problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4.31*. The midpoints of the diagonals $A C, B D, C E, \ldots$ of a convex hexagon $A B C D E F$ form a convex hexagon. Prove that its area is four times smaller than the area of the original hexagon. | 4.31. Let's denote the midpoints of the diagonals of the hexagon $A B C D E F$ as shown in Fig. 4.10. We will prove that the area of quadrilateral $A_{1} B_{1} C_{1} D_{1}$ is four times smaller than the area of quadrilateral $A B C D$. To do this, we will use the fact that the area of a quadrilateral is equal to half ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,900 |
4.32*. The diameter $P Q$ and a chord $R S$ perpendicular to it intersect at point $A$. Point $C$ lies on the circle, and point $B$ is inside the circle, such that $B C \| P Q$ and $B C = R A$. Perpendiculars $A K$ and $B L$ are dropped from points $A$ and $B$ to the line $C Q$. Prove that $S_{A C K} = S_{B C L}$.
$$
... | 4.32. Let $\alpha=\angle P Q C$. Then $2 S_{A C K}=$ $=C K \cdot A K=(A P \cos \alpha) \cdot(A Q \sin \alpha)=$ $=A R^{2} \sin \alpha \cos \alpha=B C^{2} \sin \alpha \cos \alpha=B L \cdot C L=$
$=2 S_{B C L}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,901 |
4.33*. Through a point $O$ lying inside triangle $A B C$, segments parallel to the sides are drawn. The segments $A A_{1}, B B_{1}$ and $C C_{1}$ divide triangle $A B C$ into four triangles and three quadrilaterals (Fig. 4.5). Prove that the sum of the areas of the triangles adjacent to vertices $A, B$ and $C$ is equal... | 4.33. Let $S_{a}, S_{b}$ and $S_{c}$ be the areas of the triangles adjacent to vertices $A, B$ and $C$; $S$ be the area of the fourth considered triangle. It is clear that $S_{A C C_{1}}+S_{B A A_{1}}+S_{C B B_{1}}=S_{A B C}-S+S_{a}+S_{b}+S_{c}$. Moreover, $S_{A B C}=$ $=S_{A O C}+S_{A O B}+S_{B O C}=S_{A C C_{1}}+S_{B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,902 |
4.35. The segment $M N$, parallel to the side $C D$ of the quadrilateral $A B C D$, divides its area in half (points $M$ and $N$ lie on the sides $B C$ and $A D$). The lengths of the segments drawn from points $A$ and $B$ parallel to $C D$ until they intersect with the lines $B C$ and $A D$ are $a$ and $b$. Prove that ... | 4.35. Let for definiteness the rays $A D$ and $B C$ intersect at point $O$. Then $S_{C D O}: S_{M N O}=c^{2}: x^{2}$, where $x=M N$, and $S_{A B O}: S_{M N O}=a b: x^{2}$, since $O A: O N=a: x$ and $O B: O M=b: x$. Therefore, $x^{2}-c^{2}=a b-x^{2}$, i.e., $2 x^{2}=a b+c^{2}$. | 2x^{2}=+^{2} | Geometry | proof | Yes | Yes | olympiads | false | 26,904 |
4.36. Each of the three lines divides the area of the figure in half. Prove that the part of the figure enclosed within the triangle formed by these lines has an area not exceeding $1 / 4$ of the area of the entire figure. | 4.36. Let's denote the areas of the parts of the figure, into which it is divided by the lines, as shown in Fig. 4.11. Let the area of the entire figure be denoted by \( S \). Since \( S_{3} + (S_{2} + S_{7}) = S / 2 = S_{1} + S_{6} + (S_{2} + S_{7}) \), it follows that \( S_{3} = S_{1} + S_{6} \). Adding this equality... | S_{1}\leqslantS/4 | Geometry | proof | Yes | Yes | olympiads | false | 26,905 |
4.37*. A line $l$ divides the area of a convex polygon in half. Prove that this line divides the projection of the polygon onto a line perpendicular to $l$ in a ratio not exceeding $1+\sqrt{2}$. | 4.37. Let the projection of line $l$ be denoted by $B$, and the extreme points of the projection of the polygon by $A$ and $C$. Let $C_{1}$ be the point on the polygon that projects to point $C$; line $l$ intersects the polygon at points $K$ and $L$, and $K_{1}$ and $L_{1}$ are the points on lines $C_{1} K$ and $C_{1} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,906 |
4.39*. a) Prove that any line dividing both the area and the perimeter of a triangle in half passes through the center of the inscribed circle.
b) Prove an analogous statement for any circumscribed polygon. | 4.39. a) Let the line that bisects both the area and the perimeter of triangle $ABC$ intersect sides $AC$ and $BC$ at points $P$ and $Q$ respectively. Denote the center of the inscribed circle of triangle $ABC$ by $O$, and the radius of the inscribed circle by $r$. Then, $S_{ABQOP} = r(AP + AB + BQ) / 2$ and $S_{OQCP} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,908 |
4.40*. Points $A$ and $B$ of the circle $S_{1}$ are connected by an arc of the circle $S_{2}$, which divides the area of the circle bounded by $S_{1}$ into equal parts. Prove that the arc of $S_{2}$ connecting $A$ and $B$ is longer than the diameter of $S_{1}$. | 4.40. Considering the image of the circle $S_{2}$ under symmetry with respect to the center of the circle $S_{1}$ and taking into account the equality of areas, it can be proven that the diameter $A A_{1}$ of the circle $S_{1}$ intersects $S_{2}$ at some point $K$ different from $A$, and that $A K > A_{1} K$. The circl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,909 |
4.41*. Curve G divides the square into two parts of equal area. Prove that two points $A$ and $B$ can be chosen on it such that the line $A B$ passes through the center $O$ of the square.
See also problems $6.54,6.55,16.8,18.31$
## §7. Formulas for the area of a quadrilateral | 4.41. The case when point $O$ belongs to $\Gamma$ is obvious; therefore, we will assume that $O$ does not belong to $\Gamma$. Let $\Gamma^{\prime}$ be the image of the curve $\Gamma$ under the symmetry with respect to point $O$. If the curves $\Gamma$ and $\Gamma^{\prime}$ do not intersect, then the parts into which $\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,910 |
4.42. The diagonals of quadrilateral $ABCD$ intersect at point $P$. The distances from points $A, B$, and $P$ to line $CD$ are $a, b$, and $p$. Prove that the area of quadrilateral $ABCD$ is $ab \cdot CD / 2p$. | 4.42. Let the areas of triangles $A P B, B P C, C P D$ and $D P A$ be $S_{1}, S_{2}, S_{3}$ and $S_{4}$. Then $a / p=\left(S_{3}+S_{4}\right) / S_{3}$ and $b \cdot C D / 2=S_{3}+S_{2}$, which means, $a b \cdot C D / 2 p=$ $=\left(S_{3}+S_{4}\right)\left(S_{3}+S_{2}\right) / S_{3}$. Considering that $S_{2} S_{4}=S_{1} S... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,911 |
4.43. Quadrilateral $ABCD$ is inscribed in a circle of radius $R, \varphi$ is the angle between its diagonals. Prove that the area $S$ of quadrilateral $ABCD$ is equal to $2 R^{2} \sin A \sin B \sin \varphi$. | 4.43. Applying the Law of Sines to triangles $A B C$ and $A B D$, we get $A C=2 R \sin B$ and $B D=2 R \sin A$. Therefore, $S=\frac{1}{2} A C \cdot B D \sin \varphi=$ $=2 R^{2} \sin A \sin B \sin \varphi$. | 2R^{2}\sinA\sinB\sin\varphi | Geometry | proof | Yes | Yes | olympiads | false | 26,912 |
4.44*. Prove that the area of a quadrilateral whose diagonals are not perpendicular is $\operatorname{tg} \varphi \cdot\left|a^{2}+c^{2}-b^{2}-d^{2}\right| / 4$, where $a, b, c$ and $d$ are the lengths of consecutive sides, $\varphi$ is the angle between the diagonals. | 4.44. Since the area of a quadrilateral is equal to $\left(d_{1} d_{2} \sin \varphi\right) / 2$, where $d_{1}$ and $d_{2}$ are the lengths of the diagonals, it remains to check that $2 d_{1} d_{2} \cos \varphi=\left|a^{2}+c^{2}-b^{2}-d^{2}\right|$. Let $O$ be the point of intersection of the diagonals of the quadrilate... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,913 |
4.45*. a) Prove that the area of a convex quadrilateral $ABCD$ is calculated by the formula
$$
S^{2}=(p-a)(p-b)(p-c)(p-d)-a b c d \cos ^{2}((B+D) / 2)
$$
where $p$ is the semiperimeter, and $a, b, c, d$ are the lengths of the sides.
