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32. On the plane, four points $A_{1}, A_{2}, A_{3}, A_{4}$ are given, and point $A_{4}$ is the orthocenter of triangle $A_{1} A_{2} A_{3}$. Denote the circumcircles of triangles $A_{1} A_{2} A_{3}, A_{1} A_{2} A_{4}, A_{1} A_{3} A_{4}$, and $A_{2} A_{3} A_{4}$ by $S_{4}$, $S_{3}$, $S_{2}$, and $S_{1}$, and the centers ... | 32. a) Obviously. For example, the altitudes of triangle $A_{2} A_{3} A_{4}$ are the lines $A_{1} A_{4} \perp A_{2} A_{3}, \quad A_{1} A_{3} \perp A_{2} A_{4}$, and $A_{1} A_{2} \perp A_{3} A_{4}$; the point of intersection of the altitudes is point $A_{1}$.
b) Let $A_{4}^{\prime}$ be the point symmetric to point $A_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,073 |
33. On a plane, four points \(A_{1}, A_{2}, A_{3}, A_{4}\) are given, located on the same circle \(S\). The points of intersection of the altitudes of triangles \(A_{1} A_{2} A_{3}, A_{1} A_{2} A_{4}, A_{1} A_{3} A_{4}\) and \(A_{2} A_{3} A_{4}\) are denoted by \(H_{4}, H_{3}, H_{2}\) and \(H_{1}\). Prove that
a) the ... | 33. a) Let's denote by $O^{\prime}$ the point symmetric to the center $O$ of circle $S$ with respect to the line $A_{2} A_{3}$ (see Fig. 142). Then the quadrilaterals $O O^{\prime} H_{1} A_{1}$ and $O O^{\prime} H_{1} A_{4}$ are parallelograms (see the solution to problem 32 b)). Therefore, $A_{2} H_{4} = O O^{\prime} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,074 |
35. Given a line $l$, two points $A$ and $B$ on the same side of it, and a segment $a$. Find a segment $X Y$ on the line $l$ of length $a$ such that the length of the broken line $A X Y B$ is the smallest (Fig. 42). | 35. Since the length of the segment $X Y$ is equal to $a$, it is necessary that the sum $A X + B Y$ be minimal. Suppose that the segment $X Y$ is laid out. The sliding symmetry with the axis $l$ "of the parallel translation magnitude $a$ translates point $B$ to $B'$, and point $Y$ to $X$ (Fig. 144); therefore, $B Y = B... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,076 |
37. Light reflected from a straight mirror in such a way that the angle of incidence equals the angle of reflection (i.e., according to the same law by which a billiard ball bounces off the edge of a billiard table; see problem 29). Two straight mirrors are placed on a plane, forming an angle $\alpha$ between them. Pro... | 37. First solution. Obviously, a ray immediately reflects from the side of the angle in the direction opposite to the initial one if it is perpendicular to this side; for any ray, this is not true. Suppose now that the ray $M N$ after two reflections from the sides of the angle $A B C$ goes in the direction $P Q$, oppo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,078 |
41. Given straight lines $l_{1}$ and $l_{2}$, point $A$ on line $l_{1}$ and point $B$ on line $l_{2}$. Draw a line $m$ intersecting lines $l_{1}$ and $l_{2}$ at points $X$ and $Y$ such that $A X = B Y$ and
a) line $m$ is parallel to a given line $n$;
b) line $m$ passes through a known point $M$;
c) segment $X Y$ has... | 41. First solution (based on Theorem 1, p. 60). Let the lines \( l_{1} \) and \( l_{2} \) not be parallel (Fig. 153, a). Suppose the problem is solved. According to Theorem 1, the segment \( A X \) can be rotated to an equal segment \( B Y \) such that \( A \) goes to \( B \) and \( X \) goes to \( Y \) (since \( l_{1}... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,082 |
42. Given three lines $l_{1}, l_{2}$ and $l_{3}$ and three points $A, B$ and $C$-one point on each of the lines. Draw a line $m$ intersecting lines $l_{1}, l_{2}$ and $l_{3}$ at points $X, Y$ and $Z$ such that $A X=B Y=C Z$. | 42. Let the lines $l_{1}, l_{2}$ and $l_{3}$ not all be parallel to each other, for example, $l_{3}$ is not parallel to either $l_{1}$ or $l_{2}$. Suppose the problem is solved (Fig. 155). By Theorem 1, there exists
. By the condition, point $C$ is centrally similar to point $B$ with the center of similarity $A$ and the similarity coefficient $\frac{n}{m}$, so it lies on the line $i_{1}$, which is centrally similar to $l_{1}$ with the center $A$ and the coefficient $\frac{n}{m}$, and it can ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,085 |
45. a) Given a circle $S$ and a point $A$ on it. Find the geometric locus of the midpoints of the chords of the circle passing through $A$.
