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10.83*. Prove that a triangle is acute if and only if \( p > 2R + r \). | 10.83. It is sufficient to note that $p^{2}-(2 R+r)^{2}=4 R^{2} \cos \alpha \cos \beta \cos \gamma$ (see problem 12.41, b)). | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,479 |
10.84*. Prove that triangle $ABC$ is acute-angled if and only if there exist such internal points $A_{1}, B_{1}$ and $C_{1}$ on its sides $BC, CA$, and $AB$ respectively, that $A A_{1}=B B_{1}=C C_{1}$. | 10.84. Let $\angle A \leqslant \angle B \leqslant \angle C$. If triangle $ABC$ is not acute-angled, then $C C_{1}<A C<A A_{1}$ for any points $A_{1}$ and $C_{1}$ on sides $BC$ and $AB$. Now let's prove that for an acute-angled triangle, we can choose points $A_{1}, B_{1}$, and $C_{1}$ with the required property. For th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,480 |
10.85*. Prove that triangle $ABC$ is acute-angled if and only if the lengths of its projections onto three different directions are equal.
See also problems $9.93, 10.39, 10.44, 10.48, 10.62$.
## §13. Inequalities in Triangles | 10.85. Let $\angle A \leqslant \angle B \leqslant \angle C$. Suppose first that triangle $ABC$ is acute. When the line $l$, initially parallel to $AB$, is rotated, the length of the projection of the triangle onto $l$ will first monotonically change from $c$ to $h_{b}$, then from $h_{b}$ to $a$, from $a$ to $h_{c}$, fr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,481 |
10.86. Through the point $O$ of intersection of the medians of triangle $ABC$, a line is drawn intersecting its sides at points $M$ and $N$. Prove that $N O \leqslant 2 M O$. | 10.86. Let points $M$ and $N$ lie on sides $A B$ and $A C$ respectively. Draw a line through vertex $C$ parallel to side $A B$. Let $N_{1}$ be the point of intersection of this line and line $M N$. Then $N_{1} O: M O=2$, but $N O \leqslant N_{1} O$, so $N O: M O \leqslant 2$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,482 |
10.87. Prove that if triangle $A B C$ lies inside triangle $A^{\prime} B^{\prime} C^{\prime}$, then $r_{A B C}<r_{A^{\prime} B^{\prime} C^{\prime}}$. | 10.87. The circle $S$, inscribed in triangle $ABC$, lies inside triangle $A'B'C'$. By drawing tangents to this circle, parallel to the sides of triangle $A'B'C'$, one can obtain triangle $A''B''C''$, similar to triangle $A'B'C'$, for which $S$ is the inscribed circle. Therefore, $r_{ABC}=r_{A''B''C''}<r_{A'B'C'}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,483 |
10.88. In triangle $ABC$, side $c$ is the largest, and $a$ is the smallest. Prove that $l_{c} \leqslant h_{a}$. | 10.88. The bisector $l_{c}$ divides the triangle $ABC$ into two triangles, the doubled areas of which are equal to $a l_{c} \sin (\gamma / 2)$ and $b l_{c} \sin (\gamma / 2)$. Therefore, $a h_{a}=2 S=$ $=l_{c}(a+b) \sin (\gamma / 2)$. From the condition of the problem, it follows that $a /(a+b) \leqslant 1 / 2 \leqslan... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,484 |
10.89. The medians $A A_{1}$ and $B B_{1}$ of triangle $A B C$ are perpendicular. Prove that $\operatorname{ctg} A+\operatorname{ctg} B \geqslant 2 / 3$. | 10.89. It is clear that $\operatorname{ctg} A+\operatorname{ctg} B=c / h_{c} \geqslant c / m_{c}$. Let $M$ be the point of intersection of the medians, $N$ be the midpoint of segment $A B$. Since triangle $A M B$ is right-angled, $M N=A B / 2$. Therefore, $c=2 M N=2 m_{c} / 3$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,485 |
10.91*. In an acute-angled triangle $A B C$, the bisector $A D$, the median $B M$, and the altitude $C H$ intersect at one point. Within what limits can the measure of angle $A$ vary? | 10.91. Draw a perpendicular to side $AB$ through point $B$. Let $F$ be the point of intersection of this perpendicular with the extension of side $AC$ (Fig. 10.3). We will prove that the bisector $AD$, the median $BM$, and the altitude $CH$ intersect at one point if and only if $AB = CF$. Indeed, let $L$ be the point o... | 5150' | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,487 |
10.92*. In triangle $ABC$, the sides are equal to $a, b, c$; the corresponding angles (in radians) are $\alpha, \beta, \gamma$. Prove that
$$
\frac{\pi}{3} \leqslant \frac{a \alpha + b \beta + c \gamma}{a + b + c} < \frac{\pi}{2}
$$ | 10.92. Since the larger side lies opposite the larger angle, we have $(a-b)(\alpha-\beta) \geqslant 0$, $(b-c)(\beta-\gamma) \geqslant 0$, and $(a-c)(\alpha-\gamma) \geqslant 0$. Adding these inequalities, we get
$$
2(a \alpha+b \beta+c \gamma) \geqslant a(\beta+\gamma)+b(\alpha+\gamma)+c(\alpha+\beta)=(a+b+c) \pi-a \... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,488 |
10.93*. Inside triangle $ABC$, a point $O$ is taken. Prove that $AO \sin BOC + BO \sin AOC + CO \sin AOB \leqslant p$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 10.93. Let's take points \( C_1 \) and \( B_1 \) on rays \( OB \) and \( OC \) such that \( OC_1 = OC \) and \( OB_1 = OB \). Let \( B_2 \) and \( C_2 \) be the projections of points \( B_1 \) and \( C_1 \) onto the line perpendicular to \( AO \). Then \( BO \sin AOC + CO \sin AOB = B_2 C_2 \leq BC \). By summing three... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,489 | |
10.94*. On the extension of the longest side $A C$ of triangle $A B C$ beyond point $C$, a point $D$ is taken such that $C D=C B$. Prove that angle $A B D$ is not acute. | 10.94. Since $\angle C B D=\angle C / 2$ and $\angle B \geqslant \angle A$, then $\angle A B D=\angle B + \angle C B D \geqslant (\angle A + \angle B + \angle C) / 2 = 90^{\circ}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,490 |
10.96*. On the sides $BC$, $CA$, $AB$ of triangle $ABC$, points $X$, $Y$, $Z$ are taken such that the lines $AX$, $BY$, $CZ$ intersect at a single point $O$. Prove that among the ratios $OA:OX$, $OB:OY$, $OC:OZ$, at least one is not greater than 2 and at least one is not less than 2. | 10.96. Suppose all the given ratios are less than 2. Then $S_{A B O} + S_{A O C} < 2 S_{\text{Xvo}} + 2 S_{\text{Xos}} = 2 S_{\text{OvC}}, S_{A B O} + S_{\text{OvC}} < 2 S_{A O C}$ and $S_{A O C} + S_{O B C} < 2 S_{A B O}$. Adding these inequalities, we arrive at a contradiction. Similarly, it is proved that one of the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,492 |
10.97*. Circle $S_{1}$ touches sides $A C$ and $A B$ of triangle $A B C$, circle $S_{2}$ touches sides $B C$ and $A B$, moreover, $S_{1}$ and $S_{2}$ touch each other externally. Prove that the sum of the radii of these circles is greater than the radius of the inscribed circle $S$.
