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49. For the diagonals $A D, B E$ and $C F$ of a hexagon $A B C D E F$ inscribed in a circle to intersect at one point, it is necessary and sufficient that the equality $|A B| \cdot|C D| \cdot|E F|=|B C| \cdot|D E| \cdot|F A|$ holds. | 49. Consider the triangle $A C E$, through the vertices of which the lines $A D, C F$ and $E B$ are drawn. The sines of the angles formed by these lines with the sides of the triangle $A C E$ are proportional to the chords on which they stand; therefore, the condition $R=1$ (see problem II. 44) is equivalent to the con... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,458 |
51. A circle intersects side $A B$ of triangle $A B C$ at points $C_{1}$ and $C_{2}$, side $C A$ - at points $B_{1}$ and $B_{2}$, and side $B C$ - at points $A_{1}$ and $A_{2}$. Prove that if the lines $A A_{1}, B B_{1}$ and $C C_{1}$ intersect at one point, then the lines $A A_{2}, B B_{2}$ and $C C_{2}$ also intersec... | 51. By the property of secants drawn from an external point to a circle, or by the property of segments of chords of a circle passing through one point, we have: $\left|B C_{1}\right| \cdot\left|B C_{2}\right|=\left|B A_{1}\right| \cdot\left|B A_{2}\right|,\left|C B_{1}\right| \cdot\left|C B_{2}\right|=$ $=\left|C A_{1... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,460 |
52. On the sides $AB, BC$, and $CA$ of triangle $ABC$, points $C_{1}, A_{1}$, and $B_{1}$ are taken. Let $C_{2}$ be the point of intersection of the lines $AB$ and $A_{1} B_{1}$, $A_{2}$ be the point of intersection of the lines $BC$ and $B_{1} C_{1}$, and $B_{2}$ be the point of intersection of the lines $AC$ and $A_{... | 52. Writing the equality $R=1$ (according to Ceva's and Menelaus' theorems - see problems II.44 and II.45) for the points $A_{1}, B_{1}, C_{1} ; A_{1}, B_{1}, C_{2} ; A_{1}, B_{2}, C_{1}$; $A_{2}, B_{1}, C_{1}$, we obtain that the equality $R=1$ also holds for the points $A_{2}, B_{2}, C_{2}$. Now it remains to prove t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,461 |
53. A line intersects sides $A B, B C$ and the extension of side $A C$ of triangle $A B C$ at points $D, E$ and $F$ respectively. Prove that the midpoints of segments $D C, A E$ and $B F$ lie on one line (Gauss line). | 53. Use Menelaus' theorem (see problem II. 45). As the vertices of the given triangle, take the midpoints of the sides of triangle $A B C$, on the sides and extensions of which the considered points lie. | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,462 |
54. Given a triangle $A B C$. Let's define a point $A_{1}$ on the side $B C$ as follows: $A_{1}$ is the midpoint of the side $K L$ of a regular pentagon $M K L N P$, where vertices $K$ and $L$ lie on $B C$, and vertices $M$ and $N$ lie on $A B$ and $A C$, respectively. Similarly, points $C_{1}$ and $B_{1}$ are defined ... | 54. If $a$ is the length of the side of pentagon $M K L N P, b$ is the length of the side of the pentagon with one side on $A B, c$ is the length of the side of the pentagon, one of whose sides is on $A C$, then $\frac{\left|B A_{1}\right|}{\left|C_{1} B\right|}=\frac{a}{b}$, $\frac{\left|A C_{1}\right|}{\left|B_{1} A\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,463 |
55. 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 ... | 55. Check that points $A_{1}, A_{2}, A_{3}$ and $B_{1}, B_{2}, B_{3}$ lie on the sides of triangle $O_{1} O_{2} O_{3}$ (where $O_{1}, O_{2}, O_{3}$ are the centers of the circles) or on the extensions of these sides, and the ratio of the distances from each of these points to the corresponding vertices of triangle $O_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,464 |
57. Given a triangle $A B C$, $M$ is an arbitrary point on the plane. The bisectors of two angles formed by the lines $A M$ and $B M$ intersect the line $A B$ at points $C_{1}$ and $C_{2}$ ( $C_{1}$ is on the segment $A B$ ), similarly, points $A_{1}$ and $A_{2}$, $B_{1}$ and $B_{2}$ are determined on $B C$ and $C A$. ... | 57. Let us use the equality $\frac{\sin \angle B_{1} A A_{2}}{\sin \angle A_{2} A C_{1}}=\frac{\left|A C_{1}\right|}{\left|A B_{1}\right|} \cdot \frac{\left|B_{1} A_{2}\right|}{\left|A_{2} C_{1}\right|}$. By obtaining similar equalities for other angles and multiplying them, we get our statement based on the results of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,466 |
61. Given a triangle $A B C$. On the sides $B C, C A$ and $A B$, points $A_{1}$ and $A_{2}, B_{1}$ and $B_{2}, C_{1}$ and $C_{2}$ are taken such that $A A_{1}$, $B B_{1}$ and $C C_{1}$ intersect at one point and $A A_{2}, B B_{2}$ and $C C_{2}$ also intersect at one point. Prove that: a) the points of intersection of t... | 61. Let $S$ be the point of intersection of the lines $A_{1} M, B_{1} L$, and $C_{1} K$. Applying Menelaus' theorem (problem II. $45^{*}$, note) to triangles $S M K, S K L$, and $S L M$, we get $\frac{K L_{1}}{L_{1} M} \cdot \frac{M A_{1}}{A_{1} S} \cdot \frac{S C_{1}}{C_{1} K}=-1, \frac{L M_{1}}{M_{1} K} \times$ $\tim... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,470 |
63. Given a triangle $A B C$ and a point $D$. Points $E, F$ and $G$ are located on the lines $A D, B D$ and $C D$, respectively. $K$ is the intersection point of $A F$ and $B E$, $L$ is the intersection point of $B G$ and $C F$, and $M$ is the intersection point of $C E$ and $A G$. The points $P, Q$ and $R$ are the int... | 63. Applying Ceva's theorem (problem II. 44*, note) to triangles $A B D, \quad B D C \quad$ and $C D A$, we get: $\frac{A P}{P B} \cdot \frac{B F}{F D} \cdot \frac{D E}{E A}=1$, $\frac{B Q}{Q C} \cdot \frac{C G}{G D} \cdot \frac{D F}{F B}=1, \frac{C R}{R A} \cdot \frac{A E}{E D} \cdot \frac{D G}{G C}=1$. Multiplying th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,471 |
64. Points $A$ and $A_{1}$, $B$ and $B_{1}$, $C$ and $C_{1}$ are symmetric with respect to the line $l$, and $N$ is an arbitrary point on $l$. Prove that the lines $A N, B N, C N$ intersect the lines $B_{1} C_{1}$, $C_{1} A_{1}$, $A_{1} B_{1}$ respectively at three points lying on one straight line. | 64. Let us first consider the limiting case when point $N$ is at "infinity"; then the 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 $l$). The lines parallel to $l$ and p... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,472 |
65. Let $A_{1}, A_{3}, A_{5}$ be three points on one line, and $A_{2}, A_{4}, A_{6}$ be on another. Prove that the three points where the lines $A_{1} A_{2}$ and $A_{4} A_{5}, A_{2} A_{3}$ and $A_{5} A_{6}, A_{3} A_{4}$ and $A_{6} A_{1}$ intersect pairwise lie on one line (Pappus).
