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9.1. (Yugoslavia, 81). In an acute-angled scalene triangle, a height is drawn from one vertex, a median from another, and a bisector from the third. Prove that if the lines intersect to form a triangle, it cannot be equilateral. | 9.1. Suppose the opposite, i.e., the height $A H$, the median $B M$, and the bisector $C L$ of an acute, non-equilateral triangle $A B C$ (Fig. 2), intersect to form an equilateral triangle. The points of intersection of segments $A H$ and $B M$, $B M$ and $C L$, $C L$ and $A H$ are denoted respectively as $P$, $Q$, $R... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,939 |
9.2. (Belgium, 77). Prove that if positive numbers $a, b, c \in \mathbf{R}$ are given and for each value $n \in \mathbf{N}$ there exists a triangle with sides $a^{n}, b^{n}, c^{n}$ respectively, then all these triangles are isosceles. | 9.2. Without loss of generality, we can assume that $a \leqslant b \leqslant c$. If $c>b$, then
$$
\lim _{n \rightarrow \infty} \frac{b^{n}}{c^{n}}=0, \quad \lim _{n \rightarrow \infty} \frac{a^{n}}{c^{n}}=0
$$
and for sufficiently large values of $n \in N$, the inequality $a^{n}+b^{n}>c^{n}$, necessary for the exist... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,940 |
9.3. (Sweden, 82). Find all values of $n \in \mathbf{N}$ for each of which there exists a number $m \in \mathbf{N}$, a triangle $A B C$ with sides $A B=33, A C=21, B C=n$ and points $D$, $E$ on sides $A B, A C$ respectively, satisfying the conditions $A D=D E=E C=m$. | 9.3. Let the numbers $m, n \in \mathbf{N}$ satisfy the condition of the problem. Then $m = CE < AC = 21$ and from triangle $ADE$ (Fig. 3) we have $21 - m = AE < AD + DE = 2m$, hence
$$
7 < m < 21
$$
Furthermore, since $AD = DE$, for the angle $\alpha = \angle BAC$ we find
$$
\cos \alpha = AE / (2 \cdot AD) = (21 - m... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,941 |
9.4. (Jury, SRV, 79). Find all triples of numbers $a, b, c \in \mathbf{N}$, which are the lengths of the sides of a triangle with the diameter of the circumscribed circle equal to 6.25. | 9.4. Let $a, b, c \in \mathbf{N}$ be the lengths of the sides of a triangle with a diameter $2 R=6.25$ of the circumscribed circle, area $S$, and semiperimeter $p=(a+b+c) / 2$. Since $a, b, c \leqslant 2 R$, we have
$$
a, b, c \in\{1 ; 2 ; 3 ; 4 ; 5 ; 6\}
$$
Further, we have the equalities
$$
(a b c)^{2}=(4 S \cdot ... | 5,5,6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,942 |
9.5. (New York, 78). Triangles $A B C$ and $D E F$ are inscribed in the same circle. Prove that the equality of their perimeters is equivalent to the condition
$\sin \angle A+\sin \angle B+\sin \angle C=\sin \angle D+\sin \angle E+\sin \angle F$. | 9.5. Let $R$ be the radius of the circle in which triangles $ABC$ and $DEF$ are inscribed. Then, by the Law of Sines, we have
$$
\sin \angle A + \sin \angle B + \sin \angle C = \frac{BC + AC + AB}{2R} = \frac{p_1}{2R}
$$
where $p_1$ is the perimeter of triangle $ABC$. Similarly,
$$
\sin \angle E + \sin \angle D + \s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,943 |
9.6. (SFRY, 81). A line divides a triangle into two parts of equal areas and perimeters. Prove that the center of the inscribed circle lies on this line. | 9.6. Let a line intersect the sides of triangle $ABC$ at points $K$ and $M$, lying on sides $AB$ and $AC$ respectively (Fig. 4). We will prove that the equality
$$
\frac{S_{AKM}}{S_{ABC}}=\frac{AK+AM}{AB+AC+BC}
$$
holds if and only if the line $KM$ passes through the center of the circle inscribed in triangle $ABC$. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,944 |
9.7. (Austria, 83). In triangle $A B C$, points $C^{\prime}, B^{\prime}$, and $A^{\prime}$ are taken on sides $A B, A C$, and $B C$ respectively, such that segments $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$ intersect at one point. Points $A^{\prime \prime}$, $B^{\prime \prime}$, and $C^{\prime \prime}$ are symme... | 9.7. Let the segments $A A^{\prime}, B B^{\prime}, C C^{\prime}$ intersect at point $O$ (Fig. 5)
and
$\angle A O B=\varphi$. Then we have the equalities
$$
\begin{aligned}
& 2 S_{A O R}=A O \cdot B O \sin \varphi \\
& 2 S_{A O B^{\prime}}=A O \cdot B^{\prime} O \sin \varphi \\
& 2 S_{B O A^{\prime}}=B O \cdot A^{\pr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,945 |
9.8. (Austria, 71). The medians of triangle $ABC$ intersect at point $O$. Prove that
$$
A B^{2}+B C^{2}+C A^{2}=3\left(O A^{2}+O B^{2}+O C^{2}\right)
$$ | 9.8. Let $\overrightarrow{A B}=\boldsymbol{c}, \overrightarrow{B C}=\boldsymbol{a}, \overrightarrow{C A}=\boldsymbol{b}$, then
$$
\overrightarrow{A O}=\frac{1}{3}(c-b), \overrightarrow{B O}=\frac{1}{3}(a-c), \overrightarrow{C O}=\frac{1}{3}(b-a)
$$
and
$$
\begin{aligned}
&\left(\overrightarrow{A B}^{2}+\overrightarr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,946 |
9.9. (New York, 79). Prove that if the center of gravity of a triangle coincides with the center of gravity of its boundary, then the triangle is equilateral.