b) Prove that if the quadrilateral $ABCD$ is cyclic, then $S^{2}=$ $=(p-a)(p-b)(p-c)... | 4.45. a) Let $A B=a, B C=b, C D=c$ and $A D=d$. Clearly, $S=S_{A B C}+S_{A D C}=(a b \sin B+c d \sin D) / 2$ and $a^{2}+b^{2}-2 a b \cos B=A C^{2}=c^{2}+d^{2}-2 c d \cos D$. Therefore,
\[
\begin{aligned}
& 16 S^{2}=4 a^{2} b^{2}-4 a^{2} b^{2} \cos ^{2} B+8 a b c d \sin B \sin D+4 c^{2} d^{2}-4 c^{2} d^{2} \cos ^{2} D ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,914 |
4.46. Prove that the sum of the distances from an arbitrary point inside an equilateral triangle to its sides is constant (and equal to the height of the triangle). | 4.46. From point $O$, lying inside the equilateral triangle $A B C$, drop perpendiculars $O A_{1}, O B_{1}$, and $O C_{1}$ to the sides $B C, A C$, and $A B$ respectively. Let $a$ be the length of the side of triangle $A B C$, and $h$ be the length of the height. Clearly, $S_{A B C}=S_{B C O}+S_{A C O}+S_{A B O}$. Ther... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,915 |
4.47. Prove that the length of the bisector $A D$ of triangle $A B C$ is $\frac{2 b c}{b+c} \cos \frac{\alpha}{2}$. | 4.47. Let $A D=l$. Then $2 S_{A B D}=c l \sin (\alpha / 2), 2 S_{A C D}=b l \sin (\alpha / 2)$ and $2 S_{A B C}=$ $=b c \sin \alpha$. Therefore, $c l \sin (\alpha / 2)+b l \sin (\alpha / 2)=b c \sin \alpha=2 b c \sin (\alpha / 2) \cos (\alpha / 2)$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,916 |
4.48. Inside triangle $ABC$, a point $O$ is taken; lines $AO$, $BO$, and $CO$ intersect its sides at points $A_1$, $B_1$, and $C_1$. Prove that:
a) $\frac{OA_1}{AA_1} + \frac{OB_1}{BB_1} + \frac{OC_1}{CC_1} = 1$;
b) $\frac{AC_1}{C_1B} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A} = 1$. | 4.48. a) Let the distances from points $A$ and $O$ to the line $BC$ be $h$ and $h_{1}$. Then $S_{OBC}: S_{ABC}=h_{1}: h=OA_{1}: AA_{1}$. Similarly, $S_{OAC}: S_{ABC}=OB_{1}: BB_{1}$ and $S_{OAB}: S_{ABC}=OC_{1}: CC_{1}$. Adding these equalities and considering that $S_{OBC} + S_{OAC} + S_{OAB} = S_{ABC}$, we obtain the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,917 |
4.49. Given a $(2 n-1)$-gon $A_{1} \ldots A_{2 n-1}$ and a point $O$. The lines $A_{k} O$ and $A_{n+k-1} A_{n+k}$ intersect at point $B_{k}$. Prove that the product of the ratios $A_{n+k-1} B_{k} / A_{n+k} B_{k}(k=1, \ldots, n)$ equals 1. | 4.49. It is easy to verify that the ratio of the lengths of segments $A_{n+k-1} B_{k}$ and $A_{n+k} B_{k}$ is equal to the ratio of the areas of triangles $A_{n+k-1} O A_{k}$ and $A_{k} O A_{n+k}$. By multiplying these equalities, we obtain the required result. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,918 |
4.50. Given a convex polygon $A_{1} A_{2} \ldots A_{n}$. Points $B_{1}$ and $D_{2}$ are taken on side $A_{1} A_{2}$, points $B_{2}$ and $D_{3}$ on side $A_{2} A_{3}$, and so on, such that if parallelograms $A_{1} B_{1} C_{1} D_{1}, \ldots, A_{n} B_{n} C_{n} D_{n}$ are constructed, then the lines $A_{1} C_{1}, \ldots, A... | 4.50. Since $A_{i} B_{i} C_{i} D_{i}$ is a parallelogram and point $O$ lies on the extension of its diagonal $A_{i} C_{i}$, then $S_{A_{i} B_{i} O}=S_{A_{i} D_{i} O}$, which means $A_{i} B_{i}: A_{i} D_{i}=h_{i}: h_{i-1}$, where $h_{i}$ is the distance from point $O$ to side $A_{i} A_{i+1}$. It remains to multiply thes... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,919 |
4.51. The lengths of the sides of a triangle form an arithmetic progression. Prove that the radius of the inscribed circle is one third of one of the heights of the triangle. | 4.51. Let the lengths of the sides of triangle $ABC$ be $a, b$, and $c$, with $a \leqslant b \leqslant c$. Then $2b = a + c$ and $2S_{ABC} = r(a + b + c) = 3rb$, where $r$ is the radius of the inscribed circle. On the other hand, $2S_{ABC} = h_b b$. Therefore, $r = h_b / 3$. | \frac{h_b}{3} | Geometry | proof | Yes | Yes | olympiads | false | 26,920 |
4.52. The distances from point $X$ on side $B C$ of triangle $A B C$ to the lines $A B$ and $A C$ are $d_{b}$ and $d_{c}$. Prove that $d_{b} / d_{c}=B X \cdot A C /(C X \cdot A B)$. | 4.52. It is sufficient to note that $d_{b} \cdot A B=2 S_{A X}=B X \cdot A X \sin \varphi$, where $\varphi=$ $=\angle A X B$ and $d_{c} \cdot A C=2 S_{A X C}=C X \cdot A X \sin \varphi$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,921 |
4.54*. Through a point $M$ inside the parallelogram $A B C D$, lines $P R$ and $Q S$ are drawn parallel to the sides $B C$ and $A B$ (points $P, Q, R$ and $S$ lie on the sides $A B, B C, C D$ and $D A$ respectively). Prove that the lines $B S, P D$ and $M C$ intersect at one point. | 4.54. Through the point $N$ of intersection of the lines $B S$ and $C M$, draw the lines $Q_{1} S_{1}$ and $P_{1} R_{1}$ parallel to the lines $Q S$ and $P R$ (points $P_{1}, Q_{1}, R_{1}$, and $S_{1}$ lie on the sides $A B, B C, C D$, and $D A$). Let $F$ and $G$ be the points of intersection of the lines $P R$ and $Q_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,923 |
4.55*. Prove that if no sides of a quadrilateral are parallel, then the midpoint of the segment connecting the points of intersection of opposite sides lies on the line connecting the midpoints of the diagonals (the Gauss line). | 4.55. Let $E$ and $F$ be the points of intersection of the extensions of the sides of a given quadrilateral. Denote the vertices of the quadrilateral such that $E$ is the point of intersection of the extensions of sides $A B$ and $C D$ beyond points $B$ and $C$, and $F$ is the point of intersection of the rays $B C$ an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,924 |
4.56*. In an acute-angled triangle $A B C$, the altitudes $B B_{1}$ and $C C_{1}$ are drawn, and points $K$ and $L$ are taken on the sides $A B$ and $A C$ such that $A K=B C_{1}$ and $A L=C B_{1}$. Prove that the line $A O$, where $O$ is the center of the circumcircle of triangle $A B C$, bisects the segment $K L$. | 4.56. It is sufficient to check that $S_{A K O}=S_{A L O}$, i.e., $A O \cdot A L \sin O A L=$ $=A O \cdot A K \sin O A K$. Clearly, $A L=C B_{1}=B C \cos C, \sin O A L=\cos B, A K=$ $=B C_{1}=B C \cos B$ and $\sin O A K=\cos C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,925 |
4.58*. Inside triangle $ABC$, a point $O$ is taken. Let the distances from point $O$ to the sides $BC, CA, AB$ of the triangle be denoted by $d_{a}, d_{b}, d_{c}$, and the distances from point $O$ to the vertices $A, B, C$ by $R_{a}, R_{b}, R_{c}$. Prove that:
a) $a R_{a} \geqslant c d_{c} + b d_{b}$
b) $d_{a} R_{a} ... | 4.58. First, let's prove a general statement that we will use to solve parts a)-d). Take arbitrary points \( B_{1} \) and \( C_{1} \) on the rays \( A B \) and \( A C \), and drop perpendiculars \( B_{1} K \) and \( C_{1} L \) to the line \( A O \). Since \( B_{1} C_{1} \geqslant B_{1} K + C_{1} L \), we have \( B_{1} ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 26,927 |
4.60. From the midpoint of each side of an acute-angled triangle, perpendiculars are dropped to the other two sides. Prove that the area of the hexagon bounded by them is half the area of the original triangle. | 4.60. Let $A_{1}, B_{1}$, and $C_{1}$ be the midpoints of sides $BC, CA$, and $AB$ of triangle $ABC$. The segments drawn from these points are the altitudes of triangles $AB_{1}C_{1}, A_{1}BC_{1}$, and $A_{1}B_{1}C$. Let $P, Q$, and $R$ be the points of intersection of the altitudes of these triangles, and let $O$ be t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,929 |
5.1. On the sides $BC$, $CA$, and $AB$ of triangle $ABC$, points $A_1$, $B_1$, and $C_1$ are taken such that $AC_1 = AB_1$, $BA_1 = BC_1$, and $CA_1 = CB_1$. Prove that $A_1$, $B_1$, and $C_1$ are the points of tangency of the inscribed circle with the sides. | 5.1. Let $A C_{1}=A B_{1}=x, B A_{1}=B C_{1}=y$ and $C A_{1}=C B_{1}=z$. Then $a=y+z, b=z+x$ and $c=x+y$. By subtracting the third equality from the sum of the first two, we get $z=(a+b-c) / 2$. Therefore, if triangle $A B C$ is given, the positions of points $A_{1}$ and $B_{1}$ are uniquely determined. Similarly, the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,932 |
5.2. Let $O_{a}, O_{b}$ and $O_{c}$ be the centers of the excircles of triangle $A B C$. Prove that points $A, B$ and $C$ are the feet of the altitudes of triangle $O_{a} O_{b} O_{c}$. | 5.2. The rays $C O_{a}$ and $C O_{b}$ are the bisectors of the external angles at vertex $C$, so $C$ lies on the line $O_{a} O_{b}$ and $\angle O_{a} C B = \angle O_{b} C A$. Since $C O_{c}$ is the bisector of angle $B C A$, then $\angle B C O_{c} = \angle A C O_{c}$. Adding these equalities, we get $\angle O_{a} C O_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,933 |
5.3. Prove that the side $B C$ of triangle $A B C$ is seen from the center $O$ of the inscribed circle at an angle of $90^{\circ}+\angle A / 2$, and from the center $O_{a}$ of the excircle at an angle of $90^{\circ}-\angle A / 2$. | 5.3. It is clear that $\angle B O C=180^{\circ}-\angle C B O-\angle B C O=180^{\circ}-\angle B / 2-\angle C / 2=90^{\circ}+$ $+\angle A / 2$, and $\angle B O_{a} C=180^{\circ}-\angle B O C$, since $\angle O B O_{a}=\angle O C O_{a}=90^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,934 |
5.4. Inside triangle $A B C$, a point $P$ is taken such that $\angle P A B: \angle P A C=\angle P C A: \angle P C B=\angle P B C: \angle P B A=x$. Prove that $x=1$.