b) Given a circle $S$ and three points $A, B$, and $C$ on it. Draw a chord $AX$ that is bisected by the chord $BC$. | 45. a) The desired geometric locus is obtained from the circle $S$ by a central similarity transformation with center $A$ and similarity coefficient $\frac{1}{2}$; therefore, it is a circle with diameter $A O$, where $O$ is the center of $S$ (Fig. $158, a$ ).
 Let $H$ be the point of intersection of the altitudes of triangle $ABC$, and $M$ be the point of intersection of its medians. Prove that the circle $\bar{S}$, centrally similar to the circumcircle $S$ with the center of similarity $H$ and a similarity coefficient of $1 / 2$, is peripherally similar to the circle... | 51. a) Let $A P, B Q, C R$ be the altitudes of triangle $A B C$; $A^{\prime}, B^{\prime}, C^{\prime}$ be the midpoints of its sides; $D, E, F$ be the midpoints of segments $H A, H B, H C$ (Fig. $168, a$). It is clear that points $D, E$, and $F$ lie on circle $\bar{S}$. We will now show that points $P, Q$, and $R$ also ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,092 |
58. a) Derive from the statements of problems 52 a) and 50 a) a new solution to problem 33 a) from § 1, Chapter II of the first part (p. 47).
b) Let $A_{1}, A_{2}, A_{3}, A_{4}$ be four points lying on a circle $\mathcal{S} ; O_{1}, O_{2}, O_{3}, O_{4}$ be the centers of the Euler circles (see problem 51 a)) of the tr... | 58. a) According to the theorem of problem 52 a), the quadrilaterals $A_{1} A_{3} A_{3} A_{4}$ and $M_{1} M_{2} M_{3} M_{4}$, where $M_{1}, M_{2}, M_{2}$ and $M_{4}$ are the points of intersection of the medians of triangles $A_{2} A_{3} A_{4}, A_{1} A_{3} A_{4}, A_{1} A_{2} A_{4}$ and $A_{1} A_{2} A_{3}$, are centrall... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,098 |
59. Let $A_{1}, A_{2}, A_{3}$, and $A_{4}$ be points located on the circle $S$; $H_{4}, H_{3}, H_{2}$, and $H_{1}$ be the points of intersection of the altitudes of triangles $A_{1} A_{2} A_{3}, A_{1} A_{2} A_{4}, A_{1} A_{3} A_{4}$, and $A_{2} A_{3} A_{4}$. The eight points $A_{1}, A_{2}, A_{3}, A_{4}, H_{1}, H_{2}, H... | 59. a) Consider the quadruple of points $A_{1}, A_{2}, A_{3}$, and $H_{3}$. From these four points, four triangles can be formed: $A_{1} A_{2} A_{3}, A_{1} A_{2} H_{4}, A_{1} A_{3} H_{4}$, and $A_{3} A_{3} H_{4}$. We need to prove that the Euler circles of these triangles coincide with each other. Indeed, the radii of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,099 |
60. a) In the given triangle $A B C$, inscribe a triangle $P X Y$ (point $P$ is given on side $A B$), similar to the given triangle $L M N$.
