed by the centers of the excircles of triangle $ABC$.
b) Prove that the circumcircle bisects the segment connecting the centers of the incircle and an excircle of triangle $ABC$. | 5.120. a) Let $O_{a}, O_{b}$ and $O_{c}$ be the centers of the excircles of triangle $A B C$. The vertices of triangle $A B C$ are the feet of the altitudes of triangle $O_{a} O_{b} O_{c}$ (Problem 5.2), so the nine-point circle of triangle $O_{a} O_{b} O_{c}$ passes through points $A, B$ and $C$.
b) Let $O$ be the or... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,045 |
5.121*. Prove that the Euler line of triangle $ABC$ is parallel to side $BC$ if and only if $\operatorname{tg} B \operatorname{tg} C=3$. | 5.121. Let $A A_{1}$ be the altitude, $H$ be the orthocenter. According to problem 5.47, b) $A H=2 R|\cos A|$. The medians are divided by their intersection point in the ratio 1 : 2, so the Euler line is parallel to $B C$ if and only if $A H: A A_{1}=2: 3$ and the vectors $\overrightarrow{A H}$ and $\overrightarrow{A A... | \sinB\sinC=3\cosB\cosC | Geometry | proof | Yes | Yes | olympiads | false | 27,046 |
5.122*. Prove that the segment cut off on side $A B$ of an acute-angled triangle $A B C$ by the nine-point circle is seen from its center at an angle of $2|\angle A-\angle B|$. | 5.122. Let $CD$ be the altitude, $O$ the center of the circumscribed circle, $N$ the midpoint of side $AB$, and point $E$ the midpoint of the segment connecting $C$ with the orthocenter. Then $CENO$ is a parallelogram, so $\angle NED = \angle OCH = |\angle A - \angle B|$ (see problem 2.90). Points $N, E$, and $D$ lie o... | 2|\angleA-\angleB| | Geometry | proof | Yes | Yes | olympiads | false | 27,047 |
5.123*. Prove that if the Euler line passes through the center of the inscribed circle of a triangle, then the triangle is isosceles.
Prove that if the Euler line passes through the center of the inscribed circle of a triangle, then the triangle is isosceles. | 5.123. Let $O$ and $I$ be the centers of the circumcircle and incircle of triangle $ABC$, $H$ be the orthocenter; lines $AI$ and $BI$ intersect the circumcircle at points $A_1$ and $B_1$. Suppose triangle $ABC$ is not isosceles. Then $OI: IH = OA_1: AH$ and $OI: IH = OB_1: BH$. Since $OB_1 = OA_1$, it follows that $AH ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,048 |
5.124*. The inscribed circle touches the sides of triangle $A B C$ at points $A_{1}, B_{1}$ and $C_{1}$. Prove that the Euler line of triangle $A_{1} B_{1} C_{1}$ passes through the center of the circumscribed circle of triangle $A B C$. | 5.124. Let $O$ and $I$ be the centers of the circumcircle and incircle of triangle $ABC$, and $H$ be the orthocenter of triangle $A_{1} B_{1} C_{1}$. In triangle $A_{1} B_{1} C_{1}$, draw the altitudes $A_{1} A_{2}$, $B_{1} B_{2}$, and $C_{1} C_{2}$. Triangle $A_{1} B_{1} C_{1}$ is acute-angled (for example, $\left.\an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,049 |
5.125*. In triangle $A B C$, the altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ are drawn. Let $A_{1} A_{2}, B_{1} B_{2}$, and $C_{1} C_{2}$ be the diameters of the nine-point circle of triangle $A B C$. Prove that the lines $A A_{2}, B B_{2}$, and $C C_{2}$ intersect at one point (or are parallel).
See also problems $3.... | 5.125. Let $H$ be the point of intersection of the altitudes of triangle $ABC$, $E$ and $M$ be the midpoints of segments $CH$ and $AB$ (Fig. 5.6). Then $C_{1} M C_{2} E$ is a rectangle. Let the line $CC_{2}$ intersect the line $AB$ at point $C_{3}$. We will prove that $\overline{A C_{3}}: \overline{C_{3} B} = \operator... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,050 |
5.126*. a) Prove that inside triangle $ABC$ there exists a point $P$ such that $\angle ABP = \angle CAP = \angle BCP$.
b) On the sides of triangle $ABC$, similar triangles $CA_1B$, $CAB_1$, and $C_1AB$ are constructed externally (the angles at the first vertices of all four triangles are equal, etc.). Prove that the l... | 5.126. Let's immediately solve problem b). First, we will prove that the lines \(A A_{1}\), \(B B_{1}\), and \(C C_{1}\) intersect at one point. Suppose the circumcircles of triangles \(A_{1} B C\) and \(A B_{1} C\) intersect at point \(O\). Then \(\angle(B O, O A) = \angle(B O, O C) + \angle(O C, O A) = \angle(B A_{1}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,051 |
5.127*. a) Through the Brocard point $P$ of triangle $ABC$, lines $AP$, $BP$, and $CP$ are drawn, intersecting the circumcircle at points $A_1$, $B_1$, and $C_1$. Prove that $\triangle ABC = \triangle B_1C_1A_1$.
b) Triangle $ABC$ is inscribed in circle $S$. Prove that the triangle formed by the points of intersection ... | 5.127. a) Let's prove that $\smile A B=\smile B_{1} C_{1}$, i.e., $A B=B_{1} C_{1}$. Indeed, $\smile A B=\smile A C_{1}+\smile C_{1} B$, and $\smile C_{1} B=\smile A B_{1}$, therefore $\smile A B=\smile A C_{1}+\smile A B_{1}=\smile B_{1} C_{1}$.
b) We will assume that triangles $A B C$ and $A_{1} B_{1} C_{1}$ are ins... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,052 |
5.128*. a) Let $P$ be the Brocard point of triangle $ABC$. The angle $\varphi = \angle ABP = \angle BCP = \angle CAP$ is called the Brocard angle of this triangle. Prove that $\operatorname{ctg} \varphi = \operatorname{ctg} \alpha + \operatorname{ctg} \beta + \operatorname{ctg} \gamma$.