Translate the above text into English, please keep the line breaks and format of the source text, and output the translation result directly. | 5.4. Let $A A_{1}, B B_{1}$, and $C C_{1}$ be the angle bisectors of triangle $A B C$, and $O$ be their point of intersection. Suppose $x>1$. Then $\angle P A B > \angle P A C$, i.e., point $P$ lies inside triangle $A A_{1} C$. Similarly, point $P$ lies inside triangles $C C_{1} B$ and $B B_{1} A$. But the only common ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,935 |
5.5*. Let $A_{1}, B_{1}$, and $C_{1}$ be the projections of some interior point $O$ of triangle $ABC$ onto the altitudes. Prove that if the lengths of segments $A A_{1}, B B_{1}$, and $C C_{1}$ are equal, then they are equal to $2r$.
5.6 ${ }^{*}$. An angle of measure $\alpha=\angle B A C$ rotates around its vertex $O... | 5.5. Let $d_{a}, d_{b}$ and $d_{c}$ be the distances from point $O$ to the sides $B C, C A$ and $A B$. Then $a d_{a}+b d_{b}+c d_{c}=2 S$ and $a h_{a}=b h_{b}=c h_{c}=2 S$. If $h_{a}-d_{a}=h_{b}-d_{b}=$ $=h_{c}-d_{c}=x$, then $(a+b+c) x=a\left(h_{a}-d_{a}\right)+b\left(h_{b}-d_{b}\right)+c\left(h_{c}-d_{c}\right)=6 S-2... | 2r | Geometry | proof | Yes | Yes | olympiads | false | 26,936 |
5.7*. In a non-isosceles triangle \(ABC\), a line \(MO\) is drawn through the midpoint \(M\) of side \(BC\) and the center \(O\) of the inscribed circle, intersecting the altitude \(AH\) at point \(E\). Prove that \(AE = r\). | 5.7. Let $P$ be the point of tangency of the inscribed circle with side $B C$, $P Q$ - the diameter of the inscribed circle, $R$ - the point of intersection of the lines $A Q$ and $B C$. Since $C R = B P$ (see problem 19.11, a)) and $M$ is the midpoint of side $B C$, then $R M = P M$. Moreover, $O$ is the midpoint of t... | AE=r | Geometry | proof | Yes | Yes | olympiads | false | 26,937 |
5.8*. A circle touches the sides of an angle with vertex $A$ at points $P$ and $Q$. The distances from points $P, Q$, and $A$ to a certain tangent to this circle are $u, v$, and $w$. Prove that $uv / w^2 = \sin^2(A / 2)$.
5.9*. a) On the side $AB$ of triangle $ABC$, a point $P$ is taken. Let $r, r_1$, and $r_2$ be the... | 5.8. This circle can be both an inscribed and an exscribed circle of triangle $ABC$, cut off by a tangent from the angle. Using the result of problem 3.2, in both cases it is easy to verify that $uv / w^{2} = (p-b)(p-c) \sin B \sin C / h_{a}^{2}$. It remains to note that $h_{a} = b \sin C = c \sin B$ and $(p-b)(p-c) / ... | uv/w^2=\sin^2(A/2) | Geometry | proof | Yes | Yes | olympiads | false | 26,938 |
5.10. Prove that the points symmetric to the orthocenter of triangle $ABC$ with respect to its sides lie on the circumcircle. | 5.10. Let $A_{1}, B_{1}$, and $C_{1}$ be the points symmetric to point $H$ with respect to the sides $BC, CA$, and $AB$ respectively. Since $AB \perp CH$ and $BC \perp AH$, we have $\angle(AB, BC) = \angle(CH, HA)$. Since triangle $AC_{1}H$ is isosceles, $\angle(CH, HA) = \angle(AC_{1}, C_{1}C)$. Therefore, $\angle(AB,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,939 |
5.11. From point $P$ on the arc $BC$ of the circumcircle of triangle $ABC$, perpendiculars $PX$, $PY$, and $PZ$ are dropped to $BC$, $CA$, and $AB$ respectively. Prove that $\frac{BC}{PX} = \frac{AC}{PY} + \frac{AB}{PZ}$. | 5.11. Points $X, Y$ and $Z$ lie on the same line (Problem 5.95, a)). Therefore, $S_{P Y Z}=S_{P X Z}+S_{P X Y}$. Moreover, $S_{P Y Z}=\frac{1}{2} P X \cdot P Z \sin \alpha$, since $P X \perp B C$ and $P Z \perp C A$. By substituting the other two areas in a similar manner, we get
$$
\frac{\sin \alpha}{P X}=\frac{\sin ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,940 |
5.12*. Let $O$ be the center of the circumcircle of triangle $ABC$, $I$ be the center of the incircle, and $I_{a}$ be the center of the excircle opposite to vertex $A$, which touches side $BC$. Prove that:
a) $d^{2}=R^{2}-2 R r$, where $d=O I$
b) $d_{a}^{2}=R^{2}+2 R r_{a}$, where $d_{a}=O I_{a}$. | 5.12. a) Let $M$ be the point of intersection of the line $A I$ with the circumcircle. By drawing a diameter through point $I$, we get $A I \cdot I M=(R+d)(R-d)=R^{2}-d^{2}$. Since $I M=C M$ (problem 2.4, a)), then $R^{2}-d^{2}=A I \cdot C M$. It remains to note that $A I=r / \sin (A / 2)$ and $C M=2 R \sin (A / 2)$.
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,941 |
5.13*. The continuations of the angle bisectors of triangle $ABC$ intersect the circumscribed circle at points $A_{1}, B_{1}$, and $C_{1}$; $M$ is the point of intersection of the angle bisectors. Prove that:
a) $\frac{M A \cdot M C}{M B_{1}}=2 r$
b) $\frac{M A_{1} \cdot M C_{1}}{M B}=R$. | 5.13. a) Since $B_{1}$ is the center of the circumcircle of triangle $A M C$ (see problem 2.4, a)), then $A M=2 M B_{1} \sin A C M$. It is also clear that $M C=r / \sin A C M$. Therefore, $M A \cdot M C / M B_{1}=2 r$.
b) Since $\angle M B C_{1}=\angle B M C_{1}=180^{\circ}-\angle B M C \quad$ and $\angle B C_{1} M=\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,942 |
5.14*. The lengths of the sides of triangle $ABC$ form an arithmetic progression, with $a<b<c$. The bisector of angle $B$ intersects the circumscribed circle at point $B_{1}$. Prove that the center $O$ of the inscribed circle bisects the segment $B B_{1}$. | 5.14. Let $M$ be the midpoint of side $A C$, and $N$ be the point of tangency of the inscribed circle with side $B C$. Then $B N=p-b$ (see problem 3.2), so $B N=$ $=A M$, since $p=3 b / 2$ by the condition. In addition, $\angle O B N=\angle B_{1} A M$, hence $\triangle O B N=\triangle B_{1} A M$, i.e., $O B=B_{1} A$. B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,943 |
5.15*. In triangle $A B C$, side $B C$ is the smallest. On rays $B A$ and $C A$, segments $B D$ and $C E$ are laid off, equal to $B C$. Prove that the radius of the circumscribed circle of triangle $A D E$ is equal to the distance between the centers of the inscribed and circumscribed circles of triangle $A B C$.
