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9.59*. In a square with side 1, a broken line of length $L$ is located. It is known that each point of the square is at a distance less than $\varepsilon$ from some point of this broken line. Prove that then $L \geqslant \frac{1}{2 \varepsilon}-\frac{\pi \varepsilon}{2}$. | 9.59. The geometric locus of points that are no more than $\varepsilon$ away from a given segment is shown in Fig. 9.19. The area of this figure is $\pi \varepsilon^{2}+2 \varepsilon l$, where $l$ is the length of the segment. Let's construct such figures for all $N$ segments of the given broken line. Since adjacent fi... | L\geqslant\frac{1}{2\varepsilon}-\frac{\pi\varepsilon}{2} | Geometry | proof | Yes | Yes | olympiads | false | 27,365 |
9.60*. Inside a square with side 1, there are $n^{2}$ points. Prove that there exists a broken line containing all these points, the length of which does not exceed $2 n$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 9.60. Let's divide a square into $n$ vertical strips, each containing $n$ points. We will connect the points within each strip from top to bottom, resulting in $n$ broken lines. These broken lines can be connected into one broken line in two ways (Fig. $9.20, a$ and b). Consider the segments connecting different strips... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,366 |
9.61*. Inside a square with side 100, there is a broken line $L$ with the property that any point in the square is no more than 0.5 away from $L$. Prove that there are two points on $L$ such that the distance between them is no more than 1, while the distance along $L$ between them is at least 198.
## §9. Quadrilatera... | 9.61. Let $M$ and $N$ be the endpoints of a broken line. We will walk along the broken line from $M$ to $N$. Let $A_{1}$ be the first point we encounter on the broken line that is 0.5 units away from any vertex of the square. Consider the vertices of the square adjacent to this vertex. Let $B_{1}$ be the first point af... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,367 |
9.62. In quadrilateral $A B C D$, angles $A$ and $B$ are equal, and $\angle D > \angle C$. Prove that then $A D < B C$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 9.62. Let $\angle A = \angle B$. It is sufficient to prove that if $A D \angle C$. Take a point $D_{1}$ on the side $B C$ such that $B D_{1} = A D$. Then $A B D_{1} D$ is an isosceles trapezoid. Therefore, $\angle D > \angle D_{1} D A = \angle D D_{1} B > \angle C$. | Geometry | proof | Yes | Yes | olympiads | false | 27,368 | |
9.64. Prove that if two opposite angles of a quadrilateral are obtuse, then the diagonal connecting the vertices of these angles is shorter than the other diagonal. | 9.64. Let angles $B$ and $D$ of quadrilateral $ABCD$ be obtuse. Then points $B$ and $D$ lie inside the circle with diameter $AC$. Since the distance between any two points lying inside a circle is less than its diameter, $BD < AC$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,370 |
9.66. Angle $A$ of quadrilateral $A B C D$ is obtuse; $F$ is the midpoint of side $B C$. Prove that $2 F A < B D + C D$.
保留源文本的换行和格式，直接输出翻译结果。 | 9.66. Let $O$ be the midpoint of segment $BD$. Point $A$ lies inside the circle with diameter $BD$, so $OA < BD / 2$. Moreover, $FO = CD / 2$. Therefore, $2 FA \leqslant 2 FO + 2 OA < CD + BD$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,372 |
9.67. Given a quadrilateral $A B C D$. Prove that $A C \cdot B D \leqslant A B \cdot C D+$ $+B C \cdot A D$ (Ptolemy's inequality). | 9.67. On the rays $A B, A C$ and $A D$, we lay off segments $A B^{\prime}, A C^{\prime}$ and $A D^{\prime}$ of lengths $1 / A B, 1 / A C$ and $1 / A D$. Then $A B: A C=A C^{\prime}: A B^{\prime}$, i.e., $\triangle A B C \sim \triangle A C^{\prime} B^{\prime}$. The similarity coefficient of these triangles is $1 /(A B \... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,373 |
9.68. Let $M$ and $N$ be the midpoints of sides $B C$ and $C D$ of a convex quadrilateral $A B C D$. Prove that $S_{A B C D}<4 S_{A M N}$. | 9.68. It is clear that $S_{A B C D}=S_{A B C}+S_{A C D}=2 S_{A M C}+2 S_{A N C}=2\left(S_{A M N}+\right.$ $\left.+S_{C M N}\right)$. If segment $A M$ intersects diagonal $B D$ at point $A_{1}$, then $S_{C M N}=$ $=S_{A_{1} M N}<S_{A M N}$. Therefore, $S_{A B C D}<4 S_{A M N}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,374 |
9.69. Point $P$ lies inside a convex quadrilateral $A B C D$. Prove that the sum of the distances from point $P$ to the vertices of the quadrilateral is less than the sum of the pairwise distances between the vertices of the quadrilateral. | 9.69. The diagonals $A C$ and $B D$ intersect at point $O$. Let point $P$ lie inside triangle $A O B$ for definiteness. Then $A P + B P \leqslant A O + B O < A C + B D$
(see the solution to problem 9.28) and $C P + D P < C B + B A + A D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,375 |
9.70. The diagonals divide the convex quadrilateral $A B C D$ into four triangles. Let $P$ be the perimeter of the quadrilateral $A B C D$, and $Q$ be the perimeter of the quadrilateral formed by the centers of the inscribed circles of the obtained triangles. Prove that $P Q>4 S_{A B C D}$. | 9.70. Let $r_{i}, S_{i}$ and $p_{i}$ be the radii of the inscribed circles, the areas, and the semiperimeters of the obtained triangles. Then $Q \geqslant 2 \sum r_{i}=$ $=2 \sum\left(S_{i} / p_{i}\right)>4 \sum\left(S_{i} / P\right)=4 S / P$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,376 |
9.71. Prove that the distance from one of the vertices of a convex quadrilateral to the opposite diagonal does not exceed half of this diagonal. | 9.71. Let $A C \leqslant B D$. Drop perpendiculars $A A_{1}$ and $C C_{1}$ from vertices $A$ and $C$ to the diagonal $B D$. Then $A A_{1} + C C_{1} \leqslant A C \leqslant B D$, and therefore, $A A_{1} \leqslant B D / 2$ or $C C_{1} \leqslant B D / 2$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,377 |
9.72*. The segment $K L$ passes through the intersection point of the diagonals of the quadrilateral $A B C D$, and its ends lie on the sides $A B$ and $C D$. Prove that the length of the segment $K L$ does not exceed the length of one of the diagonals. | 9.72. Draw perpendicular lines through the endpoints of the segment $K L$ and consider the projections on them of the vertices of the quadrilateral, as well as the points of intersection with them of the lines $A C$ and $B D$ (Fig. 9.22). Let us assume, for definiteness, that point $A$ lies inside the strip defined by ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,378 |
9.73*. A parallelogram $P_{2}$ is inscribed in a parallelogram $P_{1}$, and a parallelogram $P_{3}$ is inscribed in $P_{2}$, with the sides of $P_{3}$ parallel to the corresponding sides of $P_{1}$. Prove that the length of at least one side of $P_{1}$ does not exceed twice the length of the parallel side of $P_{3}$.