b) Prove that the Brocard point... | 5.128. a) Since $P C=\frac{A C \sin C A P}{\sin A P C}$ and $P C=\frac{B C \sin C B P}{\sin B P C}$, then $\frac{\sin \varphi \sin \beta}{\sin \gamma}=$ $=\frac{\sin (\beta-\varphi) \sin \alpha}{\sin \beta}$. Considering that $\sin (\beta-\varphi)=\sin \beta \cos \varphi-\cos \beta \sin \varphi$, we get $\operatorname{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,053 |
5.129*. a) Prove that the Brocard angle of any triangle does not exceed $30^{\circ}$.
b) Inside triangle $ABC$, a point $M$ is taken. Prove that one of the angles $ABM$, $BCM$, and $CAM$ does not exceed $30^{\circ}$. | 5.129. а) According to problem 10.38, а) $\operatorname{ctg} \varphi=\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma \geqslant \sqrt{3}=\operatorname{ctg} 30^{\circ}$, therefore $\varphi \leqslant 30^{\circ}$.
б) Let $P$ be the first Brocard point of triangle $A B C$. The point $M$ lies ins... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,054 |
5.130*. Let $Q$ be the second Brocard point of triangle $ABC$, $O$ the center of its circumcircle, and $A_{1}, B_{1}$, and $C_{1}$ the centers of the circumcircles of triangles $CAQ$, $ABQ$, and $BCQ$. Prove that $\triangle A_{1} B_{1} C_{1} \sim \triangle A B C$ and $O$ is the first Brocard point of triangle $A_{1} B_... | 5.130. The lines $A_{1} B_{1}, B_{1} C_{1}$, and $C_{1} A_{1}$ are the perpendicular bisectors of the segments $A Q, B Q$, and $C Q$. Therefore, for example, $\angle B_{1} A_{1} C_{1}=180^{\circ}-\angle A Q C=\angle A$. The proof for the other angles is analogous.
Moreover, the lines $A_{1} O, B_{1} O$, and $C_{1} O$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,055 |
5.131*. Let $P$ be the Brocard point of triangle $ABC$; $R_{1}, R_{2}$, and $R_{3}$ be the radii of the circumcircles of triangles $ABP$, $BCP$, and $CAP$. Prove that $R_{1} R_{2} R_{3}=R^{3}$, where $R$ is the radius of the circumcircle of triangle $ABC$. | 5.131. By the Law of Sines, $R_{1}=A B / 2 \sin A P B, R_{2}=B C / 2 \sin B P C$ and $R_{3}=$ $=C A / 2 \sin C P A$. It is also clear that $\sin A P B=\sin A, \sin B P C=\sin B$ and $\sin C P A=$ $=\sin C$. | R_{1}R_{2}R_{3}=R^{3} | Geometry | proof | Yes | Yes | olympiads | false | 27,056 |
5.132*. Let $P$ and $Q$ be the first and second Brocard points of triangle $ABC$. The lines $CP$ and $BQ$, $AP$ and $CQ$, $BP$ and $AQ$ intersect at points $A_{1}$, $B_{1}$, and $C_{1}$. Prove that the circumcircle of triangle $A_{1} B_{1} C_{1}$ passes through points $P$ and $Q$. | 5.132. Triangle $A B C_{1}$ is isosceles, with the angle at its base $A B$ equal to the Brocard angle $\varphi$. Therefore, $\angle\left(P C_{1}, C_{1} Q\right)=\angle\left(B C_{1}, C_{1} A\right)=2 \varphi$. Similarly, $\angle\left(P A_{1}, A_{1} Q\right)=\angle\left(P B_{1}, B_{1} Q\right)=\angle\left(P C_{1}, C_{1} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,057 |
5.133*. On the sides $C A, A B$, and $B C$ of an acute-angled triangle $A B C$, points $A_{1}, B_{1}$, and $C_{1}$ are taken such that $\angle A B_{1} A_{1}=\angle B C_{1} B_{1}=\angle C A_{1} C_{1}$. Prove that $\triangle A_{1} B_{1} C_{1} \sim \triangle A B C$, and that the center of the spiral similarity that transf... | 5.133. Since $\angle C A_{1} B_{1}=\angle A+\angle A B_{1} A_{1}$ and $\angle A B_{1} A_{1}=\angle C A_{1} C_{1}$, then $\angle B_{1} A_{1} C_{1}=$ $=\angle A$. Similarly, it can be proved that the other angles of triangles $A B C$ and $A_{1} B_{1} C_{1}$ are equal.