| 9.9. Let the center of gravity of the boundary of triangle $ABC$ coincide with the center of gravity of the triangle itself, i.e., with the point $O$ where its medians intersect. Denote the lengths of the sides $BC, CA, AB$ by $a, b, c$ respectively, and their midpoints (which are their centers of gravity) by $A_1, B_1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,947 |
9.10. (England, 83). Let $O$ be the center of the circumcircle of triangle $ABC$, $D$ be the midpoint of side $AB$, and $E$ be the point of intersection of the medians of triangle $ACD$. Prove that if $AB = AC$, then $OE \perp CD$. | 9.10. From the relations (Fig. 6)
$$
\begin{gathered}
\overrightarrow{O E}=\frac{1}{3} \overrightarrow{(O C}+\overrightarrow{O A}+\overrightarrow{O D})=\frac{1}{3}\left(\overrightarrow{O C}+\frac{3}{2} \overrightarrow{O A}+\frac{1}{2} \overrightarrow{O B}\right) \\
\overrightarrow{C D}=\frac{1}{2}(\overrightarrow{C A}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,948 |
9.11. (ČSSR, 72). Find all pairs of positive numbers $a, b \in \mathbf{R}$, for which there exist a right triangle $C D E$ and points $A, B$ on its hypotenuse $D E$, satisfying the conditions $\overrightarrow{D A}=\overrightarrow{A B}=\overrightarrow{B E}$ and $A C=a$, $B C=b$. | 9.11. Let $\overrightarrow{C A}=x, \overrightarrow{C B}=\boldsymbol{y}$, then (Fig. 7)
$$
\overrightarrow{D A}=\overrightarrow{B E}=\overrightarrow{A B}=y-x, \overrightarrow{C E}=2 y-x, \overrightarrow{C D}=2 x-y ;
$$
the condition $C D \perp C E$ is equivalent to the equality
$$
(2 x-y)(2 y-x)=0 \text {, or } 5(x-y... | \frac{1}{2}<\frac{}{b}<2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,949 |
9.12. (New York, 76). Find at least one right triangle with integer sides, each angle of which can be divided into three equal parts using a compass and straightedge. | 9.12. Let's choose an angle $\alpha$ that satisfies the conditions $6 \alpha < 90^{\circ}$ and $\operatorname{tg} \alpha \in Q$ (for example, the value $\operatorname{tg} \alpha = 1 / 4$ is suitable). Then each of the numbers
$$
\begin{aligned}
& \operatorname{tg} 2 \alpha = \frac{2 \operatorname{tg} \alpha}{1 - \oper... | AB=4913,AC=495,BC=4888 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,950 |
9.13. (Finland, 80). Perpendiculars drawn through the midpoints of sides $A B$ and $A C$ of triangle $A B C$ intersect the line $B C$ at points $X$ and $Y$ respectively. Prove that the equality $B C = X Y$ : a) is satisfied if
$$
\operatorname{tg} \angle B \cdot \operatorname{tg} \angle C = 3
$$
b) can also be satisf... | 9.13. In triangle $ABC$, we denote
$$
\alpha=\angle A, \beta=\angle B, \gamma=\angle C, a=BC, b=AC, c=AB
$$
$R$ is the radius of the circumscribed circle. The equality $BC=XY$ holds if and only if $\overrightarrow{BC}=\overrightarrow{YX}$ or $\overrightarrow{BC}=\overrightarrow{XY}$, which
. The altitudes of an acute-angled triangle $A B C$ intersect at point $O$, and points $B_{1}$ and $C_{1}$ are chosen on segments $O B$ and $O C$ such that
$$
\angle A B_{1} C=\angle A C_{1} B=90^{\circ}
$$
Prove that $A B_{1}=A C_{1}$. | 9.14. Let $B_{2}$ and $C_{2}$ be the feet of the altitudes dropped to sides $A C$ and $A B$ respectively (Fig. 10). Then from the relations $\triangle A B_{1} C \sim \triangle A B_{2} B_{1}, \triangle A B B_{2} \sim \triangle A C C_{2}, \triangle A C_{1} B \sim \triangle A C_{2} C_{1}$
(each of the indicated pairs of ... | AB_{1}=AC_{1} | Geometry | proof | Yes | Yes | olympiads | false | 28,952 |
9.15. (England, 81). The altitudes of triangle $ABC$ intersect at point $O$, and points $A_{1}, B_{1}, C_{1}$ are the midpoints of sides $BC, CA, AB$ respectively. The circle with center $O$ intersects line $B_{1}C_{1}$ at points $D_{1}, D_{2}$, line $C_{1}A_{1}$ at points $E_{1}, E_{2}$, and line $A_{1}B_{1}$ at point... | 9.15. Let $A_{2}, B_{2}, C_{2}$ be the feet of the altitudes dropped to the sides $B C, C A, A B$ respectively. Then the following equalities hold:
$$
A O \cdot A_{2} O = B O \cdot B_{2} O = C O \cdot C_{2} O
$$
(The first equality follows from the similarity of right triangles $A O B_{2}$ and $B O A_{2}$, and the se... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,953 |
9.16. (SFRY, 83). Inside triangle $ABC$, a point $P$ is taken, and points $M$ and $L$ are taken on sides $AC$ and $BC$ respectively, such that
$$
\angle PAC = \angle PBC, \angle PLC = \angle PMC = 90^{\circ}
$$
Prove that if $D$ is the midpoint of side $AB$, then $DM = DL$. | 9.16. Let $E$ and $F$ be the midpoints of segments $AP$ and $BP$, respectively. Then, since $DE$ and $DF$ are the midlines of triangle $APB$,
quadrilateral $DFPE$ is a parallelogram, and f... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,954 |
9.17. (SRP, 78). Find the geometric locus of points $M$, lying inside an equilateral triangle $ABC$ and satisfying the condition
$$
\angle MAB + \angle MBC + \angle MCA = 90^{\circ}
$$ | 9.17. The condition given in the problem is satisfied for all points lying on the altitudes of the equilateral triangle \(ABC\). For example,
Fig. 13
if \(M_i\) is a point on the altitude ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,955 |
9.18. (NPR, 82). The point $B_{ij} (i, j \in \{1; 2; 3\})$ denotes the point symmetric to the vertex $A_i$ of a given scalene triangle $A_1 A_2 A_3$ with respect to its angle bisector passing through the vertex $A_j$. Prove that the lines $B_{12} B_{21}, B_{13} B_{31}$, and $B_{23} B_{32}$ are parallel. | 9.18. Let $A_{1} C_{1}$ be the bisector of triangle $A_{1} A_{2} A_{3}$ (Fig. 14). Since the image of segment $A_{2} A_{3}$ under the reflection transformation relative to the line $A_{1} C_{\mathbf{i}}$ is the segment $B_{21} B_{31}$, the lines $A_{2} A_{3}$ and $B_{21} B_{31}$ intersect at point $C_{1}$. By the prope... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,956 |
9.19. (PRL, 81). The bisectors of the internal and external angles $C$ of triangle $ABC$ intersect the line $AB$ at points $L$ and $M$ respectively. Prove that if $CL = CM$, then
$$
AC^2 + BC^2 = 4R^2
$$
where $R$ is the radius of the circumscribed circle. | 9.19. Let points \( A, L, B, M \) be located on line \( AB \) in the given order (Fig. 15; the case of their arrangement in the order \( M_{2} A, L, B \) is considered similarly), then
\[
\angle L C M = \frac{1}{2} \cdot 180^{\circ} = 90^{\circ} \text{ and } \angle C L M = 45^{\circ}
\]
. Inside triangle $A B C$, a point $M$ is taken, for which $\angle M B A=30^{\circ}, \angle M A B=10^{\circ}$. Find $\angle A M C$, if $\angle A C B=80^{\circ}$ and $A C=B C$. | 9.20. Let the height $CH$ of triangle $ABC$ intersect the line
Fig. 16
$BM$ at point $E$ (Fig. 16). Then $AE = BE$ and
\[
\begin{aligned}
& \angle EAM = \angle EAB - \angle MAB = 30^\circ... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,958 |
9.21. (England, 70). On the sides $B C$ and $A C$ of triangle $A B C$, points $D$ and $E$ are chosen respectively such that $\angle B A D=50^{\circ}, \angle A B E=30^{\circ}$. Find $\angle B E D$, if
$\angle A B C=\angle A C B=50^{\circ}$ | 9.21. Let $O$ be the point of intersection of lines $A D$ and $B E$ (Fig. 17), then $\angle A O B=180^{\circ}-30^{\circ}-50^{\circ}=100^{\circ}$, $\angle B D A=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}$, $\angle C B E=50^{\circ}-30^{\circ}=20^{\circ}$, $\angle A E B=\angle C B E+\angle E C B=70^{\circ}$, $\angle C A... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,959 |
