problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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207. Through a given point $O$ inside a given angle, two rays form the given angle $\alpha$. Let one ray intersect one side of the angle at point $A$, and the other ray intersect the other side of the angle at point $B$. Find the set of feet of the perpendiculars dropped from $O$ to the line $A B$. | 207. Let $C$ be the vertex of the given angle, $\beta$ its magnitude. Drop perpendiculars $OK$ and $OL$ from $O$ to the sides of the angle (Fig. 53a). A circle can be circumscribed around quadrilateral $OKAM$. Therefore, $\widehat{KMO} = \widehat{KAO}$. Similarly, $\widehat{OML} = \widehat{OBL}$. Hence, $\widehat{KML} ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,014 |
208. In a circle, two mutually perpendicular diameters $A C$ and $B D$ are drawn. Let $P$ be an arbitrary point on the circle, and $P A$ intersects $B D$ at point $E$. A line passing through $E$ parallel to $A C$ intersects the line $P B$ at point $M$. Find the set of points $M$. | 208. Consider the quadrilateral $D E P M . \widehat{D E M}=\widehat{D P M}=$ $=90^{\circ}$, hence this quadrilateral is cyclic. Therefore, $\widehat{D M E}=\widehat{D P E}=45^{\circ}$. The desired set is the line $D C$.
. Find the set of centroids of triangles $A C D$. | 209. Let's consider the case when point $B$ lies inside the given angle. First of all, note that all resulting $\triangle B C D$ (Fig. 54) are similar to each other, since $\widehat{B C D}=\widehat{B A D}, \widehat{B D C}=$ $=\widehat{B A C}$. Therefore, if $N$ is the midpoint of $C D$, then the angles $\widehat{B N C}... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,016 |
210. One vertex of a rectangle is at a given point, two others, not lying on the same side, - on two given mutually perpendicular lines. Find the set of fourth vertices of such rectangles. | 210. If $O$ is the vertex of the angle, $A B C D$ is a rectangle (with $A$ fixed), then the points $A, B, C, D, O$ lie on the same circle. Therefore, $\widehat{C O A}=90^{\circ}$, i.e., point $C$ lies on the line perpendicular to $O A$ and passing through $O$. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,017 |
211. Let $A$ be one of the two intersection points of two given circles; through the other intersection point, an arbitrary line is drawn, intersecting one circle at point $B$ and the other at point $C$, different from the common points of these circles. Find the set of: a) centers of the circumcircles of $A B C$; b) c... | 211. Note that all resulting triangles $ABC$ are similar to each other. Consequently, if we take in each triangle a point $K$ that divides the side $BC$ in the same ratio, then since $\widehat{A K C}$ maintains a constant value, the point $K$ will trace a circle. Therefore, the point $M$, dividing $A K$ in a constant r... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,018 | |
212. Let $B$ and $C$ be two fixed points on a given circle, and $A$ be a variable point on the same circle. Find the set of the feet of the perpendiculars dropped from the midpoint of $A B$ to $A C$. | 212. Let $K$ be the midpoint of $AB$, and $M$ be the foot of the perpendicular dropped from $K$ to $AC$. All triangles $AKM$ are similar to each other (by two angles), hence all triangles $ABM$ will also be similar. It is now easy to see that the desired set is a circle with chord $BC$, and the angles subtended by this... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,019 |
214. Two circles touch each other internally at point $A$. A tangent to the smaller circle intersects the larger circle at points $B$ and $C$. Find the set of centers of circles inscribed in triangles $A B C$. | 214. Let the radii of the given circles be denoted by \( R \) and \( r (R > r) \), the point of tangency of the chord \( BC \) with the smaller circle be \( D \); \( K \) and \( L \) will be the points of intersection of the chords \( AC \) and \( AB \) with the smaller circle, and finally, \( O \) is the center of the... | \frac{r\sqrt{R}}{\sqrt{R+\sqrt{R-r}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,021 |
215. Given numbers $\alpha, \beta, \gamma$ and $k$. Let $x, y, z$ be the distances from a point $M$ inside a triangle to its sides. Prove that the set of points $M$ such that $\alpha x + \beta y + \gamma z = k$, is either empty, or a segment, or coincides with the set of all points of the triangle. | 215. Show that if $M_{1}$ and $M_{2}$ are two different points belonging to a certain set, then any point $M$ on the line segment $M_{1} M_{2}$ inside the triangle also belongs to this set. For this, denoting by $x_{1}, y_{1}, z_{1}$ the distances from $M_{1}$ to the sides of the triangle, and by $x_{2}, y_{2}, z_{2}$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,022 |
216. Find the set of points $M$ located inside a given triangle such that the distances from $M$ to the sides of the given triangle are equal to the sides of some triangle. | 216. For distances $x, y, z$ to be the sides of a triangle, it is necessary that the inequalities $x<y+z, y<z+x, z<x+y$ are satisfied. However, the set of points for which, for example, $x=y+z$, is a segment with endpoints at the feet of the angle bisectors (at the feet of the angle bisectors, two distances are equal, ... | thedesiredsetconsistsofpointslocatedinsidethetrianglewithverticesattheoftheanglebisectors | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,023 |
217. Prove that if there exists a circle tangent to the lines $A B, B C, C D$ and $D A$, then its center and the midpoints of $A C$ and $B D$ lie on the same line. | 217. Let $A B C D$ be a circumscribed quadrilateral, $O$ - the center of the inscribed circle, $M_{1}$ - the midpoint of $A C$, $M_{2}$ - the midpoint of $B D$, $r$ - the radius of the circle (the distances from $O$ to the sides are equal to $r$), $x_{1}, y_{1}, z_{1}, u_{1}$ - the distances from $M_{1}$ to $A B, B C, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,024 |
221. Given two points $A$ and $B$ and a line $l$. Find the set of centers of circles passing through $A$ and $B$ and intersecting the line $l$. | 221. If the line $A B$ is not parallel to $l$, then there exist two circles passing through $A$ and $B$ and tangent to $l$. Let their centers be $O_{1}$ and $O_{2}$. The desired set is the line $O_{1} O_{2}$, excluding the interval $\left(O_{1}, O_{2}\right)$. If $A B$ is parallel to $l$, the desired set consists of a ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,028 |
222. Given two points $O$ and $M$.
a) Determine the set of points on the plane that can serve as one of the vertices of a triangle with the center of the circumscribed circle at point $O$ and the centroid at point $M$.
b) Determine the set of points on the plane that can serve as one of the vertices of an obtuse tria... | 222. a) Let (Fig. 58) $A$ be the vertex of a certain triangle. Extend the segment $AM$ beyond $M$ by a length $|MN| = \frac{1}{2}|AM|$. The point $N$ is the midpoint of the side opposite vertex $A$; therefore, $N$ must lie inside the circumscribed circle, i.e., inside the circle with center $O$ and radius $|OA|$. Drop ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,029 |
223. A regular triangle is inscribed in a circle. Find the set of intersection points of the altitudes of all possible triangles inscribed in the same circle, two sides of which are parallel to two sides of the given regular triangle. | 223. Let (Fig. 59) $ABC$ be the original equilateral triangle, $A_{1} B_{1} C_{1}$ be an arbitrary triangle ($A_{1} C_{1} \parallel A C, A_{1} B_{1} \parallel A B$),
Fig. 59.