## § 3. Geometric Loci | 65. We will consider that these lines are parallel. This can be achieved through appropriate design or coordinate transformation (see the solution to problem II. 64). Apply Menelaus' theorem (problem II. 45) to triangle \(A_{1} A_{6} M\) (Fig. 16, \(N^{\prime} K^{\prime}\) is parallel to the given lines):
is a... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,476 |
70. On the side $A C$ of triangle $A B C$, a point $K$ is taken, and on the median $B D$ - a point $P$ such that the area of triangle $A P K$ is equal to the area of triangle $B P C$. Find the geometric locus of the points of intersection of the lines $A P$ and $B K$. | 70. Let $\varphi$ be the angle between $B D$ and $A C$; $S_{A P K}=\frac{1}{2}|A K| \times |P D| \sin \varphi, S_{B P C}=\frac{1}{2}|B P| \cdot|D C| \sin \varphi=\frac{1}{2}|B P| \cdot|A D| \sin \varphi$. Since $S_{A P K}=S_{B P C}$, then $|A K| \cdot|P D|=|B P| \cdot|A D|$, or $\frac{|A K|}{|A D|} \cdot \frac{|P D|}{|... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,477 |
71. Through the given point $O$ inside the 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 geometric locus of the feet of the perpendiculars dropped from $O$ to the line $A B$. | 71. Let $C$ be the vertex of the given angle, $\beta$ its measure. Drop perpendiculars $O K$ and $O L$ from $O$ to the sides of the angle (Fig. 17, a). Around
Fig. 17
the quadrilateral $O ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,478 |
72. 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 geometric locus of points $M$. | 72. Consider the quadrilateral $D E P M, \angle D E M=\angle D P M=90^{\circ}$, therefore, this quadrilateral is cyclic. Hence, $\angle D M E=$ $=\angle D P E=45^{\circ}$. The required geometric locus of points is the line $D C$. | DC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,479 |
73. Given an angle with its vertex at point $A$, and point $B$. An arbitrary circle passing through $A$ and $B$ intersects the sides of the angle at points $C$ and $D$ (different from $A$). Find the geometric locus of the centroids of triangles $A C D$. | 73. Consider the case when point $B$ lies inside the given angle. First, note that all resulting $\triangle B C D$ (Fig. 18) are similar to each other, since $\angle B C D=\angle B A D, \angle B D C=\angle B A C$. Therefore, if $N$ is the midpoint of $C D$, then the angles $B N C$ and $B N D$ will be constant. Let's de... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,480 |
74. One vertex of a rectangle is at a given point, two others, not belonging to the same side, - on two given mutually perpendicular lines. Find the geometric locus of the fourth vertices of such rectangles. | 74. If $O$ is the vertex of the angle, $A B C D$ is a rectangle (with $A$ fixed), then points $A, B, C, D, O$ lie on the same circle. Therefore, $\angle C O A=90^{\circ}$, 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 | 23,481 |
75. 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 geometric locus: a) of the centers of the circles circums... | 75. Note that all resulting triangles $ABC$ are similar to each other. Consequently, if in each triangle we take a point $K$ that divides the side $BC$ in the same ratio, then since $\angle AKC$ maintains a constant value, the point $K$ will trace a circle. Therefore, the point $M$, which divides $AK$ in a constant rat... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,482 |
76. Let $B$ and $C$ be two fixed points on a given circle, and $A$ be a variable point on the same circle. Find the geometric locus of the feet of the perpendiculars dropped from the midpoint of $A B$ to $A C$. | 76. 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 geometric locus of points is a circle with chord $BC$, and the ang... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,483 |
77. Find the geometric locus of the points of intersection of the diagonals of rectangles, the sides of which (or their extensions) pass through four given points of the plane. | 77. If $M, N, L$ and $K$ are given points ($M$ and $N$ on opposite sides of a rectangle, $L$ and $K$ also), $P$ is the midpoint of $M N$, $Q$ is the midpoint of $K L$, and $O$ is the point of intersection of the diagonals of the rectangle (Fig. 19), then $\angle P O Q = 90^{\circ}$. Therefore, the required geometric lo... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,484 |
78. 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 geometric locus of the centers of the circles inscribed in triangles $A B C$. | 78. Let the radii of the given circles be denoted by \( R \) and \( r (R \geqslant r) \), the point of tangency of the chord \( BC \) with the smaller circle be \( D \); let \( K \) and \( L \) be the points of intersection of the chords \( AC \) and \( AB \) with the smaller circle, and finally, let \( O \) be the cen... | \frac{r\sqrt{R}}{\sqrt{R}+\sqrt{R-r}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,485 |
79. Given two intersecting circles. Find the geometric locus of the centers of rectangles with vertices on these circles. | 79. Let \(O_{1}\) and \(O_{2}\) be the centers of the given circles, and the line \(O_{1} O_{2}\) intersects the circles at points \(A, B, C, D\) (in sequence). Consider two cases.