... | 9.73. Let's introduce such notations as in Fig. 9.23. All the parallelograms considered have a common center (Problem 1.7). The side lengths of parallelogram $P_{3}$ are $a+a_{1}$ and $b+b_{1}$, while the side lengths of parallelogram $P_{1}$ are $a+a_{1}+2 x$ and $b+b_{1}+2 y$, so we need to check that $a+a_{1}+2 x \l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,379 |
9.74. Prove that if the angles of a convex pentagon form an arithmetic progression, then each of them is greater than $36^{\circ}$. | 9.74. Let the angles of a pentagon be $\alpha, \alpha+\gamma, \alpha+2 \gamma, \alpha+3 \gamma, \alpha+4 \gamma$, where $\alpha, \gamma \geqslant 0$. Since the sum of the angles of a pentagon is $3 \pi$, we have $5 \alpha+10 \gamma=3 \pi$. From the convexity of the pentagon, it follows that all its angles are less than... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,380 |
9.75*. Let \( A B C D E \) be a convex pentagon inscribed in a circle of radius 1, such that \( A B=a, B C=b, C D=c, D E=d, A E=2 \). Prove that
$$
a^{2}+b^{2}+c^{2}+d^{2}+a b c+b c d<4
$$ | 9.75. It is clear that $4=A E^{2}=|\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D E}|^{2}=|\overrightarrow{A B}+\overrightarrow{B C}|^{2}+$ $+2(\overrightarrow{A B}+\overrightarrow{B C}, \overrightarrow{C D}+\overrightarrow{D E})+|\overrightarrow{C D}+\overrightarrow{D E}|^{2}$. Since ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,381 |
9.76*. Inside a regular hexagon with side 1, a point $P$ is taken. Prove that the distances from point $P$ to some three vertices of the hexagon are not less than 1. | 9.76. Let $B$ be the midpoint of side $A_{1} A_{2}$ of the given hexagon $A_{1} \ldots A_{6}, O$ - its center. We can assume that point $P$ lies inside triangle $A_{1} O B$. Then $P A_{3} \geqslant 1$, since the distance from point $A_{3}$ to line $B O$ is $1 ; P A_{4} \geqslant 1$ and $P A_{5} \geqslant 1$, since the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,382 |
9.77*. Prove that if the sides of a convex hexagon $A B C D E F$ are equal to 1, then the radius of the circumscribed circle of one of the triangles $A C E$ and $B D F$ does not exceed 1. | 9.77. Suppose the radii of the circumcircles of triangles $A C E$ and $B D F$ are greater than 1. Let $O$ be the center of the circumcircle of triangle $A C E$. Then $\angle A B C > \angle A O C$, $\angle C D E > \angle C O E$, and $\angle E F A > \angle E O A$, which means $\angle B + \angle D + \angle F > 2 \pi$. Sim... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,383 |
9.78*. The lengths of the sides of a convex hexagon $A B C D E F$ are less than 1. Prove that the length of one of the diagonals $A D, B E, C F$ is less than 2. | 9.78. We can assume that $A E \leqslant A C \leqslant C E$. According to problem $9.67$, $A D \cdot C E \leqslant A E \cdot C D + A C \cdot D E < A E + A C \leqslant 2 C E$, i.e., $A D < 2$. | AD<2 | Geometry | proof | Yes | Yes | olympiads | false | 27,384 |
9.79*. The heptagon $A_{1} \ldots A_{7}$ is inscribed in a circle. Prove that if the center of this circle lies inside it, then the sum of the angles at the vertices $A_{1}, A_{3}, A_{5}$ is less than $450^{\circ}$.
$$
* * *
$$ | 9.79. Since $\angle A_{1}=180^{\circ}-\smile A_{2} A_{7} / 2, \angle A_{3}=180^{\circ}-\smile A_{4} A_{2} / 2$ and $\angle A_{5}=180^{\circ}-\smile A_{6} A_{4} / 2$, then $\angle A_{1}+\angle A_{3}+\angle A_{5}=2 \cdot 180^{\circ}+$ $+\left(360^{\circ}-\smile A_{2} A_{7}-\smile A_{4} A_{2}-\smile A_{6} A_{4}\right) / 2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,385 |
9.80. a) Prove that if the lengths of the projections of a segment onto two mutually perpendicular lines are $a$ and $b$, then its length is not less than $(a+b) / \sqrt{2}$.
b) The lengths of the projections of a polygon onto the coordinate axes are $a$ and $b$. Prove that its perimeter is not less than $\sqrt{2}(a+b... | 9.80. a) It is necessary to prove that if $c$ is the hypotenuse of a right triangle, and $a$ and $b$ are its legs, then $c \geqslant (a+b) / \sqrt{2}$, i.e., $(a+b)^{2} \leqslant 2\left(a^{2}+b^{2}\right)$. Clearly, $(a+b)^{2}=\left(a^{2}+b^{2}\right)+2 a b \leqslant\left(a^{2}+b^{2}\right)+\left(a^{2}+b^{2}\right)=2\l... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,386 |
9.81*. Prove that from the sides of a convex polygon with perimeter $P$, two segments can be formed whose lengths differ by no more than $P / 3$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 9.81. Let's take a segment of length $P$ and arrange the sides of the polygon on it as follows: place the largest side at one end of the segment, the next largest side at the other end, and all the other sides between them. Since any side of the polygon is less than $P / 2$, the midpoint $O$ of the segment cannot lie o... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,387 | |
9.82*. Inside a convex polygon $A_{1} \ldots A_{n}$, a point $O$ is taken. Let $\alpha_{k}$ be the angle at vertex $A_{k}$, $x_{k} = O A_{k}$, and $d_{k}$ be the distance from point $O$ to the line $A_{k} A_{k+1}$. Prove that $\sum x_{k} \sin \left(\alpha_{k} / 2\right) \geqslant \sum d_{k}$ and $\sum x_{k} \cos \left(... | 9.82. Let $\beta_{k}=\angle O A_{k} A_{k+1}$. Then $x_{k} \sin \beta_{k}=d_{k}=x_{k+1} \sin \left(\alpha_{k+1}-\beta_{k+1}\right)$. Therefore, $2 \sum d_{k}=\sum x_{k}\left(\sin \left(\alpha_{k}-\beta_{k}\right)+\sin \beta_{k}\right)=2 \sum x_{k} \sin \left(\alpha_{k} / 2\right) \cos \left(\alpha_{k} / 2-\beta_{k}\righ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,388 |
9.83*. A regular $2 n$-gon $M_{1}$ with side length $a$ lies inside a regular $2 n$-gon $M_{2}$ with side length $2 a$. Prove that the polygon $M_{1}$ contains the center of the polygon $M_{2}$. | 9.83. Suppose that the center $O$ of polygon $M_{2}$ lies outside polygon $M_{1}$. Then there exists a side $A B$ of polygon $M_{1}$ such that polygon $M_{1}$ and point $O$ lie on opposite sides of the line $A B$. Let $C D$ be a side of polygon $M_{1}$ parallel to $A B$. The distance between the lines $A B$ and $C D$ i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,389 |
9.84*. Inside a regular polygon $A_{1} \ldots A_{n}$, a point $O$ is taken. Prove that at least one of the angles $A_{i} O A_{j}$ satisfies the inequalities $\pi(1-1 / n) \leqslant \angle A_{i} O A_{j} \leqslant \pi$. | 9.84. Let $A_{1}$ be the vertex of the polygon closest to $O$. Divide the polygon into triangles by diagonals passing through the vertex $A_{1}$. The point $O$ will be inside one of these triangles, for example, in the triangle $A_{1} A_{k} A_{k+1}$. If the point $O$ falls on the side $A_{1} A_{k}$, then $\angle A_{1} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,390 |
9.85*. Prove that for $n \geqslant 7$ there is a point inside a convex $n$-gon such that the sum of the distances from this point to the vertices is greater than the perimeter. | 9.85. Let $d$ be the length of the longest diagonal (or side) $AB$ of a given $n$-gon. Then its perimeter $P$ does not exceed $\pi d$ (Problem 13.42). Let $A_{i}^{\prime}$ be the projection of vertex $A_{i}$ onto segment $AB$. Then $\sum A A_{i}^{\prime} \geqslant n d / 2$ or $\sum B A_{i}^{\prime} \geqslant n d / 2$ (... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,391 |
9.86*. a) Convex polygons $A_{1} \ldots A_{n}$ and $B_{1} \ldots B_{n}$ are such that all their corresponding sides, except $A_{1} A_{n}$ and $B_{1} B_{n}$, are equal, and $\angle A_{2} \geqslant \angle B_{2}, \ldots, \angle A_{n-1} \geqslant \angle B_{n-1}$, with at least one of these inequalities being strict. Prove ... | 9.86. a) Suppose first that $\angle A_{i}>\angle B_{i}$, and for all other pairs of angles, equality holds. Arrange the polygons so that the vertices $A_{1}, \ldots, A_{i}$ coincide with $B_{1}, \ldots, B_{i}$. In triangles $A_{1} A_{i} A_{n}$ and $A_{1} A_{i} B_{n}$, the sides $A_{i} A_{n}$ and $A_{i} B_{n}$ are equal... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,392 |
9.87. On a segment of length 1, there are $n$ points. Prove that the sum of the distances from some point on the segment to these points is not less than $n / 2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 9.87. Let $A$ and $B$ be the endpoints of a segment; $X_{1}, \ldots, X_{n}$ be given points. Since $A X_{i} + B X_{i} = 1$, then $\sum A X_{i} + \sum B X_{i} = n$. Therefore, $\sum A X_{i} \geqslant n / 2$ or $\sum B X_{i} \geqslant n / 2$.