$O$ is the ce... | 3segmentsoflength4R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,030 |
224. Find the set of centers of all possible rectangles circumscribed about a given triangle. (We will call a rectangle circumscribed if one vertex of the triangle coincides with a vertex of the rectangle, and the other two lie on the two sides of the rectangle not containing this vertex.) | 224. If (Fig. 60) $ABC$ is a given triangle and the vertex of the circumscribed rectangle $AKLM$ coincides with $A$ - $(B$ on $KL$, $C$ on $LM$), then $L$ lies on the semicircle with diameter $BC$, and the angles $ABL$ and $ACL$ are obtuse, i.e., $L$ will have two extreme positions: $L_1$ and $L_2$, $\widehat{L_1CA} = ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,031 |
225. Given two squares with respectively parallel sides. Determine the set of points $M$ such that for any point $P$ from the first square, there is a point $Q$ from the second such that the triangle $M P Q$ is equilateral. Let the side of the first square be $a$, and the second be $b$. For what ratio between $a$ and $... | 225. If (Fig. 61) we rotate the first square around point $M$ by $60^{\circ}$, either clockwise or counterclockwise, it must fit entirely inside the second square. Conversely, for every square located inside the larger one, equal to the smaller one, with sides forming angles of $30^{\circ}$ and $60^{\circ}$ with the si... | b\geqslant\frac{}{2}(\sqrt{3}+1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,032 |
226. Inside the given triangle, find all points $M$ such that for any point $N$ lying on the boundary of the triangle, there exists a point $P$ inside or on its boundary, such that the area of triangle $M N P$ is not less than $1 / 6$ of the area of the given triangle. | 226. Such a point $M$ is unique (Fig. 62) - the centroid of the triangle (the point of intersection of the medians). It is easy to see that in this case, for any point $N$ on the boundary of the triangle, one of the vertices of the triangle can be taken as point $P$.
Let us take any other point $M_{1}$. We will assume... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,033 |
227. Given two points $A$ and $I$. Find the set of points $B$ such that there exists a triangle $A B C$ with the incenter at point $I$, and all angles of which are less than $\alpha\left(60^{\circ}<\alpha<90^{\circ}\right)$. | 227. If $A, \hat{B}, \hat{C}$ are the angles of $\triangle A B C$, then the angles of $\triangle A B I$ are $\frac{A}{2}, \frac{\hat{B}}{2}, 90^{\circ}+\frac{\hat{C}}{2}$ (Fig. 63); therefore, the desired set, depicted in Fig. 63, is a pair of triangles, two sides of which are segments of lines, and the third is an arc... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,034 |
228. Points $A, B$ and $C$ are located on the same line ( $B$ - between $A$ and $C$ ). Find the set of points $M$ such that $\operatorname{ctg} \widehat{A M B}+\operatorname{ctg} \widehat{B M C}=k$. | 228. Construct a perpendicular to $B M$ at point $M$, let $P$ be the point of intersection of this perpendicular and the one erected to the original line at point $B$. We will show that the magnitude $|P B|$ is constant. Let $\widehat{M B C}=\varphi$; through $K$ and $L$ denote the feet of the perpendiculars dropped fr... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,035 |
229. Given two points $A$ and $Q$. Find the set of points $B$ such that there exists an acute triangle $A B C$, for which $Q$ is the centroid. | 229. Extend $A Q$ beyond point $Q$ and take a point $M$ on this ray such that $|Q M|=\frac{1}{2}|A Q|$, and a point $A_{1}$ such that $\left|M A_{1}\right|=|A M|$;
$M$ is the midpoint of side $\quad B C$ of triangle $A B C ; \widehat{C B A_{1}}=\widehat{B C A}$, $\widehat{A B A_{1}}=180^{\circ}-\widehat{B A C}$.
Ther... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,036 |
231. On a plane, two rays are given. Find the set of points on the plane that are equidistant from these rays. (The distance from a point to a ray is equal to the distance from this point to the nearest point on the ray.) | 231. If the ends of the rays do not coincide, the desired set consists of parts of the following lines: the bisectors of the two angles formed by the lines containing the given rays, the perpendicular bisector of the segment connecting the ends of the rays, and two parabolas (a parabola is the set of points equidistant... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,038 |
235. Let $H$ be the orthocenter of a triangle, and $F$ be an arbitrary point on the circumcircle. Prove that the Simson line corresponding to point $F$ passes through one of the points of intersection of the line $F H$ with the nine-point circle (see problems 162, 233). | 235. Since the midpoint of $F H$ lies on the nine-point circle (see problem 162), it is sufficient to show that the Simson line corresponding to point $F$ also bisects $F H$. Let $K$ be the projection of $F$ onto one of the sides of the triangle, $D$ the foot of the altitude to that same side, $H_{1}$ the point where t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,042 |
236. Prove that on the Euler line of triangle $ABC$ there exists a point $P$ such that the distances from the centroids of triangles $ABP$, $BCP$, $CAP$ to the vertices $C$, $A$, and $B$ respectively are equal to each other (see problem 232). | 236. Show that the required property is possessed by a point $P$ on the Euler line for which $P O_{1}=O H_{1}$ (where $O$ is the center of the circumscribed circle, $H$ is the orthocenter); in this case, for each triangle, the distance from the centroid to the opposite vertex of the original triangle is $-\frac{4}{3} R... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,043 |
237. Let $P$ be a point inside triangle $ABC$ such that the angles $APB$, $BPC$, and $CPA$ are $120^{\circ}$ (assuming that the angles of triangle $ABC$ are less than $120^{\circ}$). Prove that the Euler lines of triangles $APB$, $BPC$, and $CPA$ intersect at one point (see problem 232). | 237. Let $C_{1}$ be the center of the circumcircle of $\triangle A P B$, and $C_{2}$ the point symmetric to $C_{1}$ with respect to $A B$. Similarly, for triangles $B P C$ and $C P A$, define points $A_{1}$ and $A_{2}, B_{1}$ and $B_{2}$. Since triangles $A C_{1} B, A C_{2} B, B A_{1} C, B A_{2} C, C B_{1} A, C B_{2} A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,044 |
238. A quadrilateral $A B C D$ is inscribed in a circle. Let $M$ be the point of intersection of the tangents to the circle at points $A$ and $C, N$ be the point of intersection of the tangents drawn through $B$ and $D, K$ be the point of intersection of the angle bisectors of angles $A$ and $C$ of the quadrilateral, $... | 238. A necessary and sufficient condition for the fulfillment of all four points is the equality $|A B| \cdot|C D|=|A D| \cdot|B C|$. For points a) and b), this follows from the theorem about the bisector of the interior angle of a triangle, for points c) and d) - from the result of problem 50.