9.22. (GDR, 64). On the side $BC$ of triangle $ABC$, a point $P$ is taken such that $PC = 2BP$. Find $\angle ACB$ if $\angle ABC = 45^\circ$ and $\angle APC = 60^\circ$. | 9.22. If point $C_{1}$ is symmetric to point $C$ with respect to line $A P$ (Fig. 18), then $C_{1} P = C P = 2 B P$ and $\angle C_{1} P B = 180^{\circ} - \angle A P C - \angle A P C_{1} = 180^{\circ} - 60^{\circ} - 60^{\circ} = 60^{\circ}$. Therefore, $\angle C_{1} B P = 90^{\circ}$ (since triangle $C_{1} P B$ is simil... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,960 |
9.23*. (Jury, Netherlands, 79). Inside an equilateral triangle $ABC$, points $K, L, M$ are taken such that $\angle KAB = \angle LBA = 15^{\circ}, \quad \angle MBC = \angle KCB = 20^{\circ}$, $\angle LCA = \angle MAC = 25^{\circ}$. Find the angles of triangle $KLM$.[^2]
## § 10. Circles and Circles
(see Appendix G: de... | 9.23. Let $\alpha=20^{\circ}, \beta=25^{\circ}, \gamma=15^{\circ}$ (Fig. 19). Then $\alpha, \beta, \gamma<30^{\circ}, \alpha+\beta+\gamma=60^{\circ}, \angle K A M=60^{\circ}-\angle M A C-\angle K A B=\alpha$ and similarly $\angle L B M=\beta, \angle K C L=\gamma$. Let segments $A M$ and $C L$ intersect at point $N$, li... | \angleKLM=60,\angleLKM=75,\angleKML=45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,961 |
10.1. (Brazil, 83). Prove that all points of a circle can be divided into two sets such that among the vertices of any inscribed right triangle, there are points from both sets. | 10.1. Let's divide all points on the circle into pairs of diametrically opposite points and in each pair, assign one point (any one) to the first set, and the other to the second set. Since the hypotenuse of any inscribed right triangle is the diameter of the circle, the vertices of the acute angles of such a triangle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,962 |
10.2. (New York, 75). Prove that the four distances from a point on the circle to the vertices of a square inscribed in it cannot all be rational numbers. | 10.2. Let the square $A B C D$ be inscribed in a circle with diameter 4, and let point $P$ lie on the side $A D$ (Fig. 20). Denote $\alpha=$ $=\angle A C P$. Then, if the numbers $A P=d \sin \alpha$ and $C P=d \cos x$
. The bisectors $A A_{1}, B B_{1}$, $C C_{1}$ of triangle $A B C$ intersect at point $M$. Prove that if the radii of the circles inscribed in triangles $M B_{1} A, M C_{1} A, M C_{1} B, M A_{1} B, M A_{1} C$ and $M B_{1} C$ are equal, then triangle $A B C$ is equilateral. | 10.3. Consider equal circles with centers \(O_{1}\) and \(O_{2}\), inscribed in triangles \(A C_{1} M\) and \(A B_{1} M\) respectively. Let these circles touch the segment \(A M\) at points \(K_{1}\) and \(K_{2}\) (Fig. 21). Then the right triangles \(A O_{1} K_{1}\) and \(A O_{2} K_{2}\) are equal, since
\[
O_{1} K_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,964 |
10.4. (England, 75). Seven points in a circle of unit radius are arranged so that the distance between any two of them is at least 1. Prove that one of the points coincides with the center of the circle. | 10.4. Let, contrary to the statement of the problem, none of the points $O_{1}, \ldots, \ldots, O_{7}$, arranged in the order of traversal around the center $O$ of the given circle clockwise (Fig. 22), coincides with point $O$. Since the sum of the angles $\angle O_{1} O O_{2}, \quad \angle O_{2} O O_{3}, \quad \cdots,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,965 |
10.5. (Beijing, 62). Six circles on a plane are arranged so that the center of each of them lies outside the other circles. Prove that all six circles do not have a common point. | 10.5. Let, contrary to the assertion of the problem, there exist a point $O$ belonging to all six circles. Denote by $O_{1}, O_{2}, O_{3}, O_{4}$, $O_{5}, O_{6}$ the centers of these circles in the order of traversal around point $O$ clockwise (see Fig. 22; according to the condition, point $O$ cannot be the center of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,966 |
10.6. (Austria-Poland, 78). On a plane, there are non-intersecting circles, each of which touches at least six of the other circles. Prove that the set of circles is infinite. | 10.6. Suppose that the set of circles is finite. Then the circle with center $O$ and the smallest radius $r$ touches six circles with centers $O_{1}, \ldots, O_{6}$ (arranged in clockwise order around point $O$; see Fig. 22) and radii $r_{1}, \ldots, r_{6}$, respectively. In the triangle $\mathrm{O}_{1} \mathrm{OO}_{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,967 |
10.7. (MS, 84). Prove that for any triangle $A B C$ there exist 3 circles of the same radius, one of which touches the sides $A B$ and $B C$, another the sides $B C$ and $A C$, and the third the sides $A C$ and $A B$, and all 3 circles have exactly one common point. | 10.7. Let there be a triangle $A B C$. Construct a triangle $A^{\prime} B^{\prime} C^{\prime}$ similar to it, for which the circles mentioned in the problem exist (then the existence of the corresponding circles for the original triangle $A B C$ will also be proven). For this, we draw 3 circles with centers at points $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,968 |
10.8. (MMS, Luxembourg, 80). Two circles touch each other at point $P$. A line tangent to one of them at point $A$ intersects the other at points $B$ and $C$. Prove that the line $P A$ is the bisector of angle ВРС or the adjacent angle. | 10.8. Consider a homothety with center $P$, under which a circle containing points $B$ and $C$ is transformed into another circle. Under this homothety, points $B$ and $C$ are transformed into some points $B^{\prime}$ and $C^{\prime}$, lying on the lines $B P$ and $C P$ respectively, and the line $B C$ is transformed i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,969 |
10.9. (Jury, USA, 79). A circle, the center of which lies on the side $B C$ of an isosceles triangle $A B C$, touches its equal sides $A B$ and $A C$. Prove that a segment with endpoints $P$ and $Q$, lying on the sides $A B$ and $B C$ respectively, touches the circle if and only if $B P \cdot C Q = B C^{2} / 4$. | 10.9. Let the segment $P Q$ touch the circle with center $O$ (coinciding with the midpoint of the base $B C$ of the isosceles triangle $A B C$). Denote
$\angle P B O=\angle Q C O=\alpha, \quad \angle B P O=\angle Q P O=\beta, \quad \angle C Q O=\angle P Q O=\gamma$ (Fig. 26), then, considering the quadrilateral $C B P... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,970 |
10.10. (England, 80). On the diameter $A B$ of a semicircle,
points $K$ and $L$ are taken, and on the semicircle—points $M, N$ and $C$ such that the quadrilateral $K L M N$ is a square, the area of which is equal to the area of triangle $A B C$. Prove that the center of the inscribed circle in triangle $A B C$ coincide... | 10.10. Let $a$ be the side of the square, $R$ be the radius of the semicircle with center $O$, and $r$ be the radius of the circle inscribed in triangle $ABC$,
$$
\alpha=\angle AON=45^{\circ} \text{, then we have }
$$
$$
\sin 2 \alpha=2 \sin \alpha \cos \alpha=2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}=\frac... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,971 |
10.12. (GDR, 73). Given a convex quadrilateral $A B C D$, in which $A B=A D$ and $C B=C D$. Prove that:
a) a circle can be inscribed in it;
b) a circle can be circumscribed around it if and only if $A B \perp B C$;
c) if $A B \perp B C$, then the square of the distance between the center of the inscribed circle (rad... | 10.12. a) According to the problem condition
$$
A B+C D=A D+B C
$$
therefore, a circle can be inscribed in the quadrilateral $A B C D$.