See also problems $14.26,17.16,17.1... | 10.97. Let the radii of the circles $S, S_{1}$ and $S_{2}$ be denoted by $r, r_{1}$ and $r_{2}$. Suppose triangles $A B_{1} C_{1}$ and $A_{2} B C_{2}$ are similar to triangle $A B C$, with similarity coefficients $r_{1} / r$ and $r_{2} / r$ respectively. The circles $S_{1}$ and $S_{2}$ are inscribed in triangles $A B_{... | r_{1}+r_{2}>r | Geometry | proof | Yes | Yes | olympiads | false | 27,493 |
11.1. Prove that among all triangles with a fixed angle $\alpha$ and area $S$, the triangle with the smallest length of side $BC$ is the isosceles triangle with base $BC$. | 11.1. By the Law of Cosines \(a^{2}=b^{2}+c^{2}-2 b c \cos \alpha=(b-c)^{2}+2 b c(1-\cos \alpha)=\) \((b-c)^{2}+4 S(1-\cos \alpha) / \sin \alpha\). Since the second term is constant, \(a\) is minimal if \(b=c\). | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,494 |
11.2. Prove that among all triangles $ABC$ with a fixed angle $\alpha$ and semiperimeter $p$, the triangle with the greatest area is the isosceles triangle with base $BC$. | 11.2. Let the excircle touch sides $A B$ and $A C$ at points $K$ and $L$. Since $A K=A L=p$, the excircle $S_{a}$ is fixed. The radius $r$ of the inscribed circle is maximal when it touches the circle $S_{a}$, i.e., triangle $A B C$ is isosceles. It is also clear that $S=p r$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,495 |
11.3. Prove that among all triangles with a fixed semiperimeter \( p \), the one with the greatest area is the equilateral triangle. | 11.3. According to problem 10.53, a) $S \leqslant p^{2} / 3 \sqrt{3}$, and equality is achieved only for an equilateral triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,496 |
11.4. Consider all acute-angled triangles with a given side $a$ and angle $\alpha$. What is the maximum value of the sum of the squares of the lengths of sides $b$ and $c$? | 11.4. By the Law of Cosines, $b^{2}+c^{2}=a^{2}+2 b c \cos \alpha$. Since $2 b c \leqslant b^{2}+c^{2}$ and $\cos \alpha>0$, then $b^{2}+c^{2} \leqslant a^{2}+\left(b^{2}+c^{2}\right) \cos \alpha$, i.e., $b^{2}+c^{2} \leqslant a^{2} /(1-\cos \alpha)$. Equality is achieved if $b=c$. | ^2/(1-\cos\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,497 |
11.5. Among all triangles inscribed in a given circle, find the one for which the sum of the squares of the side lengths is maximized. | 11.5. Let $O$ be the center of a circle with radius $R$; $A, B$, and $C$ be the vertices of a triangle; $\boldsymbol{a}=\overrightarrow{O A}, \boldsymbol{b}=\overrightarrow{O B}, \boldsymbol{c}=\overrightarrow{O C}$. Then $A B^{2}+B C^{2}+C A^{2}=|\boldsymbol{a}-\boldsymbol{b}|^{2}+$ $+|\boldsymbol{b}-\boldsymbol{c}|^{... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,498 |
11.6*. The perimeter of triangle $A B C$ is $2 p$. Points $M$ and $N$ are taken on sides $A B$ and $A C$ such that $M N \| B C$ and $M N$ is tangent to the inscribed circle of triangle $A B C$. Find the maximum value of the length of segment $M N$. | 11.6. Let the length of the altitude dropped to side $BC$ be denoted by $h$. Since $\triangle A M N \sim \triangle A B C$, we have $M N / B C=(h-2 r) / h$, i.e., $M N=a \left(1-\frac{2 r}{h}\right)$. Since $r=S / p=a h / 2 p$, it follows that $M N=a\left(1-a / p\right)$. The maximum value of the expression $a\left(1-a ... | \frac{p}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,499 |
11.8*. The area of triangle $ABC$ is 1. Let $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $BC, CA, AB$ respectively. Points $K, L, M$ are taken on segments $AB_{1}, CA_{1}, BC_{1}$ respectively. What is the minimum area of the common part of triangles $KLM$ and $A_{1}B_{1}C_{1}$? | 11.8. Let the intersection point of lines $K M$ and $B C$ be denoted as $T$, and the intersection points of the sides of triangles $A_{1} B_{1} C_{1}$ and $K L M$ as shown in Fig. 11.2. Then $T L: R Z = K L: K Z = L C: Z B_{1}$. Since $T L \geqslant B A_{1} = A_{1} C \geqslant L C$, it follows that $R Z \geqslant Z B_{... | \frac{1}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,501 |
11.9*. What is the minimum width that an infinite strip of paper must have so that any triangle with an area of 1 can be cut out of it? | 11.9. Since the area of an equilateral triangle with side $a$ is $a^{2} \sqrt{3} / 4$, the side of an equilateral triangle with area 1 is $2 / \sqrt[4]{3}$, and its height is $\sqrt[4]{3}$. We will prove that an equilateral triangle with area 1 cannot be cut from a strip of width less than $\sqrt[4]{3}$. Let the equila... | \sqrt[4]{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,502 |
11.10. Prove that triangles with side lengths $a, b, c$ and $a_{1}, b_{1}, c_{1}$ are similar if and only if
$$
\sqrt{a a_{1}}+\sqrt{b b_{1}}+\sqrt{c c_{1}}=\sqrt{(a+b+c)\left(a_{1}+b_{1}+c_{1}\right)}
$$ | 11.10. By squaring both sides of the given equality, it can easily be reduced to the form
$$
\left(\sqrt{a b_{1}}-\sqrt{a_{1} b}\right)^{2}+\left(\sqrt{c a_{1}}-\sqrt{c_{1} a}\right)^{2}+\left(\sqrt{b c_{1}}-\sqrt{c b_{1}}\right)^{2}=0
$$
i.e., \( a / a_{1}=b / b_{1}=c / c_{1} \). | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,503 |
11.11*. Prove that if $\alpha, \beta, \gamma$ and $\alpha_{1}, \beta_{1}, \gamma_{1}$ are the angles of two triangles, then
$$
\frac{\cos \alpha_{1}}{\sin \alpha}+\frac{\cos \beta_{1}}{\sin \beta}+\frac{\cos \gamma_{1}}{\sin \gamma} \leqslant \operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma... | 11.11. Fix angles $\alpha, \beta$ and $\gamma$. Let $A_{1} B_{1} C_{1}$ be a triangle with angles $\alpha_{1}$, $\beta_{1}$ and $\gamma_{1}$. Consider vectors $\boldsymbol{a}, \boldsymbol{b}$ and $\boldsymbol{c}$, directed along vectors $\overrightarrow{B_{1} C_{1}}, \overrightarrow{C_{1} A_{1}}$
and $\overrightarrow{A... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,504 |
11.12*. Let $a, b$ and $c$ be the lengths of the sides of a triangle with area $S$; $\alpha_{1}, \beta_{1}$ and $\gamma_{1}$ be the angles of some other triangle. Prove that $a^{2} \operatorname{ctg} \alpha_{1}+$ $+b^{2} \operatorname{ctg} \beta_{1}+c^{2} \operatorname{ctg} \gamma_{1} \geqslant 4 S$, and equality is ac... | 11.12. Let $x=\operatorname{ctg} \alpha_{1}$ and $y=\operatorname{ctg} \beta_{1}$. Then $x+y>0$ (since $\alpha_{1}+\beta_{1}<\pi$) and $\operatorname{ctg} \gamma_{1}=(1-x y) /(x+y)=\left(x^{2}+1\right) /(x+y)-x$. Therefore, $a^{2} \operatorname{ctg} \alpha_{1}+b^{2} \operatorname{ctg} \beta_{1}+c^{2} \operatorname{ctg}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,505 |
11.13*. Given a triangle with sides $a, b$, and $c$, where $a \geqslant b \geqslant c$; $x, y$, and $z$ are the angles of some other triangle. Prove that
$$
b c + c a - a b < b c \cos x + c a \cos y + a b \cos z \leqslant \left(a^{2} + b^{2} + c^{2}\right) / 2
$$
See also problem 17.21.