*) For more detailed in... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,045 |
239. Prove that 4 lines, each of which passes through the bases of two perpendiculars dropped from the vertex of an inscribed quadrilateral to the sides not containing it, intersect at one point. | 239. Let $ABCD$ be the given quadrilateral. We will assume that angles $A$ and $D$ are obtuse, and $B$ and $C$ are acute. Let the feet of the perpendiculars dropped from vertex $A$ be denoted by $M$ and $N$, and from vertex $C$ by $K$ and $L$ (see Fig. 64a), and let $R$ be the intersection point of $MN$ and $LK$. Note ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,046 |
240. $A B$ and $C D$ are two chords of a circle; $M$ is the point of intersection of the perpendiculars erected to $A B$ at point $A$ and to $C D$ at point $C; N$ is the point of intersection of the perpendiculars erected to $A B$ and $C D$ at points $B$ and $D$. Prove that the line $M N$ passes through the point of in... | 240. Let's find in what ratio $B C$ divides $M N$. This ratio is equal to the ratio
$$
\frac{S_{M C B}}{S_{C B N}}=\frac{|M C| \cdot|C B| \sin \widehat{M C B}}{|B N| \cdot|C B| \sin \widehat{N B C}}=\frac{M C \cdot \cos \widehat{B C D}}{|B N| \cos \widehat{C B A}}
$$
Similarly, the ratio in which $A D$ divides $M N$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,047 |
241. Given a circle $S$ and a tangent to it $l$. Let $N$ be the point of tangency, $NM$ be a diameter. A fixed point $A$ is taken on the line $NM$. Consider an arbitrary circle passing through $A$, with its center on $l$. Let $C$ and $D$ be the points of intersection of this circle with $l$, and $P$ and $Q$ be the poin... | 241. Since the circle with diameter $C D$ passes through a fixed point $A$ on $M N$ ($M N \perp C D$), then
$$
|C N| \cdot|N D|=|N A|^{2}(1)
$$
is a constant value. Let $K$ be the intersection point of $P Q$ with $M N$. We will show that $\frac{|N K|}{|K M|}$ is a constant value. Note that $\widehat{P N Q}=180^{\circ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,048 |
242. Given three pairwise non-intersecting circles. Let $A_{1}, A_{2}, A_{3}$ be the three points of intersection of the common internal tangents to any two of them, and $B_{1}, B_{2}, B_{3}$ be the corresponding points of intersection of the external tangents. Prove that these points lie on four lines, three on each (... | 242. Check that points $A_{1}, A_{2}, A_{3}$ and $B_{1}, B_{2}, B_{3}$ lie on the sides of triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ ( $\mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}$ - centers of circles) or on the extensions of these sides and the ratio of the distances from each of these points to the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,049 |
243. The diameter of the circle inscribed in triangle $ABC$, passing through the point of tangency with side $BC$, intersects the chord connecting the other two points of tangency at point $N$. Prove that $AN$ bisects $BC$. | 243. Let $O$ be the center of the inscribed circle, $K$ and $L$ be the points of tangency with sides $AC$ and $AB$, and the line passing through $N$ parallel to $BC$ intersects sides $AB$ and $AC$ at points $R$ and $M$. The quadrilateral $OKMN$ is inscribed ( $\widehat{ONM}=\widehat{OKM}=90^{\circ}$ ), therefore, $\wid... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,050 |
244. A circle is inscribed in triangle $ABC$. Let $M$ be the point of tangency of the circle with side $AC$, and $MK$ be a diameter. The line $BK$ intersects $AC$ at point $N$. Prove that $|AM|=|NC|$. | 244. If the sides of $\triangle A B C$ are $|B C|=a, \mid C A=b$, $|A B|=c$, then, as we know (see problem 18, section 1), $\mid M C:=$ $=\frac{a+b-c}{2}$. Draw a line through $K$ parallel to $A C$, and denote its points of intersection with $A B$ and $B C$ as $A_{1}$ and $C_{1}$. The incircle of $\triangle A B C$ is a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,051 |
245. A circle is inscribed in triangle $ABC$, $M-$ is the point of tangency of the circle with side $BC$, $MK$ is a diameter. Line $AK$ intersects the circle at point $P$. Prove that the tangent to the circle at point $P$ bisects side $BC$. | 245. Draw a line through $K$ parallel to $B C$. Let $L$ and $Q$ be the points of intersection of the tangent at point $P$ with the line $B C$ and the constructed parallel line, and let $N$ be the point of intersection of $A K$ with $B C$. Since $|C N| = |B M|$ (see problem 244), it is sufficient to prove that $|N L| = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,052 |
246. The line $l$ is tangent to the circle at point $A$, let $CD$ be a chord of the circle parallel to $l$, and $B$ be any point on the line $l$. The lines $CB$ and $DB$ intersect the circle again at points $L$ and $K$. Prove that the line $LK$ bisects the segment $AB$. | 246. Let $M$ and $N$ be the points of intersection of line $L K$ with lines $l$ and $C D$. Then $|A M|^{2}=|M L| \cdot|M K|$. From the similarity of triangles $K M B$ and $D K N$, it follows that $\frac{|M K|}{|K N|}=\frac{|M B|}{|D N|}, \quad|M K|=$ $=\frac{|K N| \cdot|M B|}{|D N|}$. From the similarity of triangles $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,053 |
247. Given two intersecting circles. Let $A$ be one of their points of intersection. From an arbitrary point lying on the extension of the common chord of the given circles, two tangents are drawn to one of them, touching it at points $M$ and $N$. Let $P$ and $Q$ be the points of intersection (different from $A$) of th... | 247. Let (Fig. 65) $B$ be the second common point of the circles,
Fig. 65. $C$ is a point on the line $A B$, from which tangents are drawn, and, finally, $K$ is the intersection point of th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,054 |
248. On the height $B D$ of triangle $A B C$ as a diameter, a circle is constructed, intersecting sides $A B$ and $B C$ at points $K$ and $L$. The lines tangent to the circle at points $K$ and $L$ intersect at point $M$. Prove that the line $B M$ bisects side $A C$. | 248. Draw a line through $M$ parallel to $A C$, until it intersects points $A_{1}$ and $C_{1}$ on lines $B A$ and $B C$. We have $\widehat{A_{1} K M}=90^{\circ}-\widehat{D K M}=90^{\circ}-\widehat{K B D}=\widehat{B A D}=\widehat{K A_{1} M}, \quad$ therefore, $\triangle K M A_{1}$ is isosceles and $\left|A_{1} M\right|=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,055 |
249. The line $l$ is perpendicular to the segment $A B$ and passes through $B$. A circle with its center on $l$ passes through $A$ and intersects $l$ at points $C$ and $D$. The tangents to the circle at points $A$ and $C$ intersect at $N$. Prove that the line $D N$ bisects the segment $A B$. | 249. Let $M$ be the intersection point of $ND$ and $AB$, and $P$ be the intersection point of the tangents to the circle at points $A$ and $D$. Since the lines $NC$, $AB$, and $PD$ are parallel, from the similarity of the corresponding triangles we get
$$
\begin{gathered}
\frac{|AM|}{|DP|}=\frac{|AN|}{|NP|}, \quad|AM|... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,056 |
250. A circle is circumscribed around $\triangle A B C$. Let $N$ be the point of intersection of the tangents to the circle passing through points $B$ and $C ; M$ - a point on the circle such that $A M \| B C, K$ - the point of intersection of $M N$ and the circle. Prove that $K A$ bisects $B C$. | 250. Let's consider that $D$ is the midpoint of $CB$ and $AD$ intersects the circle again at point $K$. We need to prove that the tangents to the circle at points $B$ and $C$ intersect on the line $MK$.