6) From the equality of triangles $A B C$ and $A D C$, it follows that $\angle B=\angle D$. Therefore, a circle can be circumscribed around the quadrilateral $A B C D$ if and only i... | R^{2}+r^{2}-r\sqrt{r^{2}+4R^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 28,973 |
10.13. (CRR, 78). Prove that the four vertices of a square cannot be located respectively on four concentric circles, the radii of which form an arithmetic progression. | 10.13. If the vertices of a square were located on concentric circles with radii $a, a+d, a+2d, a+3d$ respectively, then by Theorem 79, one of the following equalities would hold:
\[
\begin{aligned}
a^{2}+(a+d)^{2} & =(a+2d)^{2}+(a+3d)^{2} \\
a^{2}+(a+2d)^{2} & =(a+d)^{2}+(a+3d)^{2} \\
(a+d)^{2}+(a+2d)^{2} & =a^{2}+(a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,974 |
10.14. (SFRY, 83). On the arc $AB$ of the circumcircle of rectangle $ABCD$, a point $M$ is taken, distinct from vertices $A, B$. Points $P, Q, R$, and $S$ are the projections of point $M$ onto the lines $AD, AB, BC$, and $CD$ respectively. Prove that the lines $PQ$ and $RS$ are perpendicular and intersect on one of the... | 10.14. Let $L$ be a point of intersection of the circle with the line $QS$ different from $M$ (Fig. 30). Then
$$
\overrightarrow{RS}=\overrightarrow{MS}-\overrightarrow{MR}=\overrightarrow{QL}-\overrightarrow{QB}, \overrightarrow{PQ}=\overrightarrow{AQ}+\overrightarrow{MQ}
$$
from which, using Theorem 71, we get
$$
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,975 |
10.15. (England, 77). The side $B C$ of triangle $A B C$ touches the inscribed circle at point $D$. Prove that the center of the circle lies on the line passing through the midpoints of segments $B C$ and $A D$. | 10.15. Let the inscribed circle touch the sides \(AB\) and \(AC\) at points \(\boldsymbol{E}\) and \(\boldsymbol{F}\) respectively, and the exscribed circle touch the side \(CB\) at point \(L\) and the extensions of sides \(AB\) and \(AC\) at points \(M\) and \(N\) respectively (Fig. 31). Then we have
\[
\begin{aligne... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,976 |
10.16. (Austria, 72). Two circles touch each other. A regular triangle is inscribed in the larger circle, and tangents are drawn from its vertices to the smaller circle. Prove that the length of one of the three tangents is equal to the sum of the lengths of the other two. | 10.16. Let $D$ be the point of tangency of the circles, and $ABC$ be an equilateral triangle inscribed in the larger of them. Without loss of generality, we can assume that point $D$ lies on the arc $AB$ (Fig. 32). We will prove that $DC = DA + DB$. For this, we take a point $M$ on the segment $DC$ such that $AD = DM$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,977 |
10.17. (Jury, USA, 79). From a point $P$ lying on the arc $B C$ of the circumcircle of triangle $A B C$, perpendiculars $P K, P L$ and $P M$ are dropped to the lines $B C$,
$A C$ and $A B$ respectively. Prove that
$$
\frac{B C}{P K}=\frac{A C}{P L}+\frac{A B}{P M}
$$ | 10.17. On the segment $BC$ there exists a point $N$ such that $\angle PNB = \angle PCA$ (since $\angle PCB < \angle PCA = 180^\circ - \angle PBA < 180^\circ - \angle PBC$, see Fig. 33). Then we have
\[
\begin{aligned}
& \triangle BPN \sim \triangle APC \text{ and } \\
& \triangle CPN \sim \triangle APB_1
\end{aligned}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,978 |
10.18. (GDR, 70; SFRY, 72). a) Let point $O$ be the center of the inscribed circle of triangle $ABC$, and point $D$ be the point of intersection of line $AO$ with the circumcircle of triangle $ABC$ other than $A$. Prove that $DB = DC = DO$.
b) Prove that if $ABCD$ is a cyclic quadrilateral, then the centers $A_1, B_1,... | 10.18. a) In the conditions of the problem, let $\angle B A C=\alpha, \angle A B C=\beta$, $\angle B C A=\gamma$ (Fig. 34). Then we have $\angle D A B=\angle D A C=\alpha / 2$, from which $B D=D C$. Since
$$
\begin{aligned}
\angle O D C & =\angle A B C=\beta \\
\angle O C D=\angle O C B+\angle B C D & =\frac{1}{2} \an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,979 |
10.19*. (Balkan, 84). Prove that the inscribed quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is equal to the quadrilateral whose vertices $H_{1}, H_{2}, H_{3}, H_{4}$ are the points of intersection of the altitudes of the triangles
$$
A_{2} A_{3} A_{4}, A_{1} A_{3} A_{4}, A_{1} A_{2} A_{4}, A_{1} A_{2} A_{3}
$$
respectivel... | 10.19. First, let's prove that the midpoints of segments \(A_{1} H_{1}\) and \(A_{2} H_{2}\) coincide. For this, draw a line through point \(A_{3}\) perpendicular to side \(A_{3} A_{4}\), and denote by \(K\) the point of intersection of this line with the circumcircle of quadrilateral \(A_{1} A_{2} A_{3} A_{4}\) (see F... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,980 |
10.20*. a) (Jury, USA, 82). A diagonal of a given cyclic quadrilateral divides it into two triangles. Prove that the sum of the radii of the circles inscribed in these triangles does not depend on the choice of the diagonal.
b) (BHR, 78). Prove that the greatest height of an acute-angled triangle is not less than the ... | 10.20. a) Let the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ be inscribed in a circle with center $O$ and radius $R$, and the projections of point $O$ onto the chords $A_{1} A_{3}$, $A_{1} A_{2}$, $A_{2} A_{3}$, $A_{3} A_{4}$, $A_{4} A_{1}$ are their midpoints $H_{0}, H_{1}, H_{2}, H_{3}, H_{4}$, respectively. Denote
$$
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,981 |
10.21*. (SFRY, 77). On a plane, there are 100 points. Prove that there exists a finite set of circles satisfying the following three conditions: 1) any of the given points lies inside one of the circles; 2) any two points in different circles are more than 1 unit apart; 3) the sum of the diameters of all the circles is... | 10.21. Consider the following procedure, which consists of performing a certain number of steps. On the first step, cover each of the given points with a circle of diameter $1 / 200$. Suppose that after the $k$-th step $(k \in \mathbb{N})$ there exist two circles that are no more than 1 unit apart. Let $\mathrm{O}_{1}$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,982 |
10.22*. (Beijing, 63). On a plane, there are $2 n+3$ points, no three of which lie on the same line and no four of which lie on the same circle. Does there exist a circle passing through any three of these points and bounding a disk inside which lies exactly half of the remaining points? | 10.22. Let's take a line such that all points lie in one half-plane relative to it, and move it parallel until it passes through the first point $A_{\mathbf{i}}$. Then we will rotate the obtained line around point $A_{1}$ until it passes through another point $A_{2}$ for the first time. Then all other points lie in one... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,983 |
10.23*. (NPR, 78). Prove that for any convex polygon, there exists a triplet of consecutive vertices such that the circle passing through them bounds a disk that covers the entire polygon. | 10.23. Among all circles passing through any three vertices of a given polygon, two of which are adjacent, and the side connecting them is seen from the third vertex at an angle not exceeding $90^{\circ}$
. Among $n>1$ pairs of opposite sides in an inscribed $2 n$-gon, $n-1$ pairs of parallel sides are chosen. Find all values of $n$ for which the sides of the remaining pair are necessarily parallel. | 10.24. We will prove that the sought values of $n$ are all odd numbers $n>1$. Let $n$ be odd and in the $2n$-gon $A_{i} \ldots A_{2n}$, all pairs of opposite sides, except possibly the pair $A_{1} A_{n}, A_{n} A_{n+1}$, are composed of parallel segments. If (Fig. 40) $A_{1} A_{2} A_{n+1} = 180^{\circ} + \alpha$, then
... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,985 |
10.25*. (NRS, 82). On a plane, there are $n$ distinct circles of radius 1 each. Prove that at least one of them contains an arc that does not intersect with any of the other circles and has a length of at least $2 \pi / n$.