The circumcircles of triangles $A A_{1} B_{1}, B B_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,058 |
5.134*. Prove that for the Brocard angle $\varphi$ the following inequalities hold:
a) $\varphi^{3} \leqslant(\alpha-\varphi)(\beta-\varphi)(\gamma-\varphi)$;
b) $8 \varphi^{3} \leqslant \alpha \beta \gamma$ (Jaffee's inequality). | 5.134. a) Consider the function $f(x)=\ln (x / \sin x)=\ln x-\ln \sin x$. It is clear that the functions
$$
f^{\prime}(x)=\frac{1}{x}-\operatorname{ctg} x \quad \text { and } \quad f^{\prime \prime}(x)=\frac{1}{\sin ^{2} x}-\frac{1}{x^{2}}
$$
are positive for $0<x<\pi$. Therefore, the function $f(x)$ is monotonically... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,059 |
5.135*. Let vertices $B$ and $C$ of a triangle be fixed, and vertex $A$ move such that the Brocard angle $\varphi$ of triangle $A B C$ remains constant. Then point $A$ moves along a circle of radius $(a / 2) \sqrt{\operatorname{ctg}^{2} \varphi-3}$, where $a=B C$ (Neuberg circle). | 5.135. According to problem 12.44, a)
$$
\operatorname{ctg} \varphi=\frac{a^{2}+b^{2}+c^{2}}{4 S}
$$
where $S$ is the area of the triangle. Thus, for a triangle with vertices at points with coordinates $( \pm a / 2,0)$ and $(x, y)$, the Brocard angle $\varphi$ is determined by the equation
$$
\operatorname{ctg} \var... | (2)\sqrt{\operatorname{ctg}^{2}\varphi-3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,060 |
5.137. The lines $A M$ and $A N$ are symmetric with respect to the bisector of angle $A$ of triangle $A B C$ (points $M$ and $N$ lie on the line $B C$). Prove that $B M \cdot B N /(C M \cdot C N)=c^{2} / b^{2}$. In particular, if $A S-$ is the symmedian, then $B S / C S=c^{2} / b^{2}$. | 5.137. By the Law of Sines, $A B / B M = \sin A M B / \sin B A M$ and $A B / B N = \sin A N B / \sin B A N$. Therefore,
$$
\frac{A B^{2}}{B M \cdot B N} = \frac{\sin A M B \sin A N B}{\sin B A M \sin B A N} = \frac{\sin A M C \sin A N C}{\sin C A N \sin C A M} = \frac{A C^{2}}{C M \cdot C N}
$$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,062 |
5.138. Express the length of the symmedian $A S$ in terms of the side lengths of triangle $A B C$. | 5.138. Since $\angle B A S=\angle C A M$, then $B S / C M=S_{B A S} / S_{C A M}=$ $=A B \cdot A S /(A C \cdot A M)$, i.e., $A S / A M=2 b \cdot B S / a c$. It remains to note that $B S=$ $=a c^{2} /\left(b^{2}+c^{2}\right)$ and $2 A M=\sqrt{2 b^{2}+2 c^{2}-a^{2}}$ (see problems 5.137 and $\left.12.11, \mathrm{a}\right)... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,063 | |
5.139. The segment $B_{1} C_{1}$, where points $B_{1}$ and $C_{1}$ lie on the rays $A C$ and $A B$, is called antiparallel to the side $B C$ if $\angle A B_{1} C_{1}=\angle A B C$ and $\angle A C_{1} B_{1}=$ $=\angle A C B$. Prove that the symmedian $A S$ bisects any segment $B_{1} C_{1}$, antiparallel to the side $B C... | 5.139. With symmetry relative to the bisector of angle $A$, segment $B_{1} C_{1}$ transforms into a segment parallel to side $B C$, and line $A S$ transforms into line $A M$, where $M$ is the midpoint of side $B C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,064 |
5.140. The tangent at point $B$ to the circumcircle $S$ of triangle $A B C$ intersects the line $A C$ at point $K$. From point $K$, a second tangent $K D$ is drawn to the circle $S$. Prove that $B D$ is the symmedian of triangle $A B C$. | 5.140. Let's take points \( A_1 \) and \( C_1 \) on segments \( BC \) and \( BA \) such that \( A_1 C_1 \parallel BK \). Since \( \angle BAC = \angle CBK = \angle BA_1C_1 \) and \( \angle BCA = \angle BC_1A_1 \), the segment \( A_1 C_1 \) is antiparallel to side \( AC \). On the other hand, according to problem 3.31, b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,065 |
5.142*. Circle $S_{1}$ passes through points $A$ and $B$ and is tangent to line $A C$, circle $S_{2}$ passes through points $A$ and $C$ and is tangent to line $A B$. Prove that the common chord of these circles is the symmedian of triangle $A B C$. | 5.142. Let $A P$ be the common chord of the considered circles, $Q$ be the intersection point of the lines $A P$ and $B C$. Then $B Q / A B = \sin B A Q / \sin A Q B$ and $A C / C Q = \sin A Q C / \sin C A Q$. Therefore, $B Q / C Q = A B \sin B A P / A C \sin C A P$. Since $A C$ and $A B$ are tangents to the circles $S... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,067 |
5.143*. The bisectors of the external and internal angles at vertex $A$ of triangle $A B C$ intersect the line $B C$ at points $D$ and $E$. The circle with diameter $D E$ intersects the circumcircle of triangle $A B C$ at points $A$ and $X$. Prove that $A X$ is the symmedian of triangle $A B C$.
$$
* * *
$$ | 5.143. Let $S$ be the point of intersection of the lines $A X$ and $B C$. Then $A S / A B = C S / C X$ and $A S / A C = B S / B X$, which means $C S / B S = (A C / A B) \cdot (X C / X B)$. It remains to note that $X C / X B = A C / A B$ (see the solution to problem 7.16, a)). | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,068 |
5.144*. Prove that the Lemoine point of triangle $ABC$ with a right angle at $C$ is the midpoint of the altitude $CH$.