b) In the given parallelogram $A B C D$, inscribe a parallelogram, similar to the given parallelogram $K L M N$. | 60. a) Suppose that triangle $PXY$ is constructed (Fig. 174, a). Point $Y$ is obtained from $X$ using a central-similar rotation with the center at point $P$, the angle of rotation $\alpha$, equal to the angle $L$ of triangle $LMN$, and the similarity coefficient $k$, equal to the ratio of sides $\frac{LN}{LM}$ of this... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,100 |
61. Through the intersection point $A$ of two circles $S_{1}$ and $S_{2}$, an arbitrary line is drawn, intersecting the circles again at points $M_{1}$ and $M_{2}$; then, through the centers $O_{1}$ and $O_{2}$ of the circles, lines $O_{1} J$ and $O_{2} J$ are drawn parallel to the tangents $M_{1} N$ and $M_{2} N$ to t... | 61. Connect the second intersection point $B$ of $S_{1}$ and $S_{2}$ with points $M_{1}$ and $M_{2}$, $O_{1}$ and $O_{2}$ (Fig. 175). Triangle $B M_{1} M_{2}$ will be similar to $\triangle B O_{2} O_{2}$ (since $\angle B O_{1} O_{2}=\frac{1}{2} \angle B O_{1} A=\angle B M_{1} A$, $\angle B O_{8} O_{2}=\frac{1}{2} \angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,101 |
67. Given a line $l$, a point $A$ on it, and circles $S_{1}$ and $S_{2}$. Construct a triangle $A B C$ in which the line $l$ is the angle bisector of angle $A$, vertices $B$ and $C$ lie on circles $S_{1}$ and $S_{2}$ respectively, and the ratio of sides $A B$ and $A C$ is a given value $m: n$. | 67. Let triangle $ABC$ be constructed (Fig. 180). Point $B$ is translated to point $C$ by a central similarity with center $A$, axis $l$, and similarity coefficient $\frac{n}{m}$. Therefore, point $C$ lies simultaneously on the circle $S_{\mathrm{z}}$ and on the circle $S_{1}^{\prime}$, obtained from $S_{1}$ by this ce... | two,one,ornosolutions | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,107 |
68. Construct a quadrilateral $ABCD$, the diagonal of which is the bisector of angle $A$, given:
a) sides $AB$ and $AD$, diagonal $AC$, and the difference between angles $B$ and $D$;
b) sides $BC$ and $CD$, the ratio of sides $AB$ and $AD$, and the difference between angles $B$ and $D$;
c) sides $AB$ and $AD$, diago... | 68. Suppose that the quadrilateral $A B C D$ is constructed. The central similarity with center $A$, axis $A C$ and similarity coefficient $\frac{A B}{A D}$ translates the triangle $A D C$ into the triangle $A B C^{\prime}$ (Fig. 181).
a) $A C$ and $A C^{\prime}=A C \cdot \frac{A B}{A D}$ are known to us; therefore, w... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,108 |
74. a) The hypotenuse of a right triangle slides with its ends along two mutually perpendicular lines. Find the geometric locus of the vertices of the right angle.
b) The larger side of an isosceles triangle with an obtuse angle of $120^{\circ}$ slides with its ends along the sides of an angle of $60^{\circ}$. Find th... | 74. a) Points $C_{1}$ and $C_{2}$ in Fig. 188, a are located on the circumcircle $S$ of triangle $ABO$. According to Theorem 2 (pages $119-120$), when segment $AB$ slides with its endpoints
... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,114 |
76. Through the intersection point $A$ of circles $S_{1}$ and $S_{8}$, an arbitrary line $l$ and a fixed line $l_{0}$ are drawn, intersecting $S_{1}$ and $S_{2}$ again at points $M_{1}, M_{2} \| N_{1}, N_{2}$; let $M_{1} M_{2} P$ be an equilateral triangle constructed on the segment $M_{1} M_{2}$, and $Q$ be the point ... | 76. a) Consider the quadrilateral $B M_{1} M_{2} P$, where $B$ is the second intersection point of $S_{1}$ and $S_{2}$. When $M_{1}^{2} M_{3}$ rotates around $A$, the triangle $B M_{1} M_{8}$ remains similar to itself (see the solution to problem 61); therefore, the quadrilateral $B M_{1} M_{2} P$ also remains similar ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,116 |