## § 11. Polygons
(see Appendix G: definitions $1,2,35,37$; theorems $2,65,70,73,74,80$ ) | 10.25. For \( n=1 \), the full arc of the only circle has a length of \( 2 \pi \geqslant 2 \pi / n \), i.e., the statement of the problem is true. Let \( n \geqslant 2 \). Suppose first that the centers of all \( n \) circles lie on one straight line. Choose the center such that all other centers lie on one ray relativ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,986 |
11.1. (SFRY, 8I; Italy, 82). Prove that if for some point $O$, lying inside the quadrilateral $A B C D$, the areas of the triangles $A B O, B C O, C D O$, $D A O$ are equal, then this point lies on at least one of the diagonals $A C$ or $B D$. | 11.1. Let diagonal $A C$ intersect lines $O B$ and $O D$ at points $P$ and $Q$ respectively (Fig. 42). Since the areas of triangles $A O B$ and $C O B$ are equal, their heights to the common side $O B$ are also equal, and
. The areas of quadrilaterals $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are $S$ and $S^{\prime}$ respectively. Prove that if there exists a point 0 inside quadrilateral $A B C D$ such that
$$
\overrightarrow{O A}=\overrightarrow{A^{\prime} B^{\prime}}, \overrightarrow{O B}=\overrigh... | 11.2. Parallelograms $O A M B$ and $A^{\prime} B^{\prime} C^{\prime} N$, constructed on vectors $\overrightarrow{O A}, \overrightarrow{O B}$ and $\overrightarrow{A^{\prime} B^{\prime}}, \overrightarrow{B^{\prime} C^{\prime}}$ (Fig. 43), are equal, therefore we have
$$
S_{A O B}=(1 / 2) S_{O A M B}=(1 / 2) S_{A^{\prime... | 2S^{\} | Geometry | proof | Yes | Yes | olympiads | false | 28,988 |
11.3. (SFROI, 72). Eight lines connecting the vertices of a parallelogram to the midpoints of non-adjacent sides, intersecting, form an octagon. Prove that its area is one-sixth of the area of the parallelogram. | 11.3. Let $A_{1}, B_{1}, C_{1}, D_{1}$ be the midpoints of the sides $A B, B C, C D, D A$ of parallelogram $A B C D$ with area $S$, and let $K, K_{1}, K_{2}$ be the points of intersection of line $A C_{1}$ with lines $B D_{1}, A_{1} D, C D_{1}$, respectively. Similarly, we obtain points $L, L_{1}, L_{2}, M, M_{1}, M_{2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,989 |
11.4. (SFRY, 70). The diagonals of a convex pentagon $A B C D E$, intersecting, form a pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ and a five-pointed star.
a) Find the sum of the angles of this star at the vertices $A$, $B, C, D, E$.
b) Find the ratio of the area of the pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ to the ar... | 11.4. b) The sum of the interior angles of the pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ and the decagon $A A_{1} B B_{1} C C_{1} D D_{1} E E_{1}$ are respectively $3 \cdot 180^{\circ}$ and 8 $\cdot 180^{\circ}$. Moreover, each angle of this pentagon, together with the angle of the decagon at the same vertex, sums to $3... | \frac{7-3\sqrt{5}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 28,990 |
11.5. (Jury, France, 79). The corresponding sides of quadrilaterals $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are equal. Prove that one of the following two statements holds:
a) $B D \perp A C$ and $B^{\prime} D^{\prime} \perp A^{\prime} C^{\prime}$;
b) the points $M$ and $M^{\prime}$ of intersection ... | 11.5. Let in quadrilateral $A B C D$ point $O$ be the midpoint of segment $B D$, and points $K$ and $L$ be the projections of vertices $A$ and $C$ onto line $B D$ respectively; in this case, $\overrightarrow{O K}=x \overrightarrow{B D}, \overrightarrow{O L}=y \overrightarrow{B D}$ (corresponding elements in quadrilater... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,991 |
11.6. (Austria, 73). Prove that if all angles of a convex octagon are equal, and the ratio of the lengths of any two adjacent sides is rational, then the opposite sides of this octagon are equal. | 11.6. Without loss of generality, we can assume that the lengths of the sides of the given octagon $A_{1} A_{2} \ldots A_{8}$ are rational numbers (otherwise, we will prove the required statement for a similar octagon $A_{1}^{\prime} A_{2}^{\prime} \ldots A_{\mathbf{8}}^{\prime}$, where $A_{1}^{\prime} A_{2}^{\prime}=1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,992 |
11.7. (SFRY, 76). On a plane, a regular hexagon with side $a$ is drawn. For any value of $n \in \mathbf{N}$, greater than 1, construct a segment of length $a / n$ using only a straightedge. | 11.7. Given a regular hexagon $A_{0} A_{1} B_{2} C_{2} C_{1} B_{0}$ (Fig. 48), we will construct the following points using a ruler: $A_{2}$ and $A_{8}$ at the intersection of line $A_{0} A_{1}$ with lines $C_{2} B_{2}$ and $C_{1} B_{2}$, respectively,
. Prove that if for the angles of an equilateral convex pentagon $A B C D E$ the inequalities $\angle A \geqslant \angle B \geqslant \angle C \geqslant \angle D \geqslant \angle E$ are satisfied, then this pentagon is regular. | 11.8. Since
$$
\begin{aligned}
A C=2 \cdot A B \sin ( & \angle B / 2) \geqslant \\
& \geqslant 2 \cdot C D \sin (\angle D / 2)=C E
\end{aligned}
$$
(Fig. 49), from triangle $A C E$ we have $\angle A E C \geqslant \angle E A C$. On the other hand, we obtain
$$
\begin{aligned}
& \angle E A C=\angle A-\left(180^{\circ}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,994 |
11.9. (GDR, 74; GDR, 79). a) Prove that if the vertices of one convex $n$-gon lie within another equal $n$-gon, then the vertices of these $n$-gons coincide.
b) Is the statement in a) true for non-convex polygons?
c) Is the statement in a) false for any non-convex polygon? | 11.9. a) If the vertices of the first polygon lie within the second, equal convex polygon, then the first polygon is entirely contained within the second. From the equality of their areas, it follows that they coincide.
0) Incorrect (see Fig. 50, which shows equal non-convex quadrilaterals \(ABCE\) and \(ACDE\)).
b) ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,995 |
11.10. (Jury, Belgium, 79). Can any correct $2 n$-gon be divided into rhombuses? | 11.10. We will prove a more general statement: any equilateral 2n-gon, the opposite sides of which are parallel, can be divided into rhombuses. For $n=2$, the statement is true, as an equilateral quadrilateral is already a rhombus. Suppose the statement is proven for some value $n \geqslant 2$ and we have a $2(n+1)$-go... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,996 |
11.11. (GDR, 64). Outside the rhombus $A B C D$ with a given side $a$, a point $O$ is taken, which is at a given distance $b>a$ from each of the vertices $A$ and $C$; Prove that the product $O B \cdot O D$ does not depend on the size of the angle $B A D$.