保留源文本的换行和格式，直接输出翻译结果。 | 5.144. Let $L, M$ and $N$ be the midpoints of segments $CA, CB$ and $CH$. Since $\triangle BAC \sim \triangle CAH$, then $\triangle BAM \sim \triangle CAN$, and therefore, $\angle BAM = \angle CAN$. Similarly, $\angle ABL = \angle CBN$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,069 |
5.145*. Through a point $X$ lying inside triangle $A B C$, three segments are drawn, antiparallel to its sides (see problem 5.139). Prove that these segments are equal if and only if $X$ is the Lemoine point. | 5.145. Let $B_{1} C_{1}, C_{2} A_{2}$ and $A_{3} B_{3}$ be given segments. Then the triangles $A_{2} X A_{3}, B_{1} X B_{3}$ and $C_{1} X C_{2}$ are isosceles; let the lengths of their legs be $a, b$ and $c$. The line $A X$ bisects the segment $B_{1} C_{1}$ if and only if this line contains the symmedian. Therefore, if... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,070 |
5.146*. Let $A_{1}, B_{1}$ and $C_{1}$ be the projections of the Lemoine point $K$ onto the sides of triangle $A B C$. Prove that $K$ is the centroid of triangle $A_{1} B_{1} C_{1}$. | 5.146. Let $M$ be the point of intersection of the medians of triangle $ABC$; $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ be the distances from points $K$ and $M$ to the sides of the triangle. Since points $K$ and $M$ are isogonal conjugates, we have $a_{1} a_{2}=b_{1} b_{2}=c_{1} c_{2}$; moreover, $a a_{2}=$ $=b b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,071 |
5.147*. Let $A_{1}, B_{1}$ and $C_{1}$ be the projections of the Lemoine point $K$ of triangle $A B C$ onto the sides $B C, C A$ and $A B$. Prove that the median $A M$ of triangle $A B C$ is perpendicular to the line $B_{1} C_{1}$. | 5.147. The medians of triangle \(A_{1} B_{1} C_{1}\) intersect at point \(K\) (problem 5.146), therefore, the sides of triangle \(A B C\) are perpendicular to the medians of triangle \(A_{1} B_{1} C_{1}\). After a \(90^{\circ}\) rotation, the sides of triangle \(A B C\) will be parallel to the medians of triangle \(A_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,072 |
5.148*. Lines $A K, B K$ and $C K$, where $K$ is the Lemoine point of triangle $A B C$, intersect the circumcircle at points $A_{1}, B_{1}$ and $C_{1}$. Prove that $K$ is the Lemoine point of triangle $A_{1} B_{1} C_{1}$. | 5.148. Let $A_{2}, B_{2}$ and $C_{2}$ be the projections of point $K$ onto the lines $B C, C A$ and $A B$. Then $\triangle A_{1} B_{1} C_{1} \sim \triangle A_{2} B_{2} C_{2}$ (Problem 5.111) and $K$ is the centroid of triangle $A_{2} B_{2} C_{2}$ (Problem 5.146). Therefore, the similarity transformation that maps trian... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,073 |
5.149*. Prove that the lines connecting the midpoints of the sides of a triangle with the midpoints of the corresponding altitudes intersect at the Lemoine point.
See also problems $11.22,19.55,19.56$.
## Problems for independent solving | 5.149. Let $K$ be the Lemoine point of triangle $ABC$; $A_{1}, B_{1}$, and $C_{1}$ be the projections of point $K$ onto the sides of triangle $ABC$; $L$ be the midpoint of segment $B_{1}C_{1}$; $N$ be the intersection point of line $KL$ and median $AM$; $O$ be the midpoint of segment $AK$ (Fig. 5.7). Points $B_{1}$ and... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,074 |
6.1. Prove that if the center of the circle inscribed in a quadrilateral coincides with the point of intersection of the diagonals, then this quadrilateral is a rhombus. | 6.1. Let $O$ be the center of the inscribed circle and the point of intersection of the diagonals of quadrilateral $ABCD$. Then $\angle ACB = \angle ACD$ and $\angle BAC = \angle CAD$. Therefore, triangles $ABC$ and $ADC$ are equal, as side $AC$ is common to both. Consequently, $AB = DA$. Similarly, $AB = BC = CD = DA$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,075 |
6.2. Quadrilateral $ABCD$ is circumscribed around a circle with center $O$. Prove that $\angle AOB + \angle COD = 180^{\circ}$. | 6.2. It is clear that $\angle A O B=180^{\circ}-\angle B A O-\angle A B O=180^{\circ}-(\angle A+\angle B) / 2$ and $\angle C O D=$ $=180^{\circ}-(\angle C+\angle D) / 2$. Therefore, $\angle A O B+\angle C O D=360^{\circ}-(\angle A+\angle B+\angle C+$ $+\angle D) / 2=180^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,076 |
6.3. Prove that if there exists a circle that touches all sides of a convex quadrilateral $A B C D$, and a circle that touches the extensions of all its sides, then the diagonals of such a quadrilateral are perpendicular. | 6.3. Consider two circles that touch the sides of a given quadrilateral and their extensions. The lines containing the sides of the quadrilateral are common internal and external tangents to these circles. The line connecting the centers of the circles contains a diagonal of the quadrilateral, and, moreover, it is the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,077 |
6.4. A circle cuts equal chords on all four sides of a quadrilateral. Prove that a circle can be inscribed in this quadrilateral. | 6.4. Let $O$ be the center of the given circle, $R$ its radius, and $a$ the length of the chords cut by the circle on the sides of the quadrilateral. Then the distances from point $O$ to the sides of the quadrilateral are $\sqrt{R^{2}-a^{2} / 4}$, i.e., it is equidistant from the sides of the quadrilateral and is the c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,078 |
6.5. Prove that if a circle can be inscribed in a quadrilateral, then the center of this circle lies on the same line as the midpoints of the diagonals. | 6.5. For a parallelogram, the statement of the problem is obvious, so we can assume that the lines $A B$ and $C D$ intersect. Let $O$ be the center of the inscribed circle of the quadrilateral $A B C D ; M$ and $N$ be the midpoints of the diagonals $A C$ and $B D$. Then $S_{A N B}+S_{C N D}=S_{A M B}+S_{C M D}=S_{A O B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,079 |
6.6. Quadrilateral $A B C D$ is circumscribed around a circle with center $O$. In triangle $A O B$, the altitudes $A A_{1}$ and $B B_{1}$ are drawn, and in triangle $C O D$, the altitudes $C C_{1}$ and $D D_{1}$ are drawn. Prove that the points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ lie on the same line. | 6.6. Let the inscribed circle touch the sides $D A, A B$ and $B C$ at points $M, H$ and $N$ respectively. Then $O H$ is the altitude of triangle $A O B$, and under symmetry with respect to the lines $A O$ and $B O$, point $H$ maps to points $M$ and $N$ respectively. Therefore, according to problem 1.58, points $A_{1}$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,080 |