. It is sufficient to prove that the path along the arc of the circle is no more than 1.6 times longer than the path along the straight line. The ratio of the length of an arc of angular measure $2 \var... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,394 |
9.89*. In a certain forest, the distance between any two trees does not exceed the difference in their heights. All trees have a height of less than $100 \mathrm{m}$. Prove that this forest can be fenced with a fence of length $200 \mathrm{m}$. | 9.89. Let trees of heights $a_{1}>a_{2}>\ldots>a_{n}$ grow at points $A_{1}, \ldots, A_{n}$. Then by the condition $A_{1} A_{2} \leqslant\left|a_{1}-a_{2}\right|=a_{1}-a_{2}, \ldots, A_{n-1} A_{n} \leqslant a_{n-1}-a_{n}$. Therefore, the length of the broken line $A_{1} A_{2} \ldots A_{n}$ does not exceed ( $a_{1}-a_{2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,395 |
9.90*. A polygon (not necessarily convex), cut out of paper, is folded along some line and the two halves are glued together. Can the perimeter of the resulting polygon be greater than the perimeter of the original?
$$
* * *
$$ | 9.90. Let's highlight the parts in the resulting polygon where the gluing occurred (in Fig. 9.27, these parts are shaded). All sides that do not belong to the shaded polygons are part of the perimeter of both the original and the resulting polygons. As for the shaded polygons, their sides lying on the fold line are par... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,396 |
9.91. Prove that a closed broken line of length 1 can be placed in a circle of radius 0.25. | 9.91. Let's take two points $A$ and $B$ on a broken line, dividing its perimeter in half. Then $A B \leqslant 1 / 2$. We will prove that all points of the broken line lie inside a circle of radius $1 / 4$
. Therefore, $P / 4=R_{1} \geqslant R$.
Second solution. If $02 x / \pi$. Therefore, $a+b+$ $+c=2 R(\sin \alpha+\si... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,399 |
10.2. The medians $A A_{1}$ and $B B_{1}$ of triangle $A B C$ intersect at point $M$. Prove that if the quadrilateral $A_{1} M B_{1} C$ is cyclic, then $A C = B C$. | 10.2. Suppose, for example, that $a>b$. Then $m_{a}<m_{b}$ (Problem 10.1). And since the quadrilateral $A_{1} M B_{1} C$ is circumscribed, $\frac{a}{2}+\frac{m_{b}}{3}=\frac{b}{2}+\frac{m_{a}}{3}$, i.e., $(a-b) / 2=$ $=\left(m_{a}-m_{b}\right) / 3$. This leads to a contradiction. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,401 |
10.4. a) Prove that if $a, b, c$ are the lengths of the sides of any triangle, then $a^{2}+b^{2} \geqslant c^{2} / 2$.
b) Prove that $m_{a}^{2}+m_{b}^{2} \geqslant 9 c^{2} / 8$. | 10.4. a) Since \( c \leqslant a + b \), then \( c^{2} \leqslant (a + b)^{2} = a^{2} + b^{2} + 2ab \leqslant 2\left(a^{2} + b^{2}\right) \).
b) Let \( M \) be the point of intersection of the medians of triangle \( ABC \). According to problem a), \( M A^{2} + M B^{2} \geqslant A B^{2} / 2 \), i.e., \( \frac{4 m_{a}^{2... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,403 |
10.5*. a) Prove that $m_{a}^{2}+m_{b}^{2}+m_{c}^{2} \leqslant 27 R^{2} / 4$.
b) Prove that $m_{a}+m_{b}+m_{c} \leqslant 9 R / 2$. | 10.5. a) Let $M$ be the point of intersection of the medians, and $O$ be the center of the circumscribed circle of triangle $ABC$. Then $AO^{2}+BO^{2}+CO^{2}=(\overrightarrow{AM}+\overrightarrow{MO})^{2}+$ $(\overrightarrow{BM}+\overrightarrow{MO})^{2}+(\overrightarrow{CM}+\overrightarrow{MO})^{2}=AM^{2}+BM^{2}+CM^{2}+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,404 |
10.6*. Prove that $\left|a^{2}-b^{2}\right| /(2 c)<m_{c} \leqslant\left(a^{2}+b^{2}\right) /(2 c)$. | 10.6. Heron's formula can be rewritten as $16 S^{2}=2 a^{2} b^{2}+2 a^{2} c^{2}+$ $+2 b^{2} c^{2}-a^{4}-b^{4}-c^{4}$. Since $m_{c}^{2}=\left(2 a^{2}+2 b^{2}-c^{2}\right) / 4$ (problem 12.11, a) $)$, the inequalities $m_{c}^{2} \leqslant\left(\left(a^{2}+b^{2}\right) / 2 c\right)^{2}$ and $m_{c}^{2}>\left(\left(a^{2}-b^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,405 |
10.7*. Let $x=a b+b c+c a, x_{1}=m_{a} m_{b}+m_{b} m_{c}+m_{c} m_{a}$. Prove that $9 / 20<x_{1} / x<5 / 4$.
See also problems $9.1,10.74,10.76,17.17$.