Consider the quadrilateral $CMBK$. For the tangents to the circle at points $C$ and $B$ to intersect on the diagonal... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,057 |
252. Given a semicircle with diameter $A B . C$-point on the semicircle, $D$-the foot of the perpendicular
dropped from $C$ to $A B$. Consider three circles: the first circle, with center $O_{1}$, touches segments $A D, D C$ and arc $\breve{A C}$; the second, with center $O_{2}$, touches segments $D B, D C$ and arc $\w... | 252. Let (Fig. 67) $O$ be the midpoint of $AB$, $N_{1}$ and $N_{2}$ be the points of tangency of circles $O_{1}$ and $O_{2}$ with $AB$, $O_{3}$ be the midpoint of $O_{1}O_{2}$, $N_{3}$ be the foot of the perpendicular dropped from $O_{3}$ to $AB$, $a$ and $b$ be the legs of the triangle ($|CB|=a, |AC|=b$), $c$ be the h... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,058 |
253. Let $A B C D E F$ be a cyclic hexagon. Denote by $K$ the intersection point of $A C$ and $B F$, and by $L$ the intersection point of $C E$ and $F D$. Prove that the diagonals $A D, B E$ and the line $K L$ intersect at one point (Pascal). | 253. Let (Fig. 68) $M$ be the point of intersection of $A D$ and $K L$;
$$
\frac{|K M|}{|M L|}=\frac{S_{A K D}}{S_{A L D}}=\frac{\frac{1}{2}|A K| \cdot|A D| \sin \widehat{K A D}}{\frac{1}{2}|D L| \cdot|A D| \sin \widehat{A D L}}=\frac{|A K| \cdot|C D|}{|D L| \cdot|A F|}
$$
(We used the fact that the sines of inscribe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,059 |
254. $A B C D$ is a cyclic quadrilateral. The perpendicular to $B A$, erected at point $A$, intersects the line $C D$ at point $M$; the perpendicular to $D A$, erected at point $A$, intersects the line $B C$ at point $N$. Prove that $M N$ passes through the center of the circle. | 254. Let $L$ and $P$ be the points of intersection of the lines $A M$ and $A N$ with the circle. As follows from problem 253, the lines $B L$, $D P$, and $M N$ intersect at one point. But $B L$ and $D P$ are diameters, intersecting at the center of the circle, hence $M N$ passes through the center of the circle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,060 |
255. On each side of a triangle, two points are taken such that all six segments connecting each point with the opposite vertex are equal to each other. Prove that the midpoints of these six segments lie on a single circle. | 255. We will prove that the center of the desired circle coincides with the orthocenter (the point of intersection of the altitudes). Let $B D$ be the altitude, $H$ be the point of intersection of the altitudes, and $K$ and $L$ be the midpoints of the constructed segments emanating from vertex $B, |B K|=|B L|=l, M$ be ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,061 |
256. In triangle $ABC$, on the rays $AB$ and $CB$, segments $|AM|=|CN|=p$ are laid out, where $p$ is the semiperimeter of the triangle (with $B$ lying between $A$ and $M$ and between $C$ and $N$). Let $K$ be the point on the circumcircle of $\triangle ABC$ that is diametrically opposite to $B$. Prove that the perpendic... | 256. Let (Fig. 70) the lengths of the sides of triangle $ABC$ be: $|BC|=a, |CA|=b, |AB|=c$. Draw lines through the center of the inscribed circle parallel to $AB$ and $BC$, intersecting $AK$ and $KC$ at points $P$ and $Q$; in triangle $OPQ$ we have $\widehat{POQ}=\hat{B}$, $|OQ|=\frac{a+b-c}{2}=p-c$, $|OP|=\frac{b+c-a}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,062 |
257. From a certain point on the circumcircle of an equilateral triangle $ABC$, lines parallel to $BC$, $CA$, and $AB$ are drawn and intersect $CA$, $AB$, and $BC$ at points $M$, $N$, and $Q$. Prove that $M$, $N$, and $Q$ lie on the same line. | 257. Let, for definiteness, $P$ lie on the arc $\widehat{A C}$ (Fig. 71). Points $A$, $M, P, N$ lie on the same circle, hence $\widehat{N M P}=\widehat{N A P}$. Similarly, $P, M$, $Q, C$ lie on the same circle, $\widehat{P M Q}=$ $=180^{\circ}-\widehat{P C Q}=180^{\circ}-\widehat{P A N}=180^{\circ}-\widehat{P M N}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,063 |
258. Points $A$ and $A_{1}, B$ and $B_{1}, C$ and $C_{1}$ are symmetric with respect to line $l, N$ is an arbitrary point on $l$. Prove that the lines $A N, B N, C N$ intersect the lines $B_{1} C_{1}, A_{1} C_{1}, A_{1} B_{1}$ respectively at three points lying on one line. | 258. Let's first consider the limiting case when point $N$ is at "infinity"; in this case, lines $A N$, $B N$, and $C N$ are parallel to line $l$. Let the distances from points $A, B$, and $C$ to line $l$ be $a, b$, and $c$ (for convenience, assume that $A, B$, and $C$ are on the same side of $l$). Lines parallel to $l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,064 |
259. Through the intersection point of the altitudes of a triangle, two mutually perpendicular lines are drawn. Prove that the midpoints of the segments cut off by these lines on the sides of the triangle (on the lines forming the triangle) lie on one straight line. | 259. Let the mutually perpendicular lines-axes $x=0$ and $y=0$ of a rectangular coordinate system, the altitudes of the triangle lie on the lines $y=k_{i} x(i=1,2,3)$, the sides of the triangle should have angular coefficients $-\frac{1}{k_{i}}$, and from the condition of the vertices' ( $x_{i}, y_{i}$ ) belonging to t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,065 |
260. Prove that three lines, symmetric to an arbitrary line passing through the orthocenter of a triangle with respect to the sides of the triangle, intersect at one point. | 260. Let (Fig. 72) $ABC$ be a given triangle, and $H$ be the point of intersection of its altitudes. Note that the points symmetric to $H$ with respect to its sides
Fig. 72. lie on the cir... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,066 |
261. Two equal non-intersecting circles are given. On two common internal tangents, take two arbitrary points $F$ and $F^{\prime}$. From both points, one more tangent can be drawn to each circle. Let the tangents drawn from points $F$ and $F^{\prime}$ to one circle meet at point $A$, and to the other at point $B$. It i... | 261. Consider the general case of arbitrary circles. Let points $F$ and $F^{\prime}$ be located as shown in Fig. 73. The notation is clear from the figure. We will prove that there exists a circle inscribed in the quadrilateral $A K B M$, after which we will use the result of problem 242. For this, it is sufficient to ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,067 |
262. Given a triangle $A B C$; $A A_{1}, B B_{1}$, and $C C_{1}$ are its altitudes. Prove that the Euler lines of triangles $A B_{1} C_{1}, A_{1} B C_{1}, A_{1} B_{1} C$ intersect at a point $P$ on the nine-point circle, such that one of the segments $P A_{1}, P B_{1}, P C_{1}$ is equal to the sum of the other two segm... | 262. Let $H$ be the point of intersection of the altitudes of triangle $ABC$, and let $A_{2}, B_{2}, C_{2}$ be the midpoints of segments $AH, BH, CH$. Note that triangles $AB_{1}C_{1}, A_{1}BC_{1}, A_{1}B_{1}C$ are similar to each other (corresponding vertices are denoted by the same letters), and $A_{2}, B_{2}$, and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,068 |
263. Prove that if in triangle $ABC$ angle $B$ is obtuse and $|AB|=\frac{|AC|}{2}$, then $\hat{C}>\frac{1}{2} \hat{A}$. | 263. Let $D$ be the midpoint of $AC$. Construct a perpendicular to $AC$ at $D$ and denote by $M$ the point of intersection of this perpendicular with $BC$. $\triangle AMC$ is isosceles, so $\widehat{MAC} = \widehat{BCA}$. By the given condition, $\triangle ABD$ is also isosceles, $\widehat{ABD} = \widehat{BDA}$, $\wide... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,069 |
264. Prove that the circumcircle of a triangle cannot pass through the center of an excircle. (An excircle is tangent to one side of the triangle and the extensions of the other two sides. For each triangle, there are three excircles.) | 264. If the circle touches the extensions of sides $A B$ and $A C$ and its center is $O$, it is easy to find that $\widehat{B O C}=180^{\circ}-\left(90^{\circ}-\frac{\hat{B}}{2}\right)-$ $-\left(90^{\circ}-\frac{\hat{C}}{2}\right)=\frac{\hat{B}+\hat{C}}{2}=90^{\circ}-\frac{\hat{A}}{2}$. Therefore, $\widehat{B O C}+\wid... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,070 |
265. In a triangle, from vertex $A$ come out the median, bisector, and altitude. Which angle is larger: between the median and the bisector or between the bisector and the altitude, if angle $\hat{A}$ equals $\alpha$? | 265. Let $A D$ be the altitude, $A L$ the angle bisector, and $A M$ the median. Extend the angle bisector to intersect the circumcircle of the triangle at point $A_{1}$. Since $M A_{1} \| A D$, then $\widehat{M A_{1} A}=L \widehat{A D}$