## §... | 5.15. Let $O$ and $O_{1}$ be the centers of the inscribed and circumscribed circles of triangle $ABC$. Consider a circle with radius $d = OO_{1}$ centered at $O$. In this circle, draw chords $O_{1}M$ and $O_{1}N$, parallel to sides $AB$ and $AC$ respectively. Let $K$ be the point of tangency of the inscribed circle wit... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,944 |
5.16. In triangle $ABC$, angle $C$ is a right angle. Prove that $r=(a+$ $+b-c) / 2$ and $r_{c}=(a+b+c) / 2$. | 5.16. Let the inscribed circle touch side $A C$ at point $K$, and the escribed circle touch the extension of side $A C$ at point $L$. Then $r = C K$ and $r_{c} = C L$. It remains to use the result of problem 3.2. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,945 |
5.17. Let $M$ be the midpoint of side $AB$ of triangle $ABC$. Prove that $CM = AB / 2$ if and only if $\angle ACB = 90^{\circ}$. | 5.17. Since $A B / 2=A M=B M$, then $C M=A B / 2$ if and only if point $C$ lies on the circle with diameter $A B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,946 |
5.18. Given a trapezoid $A B C D$ with base $A D$. The bisectors of the external angles at vertices $A$ and $B$ intersect at point $P$, and at vertices $C$ and $D$ at point $Q$. Prove that the length of segment $P Q$ is equal to half the perimeter of the trapezoid. | 5.18. Let $M$ and $N$ be the midpoints of sides $A B$ and $C D$. Triangle $A P B$ is right-angled, so $P M = A B / 2$ and $\angle M P A = \angle P A M$, which means $P M \| A D$. Similar reasoning shows that points $P, M, N$, and $Q$ lie on the same line and $P Q = P M + M N + N Q = (A B + (B C + A D) + C D) / 2$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,947 |
5.19. In an isosceles triangle \(ABC\) with base \(AC\), the bisector \(CD\) is drawn. A line passing through point \(D\) perpendicular to \(DC\) intersects \(AC\) at point \(E\). Prove that \(EC = 2AD\). | 5.19. Let $F$ be the point of intersection of lines $D E$ and $B C$; $K$ be the midpoint of segment $E C$. Segment $C D$ is the bisector and altitude of triangle $E C F$, so $E D = D F$, which means $D K \| F C$. The median $D K$ of the right triangle $E D C$ is half the length of its hypotenuse $E C$ (problem 5.17), s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,948 |
5.20. On the median $B M$ and the bisector $B K$ of triangle $A B C$ (or on their extensions), points $D$ and $E$ are taken such that $D K \| A B$ and $E M \| B C$. Prove that $E D \perp B K$.
On the median $B M$ and the bisector $B K$ of triangle $A B C$ (or on their extensions), points $D$ and $E$ are taken such tha... | 5.20. The line $E M$ passes through the midpoint of side $A B$, so it passes through the midpoint $O$ of segment $D K$. In addition, $\angle E K O=\angle A B K=\angle K B C=$ $=\angle K E O$. Therefore, $O E=O K=O D$. According to problem $5.17 \angle D E K=90^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,949 |
5.21. The sum of the angles at the base of a trapezoid is $90^{\circ}$. Prove that the segment connecting the midpoints of the bases is equal to half the difference of the bases. | 5.21. Let the sum of the angles at the base $A D$ of trapezoid $A B C D$ be $90^{\circ}$. Denote the point of intersection of lines $A B$ and $C D$ as $O$. The point $O$ lies on the line passing through the midpoints of the bases. Draw a line $C K$ through point $C$, parallel to this line, and a line $C E$, parallel to... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,950 |
5.22. In a right triangle $A B C$, the height $C K$ is drawn from the vertex of the right angle $C$, and in triangle $A C K$ - the bisector $C E$. Prove that $C B=B E$. | 5.22. It is clear that $\angle C E B=\angle A+\angle A C E=\angle B C K+\angle K C E=\angle B C E$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,951 |
5.23. In triangle $ABC$ with a right angle at $C$, the altitude $CD$ and the angle bisector $CF$ are drawn; $DK$ and $DL$ are the angle bisectors of triangles $BDC$ and $ADC$. Prove that $CLFK$ is a square. | 5.23. Segments $C F$ and $D K$ are the bisectors of similar triangles $A C B$ and $C D B$, therefore $A B: F B = C B: K B$. Consequently, $F K \| A C$. Similarly, it can be proven that $L F \| C B$. Therefore, $C L F K$ is a rectangle, in which the diagonal $C F$ is the bisector of angle $L C K$, i.e., it is a square. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,952 |
5.24*. On the hypotenuse $AB$ of a right triangle $ABC$, a square $ABPQ$ is constructed externally. Let $\alpha=\angle ACQ, \beta=\angle QCP$ and $\gamma=$ $=\angle PCB$. Prove that $\cos \beta=\cos \alpha \cos \gamma$.
See also problems $1.40,1.43,2.39,2.66,2.67,3.38,5.32,5.39,5.42,5.68$, $5.144,6.80,11.14$.
## §3. ... | 5.24. Since $\frac{\sin ACQ}{AQ} = \frac{\sin AQC}{AC}$, then $\frac{\sin \alpha}{a} = \frac{\sin (180^{\circ} - \alpha - 90^{\circ} - \varphi)}{a \cos \varphi} = \frac{\cos (\alpha + \varphi)}{a \cos \varphi}$, where $a$ is the side of the square $ABPQ$, $\varphi = \angle CAB$. Therefore, $\operatorname{ctg} \alpha = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,953 |
5.25. From a point $M$ inside an equilateral triangle $ABC$, perpendiculars $MP$, $MQ$, and $MR$ are dropped to the sides $AB$, $BC$, and $CA$ respectively. Prove that $AP^2 + BQ^2 + CR^2 = PB^2 + QC^2 + RA^2$ and $AP + BQ + CR = PB + QC + RA$. | 5.25. By the Pythagorean theorem $A P^{2}+B Q^{2}+C R^{2}=\left(A M^{2}-P M^{2}\right)+\left(B M^{2}-Q M^{2}\right)+$ $+\left(C M^{2}-R M^{2}\right)$ and $P B^{2}+Q C^{2}+R A^{2}=\left(B M^{2}-P M^{2}\right)+\left(C M^{2}-Q M^{2}\right)+\left(A M^{2}-R M^{2}\right)$. These expressions are equal.
Since $A P^{2}+B Q^{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,954 |
5.26. Points $D$ and $E$ divide the sides $AC$ and $AB$ of an equilateral triangle $ABC$ in the ratios $AD:DC = BE:EA = 1:2$. The lines $BD$ and $CE$ intersect at point $O$. Prove that $\angle AOC = 90^{\circ}$.
$$
* * *
$$ | 5.26. Let point $F$ divide segment $B C$ in the ratio $C F: F B=1: 2 ; P$
and $Q$ be the points of intersection of segment $A F$ with $B D$ and $C E$ respectively. It is clear that triangle $O P Q$ is equilateral. Using the result of problem 1.3, it is easy to verify that $A P: P F=3: 4$ and $A Q: Q F=6: 1$. Therefore,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,955 |
5.27. A circle divides each side of a triangle into three equal parts. Prove that this triangle is equilateral. | 5.27. Let $A$ and $B$, $C$ and $D$, $E$ and $F$ be the points of intersection of the circle with the sides $P Q$, $Q R$, $R P$ of triangle $P Q R$. Consider the median $P S$. It connects the midpoints of the parallel chords $F A$ and $D C$ and is therefore perpendicular to them. Hence, $P S$ is the altitude of triangle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,956 |
5.28. Prove that if the point of intersection of the altitudes of an acute-angled triangle divides the altitudes in the same ratio, then the triangle is equilateral. | 5.28. Let $H$ be the point of intersection of the altitudes $A A_{1}, B B_{1}$ and $C C_{1}$ of triangle $A B C$. By the condition, $A_{1} H \cdot B H = B_{1} H \cdot A H$. On the other hand, since points $A_{1}$ and $B_{1}$ lie on the circle with diameter $A B$, then $A H \cdot A_{1} H = B H \cdot B_{1} H$. Therefore,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,957 |
5.29. a) Prove that if $a+h_{a}=b+h_{b}=c+h_{c}$, then triangle $A B C$ is equilateral.
b) In triangle $A B C$, three squares are inscribed: one has two vertices on side $A C$, another on $B C$, and the third on $A B$. Prove that if all three squares are equal, then triangle $A B C$ is equilateral. | 5.29. a) Suppose that triangle $A B C$ is irregular, for example, $a \neq b$. Since $a+h_{a}=a+b \sin \gamma$ and $b+h_{b}=b+a \sin \gamma$, we have $(a-b)(1-\sin \gamma)=0$. Therefore, $\sin \gamma=1$, i.e., $\gamma=90^{\circ}$. But then $a \neq c$, and similar reasoning shows that $\beta=90^{\circ}$. This leads to a ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,958 |
5.30. In triangle $A B C$, a circle is inscribed, touching its sides at points $A_{1}, B_{1}, C_{1}$. Prove that if triangles $A B C$ and $A_{1} B_{1} C_{1}$ are similar, then triangle $A B C$ is equilateral. | 5.30. If $\alpha, \beta$ and $\gamma$ are the angles of triangle $A B C$, then the angles of triangle $A_{1} B_{1} C_{1}$ are $(\beta+\gamma) / 2, (\gamma+\alpha) / 2$ and $(\alpha+\beta) / 2$. Let's assume for definiteness that $\alpha \geqslant \beta \geqslant \gamma$. Then $(\alpha+\beta) / 2 \geqslant (\alpha+\gamm... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,959 |
5.31. The radius of the inscribed circle of a triangle is 1, and the lengths of the altitudes are integers. Prove that the triangle is equilateral.