## §2. Altitudes | 10.7. Let $y=a^{2}+b^{2}+c^{2}$ and $y_{1}=m_{a}^{2}+m_{b}^{2}+m_{c}^{2}$. Then $3 y=4 y_{1}$ (problem 12.11, b)), $y<2 x$ (problem 9.7) and $2 x_{1}+y_{1}<2 x+y$, since $(m_{a}+m_{b}+m_{c})^{2}<(a+b+c)^{2}$ (see problem 9.2). Adding the inequality $8 x_{1}+4 y_{1}<8 x+4 y$ to the equality $3 y=4 y_{1}$, we get $8 x_{1... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,406 |
10.8. Prove that in any triangle, the sum of the lengths of the altitudes is less than the perimeter. | 10.8. It is clear that $h_{a} \leqslant b, h_{b} \leqslant c, h_{c} \leqslant a$, and at least one of these inequalities is strict. Therefore, $h_{a}+h_{b}+h_{c}<a+b+c$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,407 |
10.9. Two altitudes of a triangle are greater than 1. Prove that its area is greater than $1 / 2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 10.9. Let $h_{a}>1$ and $h_{b}>1$. Then $a \geqslant h_{b}>1$. Therefore $S=a h_{a} / 2>1 / 2$. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,408 |
10.10. In triangle $A B C$, the altitude $A M$ is not less than $B C$, and the altitude $B H$ is not less than $A C$. Find the angles of triangle $A B C$. | 10.10. According to the condition $B H \geqslant A C$, and since the perpendicular is shorter than the slant, then $B H \geqslant A C \geqslant A M$. Similarly $A M \geqslant B C \geqslant B H$. Therefore $B H=A M=A C=$ $=B C$. Since $A C=A M$, the segments $A C$ and $A M$ coincide, i.e., $\angle C=90^{\circ}$, and sin... | 45,45,90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,409 |
10.11. Prove that $\frac{1}{2 r}<\frac{1}{h_{a}}+\frac{1}{h_{b}}<\frac{1}{r}$. | 10.11. It is clear that $\frac{1}{h_{a}}+\frac{1}{h_{b}}=\frac{a+b}{2 S}=\frac{a+b}{(a+b+c) r}$ and $a+b+c<2(a+b)<2(a+b+c)$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,410 |
10.12. Prove that $h_{a}+h_{b}+h_{c} \geqslant 9 r$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 10.12. Since $a h_{a}=2 S=r(a+b+c)$, then $h_{a}=r \quad 1+\frac{b}{a}+\frac{c}{a}$. By adding such equalities for $h_{a}, h_{b}$ and $h_{c}$ and using the inequality $\frac{x}{y}+\frac{y}{x} \geqslant 2$, we will obtain the required. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,411 | |
10.13. Let $a<b$. Prove that $a+h_{a} \leqslant b+h_{b}$. | 10.13. Since $h_{a}-h_{b}=2 S(1 / a-1 / b)=2 S(b-a) / a b$ and $2 S \leqslant a b$, then $h_{a}-h_{b} \leqslant b-a$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,412 |
10.14*. Prove that $h_{a} \leqslant \sqrt{r_{b} r_{c}}$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 10.14. According to problem $12.21 \frac{2}{h_{a}}=\frac{1}{r_{b}}+\frac{1}{r_{c}}$. In addition, $\frac{1}{r_{b}}+\frac{1}{r_{c}} \geqslant 2 / \sqrt{r_{b} r_{c}}$. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,413 | |
10.15*. Prove that $h_{a} \leqslant(a / 2) \operatorname{ctg}(\alpha / 2)$. | 10.15. Since $2 \sin \beta \sin \gamma = \cos (\beta - \gamma) - \cos (\beta + \gamma) \leqslant 1 + \cos \alpha$, then
$$
\frac{h_{a}}{a} = \frac{\sin \beta \sin \gamma}{\sin \alpha} \leqslant \frac{1 + \cos \alpha}{2 \sin \alpha} = \frac{1}{2} \operatorname{ctg} \frac{\alpha}{2}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,414 |
10.16*. Let $a \leqslant b \leqslant c$. Prove that then $h_{a}+h_{b}+$ $+h_{c} \leqslant 3 b\left(a^{2}+a c+c^{2}\right) /(4 p R)$.
See also problems $10.28,10.55,10.74,10.80$.
## §3. Bisectors | 10.16. Since $b / 2 R=\sin \beta$, after multiplying by $2 p$ we transition to the inequality $(a+b+c)\left(h_{a}+h_{b}+h_{c}\right) \leqslant 3 \sin \beta\left(a^{2}+a c+c^{2}\right)$. Subtracting $6 S$ from both sides, we get $a\left(h_{b}+h_{c}\right)+b\left(h_{a}+h_{c}\right)+c\left(h_{a}+h_{b}\right) \leqslant 3 \... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,415 |
10.17*. Prove that $l_{a} \leqslant \sqrt{p(p-a)}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 10.17. According to problem 12.35, a) $l_{a}^{2}=4 b c p(p-a) /(b+c)^{2}$. In addition, $4 b c \leqslant(b+$ $+c)^{2}$. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,416 | |
10.18*. Prove that $h_{a} / l_{a} \geqslant \sqrt{2 r / R}$. | 10.18. It is clear that $h_{a} / l_{a}=\cos ((\beta-\gamma) / 2)$. According to problem 12.36, a)
$2 r / R=8 \sin (\alpha / 2) \sin (\beta / 2) \sin (\gamma / 2)=4 \sin (\alpha / 2)[\cos ((\beta-\gamma) / 2)-$
$$
-\cos ((\beta+\gamma) / 2)]=4 x(q-x), \quad \text { where } \quad x=\sin (\alpha / 2) \quad \text { and }... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,417 |
10.19*. Prove that: a) $l_{a}^{2}+l_{b}^{2}+l_{c}^{2} \leqslant p^{2}$; b) $l_{a}+l_{b}+l_{c} \leqslant \sqrt{3} p$. | 10.19. a) According to problem $10.17 l_{a}^{2} \leqslant p(p-a)$. Adding three similar inequalities, we obtain the required result.
b) For any numbers $l_{a}, l_{b}$, and $l_{c}$, the inequality $\left(l_{a}+l_{b}+l_{c}\right)^{2} \leqslant 3\left(l_{a}^{2}+\right.$ $\left.+l_{b}^{2}+l_{c}^{2}\right)$ holds. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,418 |
10.20*. Prove that $l_{a}+l_{b}+m_{c} \leqslant \sqrt{3} p$.
See also problems $6.41,10.75,10.74,10.95$.
## §4. Lengths of sides | 10.20. It is sufficient to prove that $\sqrt{p(p-a)}+\sqrt{p(p-b)}+m_{c} \leqslant \sqrt{3 p}$. We can assume that $p=1$; let $x=1-a$ and $y=1-b$. Then $m_{c}^{2}=\left(2 a^{2}+2 b^{2}-c^{2}\right) / 4=$ $=1-(x+y)+(x-y)^{2} / 4=m(x, y)$. Consider the function $f(x, y)=\sqrt{x}+$ $+\sqrt{y}+\sqrt{m(x, y)}$. We need to p... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,419 |
10.21. Prove that $\frac{9 r}{2 S} \leqslant \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{9 R}{4 S}$. | 10.21. It is clear that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\left(h_{a}+h_{b}+h_{c}\right) / 2 S$. Moreover, $9 r \leqslant h_{a}+h_{b}+h_{c}$ (problem 10.12) and $h_{a}+h_{b}+h_{c} \leqslant m_{a}+m_{b}+m_{c} \leqslant 9 R / 2$ (problem 10.5, b)). | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,420 |
10.22*. Prove that $2 b c \cos \alpha /(b+c)<b+c-a<2 b c / a$. | 10.22. First, let's prove that \( b+c-a < \frac{2bc}{a} \). Let \( 2x = b+c-a \), \( 2y = a+c-b \), and \( 2z = a+b-c \). We need to prove that \( 2x < \frac{2(x+y)(x+z)}{y+z} \), i.e., \( xy + xz < xy + xz + x^2 + yz \). The last inequality is obvious.