78. Given a triangle $ABC$ and some point $O$. Through $O$, lines $l_{1}, l_{2}$, and $l_{3}$ are drawn, the angles between which are equal to the angles of the triangle; let $\bar{A}$, $\bar{B}$, $\bar{C}$ be the points of intersection of these lines with the corresponding sides of $ABC$ (see Fig. 95; note the directi... | 78. Let $\bar{A} \bar{B} \bar{C}$ and $\overrightarrow{A^{\prime} B^{\prime}} \bar{C}$ be two positions of the triangle $\bar{A} \bar{B} \bar{C}$ (Fig. 192). Since the angle between $\overline{O A}$ and $\overline{O B}$ (or $\overline{O A^{\prime}}$ and $\overline{O B^{\prime}}$) is equal to the angle between $\bar{A} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,118 |
79. Construct a quadrilateral $A B C D$, similar to a given one (for example, a square),
a) the vertices of which lie on four given lines;
b) the sides of which pass through four given points;
c) the sides $B C, C D$ and the diagonal $B D$ of which pass through three given points, and the vertex $A$ lies on a given ... | 79. a) Let $l_{1}, l_{2}, l_{3}, l_{4}$ be four given lines. Consider all possible quadrilaterals $\bar{A} \bar{B} \bar{C} \bar{D}$, similar to a given one, with three vertices $\bar{A}, \bar{B}$, and $\bar{C}$ lying on the lines $l_{1}, l_{3}$, and $l_{3}$; by specifying vertex $\vec{A}$ or the direction of side $\vec... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,119 |
81. Let's rotate all sides of triangle $A B C$ around the midpoints of the sides by the same angle $\alpha$ (and in the same direction); let $A^{\prime} B^{\prime} C^{\prime}$ be the triangle formed by the rotated lines. Find the geometric locus of the points of intersection of the altitudes, angle bisectors, and media... | 81. When the angle α of triangle $A^{\prime} B^{\prime} C^{\prime}$ changes, it remains similar to itself (and to triangle $A B C$), and its sides always pass through fixed points - the midpoints of the sides of $A B C$; therefore, all its points (and, in particular, the points of intersection of the altitudes, angle b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,121 |
82. Let $M, K$ and $L$ be three points lying on the sides $AB, BC$ and $AC$ of triangle $ABC$. Prove that
a) the circles $S_{1}, S_{2}$ and $S_{3}$, circumscribed around triangles $LMA, M KB$ and $KLC$, intersect at one point;
b) the triangle formed by the centers of circles $S_{1}$, $S_{2}$ and $S_{3}$ is similar to... | 82. a) If triangle $K L M$ changes, remaining similar to itself, such that its vertices $K, L$, and $M$ slide along the sides $B C, C A$, and $A B$ of triangle $A B C$, then all positions of triangle $K L M$ have a common center of rotation $O$,
 triangles $A B C$ and $A_{1} B_{1} C_{1}$ are inscribed in a circle $S$; let $\bar{A}, \bar{B}$ and $\bar{C}$ be the points of intersection of their corresponding sides (Fig. 98). Prove that
a) triangle $\bar{A} \bar{B} \bar{C}$ is similar to triangles $A B C$ and $A_{1} B_{1}... | 83. a) Triangle $A_{1} B_{1} C_{1}$ is obtained from triangle $A B C$ by rotating around point $O$ by some angle $\alpha$. Therefore, $\angle A \bar{B} A_{1} = \angle A \bar{C} A_{1} = \angle A O A_{1} = \alpha$, so the points $A, A_{1}, \bar{B}, \bar{C}$, and $O$ lie on one circle (Fig. 194). Similarly, it can be prov... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,123 |
85. Prove that the bases of the perpendiculars dropped from some point $P$ to the sides of triangle $ABC$ lie on one straight line (the Simson line, fig. 99) if and only if the point $P$ lies on the circumcircle of $ABC$. | 85. First solution. If the bases of the perpendiculars dropped from point $P$ to the sides of $\triangle ABC$ lie on the same line $l_{1}$, then the lines $\bar{l}_{1}, \bar{l}_{2}$, and $\bar{l}_{3}$, which are symmetric to the line $l$ with respect to the sides of $\triangle ABC$, are centrally similar to $l$ with th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,125 |
87. On the plane, four lines are given, no two of which are parallel and no three intersect at the same point. Prove that the points of intersection of the altitudes of the four triangles formed by these lines lie on one line.