Answer: if $\alpha$ $>90^{\circ}$ - the opposite; if $\alpha=90^{\circ}$, the angl... | if\\alpha>90\-\the\opposite;\if\\alpha=90,\the\angles\\equal | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,071 |
266. Prove that if the medians drawn from vertices $B$ and $C$ of triangle $ABC$ are perpendicular, then $\operatorname{ctg} \hat{B} + \operatorname{ctg} \hat{C} \geqslant 2 / 3$. | 266. If $A D$ is the altitude, $A N$ is the median, and $M$ is the point of intersection of the medians, then
$\operatorname{ctg} \hat{B}+\operatorname{ctg} \hat{C}=\frac{|D B|}{|A D|}+\frac{|C D|}{|A D|}=\frac{|C B|}{|A D|} \geqslant \frac{|C B|}{|A N|}=\frac{|C B|}{3|M N|}=\frac{2}{3}$. | \frac{2}{3} | Geometry | proof | Yes | Yes | olympiads | false | 25,072 |
268. From an external point $A$ to a circle, two tangents $A B$ and $A C$ are drawn, and the midpoints of these tangents $D$ and $E$ are connected by a line $D E$. Prove that this line does not intersect the circle. | 268. If $|O A|=a, R$ is the radius of the circle, and $K$ is the intersection point of $O A$ and $D E$, it is easy to find that
$$
|O K|=a-\frac{a^{2}-R^{2}}{2 a}=\frac{a^{2}+R^{2}}{2 a}>R
$$ | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,074 |
270. In triangle $A B C$, the angles are related by $3 \hat{A}-\hat{C}<\pi$. Angle $B$ is divided into four equal parts by lines intersecting side $A C$. Prove that the third segment, counting from vertex $A$, into which side $A C$ is divided, is less than $|A C| / 4$. | 270. Let $K, L$ and $M$ be the points of intersection of the drawn lines with $AC$. Denote $|AC|=b, |BC|=a, |AB|=c, |BL\rangle=l$. By the angle bisector theorem for the internal angle, we find $|LC|=\frac{ba}{a+c}$; applying this theorem again for $\triangle BCL$, we get
$$
|LM|=|LC| \frac{|BL|}{|BL|+|BC|}=\frac{ba}{a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,076 |
271. Let $a, b, c, d$ be the lengths of the consecutive sides of a quadrilateral. Prove that if $S$ is its area, then $S \leqslant (ac + bd) / 2$, and equality holds only for a cyclic quadrilateral with perpendicular diagonals. | 271. If $A B C D$ is the given quadrilateral, then take the quadrilateral $A B_{1} C D$, where $B_{1}$ is symmetric to $B$ with respect to the perpendicular bisector of the diagonal $A C$. Obviously, the areas of $A B C D$ and $A B_{1} C D$ are equal, and the sides of $A B_{1} C D$ in order are $b$, $a, c, d$. The ineq... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,077 |
272. Prove that if the lengths of the bisectors of a triangle are less than 1, then its area is less than $\sqrt{3} / 3$. | 272. Consider two cases:
1) The given triangle $A B C$ is acute-angled. Let $\hat{B}$ be the largest angle: $60^{\circ} \leqslant \hat{B}<90^{\circ}$. Since the angle bisectors of angles $A$ and $C$ are less than 1, the heights of these angles $h_{A}$ and $h_{C}$ are also less than 1. We have
$$
S_{A B C}=\frac{h_{A} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,078 |
273. Prove that triangle $ABC$ will be acute, right, or obtuse depending on whether the expression $a^{2}+b^{2}+c^{2}-8 R^{2}$ is positive, zero, or negative ( $a, b, c$ are the sides of the triangle, $R$ - the radius of the circumscribed circle). | 273. Let $c$ be the largest side, opposite vertex $C$. If $a^{2}+b^{2}+c^{2}-8 R^{2}>0$, then $a^{2}+b^{2}>8 R^{2}-c^{2}>c^{2}$ (since $c<2 R$), i.e., the triangle is acute-angled. Conversely, let the triangle be acute-angled; then $a^{2}+b^{2}+c^{2}=2 m_{c}^{2}+\frac{3}{2} c^{2}\left(m_{c}\right.$ - the median to side... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,079 |
274. Prove that if the lengths of the sides of a triangle are related by the inequality $a^{2}+b^{2}>5 c^{2}$, then $c$ is the length of the smallest side. | 274. Suppose the opposite, for example, that $c \geqslant a$; then $2 c \geqslant$ $\geqslant c+a>b$; squaring the inequalities and adding them, we get $5 c^{2}>a^{2}+b^{2}-$ a contradiction. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,080 |
275. In triangle $ABC$, angle $B$ is the middle in size: $\hat{A}<\hat{B}<\hat{C} ; I$ is the incenter,
O is the circumcenter, $H$ is the orthocenter. Prove that $I$ lies inside $\triangle B O H$. | 275. It is not difficult to prove that the bisector of angle $B$ is also the bisector of angle $O B H$ (the same is true for angles $O A H, O C H$). Figure 76, b shows the cases of an acute triangle (a) and an obtuse triangle (b). The solution is the same in both cases.
|B C| \leqslant(|A B|-$ $-|A C|)|M C|$ | 277. Draw a line through $M$ parallel to $A C$, intersecting $A B$ at point $K$. It is easy to find
$$
\left|A K_{i}=C M\right| \frac{|A B|}{|C B|}, \quad|M K|=|M B| \frac{|A C|}{|C B|} .
$$
Since $|A M| \leqslant|A K|+|K M|$, by substituting $|A K|$ and $|K M|$, we get
$$
\begin{gathered}
|A M| \leqslant \frac{|C M... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,083 |
278. Let $a, b, c$ be the lengths of the sides of triangle $ABC$, and $M$ be an arbitrary point on the plane. Find the minimum of the expression
$$
|M A|^{2}+|M B|^{2}+|M C|^{2}
$$ | 278. The minimum is $\frac{a^{2}+b^{2}+c^{2}}{3}$ and is achieved if $M-$ is the centroid of $\triangle A B C$. (This can be proven, for example, using the coordinate method or by using Leibniz's theorem - see problem 153.) | \frac{^{2}+b^{2}+^{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,084 |
279. The sides of an angle of measure $\alpha$ are the cushions of a billiard table. What is the maximum number of bounces a billiard ball (neglecting the size of the ball) can make off the cushions? | 279. To "straighten" the path of the ball, instead of "reflecting" the ball off the cushion, we will reflect the billiard table itself mirror-like relative to this cushion. We will obtain a system of rays with a common vertex, any two adjacent rays forming an angle $\alpha$. The maximum number of rays in the system tha... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,085 | |
280. Four villages are located at the vertices of a square with a side length of 2 km. The villages are connected by roads in such a way that it is possible to travel from any village to any other. Can the total length of the roads be less than 5.5 km? | 280. If the roads are built as shown in Fig. $77$ (A, B, C, and D are villages, roads are solid lines), then their total length will be $\frac{8 \sqrt{\overline{3}}}{3}+\left(2-\frac{2 \sqrt{\overline{3}}}{3}\right)=2+2 \sqrt{\overline{3}}<5.5$. It can be shown that the specified layout of the roads realizes the minimu... | 2+2\sqrt{3}<5.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,086 |
281. Point $A$ is located between two parallel lines at distances $a$ and $b$ from them. This point serves as the vertex of an angle of measure $\alpha$ of all possible triangles, the other two vertices of which lie on the given lines. Find the minimum value of the area of such triangles. | 281. If one of the sides of the triangle passing through $A$ forms an angle $\varphi$ with a line perpendicular to the given parallel lines, then the other side will form an angle $180^{\circ}-\varphi-\alpha$; finding these sides, we get that the area of the triangle will be
$$
-\frac{a b \sin \alpha}{2 \cos \varphi \... | \operatorname{ctg}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,087 |
282. Given a circle of radius $R$, $O$-its center, $AB$-its diameter, point $M$-on the radius $OA$, such that $\frac{|AM|}{|MO|}=k$. Through $M$ a random chord $CD$ is drawn. What is the maximum value of the area of the quadrilateral $ACBD$? | 282. Since $\frac{S_{A C D}}{S_{O C D}}=\frac{|A M|}{|M O|}=k, \frac{S_{B C D}}{S_{O C D}}=\frac{|B M|}{|O M|}=k$. 2, then $S_{A C B D}=S_{A C D}+S_{B C D}=2(k+1) S_{O C D}$. Therefore, the area of $A C B D$ will be the largest when the area of triangle $C O D$ is the largest. But triangle $C O D$ is isosceles, with th... | if\k\leqslant\sqrt{2}-1,\then\S_{\max}=(k+1)R^{2};\if\k>\sqrt{2}-1,\then\S_{\max}=\frac{2\sqrt{k(k+2)}}{k+1}R^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,088 |
283. The vertex of an angle of measure $\alpha$ is at point $O$. $A$ is a fixed point inside the angle. Points $M$ and $N$ are taken on the sides of the angle such that $\widehat{M A N}=\beta(\alpha+\beta<$ $<180^{\circ}$ ). Prove that if $|A M|=|A N|$, then the area of quadrilateral OMAN reaches a maximum (among all p... | 283. Let $M_{1}$ and $N_{1}$ be two other points on the sides of the angle (Fig. 78). Then $\quad{\widehat{M_{1} A N_{1}}}_{1}=\beta, \quad \widehat{A M_{1} M}=360^{\circ}-\alpha-\beta-\widehat{O N_{1} A}>$ $>180^{\circ}-\widehat{N_{1} A}=\widehat{A N_{1} N}$. From this, considering that $\widehat{M_{1}} \widehat{A M}=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,089 |
284. Given the result of the previous problem, solve the following. Inside an angle with vertex $O$, a point $A$ is taken.