Since \( 2bc \cos \alpha = b^2 + c^2 - a^2 = (b+c-a)(b+c+a) - 2bc... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,421 |
10.23*. Prove that if $a, b, c$ are the lengths of the sides of a triangle with a perimeter of 2, then $a^{2}+b^{2}+c^{2}<2(1-a b c)$. | 10.23. According to problem $12.30 a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+a c)=$ $=4 p^{2}-2 r^{2}-2 p^{2}-8 r R=2 p^{2}-2 r^{2}-8 R r$ and $a b c=4 p r R$. Therefore, we need to prove the inequality $2 p^{2}-2 r^{2}-8 r R<2(1-4 p r R)$, where $p=1$. This inequality is obvious. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,422 |
10.24*. Prove that $20 R r-4 r^{2} \leqslant a b+b c+c a \leqslant 4(R+r)^{2}$.
## §5. Radii of Circumscribed, Inscribed, and Externally Tangent Circles | 10.24. According to problem $12.30 \, ab+bc+ca=r^{2}+p^{2}+4 R r$. In addition, $16 R r-5 r^{2} \leqslant p^{2} \leqslant 4 R^{2}+4 R r+3 r^{2}$ (problem 10.34). | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,423 |
10.25. Prove that $r r_{c} \leqslant c^{2} / 4$. | 10.25. Since $r(\operatorname{ctg} \alpha+\operatorname{ctg} \beta)=c=r_{c}(\operatorname{tg} \alpha+\operatorname{tg} \beta)$, then
$$
c^{2}=r r_{c} \quad 2+\frac{\operatorname{tg} \alpha}{\operatorname{tg} \beta}+\frac{\operatorname{tg} \beta}{\operatorname{tg} \alpha} \geqslant 4 r r_{c}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,424 |
10.26*. Prove that $r / R \leqslant 2 \sin (\alpha / 2)(1-\sin (\alpha / 2))$.
10.27*. Prove that $6 r \leqslant a+b$. | 10.26. It is sufficient to use the results of problems 12.36, a) and 10.45. Note also that $x(1-x) \leqslant 1 / 4$, i.e., $r / R \leqslant 1 / 2$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,425 |
10.28*. Prove that $\frac{r_{a}}{h_{a}}+\frac{r_{b}}{h_{b}}+\frac{r_{c}}{h_{c}} \geqslant 3$. | 10.28. Since $\frac{2}{h_{a}}=\frac{1}{r_{a}}+\frac{1}{r_{c}}$ (Problem 12.21), then $\frac{r_{a}}{h_{a}}=\frac{r_{a}}{r_{b}}+\frac{r_{a}}{r_{c}} \quad / 2$. Let's write similar equalities for $r_{b} / h_{b}$ and $r_{c} / h_{c}$ and add them. Considering that $\frac{x}{y}+\frac{y}{x} \geqslant 2$, we obtain the require... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,426 |
10.29*. Prove that $27 R r \leqslant 2 p^{2} \leqslant 27 R^{2} / 2$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 10.29. Since $R r=R S / p=a b c / 4 p$ (see problem 12.1), we arrive at the inequality $27 a b c \leqslant 8 p^{3}=(a+b+c)^{3}$.
Since $(a+b+c)^{2} \leqslant 3\left(a^{2}+b^{2}+c^{2}\right)$ for any numbers $a, b$ and $c$, then $p^{2} \leqslant 3\left(a^{2}+b^{2}+c^{2}\right) / 4=m_{a}^{2}+m_{b}^{2}+m_{c}^{2}$ (see pr... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,427 | |
10.30*. Let $O$ be the center of the inscribed circle of triangle $ABC$, and suppose $OA \geqslant OB \geqslant OC$. Prove that $OA \geqslant 2r$ and $OB \geqslant r \sqrt{2}$. | 10.30. Since $O A=r / \sin (A / 2), O B=r / \sin (B / 2)$ and $O C=r / \sin (C / 2)$, and the angles $\angle A / 2, \angle B / 2$ and $\angle C / 2$ are acute, then $\angle A \leqslant \angle B \leqslant \angle C$. Therefore, $\angle A \leqslant 60^{\circ}$ and $\angle B \leqslant 90^{\circ}$, which means $\sin (A / 2)... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,428 |
10.31*. Prove that the sum of the distances from any point inside a triangle to its vertices is not less than $6 r$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 10.31. If $\angle C \geqslant 120^{\circ}$, then the sum of the distances from any point inside the triangle to its vertices is not less than $a+b$ (Problem 12.21); moreover, $a+b \geqslant 6 r$ (Problem 12.27).
If all angles of the triangle are less than $120^{\circ}$, then at the point of minimum of the sum of the d... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 27,429 | |
10.32*. Prove that $3\left(\frac{a}{r_{a}}+\frac{b}{r_{b}}+\frac{c}{r_{c}}\right) \geqslant 4\left(\frac{r_{a}}{a}+\frac{r_{b}}{b}+\frac{r_{c}}{c}\right)$. | 10.32. Let $\alpha=\cos (A / 2), \beta=\cos (B / 2)$ and $\gamma=\cos (C / 2)$. According to problem 12.17, b) $a / r_{a}=\alpha / \beta \gamma, b / r_{b}=\beta / \gamma \alpha$ and $c / r_{c}=\gamma / \alpha \beta$. Therefore, after multiplying by $\alpha \beta \gamma$, the required inequality can be rewritten as $3\l... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,430 |
10.33*. Prove that
a) $5 R-r \geqslant \sqrt{3} p$
b) $4 R-r_{a} \geqslant(p-a)\left[\sqrt{3}+\left(a^{2}+(b-c)^{2}\right) /(2 S)\right]$. | 10.33. a) Adding the equality $4 R+r=r_{a}+r_{b}+r_{c}$ (Problem 12.24) to the inequality $R-2 r \geqslant 0$ (Problem 10.26), we get
\[
\begin{aligned}
5 R-r \geqslant r_{a}+r_{b}+r_{c} & =p r\left((p-a)^{-1}+(p-b)^{-1}+(p-c)^{-1}\right)= \\
& =p\left(a b+b c+c a-p^{2}\right) / S=p\left(2(a b+b c+c a)-a^{2}-b^{2}-c^{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,431 |
10.37*. a) $\sin \alpha + \sin \beta + \sin \gamma \leqslant 3 \sqrt{3} / 2$;
б) $\cos (\alpha / 2) + \cos (\beta / 2) + \cos (\gamma / 2) \leqslant 3 \sqrt{3} / 2$.
a) $\sin \alpha + \sin \beta + \sin \gamma \leqslant 3 \sqrt{3} / 2$;
b) $\cos (\alpha / 2) + \cos (\beta / 2) + \cos (\gamma / 2) \leqslant 3 \sqrt{3}... | 10.37. a) It is clear that $\sin \alpha + \sin \beta + \sin \gamma = p / R$. Moreover, $p \leqslant 3 \sqrt{3} R / 2$ (Problem 10.29).
b) Follows from a) (see the remark). | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,433 |
10.38*. a) $\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma \geqslant \sqrt{3}$
b) $\operatorname{tg}(\alpha / 2)+\operatorname{tg}(\beta / 2)+\operatorname{tg}(\gamma / 2) \geqslant \sqrt{3}$. | 10.38. a) According to problem 12.44, a) $\operatorname{ctg} \alpha+\operatorname{ctg} \beta+\operatorname{ctg} \gamma=\left(a^{2}+b^{2}+c^{2}\right) / 4 S$. In addition, $a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S$ (problem 10.53, b)).