6.7. The angles at the base $A D$ of the trapezoid $A B C D$ are equal to $2 \alpha$ and $2 \beta$. Prove that the trapezoid is circumscribed if and only if $B C / A D=\operatorname{tg} \alpha \operatorname{tg} \beta$. | 6.7. Let $r$ be the distance from the intersection point of the angle bisectors of angles $A$ and $D$ to the base $AD$, and $r'$ be the distance from the intersection point of the angle bisectors of angles $B$ and $C$ to the base $BC$. Then $AD = r(\operatorname{ctg} \alpha + \operatorname{ctg} \beta)$ and $BC = r'(\op... | BC/AD=\operatorname{tg}\alpha\cdot\operatorname{tg}\beta | Geometry | proof | Yes | Yes | olympiads | false | 27,081 |
6.9*. Given a convex quadrilateral $A B C D$. The rays $A B$ and $C D$ intersect at point $P$, and the rays $B C$ and $A D$ intersect at point $Q$. Prove that the quadrilateral $A B C D$ is tangential if and only if one of the following conditions is satisfied: $A B + C D = B C + A D, A P + C Q = A Q + C P$ or $B P + B... | 6.9. First, let's prove that if quadrilateral $A B C D$ is circumscribed, then all conditions are satisfied. Let $K, L, M$ and $N$ be the points of tangency of the inscribed circle with the sides $A B, B C, C D$ and $D A$. Then $A B+C D=$ $=A K+B K+C M+D M=A N+B L+C L+$ $+D N=B C+A D, A P+C Q=A K+P K+$ $+Q L-C L=A N+P ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,083 |
6.10*. Through the intersection points of the extensions of the sides of a convex quadrilateral $A B C D$, two lines are drawn, dividing it into four quadrilaterals. Prove that if the quadrilaterals adjacent to vertices $B$ and $D$ are circumscribed, then the quadrilateral $A B C D$ is also circumscribed. | 6.10. Let rays $A B$ and $D C$ intersect at point $P$, rays $B C$ and $A D-$ at point $Q$; the given lines passing through points $P$ and $Q$ intersect at point $O$. According to problem $6.9$, $B P+B Q=O P+O Q$ and $O P+O Q=D P+D Q$. Therefore, $B P+B Q=D P+D Q$, which means that quadrilateral $A B C D$ is circumscrib... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,084 |
6.11*. On the side $B C$ of triangle $A B C$, points $K_{1}$ and $K_{2}$ are taken. Prove that the common external tangents to the inscribed circles of triangles $A B K_{1}$ and $A C K_{2}$ are the same as the common external tangents to the inscribed circles of triangles $A B K_{2}$ and $A C K_{1}$, and they intersect... | 6.11. Let $O$ be the point of intersection of the common external tangents to the incircles of triangles $A B K_{1}$ and $A C K_{2}$ (Fig. 6.5). Draw a tangent $l$ from point $O$ to the incircle of the triangle formed by the lines $A K_{1}, A K_{2}$, and the tangent to the incircles of triangles $A B K_{1}$ and $A C K_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,085 |
6.12*. Through each of the intersection points of the extensions of the sides of a convex quadrilateral $ABCD$, two lines are drawn. These lines divide the quadrilateral into nine quadrilaterals.
a) Prove that if three of the quadrilaterals adjacent to the vertices $A, B, C, D$ are circumscribed, then the fourth quadr... | 6.12. a) We associate circles $(x-a)^{2}+(y-b)^{2}=r^{2}$ with a given orientation (direction of traversal) to the point with coordinates $(a, b, \pm r)$, where the sign before $r$ corresponds to the orientation of the circle. Consider a pair of intersecting lines with given orientations (directions). It is easy to ver... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,086 |
6.13*. Circles $S_{1}$ and $S_{2}, S_{2}$ and $S_{3}, S_{3}$ and $S_{4}, S_{4}$ and $S_{1}$ touch each other externally. Prove that the four common tangents (at the points of tangency of the circles) either intersect at one point or touch one circle. | 6.13. Orientations can be chosen consistently on circles and tangents (Fig. 6.6). Let $A_{i}$ be the point of intersection of the tangents to the circle $S_{i}$. The orientations of the tangents define the orientation of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ (this quadrilateral may be non-convex). From the equali... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,087 |
6.14*. Prove that the point of intersection of the diagonals of a circumscribed quadrilateral coincides with the point of intersection of the diagonals of the quadrilateral whose vertices are the points of tangency of the sides of the original quadrilateral with the inscribed circle.
$$
* * *
$$ | 6.14. Let the sides $A B, B C, C D, D A$ of the quadrilateral $A B C D$ touch the inscribed circle at points $E, F, G, H$ respectively.
First, we will show that the lines $F H, E G$ and $A C$ intersect at one point. Denote the points where the lines $F H$ and $E G$ intersect the line $A C$ as $M$ and $M^{\prime}$ resp... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,088 |
6.15*. Quadrilateral $A B C D$ is inscribed; $H_{c}$ and $H_{d}$ are the orthocenters of triangles $A B D$ and $A B C$. Prove that $C D H_{c} H_{d}$ is a parallelogram. | 6.15. Segments $C H_{d}$ and $D H_{c}$ are parallel, as they are perpendicular to line $B C$. Moreover, since $\angle B C A=\angle B D A=\varphi$, the lengths of these segments are equal to $A B|\operatorname{ctg} \varphi|$ (see problem 5.47, b) ). | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,089 |
6.16*. Quadrilateral $A B C D$ is inscribed. Prove that the centers of the inscribed circles of triangles $A B C, B C D, C D A$ and $D A B$ form a rectangle. | 6.16. Let $O_{a}, O_{b}, O_{c}$ and $O_{d}$ be the centers of the inscribed circles of triangles $B C D, A C D, A B D$ and $A B C$ respectively. Since $\angle A D B=\angle A C B$, then $\angle A O_{c} B=90^{\circ}+(\angle A D B / 2)=90^{\circ}+(\angle A C B / 2)=\angle A O_{d} B$ (see problem 5.3). Therefore, the quadr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,090 |
6.17*. The extensions of the sides of a quadrilateral $A B C D$, inscribed in a circle with center $O$, intersect at points $P$ and $Q$, and its diagonals intersect at point $S$.