Taking a point $O$ outside the rhombus $A B C D$ with a given s... | 11.11. Since each of the points $O, D, B$ is equidistant from vertices $A$ and $C$, these points lie on the same line. Consider a circle with center $A$ and radius $a$ (Fig. 53). By Theorem 73, the product $O B \cdot O D$ is equal to the square of the length of the tangent $O E$ drawn
. Outside the parallelogram $A B C D$, a point $P$ is taken such that $\angle P A B = \angle P C B$, and vertices $A$ and $C$ lie in different half-planes relative to the line $P B$. Prove that $\angle A P B = \angle D P C$. | 11.12. Let's take a point $Q$ such that the quadrilateral $Q P A B$ is a parallelogram (Fig. 54). Then $Q P D C$ is also a parallelogram, because
$$
C D=B A=Q P, C D\|B A\| Q P
$$
Since the vertices of the equal angles $P Q B$ and $P C B$ lie on the same side relative to the line $P B$, the points $Q, P, B, C$ lie on... | proof | Geometry | proof | Yes | Yes | olympiads | false | 28,998 |
11.13. (Kuri, Belgium, 83). In the hexagon $A B C D E F$, the angles at vertices $A, C, E$ are equal and do not exceed $180^{\circ}$, and
$$
\angle A B F=\angle C B D, \angle A F B=\angle E F D
$$
Prove that if point $A^{\prime}$ is symmetric to vertex $A$ with respect to the diagonal $B F$ and does not lie on the li... | 11.13. Note that triangles $A^{\prime} E F$ and $B D F$ are similar. Indeed, we have (Fig. 55)
$$
\angle E F D = \angle B F A = \angle B F A^{\prime}, \angle F E D = \angle F A B = \angle F A^{\prime} B,
$$
. The vertices of the convex pentagon $A B C D E$ are arranged so that triangles $A B C$ and $C D E$ are equilateral. Prove that if point $O$ is the center of triangle $A B C$, and points $M$ and $N$ are the midpoints of sides $B D$ and $A E$ respectively, then triangles $O M E$ and $O N D$ are similar. | 11.14. Let $P$ and $Q$ be the midpoints of segments $B C$ and $A C$ respectively. Note that if triangle $O P M$ is rotated around point $O$ by $60^{\circ}$ (clockwise; see Fig. 56), and then a homothety with center $O$ and coefficient 2 is applied to it, it will transform into triangle $O C E$. Indeed, since
$$
\angle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,000 |
11.15. (Jury, USSR, 82). In a convex pentagon $A B C D E$, the angles at vertices $B, E$ are right angles and $\angle B A C = \angle E A D$. Prove that if diagonals $B D$ and $C E$ intersect at point $O$, then lines $A O$ and $B E$ are perpendicular. | 11.15. Let the extension of the perpendicular $A H$ to the line $B E$ intersect the lines $C E$ and $B D$ at points $P$ and $Q$ respectively. We will prove that $A P = A Q$, from which the equalities $P = Q = 0$ will follow. Drop the perpendicular $C K$ to the line $B E$ (Fig. 57), then from the similarity of right tri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,001 |
11.16. (England, 66). Find the number of sides of a regular polygon if for four of its consecutive vertices \( A, B, C, D \) the equality
\[
\frac{1}{A B}=\frac{1}{A C}+\frac{1}{A D}
\]
is satisfied. | 11.16. Let a circle with center 0 and radius $R$ be circumscribed around a polygon (Fig. 58). Denote $\alpha=\angle A O B$, then $0<\alpha<$ $<120^{\circ}$ and
$A B=2 R \sin (\alpha / 2), A C=2 R \sin \alpha$,
$$
A D=2 R \sin (3 \alpha / 2)
$$
from which we have
$$
\frac{1}{\sin (\alpha / 2)}=\frac{1}{\sin \alpha}+... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,002 |
11.17*. (England, 71). On the circle inscribed in a regular $2 n$-gon, points $A$ and $B$ are taken. Prove that if the diagonals connecting opposite vertices of the $2 n$-gon are seen from point $A$ at angles $\alpha_{i}, \ldots, \alpha_{n}$, and from point $B$ at angles $\beta_{i}, \ldots, \beta_{n}$ respectively, the... | 11.17. Let point 0 be the center of a regular $2n$-gon $C_{0} C_{1} \ldots C_{2n-1}$, and $R$ be the radius of the circumscribed circle around it. Take
Fig. 59
an arbitrary point $A$ at a ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,003 |
11.18*. (PRC, 78). The vertices of a convex polygon with an odd number of sides are colored such that any two adjacent vertices have different colors. Prove that for any coloring satisfying this condition, the polygon can be divided into triangles by non-intersecting diagonals such that the ends of each diagonal have d... | 11.18. We will prove the statement by induction on the number $n$ of sides of the polygon. For $n=3$, the statement is true, since a triangle has no diagonals. Suppose it has already been proven for some odd value $n \geqslant 3$ and consider a convex $(n+2)$-gon, the vertices of which are colored in the manner specifi... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,004 |
12.1. (New York, 80). The medians of an equilateral triangle \(ABC\) divide it into four triangles \(ADE, BDF, DEF\), and \(CEF\), on the sides of which midpoints \(K, L, M, N, O, P, Q, R, S\) are marked. Each of the 15 points obtained is colored in one of two colors. Prove that there will be 3 points of the same color... | 12.1. We will prove a stronger statement: among 10 points $A$, $D$, $E$, $K$, $L$, $M$, $N$, $O$, $P$, $Q$, arranged as shown in Fig. 60, there will be 3 points of the same color that are vertices of an equilateral triangle. Suppose this is not the case. Then, without loss of generality, we can assume that point $O$ is... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,005 |
12.2. (CIIIA, 81). On a plane, an angle of $180^{\circ} / n$ is drawn, where the number $n \in \mathbf{N}$ is not divisible by 3. Prove that this angle can be divided into 3 equal angles using a compass and a straightedge. | 12.2. If $n=3 k+1\left(k \in \mathrm{Z}^{+}\right)$, then using a compass and straightedge, we construct the angle
$$
60^{\circ}-k \cdot \frac{180^{\circ}}{n}=\frac{(3 k+1) \cdot 180^{\circ}-3 k \cdot 180^{\circ}}{3 n}=\frac{1}{3} \cdot \frac{180^{\circ}}{n},
$$
and if $n=3 k-1$ ( $k \in \mathrm{N}$ ), then we constr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,006 |
12.3. (SFRY, 72). Prove that any two diameters of a convex set in the plane have at least one common point. | 12.3. Suppose the opposite. Let two non-intersecting segments be diameters of a convex set. There are two cases: 1) neither segment intersects the extension of the other; 2) one segment intersects the extension of the other. In the first case, consider the convex quadrilateral \(ABCD\), where sides \(AB\) and \(CD\) ar... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,007 |
12.4. (GDR, 82). Can any convex quadrilateral be divided by a broken line into two parts, the diameter of each of which is less than the diameter of the original quadrilateral? | 12.4. The answer to the question of the problem is negative. Indeed, consider a convex quadrilateral $ABCD$ in which $AB=AC=BC=d, BD<d$ (Fig. 63). Its diameter is equal to $d$. On the other hand, at least two of the three vertices $A, B, C$ fall into one of the two parts into which the broken line divides this quadrila... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,008 |
12.5. (GDR, 72). Prove that if $n$ points on a plane are arranged such that any line passing through two of them contains at least one more of the points, then all of them lie on one line. | 12.5. Suppose not all points lie on the same line. Draw all possible lines through all pairs of points and consider all non-zero distances between the points and the drawn lines. Since there are a finite number of these distances, there will be a point \( A \) and a line \( l \), the distance between which is minimal. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,009 |
12.6. (Czechoslovakia, 68). Segments $A B$ and $C D$ are equal and not parallel. Find the geometric locus of points $O$ that have the following property: the segment symmetric to segment $A B$ with respect to point $O$ is symmetric to segment $C D$ with respect to some line. | 12.6. Let $A^{\prime} B^{\prime}$ denote the image of the segment $A B$ under the symmetry with respect to the desired point $O$. Consider an arbitrary line $l$ such that the image $C^{\prime} D^{\prime}$ of the segment $C D$ under the symmetry with respect to $l$ is parallel to $A^{\prime} B^{\prime}$.
. Prove that if a set on the plane has more than one center of symmetry, then it has infinitely many. | 12.7. Let the set $M$ have two distinct centers of symmetry $O_{1}$ and $O_{2}$. Then the point $O_{3}$, symmetric to the point $O_{1}$ with respect to the point $O_{2}$, is also a center of symmetry of the set $M$. Indeed,
if we denote by $C_{O}(A)$ the point symmetric to the point $A$ with respect to the point $O$, t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,011 |
12.8. (Belgium, 78). Prove that the union $L$ of the axes of symmetry of a set $M$ in the plane is contained in the union of the axes of symmetry of the set $L$.