See also problems $1.29,1.45,1.46,1.50$ b), $1.59,2.13,2.15,2.18,2.19,2.37$, $2.45,2.55,4.46,5.59,5.60,6.80,7.16$ b), $7.17,7.22,7.38,7.45,10.3,10.77,11.3$, $11.5,14.21$ a... | 5.31. In any triangle, the height is greater than the diameter of the inscribed circle. Therefore, the lengths of the heights are integers greater than 2, i.e., all of them are at least 3. Let $S$ be the area of the triangle, $a$ be its largest side, and $h$ be the corresponding height.
Suppose the triangle is irregul... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,960 |
5.32. In triangle $A B C$ with angle $A$ equal to $120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}$ and $C C_{1}$ are drawn. Prove that triangle $A_{1} B_{1} C_{1}$ is a right triangle. | 5.32. Since the exterior angle at vertex $A$ of triangle $A B A_{1}$ is $120^{\circ}$ and $\angle A_{1} A B_{1}=60^{\circ}$, then $A B_{1}$ is the bisector of this exterior angle. In addition, $B B_{1}$ is the bisector of the interior angle at vertex $B$, so $A_{1} B_{1}$ is the bisector of angle $A A_{1} C$. Similarly... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,961 |
5.33. In triangle $A B C$ with angle $A$ equal to $120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}$ and $C C_{1}$ intersect at point $O$. Prove that $\angle A_{1} C_{1} O=30^{\circ}$. | 5.33. According to the solution of the previous problem, ray $A_{1} C_{1}$ is the bisector of angle $A A_{1} B$. Let $K$ be the point of intersection of the angle bisectors of triangle $A_{1} A B$. Then $\angle C_{1} K O=\angle A_{1} K B=90^{\circ}+\angle A / 2=120^{\circ}$. Therefore, $\angle C_{1} K O+\angle C_{1} A ... | 30 | Geometry | proof | Yes | Yes | olympiads | false | 26,962 |
5.34. a) Prove that if angle $A$ of triangle $ABC$ is $120^{\circ}$, then the circumcenter and the orthocenter are symmetric with respect to the bisector of the external angle $A$.
b) In triangle $ABC$, angle $A$ is $60^{\circ}$; $O$ is the circumcenter, $H$ is the orthocenter, $I$ is the incenter, and $I_{a}$ is the ... | 5.34. a) Let $S$ be the circumcircle of triangle $ABC$, and $S_{1}$ be the circle symmetric to $S$ with respect to the line $BC$. The orthocenter $H$ of triangle $ABC$ lies on the circle $S_{1}$ (Problem 5.10), so it is sufficient to check that the center $O$ of circle $S$ also belongs to $S_{1}$ and that the external ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,963 |
5.35. In triangle $ABC$, angle $A$ is equal to $120^{\circ}$. Prove that a triangle can be formed from segments of lengths $a, b, b+c$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 5.35. Construct an equilateral triangle $A B_{1} C$ externally on the side $A C$ of triangle $A B C$. Since $\angle A=120^{\circ}$, point $A$ lies on the segment $B B_{1}$. Therefore, $B B_{1}=b+c$ and, moreover, $B C=a$ and $B_{1} C=b$, i.e., triangle $B B_{1} C$ is the desired one. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,964 | |
5.36*. In an acute-angled triangle $ABC$ with angle $A$ equal to $60^{\circ}$, the altitudes intersect at point $H$.
a) Let $M$ and $N$ be the points of intersection of the perpendicular bisectors of segments $BH$ and $CH$ with sides $AB$ and $AC$ respectively. Prove that points $M, N$, and $H$ lie on the same line.
... | 5.36. a) Let \( M_{1} \) and \( N_{1} \) be the midpoints of segments \( B H \) and \( C H \), and \( B B_{1} \) and \( C C_{1} \) be the altitudes. Right triangles \( A B B_{1} \) and \( B H C_{1} \) have a common acute angle at vertex \( B \), so \( \angle C_{1} H B = \angle A = 60^{\circ} \). Since triangle \( B M H... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,965 |
5.37*. In triangle $A B C$, the angle bisectors $B B_{1}$ and $C C_{1}$ are drawn. Prove that if $\angle C C_{1} B_{1}=30^{\circ}$, then either $\angle A=60^{\circ}$ or $\angle B=120^{\circ}$.
See also problem 2.33.
## §5. Integer Triangles | 5.37. Since $\angle B B_{1} C=\angle B_{1} B A+\angle B_{1} A B>\angle B_{1} B A=\angle B_{1} B C$, then $B C>B_{1} C$. Therefore, the point $K$, symmetric to $B_{1}$ with respect to the bisector $C C_{1}$, lies on the side $B C$, not on its extension. Since $\angle C C_{1} B=30^{\circ}$, then $\angle B_{1} C_{1} K=60^... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,966 |
5.38. The lengths of the sides of a triangle are consecutive integers. Find these numbers, given that one of the medians is perpendicular to one of the angle bisectors. | 5.38. Let $B M$ be the median, $A K$ the bisector of triangle $A B C$, and $B M \perp A K$. The line $A K$ is both a bisector and an altitude of triangle $A B M$, so $A M = A B$, i.e., $A C = 2 A M = 2 A B$. Therefore, $A B = 2$, $B C = 3$, and $A C = 4$. | AB=2,BC=3,AC=4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,967 |
5.39. The lengths of all sides of a right-angled triangle are integers, and the greatest common divisor of these numbers is 1. Prove that its legs are $2 m n$ and $m^{2}-n^{2}$, and the hypotenuse is $m^{2}+n^{2}$, where $m$ and $n$ are natural numbers.
A right-angled triangle, the lengths of whose sides are integers,... | 5.39. Let $a$ and $b$ be the legs, and $c$ the hypotenuse of a given triangle. If the numbers $a$ and $b$ are odd, then $a^{2}+b^{2}$ gives a remainder of 2 when divided by 4 and cannot be a perfect square. Therefore, one of the numbers $a$ and $b$ is even, and the other is odd; let $a=2 p$ for definiteness. The number... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,968 |
5.40*. The radius of the inscribed circle of a triangle is 1, and the lengths of its sides are integers. Prove that these numbers are $3,4,5$. | 5.40. Let $p$ be the semiperimeter of a triangle, and $a, b, c$ be the lengths of its sides. According to Heron's formula, $S^{2}=p(p-a)(p-b)(p-c)$. On the other hand, $S^{2}=p^{2} r^{2}=p^{2}$, since $r=1$. Therefore, $p=(p-a)(p-b)(p-c)$. If we introduce the unknowns $x=p-a, y=p-b, z=p-c$, then this equation can be re... | 3,4,5 | Geometry | proof | Yes | Yes | olympiads | false | 26,969 |
5.41*. Provide an example of a cyclic quadrilateral with pairwise distinct integer side lengths, for which the lengths of the diagonals, the area, and the radius of the circumscribed circle are integers (Brahmagupta). | 5.41. Let $a_{1}$ and $b_{1}, a_{2}$ and $b_{2}$ be the legs of two
Fig. 5.3 different Pythagorean triangles, $c_{1}$ and $c_{2}$ - their hypotenuses. Take two perpendicular lines and mark ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,970 |
5.42*. a) Indicate two right-angled triangles from which a triangle can be formed, the lengths of the sides and the area of which are integers.
b) Prove that if the area of a triangle is an integer and the lengths of the sides are consecutive natural numbers, then this triangle can be formed from two right-angled tria... | 5.42. a) The lengths of the hypotenuses of right triangles with legs 5 and 12, 9 and 12 are 13 and 15. By attaching the equal legs of these triangles to each other, we obtain a triangle with an area of $12(5+$
$+9) / 2=84$.
b) Suppose first that the length of the smallest side of the given triangle is an even number,... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,971 |
5.43*. a) In triangle $A B C$, the lengths of the sides of which are rational numbers, the height $B B_{1}$ is drawn. Prove that the lengths of the segments $A B_{1}$ and $C B_{1}$ are rational numbers.