The line $O A$ forms angles $\varphi$ and $\psi$ with the sides of the angle. Find points $M$ and $N$ on the sides of the angle such that $\widehat{M A N}=\beta\left(\varphi+\psi+\beta<180^{\circ}... | 284. Given the result of the previous problem, we need to determine under what conditions points $M$ and $N$ can be found on the sides of the angle such that $\widehat{M A N}=\beta$ and $|M A|=|A N|$. Describe a circle around triangle $M O N$ (Fig. 79). Since $\varphi+\psi+\beta\widehat{L M N}=\widehat{L O N}$ and $\wi... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,090 |
286. Let $A B C D$ be a cyclic quadrilateral. The diagonal $A C$ is equal to $a$ and forms angles $\alpha$ and $\beta$ with the sides $A B$ and $A D$. Prove that the area of the quadrilateral is bounded by the values
$$
\frac{a^{2} \sin (\alpha+\beta) \sin \beta}{2 \sin \alpha} \text { and } \frac{a^{2} \sin (\alpha+\... | 286. Let $\sin \alpha \geqslant \sin \beta$ for definiteness; take a point $K$ on the extension of $AB$ such that $\widehat{B K C}=\beta$ (Fig. 81); since $\widehat{C} B K = \widehat{A D C}$ (since $A B C D$ is cyclic), $\triangle K B C$ is similar to $\triangle A C D$,
$. Prove that if among these quadrilaterals... | 287. Consider another position of points $M_{1}$ and $N_{1}\left(\widehat{M_{1} A N_{1}}=\beta\right)$ and show, taking into account the condition (that $\alpha+\beta>180^{\circ}$), that the "added" triangle has a greater area than the triangle whose area is reduced (analogous to the solution of problem 283). | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,093 |
288. Inside an angle with vertex $O$, there is a point $A$ such that $O A$ forms angles $\varphi$ and $\psi$ with the sides of the given angle. Find points $M$ and $N$ on the sides of the angle such that $\widehat{M A N}=\beta\left(\varphi+\psi+\beta>180^{\circ}\right)$ and the area of the quadrilateral $O M A N$ is mi... | 288. Given the result of problem 287, reasoning in the same way as in problem 279, we get that if $\varphi>90^{\circ}-\frac{\beta}{2}$ and $\psi>90^{\circ}-\frac{\beta}{2}$, a quadrilateral of the smallest area exists and for it $|M A|=|A N|$. If this condition is not satisfied, the desired quadrilateral degenerates (o... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,094 |
290. Find the radius of the largest circle that can be covered by three circles of radius $R$. Solve the problem in the general case when the radii are $R_{1}, R_{2}, R_{3}$. | 290. The radius of the largest circle is equal to the radius of the circumscribed circle around an equilateral triangle with side $2 R$, i.e., $2 R / \sqrt{3}$. (Let's take such a triangle and construct circles on its sides as diameters.) For any circle with a larger radius, if it were covered by the given circles, the... | \frac{2R}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,096 |
292. What is the maximum area of a regular triangle that can be covered by three regular triangles with a side length of 1? | 292. First, note that the side of the smallest equilateral triangle covering a rhombus with side $a$ and acute angle $60^{\circ}$ is $2a$. Indeed, if the vertices of the acute angles $M$ and $N$ of the rhombus are on the sides $AB$ and $BC$ of the equilateral triangle $ABC$ and $\widehat{BNM}=\alpha, 90^{\circ} \geqsla... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,098 |
293. In triangle $ABC$, points $M$ and $N$ are taken on sides $AC$ and $BC$, and point $L$ is taken on segment $MN$. Let the areas of triangles $ABC$, $AML$, and $BNL$ be $S$, $P$, and $Q$ respectively. Prove that
$$
\sqrt[3]{S} \geqslant \sqrt[3]{P}+\sqrt[3]{Q}
$$ | 293. Let the ratios $\frac{|A M|}{|M C|}, \frac{|C N|}{|N B|}$ and $\frac{|M L|}{|L M|}$ be denoted by $\alpha, \beta$ and $\gamma$. Then we will have (see the solution of problem 35) $\frac{P}{Q}=\alpha \beta \gamma$, $S=Q(\alpha+1)(\beta+1)(\gamma+1)$. Then we will use the inequality
$$
(\alpha+1)(\beta+1)(\gamma+1)... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 25,099 |
1. In space, two triangles $A B C$ and $A_{1} B_{1} C_{1}$ are given. Prove that the greatest of all distances $M M_{1}$, where $M \in A B C^{1}$, and $M_{1} \in A_{1} B_{1} C_{1}$, is the largest of the nine segments $A A_{1}, A B_{1}, A C_{1}, B A_{1}, B B_{1}$, $B C_{1}, C A_{1}, C B_{1}$, and $C C_{1}$. | 1. Let the greatest of the nine distances between the vertices of two triangles be equal to $a$, and let $M$ be an arbitrary point of triangle $ABC$ and $M_{1}$ be an arbitrary point of triangle $A_{1} B_{1} C_{1}$ (Fig. 59). We will prove that $M M_{1} \leqslant a$. Draw an arbitrary line through $M$
 half the difference of the distances $A A_... | 2. Draw through the midpoint $M_{1}$ of the segment $A_{1} B_{1}$ the segments $M_{1} A^{\prime} \# A_{1} A$ and $M_{1} B^{\prime} \# B_{1} B$ (Fig. $60, a$). In this case, the quadrilateral $M_{1} A_{1} A A^{\prime}$ will be a parallelogram ${ }^{1}$); therefore, $A A^{\prime} \# A_{1} M_{1}$. Similarly, the quadrilat... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,101 |
4. Let the convex quadrilateral $A B C D$ be such that $A C+C D \geqslant A B+B D$. Prove that in this case, vertex $A$ of the quadrilateral is closer to vertex $B$ than to vertex $C$.