a) The rectangle \(K L M N\) is positioned such that the opposite vertices \(K, M\) lie on one circle, and \(L\) and \(N\) lie on the othe... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,486 |
81. Through a point lying at an equal distance from two given parallel lines, a line is drawn intersecting these lines at points $M$ and $N$. Find the geometric locus of the vertices $P$ of equilateral triangles $M N P$. | 81. The desired geometric locus of points - two lines, perpendicular to the given lines. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,488 |
82. Given two points $\boldsymbol{A}$ and $\boldsymbol{B}$ and a line $l$. Find the geometric locus of the centers of circles passing through $A$ and $B$ and intersecting the line $l$. | 82. 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 geometric locus of points is the line $O_{1} O_{2}$, excluding the interval $\left(O_{1} O_{2}\right)$. If $A B$ is parallel to $l$, then the d... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,489 |
83. Given two points $O$ and $M$. Determine: a) the geometric locus of points on the plane that can serve as one of the vertices of a triangle with the circumcenter at point $O$ and the centroid at point $M$; b) the geometric locus of points on the plane that can serve as one of the vertices of an obtuse triangle with ... | 83. a) Let $A$ (Fig. 21) be the vertex of a certain triangle. Extend the segment $A M$ beyond $M$ by $|M N|=\frac{1}{2}|A M|$. The point
Fig. 21 $N$ is the midpoint of the side opposite ve... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,490 |
84. A regular triangle is inscribed in a circle. Find the geometric locus of the points of intersection 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. | 84. Let $A B C$ (Fig. 22) be the original equilateral triangle, and $A_{1} B_{1} C_{1}$ be an arbitrary triangle such that $A_{1} C_{1} \parallel A C$ and $A_{1} B_{1} \parallel A B$,
Fig. ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,491 |
85. Find the geometric locus of the centers of all possible rectangles circumscribed around a given triangle. (A rectangle will be called 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 that do not contain this vertex.) | 85. If $ABC$ (Fig. 23) is a given triangle and the vertex of the circumscribed rectangle $AKLM$ coincides with $A (B$ on $KL, C$ on $LM)$, then $L$ belongs to the semicircle with diameter $BC$, and the angles $ABL$ and $ACL$ are obtuse, i.e., $L$ will have two extreme positions: $L_{1}$ and $L_{2}$, $\angle L_{1}CA = \... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,492 |
86. Given two squares with respectively parallel sides. Determine the geometric locus of points $M$ such that for any point $P$ from the first square, there exists a point $Q$ from the second such that the triangle $M P Q$ is equilateral. Let the sides of the first square be $a$, and the second be $b$. For what ratio b... | 86. If (Fig. 24) the first square is rotated around point $M$ by $60^{\circ}$, either clockwise or counterclockwise, it must fit entirely inside the second square. Conversely, for each square located
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,493 |
88. Given two points $\boldsymbol{A}$ and I. Find the geometric locus of points $B$ such that there exists a triangle $A B C$ with the incenter at point $I$, all angles of which are less than $\alpha$ ( $60^{\circ}<\alpha<$ $<90^{\circ}$ ). | 88. If $A, B, C$ are the angles of $\triangle A B C$, then the angles of $\triangle A B I$ are $\frac{A}{2}, \frac{B}{2}$, $90^{\circ}+\frac{C}{2}$ (Fig. 25); therefore, the required geometric locus of points is a pair of triangles, two sides of which are segments of lines, and the third is an arc, which is part of a s... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,495 |
89. Points $A, B$ and $C$ are located on the same line ( $B$ - between $A$ and $C$ ). Find the geometric locus of points $M$ such that $\operatorname{ctg} \angle A M B+\operatorname{ctg} \angle B M C=k$. | 89. Draw a perpendicular to $B M$ at point $M$; let $P$ be the point of intersection of this perpendicular and the perpendicular erected to the original line at point $B$. We will show that the magnitude $|P B|$ is constant. Let $\angle M B C=\varphi$; through $K$ and $L$ denote the feet of the perpendiculars dropped f... | |PB|=\frac{k|BA|\cdot|BC|}{|BA|+|BC|} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,496 |
90. Given two points $A$ and $Q$. Find the geometric locus of points $B$ such that there exists an acute triangle $A B C$, for which $Q$ is the centroid. | 90. 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 $B C$ of triangle $A B C ; \quad \angle C B A_{1}=\angle B C A$, $\angle A B A_{1}=180^{\circ}-\angle B A C$.
Therefore, if we c... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,497 |
92. On a plane, two rays are given. Find the geometric locus 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.) | 92. If the ends of the rays do not coincide, the sought geometric locus of points 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 geome... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,499 |
93. Given an angle and a circle inscribed in this angle with its center at point $O$. An arbitrary tangent line to the circle intersects the sides of the angle at points $M$ and $N$. Find the geometric locus of the centers of the circumcircles of triangles $MON$. | 93. Let $A$ be the vertex of an angle. It can be proven that the center of the circle circumscribed around $\triangle M O N$ coincides with the point of intersection of the bisector $A O$ and the circle circumscribed around $A M N$. Let $\alpha$ be the measure of the angle, $r$ be the radius of the circle, and $K$ be t... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,500 |
94. Given two circles, points $A$ and $B$ are taken on each, equidistant from the midpoint of the segment connecting their centers. Find the geometric locus of the midpoints of segments $A B$. | 94. Let's denote: $O_{1}, O_{2}$ - the centers of the circles, $r_{1}, r_{2}$ - their radii, $M$ - the midpoint of $A B$, $O$ - the midpoint of $O_{1} O_{2}$. We have (by the formula for the length of the median, problem I.11) $\left|O_{1} M\right|^{2}=\frac{1}{4}\left(2 r_{1}^{2}+2\left|O_{1} B\right|^{2}-|A B|^{2}\ri... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,501 |
95. Given a segment $A B$. Take an arbitrary point $M$ on $A B$ and consider two squares $A M C D$ and $M B E F$, located on the same side of $A B$. Describe circles around these squares and denote by $N$ their point of intersection, different from $M$. Prove that: a) $A F$ and $B C$ intersect at $N$; b) $M N$ passes t... | 95. a) Since $\angle F N B=90^{\circ}, \angle C N M=135^{\circ}, \angle F N M=45^{\circ}$ (assuming $|A M|>|M B|$), then $\angle F N C=90^{\circ}$ and $C, N$ and $B$ are collinear, etc.