b) The lengths of the sides and diagonals of a convex quadrilateral are rational numbers. Prove that the diagonals d... | 5.43. a) Since $A B^{2}-A B_{1}^{2}=B B_{1}^{2}=B C^{2}-\left(A C \pm A B_{1}\right)^{2}$, then $A B_{1}=$ $= \pm\left(A B^{2}+A C^{2}-B C^{2}\right) / 2 A C$.
b) Let the diagonals $A C$ and $B D$ intersect at point $O$. We will prove, for example, that the number $q=B O / O D$ is rational (then the number $O D=B D /(... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,972 |
5.44. Triangles $A B C$ and $A_{1} B_{1} C_{1}$ are such that their corresponding angles are equal or sum up to $180^{\circ}$. Prove that in fact all corresponding angles are equal. | 5.44. In triangles $A B C$ and $A_{1} B_{1} C_{1}$, there cannot be two pairs of corresponding angles that sum up to $180^{\circ}$, because otherwise their total sum would be $360^{\circ}$ and the third angles of the triangles would have to be zero. Now suppose that the angles of the first triangle are $\alpha, \beta$,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,973 |
5.45. Inside triangle $A B C$, a point $O$ is taken arbitrarily, and points $A_{1}, B_{1}$ and $C_{1}$ are constructed, which are symmetric to $O$ with respect to the midpoints of sides $B C, C A$ and $A B$. Prove that $\triangle A B C=\triangle A_{1} B_{1} C_{1}$ and the lines $A A_{1}, B B_{1}$ and $C C_{1}$ intersec... | 5.45. It is clear that $\overrightarrow{A_{1} C}=\overrightarrow{B O}$ and $\overrightarrow{C B_{1}}=\overrightarrow{O A}$, therefore $\overrightarrow{A_{1} B_{1}}=\overrightarrow{B A}$. Similarly, $\overrightarrow{B_{1} C_{1}}=\overrightarrow{C B}$ and $\overrightarrow{C_{1} A_{1}}=\overrightarrow{A C}$, i.e., $\trian... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,974 |
5.46. Through the point $O$ of intersection of the angle bisectors of triangle $ABC$, lines parallel to its sides are drawn. The line parallel to $AB$ intersects $AC$ and $BC$ at points $M$ and $N$, and the lines parallel to $AC$ and $BC$ intersect $AB$ at points $P$ and $Q$. Prove that $MN = AM + BN$ and the perimeter... | 5.46. Since $\angle M A O=\angle P A O=\angle A O M$, then $A M O P-$ is a rhombus. Similarly, $B N O Q$ is a rhombus. Therefore, $M N=M O+O N=A M+B N$ and $O P+P Q+$ $+Q O=A P+P Q+Q B=A B$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,975 |
5.47. a) Prove that the altitudes of a triangle intersect at one point.
b) Let $H$ be the point of intersection of the altitudes of triangle $ABC$, $R$ be the radius of the circumscribed circle. Prove that $A H^{2}+B C^{2}=4 R^{2}$ and $A H=$ $=B C|\operatorname{ctg} \alpha|$. | 5.47. a) Draw lines through the vertices of triangle $A B C$ parallel to its opposite sides. As a result, we obtain triangle $A_{1} B_{1} C_{1}$, the midpoints of whose sides are points $A, B$, and $C$. The altitudes of triangle $A B C$ are the perpendicular bisectors of the sides of triangle $A_{1} B_{1} C_{1}$, so th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,976 |
5.48. Let $x=\sin 18^{\circ}$. Prove that $4 x^{2}+2 x=1$.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | 5.48. Let $A D$ be the bisector of isosceles triangle $A B C$ with base $A B$ and angle $36^{\circ}$ at vertex $C$. Then triangle $A C D$ is isosceles and $\triangle A B C \sim \triangle B D A$. Therefore, $C D=A D=A B=2 x B C$ and $D B=2 x A B=4 x^{2} B C$, hence $B C=C D+D B=\left(2 x+4 x^{2}\right) B C$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,977 | |
5.49*. Prove that the projections of vertex $A$ of triangle $A B C$ onto the bisectors of the external and internal angles at vertices $B$ and $C$ lie on the same line. | 5.49. Let $B_{1}$ and $B_{2}$ be the projections of point $A$ onto the bisectors of the internal and external angles at vertex $B$, and let $M$ be the midpoint of side $AB$. Since the bisectors of the internal and external angles are perpendicular, $A B_{1} B B_{2}$ is a rectangle, and its diagonal $B_{1} B_{2}$ passes... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,978 |
5.50*. Prove that if in a triangle two angle bisectors are equal, then it is isosceles.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 5.50. Suppose the angle bisectors of angles $A$ and $B$ are equal, but $a > b$. Then $\cos \frac{A}{2} \frac{1}{c} + \frac{1}{a}$, i.e., $\frac{b c}{b+c} < \frac{a c}{a+c}$. Multiplying the obtained inequalities, we arrive at a contradiction, since $l_{a} = 2 b c \cos (A / 2) / (b+c)$ and $l_{b} = 2 a c \cos (B / 2) / ... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,979 | |
5.51*. a) In triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$, the sides $A C$ and $A^{\prime} C^{\prime}$, the angles at vertices $B$ and $B^{\prime}$, and the angle bisectors of angles $B$ and $B^{\prime}$ are equal. Prove that these triangles are congruent (more precisely, $\triangle A B C=\triangle A^{\prim... | 5.51. a) According to problem 4.47, the length of the bisector of angle $B$ in triangle $A B C$ is $2 a c \cos (B / 2) /(a+c)$, so it is sufficient to verify that the system of equations $a c /(a+c)=p, a^{2}+c^{2}-2 a c \cos B=q$ has (up to the permutation of numbers $a$ and $c$) a unique positive solution. Let $a+c=u$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,980 |
5.52*. Prove that a line divides the perimeter and area of a triangle in equal ratios if and only if it passes through the center of the inscribed circle. | 5.52. Let points $M$ and $N$ lie on sides $A B$ and $A C$. If $r_{1}$ is the radius of the circle with center on segment $M N$, touching sides $A B$ and $A C$, then $S_{A M N}=$ $=q r_{1}$, where $q=(A M+A N) / 2$. The line $M N$ passes through the center of the inscribed circle if and only if $r_{1}=r$, i.e., $S_{A M ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,981 |
5.53*. Point $E$ is the midpoint of the arc $AB$ of the circumcircle of triangle $ABC$, on which point $C$ lies; $C_{1}$ is the midpoint of side $AB$. A perpendicular $EF$ is dropped from point $E$ to $AC$. Prove that:
a) the line $C_{1}F$ bisects the perimeter of triangle $ABC$;
b) three such lines, constructed for ... | 5.53. a) Let's take a point \( B' \) on the extension of segment \( AC \) beyond point \( C \) such that \( CB' = CB \). Triangle \( BCB' \) is isosceles, so \( \angle AEB = \angle ACB = 2 \angle CBB' \), which means \( E \) is the center of the circumscribed circle of triangle \( ABB' \). Therefore, point \( F \) bise... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,982 |
5.54*. On the sides $AB$ and $BC$ of an acute-angled triangle $ABC$, squares $AB C_{1} D_{1}$ and $A_{2} B C D_{2}$ are constructed externally. Prove that the point of intersection of the lines $A D_{2}$ and $C D_{1}$ lies on the altitude $BH$. | 5.54. Let $X$ be the intersection point of the lines $A D_{2}$ and $C D_{1}; M, E_{1}$ and $E_{2}$ be the projections of points $X, D_{1}$ and $D_{2}$ onto the line $A C$. Then $C E_{2}=C D_{2} \sin \gamma=a \sin \gamma$ and $A E_{1}=$ $=c \sin \alpha$. Since $a \sin \gamma=c \sin \alpha$, then $C E_{2}=A E_{1}=q$. The... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,983 |
5.55*. On the sides of triangle $A B C$, squares are constructed externally with centers $A_{1}, B_{1}$ and $C_{1}$. Let $a_{1}, b_{1}$ and $c_{1}$ be the lengths of the sides of triangle $A_{1} B_{1} C_{1}$, $S$ and $S_{1}$ be the areas of triangles $A B C$ and $A_{1} B_{1} C_{1}$. Prove that:
a) $a_{1}^{2}+b_{1}^{2}... | 5.55. a) By the cosine theorem, $B_{1} C_{1}^{2}=A C_{1}^{2}+A B_{1}^{2}-2 A C_{1} \cdot A B_{1} \cdot \cos \left(90^{\circ}+\alpha\right)$, i.e., $a_{1}^{2}=\frac{c^{2}}{2}+\frac{b^{2}}{2}+b c \sin \alpha=\frac{b^{2}+c^{2}}{2}+2 S$. Writing similar equalities for $b_{1}^{2}$ and $c_{1}^{2}$ and adding them, we obtain ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,984 |
5.56*. On the sides $AB$, $BC$, and $CA$ of triangle $ABC$ (or on their extensions), points $C_1$, $A_1$, and $B_1$ are taken such that $\angle(C C_1, AB) = \angle(A A_1, BC) = \angle(B B_1, CA) = \alpha$. The lines $A A_1$ and $B B_1$, $B B_1$ and $C C_1$, $C C_1$ and $A A_1$ intersect at points $C'$, $A'$, $B'$ respe... | 5.56. First, let's prove that point \( B' \) lies on the circumcircle of triangle \( AHC \), where \( H \) is the orthocenter of triangle \( ABC \). \(\angle (AB', B'C) = \angle (AA_1, CC_1) = \angle (AA_1, BC) + \angle (BC, AB) + \angle (AB, CC_1) = \angle (BC, AB)\). From the solution to problem 5.10, \(\angle (BC, A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,985 |
5.58*. On the sides of triangle $A B C$, points $A_{1}, B_{1}$, and $C_{1}$ are taken such that $A B_{1}: B_{1} C=c^{n}: a^{n}, B C_{1}: C_{1} A=a^{n}: b^{n}$, and $C A_{1}: A_{1} B=b^{n}: c^{n}$ (where $a, b$, and $c$ are the lengths of the sides of the triangle). The circumcircle of triangle $A_{1} B_{1} C_{1}$ inter... | 5.58. Let $a_{1}=B A_{1}, a_{2}=A_{1} C, b_{1}=C B_{1}, b_{2}=B_{1} A, c_{1}=A C_{1}$ and $c_{2}=C_{1} B$. The products of the lengths of the segments of the secants passing through one point are equal, so $a_{1}\left(a_{1}+x\right)=c_{2}\left(c_{2}-z\right)$, i.e., $a_{1} x+c_{2} z=c_{2}^{2}-a_{1}^{2}$. Similarly, we ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,986 |
5.59*. In triangle $ABC$, trisectors (rays that divide angles into three equal parts) are drawn. The trisectors of angles $B$ and $C$ closest to side $BC$ intersect at point $A_{1}$; similarly, we define points $B_{1}$ and $C_{1}$ (Fig. 5.1). Prove that triangle $A_{1} B_{1} C_{1}$ is equilateral (Morley's theorem).