## §2. Extreme Points of a Tr... | 11.13. Let $f=b c \cos x+c a \cos y+a b \cos z$. Since $\cos x=-\cos y \cos z+ \sin y \sin z$, then $f=c(a-b \cos z) \cos y+b c \sin y \sin z+a b \cos z$. Consider a triangle, the lengths of two sides of which are $a$ and $b$, and the angle between them is $z$; let $\xi$ and $\eta$ be the angles opposite the sides $a$ ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,506 |
11.14. On the hypotenuse $AB$ of a right triangle $ABC$, a point $X$ is taken; $M$ and $N$ are its projections on the legs $AC$ and $BC$.
a) For which position of point $X$ will the length of segment $MN$ be the smallest?
b) For which position of point $X$ will the area of quadrilateral $CMXN$ be the largest? | 11.14. a) Since $C M X N$ is a rectangle, then $M N = C X$. Therefore, the length of the segment $M N$ will be the smallest if $C X$ is the height.
b) Let $S_{A B C} = S$. Then $S_{A M X} = A X^{2} \cdot S / A B^{2}$ and $S_{B N X} = B X^{2} \cdot S / A B^{2}$. Since $A X^{2} + B X^{2} \geqslant A B^{2} / 2$ (with equ... | )CXistheheight.\quadb)XisthemidpointofsideAB | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,507 |
11.15. From a point $M$, lying on the side $A B$ of an acute-angled triangle $A B C$, perpendiculars $M P$ and $M Q$ are dropped to the sides $B C$ and $A C$. For what position of point $M$ is the length of the segment $P Q$ minimal? | 11.15. Points $P$ and $Q$ lie on the circle constructed on segment $C M$ as its diameter. In this circle, the constant angle $C$ subtends the chord $P Q$, so the length of the chord $P Q$ will be minimal if the diameter $C M$ of the circle is minimal, i.e., $C M$ is the altitude of triangle $A B C$. | CMisthealtitudeoftriangleABC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,508 |
11.16. Given a triangle $A B C$. Find a point $M$ on the line $A B$ such that the sum of the radii of the circumcircles of triangles $A C M$ and $B C M$ is minimized. | 11.16. By the Law of Sines, the radii of the circumcircles of triangles $A C M$ and $B C M$ are $A C /(2 \sin A M C)$ and $B C /(2 \sin B M C)$, respectively. It is easy to verify that $\sin A M C=\sin B M C$. Therefore, $A C /(2 \sin A M C)+$ $+B C /(2 \sin B M C)=(A C+B C) /(2 \sin B M C)$. The last expression will b... | CM\perpAB | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,509 |
11.17. From a point $M$ on the circumcircle of triangle $A B C$, perpendiculars $M P$ and $M Q$ are dropped to the lines $A B$ and $A C$. For what position of point $M$ is the length of segment $P Q$ maximized? | 11.17. Points $P$ and $Q$ lie on the circle with diameter $A M$, so $P Q=$ $=A M \sin P A Q=A M \sin A$. Therefore, the length of segment $P Q$ is maximized when $A M$ is the diameter of the circumscribed circle. | AM | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,510 |
11.18*. Inside triangle $ABC$, a point $O$ is taken. Let $d_{a}, d_{b}, d_{c}$ be the distances from it to the lines $BC, CA, AB$. For which position of point $O$ will the product $d_{a} d_{b} d_{c}$ be the greatest? | 11.18. It is clear that $2 S_{A B C}=a d_{a}+b d_{b}+c d_{c}$. Therefore, the product $\left(a d_{a}\right)\left(b d_{b}\right)\left(c d_{c}\right)$ will be the largest if $a d_{a}=b d_{b}=c d_{c}$ (see p. 217). Since the value $a b c$ is constant, the product $\left(a d_{a}\right)\left(b d_{b}\right)\left(c d_{c}\righ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,511 |
11.19*. Points $A_{1}, B_{1}$ and $C_{1}$ are taken on the sides $B C, C A$ and $A B$ of triangle $A B C$, such that segments $A A_{1}, B B_{1}$ and $C C_{1}$ intersect at a single point $M$. For what position of point $M$ is the value of $\frac{M A_{1}}{A A_{1}} \cdot \frac{M B_{1}}{B B_{1}} \cdot \frac{M C_{1}}{C C_{... | 11.19. Let $\alpha=M A_{1} / A A_{1}, \beta=M B_{1} / B B_{1}$ and $\gamma=M C_{1} / C C_{1}$. Since $\alpha+\beta+$ $+\gamma=1$ (see problem 4.48, a)), then $\sqrt[3]{\alpha \beta \gamma} \leqslant(\alpha+\beta+\gamma) / 3=1 / 3$, and equality is achieved when $\alpha=\beta=\gamma=1 / 3$, i.e., $M$ is the point of int... | M | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,512 |
11.20*. From a point $M$ inside a given triangle $ABC$, perpendiculars $M A_{1}, M B_{1}, M C_{1}$ are dropped to the lines $BC, CA, AB$. For which points $M$ inside the given triangle $ABC$ does the value $a / M A_{1} + b / M B_{1} + c / M C_{1}$ attain its minimum? | 11.20. Let \( x = M A_{1}, y = M B_{1} \) and \( z = M C_{1} \). Then \( a x + b y + c z = 2 S_{B M C} + 2 S_{A M C} + 2 S_{A M B} = 2 S_{A B C} \). Therefore, \(\frac{a}{x} + \frac{b}{y} + \frac{c}{z} \cdot 2 S_{A B C} = \frac{a}{x} + \frac{b}{y} + \frac{c}{z} \quad (a x + b y + c z) = a^{2} + b^{2} + c^{2} + a b \qua... | MistheincenteroftriangleABC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,513 |
11.21*. Given a triangle $A B C$. Find a point $O$ inside it for which the sum of the lengths of the segments $O A, O B, O C$ is minimal. (Pay attention to the case when one of the angles of the triangle is greater than $120^{\circ}$.) | 11.21. Suppose first that all angles of triangle \(ABC\) are less than \(120^\circ\). Then there exists a point \(O\) inside it from which all sides are seen at an angle of \(120^\circ\). Draw lines through the vertices \(A\), \(B\), and \(C\) perpendicular to the segments \(OA\), \(OB\), and \(OC\). These lines form a... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,514 |
11.22*. Find a point \( O \) inside triangle \( ABC \) for which the sum of the squares of the distances from it to the sides of the triangle is minimal.
See also problem 18.22 a).
## §3. Angle | 11.22. Let the distances from point $O$ to sides $BC$, $CA$, and $AB$ be $x$, $y$, and $z$ respectively. Then $ax + by + cz = 2(S_{BOC} + S_{COA} + S_{AOB}) = 2S_{ABC}$. It is also clear that $x : y : z = (S_{BOC} / a) : (S_{COA} / b) : (S_{AOB} / c)$.
The equation $ax + by + cz = 2S$ defines a plane in three-dimensio... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,515 |
11.23. On one side of an acute angle, points $A$ and $B$ are given. Construct a point $C$ on the other side of the angle from which the segment $A B$ is seen at the largest angle. | 11.23. Let $O$ be the vertex of the given angle. Point $C$ is the point of tangency of the side of the angle with a circle passing through points $A$ and $B$, i.e., $O C^{2}=O A \cdot O B$. To find the length of the segment $O C$, it is sufficient to draw a tangent to any circle passing through points $A$ and $B$. | OC^{2}=OA\cdotOB | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,516 |
11.24. Given an angle $X A Y$ and a point $O$ inside it. Draw a line through point $O$ that cuts off a triangle of the smallest area from the given angle. | 11.24. Consider the angle \(X^{\prime} A^{\prime} Y^{\prime}\), symmetric to the angle \(X A Y\) with respect to point \(O\). Let \(B\) and \(C\) be the points of intersection of the sides of these angles. Denote the points of intersection of the line passing through point \(O\) with the sides of the angles \(X A Y\) a... | BC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,517 |
11.25. Through a given point $P$, lying inside the angle $A O B$, draw a line $M N$ such that the value of $O M+O N$ is minimized (points $M$ and $N$ lie on the sides $O A$ and $O B$). | 11.25. Let's take points $K$ and $L$ on sides $O A$ and $O B$ such that $K P \| O B$ and $L P \| O A$. Then $K M: K P = P L: L N$, and therefore, $K M + L N \geqslant 2 \sqrt{K M \cdot L N} = 2 \sqrt{K P \cdot P L} = 2 \sqrt{O K \cdot O L}$, with equality achieved when $K M = L N = \sqrt{O K \cdot O L}$. It is also cle... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,518 |
11.26. Given an angle $X A Y$ and a circle inside it. Construct a point on the circle such that the sum of its distances from the lines $A X$ and $A Y$ is minimal. | 11.26. On the rays $A X$ and $A Y$, equal segments $A B$ and $A C$ are laid out. If point $M$ lies on segment $B C$, then the sum of the distances from it to the lines $A B$ and $A C$ is equal to $2\left(S_{A B M}+S_{A C M}\right) / A B=2 S_{A B C} / A B$. Therefore, the sum of the distances from a point to the lines $... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,519 | |
11.27*. Inside the acute angle $B A C$ there is a point $M$. Construct points $X$ and $Y$ on the sides $B A$ and $A C$ respectively, such that the perimeter of triangle $X Y M$ is minimized. | 11.27. Let points \( M_{1} \) and \( M_{2} \) be symmetric to \( M \) with respect to lines \( AB \) and \( AC \). Since \( \angle BAM_{1} = \angle BAM \) and \( \angle CAM_{2} = \angle CAM \), it follows that \( \angle M_{1}AM_{2} = 2 \angle BAC < 180^\circ \). Therefore, the segment \( M_{1}M_{2} \) intersects rays \... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,520 |
11.28*. Given angle $X A Y$. The ends $B$ and $C$ of segments $B O$ and $C O$ of length 1 move along rays $A X$ and $A Y$. Construct quadrilateral $A B O C$ of maximum area.