b) Consider the isosceles right triangle $A B K$ with hypotenuse $A B$ ( $K$ is on the opposite side of $A B$ from the squares). The ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,502 |
96. Given a circle and a point $A$. Let $M$ be an arbitrary point on the circle. Find the geometric locus of the points of intersection of the perpendicular bisector of the segment $A M$ and the tangent to the circle passing through $M$. | 96. Let $N$ be the point of intersection of the perpendicular bisector and the tangent, $O$ be the center of the circle, and $R$ be its radius. We have: $|O N|^{2}-$ $-|N A|^{2}=R^{2}+|M N|^{2}-|N A|^{2}=R^{2}$. Therefore, the required geometric locus of points is a line perpendicular to $O A$ (Problem II.1). | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,503 |
97. Two circles touch each other at point $A$. One line passing through $A$ intersects these circles again at points $B$ and $C$, another - at points $B_{1}$ and $C_{1}$ ( $B$ and $B_{1}$ - on the same circle). Find the geometric locus of the points of intersection of the circles circumscribed around triangles $A B_{1}... | 97. If $O_{1}$ and $O_{2}$ are the centers of the given circles, $Q_{1}$ and $Q_{2}$ are the centers of the circumcircles of triangles $A B C_{1}$ and $A B_{1} C$, then $O_{1} Q_{1} O_{2} Q_{2}$ is a parallelogram. The line $Q_{1} Q_{2}$ passes through the midpoint of the segment $O_{1} O_{2}$ (point $D$). The second p... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,504 |
98. Find the geometric locus of the vertices of right angles of all possible isosceles right triangles, the ends of whose hypotenuses lie on two given circles. | 98. Let $O_{1}$ and $O_{2}$ be the centers of the given circles, and $r_{1}$ and $r_{2}$ their radii. Consider two isosceles right triangles with hypotenuse $O_{1} O_{2}-O_{1} O_{2} O$ and $O_{1} O_{2} O^{\prime}$. The desired geometric locus of points is two annuli with centers at $O$ and $O^{\prime}$ and radii: the o... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,505 |
99. The sides of the given triangle are the diagonals of three parallelograms. The sides of these parallelograms are parallel to two lines $-l$ and $p$. Prove that the three diagonals of these parallelograms, different from the sides of the triangle, intersect at one point $M$. Find the geometric locus of points $M$ if... | 99. The union of three constructed parallelograms represents a parallelogram circumscribed around a given triangle, divided into four smaller ones. It is not difficult to express the ratios in which each of the considered diagonals is divided by another diagonal in terms of the segments of the sides of the larger paral... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,506 |
100. Let $B$ and $C$ be two fixed points on a circle, $A$ be an arbitrary point on this circle; $H$ be the intersection point of the altitudes of triangle $A B C$, and $M$ be the projection of $H$ onto the bisector of angle $B A C$. Find the geometric locus of points $M$. | 100. We will prove that $\frac{|A M|}{|A D|}=|\cos \angle B A C|$, where $D$ is the intersection point of $A M$ with the circle. Let $O$ be the center of the circle, $P$ the midpoint of $B C$, and $K$ the midpoint of $A H$. Triangles $D O A$ and $M K A$ are similar. Therefore, $\frac{|M A|}{|A D|}=\frac{|A K|}{|D O|}=\... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,507 |
101. Given a triangle $A B C$. Let $D$ be an arbitrary point on the line $B C$. Lines passing through $D$ parallel to $A B$ and $A C$ intersect $A C$ and $A B$ at points $E$ and $F$. Find the geometric locus of the centers of circles passing through points $D$, $E$, and $F$. | 101. Let $B_{0}$ and $C_{0}$ be the midpoints of sides $A C$ and $A B$, $B B_{1}$ and $C C_{1}$ be the altitudes, $K$ be the midpoint of $D E$ (Fig. 26), $G K$ and $C_{0} N$ be perpendicular to $A B$, and $B_{0} M$ be perpendicular to $A C$. Then $\frac{|M L|}{|N M|}=\frac{\left|G C_{1}\right|}{\left|C_{0} C_{1}\right|... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,508 |
102. Given $A B C$ - an equilateral triangle. Find the geometric locus of points $M$ inside this triangle such that $\angle M A B + \angle M B C + \angle M C A = \pi / 2$. | 102. It is obvious that any point of any height of triangle $ABC$ belongs to the desired geometric locus. We will show that there are no other points. Take a point $M$ that does not lie on the heights of triangle $ABC$. Let the line $BM$ intersect the heights dropped from vertices $A$ and $C$ at points $M_{1}$ and $M_{... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,509 |
103. Inside a triangle, a point $M$ is taken such that there exists a line $l$ passing through $M$ and dividing the given triangle into two parts in such a way that when one part is reflected across $l$, it lies inside or on the boundary of the other part. Find the geometric locus of points $M$.
## § 4. Triangle. Tria... | 103. Note that if any line $l$ passing through $M$ has the required property, then there exists either a line $l_{1}$ passing through $M$ and one of the vertices of the triangle, or a line $l_{2}$ passing through $M$ and perpendicular to one of the sides of the triangle, also possessing this property. Indeed, let the l... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,510 |
115. The circle inscribed in triangle $ABC$ touches sides $AB$ and $AC$ at points $C_{1}$ and $B_{1}$, and the circle that touches side $BC$ and the extensions of $AB$ and $AC$ touches the lines $AB$ and $AC$ at points $C_{2}$ and $B_{2}$. Let $D$ be the midpoint of $BC$. Line $AD$ intersects lines $B_{1} C_{1}$ and $B... | 115. Draw a line through $D$ perpendicular to the bisector of angle $A$, and denote the points of its intersection with $A B$ and $A C$ as $K$ and $M$, and prove that $|A K|=|A M|=\frac{b+c}{2}$. Since $\left|A C_{1}\right|=\left|A B_{1}\right|=p-a$, $\left|A C_{2}\right|=\left|B C_{2}\right|=p$ ( $p$ - the semiperimet... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,518 |
116. In triangle $ABC$, the bisector of the internal angle $AD$ is drawn. Construct the tangent $l$ to the circumscribed circle at point $A$. Prove that the line drawn through $D$ parallel to $l$ is tangent to the inscribed circle of triangle $ABC$. | 116. Prove that $l$ forms the same angles with $A D$ as the line $B C$, which is tangent to our circle. It follows from this that the other tangent to the circle passing through $D$ will be parallel to $l$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,519 |
117. In triangle $ABC$, a line is drawn intersecting sides $AC$ and $BC$ at points $M$ and $N$ such that $|MN| = |AM| + |BN|$. Prove that all such lines are tangent to the same circle. | 117. Let's construct a circle that is tangent to the lines $M N, A C$ and $B C$ in such a way that the points of tangency $P$ and $Q$ with the lines $A C$ and $B C$ are outside the segments $C M$ and $C N$ (this will be the excircle of triangle $M C N$). If $R$ is the point of tangency with $M N$, then $|M P|=|M R|,|N ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,520 |