b) Follows from a) (see remark). | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,434 |
10.39*. a) $\operatorname{ctg}(\alpha / 2)+\operatorname{ctg}(\beta / 2)+\operatorname{ctg}(\gamma / 2) \geqslant 3 \sqrt{3}$.
b) For an acute-angled triangle
$$
\operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg} \gamma \geqslant 3 \sqrt{3}
$$ | 10.39. a) According to problem 12.44, a) $\operatorname{ctg}(\alpha / 2)+\operatorname{ctg}(\beta / 2)+\operatorname{ctg}(\gamma / 2)=p / r$. In addition, $p \geqslant 3 \sqrt{3} r$ (problem 10.53, a)).
b) Follows from a) (see the remark). For an obtuse triangle, $\operatorname{tg} \alpha+\operatorname{tg} \beta+$ $+\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,435 |
10.40*. a) $\sin (\alpha / 2) \sin (\beta / 2) \sin (\gamma / 2) \leqslant 1 / 8$;
б) $\cos \alpha \cos \beta \cos \gamma \leqslant 1 / 8$. | 10.40. a) According to problem 12.36, a) $\sin (\alpha / 2) \sin (\beta / 2) \sin (\gamma / 2)=r / 4 R$. In addition, $r \leqslant R / 2$ (problem 10.26).
b) For an acute triangle, it follows from a) (see the remark). For an obtuse triangle, $\cos \alpha \cos \beta \cos \gamma<0$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,436 |
10.41*. a) $\sin \alpha \sin \beta \sin \gamma \leqslant 3 \sqrt{3} / 8$
b) $\cos (\alpha / 2) \cos (\beta / 2) \cos (\gamma / 2) \leqslant 3 \sqrt{3} / 8$. | 10.41. a) Since $\sin x=2 \sin (x / 2) \cos (x / 2)$, using the results of problems 12.36, a) and 12.36, b), we get $\sin \alpha \sin \beta \sin \gamma=p r / 2 R^{2}$. Moreover, $p \leqslant 3 \sqrt{3} R / 2$ (problem 10.29) and $r \leqslant R / 2$ (problem 10.26).
b) Follows from a) (see the remark). | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,437 |
10.42*. a) $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma \geqslant 3 / 4$.
b) For an obtuse triangle
$$
\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma>1
$$ | 10.42. According to problem 12.39, b) $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1-2 \cos \alpha \cos \beta \cos \gamma$. It remains to note that $\cos \alpha \cos \beta \cos \gamma \leqslant 1 / 8$ (problem 10.40, b)), and for an obtuse triangle $\cos \alpha \cos \beta \cos \gamma<0$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,438 |
10.43*. a) $\cos \alpha \cos \beta+\cos \beta \cos \gamma+\cos \gamma \cos \alpha \leqslant 3 / 4$. | 10.43. It is clear that $2(\cos \alpha \cos \beta + \cos \beta \cos \gamma + \cos \gamma \cos \alpha) = (\cos \alpha + \cos \beta + \cos \gamma)^2 - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma$. It remains to note that $\cos \alpha + \cos \beta + \cos \gamma \leqslant 3 / 2$ (problem 10.36, a)) and $\cos^2 \alpha + \c... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,439 |
10.44*. For an acute triangle
$$
\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma \leqslant \sin (\alpha+\beta)+\sin (\beta+\gamma)+\sin (\gamma+\alpha)
$$
## §7. Inequalities for Angles of a Triangle | 10.44. Let the extensions of the angle bisectors of an acute-angled triangle $ABC$ with angles $\alpha, \beta$, and $\gamma$ intersect the circumcircle at points $A_{1}, B_{1}$, and $C_{1}$. Then $S_{ABC}=R^{2}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) / 2$ and $S_{A_{1} B_{1} C_{1}}=R^{2}(\sin (\alpha+\beta)+\sin (\be... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,440 |
10.45. Prove that $1-\sin (\alpha / 2) \geqslant 2 \sin (\beta / 2) \sin (\gamma / 2)$. | 10.45. It is clear that $2 \sin (\beta / 2) \sin (\gamma / 2)=\cos ((\beta-\gamma) / 2)-\cos ((\beta+\gamma) / 2) \leqslant$ $\leqslant 1-\sin (\alpha / 2)$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,441 |
10.46. Prove that $\sin (\gamma / 2) \leqslant c /(a+b)$. | 10.46. Drop perpendiculars $A A_{1}$ and $B B_{1}$ from vertices $A$ and $B$ to the bisector of angle $A C B$. Then $A B \geqslant A A_{1}+B B_{1}=b \sin (\gamma / 2)+a \sin (\gamma / 2)$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,442 |
10.47*. Prove that if $a+b<3 c$, then $\operatorname{tg}(\alpha / 2) \operatorname{tg}(\beta / 2)<1 / 2$. | 10.47. According to problem $12.32 \operatorname{tg}(\alpha / 2) \operatorname{tg}(\beta / 2)=(a+b-c) /(a+b+c)$. And since $a+b<3 c$, then $a+b-c<(a+b+c) / 2$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,443 |
10.49*. Prove that $\cos 2 \alpha+\cos 2 \beta-\cos 2 \gamma \leqslant 3 / 2$. | 10.49. First, note that $\cos 2 \gamma - \cos (\pi - \alpha - \beta) = \cos 2 \alpha \cos 2 \beta - \sin 2 \alpha \sin 2 \beta$. Therefore, $\cos 2 \alpha + \cos 2 \beta - \cos 2 \gamma = \cos 2 \alpha + \cos 2 \beta - \cos 2 \alpha \cos 2 \beta + \sin 2 \alpha \sin 2 \beta$. Since $a \cos \varphi + b \sin \varphi \leq... | \frac{3}{2} | Inequalities | proof | Yes | Yes | olympiads | false | 27,445 |
10.51. The inscribed circle touches the sides of triangle $ABC$ at points $A_{1}, B_{1}$ and $C_{1}$. Prove that triangle $A_{1} B_{1} C_{1}$ is acute-angled. | 10.51. If the angles of triangle $A B C$ are $\alpha, \beta$ and $\gamma$, then the angles of triangle $A_{1} B_{1} C_{1}$ are $(\beta+\gamma) / 2, (\gamma+\alpha) / 2$ and $(\alpha+\beta) / 2$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,447 |
10.53. Prove that:
a) $3 \sqrt{3} r^{2} \leqslant S \leqslant p^{2} / 3 \sqrt{3}$
b) $S \leqslant\left(a^{2}+b^{2}+c^{2}\right) / 4 \sqrt{3}$. | 10.53. a) It is clear that $S^{2} / p=(p-a)(p-b)(p-c) \leqslant((p-a+p-b+p-c) / 3)^{3}=p^{3} / 27$. Therefore, $p r=S \leqslant p^{2} / 3 \sqrt{3}$, i.e., $r \leqslant p / 3 \sqrt{3}$. Multiplying the last inequality by $r$, we get the required result.
b) Since $(a+b+c)^{2} \leqslant 3\left(a^{2}+b^{2}+c^{2}\right)$, ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,449 |
10.54*. Prove that $a^{2}+b^{2}+c^{2}-(a-b)^{2}-(b-c)^{2}-(c-a)^{2} \geqslant 4 \sqrt{3} S$. | 10.54. Let $x=p-a, y=p-b, z=p-c$. Then $\left(a^{2}-(b-c)^{2}\right)+\left(b^{2}-(a-c)^{2}\right)+$ $+\left(c^{2}-(a-b)^{2}\right)=4(p-b)(p-c)+4(p-a)(p-c)+4(p-a)(p-b)=4(y z+z x+x y)$
and
$$
4 \sqrt{3} S=4 \sqrt{3 p(p-a)(p-b)(p-c)}=4 \sqrt{3(x+y+z) x y z} .