untranslated part:
(Бельгия, 78). Доказать, что объединение $L$ осей симметрии множества $M$ на плоскости содержится в объединении осей симметрии множества ... | 12.8. Let the set $M$ have axes of symmetry $l_{0}$ and $l_{1}$ (not necessarily distinct). Then the line $l_{2}$, symmetric to the line $l_{1}$ with respect to the line $l_{0}$, is also an axis of symmetry of the set $M$. Indeed, if $S_{l}(A)$ denotes the point symmetric to the point $A$ with respect to the line $l$, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,012 |
12.9. (IMO, GDR, 79). A set on the plane has two axes of symmetry intersecting at an angle $\alpha$, where the number $\alpha / \pi$ is irrational. Prove that if this set contains more than one point, then it contains infinitely many points. | 12.9. Let the axes of symmetry $l_{0}$ and $l_{1}$ of the set $M$ intersect at point $O$, and the axis $l_{0}$, when rotated clockwise around point $O$ by an angle $\alpha$, transitions to the axis $\boldsymbol{l}_{1}$. Then, if we denote by $S_{l}(A)$ the point symmetric to point $A$ with respect to the line $l$, then... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,013 |
12.10. (SFRY, 76). Find all values of $n \in \mathbf{N}$, greater than 2, for which it is possible to select $n$ points on a plane such that any two of them are vertices of an equilateral triangle, the third vertex of which is also one of the selected points. | 12.10. We will prove that the condition of the problem is satisfied only by the value $n=3$ (in which case the points can be placed at the vertices of an equilateral triangle). Indeed, suppose that it is possible to arrange $n \geqslant 4$ points in the manner specified in the problem. We select two points $A$ and $B$,... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,014 |
12.11. (CSSR, 80). The set $M$ is obtained from the plane by removing three distinct points $A, B$, and $C$. Find the smallest number of convex sets whose union is the set $M$. | 12.11. Let points $A, B$, and $C$ initially lie on the same line. Then these points divide the line into 4 intervals, and no points from different intervals can lie in the same convex set. Therefore, the number of required sets cannot be less than 4. The number 4 is achieved if the set $M$ is divided into parts as show... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,015 |
12.12. (Jury, SRP, 79). Prove that for any value of $n \in \mathbf{N}$, greater than some number $n_{0}$, the entire plane can be divided into $n$ parts by drawing several lines, among which there must be intersecting ones. Find the smallest such value of $n_{0}$. | 12.12. Among the lines drawn according to the condition, there are necessarily intersecting lines that already divide the plane into 4 parts. If another line is drawn, then, as a simple case analysis shows, the number of parts will increase by at least 2. Therefore, it is impossible to get exactly 5 parts, from which i... | 5 | Geometry | proof | Yes | Yes | olympiads | false | 29,016 |
12.13. (Czechoslovakia, 82). On the coordinate plane, find a convex set that contains infinitely many points with both integer coordinates, but in the intersection with any line contains only a finite (or empty) set of such points. | 12.13. A set that satisfies the condition of the problem, for example, is a strip on the coordinate plane
$$
M=\{(x ; y) \mid \sqrt{2} x-1<y<\sqrt{2} x\}
$$
Indeed, this set contains infinitely many integer points of the form ( $x ;[\sqrt{2} x]$ ) for $x \in \mathbf{Z}$. On the other hand, any line $y=k x+b$ in the c... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,017 |
12.14. (GDR, 74). Find all pairs of non-zero vectors $\boldsymbol{x}, \boldsymbol{y}$, for which the sequence of numbers
$$
a_{n}=|\boldsymbol{x}-n \boldsymbol{y}|(n \in \mathbf{N})
$$
is: a) increasing; b) decreasing. | 12.14. Since $a_{n} \geqslant 0$ for all values of $n$, the increase or decrease of the sequence $\left\{a_{n}\right\}$ is equivalent to the increase or decrease of the sequence of numbers
$$
a_{n}^{2}=(x-n y)^{2}=x^{2}-2 n x y+n^{2} y^{2}
$$
which represents the sequence of values of a quadratic trinomial with a pos... | 3|y|>2|x|\cos\varphi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,018 |
12.15. (SFRY, 73). Several points on the plane are arranged so that the distance between any two of them is greater than 2. Prove that any set of area less than $\pi$ can be parallel translated along the plane by a vector of length less than 1 so that it does not contain any of the points. | 12.15. Let $M$ be a set of area less than $\pi$, and $U_{\mathbf{i}}, \ldots, U_{n}$ be unit radius circles centered at given points $A_{\mathbf{i}}, \ldots, A_{n}$ respectively,
$$
V_{i}=U_{i} \cap M \quad(i=1, \ldots, n)
$$
Since the distances between the centers of the circles are greater than 2, the circles do no... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,019 |
12.16. (Belgium, 77). Inside a circle of radius $n \in \mathbf{N}$, there are $4 n$ segments each of length 1. Prove that if a certain line is given, then there exists another line, either parallel or perpendicular to it, that intersects at least two segments. | 12.16. Note that the sum of the lengths of the projections of each segment onto a given line $l$ and a line $l^{\prime}$ perpendicular to it is not less than 1. Indeed, if a vector $\boldsymbol{a}$ of length 1 is parallel to some segment, and vectors $\boldsymbol{x}$ and $\boldsymbol{y}$ are the projections of vector $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,020 |
12.17*. (CSSR, 73). In a square with side 50, a broken line is located. Prove that if the distance from any point of the square to at least one point of the broken line is no more than 1, then the length of the broken line is greater than 1248. | 12.17. Let $U(M)$ denote the union of all circles of radius 1, the centers of which belong to a set $M$ on the plane. We will prove by induction on $n \in \mathbb{N}$ that for any broken line $A_{0} A_{1} \ldots A_{n}$, the inequality
$$
S_{U\left(A_{0} \ldots A_{n}\right)} \leqslant 2 \sum_{i=1}^{n} A_{i-1} A_{i} + \... | 1248 | Geometry | proof | Yes | Yes | olympiads | false | 29,021 |
12.18*. (Jury, CSSR, 79). On a line, there are $n^{2}+1(n \in \mathrm{N})$ segments. Prove that either among them one can select $n+1$ segments that do not intersect each other, or there exists a common point for some $n+1$ segments. | 12.18. Let's choose a direction on the number line and say that one segment is to the left of another if the left end of the first segment is to the left of (more precisely, not to the right of) the left end of the second. We will assign one of the $n$ numbers $1,2, \ldots, n$ to each segment as follows. On the first s... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,022 |
12.19*. (USA, 83). On a line, there are several sets, each of which is the union of two segments. Prove that if any three of these sets have a common point, then there exists a point belonging to at least half of all sets. | 12.19. To each set $A_{i}(i=1, \ldots, n)$, representing the union of two segments of a line (where the left and right directions are distinguished), we will correspond a segment $B_{i}$, the left end of which coincides with the leftmost point of the set $A_{i}$, and the right end with the rightmost. Since $B_{i} \sups... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,023 |
12.20*. (SFRY, 75). On a plane, $n+4$ points are marked, four of which are located at the vertices of a square, and the remaining $n$ lie inside this square. Any of the marked points are allowed to be connected by segments, provided that no constructed segment contains marked points other than its endpoints, and no two... | 12.20. Suppose that for $n+4$ points, a certain network of segments has already been constructed, satisfying the condition of the problem, and no more segments can be drawn (such networks, called maximal in the future, necessarily exist, since the number of all possible segments is limited by the number $C_{n+4}^{2}$).... | 3n+5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,024 |
13.1. (England, 67). The lengths $a$ and $b$ of two sides of a triangle satisfy the condition $a > b$, and the lengths of the corresponding altitudes are $h_{a}$ and $h_{b}$. Prove the inequality
$$
a + h_{a} \geqslant b + h_{b}
$$
Determine when equality is achieved. | 13.1. Note that
$$
2 S=a b \sin \gamma, then
$$
$$
\left(a+h_{a}\right)-\left(b+h_{b}\right)=\left(a+\frac{2 S}{a}\right)-\left(b+\frac{2 S}{b}\right)=(a-b)\left(1-\frac{2 S}{a b}\right) \geqslant 0
$$
and equality is achieved if and only if $2 S=a b$, i.e., when the angle between the given sides is a right angle. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,025 |
13.3. (SFRY, 75). Prove that if the midpoints of the sides of a convex $n$-gon (where $n \geqslant 4$) are connected in sequence, then the area of the resulting polygon will be no less than half the area of the original.