118. Prove that the points symmetric to the center of the circle circumscribed around a triangle with respect to the midpoints of its medians lie on the heights of the triangle. | 118. If $O$ is the center of the circle circumscribed around $\triangle A B C, D$ is the midpoint of $C B, H$ is the point of intersection of the altitudes, $L$ is the midpoint of $A H$, then $|A L|=|O D|$ and, since $A L \| O D$, $O L$ bisects $A D$, i.e., $L$ is symmetric to $O$ with respect to the midpoint of $A D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,521 |
119. Prove that if the height of a triangle is $\sqrt{2}$ times the radius of the circumscribed circle, then the line connecting the bases of the perpendiculars dropped from the base of this height to the sides containing it passes through the center of the circumscribed circle. | 119. Let $B D$ be the height of the triangle, and $|B D|=R \sqrt{2}$, where $R$ is the radius of the circumscribed circle. $K$ and $M$ are the feet of the perpendiculars dropped from $D$ to $A B$ and $B C$, and $O$ is the center of the circumscribed circle. If angle $C$ is acute, then $\angle K B O=90^{\circ}-\angle C$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,522 |
120. Let $A B C$ be a right triangle ( $\angle C=90^{\circ}$ ), $C D$ be the altitude, $K$ be a point on the plane such that $|A K|=|A C|$. Prove that the diameter of the circumcircle of triangle $A B K$, passing through vertex $A$, is perpendicular to the line $D K$. | 120. Note that $\triangle A D K$ is similar to $\triangle A B K$, since $|A K|^{2}=$ $=|A C|^{2}=|A D| \cdot|A B|$. If $O$ is the center of the circumcircle of $\triangle A B K$, then $\angle O A D+\angle A D K=90^{\circ}-\angle A K B+\angle A D K=90^{\circ}$ (it was assumed that $\angle A K B$ is acute; if $\angle A K... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,523 |
122. Two circles pass through the vertex of an angle and a point lying on the bisector. Prove that the segments of the sides of the angle, enclosed between the circles, are equal. | 122. If $O$ is the vertex of the angle, $A$ is a point on the bisector, $B_{1}$ and $B_{2}$ are the points of intersection with the sides of the angle of one circle, $C_{1}$ and $C_{2}$ ( $B_{1}$ and $C_{1}$ are on the same side) are the points of intersection of another circle, then $\triangle A B_{1} C_{1}=\triangle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,525 |
130. Given a right triangle $A B C$; angle $C$ is right, $O$ is the center of the inscribed circle, $M$ is the point of tangency of the inscribed circle with the hypotenuse, the circle with center at $M$, passing through $O$, intersects the angle bisectors of angles $A$ and $B$ at points $K$ and $L$, different from $O$... | 130. If $K N$ is the perpendicular from $K$ to $A B, \angle C A B=\alpha$, then
\[
\begin{aligned}
& \frac{|K N|}{|O M|}=\frac{|A K|}{|A O|}=\frac{|A O|-|K O|}{|A O|}=\frac{|A O|-2|O M| \sin \frac{\alpha}{2}}{|A O|}= \\
& =\frac{|A O|-2|A O| \sin ^{2} \frac{\alpha}{2}}{|A O|}=\cos \alpha=\frac{|C D|}{|C B|}. \text{ Si... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,532 |
131. Prove that in triangle $ABC$, the bisector of angle $A$, the midline parallel to $AC$, and the line connecting the points of tangency of the inscribed circle with sides $CB$ and $CA$, intersect at one point. | 131. Let $C_{1}$ and $A_{1}$ be the midpoints of $A B$ and $B C$, $B^{\prime}$ and $A^{\prime}$ - the points of tangency of the inscribed circle with $A C$ and $B C$. Suppose, for definiteness, that $c \geqslant b$ (where $c$ and $b$ are the sides of $\triangle A B C$), then the angle bisector of $\angle A$ intersects ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,533 |
132. Prove that three lines, passing respectively through the bases of two altitudes of a triangle, the ends of two of its angle bisectors, and through two points of tangency of the inscribed circle with its sides (all points are located on two sides of the triangle), intersect at one point. | 132. Consider an angle with vertex $A$. On one side of the angle, three points $-B_{1}, B_{2}, B_{3}$ are taken, and on the other side, $C_{1}, C_{2}, C_{3}$. From Menelaus' theorem (Problem II. 45, note), it follows that for the lines $B_{1} C_{1}, B_{2} C_{2}, B_{3} C_{3}$ to intersect at one point, it is necessary a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,534 |
133. On the sides $B C, C A$ and $A B$ of triangle $A B C$, points $A_{1}, B_{1}$ and $C_{1}$ are taken such that the lines $A A_{1}, B B_{1}$ and $C C_{1}$ intersect at one point. Prove that if $A A_{1}$ is the bisector of angle $B_{1} A_{1} C_{1}$, then $A A_{1}$ is the altitude of triangle $A B C$. | 133. Draw a line through $A$ parallel to $B C$, and denote by $K$ and $L$ the points of its intersection with $A_{1} C_{1}$ and $A_{1} B_{1}$, respectively. We have: $\frac{|K A|}{\left|B A_{1}\right|}=\frac{\left|A C_{1}\right|}{\left|C_{1} B\right|}, \frac{\left|C B_{1}\right|}{\left|B_{1} A\right|}=\frac{\left|A_{1}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,535 |
134. On the sides $B C, C A$ and $A B$ of triangle $A B C$, points $A_{1}, B_{1}$ and $C_{1}$ are taken respectively such that $\angle A A_{1} C = \angle B B_{1} A = \angle C C_{1} B$ (angles are measured in the same direction). Prove that the center of the circle circumscribed around the triangle bounded by the lines ... | 134. Let $K$ be the point of intersection of $A A_{1}$ and $B B_{1}$, and $H$ be the point of intersection of the altitudes of triangle $A B C$. Points $A, K, H$, and $B$ lie on the same circle (angles $A K B$ and $A H B$ are either equal or their sum is $180^{\circ}$, depending on whether points $K$ and $H$ are on the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,536 |
135. The vertices of triangle $A_{1} B_{1} C_{1}$ are located on the lines $B C, C A$ and $A B\left(A_{1}\right.$ - on $B C, B_{1}$ - on $C A, C_{1}$ - on $\left.A B\right)$. Prove that if triangles $A B C$ and $A_{1} B_{1} C_{1}$ are similar (corresponding vertices are $A$ and $A_{1}, B$ and $B_{1}, C$ and $C_{1}$ ), ... | 135. Let $H$ be the orthocenter of triangle $A_{1} B_{1} C_{1}$. Points $A_{1}, H, B_{1}$, and $C$ lie on the same circle, and points $B_{1}, H, C_{1}$, and $A$ also lie on the same circle, with the radii of these circles being equal. Angles $H B_{1} C$ and $H B_{1} A$ are either equal or supplementary to $180^{\circ}$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,537 |
136. 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 the same circle. | 136. 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 originating from vertex $B, |B K|=|B L|=l, M$ b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,538 |
137. 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 $ABC$ that is diametrically opposite to $B$. Prove that the perpendicular drop... | 137. Let (Fig. 28): \( |BC| = a, |CA| = b, |AB| = c \). 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: \( \angle POQ = \angle ABC, \quad |OQ| = p - c, \quad |OP| = p - a \), where \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,539 |