## §4. Quadrilaterals | 11.28. The quadrilateral $ABOC$ of the greatest area is convex. Among all triangles $ABC$ with a fixed angle $A$ and side $BC$, the one with the greatest area is the isosceles triangle with base $BC$. Therefore, among all considered quadrilaterals $ABOC$ with a fixed diagonal $BC$, the one with the greatest area is the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,521 |
11.29. Inside a convex quadrilateral, find a point for which the sum of the distances to the vertices is the smallest. | 11.29. Let $O$ be the point of intersection of the diagonals of a convex quadrilateral $ABCD$, and $O_{1}$ be any other point. Then $A O_{1}+C O_{1} \geqslant A C=A O+C O$ and $B O_{1}+D O_{1} \geqslant B D=B O+D O$, and at least one of these inequalities is strict. Therefore, $O$ is the desired point. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,522 |
11.30. The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$. What is the smallest area that this quadrilateral can have if the area of triangle $AOB$ is 4 and the area of triangle $COD$ is 9? | 11.30. Since $S_{A O B}: S_{B O C}=A O: O C=S_{A O D}: S_{D O C}$, then $S_{B O C} \cdot S_{A O D}=$ $=S_{A O B} \cdot S_{D O C}=36$. Therefore, $S_{B O C}+S_{A O D} \geqslant 2 \sqrt{S_{B O C} \cdot S_{A O D}}=12$, and equality is achieved if $S_{B O C}=S_{A O D}$, i.e., $S_{A B C}=S_{A B D}$, from which $A B \| C D$.... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,523 |
11.31. Trapezoid $ABCD$ with base $AD$ is divided by diagonal $AC$ into two triangles. A line $l$, parallel to the base, cuts these triangles into two triangles and two quadrilaterals. At what position of line $l$ is the sum of the areas of the resulting triangles minimal? | 11.31. Let $S_{0}$ and $S$ be the considered sums of the areas of triangles for the line $l_{0}$, passing through the intersection point of the diagonals of the trapezoid, and for some other line $l$. It is easy to verify that $S=S_{0}+s$, where $s$ is the area of the triangle formed by the diagonals $A C$ and $B D$ an... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,524 |
11.32. The area of a trapezoid is 1. What is the smallest value that the largest diagonal of this trapezoid can have? | 11.32. Let the lengths of the diagonals of the trapezoid be denoted by \(d_{1}\) and \(d_{2}\), the lengths of their projections on the base by \(p_{1}\) and \(p_{2}\), the lengths of the bases by \(a\) and \(b\), and the height by \(h\). Let \(d_{1} \geqslant d_{2}\) for definiteness. Then \(p_{1} \geqslant p_{2}\). I... | \sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,525 |
11.33*. On the base $AD$ of trapezoid $ABCD$, a point $K$ is given. Find a point $M$ on the base $BC$ such that the area of the common part of triangles $AMD$ and $BKC$ is maximized. | 11.33. We will prove that the desired point is point $M$, which divides side $B C$ in the ratio $B M: M C=A K: K D$. Let the points of intersection of segments $A M$ and $B K$, $D M$ and $C K$ be denoted as $P, Q$ respectively. Then $K Q: Q C=K D: M C=$ $=K A: M B=K P: P B$, i.e., line $P Q$ is parallel to the bases of... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,526 |
11.34*. Prove that among all quadrilaterals with fixed side lengths, the one with the largest area is the inscribed quadrilateral.
See also problems $9.35,15.3$ b).
## §5. Polygons | 11.34. According to problem 4.45, a)
$$
S^{2}=(p-a)(p-b)(p-c)(p-d)-a b c d \cos ^{2}((B+D) / 2)
$$
This value is maximal when $\cos ((B+D) / 2)=0$, i.e., $\angle B+\angle D=180^{\circ}$.
 | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,527 |
11.35. A polygon has a center of symmetry $O$. Prove that the sum of the distances to the vertices is minimal for the point $O$.
保留源文本的换行和格式，直接输出翻译结果。 | 11.35. If $A$ and $A^{\prime}$ are vertices of a polygon symmetric with respect to point $O$, then the sum of the distances to points $A$ and $A^{\prime}$ is the same for all points on the segment $A A^{\prime}$, and for all other points it is greater. Point $O$ belongs to all such segments. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,528 |
11.36. Among all polygons inscribed in a given circle, find the one for which the sum of the squares of the side lengths is maximized. | 11.36. If in triangle $ABC$ angle $B$ is obtuse or right, then by the cosine theorem $AC^{2} \geqslant AB^{2} + BC^{2}$. Therefore, if in a polygon the angle at vertex $B$ is not acute, then by removing vertex $B$, we obtain a polygon with no less sum of the squares of the side lengths. Since any $n$-sided polygon with... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,529 |
11.37*. Given a convex polygon $A_{1} \ldots A_{n}$. Prove that the point of the polygon for which the sum of the distances from it to all vertices is maximal is a vertex.
See also problem 6.72.
## §6. Various Problems | 11.37. If point $X$ divides a certain segment $P Q$ in the ratio $\lambda:(1-\lambda)$, then $\overrightarrow{A_{i} X}=(1-\lambda) \overrightarrow{A_{i} P}+\lambda \overrightarrow{A_{i} Q}$, and therefore $A_{i} X \leqslant(1-\lambda) A_{i} P+\lambda A_{i} Q$. Consequently, $f(X)=\sum A_{i} X \leqslant(1-\lambda) \sum ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,530 |
11.38. Inside a circle with center $O$, a point $A$ is given. Find the point $M$ on the circle for which the angle $O M A$ is maximal. | 11.38. The geometric locus of points $X$, for which the angle $O X A$ is constant, consists of two arcs of circles $S_{1}$ and $S_{2}$ symmetric with respect to the line $O A$. Consider the case when the diameter of the circles $S_{1}$ and $S_{2}$ is equal to the radius of the original circle, i.e., these circles touch... | M_{1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,531 |
11.39. On a plane, a line $l$ and points $A$ and $B$ are given, lying on opposite sides of it. Construct a circle passing through points $A$ and $B$ such that the line $l$ cuts off the shortest chord on it. | 11.39. Let the intersection point of line $l$ and segment $A B$ be denoted as $O$. Consider an arbitrary circle $S$ passing through points $A$ and $B$. It intersects $l$ at some points $M$ and $N$. Since $M O \cdot N O = A O \cdot B O$ is a constant, we have
$$
M N = M O + N O \geqslant 2 \sqrt{M O \cdot N O} = 2 \sqr... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,532 | |
11.40. Given a line $l$ and points $P$ and $Q$, lying on the same side of it. On the line $l$, we take a point $M$ and in the triangle $P Q M$ we draw the altitudes $P P^{\prime}$ and $Q Q^{\prime}$. For what position of point $M$ is the length of the segment $P^{\prime} Q^{\prime}$ minimal? | 11.40. Let's construct a circle with diameter $P Q$. If this circle intersects the line $l$, then any of the intersection points is the one we are looking for, since in this case $P^{\prime}=Q^{\prime}$. If the circle does not intersect the line $l$, then for any point $M$ on the line $l$, the angle $P M Q$ is acute an... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,533 |
11.41. Points $A, B$ and $O$ do not lie on the same line. Draw a line $l$ through point $O$ such that the sum of the distances from it to points $A$ and $B$ is: a) greatest; b) smallest. | 11.41. Let the sum of the distances from points $A$ and $B$ to the line $l$ be $2h$. If the line $l$ intersects the segment $AB$ at point $X$, then $S_{AOB} = h \cdot OX$, so the value of $h$ is extremal when the value of $OX$ is extremal, i.e., the line $OX$ corresponds to a side or altitude of triangle $AOB$. If the ... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,534 | |
11.42. If five points are given on a plane, then by considering all possible triples of these points, one can form 30 angles. Let the smallest of these angles be $\alpha$. Find the maximum value of $\alpha$.