a) The distances from points $P, Q$, and $S$ to point $O$ are $p, q$, and $s$, and the radius of the circumscribed circle is $R$. Find the l... | 6.17. a) Let rays $A B$ and $D C$ intersect at point $P$, and rays $B C$ and $A D$ intersect at point $Q$. We need to prove that the point $M$, where the circumcircles of triangles $C B P$ and $C D Q$ intersect, lies on the segment $P Q$. Indeed, $\angle C M P + \angle C M Q = \angle A B C + \angle A D C = 180^{\circ}$... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,091 |
6.18*. Diagonal $AC$ divides quadrilateral $ABCD$ into two triangles, the inscribed circles of which touch diagonal $AC$ at one point. Prove that the inscribed circles of triangles $ABD$ and $BCD$ also touch diagonal $BD$ at one point, and the points of their tangency with the sides of the quadrilateral lie on one circ... | 6.18. Let the inscribed circles of triangles $ABC$ and $ACD$ touch the diagonal $AC$ at points $M$ and $N$ respectively. Then $AM = (AC + AB - BC) / 2$ and $AN = (AC + AD - CD) / 2$ (see problem 3.2). Points $M$ and $N$ coincide if and only if $AM = AN$, i.e., $AB + CD = BC + AD$. Therefore, if points $M$ and $N$ coinc... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,092 |
6.19*. Prove that the projections of the intersection point of the diagonals of a cyclic quadrilateral onto its sides are the vertices of a circumscribed quadrilateral, provided that they do not fall on the extensions of the sides. | 6.19. Let $O$ be the point of intersection of the diagonals $AC$ and $BD$; $A_1, B_1, C_1$, and $D_1$ are its projections on the sides $AB, BC, CD$, and $DA$. Points $A_1$ and $D_1$ lie on the circle with diameter $AO$, so $\angle OA_1D_1 = \angle OAD_1$. Similarly, $\angle OA_1B_1 = \angle OBB_1$. Since $\angle CAD = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,093 |
6.20*. Prove that if the diagonals of a quadrilateral are perpendicular, then the projections of the point of intersection of the diagonals onto the sides are the vertices of an inscribed quadrilateral.
See also problems $2.71-2.79,13.33,13.34,16.4$.
## §2. Quadrilaterals | 6.20. Let us use the notations of Fig. 6.7. The condition for the inscribed quadrilateral $A_{1} B_{1} C_{1} D_{1}$ is equivalent to $(\alpha+\beta)+(\gamma+\delta)=180^{\circ}$, and the perpendicularity of the diagonals $A C$ and $B D$ is equivalent to $\left(\alpha_{1}+\delta_{1}\right)+\left(\beta_{1}+\gamma_{1}\rig... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,094 |
6.21. The angle between sides $A B$ and $C D$ of quadrilateral $A B C D$ is equal to $\varphi$. Prove that $A D^{2}=A B^{2}+B C^{2}+C D^{2}-2(A B \cdot B C \cos B+$ $+B C \cdot C D \cos C+C D \cdot A B \cos \varphi)$ | 6.21. By the cosine theorem, \( A D^{2}=A C^{2}+C D^{2}-2 A C \cdot C D \cdot \cos A C D \) and \( A C^{2}=A B^{2}+B C^{2}-2 A B \cdot B C \cos B \). Since the length of the projection of segment \( A C \) onto line \( l \), perpendicular to \( C D \), is equal to the sum of the lengths of the projections of segments \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,095 |
6.22. In quadrilateral $A B C D$, sides $A B$ and $C D$ are equal, and rays $A B$ and $D C$ intersect at point $O$. Prove that the line connecting the midpoints of the diagonals is perpendicular to the bisector of angle $A O D$. | 6.22. Let $\angle A O D=2 \alpha$. Then the distances from point $O$ to the projections of the midpoints of diagonals $A C$ and $B D$ onto the bisector of angle $A O D$ are $\cos \alpha(O A+O C) / 2$ and $\cos \alpha(O B+O D) / 2$ respectively. Since $O A+O C=A B+O B+O C=C D+O B+O C=O B+O D$, these projections coincide... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,096 |
6.23. On the sides $B C$ and $A D$ of quadrilateral $A B C D$, points $M$ and $N$ are taken such that $B M: M C = A N: N D = A B: C D$. The rays $A B$ and $D C$ intersect at point $O$. Prove that the line $M N$ is parallel to the bisector of angle $A O D$. | 6.23. Complete triangles $A B M$ and $D C M$ to parallelograms $A B M M_{1}$ and $D C M M_{2}$. Since $A M_{1}: D M_{2}=B M: M C=A N: D N$, then $\triangle A N M_{1} \sim \triangle D N M_{2}$. Therefore, point $N$ lies on the segment $M_{1} M_{2}$ and $M M_{1}: M M_{2}=A B: C D=A N: N D=M_{1} N: M_{2} N$, i.e., $M N$ i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,097 |
6.25. Two different parallelograms $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$ with respectively parallel sides are inscribed in the quadrilateral $P Q R S$ (points $A$ and $A_{1}$ lie on side $P Q, B$ and $B_{1}$ - on $Q R$, and so on). Prove that the diagonals of the quadrilateral are parallel to the sides of the paralle... | 6.25. Let $A B>A_{1} B_{1}$ for definiteness. When the triangle $S D_{1} C_{1}$ is translated by the vector $\overrightarrow{C B}$, it transforms into $S^{\prime} D_{1}^{\prime} C_{1}^{\prime}$, and the segment $C D$ transforms into $B A$. Since $Q A_{1}: Q A=A_{1} B_{1}: A B=S^{\prime} D_{1}^{\prime}: S^{\prime} A$, i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,099 |
6.26. The midpoints $M$ and $N$ of the diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ do not coincide. The line $MN$ intersects the sides $AB$ and $CD$ at points $M_1$ and $N_1$. Prove that if $MM_1 = NN_1$, then $AD \parallel BC$. | 6.26. Suppose that lines $A D$ and $B C$ are not parallel. Let $M_{2}, K, N_{2}$ be the midpoints of sides $A B, B C, C D$ respectively. If $M N \| B C$, then $B C \| A D$, since $A M=M C$ and $B N=N D$. Therefore, we will assume that lines $M N$ and $B C$ are not parallel, i.e., $M_{1} \neq M_{2}$ and $N_{1} \neq N_{2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,100 |