138. From a certain point on the circumcircle of an equilateral triangle $ABC$, lines parallel to $BC$, $CA$, and $AB$ are drawn, intersecting $CA$, $AB$, and $BC$ at points $M$, $N$, and $Q$ respectively. Prove that $M$, $N$, and $Q$ lie on the same line. | 138. Let $P$ lie on the arc $A C$ for definiteness. Points $A, M$, $P, N$ lie on the same circle, so $\angle N M P=\angle N A P$. Similarly, $P, M, Q, C$ lie on the same circle, $\angle P M Q=180^{\circ}-\angle P C Q=$ $=180^{\circ}-\angle P A N=180^{\circ}-\angle P M N$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,540 |
139. 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. | 139. Let $A B C$ be the given triangle (Fig. 29), and $H$ be the point of intersection of its altitudes. Note that the points symmetric to $H$ with respect to its sides lie on the circumcircle of triangle $A B C$ (see problem II.107). If $H_{1}$ is the point symmetric to $H$ with respect to
$. | 140. Let points $A, B, C$ and $M$ in the Cartesian coordinate system have coordinates respectively $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right),(x, y)$,
F... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,542 |
141. Let $A B C$ be an equilateral triangle with side $a$, and $M$ be some point on the plane at a distance $d$ from the center of the triangle $A B C$. Prove that the area of the triangle with sides equal to the segments $M A, M B$, and $M C$ is given by the formula $S=\frac{\sqrt{3}}{12}\left|a^{2}-3 d^{2}\right|$. | 141. Consider the case when point $M$ (Fig. 30) lies inside triangle $ABC$. Rotate triangle $ABM$ around $A$ by an angle of $60^{\circ}$ so that $B$ moves to $C$. We obtain triangle $AM_1C$, which is equal to $\triangle ABM$, and $\triangle AMM_1$ is equilateral. Therefore, the sides of $\triangle CMM_1$ are equal to t... | \frac{\sqrt{3}}{12}(^2-3d^2) | Geometry | proof | Yes | Yes | olympiads | false | 23,543 |
143. Given a triangle $A B C$. On the rays $A B$ and $C B$, segments $A K$ and $C M$ are laid off, equal to $A C$. Prove that the radius of the circumcircle of triangle $B K M$ is equal to the distance between the centers of the inscribed and circumscribed circles of triangle $A B C$, and that the line $K M$ is perpend... | 143. Let (Fig. $31, a$) $O$ be the center of the circumscribed circle, and $I$ be the center of the inscribed circle. Drop perpendiculars from $O$ and $I$ to $AB$ and $BC$; $ON, OP$,
$a$
!... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,545 |
145. Prove that if the lengths of the sides of a triangle form an arithmetic progression, then: a) the radius of the inscribed circle is equal to $1 / 3$ of the height dropped to the middle side; b) the line connecting the centroid of the triangle with the center of the inscribed circle is parallel to the middle side; ... | 145. Let the sides of the triangle be $a, b$, and $c$, with $b = (a + c) / 2$.
a) From the equality $p r = \frac{1}{2} b h_{b}$ (where $p$ is the semiperimeter, $r$ is the radius of the inscribed circle, and $h_{b}$ is the height dropped to side $b$), we get: $\frac{1}{2}(a + b + c) = \frac{1}{2} b h_{b}$; but $a + c =... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,547 |
146. Let $K$ be the midpoint of side $BC$ of triangle $ABC$, and $M$ be the foot of the altitude dropped to $BC$. The incircle of triangle $ABC$ touches side $BC$ at point $D$; the excircle opposite to $A$, which touches the extensions of $AB$ and $AC$ and side $BC$, touches $BC$ at point $E$. The common tangent to the... | 146. Let $N$ be the point of intersection of the common tangent with $B C$. It is sufficient to check that $|F N| \cdot|N G|=|K N| \cdot|N M|=|D N| \cdot|N E|$. All segments are easily calculated, since $|B D|=|C E|=p-b$, $|D E|=|b-c|, \frac{|D N|}{|N E|}=\frac{r}{r_{a}}=\frac{p-a}{p}\left(r_{a}-\right.$ radius of the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,548 |
147. Prove that the centroid of a triangle, the orthocenter, and the circumcenter lie on the same line (Euler's line). | 147. Draw lines through the vertices of triangle $A B C$ parallel to the opposite sides. They form $\triangle A_{1} B_{1} C_{1}$, similar to $\triangle A B C$; it is obtained from $\triangle A B C$ by a homothety with the center at the common centroid of $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$, and the coef... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,549 |
148. Which sides does the Euler line intersect in an acute and obtuse triangle? | 148. In an acute triangle, the Euler line intersects the largest and the smallest sides. In an obtuse triangle - the largest and the middle. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,550 |
149. Let $K$ be the point symmetric to the center of the circumcircle of $\triangle ABC$ with respect to the side $BC$. Prove that the Euler line of triangle $ABC$ bisects the segment $AK$. | 149. Show that the required property is possessed by a point $P$ on the Euler line for which $|P O|=|O H|$ (where $O$ is the circumcenter and $H$ is the orthocenter); in this case, for each triangle, the distance from the centroid to the opposite vertex of the original triangle is $\frac{4}{3} R$, where $R$ is the radi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,551 |
150. 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. | 150. Let $C_{1}$ be the center of the circumcircle of $\triangle A P B$, and $C_{2}$ be 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... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,552 |
155. Let $M$ be a point on the circumcircle of triangle $ABC$. A line passing through $M$ and perpendicular to $BC$ intersects the circle again at point $N$. Prove that the Simson line corresponding to point $M$ is parallel to the line $AN$. | 155. The distance between the projections of $M$ on $A C$ and $B C$ is $|C M| \sin C$. If $K$ and $L$ are the projections of $M$ on $A B$ and $B C$, then the projection of $A B$ on the line $K L$ (which is the Simson line) is $|A B||\cos \angle B K L|=$ $=|A B||\cos \angle \dot{B} M L|=|A B| \sin \angle C B M=|C M| \si... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,556 |
157. Let $A A_{1}, B B_{1}, C C_{1}$ be the altitudes of triangle $A B C$. The lines $A A_{1}, B B_{1}, C C_{1}$ intersect the circumcircle of triangle $A B C$ again at points $A_{2}, B_{2}, C_{2}$, respectively. The Simson lines corresponding to points $A_{2}, B_{2}, C_{2}$ form triangle $A_{3} B_{3} C_{3}$ (where $A_... | 157. Prove that the Simson line corresponding to $A_{1}$ is perpendicular to $B_{1} C_{1}$ (the same applies to other points). Further, it can be proven that the Simson line corresponding to point $A_{1}$ passes through the midpoint of $A_{1} H$, where $H$ is the orthocenter of triangle $A B C$ (see also the solution t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,558 |
160. Prove that the midpoints of the sides of a triangle, the feet of the altitudes, and the midpoints of the segments of the altitudes from the vertices to their point of intersection lie on one circle - the "nine-point circle" (E i l e r).