If five points are given on a plane, then by considering all possible triples of these points, one can form 30 ... | 11.42. First, assume that the points are the vertices of a convex pentagon. The sum of the angles of a pentagon is $540^{\circ}$, so one of its angles does not exceed $540^{\circ} / 5=108^{\circ}$. The diagonals divide this angle into three angles, so one of them does not exceed $108^{\circ} / 3=36^{\circ}$. In this ca... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,535 |
11.43*. In a city, there are 10 streets parallel to each other, and 10 streets intersecting them at right angles. What is the minimum number of turns a closed bus route can have, passing through all intersections? | 11.43. A closed route passing through all intersections can have 20 turns (Fig. 11.8). It remains to prove that such a route cannot have fewer than 20 turns. After each turn, there is a transition from a horizontal street to a vertical one or vice versa. Therefore, the number of horizontal segments in a closed route is... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 27,536 |
11.44*. What is the maximum number of cells on an $8 \times 8$ chessboard that can be cut by a single straight line? | 11.44. A line can intersect 15 cells (Fig. 11.9). We will now prove that a line cannot intersect more than 15 cells. The number of cells intersected by a line is one less than the number of points of intersection with the segments that form the sides of the cells. Inside the square, there are 14 such segments. Therefor... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,537 |
11.45*. What is the maximum number of points that can be placed on a segment of length 1 so that on any segment of length $d$ contained in this segment, there are no more than $1+1000 d^{2}$ points?
See also problems $15.1,17.20$.
## §7. Extremal Properties of Regular Polygons | 11.45. First, let's prove that it is impossible to place 33 points in such a way. Indeed, if 33 points are on a segment of length 1, then the distance between some two of them does not exceed \(1 / 32\). The segment with endpoints at these points contains two points, but it should contain no more than \(1 + 1000 / 32^2... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 27,538 |
11.46*. a) Prove that among all $n$-gons circumscribed about a given circle, the one with the smallest area is the regular $n$-gon.
b) Prove that among all $n$-gons circumscribed about a given circle, the one with the smallest perimeter is the regular $n$-gon. | 11.46. a) Let an irregular $n$-sided polygon be circumscribed around a circle $S$. We will circumscribe a regular $n$-sided polygon around this circle, and then circumscribe a circle $S_{1}$ around it (Fig. 11.10). We will prove that the area of the part of the irregular $n$-sided polygon contained within $S_{1}$ is gr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,539 |
11.47*. Triangles $A B C_{1}$ and $A B C_{2}$ have a common base $A B$ and $\angle A C_{1} B=\angle A C_{2} B$. Prove that if $\left|A C_{1}-C_{1} B\right|<\left|A C_{2}-C_{2} B\right|$, then:
a) the area of triangle $A B C_{1}$ is greater than the area of triangle $A B C_{2}$;
b) the perimeter of triangle $A B C_{1}... | 11.47. The sides of triangle $ABC$ are proportional to $\sin \alpha, \sin \beta$, and $\sin \gamma$. If angle $\gamma$ is fixed, then the magnitude $|\sin \alpha - \sin \beta| = 2|\sin((\alpha - \beta) / 2) \sin(\gamma / 2)|$ is greater the larger the magnitude $\varphi = |\alpha - \beta|$. It is worth noting that the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,540 |
12.1. Prove that the area $S$ of a triangle is equal to $a b c / 4 R$. | 12.1. By the Law of Sines, $\sin \gamma = c / 2R$, therefore $S = (ab \sin \gamma) / 2 = abc / 4R$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,541 |
12.2. Point $D$ lies on the base $A C$ of isosceles triangle $A B C$. Prove that the radii of the circumscribed circles of triangles $A B D$ and $C B D$ are equal. | 12.2. The radii of the circumscribed circles of triangles $A B D$ and $C B D$ are equal to $A B / 2 \sin A D B$ and $B C / 2 \sin B D C$. It remains to note that $A B=B C$ and $\sin A D B=$ $=\sin B D C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,542 |
12.3. Express the area of triangle $ABC$ in terms of the length of side $BC$ and the measures of angles $B$ and $C$. | 12.3. By the Law of Sines $b=a \sin \beta / \sin \alpha=a \sin \beta / \sin (\beta+\gamma)$, therefore $S=$ $=a b \sin \gamma / 2=a^{2} \sin \beta \sin \gamma / 2 \sin (\beta+\gamma)$. | ^{2}\sin\beta\sin\gamma/2\sin(\beta+\gamma) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,543 |
12.4. Prove that
$$
\frac{a+b}{c}=\cos \frac{\alpha-\beta}{2} / \sin \frac{\gamma}{2}, \quad \text { and } \quad \frac{a-b}{c}=\sin \frac{\alpha-\beta}{2} / \cos \frac{\gamma}{2} .
$$ | 12.4. By the Law of Sines $(a+b) / c=(\sin \alpha+\sin \beta) / \sin \gamma$. Moreover, $\sin \alpha+$ $+\sin \beta=2 \sin ((\alpha+\beta) / 2) \cos ((\alpha-\beta) / 2)=2 \cos (\gamma / 2) \cos ((\alpha-\beta) / 2)$ and $\sin \gamma=$ $=2 \sin (\gamma / 2) \cos (\gamma / 2)$. The second equality is proved similarly. | proof | Algebra | proof | Yes | Yes | olympiads | false | 27,544 |
12.5. In an acute-angled triangle $A B C$, the altitudes $A A_{1}$ and $C C_{1}$ are drawn. Points $A_{2}$ and $C_{2}$ are symmetric to $A_{1}$ and $C_{1}$ with respect to the midpoints of sides $B C$ and $A B$. Prove that the line connecting vertex $B$ with the center $O$ of the circumscribed circle bisects the segmen... | 12.5. In triangle $A_{2} B C_{2}$, the lengths of sides $A_{2} B$ and $B C_{2}$ are $b \cos \gamma$ and $b \cos \alpha$; line $B O$ divides angle $A_{2} B C_{2}$ into angles $90^{\circ}-\gamma$ and $90^{\circ}-\alpha$. Let line $B O$ intersect segment $A_{2} C_{2}$ at point $M$. By the Law of Sines, $A_{2} M=$ $=A_{2} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,545 |
12.6*. Through point $S$ lines $a, b, c$ and $d$ are drawn; line $l$ intersects them at points $A, B, C$ and $D$. Prove that the value of $A C \cdot B D /(B C \cdot A D)$ does not depend on the choice of line $l$. | 12.6. Let $\alpha=\angle(a, c), \beta=\angle(c, d)$ and $\gamma=\angle(d, b)$. Then
$$
(A C / A S) /(B C / B S)=\sin \alpha / \sin (\beta+\gamma), \quad(B D / B S) /(A D / A S)=\sin \gamma / \sin (\alpha+\beta)
$$
Therefore
$$
(A C \cdot B D) /(B C \cdot A D)=\sin \alpha \sin \gamma / \sin (\alpha+\beta) \sin (\beta... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,546 |
12.7*. Given lines $a$ and $b$, intersecting at point $O$, and an arbitrary point $P$. A line $l$, passing through point $P$, intersects lines $a$ and $b$ at points $A$ and $B$. Prove that the value of $(A O / O B) / (P A / P B)$ does not depend on the choice of line $l$. | 12.7. Since $O A / P A=\sin O P A / \sin P O A$ and $O B / P B=\sin O P B / \sin P O B$, then $(O A / O B) /(P A / P B)=\sin P O B / \sin P O A$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,547 |
12.9. Two similar isosceles triangles have a common vertex. Prove that the projections of their bases onto the line connecting the midpoints of the bases are equal. | 12.9. Let $O$ be the common vertex of the given triangles, $M$ and $N$ be the midpoints of the bases, and $k$ be the ratio of the lengths of the bases to the heights. The projections of the bases of the given triangles onto the line $M N$ are $k \cdot O M \sin O M N$ and $k \cdot O N \sin O N M$. It remains to note tha... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,549 |
12.10*. On a circle with diameter $A B$, points $C$ and $D$ are taken. The line $C D$ and the tangent to the circle at point $B$ intersect at point $X$. Express $B X$ in terms of the radius of the circle $R$ and the angles $\varphi=\angle B A C$ and $\psi=\angle B A D$.