In another connection, this problem will be presented in $\$ 2$ of Chapter I of the third pa... | 87. The circumcircles of the four triangles formed by our four lines intersect at one point $P$ (problems 64, 86 a)); the feet of the perpendiculars dropped from point $P$ to the lines $l_{1}, l_{2}, l_{3}$ and $l_{4}$ lie on the same line $m$ (problem 85). The points $H_{1}, H_{2}, H_{3}$ and $H_{4}$ of intersection o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,127 |
92. Let $A_{1} B_{1} C_{1}$ and $A_{2} B_{2} C_{2}$ be triangles inscribed in triangle $A B C$ and similar to this triangle (the order of the letters indicates the correspondence of the sides), and such that points $A_{1}, B_{1}$, and $C_{1}$ lie on the sides $A B, B C$, and $C A$ of triangle $A B C$, respectively, and... | 92. a) Triangle $A_{1} B_{1} C_{1}$ can be obtained from triangle $A B C$ by rotating around point $O_{2}$ by the angle $A O_{1} A_{1}$, followed by a central similarity transformation with the coefficient $\frac{O_{1} A_{1}}{O_{1} A}$; therefore, the angle between the lines $A B$ and $A_{1} B_{1}$ is equal to the angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,132 |
93. a) Let $A_{1}, B_{1}, C_{1}$ and $A_{2}, B_{2}, C_{2}$ be the projections of the first and second centers of rotation of triangle $A B C$ onto the sides of the triangle. Prove that triangles $A_{1} B_{1} C_{1}$ and $A_{2} B_{2} C_{2}$ are both similar to triangle $A B C$ and equal to each other, and that the six po... | 93. a) Since in right triangles $A A_{1} O_{1}$, $B B_{1} O_{1}$, and $C C_{1} O_{1}$ the angles $O_{1} A A_{1}$, $O_{1} B B_{1}$, and $O_{1} C C_{1}$ are equal (see problem 90 a)), these triangles are similar (Fig. 207). Therefore, $\angle A O_{1} A_{1} = \angle B O_{1} B_{1} = \angle C O_{1} C_{1}$ and $\frac{O_{1} A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,133 |
94. Let $O_{1}$ be one of the rotation centers of triangle $ABC$; $A^{\prime}, B^{\prime}, C^{\prime}$ are the points of intersection of the lines $AO_{1}, BO_{1}$, $CO_{1}$ with the circumcircle of triangle $ABC$. Prove that
a) triangle $A^{\prime} B^{\prime} C^{\prime}$ is equal to triangle $ABC$;
b) the six triang... | 94. a) Let's denote the angle $\angle A^{\prime} A B$ as $\alpha$. Since $\angle A^{\prime} A B = \angle B^{\prime} B C = \angle C^{\prime} C A = \alpha$, then $-B A^{\prime} = -C B^{\prime} = -A C^{\prime} = 2\alpha$ and triangle $C^{\prime} A^{\prime} B^{\prime}$ is obtained from triangle $A B C$ by a rotation of ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,134 |
98. a) Given a line $l$ and two points $A$ and $B$ on the same side of this line. Find a point $X$ on the line $l$ such that the sum $A X + B X$ has the minimum possible value.
b) Given a line $l$ and two points $A$ and $B$ on opposite sides of this line. Find a point $X$ on the line $l$ such that the difference $A X ... | 98. a) Let $B^{\prime}$ be the point symmetric to $B$ with respect to the line $l$ (Fig. $214, a$). If $X^{\prime \prime}$ is any point on $l$, then $A X^{\prime \prime} + X^{\prime \prime} B = A X^{\prime} + X^{\prime} B^{\prime}$. Therefore, the sum $A X^{\prime} + B X^{\prime \prime}$ will be the smallest when the s... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,138 |
99. a) In the given triangle $A B C$, inscribe a triangle, one vertex of which coincides with a given point $P$ on side $A B$ and the perimeter of which has the smallest possible value.
b) In the given triangle $A B C$, inscribe a triangle of the smallest possible perimeter. | 99. a) First solution. Let \( P X Y \) be an arbitrary triangle inscribed in \( A B C \), one of whose vertices coincides with the point \( P \). Reflect the triangle \( A B C \) together with the triangle \( P X Y \) across the line \( B C \); the resulting triangle \( A^{\prime} B C \) and the inscribed triangle \( P... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,139 |
101. Inscribed in the triangle $A B C$ is a triangle whose sides are perpendicular to the radii $O A, O B, O C$ of the circumscribed circle. Derive from this construction another solution to problem 99 b) for an acute-angled triangle.