Does the statement of this problem remain valid if the quadrilateral $A B C D$ is not convex? | 4. We will prove the required statement by contradiction. If the convex quadrilateral $ABCD$ (Fig. 61, $\partial$) is such that $AB \geqslant AC$, then $\angle BCA \geqslant \angle ABC$; but in this case, obviously,
$$
\angle BCD > \angle BCA \geqslant \angle ABC > \angle DBC
$$
and considering $BD > CD$. Adding this... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,103 |
5. On the plane, four points $A, B, C$, and $D$ are given. Prove that:
a) if points $B$ and $C$ belong to the segment $A D$ and are symmetric with respect to the midpoint $S$ of this segment, then the sum of the distances from any point $M$ on the plane to points $A$ and $D$ is not less than the sum of the distances f... | 5. a) Let $N$ be the point symmetric to point $M$ with respect to the common midpoint $S$ of segments $AD$ and $BC$ (Fig. 62). In this case, quadrilaterals $AMDN$ and $BMCN$ are parallelograms, with the second one lying inside the first. If one convex quadrilateral lies inside another, then the perimeter of the former ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,104 |
6. a) The radii of two circles \( S_1 \) and \( S_2 \) are \( r_1 \) and \( r_2 \), and the distance between their centers \( O_1 \) and \( O_2 \) is \( d \). What is the minimum distance between the points of the circles? What is the maximum distance between their points?
b) Answer the same question with the circles ... | 6. a) Let the points of intersection of the circles $S_{1}$ and $S_{2}$ with their line of centers $O_{1} O_{2}$ be denoted by $A_{1}, B_{1}$ and $A_{2}, B_{2}$, with the point $O_{2}$ lying on the ray $O_{1} B_{1}$, and the point $O_{1}$ on the ray $O_{2} B_{2}$ (Fig. $63, a-c$; if the circles $S_{1}$ and $S_{2}$ are ... | r_1+r_2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,105 |
7. a) Let $T_{1}$ and $T_{2}$ be the contours of two equilateral triangles parallel to each other on a plane, with a common center and sides of lengths 1 and 2, respectively. What is the distance $\rho\left(T_{1}, T_{2}\right)$ from $T_{1}$ to $T_{2}$? What is the distance $\rho\left(T_{2}, T_{1}\right)$ from $T_{2}$ t... | 7. a) If $T_{1}$ is the smaller of the two triangles (Fig. 64), then the distance from point $A_{1}$ of triangle $T_{1}$ (understood as a line, not as part of the plane ${ }^{1}$)) to the nearest point of triangle $T_{2}$ will be the same for all
 Prove that the sum of the distances from $\Phi_{1}$ to $\Phi_{2}$ and from $\Phi_{2}$ to $\Phi_{3}$ is not less than the distance from $\Phi_{1}$ to $\Phi_{3}$:
$$
\rho\left(\Phi_{1}, \Phi_{2}\right)+\rho\left(\Phi_{2}, \Phi_{3}\right) \geqsla... | 8. Let $\rho\left(\Phi_{1}, \Phi_{2}\right)=a, \rho\left(\Phi_{2}, \Phi_{3}\right)=b$, and let $A$ be an arbitrary point of figure $\Phi_{1}$ (see Fig. $65, a$, where $\Phi_{1}, \Phi_{2}$, and $\Phi_{3}$ are closed curves). Since the greatest of the distances from a point of figure $\Phi_{1}$ to the nearest point of fi... | proof | Other | proof | Yes | Yes | olympiads | false | 25,107 |
11. The distances from some point $M$ to the vertices $A$ and $B$ of an equilateral triangle $ABC$ are $MA=2$ and $MB=3$. What can the distance $MC$ from point $M$ to the third vertex of the triangle be? | 11. If $MA$, $MB$, and $MC$ are the distances from an arbitrary point $M$ on the plane to the vertices of an equilateral triangle $ABC$, then the segment $CM$ is not greater than the sum of the distances $AM$ and $BM$ and not less than their difference (see, for example, problem 134b in the book by D. O. Shklyarsky, N.... | 1\leqslantMC\leqslant5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,109 |
12. Let $A_{1}, A_{2}, \ldots, A_{1000}$ be any 1000 points on the plane. Prove that on any circle of radius 1, there exists a point $M$ such that the sum of the distances from $M$ to the points $A_{1}, A_{2}, \ldots, A_{1000}$ is not less than 1000. | 12. Let $M_{1}$ and $M_{2}$ be two arbitrary diametrically opposite points on the circle $S$ with radius 1. In this case, $M_{1} M_{2}=2$, and therefore,
$$
\begin{gathered}
M_{1} A_{1}+M_{2} A_{1} \geqslant M_{1} M_{2}=2 \\
M_{1} A_{2}+M_{2} A_{2} \geqslant 2, \ldots, M_{1} A_{1000}+M_{2} A_{1000} \geqslant 2
\end{ga... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,110 |
14. On a plane, there are two points $A$ and $B$, the distance between which is $d$. Construct a square such that these points lie on its boundary, so that the sum of the distances from point $A$ to the vertices of the square is the smallest possible. What is this sum? | 14. We assert that the desired figure is the square $A A_{1} B B_{1}$, whose diagonal is the segment $A B$; the sum of the distances from $A$ to the vertices of this square is
$A B + A A_{1} + A B_{1} = A B (1 + \sqrt{2}) = (1 + \sqrt{2}) d$.
Indeed, let $C D E F$ be an arbitrary square with side $h$, on the boundary... | (1+\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,112 |
15. Through one point $O$ pass four lines $l_{1}, l_{2}, l_{3}$, and $l_{4}$ (the lines are numbered in the clockwise order; see Fig. 5). Through an arbitrary point $A_{1}$ on line $l_{1}$, a line $A_{1} A_{2} \| l_{4}$ is drawn until it intersects $l_{2}$ at point $A_{2}$; through point $A_{2}$, a line $A_{2} A_{3} \|... | 15. Extend segments $A_{2} A_{3}$ to intersect with $l_{4}$ at point $C$ and $B A_{6}$ to intersect with $A_{2} C$ at point $D$ (Fig. $70, a$). If we denote $O A_{1}=x, A_{2} A_{3}=\dot{y}$, and $O B=z$, then, since $A_{2} C=A_{1} O$ (because
 \cdot 180^{\circ} \leqslant \angle A M B \leqslant 180^{\circ}
$$ | 17. Let $AB$ be the closest to the considered point $M$ among all diagonals and sides of our regular $n$-gon. If point $M$ lies on the diagonal $AB$, then $\angle AOB = 180^{\circ}$; if it does not lie on $AB$, then since point $M$ lies inside the polygon, there will be a neighboring vertex $C$ of $B$ such that $C$ and... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,115 |
18. In an acute-angled triangle $ABC$, the bisector $AD$, the median $BM$, and the altitude $CH$ intersect at one point.
a) Prove that the angle $A$ of the triangle is greater than $45^{\circ}$.