$$
Thus, we need to prove the inequality $x y+y z+z x \geqsl... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,450 |
10.55*. Prove that
a) $S^{3} \leqslant(\sqrt{3} / 4)^{3}(a b c)^{2}$
b) $\sqrt[3]{h_{a} h_{b} h_{c}} \leqslant \sqrt[4]{3} \sqrt{S} \leqslant \sqrt[3]{r_{a} r_{b} r_{c}}$. | 10.55. a) Multiplying three equalities of the form $S=(a b \sin \gamma) / 2$, we get $S^{3}=$ $=\left((a b c)^{2} \sin \gamma \sin \beta \sin \alpha\right) / 8$. It remains to use the result of problem 10.41.
b) Since $\left(h_{a} h_{b} h_{c}\right)^{2}=(2 S)^{6} /(a b c)^{2}$ and $(a b c)^{2} \geqslant(4 / \sqrt{3})^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,451 |
10.56*. On the sides $BC$, $CA$, and $AB$ of triangle $ABC$, points $A_1$, $B_1$, and $C_1$ are taken, such that $AA_1$, $BB_1$, and $CC_1$ intersect at one point. Prove that $S_{A_1B_1C_1} / S_{ABC} \leqslant 1 / 4$. | 10.56. Let $p = \frac{B A_{1}}{B C}$, $q = \frac{C B_{1}}{C A}$, and $r = \frac{A C_{1}}{A B}$. Then $\frac{S_{A_{1} B_{1} C_{1}}}{S_{A B C}} = 1 - p(1 - r) - q(1 - p) - r(1 - q) = 1 - (p + q + r) + (pq + qr + rp)$. By Ceva's theorem (problem 5.77), $pqr = (1 - p)(1 - q)(1 - r)$, i.e., $2pqr = 1 - (p + q + r) + (pq + q... | \frac{S_{A_1B_1C_1}}{S_{ABC}}\leq\frac{1}{4} | Geometry | proof | Yes | Yes | olympiads | false | 27,452 |
10.57*. On the sides $BC$, $CA$, and $AB$ of triangle $ABC$, arbitrary points $A_1$, $B_1$, and $C_1$ are taken. Let $a = S_{AB_1C_1}$, $b = S_{A_1BC_1}$, $c = S_{A_1B_1C}$, and $u = S_{A_1B_1C_1}$. Prove that
$$
u^{3} + (a + b + c) u^{2} \geqslant 4abc
$$ | 10.57. We can assume that the area of triangle $ABC$ is equal to 1. Then $a + b + c = 1 - u$, so the given inequality can be rewritten as $u^2 \geq 4abc$. Let $x = BA_1 / BC$, $y = CB_1 / CA$, and $z = AC_1 / AB$. Then $u = 1 - (x + y + z) + xy + yz + zx$ and $abc = xyz(1 - x)(1 - y)(1 - z) = v(u - v)$, where $v = xyz$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,453 |
10.58*. On the sides $BC$, $CA$, and $AB$ of triangle $ABC$, points $A_1$, $B_1$, and $C_1$ are taken. Prove that the area of one of the triangles $AB_1C_1$, $A_1BC_1$, $A_1B_1C$ does not exceed:
a) $S_{ABC} / 4$
b) $S_{A_1B_1C_1}$.
See also problems $9.33, 9.37, 9.40, 10.9, 20.1, 20.7$.
## §9. Opposite the larger ... | 10.58. a) Let $x = \frac{B A_{1}}{B C}$, $y = \frac{C B_{1}}{C A}$, and $z = \frac{A C_{1}}{A B}$. We can assume that the area of triangle $A B C$ is 1. Then, $S_{A B_{1} C_{1}} = z(1 - y)$, $S_{A_{1} B C_{1}} = x(1 - z)$, and $S_{A_{1} B_{1} C} = y(1 - x)$. Since $x(1 - x) \leq \frac{1}{4}$, $y(1 - y) \leq \frac{1}{4}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,454 |
10.62*. In an acute-angled triangle $A B C$, the greatest height $A H$ is equal to the median $B M$. Prove that $\angle B \leqslant 60^{\circ}$. | 10.62. Let point $B_{1}$ be symmetric to $B$ with respect to point $M$. Since the height dropped from point $M$ to side $B C$ is half of $A H$, i.e., half of $B M$, then $\angle M B C=30^{\circ}$. Since $A H$ is the largest of the heights, then $B C$ is the smallest of the sides. Therefore, $A B_{1}=B C \leqslant A B$,... | \angleB\leqslant60 | Geometry | proof | Yes | Yes | olympiads | false | 27,458 |
10.64. a) Inside triangle $A B C$, there is a segment $M N$. Prove that the length of $M N$ does not exceed the length of the largest side of the triangle.
b) Inside a convex polygon, there is a segment $M N$. Prove that the length of $M N$ does not exceed the length of the largest side or the largest diagonal of this ... | 10.64. We will conduct the proof for the general case immediately. Let the line $M N$ intersect the sides of the polygon at points $M_{1}$ and $N_{1}$. It is clear that $M N \leqslant M_{1} N_{1}$. Suppose point $M_{1}$ lies on side $A B$, and point $N_{1}$ lies on $P Q$. Since $\angle A M_{1} N_{1} + \angle B M_{1} N_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,460 |
10.65*. Inside the sector $A O B$ of a circle with radius $R = A O = B O$, there lies a segment $M N$. Prove that $M N \leqslant R$ or $M N \leqslant A B$. (It is assumed that $\angle A O B < 180^{\circ}$.) | 10.65. A segment can be extended to intersect with the boundary of the sector, as this will only increase its length. Therefore, we can assume that points \( M \) and \( N \) lie on the boundary of the sector. There are three possible cases.
1. Points \( M \) and \( N \) lie on the arc of the circle. Then \( MN = 2R \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,461 |
10.66*. A circle is inscribed in an angle with vertex $A$, touching the sides of the angle at points $B$ and $C$. In the region bounded by segments $A B, A C$ and the smaller arc $B C$, there is a segment. Prove that its length does not exceed $A B$.
保留源文本的换行和格式，直接输出翻译结果如下：
10.66*. A circle is inscribed in an angle w... | 10.66. If this segment does not have common points with the circle, then using a homothety with center $A$ (and a coefficient greater than 1) it can be transformed into a segment that has a common point $X$ with the arc $A B$ and lies within our region. Draw the tangent $D E$ to the circle through point $X$ (points $D$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,462 |
10.67*. Inside a circle, there is a convex pentagon. Prove that at least one of its sides is not greater than the side of a regular pentagon inscribed in this circle. | 10.67. Suppose first that the center $O$ of the circle lies inside the given pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$. Consider the angles $A_{1} O A_{2}, A_{2} O A_{3}, \ldots, A_{5} O A_{1}$. The sum of these five angles is $2 \pi$, so one of them, for example $A_{1} O A_{2}$, does not exceed $2 \pi / 5$. Then the se... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,463 |
10.68*. Given a triangle $ABC$ with sides $a>b>c$ and an arbitrary point $O$ inside it. Let the lines $AO, BO, CO$ intersect the sides of the triangle at points $P, Q, R$. Prove that $OP + OQ + OR < a$.