## §2. The Law of Cosines | 12.10. By the Law of Sines, $B X / \sin B D X = B D / \sin B X D = 2 R \sin \psi / \sin B X D$. Moreover, $\sin B D X = \sin B D C = \sin \varphi$; the measure of angle $B X D$ is easily calculated: if points $C$ and $D$ lie on the same side of $A B$, then $\angle B X D = \pi - \varphi - \psi$, and if on opposite sides... | BX=2R\sin\varphi\sin\psi/\sin|\varphi\\psi| | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,550 |
12.11. Prove that:
a) $m_{a}^{2}=\left(2 b^{2}+2 c^{2}-a^{2}\right) / 4$
b) $m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=3\left(a^{2}+b^{2}+c^{2}\right) / 4$. | 12.11. a) Let $A_{1}$ be the midpoint of segment $BC$. By adding the equalities $AB^{2} = A A_{1}^{2} + A_{1} B^{2} - 2 A A_{1} \cdot B A_{1} \cos B A_{1} A$ and $AC^{2} = A A_{1}^{2} + A_{1} C^{2} - 2 A A_{1} \cdot A_{1} C \cos C A_{1} A$ and considering that $\cos B A_{1} A = -\cos C A_{1} A$, we obtain the required ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,551 |
12.12. Prove that $4 S=\left(a^{2}-(b-c)^{2}\right) \cot(\alpha / 2)$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 12.12. By the cosine theorem $a^{2}-(b-c)^{2}=2 b c(1-\cos \alpha)=4 S(1-\cos \alpha) / \sin \alpha=$ $=4 S \operatorname{tg}(\alpha / 2)$. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,552 | |
12.13. Prove that $\cos ^{2}(\alpha / 2)=p(p-a) / b c$ and $\sin ^{2}(\alpha / 2)=$ $=(p-b)(p-c) / b c$. | 12.13. By the cosine theorem, $\cos \alpha=\left(b^{2}+c^{2}-a^{2}\right) / 2 b c$. It remains to use the formulas $\cos ^{2}(\alpha / 2)=(1+\cos \alpha) / 2$ and $\sin ^{2}(\alpha / 2)=(1-\cos \alpha) / 2$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,553 |
12.14. The lengths of the sides of a parallelogram are $a$ and $b$, and the lengths of the diagonals are $m$ and $n$. Prove that $a^{4}+b^{4}=m^{2} n^{2}$ if and only if the acute angle of the parallelogram is $45^{\circ}$. | 12.14. Let $\alpha$ be the angle at the vertex of the parallelogram. By the cosine theorem, $m^{2}=a^{2}+b^{2}+2 a b \cos \alpha$ and $n^{2}=a^{2}+b^{2}-2 a b \cos \alpha$. Therefore, $m^{2} n^{2}=\left(a^{2}+\right.$ $\left.+b^{2}\right)^{2}-(2 a b \cos \alpha)^{2}=a^{4}+b^{4}+2 a^{2} b^{2}\left(1-2 \cos ^{2} \alpha\r... | \cos^{2}\alpha=1/2 | Geometry | proof | Yes | Yes | olympiads | false | 27,554 |
12.15. Prove that the medians $A A_{1}$ and $B B_{1}$ of triangle $A B C$ are perpendicular if and only if $a^{2}+b^{2}=5 c^{2}$. | 12.15. Let $M$ be the point of intersection of the medians $A A_{1}$ and $B B_{1}$. The angle $A M B$ is a right angle if and only if $A M^{2}+B M^{2}=A B^{2}$, i.e., $4\left(m_{a}^{2}+m_{b}^{2}\right) / 9=c^{2}$. According to problem $12.11 m_{a}^{2}+m_{b}^{2}=\left(4 c^{2}+a^{2}+b^{2}\right) / 4$. | ^{2}+b^{2}=5c^{2} | Geometry | proof | Yes | Yes | olympiads | false | 27,555 |
12.16*. Let $O$ be the center of the circumscribed circle of (non-equilateral) triangle $ABC$, and $M$ be the point of intersection of the medians. Prove that the line $OM$ is perpendicular to the median $CC_{1}$ if and only if $a^{2} + b^{2} = 2c^{2}$.