b) Specify all possible values that $\angle A$ can have under the conditions of the problem. | 18. a) It is not difficult to construct $\triangle ABC$ with a given (acute) angle $\angle A = \alpha$ such that the bisector $AD$, the altitude $CH$, and the median $BM$ intersect at one point $Q$: it is sufficient to lay off on side $AU$ of angle $UAV = \alpha$ an arbitrary segment $AC$, drop a perpendicular $CH$ (th... | 5150' | Geometry | proof | Yes | Yes | olympiads | false | 25,116 |
20. In the ring bounded by concentric circles of radii $R$ and $r$ with center $O$, two points $A$ and $B$ are taken; the distance $AB$ is equal to 1. What is the smallest value that the angle $AOB$ can have? | 20. It is clear that if $R<\frac{1}{2}$, then the required segment $A B$ does not exist; therefore, we can assume that $R \geqslant \frac{1}{2}$ (and $R \geqslant r$). If the segment $A B$ is moved parallel, say in the direction of the perpendicular dropped from $O$ to $A B$, moving $A B$ away from $O$, then $\angle A ... | \\alpha=\begin{cases}2\arcsin\frac{1}{2R}& | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,118 |
21. How to draw $n$ rays starting from one point on a plane so that the sum of all possible pairwise angles between these rays is the greatest possible? (Here, the angle between two rays is understood to be no more than $180^{\circ}$.) What is this greatest possible value of the sum of the angles? | 21. Let $l_{1}, l_{2}, \ldots, l_{n}$ be some $n$ rays emanating from a point $O$. If rays $l_{1}$ and $l_{2}$ are directed in opposite directions, then $\angle\left(l_{1}, l_{2}\right)=180^{\circ}$ and for any third ray $l_{3}$, the sum $\angle\left(l_{1}, l_{3}\right)+\angle\left(l_{2}, l_{3}\right)=180^{\circ}$ (reg... | k^{2}\cdot180\quad | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,119 |
22. Let $a_{1}, a_{2}, \ldots, a_{k}$ be some $k$ positive numbers. On the ray $O X$, a segment $O X_{1}=x_{1}$ is laid out, where $x_{1}$ is one of the numbers $a_{1}, a_{2}, \ldots, a_{h}$; then on the ray $X_{1} Y \perp O X_{1}$, a segment $X_{1} X_{2}=x_{2}$ is laid out, where
 $\frac{360^{\circ}}{14}=25 \frac{5}{7}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,121 |
26. a) On a plane, four points are given. Prove that from these points, one can choose three that form the vertices of a right or obtuse triangle; however, the arrangement of the points can be such that no three of them form the vertices of an obtuse triangle ${ }^{1}$ ).
b) On a plane, five points are given. Prove th... | 26. Let's start again with constructing a convex polygon $T$ (the convex hull of our points; see the beginning of the solution to problem 25), inside and on the boundary of which all our points are located, and the vertices coincide with some of them. Next, if point $D$ lies inside $\triangle A B C \equiv T$ (Fig. 73, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,124 |
28. Let $A, B, C, D$ be any four points on a plane; $\alpha=\alpha_{1}$ and $\alpha_{2}$ be the two smallest of the 12 angles formed by these points, and $\beta=\beta_{1}$ and $\beta_{2}$ be the two largest angles (so that $\alpha=. \alpha_{1} \leqslant \alpha_{2} \leqslant \ldots \leqslant \leqslant \beta_{2} \leqslan... | 28. a) If all our points belong to a segment, then the two smallest angles are equal to 0 (i.e., $\alpha_{2} \equiv 0$; compare with problem 29); if point $D$ belongs to $\triangle A B C$, then two angles of the triangle, say, $A$ and $B$, are acute and smaller than the "parts" $B A D$ and $C A D$, respectively $A B D$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,126 |
30. Let $A, B, C, D$ be any four points in space; the 12 angles formed by these points $(\angle B A C, \angle A B C, \angle A C B, \ldots, \angle B D C)$ we denote by $\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{A}_{12}$. Prove that
a) the smallest of the angles $A_{1}, A_{2}, \ldots, A_{12}$ is always no more th... | 30. The statement of the problem follows from the fact that e v e r y triangle has an angle not exceeding $60^{\circ}$ and an angle not less than $60^{\circ}$ (cf. above, pp. 22-23; in this case, if the tetrahedron (triangular pyramid) $A B C D$ is regular, i.e., all its faces are equilateral triangles, then none of th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,128 |
31. For what number $n$ of points $M_{1}, M_{2}, \ldots, M_{n}$
a) in the plane;
b) in space
can they be arranged so that no angle
${ }^{1}$ ) See, for example, B. A. Rosenfeld, I. M. Yaglom, Multidimensional Spaces, Encyclopedia of Elementary Mathematics (EEM), vol. V (geometry), Moscow, "Nauka", 1966, pp. 349-392... | 31. a) Consider a convex polygon $T$ with vertices at some of our points, containing all other points inside itself (see the beginning of the solution to problem 25). In this case,
$1^{\circ}$ If the number of vertices of $T$ is greater than four, then at least one of its angles is obtuse (see problem 24); this means ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,129 |
35*. What is the greatest number of rays in space that form pairwise obtuse angles? | 35. It is easy to see that it is possible to draw four rays, each pair of which forms obtuse angles: for example, it is sufficient to connect the center $O$ of a regular tetrahedron $ABCD$ with all its vertices (it is clear, for example, that $\angle AOB > \angle AO_1B = 90^\circ$, where $O_1$ is the center of the face... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 25,133 |
36. a) From one point in space, three rays emanate. Prove that at least one of the pairwise angles between these rays does not exceed $120^{\circ}$; however, the rays can be drawn such that no pairwise angle between them is less than $120^{\circ}$.
b) From one point in space, four rays emanate. Prove that at least one... | 36. a) Let $O A, O B$ and $O C$ be given rays. Since
$$
\angle A O B + \angle B O C + \angle C O A \leqslant 360^{\circ}
$$
where equality holds only if all rays lie in the same plane (the sum of the planar angles of a trihedral angle $\leqslant 360^{\circ}$), at least one of these angles $\leqslant \frac{360^{\circ}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,134 |
37. Prove that no triangle inscribed in a convex polygon \( M \) can have a larger area than the largest (by area) of all triangles whose vertices coincide with any three vertices of \( M \).
The theorem of problem 37 indicates that to solve the following problem: inscribe a triangle of the largest possible area in a ... | 37. If $X Y Z$ is a triangle inscribed in a given polygon $M$ (Fig. $85, a$), and $P P_{1}$ is a side of $M$ on which the vertex $X$ of the triangle lies, then either $\triangle P Y Z, \triangle X Y Z$, and $\triangle P_{1} Y Z$ are equal in area
 is inscribed in triangle \( ABC \) (see Fig. 9). Prove that the area of at least one of the triangles \( APR, BPQ \), and \( CRQ \) does not exceed
a) \(\frac{1}{4}\) of the area of triangle \( ABC \)
b) the area of triangle \( PQR \).
It is interesting to note that the perimeter of at least ... | 38. a) The areas of triangles having a common angle are in the ratio of the products of the lengths of the sides enclosing this angle (since \( S_{A B C} = \frac{1}{2} a b \sin C \)); therefore, in the notation of Fig. 9 (p. 37)
\[
\frac{S_{A P R}}{S_{A B C}} = \frac{A P \cdot A R}{A B \cdot A C} ; \quad \frac{S_{B P ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,136 |
39. On the extensions of the sides of angles $A, B, C$ of triangle $A B C$, segments equal to the opposite sides are laid out (Fig. 10). Show that the area of the resulting hexagon $\geqslant 13 S$, where $S$ is the area of triangle $A B C$. Can this estimate be improved?
), then (see Fig. 10 on p. 37, where it is assumed that $B C=A U=A V=a, C A=B W=B X=b, A B=C Y=$ $=C Z=\dot{c})$
$$
\frac{S_{A X V}}{S_{A B C}}... | S_{UVWXYZ}\geqslant13S | Geometry | proof | Yes | Yes | olympiads | false | 25,137 |
40. Prove that in any convex hexagon, there exists a diagonal that cuts off a triangle with an area $\leqslant \frac{1}{6} S$, where $S$ is the area of the hexagon.
Can the estimate contained in this problem be improved? | 40. Let the points of intersection of the "main" diagonals $A D, B E$ and $C F$ of the hexagon $A B C D E F$ be denoted by $P, Q$ and $R$ (Fig. 87). In this case,
$S_{A Q B}+S_{B Q C}+S_{C R D}+S_{D R E}+S_{E P F}+S_{F P A} \leqslant$
$$
\leqslant S_{A B C D E F}-S_{P Q R} \leqslant S
$$
therefore, the area of at le... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25,138 |
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