## §11. Inequalities for Right Triangles
In all problems of this paragraph, $ABC$ is a right triangle with a right ... | 10.68. Let's take points \(A_1\) and \(A_2\) on sides \(BC\), \(CA\), and \(AB\) such that \(B_1 C_2 \parallel BC\), \(C_1 A_2 \parallel CA\), and \(A_1 B_2 \parallel AB\) (Fig. 10.1). In triangles \(A_1 A_2 O\), \(B_1 B_2 O\), and \(C_1 C_2 O\), the largest sides are \(A_1 A_2\), \(B_1 O\), and \(C_2 O\) respectively.... | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 27,464 | |
10.69. Prove that $c^{n}>a^{n}+b^{n}$ for $n>2$. | 10.69. Since $c^{2}=a^{2}+b^{2}$, then $c^{n}=\left(a^{2}+\right.$ $\left.+b^{2}\right) c^{n-2}=a^{2} c^{n-2}+b^{2} c^{n-2}>a^{n}+b^{n}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,465 |
10.72*. Prove that $c / r \geqslant 2(1+\sqrt{2})$. | 10.72. It is clear that $a+b \geqslant 2 \sqrt{a b}$ and $c^{2}=a^{2}+b^{2} \geqslant 2 a b$. Therefore
$$
\frac{c^{2}}{r^{2}}=\frac{(a+b+c)^{2} c^{2}}{a^{2} b^{2}} \geqslant \frac{(2 \sqrt{a b}+\sqrt{2 a b})^{2} \cdot 2 a b}{a^{2} b^{2}}=4(1+\sqrt{2})^{2}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,468 |
10.73*. Prove that $m_{a}^{2}+m_{b}^{2}>29 r^{2}$.
## §12. Inequalities for Acute Triangles | 10.73. According to problem 12.11, a) $m_{a}^{2}+m_{b}^{2}=\left(4 c^{2}+a^{2}+b^{2}\right) / 4=5 c^{2} / 4$. In addition, $5 c^{2} / 4 \geqslant 5(1+\sqrt{2})^{2} r^{2}=(15+10 \sqrt{2}) r^{2}>29 r^{2}$ (see problem 10.72). | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,469 |
10.74. Prove that for an acute-angled triangle
$$
\frac{m_{a}}{h_{a}}+\frac{m_{b}}{h_{b}}+\frac{m_{c}}{h_{c}} \leqslant 1+\frac{R}{r}
$$ | 10.74. Let $O$ be the center of the circumscribed circle, $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $B C, C A, A B$ respectively. Then $m_{a}=A A_{1} \leqslant A O+O A_{1}=R+O A_{1}$. Similarly, $m_{b} \leqslant R+O B_{1}$ and $m_{c} \leqslant R+O C_{1}$. Therefore,
$$
\frac{m_{a}}{h_{a}}+\frac{m_{b}}{h_{b}}+\fr... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,470 |
10.75. Prove that for an acute-angled triangle
$$
\frac{1}{l_{a}}+\frac{1}{l_{b}}+\frac{1}{l_{c}} \leqslant \sqrt{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) .
$$ | 10.75. According to problem $4.47 \frac{1}{b}+\frac{1}{c}=\frac{2 \cos (\alpha / 2)}{l_{a}} \geqslant \frac{\sqrt{2}}{l_{a}}$. By adding three similar inequalities, we obtain the required result. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,471 |
10.76. Prove that if a triangle is not obtuse, then $m_{a}+m_{b}+$ $+m_{c} \geqslant 4 R$. | 10.76. Let's denote the point of intersection of the medians by $M$, and the center of the circumscribed circle by $O$. If triangle $ABC$ is not obtuse, then point $O$ lies inside it (or on its side); for definiteness, let's assume it lies inside triangle $AMB$. Then $AO + BO \leq AM + BM$, i.e., $2R \leq 2m_a/3 + 2m_b... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,472 |
10.77*. Prove that if in an acute-angled triangle $h_{a}=l_{b}=$ $=m_{c}$, then this triangle is equilateral. | 10.77. In any triangle $h_{b} \leqslant l_{b} \leqslant m_{b}$ (see problem 2.68), therefore $h_{a}=$ $=l_{b} \geqslant h_{b}$ and $m_{c}=l_{b} \leqslant m_{b}$. Consequently, $a \leqslant b$ and $b \leqslant c$ (see problem 10.1), i.e., $c-$ is the largest side, and $\gamma$ is the largest angle.
From the equality $h... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,473 |
10.78*. In an acute-angled triangle $A B C$, the altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ are drawn. Prove that the perimeter of triangle $A_{1} B_{1} C_{1}$ does not exceed half the perimeter of triangle $A B C$. | 10.78. According to problem 1.60, the ratio of the perimeters of triangles $A_{1} B_{1} C_{1}$ and $A B C$ is $r / R$. Moreover, $r \leqslant R / 2$ (problem 10.26).
Remark. Using the result of problem 12.72, it is easy to verify that $S_{A_{1} B_{1} C_{1}} / S_{A B C}=r_{1} / 2 R_{1} \leqslant 1 / 4$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,474 |
10.79*. Let $\angle A < \angle B < \angle C < 90^{\circ}$. Prove that the center of the inscribed circle of triangle $ABC$ lies inside triangle $BOH$, where $O$ is the center of the circumscribed circle, and $H$ is the orthocenter. | 10.79. Let $A A_{1}$ and $B B_{1}$ be the angle bisectors of triangles $O A H$ and $O B H$. According to problem 2.1, they are the angle bisectors of angles $A$ and $B$, i.e., the incenter of the inscribed circle is the point of intersection of the lines $A A_{1}$ and $B B_{1}$. From the inequality $A C > B C$, it foll... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,475 |
10.80*. Let $h$ be the greatest height of an acute-angled triangle. Prove that $r+R \leqslant h$. | 10.80. Let $90^{\circ} \geqslant \alpha \geqslant \beta \geqslant \gamma$. Then $C H-$
Fig. 10.2 is the largest height. The centers of the inscribed and circumscribed circles are denoted b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,476 |
10.81*. On the sides $B C, C A$, and $A B$ of an acute-angled triangle $A B C$, points $A_{1}, B_{1}$, and $C_{1}$ are taken. Prove that
$$
2\left(B_{1} C_{1} \cos \alpha+C_{1} A_{1} \cos \beta+A_{1} B_{1} \cos \gamma\right) \geqslant a \cos \alpha+b \cos \beta+c \cos \gamma
$$ | 10.81. Let $B_{2} C_{2}$ be the projection of the segment $B_{1} C_{1}$ onto the side $B C$. Then $B_{1} C_{1} \geqslant B_{2} C_{2}=B C-B C_{1} \cos \beta-C B_{1} \cos \gamma$. Similarly, $A_{1} C_{1} \geqslant A C-A C_{1} \cos \alpha-C A_{1} \cos \gamma$ and $A_{1} B_{1} \geqslant A B-A B_{1} \cos \alpha-B A_{1} \cos... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 27,477 |
10.82*. Prove that a triangle with sides $a, b$ and $c$ is acute-angled if and only if $a^{2}+b^{2}+c^{2}>8 R^{2}$. | 10.82. Since $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+2 \cos \alpha \cos \beta \cos \gamma=1$ (problem 12.39, b)), the triangle $A B C$ is acute-angled if and only if $\cos ^{2} \alpha+\cos ^{2} \beta+$ $+\cos ^{2} \gamma<1$. Multiplying both sides of the last inequality by $4 R^{2}$, we obtain the required r... | proof | Geometry | proof | Yes | Yes | olympiads | false | 27,478 |
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