## §3. Inscribed, Circumscribed, and Excircle; Their Radii | 12.16. Let $m=C_{1} M$ and $\varphi=\angle C_{1} M O$. Then $O C_{1}^{2}=C_{1} M^{2}+$ $+O M^{2}-2 O M \cdot C_{1} M \cos \varphi$ and $B O^{2}=C O^{2}=O M^{2}+M C^{2}+2 O M \cdot C M \cos \varphi=$ $=O M^{2}+4 C_{1} M^{2}+4 O M \cdot C_{1} M \cos \varphi$. Therefore, $B C_{1}^{2}=B O^{2}-O C_{1}^{2}=3 C_{1} M^{2}+$ $+... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,556 |
12.17. Prove that:
a) $a=r(\operatorname{ctg}(\beta / 2)+\operatorname{ctg}(\gamma / 2))=r \cos (\alpha / 2) /(\sin (\beta / 2) \sin (\gamma / 2))$;
b) $a=r_{a}(\operatorname{tg}(\beta / 2)+\operatorname{tg}(\gamma / 2))=r_{a} \cos (\alpha / 2) /(\cos (\beta / 2) \cos (\gamma / 2))$;
c) $p-b=r \operatorname{ctg}(\be... | 12.17. Let the inscribed circle touch side $B C$ at point $K$, and the excircle touch at point $L$. Then $B C = B K + K C = r \operatorname{ctg}(\beta / 2) + r \operatorname{ctg}(\gamma / 2)$ and $B C = B L + L C = r_{a} \operatorname{ctg} L B O_{a} + r_{a} \operatorname{ctg} L C O_{a} = r_{a} \operatorname{tg}(\beta /... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,557 |
12.18. Prove that:
a) $r p=r_{a}(p-a), r r_{a}=(p-b)(p-c)$ and $r_{b} r_{c}=p(p-a)$;
b) $S^{2}=p(p-a)(p-b)(p-c) \quad$ (Heron's formula);
c) $S^{2}=r r_{a} r_{b} r_{c}$. | 12.18. a) According to problem $12.17 p=r_{a} \operatorname{ctg}(\alpha / 2)$ and $r \operatorname{ctg}(\alpha / 2)=p-a ; r \operatorname{ctg}(\beta / 2)=$ $=p-b$ and $r_{a} \operatorname{tg}(\beta / 2)=p-c ; r_{c} \operatorname{tg}(\beta / 2)=p-a$ and $r_{b} \operatorname{ctg}(\beta / 2)=p$. By multiplying these pairs... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,558 |
12.21. Prove that $\frac{2}{h_{a}}=\frac{1}{r_{b}}+\frac{1}{r_{c}}$. | 12.21. According to problem 12.18, a) $1 / r_{b}=(p-b) / p r$ and $1 / r_{c}=(p-c) / p r$. Therefore, $\frac{1}{r_{b}}+\frac{1}{r_{c}}=a / p r=a / S=2 / h_{a}$. | \frac{2}{h_{}}=\frac{1}{r_{b}}+\frac{1}{r_{}} | Geometry | proof | Yes | Yes | olympiads | false | 27,561 |
12.22. Prove that $\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{1}{r_{a}}+\frac{1}{r_{b}}+\frac{1}{r_{c}}=\frac{1}{r}$. | 12.22. It is easy to verify that $1 / h_{a}=a / 2 p r$ and $1 / r_{a}=(p-a) / p r$. Adding similar equalities, we obtain the required result. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,562 |
12.23. Prove that
$$
\frac{1}{(p-a)(p-b)}+\frac{1}{(p-b)(p-c)}+\frac{1}{(p-c)(p-a)}=\frac{1}{r^{2}}
$$ | 12.23. According to problem 12.18, a) $1 /((p-b)(p-c))=1 / r r_{a}$. It remains to add similar equalities and use the result of problem 12.22. | proof | Algebra | proof | Yes | Yes | olympiads | false | 27,563 |
12.24. Prove that $r_{a}+r_{b}+r_{c}=4 R+r$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 12.24. According to problem $12.14 S R=a b c$. It is also clear that
$$
\begin{aligned}
a b c= & p(p-b)(p-c)+p(p-c)(p-a)+p(p-a)(p-b)- \\
& -(p-a)(p-b)(p-c)=\frac{S^{2}}{(p-a)}+\frac{S^{2}}{(p-b)}+\frac{S^{2}}{(p-c)}-\frac{S^{2}}{p}=S\left(r_{a}+r_{b}+r_{c}-r\right)
\end{aligned}
$$ | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,564 | |
12.25. Prove that $r_{a} r_{b}+r_{b} r_{c}+r_{c} r_{a}=p^{2}$. | 12.25. According to problem $12 \cdot 18$, a) $r_{a} r_{b}=p(p-c), r_{b} r_{c}=p(p-a)$ and $r_{c} r_{a}=p(p-b)$. Adding these equalities, we get the required result. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,565 |
12.26. Prove that $\frac{1}{r^{3}}-\frac{1}{r_{a}^{3}}-\frac{1}{r_{b}^{3}}-\frac{1}{r_{c}^{3}}=\frac{12 R}{S^{2}}$.
12.27* . Prove that $a(b+c)=\left(r+r_{a}\right)\left(4 R+r-r_{a}\right)$ and $a(b-c)=$ $=\left(r_{b}-r_{c}\right)\left(4 R-r_{b}-r_{c}\right)$. | 12.26. Since $S=r p=r_{a}(p-a)=r_{b}(p-b)=r_{c}(p-c)$, the expression on the left side equals $\left(p^{3}-(p-a)^{3}-(p-b)^{3}-(p-c)^{3}\right) / S^{3}=3 a b c / S^{3}$. It remains to note that $a b c / S=4 R$ (problem 12.1). | \frac{12R}{S^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 27,566 |
12.28*. Let $O$ be the center of the inscribed circle of triangle $ABC$. Prove that $\frac{O A^{2}}{b c}+\frac{O B^{2}}{a c}+\frac{O C^{2}}{a b}=1$. | 12.28. Since $O A=r / \sin (\alpha / 2)$ and $b c=2 S / \sin \alpha$, then $O A^{2} / b c=r^{2} \operatorname{ctg}(\alpha / 2) / S=$ $=r(p-a) / S$ (see problem $12 \cdot 17$, b). It remains to note that $r(p-a+p-b+p-c)=$ $=r p=S$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,567 |
12.29*. a) Prove that if for some triangle $p=2 R+r$, then this triangle is a right triangle.
b) Prove that if $p=2 R \sin \varphi+r \operatorname{ctg}(\varphi / 2)$, then $\varphi$ is one of the angles of the triangle (it is assumed that $0<\varphi<\pi$).
## §4. Lengths of sides, heights, and angle bisectors | 12.29. Let's solve problem b) immediately, which is a special case of problem a). Since $\operatorname{ctg}(\varphi / 2)=\sin \varphi /(1-\cos \varphi)$, we have $p^{2}(1-x)^{2}=\left(1-x^{2}\right)(2 R(1-x)+r)^{2}$, where $x=\cos \varphi$. The root $x_{0}=1$ of this equation does not interest us, as in this case $\ope... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,568 |
12.30. Prove that $a b c=4 p r R$ and $a b+b c+c a=r^{2}+p^{2}+4 r R$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 12.30. It is clear that $2 p r=2 S=a b \sin \gamma=a b c / 2 R$, i.e., $4 p r R=a b c$. To prove the second equality, we will use Heron's formula: $S^{2}=$ $=p(p-a)(p-b)(p-c)$, i.e., $p r^{2}=(p-a)(p-b)(p-c)=p^{3}-p^{2}(a+b+c)+$ $+p(a b+b c+c a)-a b c=-p^{3}+p(a b+b c+c a)-4 p r R$. Dividing by $p$, we obtain the requi... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,569 | |
12.34. Prove that
$h_{a}=2(p-a) \cos (\beta / 2) \cos (\gamma / 2) / \cos (\alpha / 2)=$
$$
=2(p-b) \sin (\beta / 2) \cos (\gamma / 2) / \sin (\alpha / 2)
$$ | 12.34. Since $a h_{a}=2 S=2(p-a) r_{a}$ and $r_{a} / a=\cos (\beta / 2) \cos (\gamma / 2) / \cos (\alpha / 2)$ (Problem 12.17, b)), then $h_{a}=2(p-a) \cos (\beta / 2) \cos (\gamma / 2) / \cos (\alpha / 2)$. Considering that $(p-a) \operatorname{ctg}(\beta / 2)=r_{c}=(p-b) \operatorname{ctg}(\alpha / 2)$ (Problem 12.17... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,573 |
12.35. Prove that the length of the angle bisector $l_{a}$ can be calculated using the following formulas:
a) $l_{a}=\sqrt{4 p(p-a) b c /(b+c)^{2}}$
b) $l_{a}=2 b c \cos (\alpha / 2) /(b+c)$
c) $l_{a}=2 R \sin \beta \sin \gamma / \cos ((\beta-\gamma) / 2)$
d) $l_{a}=4 p \sin (\beta / 2) \sin (\gamma / 2) /(\sin \be... | 12.35. a) Let the extension of the bisector $A D$ intersect the circumcircle of triangle $A B C$ at point $M$. Then $A D \cdot D M = B D \cdot D C$ and, since $\triangle A B D \sim \triangle A M C, A B \cdot A C = A D \cdot A M = A D(A D + D M) = A D^{2} + B D \cdot D C$. Moreover, $B D = a c / (b + c)$ and $D C = a b ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,574 |
12.36. a) $\sin (\alpha / 2) \sin (\beta / 2) \sin (\gamma / 2)=r / 4 R$;
b) $\operatorname{tg}(\alpha / 2) \operatorname{tg}(\beta / 2) \operatorname{tg}(\gamma / 2)=r / p$;
c) $\cos (\alpha / 2) \cos (\beta / 2) \cos (\gamma / 2)=p / 4 R$. | 12.36. a) Let $O$ be the center of the inscribed circle, $K$ be the point of tangency of the inscribed circle with side $AB$. Then
$2 R \sin \gamma = AB = AK + KB = r(\operatorname{ctg}(\alpha / 2) + \operatorname{ctg}(\beta / 2)) = $
$= r \sin ((\alpha + \beta) / 2) (\sin (\alpha / 2) \sin (\beta / 2))$
Considering... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,575 |
12.37. a) $\cos (\alpha / 2) \sin (\beta / 2) \sin (\gamma / 2)=(p-a) / 4 R$;
b) $\sin (\alpha / 2) \cos (\beta / 2) \cos (\gamma / 2)=r_{a} / 4 R$. | 12.37. a) Multiplying the equalities
$$
r \cos (\alpha / 2) \sin (\alpha / 2)=p-a, \quad \sin (\alpha / 2) \sin (\beta / 2) \sin (\gamma / 2)=r / 4 R
$$
(see problems 12.17, b) and 12.36, a)), we obtain the required result.