Translate the above text into English, please retain the line breaks and format of t... | 101. If $D, E, F$ are the feet of the altitudes of triangle $ABC$, then $\triangle ACD \sim \triangle BCE$ (see Fig. 220); therefore, $\frac{CE}{CD} = \frac{CB}{CA}$. Consequently, $\triangle ABC \approx \triangle DEC$ " $\angle CED = \angle CBA$. Let $MN$ be the tangent to the circumcircle at point $C$. Clearly, $\ang... | DE+EF+FD\leqD'E'+E'F'+F'D' | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,141 |
102. Inscribed in this triangle $A B C$ is a triangle $D E F$ such that the value of $a \cdot E F + b \cdot F D + c \cdot D E$, where $a, b, c$ are given positive numbers, is as small as possible. | 102. First, let's find a point whose distances to the vertices $A, B$, and $C$ of a triangle are in the ratio of the numbers $a, b$, and $c$. Constructing such a point is easy if we use the fact that the geometric locus of points whose ratio of distances to two given points is known is a circle (see footnote on page 10... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,142 |
103. In the plane of triangle $ABC$, find the point $M$ for which the sum of the distances to the vertices of the triangle is the smallest. | 103. The desired point $M$ does not lie outside the triangle $ABC$, since in the opposite case it is easy to indicate a point $M''$ such that $AM' + BM' + CM' < AM + BM + CM$ (Fig. 222, a).
... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,143 | |
104. a) Construct an equilateral triangle around the given triangle $ABC$ such that the perpendiculars erected at points $A, B, C$ to the sides of the equilateral triangle intersect at one point. Derive from this construction another solution to problem 103.
b) Inscribed in the given triangle $ABC$ an equilateral tria... | 104. a) If $D E F$ is an equilateral triangle circumscribed around triangle $A B C$, and $A M, B M, C M$ are perpendiculars to the sides of triangle $D E F$, then, obviously, $\angle A M B = \angle B M C = \angle C M A = 120^{\circ}$ (Fig. 223, a). From this, it follows that $M$ is the intersection point of the arcs of... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,144 |
105. a) Let $ABC$ be an equilateral triangle, and $M$ be an arbitrary point in the plane of this triangle. Prove that $MA + MC \geqslant MB$. In what case is $MA + MC = MB$?
b) Derive from the statement of part a) another solution to problem 103. | 105. a) First solution. Rotate triangle $CAM$ around point $A$ by $60^{\circ}$ to position $ABM'$ (Fig. 225, a). Then $MM' = AM' = AM$, and $CM = BM'$. But $BM \leq BM' + MM'$, and therefore,
$$
BM \leq AM + CM
$$
Equality $BM = BM' + MM'$ holds only if $M'$ lies on segment $BM$. Since $\angle AMM' = 60^{\circ}$, in ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,145 |
3-ча 1. Determine the ratio of two numbers if the ratio of their arithmetic mean to geometric mean is $25: 24$. | Let $x$ and $y$ be the required numbers. According to the condition, $\frac{x+y}{2}: \sqrt{x y}=25: 24$, i.e., $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac{25}{12}$. Let $q=\sqrt{x / y}$. Then $q+\frac{1}{q}=\frac{25}{12}$, i.e., $q^{2}-\frac{25}{12} q+1=0$. Solving this quadratic equation, we find $q_{1}=4 / 3$ and $q... | x:16:9or9:16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,147 |
3-2. Construct a triangle given two sides $a$ and $b$ and the bisector $m$ of the angle between them. | Solution 2. Let $A B C$ be the desired triangle, and $C D$ its bisector. Take a point $E$ on the line $A C$ such that $B E \| C D$. Then the angles at the side $B E$ of triangle $B C E$ are equal, so $E C=C B=a$. Moreover, $E B: C D=E C: C A$, so $E B=\frac{m(a+b)}{b}$. In triangle $B C E$, we know the lengths of all s... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,148 |
3-3. A pyramid, all lateral edges of which are inclined to the base plane at an angle $\varphi$, has an isosceles triangle with an angle $\alpha$ between the equal sides as its base. Determine the dihedral angle at the edge connecting the vertex of the pyramid with the vertex of the angle $\alpha$. | Solution 3. Let $ABC$ be the base of the given pyramid ($\angle BAC = \alpha$), and $S$ be the vertex of the pyramid. The lateral edges are inclined at equal angles, so they are equal. From the congruence of triangles $ASB$ and $ASC$, it follows that the feet of the perpendiculars dropped from points $B$ and $C$ to the... | \tan\frac{\theta}{2}=\frac{\tan\frac{\alpha}{2}}{\sin\varphi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,149 |