Translate the above text into English, please retain the original text's line... | 160. Our statement follows from the fact that $D$ lies on the nine-point circle, and this circle is homothetic to the circumcircle with center $H$ and coefficient $1 / 2$. (see problem II.160). | Geometry | math-word-problem | Yes | Yes | olympiads | false | 23,561 | |
163. The height dropped from vertex $A$ of triangle $ABC$ intersects the circumscribed circle at point $A_{1}$. Prove that the distance from the center of the nine-point circle to side $BC$ is equal to $\frac{1}{4}\left|A A_{1}\right|$. | 163. Let $M_{0}$ be the midpoint of $H P$, $A_{0}$ be the midpoint of $H A$, and $A_{0}, A_{1}$, and $M_{0}$ lie on the nine-point circle. Therefore, $M$ also lies on this circle, since from the problem statement it follows that $\left|M_{0} H\right| \cdot|H M|=\left|A_{0} H\right| \cdot\left|H A_{1}\right|$ and $H$ is... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,564 |
164. In triangle $A B C$, $A A_{1}$ is the altitude, $H$ is the orthocenter. Let $P$ be an arbitrary point on the circumcircle of triangle $A B C$, and $M$ be a point on the line $H P$ such that $|H P| \cdot|H M|=\left|H A_{1}\right| \cdot|H A|(H$ is on the segment $M P$ if triangle $A B C$ is acute and outside it if i... | 164. We will prove that $M$ and $N$ are on the corresponding midlines of triangle $ABC$. If $P$ is the midpoint of $AB$, then $\angle MPA = 2 \angle ABM = \angle ABC = \angle APL$. Let, for definiteness, $ABC$ be an acute triangle, $\angle C \geqslant \angle A$, then $\angle MNK = 180^{\circ} - \angle KNB = \angle KCB ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,565 |
166. 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 II. 153, II. 159). | 166. Since the midpoint of $F H$ lies on the nine-point circle (see problem II.160), 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 some side of the triangle, $D$ the foot of the altitude drawn to the same side, $H_{1}$ the intersection... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,566 |
167. Let $l$ be an arbitrary line passing through the center of the circumcircle of triangle $ABC$, and let $A_1$, $B_1$, and $C_1$ be the projections of $A$, $B$, and $C$ onto $l$. Draw a line through $A_1$ perpendicular to $BC$, a line through $B_1$ perpendicular to $AC$, and a line through $C_1$ perpendicular to $AB... | 167. In Fig. $32 O$ is the center of the circumscribed circle, $A_{1}, B_{1}, C_{1}$ are the midpoints of the sides, $L$ and $K$ are the projections of $A$ and $B$ onto $l$, $M$ is the intersection point of the lines passing through $L$ and $K$ perpendicular to $BC$ and $CA$. The triangle $ABC$, for definiteness, is an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,567 |
168. 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 $P A, P B, P C$ is equal to the sum of the other two segments (Victor... | 168. Let $H$ be the orthocenter 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 $C_{2}$ are the centers of t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,568 |
169. Prove that three circles, each passing through a vertex of a triangle, the foot of the altitude dropped from this vertex, and tangent to the radius of the circumcircle of the triangle drawn to this vertex, intersect at two points located on the Euler line of the given triangle. | 169. Let $A B C$ be a given triangle, and $A_{1}, B_{1}, C_{1}$ the midpoints of the corresponding sides. Prove that the circle passing through vertex $A$ and satisfying the condition of the problem passes through the points of intersection of the internal and external bisectors of angle $A$ with the midline $B_{1} C_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,569 |
170. Consider three circles, each passing through one vertex of a triangle and the bases of the two angle bisectors - internal and external, emanating from this vertex (these circles are known as Apollonian circles). Prove that: a) these three circles intersect at two points ($M_{1}$ and $M_{2}$); b) the line $M_{1} M_... | 170. a) Similarly to how it was done in the previous problem, it can be proven that these three circles intersect at two points $M_{1}$ and $M_{2}$, and that $\left|A M_{1}\right|:\left|B M_{1}\right|:\left|C M_{1}\right|=b c: a c: a b$ (also for $\left.M_{2}\right)$.
b) Follows from a) and problem II.14.