!... | 5.59. Let in the original triangle $\angle A=3 \alpha, \angle B=3 \beta$ and $\angle C=3 \gamma$. Take an equilateral triangle $A_{2} B_{2} C_{2}$ and construct isosceles triangles $A_{2} B_{2} R, B_{2} C_{2} P$ and $C_{2} A_{2} Q$ on its sides as bases with angles at the bases $60^{\circ}-\gamma, 60^{\circ}-\alpha, 60... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,987 |
5.60*. On the sides of an equilateral triangle $A B C$, isosceles triangles $A_{1} B C, A B_{1} C$, and $A B C_{1}$ are constructed inwardly with angles $\alpha, \beta$, and $\gamma$ at the bases, respectively, such that $\alpha + \beta + \gamma = 60^{\circ}$. The lines $B C_{1}$ and $B_{1} C$ intersect at point $A_{2}... | 5.60. Point $A_{1}$ lies on the bisector of angle $B A C$, so point $A$ lies on the extension of the bisector of angle $B_{2} A_{1} C_{2}$. Moreover, $\angle B_{2} A C_{2}=\alpha=$ $=\left(180^{\circ}-\angle B_{2} A_{1} C_{2}\right) / 2$. Therefore, $A$ is the center of the excircle of triangle $B_{2} A_{1} C_{2}$ (see... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,988 |
5.61*. A circle of radius $u_{a}$ is inscribed in angle $A$ of triangle $A B C$, and a circle of radius $u_{b}$ is inscribed in angle $B$; these circles touch each other externally. Prove that the radius of the circumscribed circle of the triangle with sides $a_{1}=\sqrt{u_{a} \operatorname{ctg}(\alpha / 2)}, b_{1}=\sq... | 5.61. The length of the common tangent to the given circles is $2 \sqrt{u_{a} u_{b}}$, therefore
$$
u_{a} \operatorname{ctg} \frac{\alpha}{2}+2 \sqrt{u_{a} u_{b}}+u_{b} \operatorname{ctg} \frac{\beta}{2}=c
$$
i.e., $a_{1}^{2}+2 a_{1} b_{1} \sqrt{\operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \frac{\beta}{2}}+b_... | \sqrt{p} | Geometry | proof | Yes | Yes | olympiads | false | 26,989 |
5.62*. Circle $S_{1}$ is inscribed in angle $A$ of triangle $ABC$; circle $S_{2}$ is inscribed in angle $B$ and is tangent to $S_{1}$ (externally); circle $S_{3}$
is inscribed in angle $C$ and is tangent to $S_{2}$; circle $S_{4}$ is inscribed in angle $A$ and is tangent to $S_{3}$ and so on. Prove that circle $S_{7}$ ... | 5.62. Let $u_{1}$ and $u_{2}$ be the radii of the circles $S_{1}$ and $S_{2}$. According to problem 5.61, the radius of the circumcircle of a triangle with sides $\tilde{u}_{1}=\sqrt{u_{1} \operatorname{ctg} \frac{\alpha}{2}}, \tilde{u}_{2}=\sqrt{u_{2} \operatorname{ctg} \frac{\beta}{2}}, \tilde{c}=\sqrt{c}$ is $\sqrt{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,990 |
5.63*. Circle $S_{1}$ is inscribed in angle $A$ of triangle $ABC$. From vertex $C$ a tangent (different from $CA$) is drawn to it, and a circle $S_{2}$ is inscribed in the resulting triangle with vertex $B$. From vertex $A$ a tangent is drawn to $S_{2}$, and a circle $S_{3}$ is inscribed in the resulting triangle with ... | 5.63. Let $r_{i}$ be the radius of the circle $S_{i}$, and $h_{i}$ be the height of triangle $ABC$, dropped from vertex $A$ when $i=3k+1$, from vertex $B$ when $i=3k+2$, and from vertex $C$ when $i=3k$. The formula from problem 5.9 a) can be written as
$$
\frac{r}{r_{i}}-1 \quad \frac{r}{r_{i+1}}-1 \quad=1-\frac{2 r}{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,991 |
5.64*. On the sides $BC$, $CA$, and $AB$ of triangle $ABC$ (or on their extensions), points $A_{1}$, $B_{1}$, and $C_{1}$ are taken respectively. Prove that points $A_{1}$, $B_{1}$, and $C_{1}$ lie on one straight line if and only if
$$
\frac{\overline{B A_{1}}}{\overline{C A_{1}}} \cdot \frac{\overline{C B_{1}}}{A B_... | 5.64. Let the projection onto a line perpendicular to the line $A_{1} B_{1}$ map points $A, B$, and $C$ to $A^{\prime}, B^{\prime}$, and $C^{\prime}$, point $C_{1}$ to $Q$, and the two points $A_{1}$ and $B_{1}$ to a single point $P$. Since $\overline{A_{1} B}: \overline{A_{1} C}=\overline{P B^{\prime}}: \overline{P C^... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,992 |
5.66*. Circle $S$ touches circles $S_{1}$ and $S_{2}$ at points $A_{1}$ and $A_{2}$. Prove that the line $A_{1} A_{2}$ passes through the point of intersection of the common external or common internal tangents to circles $S_{1}$ and $S_{2}$. | 5.66. Let $O, O_{1}$ and $O_{2}$ be the centers of circles $S, S_{1}$ and $S_{2}$; $X$ be the intersection point of the lines $O_{1} O_{2}$ and $A_{1} A_{2}$. Applying Menelaus' theorem to triangle $O O_{1} O_{2}$ and points $A_{1}, A_{2}$ and $X$, we get $\frac{O_{1} X}{O_{2} X} \cdot \frac{O_{2} A_{2}}{O A_{2}} \cdot... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,994 |
5.68*. From the vertex $C$ of the right angle of triangle $A B C$, the height $C K$ is dropped, and in triangle $A C K$ the bisector $C E$ is drawn. A line passing through point $B$ parallel to $C E$ intersects $C K$ at point $F$. Prove that the line $E F$ bisects the segment $A C$. | 5.68. Since $\angle B C E=90^{\circ}-\angle B / 2$, then $\angle B C E=\angle B E C$, which means $B E=B C$. Therefore, $C F: K F=B E: B K=B C: B K$ and $A E: K E=C A: C K=B C: B K$. Let the line $E F$ intersect $A C$ at point $D$. By Menelaus' theorem $\frac{A D}{C D} \cdot \frac{C F}{K F} \cdot \frac{K E}{A E}=$ $=1$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,996 |
5.70*. Lines $A A_{1}, B B_{1}, C C_{1}$ intersect at a single point $O$. Prove that the points of intersection of lines $A B$ and $A_{1} B_{1}, B C$ and $B_{1} C_{1}, A C$ and $A_{1} C_{1}$
lie on the same line (Desargues). | 5.70. Let $A_{2}, B_{2}, C_{2}$ be the points of intersection of the lines $B C$ and $B_{1} C_{1}, A C$ and $A_{1} C_{1}, A B$ and $A_{1} B_{1}$. Apply Menelaus' theorem to the following triangles and points on their sides: $O A B$ and $\left(A_{1}, B_{1}, C_{2}\right), O B C$ and $\left(B_{1}, C_{1}, A_{2}\right), O A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,998 |
5.71*. Points $A_{1}, B_{1}$, and $C_{1}$ are taken on one line, and points $A_{2}, B_{2}$, and $C_{2}$ are taken on another. The lines $A_{1} B_{2}$ and $A_{2} B_{1}$, $B_{1} C_{2}$ and $B_{2} C_{1}$, $C_{1} A_{2}$ and $C_{2} A_{1}$ intersect at points $C, A$, and $B$ respectively. Prove that points $A, B$, and $C$ li... | 5.71. Consider the triangle $A_{0} B_{0} C_{0}$ formed by the lines $A_{1} B_{2}, B_{1} C_{2}$, and $C_{1} A_{2}$ (where $A_{0}$ is the intersection point of the lines $A_{1} B_{2}$ and $A_{2} C_{1}$, and so on), and apply Menelaus' theorem to the following five sets of points: $\left(A, B_{2}, C_{1}\right),\left(B, C_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,999 |