3-ча 1. A railway train passes by an observer in $t_{1}$ seconds, and at the same speed, it passes through a bridge of length $a$ meters in $t_{2}$ seconds. Find the length and speed of the train. | Solution 2. Let the length of the train be $l$ m. To pass over the bridge, the front carriage must enter the bridge, travel $a$ m on it, and then travel another $l$ m so that the last carriage leaves the bridge. As a result, we get the system of equations $l=t_{1} v, l+a=t_{2} v$. Solving it, we find $l=\frac{t_{1} a}{... | =\frac{t_{1}}{t_{2}-t_{1}},v=\frac{}{t_{2}-t_{1}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,150 |
3-2. Construct a square, three vertices of which lie on three given parallel lines. | Solution 2. Let $a, b, c$ be the given lines, with line $b$ lying between $a$ and $c$. Suppose the vertices $A, B, C$ of the square $ABCD$ lie on lines $a, b, c$ respectively.
First solution. From the fact that $\angle ABC=90^{\circ}$ and $AB=BC$, the following construction follows. Take an arbitrary point $B$ on line... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,151 |
3-3. Find the volume of a regular quadrilateral pyramid, the sides of the base of which are $a$, and the planar angles at the vertex are equal to the angles of inclination of the lateral edges to the base plane. | Solution 3. Let the length of the lateral edge be $b$, and the height of the pyramid be $h$. The desired volume $V$ is equal to $a^{2} h / 3$. Let the planar angles at the vertex and the angles of inclination of the lateral edges to the base plane be $\alpha$. Then $b \sin \alpha=h, b \cos \alpha=\frac{\sqrt{2}}{2} a$ ... | \frac{1}{3}^{3}\cos\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,152 |
3-ча 1. Construct two progressions: an arithmetic and a geometric one, each consisting of four terms; in such a way that if the corresponding terms of both progressions are added, the resulting numbers should be: $27, 27, 39, 87$. | Let $a, a+d, a+2 d, a+3 d$ be the desired arithmetic progression, and $b, b q, b q^{2}, b q^{3}$ be the desired geometric progression. According to the problem,
\[
\begin{aligned}
a+b & =27 \\
a+d+b q & =27 \\
a+2 d+b q^{2} & =39 \\
a+3 d+b q^{3} & =87
\end{aligned}
\]
Subtract the first equation from the second, the... | \begin{pmatrix}24,&18,&12,&6\\3,&9,&27,&81\end{pmatrix} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,153 |
3-2. Prove: if the sides of a triangle form an arithmetic progression, then the radius of the inscribed circle is $\frac{1}{3}$ of one of the heights. | Solution 2. Let $a, a+d, a+2d$ be the sides of a triangle, $h$ be the height dropped to the side $a+d$. The area of the triangle, on the one hand, is equal to $\frac{(a+d) h}{2}$, and on the other hand, it is equal to the product of half the perimeter and the radius of the inscribed circle: $\frac{a+(a+d)+(a+2d)}{2} r$... | \frac{1}{3} | Geometry | proof | Yes | Yes | olympiads | false | 26,154 |
3-3. The height of a truncated cone is equal to the radius of its larger base; the perimeter of a regular hexagon circumscribed around the smaller base is equal to the perimeter of an equilateral triangle inscribed in the larger base. Determine the angle of inclination of the cone's generatrix to the base plane. | Solution 3. Let $R$ be the radius of the larger base circle, and $r$ be the radius of the smaller base circle. The perimeter of the regular hexagon circumscribed around the smaller base is $\frac{12 r}{\sqrt{3}}$. The perimeter of the regular hexagon inscribed in the larger base is $3 R \sqrt{3}$. According to the cond... | \varphi=\arctan(4) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,155 |
3-ча 1. Solve the system of equations:
$$
\left\{\begin{aligned}
x^{2}+y^{2}-2 z^{2} & =2 a^{2} \\
x+y+2 z & =4\left(a^{2}+1\right) \\
z^{2}-x y & =a^{2}
\end{aligned}\right.