6.27*. Prove that two quadrilaterals are similar if and only if their four corresponding angles are equal and the angles between the corresponding diagonals are equal. | 6.27. A similarity transformation can align one pair of corresponding sides of quadrilaterals, so it is sufficient to consider quadrilaterals $A B C D$ and $A B C_{1} D_{1}$, where points $C_{1}$ and $D_{1}$ lie on the rays $B C, A D$ and $C D \| C_{1} D_{1}$. Let the points of intersection of the diagonals of quadrila... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,101 |
6.28*. Quadrilateral $A B C D$ is convex; points $A_{1}, B_{1}, C_{1}$ and $D_{1}$ are such that $A B\left\|C_{1} D_{1}, A C\right\| B_{1} D_{1}$ and so on for all pairs of vertices. Prove that quadrilateral $A_{1} B_{1} C_{1} D_{1}$ is also convex, and $\angle A+\angle C_{1}=180^{\circ}$. | 6.28. Any quadrilateral is determined up to similarity by the directions of its sides and diagonals, so it is sufficient to construct one example of a quadrilateral \(A_{1} B_{1} C_{1} D_{1}\) with the required directions of its sides and diagonals. Let \(O\) be the point of intersection of the diagonals \(A C\) and \(... | \angleA+\angleC_{1}=180 | Geometry | proof | Yes | Yes | olympiads | false | 27,102 |
6.29*. Perpendiculars are dropped from the vertices of a convex quadrilateral onto its diagonals. Prove that the quadrilateral formed by the bases of the perpendiculars is similar to the original quadrilateral. | 6.29. Let $O$ be the point of intersection of the diagonals of quadrilateral $ABCD$. Without loss of generality, we can assume that $\alpha=\angle AOB<90^{\circ}$. Drop perpendiculars $AA_1, BB_1, CC_1, DD_1$ to the diagonals of quadrilateral $ABCD$. Since $OA_1=OA \cos \alpha, OB_1=OB \cos \alpha, OC_1=OC \cos \alpha,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,103 |
6.30*. A convex quadrilateral is divided by its diagonals into four triangles. Prove that the line connecting the points of intersection of the medians of two opposite triangles is perpendicular to the line connecting the points of intersection of the altitudes of the other two triangles. | 6.30. Let the diagonals of quadrilateral $A B C D$ intersect at point $O ; H_{a}$ and $H_{b}$ be the orthocenters of triangles $A O B$ and $C O D ; K_{a}$ and $K_{b}$ be the midpoints of sides $B C$ and $A D ; P$ be the midpoint of diagonal $A C$. The points of intersection of the medians of triangles $A O D$ and $B O ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,104 |
6.31*. The diagonals of the circumscribed trapezoid $A B C D$ with bases $A D$ and $B C$ intersect at point $O$. The radii of the inscribed circles of triangles $A O D, A O B, B O C$ and $C O D$ are $r_{1}, r_{2}, r_{3}$ and $r_{4}$ respectively. Prove that $\frac{1}{r_{1}}+\frac{1}{r_{3}}=\frac{1}{r_{2}}+\frac{1}{r_{4... | 6.31. Let $S=S_{A O D}, x=A O, y=D O, a=A B, b=B C, c=C D, d=D A ; k-$ the similarity coefficient of triangles $B O C$ and $A O D$. Then
$$
\begin{aligned}
& 2 \frac{1}{r_{1}}+\frac{1}{r_{3}}=\frac{d+x+y}{S}+\frac{k d+k x+k y}{k^{2} S} \\
& 2 \frac{1}{r_{2}}+\frac{1}{r_{4}}=\frac{a+x+k y}{k S}+\frac{c+k x+y}{k S}
\end... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,105 |
6.32*. A circle of radius $r_{1}$ touches the sides $D A, A B$ and $B C$ of a convex quadrilateral $A B C D$, a circle of radius $r_{2}$ touches the sides $A B, B C$ and $C D$; similarly, $r_{3}$ and $r_{4}$ are defined. Prove that $\frac{A B}{r_{1}}+\frac{C D}{r_{3}}=\frac{B C}{r_{2}}+\frac{A D}{r_{4}}$. | 6.32. It is easy to verify that $A B=r_{1}(\operatorname{ctg}(A / 2)+\operatorname{ctg}(B / 2))$ and $C D=$ $=r_{3}(\operatorname{ctg}(C / 2)+\operatorname{ctg}(D / 2))$. Therefore, $A B / r_{1}+C D / r_{3}=\operatorname{ctg}(A / 2)+\operatorname{ctg}(B / 2)+$ $+\operatorname{ctg}(C / 2)+\operatorname{ctg}(D / 2)=B C /... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,106 |
6.33*. For a convex quadrilateral $A B C D$, it is known that the radii of the circles inscribed in triangles $A B C, B C D, C D A$ and $D A B$ are equal to each other. Prove that $A B C D$ is a rectangle. | 6.33. Complete triangles $A B D$ and $D B C$ to parallelograms $A B D A_{1}$ and $D B C C_{1}$. The segments connecting point $D$ to the vertices of parallelogram $A C C_{1} A_{1}$ divide it into four triangles equal to triangles $D A B, C D A, B C D$, and $A B C$, so the radii of the inscribed circles of these triangl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,107 |
6.34*. Given a convex quadrilateral $A B C D ; A_{1}, B_{1}, C_{1}$ and $D_{1}$ are the centers of the circumscribed circles of triangles $B C D, C D A, D A B$ and $A B C$. Similarly, for quadrilateral $A_{1} B_{1} C_{1} D_{1}$, points $A_{2}, B_{2}, C_{2}$ and $D_{2}$ are defined. Prove that quadrilaterals $A B C D$ a... | 6.34. Points $C_{1}$ and $D_{1}$ lie on the perpendicular bisector of segment $A B$, so $A B \perp C_{1} D_{1}$. Similarly, $C_{1} D_{1} \perp A_{2} B_{2}$, which means $A B \| A_{2} B_{2}$. Similarly, it can be proven that the corresponding sides and diagonals of quadrilaterals $A B C D$ and $A_{2} B_{2} C_{2} D_{2}$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,108 |
6.35*. Circles, the diameters of which are the sides $A B$ and $C D$ of a convex quadrilateral $A B C D$, touch the sides $C D$ and $A B$ respectively. Prove that $B C \| A D$. | 6.35. Let $M$ and $N$ be the midpoints of sides $A B$ and $C D$. Drop a perpendicular $D P$ from point $D$ to line $M N$, and a perpendicular $M Q$ from point $M$ to $C D$. Then, $Q$ is the point of tangency of line $C D$ and the circle with diameter $A B$. The right triangles $P D N$ and $Q M N$ are similar, so $D P =... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,109 |
6.36*. Four lines define four triangles. Prove that the orthocenters of these triangles lie on one line.