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。 | 13.3. Let points $B_{1}, B_{2}, \ldots, B_{n}$ be the midpoints of the sides $A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{n} A_{1}$ of a convex $n$-gon
Fig. 76
$A_{1} A_{2} \ldots A_{n}$ with are... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 29,027 |
13.4. (Austria-Poland, 78). A parallelogram is inscribed in a regular hexagon, the center of symmetry of which coincides with the center of the hexagon. Prove that the area of the parallelogram does not exceed $2 / 3$ of the area of the hexagon. | 13.4. Let the parallelogram $ABCD$ be inscribed in a regular hexagon $M$, the center $O$ of which is the point of intersection of the diagonals of the parallelogram. We choose vertices $E$ and $F$ of the hexagon, lying in the same half-plane with point $B$ relative to the line $OA$ and satisfying the inequalities $\ang... | S_{ABCD}\leq(2/3)S_{M} | Geometry | proof | Yes | Yes | olympiads | false | 29,028 |
13.5. (SFRY, 77). Prove that the area of any square lying within a triangle does not exceed half the area of this triangle. | 13.5. Let's prove a more general statement: the area of any parallelogram $KLMN$ lying inside triangle $ABC$ does not exceed half the area of this triangle. Note that each of the lines $KL$ and $MN$ intersects two sides of triangle $ABC$ (possibly at its vertices), and therefore, at least two of the four intersection p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,029 |
13.6. (Beijing, 64). In triangle $A B C$, where the angle at vertex $A$ is not acute, a square $B_{1} C_{1} D E$ is inscribed (side $D E$ lies on segment $B C$, and vertices $B_{1}$ and $C_{1}$ lie on segments $A B$ and $A C$ respectively). Then, in triangle $A B_{1} C_{1}$, a square $B_{2} C_{2} D_{1} E_{1}$ is inscri... | 13.6. Since by condition $\angle B A C \geqslant 90^{\circ}$, point $A$ lies within the circle with diameter $B C$ and center $O$ on it. Therefore, for the height $A H$ of triangle $A B C$, we have
$$
A H \leqslant A O \leqslant B O = B C / 2
$$
(Fig. 79), i.e., $B C \geqslant 2 A H$. From this and the similarity of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,030 |
13.7. (CSSR, 75). Prove that any acute-angled triangle of area 1 can be placed in a right-angled triangle of area no more than $\sqrt{3}$. | 13.7. Let in an acute-angled triangle $ABC$ the angle at vertex $A$ be the largest. We draw a circle with center at the midpoint $M$ of side $BC$ and radius $R=MA$, which intersects the line $BC$ at points $D$ and $E$ (Fig. 80). Then the angle $DAE$ is a right angle and
$$
a=MB=MC<R
$$
(otherwise $MB \geqslant MD, MC... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,031 |
13.8. (Jury, 81). Prove that if points $A$, $B, C, D, E$ are sequentially located on a semicircle of radius 1, then the following inequality holds:
$$
A B^{2}+B C^{2}+C D^{2}+D E^{2}+A B \cdot B C \cdot C D+B C \cdot C D \cdot D E<4 .
$$ | 13.8. Let $A B=a, B C=b, C D=c, D E=d, A C=x, C E=y$, $\angle C A E=\alpha, \angle A E C=\beta$. Without loss of generality, we can assume that points $A$ and $E$ are the endpoints of a diameter of a semicircle (Fig. 81), for if this is not the case, they can be moved to the specified endpoints, after which the express... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,032 |
13.9. (CSSR, 83). Prove that for any point 0 lying on the side $A B$ of triangle $A B C$ and not coinciding with its vertices, the inequality
$$
O C \cdot A B<O A \cdot B C+O B \cdot A C
$$
holds. | 13.9. Since
$$
\overrightarrow{A O}=x \cdot \overrightarrow{A B} \text { and } \overrightarrow{O B}=(1-x) \cdot \overrightarrow{A B}
$$
where \( x \in (0 ; 1) \), we have (Fig. 82)
$$
\begin{aligned}
O C & =|\overrightarrow{C A}+\overrightarrow{A O}|=|\overrightarrow{C A}+x(\overrightarrow{C B}-\overrightarrow{C A})... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,033 |
13.10. (GDR, 62). Prove that for any convex quadrilateral, the ratio of the greatest of the distances between its vertices to the smallest of them is not less than $\sqrt{2}$ | 13.10. Let $M$ be the greatest and $m$ the least of the distances between the vertices of a quadrilateral. Since at least one of its angles, say angle $ABC$, is not acute, by the cosine rule we have
$$
M^{2} \geqslant A C^{2} \geqslant A B^{2}+B C^{2} \geqslant m^{2}+m^{2}=2 m^{2}
$$
from which
$$
M \geqslant \sqrt{... | M/\geqslant\sqrt{2} | Geometry | proof | Yes | Yes | olympiads | false | 29,034 |
13.11. (Austria, 75). Prove that if 6 distinct points are located on a plane, then the ratio of the greatest of the distances between these points to the smallest of them is not less than $\sqrt{3}$. | 13.11. First of all, note that if among the given points, we can always choose three points \( A, B, C \) such that \( 120^{\circ} \leqslant \angle A B C \), then the statement of the problem is satisfied. Indeed, let \( M \) be the largest and \( m \) be the smallest of the distances between the points. Then, by the c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,035 |
13.12. (New York, 77-79). Prove that if \(a, b\), and \(c\) are the lengths of the sides of a triangle, \(P\) is its perimeter, and \(S\) is its area, then the following inequalities hold:
1) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{9}{P}\).
2) \(a^{2}+b^{2}+c^{2} \geqslant \frac{P^{2}}{3}\).
3) \(P^{2} \... | 13.12. 1) By the mean value theorem, we have
$(a b+b c+c a)(a+b+c)=a^{2} b+b^{2} a+c^{2} a+a^{2} c+b^{2} c+c^{2} b+3 a b c \geqslant$
$\geqslant 6 a b c+3 a b c=9 a b c$,
from which the inequality follows
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geqslant \frac{9}{P}
$$
2) The inequality
$$
a^{2}+b^{2}+c^{2} \geqsl... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,036 |
13.13. (SFRY, 75). Prove that for any point $O$ lying inside triangle $ABC$ with semiperimeter $p$, the following inequality holds:
$$
O A \cdot \cos \frac{\angle B A C}{2} + O B \cdot \cos \frac{\angle A B C}{2} + O C \cdot \cos \frac{\angle A C B}{2} \geqslant p
$$
Determine when equality is achieved. | 13.13. Let $\alpha_{1}=\angle O A C, \quad \alpha_{2}=\angle O A B, \quad \beta_{1}=\angle O B A$, $\beta_{2}=\angle O B C, \quad \gamma_{1}=\angle O C B, \quad \gamma_{2}=\angle O C A, \quad \alpha=\alpha_{1}+\alpha_{2}, \quad \beta=\beta_{1}+\beta_{2}$,
. Prove that for the angles $\alpha, \beta, \gamma$ of any triangle, the inequality
$$
\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma \geqslant 3 / 4
$$
holds. Determine when equality is achieved. Prove that the quantity $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma$ does not attain a maxi... | 13.15. Let
$$
f(\alpha, \beta, \gamma)=\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma
$$
where $\alpha, \beta, \gamma$ are the angles of a triangle. Then $\alpha+\beta+\gamma=180^{\circ}$, hence,
$4\left(f(\alpha, \beta, \gamma)-\frac{3}{4}\right)=4\left(\frac{1+\cos 2 \alpha}{2}+\frac{1+\cos 2 \beta}{2}+\frac{1+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,038 |
13.16. (SFRY, 76). Prove that for the angles $\alpha, \boldsymbol{\beta}, \gamma$ of any non-obtuse triangle, the inequality
$$
\sin \alpha+\sin \beta+\sin \gamma>\cos \alpha+\cos \beta+\cos \gamma
$$
holds. | 13.16. For the angles $\alpha, \beta, \gamma \leq 90^{\circ}$ of a triangle, we have
$\cos \alpha + \cos \beta + \cos \gamma = \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} + \cos \frac{\alpha + \gamma}{2} \cos \frac{\alpha - \gamma}{2} +$
$$
\begin{aligned}
+ \cos \frac{\beta + \gamma}{2} \cos \frac{\b... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,039 |
13.17. (Jury, SRV, 77). Prove that for any triangle with side lengths $a, b, c$ and area $S$ the following inequality holds:
$$
\frac{a b + a c + b c}{4 S} \geqslant \sqrt{3}
$$ | 13.17. Let in triangle $ABC$ the equalities
$$
\begin{aligned}
AB=c, & AC=b, \quad BC=a \text { and } \\
& \angle BAC=\alpha
\end{aligned}
$$
be satisfied.
Consider the arc $BAC$ of the circumcircle of triangle $ABC$,
![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-238.jpg?height=480&width=485&... | \sqrt{3} | Inequalities | proof | Yes | Yes | olympiads | false | 29,040 |
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