c) Prove th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,570 |
171. A line symmetric to the median of a triangle with respect to the bisector of the same angle is called a symmedian. Let the symmedian from vertex $B$ of triangle $A B C$ intersect $A C$ at point $K$. Prove that $|A K|:|K C| = |A B|^{2}:|B C|^{2}$. | 171. Let's take point $A_{1}$ on $BC$ and point $C_{1}$ on $BA$ such that $\left|B A_{1}\right| = |B A|$, $\left|B C_{1}\right| = |B C|$ (triangle $\triangle A_{1} B C_{1}$ is symmetric to $\triangle A B C$ with respect to the bisector of angle $B$). Clearly, $B K$ bisects $A_{1} C_{1}$. We construct two parallelograms... | \frac{|AK|}{|KC|}=\frac{|AB|^{2}}{|BC|^{2}} | Geometry | proof | Yes | Yes | olympiads | false | 23,571 |
172. Let $D$ be an arbitrary point on side $BC$, and let $E$ and $F$ be points on $AC$ and $AB$ such that $DE$ is parallel to $AB$, and $DF$ is parallel to $AC$. The circle passing through $D$, $E$, and $F$ intersects $BC$, $CA$, and $AB$ again at points $D_{1}$, $E_{1}$, and $F_{1}$, respectively. Let $M$ and $N$ be t... | 172. We have (Fig. 34) $\angle F E_{1} A = \angle E D F = \angle A$; therefore, $|A F| = |E_{1} F|, \quad \angle F E_{1} N = \angle F D B = \angle C, \quad \angle E_{1} F N = \angle A. \quad$ Consequently, $\triangle E_{1} F N$ is similar to $\triangle A B C, \frac{|A F|}{|F N|} = \frac{|E_{1} F|}{|F N|} = \frac{|A C|}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,572 |
174. Given a trapezoid $A B C D$, in which the lateral side $C D$ is perpendicular to the bases $A D$ and $B C$. A circle with diameter $A B$ intersects $A D$ at point $P$ ( $P$ is different from $A$ ). The tangent to the circle at point $P$ intersects $C D$ at point $M$. From $M$, a second tangent to the circle is dra... | 174. Let $N$ be the intersection of $B Q$ and $C D$, $O$ be the center of the circle, and $R$ be its radius. Note that $\angle N B C=\frac{1}{2} \angle P M Q$. (If $Q$ is on the segment $N B$, then $\angle N B C=90^{\circ}-\angle Q B P=90^{\circ}-\frac{1}{2} \angle Q O P=\frac{1}{2} \angle P M Q$.) Therefore, triangles... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,574 |
175. Let $M$ and $N$ be the projections of the orthocenter of triangle $ABC$ onto the internal and external angle bisectors of $\angle B$. Prove that the line $MN$ bisects the side $AC$. | 175. Let $H$ be the orthocenter, $O$ be the circumcenter, and $B_{1}$ be the midpoint of $C A$. The line $M N$ passes through the midpoint of $B H-$ point $K,|B K|=\left|B_{1} O\right|$. Prove that the line $M N$ is parallel to $O B$ (if $\angle C>\angle A$, then $\angle M K N=2 \angle M B N=\angle C-\angle A=\angle O ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,575 |
176. Given a circle and two points $A$ and $B$ on it. The tangents to the circle passing through $A$ and $B$ intersect at point $C$. A circle passing through $C$ touches the line $A B$ at point $B$ and intersects the given circle again at point $M$. Prove that the line $A M$ bisects the segment $C B$. | 176. Let the line $A M$ intersect the circle passing through $B, C$ and $M$ again at point $D$. Then $\angle M D B = \angle M B A = \angle M A C$, $\angle M D C = \angle M B C = \angle M A B$. Therefore, $A B D C$ is a parallelogram. | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,576 |
177. From a point $A$ located outside a circle, two tangents $A M$ and $A N$ ($M$ and $N$ - points of tangency) and a secant intersecting the circle at points $K$ and $L$ are drawn. Draw an arbitrary line $l$ parallel to $A M$. Let $K M$ and $L M$ intersect $l$ at points $P$ and $Q$. Prove that the line $M N$ bisects t... | 177. From the solution of problem I. 234, it follows that $\frac{|L M|}{|M K|}=\frac{|L N|}{|N K|}$. We can assume that $l$ passes through $N$. Apply the Law of Sines to $\triangle N K P$. Replace the ratio of sines with the ratio of the corresponding chords. We will have: $\quad|N P|=\frac{|N K| \sin \angle N K P}{\si... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,577 |
178. 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 $BC$. | 178. 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 cyclic $\quad(\angle ONM = \angle OKM = 90^{\circ})$; therefore, $\angle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,578 |
179. A circle is inscribed in triangle $A B C$. Let $M$ be the point of tangency of the circle with side $A C$, and $M K$ be a diameter. Line $B K$ intersects $A C$ at point $N$. Prove that $|A M|=|N C|$. | 179. If $|B C|=a,|C A|=b,|A B|=c$, then, as is known (see problem I.18), $|M C|=\frac{a+b-c}{2}$. Draw a line through $K$ parallel to $A C$; denote its points of intersection with $A B$ and $B C$ as $A_{1}$ and $C_{1}$, respectively. The incircle of $\triangle A B C$ is an excircle (touching $A_{1} C_{1}$ and the exten... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,579 |
180. 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 $BC$. | 180. 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 II.179), it is sufficient to prove that $|N L|=|... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,580 |
181. 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 $AB$. | 181. 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 $|M K|=\frac{|K N| \cdot|M B|}{|D N|}$. From the similarity of triangles $C N L$ and $M L B$, it follows that $|M L|=\frac{|... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,581 |
182. 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... | 182. Let (Fig. 35) $B$ be the second common point of the circles, $C$ be a point on the line $A B$ from which tangents are drawn, and, finally, $K$ be the intersection point of the lines $M N$ and $P Q$. Using the Law of Sines and the result from problem 1.234, we get:
Fig. 36 Consider the parallelogram $A_{1} M O N$ ( $M$ and $N$ on $A_{1} B_{1}$ and $A_{1} C_{1}$ ). Since $A_{1} O$ f... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,587 |
188. Let points $A_{1}, B_{1}, C_{1}$ be symmetric to some point $P$ with respect to the sides $B C, C A$, and $A B$ of triangle $A B C$. Prove that: a) the circumcircles of triangles $A_{1} B C, A B_{1} C, A B C_{1}$ have a common point; b) the circumcircles of triangles $A_{1} B_{1} C, A_{1} B C_{1}$, $A B_{1} C_{1}$... | 188. The statements of the problem follow from the fact: if on each side of a triangle circles are constructed in such a way that the sum of the angular magnitudes of their arcs (located on the same side as the triangle) is $2 \pi$, then these circles have a common point. | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,588 |
190. The perpendicular erected to side $A B$ of triangle $A B C$ at its midpoint $D$ intersects the circumcircle of triangle $A B C$ at point $E$ (points $C$ and $E$ are on the same side of $A B$), and $F$ is the projection of $E$ onto $A C$. Prove that the line $D F$ bisects the perimeter of triangle $A B C$ and that ... | 190. Let's take point $M$ on the extension of $AC$ beyond point $C$ such that $|CM|=|CB|$; then $E$ is the center of the circle circumscribed around $AMB$ $(|AE|=|BE|, \angle AEB=\angle ACB=2 \angle AMB)$. From this, it follows that $F$ is the midpoint of $AM$, and $DF$ divides the perimeter of $\triangle ABC$ in half.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,590 |
191. Prove that a line dividing the perimeter and the area of a triangle in the same ratio passes through the center of the inscribed circle. | 191. Let a line intersect sides $A C$ and $A B$ of triangle $A B C$ at points $M$ and $N$. Denote: $|A M| + |A N| = 2 l$. The radius of the circle with center on $M N$, touching $A C$ and $A B$, is $\frac{S_{A M N}}{l}$, and by the condition $\frac{S_{A M N}}{l} = \frac{S_{A B C}}{p} = r$, where $p$ and $r$ are the sem... | proof | Geometry | proof | Yes | Yes | olympiads | false | 23,591 |
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