## §3. Ptolemy's Theorem | 6.36. It is sufficient to check that the orthocenters of any three of the given four triangles lie on one straight line. Let some straight line intersect the lines $B_{1} C_{1}, C_{1} A_{1}$, and $A_{1} B_{1}$ at points $A, B$, and $C$ respectively; $A_{2}, B_{2}$, and $C_{2}$ are the orthocenters of triangles $A_{1} B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,110 |
6.37*. Quadrilateral $A B C D$ is inscribed. Prove that $A B \cdot C D +$ $+A D \cdot B C=A C \cdot B D$ (Ptolemy). | 6.37. Let's take a point \( M \) on the diagonal \( BD \) such that \( \angle MCD = \angle BCA \). Then \( \triangle ABC \sim \triangle DMC \), since angles \( BAC \) and \( BDC \) subtend the same arc. Therefore, \( AB \cdot CD = AC \cdot MD \). Since \( \angle MCD = \angle BCA \), then \( \angle BCM = \angle ACD \) a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,111 |
6.38*. Quadrilateral $ABCD$ is inscribed. Prove that
$$
\frac{AC}{BD}=\frac{AB \cdot AD + CB \cdot CD}{BA \cdot BC + DA \cdot DC}
$$ | 6.38. Let $S$ be the area of quadrilateral $ABCD$, and $R$ be the radius of its circumscribed circle. Then $S=S_{ABC}+S_{ADC}=AC(AB \cdot BC+AD \cdot DC) / 4R$ (see problem 12.1). Similarly, $S=BD(AB \cdot AD+BC \cdot CD) / 4R$. By equating these expressions for $S$, we obtain the required result. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,112 |
6.39*. Let $\alpha=\pi / 7$. Prove that $\frac{1}{\sin \alpha}=\frac{1}{\sin 2 \alpha}+\frac{1}{\sin 3 \alpha}$. | 6.39. Let a regular heptagon $A_{1} \ldots A_{7}$ be inscribed in a circle. Applying Ptolemy's theorem to the quadrilateral $A_{1} A_{3} A_{4} A_{5}$, we get $A_{1} A_{3} \cdot A_{5} A_{4} + A_{3} A_{4} \cdot A_{1} A_{5} = A_{1} A_{4} \cdot A_{3} A_{5}$, i.e., $\sin 2 \alpha \sin \alpha + \sin \alpha \sin 3 \alpha = \s... | proof | Algebra | proof | Yes | Yes | olympiads | false | 27,113 |
6.40*. The distances from the center of the circumscribed circle of an acute-angled triangle to its sides are $d_{a}, d_{b}$, and $d_{c}$. Prove that $d_{a}+d_{b}+d_{c}=R+r$. | 6.40. Let $A_{1}, B_{1}$ and $C_{1}$ be the midpoints of sides $B C, C A$ and $A B$. By Ptolemy's theorem, $A C_{1} \cdot O B_{1} + A B_{1} \cdot O C_{1} = A O \cdot B_{1} C_{1}$, where $O$ is the center of the circumscribed circle. Therefore, $c d_{b} + b d_{c} = a R$. Similarly, $a d_{c} + c d_{a} = b R$ and $a d_{b}... | d_{}+d_{b}+d_{}=R+r | Geometry | proof | Yes | Yes | olympiads | false | 27,114 |
6.41*. The bisector of angle $A$ of triangle $A B C$ intersects the circumscribed circle at point $D$. Prove that $A B+A C \leqslant 2 A D$. | 6.41. By Ptolemy's theorem $A B \cdot C D+A C \cdot B D=A D \cdot B C$. Given that $C D=$ $=B D \geqslant B C / 2$, we obtain the required result. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,115 |
6.42*. On the arc $C D$ of the circumscribed circle of the square $A B C D$, a point $P$ is taken. Prove that $P A+P C=\sqrt{2} P B$.
保留源文本的换行和格式，直接输出翻译结果。 | 6.42. Applying Ptolemy's theorem to the quadrilateral $A B C P$ and reducing by the side length of the square, we obtain the required result. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,116 |
6.43*. Given a parallelogram $A B C D$. A circle passing through point $A$ intersects segments $A B, A C$ and $A D$ at points $P, Q$ and $R$ respectively. Prove that $A P \cdot A B=A R \cdot A D=A Q \cdot A C$. | 6.43. Applying Ptolemy's theorem to the quadrilateral $A P Q R$, we get $A P \cdot R Q + A R \cdot Q P = A Q \cdot P R$. Since $\angle A C B = \angle R A Q = \angle R P Q$ and $\angle R Q P = 180^{\circ} - \angle P A R = \angle A B C$, then $\triangle R Q P \sim \triangle A B C$, which means $R Q : Q P : P R = A B : B ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,117 |
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