b) According to problem $12 \cdot 17$, b) $r_{a} \operatorname{tg}(\gamma / 2)=p-b=r \operato... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,576 |
12.39. a) $\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma+4 \cos \alpha \cos \beta \cos \gamma+1=0$;
b) $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+2 \cos \alpha \cos \beta \cos \gamma=1$. | 12.39. a) Adding the equalities $\cos 2 \alpha + \cos 2 \beta = 2 \cos (\alpha + \beta) \cos (\alpha - \beta) = -2 \cos \gamma \cos (\alpha - \beta)$ and $\cos 2 \gamma = 2 \cos^2 \gamma - 1 = -2 \cos \gamma \cos (\alpha + \beta) - 1$ and considering that $\cos (\alpha + \beta) + \cos (\alpha - \beta) = 2 \cos \alpha \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 27,578 |
12.40. $\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma=4 \sin \alpha \sin \beta \sin \gamma$. | 12.40. By adding the equalities $\sin 2 \alpha + \sin 2 \beta = 2 \sin (\alpha + \beta) \cos (\alpha - \beta) = 2 \sin \gamma \times \cos (\alpha - \beta)$ and $\sin 2 \gamma = 2 \sin \gamma \cos \gamma = -2 \sin \gamma \cos (\alpha + \beta)$ and considering that $\cos (\alpha - \beta) - \cos (\alpha + \beta) = 2 \sin ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 27,579 |
12.41. a) $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=\left(p^{2}-r^{2}-4 r R\right) / 2 R^{2}$.
b) $4 R^{2} \cos \alpha \cos \beta \cos \gamma=p^{2}-(2 R+r)^{2}$. | 12.41. a) It is clear that $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=\left(a^{2}+b^{2}+c^{2}\right) / 4 R$ and $a^{2}+b^{2}+c^{2}=$ $=(a+b+c)^{2}-2(a b+b c+c a)=4 p^{2}-2\left(r^{2}+p^{2}+4 r R\right)$ (see problem 12.30).
b) According to problem 12.39, b) $2 \cos \alpha \cos \beta \cos \gamma=\sin ^{2} \alph... | Geometry | proof | Yes | Yes | olympiads | false | 27,580 | |
12.42. $a b \cos \gamma + b c \cos \alpha + c a \cos \beta = \left(a^{2} + b^{2} + c^{2}\right) / 2$. | 12.42. The Law of Cosines can be rewritten as $a b \cos \gamma=\left(a^{2}+b^{2}-c^{2}\right) / 2$. By adding three similar equations, we obtain the required result. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,581 |
12.43. $\frac{\cos ^{2}(\alpha / 2)}{a}+\frac{\cos ^{2}(\beta / 2)}{b}+\frac{\cos ^{2}(\gamma / 2)}{c}=\frac{p}{4 R r}$.
## §6. Tangents and Cotangents of the Angles of a Triangle
Let $\alpha, \beta$ and $\gamma$ be the angles of triangle $ABC$. The problems in this section require proving the relationships specified... | 12.43. According to problem $12.13 \cos ^{2}(\alpha / 2) / a=p(p-a) / a b c$. It remains to note that $p(p-a)+p(p-b)+p(p-c)=p^{2}$ and $a b c=4 S R=4 p r R$. | \frac{p}{4Rr} | Geometry | proof | Yes | Yes | olympiads | false | 27,582 |
12.44. a) $\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma=\left(a^{2}+b^{2}+c^{2}\right) / 4 S$
b) $a^{2} \operatorname{ctg} \alpha+b^{2} \operatorname{ctg} \beta+c^{2} \operatorname{ctg} \gamma=4 S$. | 12.44. a) Since \( b c \cos \alpha = 2 S \operatorname{ctg} \alpha \), then \( a^{2} = b^{2} + c^{2} - 4 S \operatorname{ctg} \alpha \). Adding three similar equations, we obtain the required result.
b) For an acute triangle, \( a^{2} \operatorname{ctg} \alpha = 2 R^{2} \sin 2 \alpha = 4 S_{\text{vOC}} \), where \( O ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,583 |
12.45. a) $\operatorname{ctg}(\alpha / 2)+\operatorname{ctg}(\beta / 2)+\operatorname{ctg}(\gamma / 2)=p / r$;
b) $\operatorname{tg}(\alpha / 2)+\operatorname{tg}(\beta / 2)+\operatorname{tg}(\gamma / 2)=\left(\frac{a}{r_{a}}+\frac{b}{r_{b}}+\frac{c}{r_{c}}\right) / 2$. | 12.45. According to problem $12.17 \operatorname{ctg}(\alpha / 2)+\operatorname{ctg}(\beta / 2)=c / r$ and $\operatorname{tg}(\alpha / 2)+\operatorname{tg}(\beta / 2)=$ $=c / r_{c}$. It remains to add such equalities for all pairs of angles of the triangle. | Geometry | proof | Yes | Yes | olympiads | false | 27,584 | |
12.46. $\operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg} \gamma=\operatorname{tg} \alpha \operatorname{tg} \beta \operatorname{tg} \gamma$.
12.46. $\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma$. | 12.46. It is clear that $\operatorname{tg} \gamma=-\operatorname{tg}(\alpha+\beta)=-(\operatorname{tg} \alpha+\operatorname{tg} \beta) /(1-\operatorname{tg} \alpha \operatorname{tg} \beta)$. Multiplying both sides by $1-\operatorname{tg} \alpha \operatorname{tg} \beta$, we obtain the required result. | proof | Algebra | proof | Yes | Yes | olympiads | false | 27,585 |
12.48. a) $\operatorname{ctg} \alpha \operatorname{ctg} \beta+\operatorname{ctg} \beta \operatorname{ctg} \gamma+\operatorname{ctg} \alpha \operatorname{ctg} \gamma=1$;
b) $\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma-\operatorname{ctg} \alpha \operatorname{ctg} \beta \operatorname{ctg}... | 12.48. a) Multiply both sides of the equality by $\sin \alpha \sin \beta \sin \gamma$. The further steps of the proof are as follows: $\cos \gamma(\sin \alpha \cos \beta+\sin \beta \cos \alpha)+\sin \gamma(\cos \alpha \cos \beta-\sin \alpha \sin \beta)=$ $=\cos \gamma \sin (\alpha+\beta)+\sin \gamma \cos (\alpha+\beta)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 27,587 |
12.51. Prove that if $\frac{1}{b}+\frac{1}{c}=\frac{1}{l_{a}}$, then $\angle A=120^{\circ}$. | 12.51. According to problem $4.47 \frac{1}{b}+\frac{1}{c}=2 \cos (\alpha / 2) / l_{a}$, therefore $\cos (\alpha / 2)=1 / 2$, i.e. $\alpha=120^{\circ}$. | \alpha=120 | Geometry | proof | Yes | Yes | olympiads | false | 27,590 |
12.52. In triangle $A B C$, the height $A H$ is equal to the median $B M$. Find the angle $M B C$.
Fig. 12.2 | 12.52. Drop a perpendicular $M D$ from point $M$ to line $B C$. Then $M D = A H / 2 = B M / 2$. In the right triangle $B D M$, the leg $M D$ is equal to half of the hypotenuse $B M$. Therefore, $\angle M B C = \angle M B D = 30^{\circ}$. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,591 |
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