5.72*. On the sides $AB$, $BC$, and $CD$ of quadrilateral $ABCD$ (or on their extensions), points $K$, $L$, and $M$ are taken. Lines $KL$ and $AC$ intersect at point $P$, and lines $LM$ and $BD$ intersect at point $Q$. Prove that the point of intersection of lines $KQ$ and $MP$ lies on line $AD$. | 5.72. Let $N$ be the intersection point of lines $A D$ and $K Q$, and $P^{\prime}$ be the intersection point of lines $K L$ and $M N$. By applying Desargues' theorem to triangles $K B L$ and $N D M$, we obtain that points $P^{\prime}, A$, and $C$ lie on the same line. Therefore, $P^{\prime}=P$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,000 |
5.73*. The extensions of sides $A B$ and $C D$ of quadrilateral $A B C D$ intersect at point $P$, and the extensions of sides $B C$ and $A D$ intersect at point $Q$. A line through point $P$ intersects sides $B C$ and $A D$ at points $E$ and $F$. Prove that the points of intersection of the diagonals of quadrilaterals ... | 5.73. It is sufficient to apply Desargues' theorem to triangles $A E D$ and $B F C$ and Pappus' theorem to the triples of points $(B, E, C)$ and $(A, F, D)$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,001 |
5.77*. Given a triangle $A B C$. On the lines $A B, B C$ and $C A$, points $C_{1}, A_{1}$ and $B_{1}$ are taken, such that $k$ of them lie on the sides of the triangle and $3-k$ on the extensions of the sides. Let
$$
R=\frac{B A_{1}}{C A_{1}} \cdot \frac{C B_{1}}{A B_{1}} \cdot \frac{A C_{1}}{B C_{1}}
$$
Prove that:
... | 5.77. a) This problem is a reformulation of problem 5.64, since the number $\overline{B A_{1}}: \overline{C A_{1}}$ is negative if point $A_{1}$ lies on the segment $B C$, and positive if it lies outside the segment $B C$.
b) Suppose first that the lines $A A_{1}, B B_{1}$, and $C C_{1}$ intersect at point $M$. Any th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,003 |
5.78*. The incircle (or excircle) of triangle $ABC$ touches the lines $BC, CA$, and $AB$ at points $A_{1}, B_{1}$, and $C_{1}$. Prove that the lines $AA_{1}, BB_{1}$, and $CC_{1}$ intersect at one point.
The point of intersection of the lines connecting the vertices of the triangle to the points of tangency of the inc... | 5.78. It is clear that $A B_{1}=A C_{1}, B A_{1}=B C_{1}$ and $C A_{1}=C B_{1}$, and in the case of the inscribed circle, three points lie on the sides of triangle $ABC$, while in the case of the excircle, one point lies on the side. It remains to use Ceva's theorem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,004 |
5.79*. Prove that the lines connecting the vertices of a triangle with the points of tangency of the excircles with the sides intersect at one point (the Nagel point).
Translate the above text into English, please retain the line breaks and format of the source text, and output the translation result directly. | 5.79. Let the excircles touch the sides $B C, C A$ and $A B$ at points $A_{1}, B_{1}$ and $C_{1}$. Then
$$
\frac{B A_{1}}{C A_{1}} \frac{C B_{1}}{A B_{1}} \frac{A C_{1}}{B C_{1}}=\frac{p-c}{p-b} \frac{p-a}{p-c} \frac{p-b}{p-a}=1
$$ | Algebra | proof | Yes | Yes | olympiads | false | 27,005 | |
5.80*. Prove that the altitudes of an acute-angled triangle intersect at one point.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
However, it seems the instruction is already in English, so here is the translation of the... | 5.80. Let $A A_{1}, B B_{1}$ and $C C_{1}$ be the altitudes of triangle $A B C$. Then
$$
\frac{A C_{1}}{C_{1} B} \cdot \frac{B A_{1}}{A_{1} C} \cdot \frac{C B_{1}}{B_{1} A}=\frac{b \cos A}{a \cos B} \cdot \frac{c \cos B}{b \cos C} \cdot \frac{a \cos C}{c \cos A}=1
$$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,006 |
5.81*. Lines $A P, B P$ and $C P$ intersect the sides of triangle $A B C$ (or their extensions) at points $A_{1}, B_{1}$ and $C_{1}$. Prove that:
a) the lines passing through the midpoints of sides $B C, C A$ and $A B$ parallel to lines $A P, B P$ and $C P$, respectively, intersect at one point;
b) the lines connecti... | 5.81. Let $A_{2}, B_{2}$ and $C_{2}$ be the midpoints of sides $B C, C A$ and $A B$. The lines considered pass through the vertices of triangle $A_{2} B_{2} C_{2}$, and in problem a) they divide its sides in the same ratios as lines $A P, B P$ and $C P$ divide the sides of triangle $A B C$, while in problem b) they div... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,007 |
5.82*. On the sides $BC$, $CA$, and $AB$ of triangle $ABC$, points $A_1$, $B_1$, and $C_1$ are taken such that the segments $AA_1$, $BB_1$, and $CC_1$ intersect at one point. The lines $A_1B_1$ and $A_1C_1$ intersect the line passing through vertex $A$ parallel to side $BC$ at points $C_2$ and $B_2$ respectively. Prove... | 5.82. Since $\triangle A C_{1} B_{2} \sim \triangle B C_{1} A_{1}$ and $\triangle A B_{1} C_{2} \sim \triangle C B_{1} A_{1}$, then $A B_{2} \cdot C_{1} B=$ $=A C_{1} \cdot B A_{1}$ and $A C_{2} \cdot C B_{1}=A_{1} C \cdot B_{1} A$. Therefore
$$
\frac{A B_{2}}{A C_{2}}=\frac{A C_{1}}{C_{1} B} \cdot \frac{B A_{1}}{A_{1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,008 |
5.83*. a) Let $\alpha, \beta$, and $\gamma$ be arbitrary angles, such that the sum of any two of them is less than $180^{\circ}$. On the sides of triangle $ABC$, triangles $A_{1}BC, AB_{1}C$, and $ABC_{1}$ are constructed externally, having angles $\alpha, \beta$, and $\gamma$ at vertices $A, B$, and $C$. Prove that th... | 5.83. Let the lines $A A_{1}, B B_{1}$, and $C C_{1}$ intersect the lines $B C, C A$, and $A B$ at points $A_{2}, B_{2}$, and $C_{2}$.
a) If $\angle B+\beta<180^{\circ}$ and $\angle C+\gamma<180^{\circ}$, then
\[
\frac{B A_{2}}{A_{2} C}=\frac{S_{A B A_{1}}}{S_{A C A_{1}}}=\frac{A B \cdot B A_{1} \sin (B+\beta)}{A C \c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,009 |
5.86*. On the sides $BC$, $CA$, $AB$ of triangle $ABC$, points $A_{1}$, $B_{1}$, $C_{1}$ are taken. Prove that
$$
\frac{A C_{1}}{C_{1} B} \cdot \frac{B A_{1}}{A_{1} C} \cdot \frac{C B_{1}}{B_{1} A}=\frac{\sin A C C_{1}}{\sin C_{1} C B} \cdot \frac{\sin B A A_{1}}{\sin A_{1} A C} \cdot \frac{\sin C B B_{1}}{\sin B_{1} ... | 5.86. Applying the Law of Sines to triangles $A C C_{1}$ and $B C C_{1}$, we get $\frac{A C_{1}}{C_{1} C}=\frac{\sin A C C_{1}}{\sin A}$ and $\frac{C C_{1}}{C_{1} B}=\frac{\sin B}{\sin C_{1} C B}$, i.e., $\frac{A C_{1}}{C_{1} B}=\frac{\sin A C C_{1}}{\sin C_{1} C B} \cdot \frac{\sin B}{\sin A}$. Similarly, $\frac{B A_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,012 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.