$$ | Solve 1. Let's write these equations as follows:
$$
\left\{\begin{aligned}
x^{2}+y^{2} & =2 z^{2}+2 a^{2} \\
x+y & =4\left(a^{2}+1\right)-2 z \\
-xy & =a^{2}-z^{2}
\end{aligned}\right.
$$
Square the second equation, add to it the third equation multiplied by 2, and subtract the first equation. The result is:
$$
0=16... | ^{2}\+1,\quad^{2}\+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,156 |
3-2. In triangle $ABC$, from an arbitrary point $D$ on side $AB$, two lines parallel to sides $AC$ and $BC$ are drawn, intersecting $BC$ and $AC$ at points $F$ and $G$ respectively. Prove that the sum of the circumferences of the circumcircles of triangles $ADG$ and $BDF$ is equal to the circumference of the circumcirc... | Solution 2. The radii of the circumscribed circles of similar triangles $A D G, D B F$ and $A B C$ are proportional to the corresponding sides:
$$
\frac{R_{1}}{A D}=\frac{R_{2}}{D B}=\frac{R}{A B}
$$
therefore, $\frac{R_{1}+R_{2}}{A D+D B}=\frac{R}{A B}$, which means, $R_{1}+R_{2}=R$. Multiplying this equality by $2 ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,157 |
3-3. The lateral surface development of a cone represents a sector with an angle of $120^{\circ}$; a triangular pyramid is inscribed in the cone, the angles of the base of which form an arithmetic progression with a difference of $15^{\circ}$. Determine the angle of inclination to the base plane of the smallest of the ... | Solution 3. Let $l$ be the length of the generatrix of the cone. The length of the circumference of the base is equal to the length of the arc of the development, so the radius of the base circle is $l / 3$. Considering that the sum of the angles of any triangle is $180^{\circ}$, we get that the base of the pyramid is ... | \cos\varphi=\frac{1}{\sqrt{17}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,158 |
3-cha 3. In two different planes lie two triangles: $A B C$ and $A_{1} B_{1} C_{1}$. The line $A B$ intersects with the line $A_{1} B_{1}$, the line $B C$ with the line $B_{1} C_{1}$, and the line $C A$ with the line $C_{1} A_{1}$. Prove that the lines $A A_{1}, B B_{1}$, and $C C_{1}$ either all intersect at one point... | Solution 3. Consider the planes $A B A_{1} B_{1}$, $B C B_{1} C_{1}$, and $A C A_{1} C_{1}$. The intersection of the first two planes is the line $B B_{1}$. If the third plane intersects the line $B B_{1}$ at some point, then this point is both a point of intersection of the three specified planes and a point of inters... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,161 |
3-ча 1. How many real solutions does the system of two equations with three unknowns have:
$$
\left\{\begin{aligned}
x+y & =2 \\
x y-z^{2} & =1 ?
\end{aligned}\right.
$$ | Sol 1. From the second equation, it follows that $x y \geqslant 1$. The numbers $x$ and $y$ cannot both be negative, since their sum is 2. Therefore, the numbers $x$ and $y$ are positive and $x+y \geqslant 2 \sqrt{x y} \geqslant 2$, and the equality $x+y=2$ is possible only when $x=y=1$. In this case, $z=0$.
Part 2. S... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,162 |
3-ча 1. Six different colors are chosen; it is required to paint 6 faces of a cube, each in a special color from the chosen ones. In how many geometrically distinct ways can this be done? Geometrically distinct are two such colorings that cannot be made to coincide with each other by rotating the cube around its center... | Solution 1. A cube can be rotated so that the face painted the first color takes the specified position. There are 5 different options for painting the opposite face; different colorings of the opposite face give geometrically different colorings of the cube.
Among the remaining four faces, one can choose the face pai... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,163 |
3-ча 3. We will denote by $M(a, b)$ the least common multiple of two numbers $a$ and $b$, and $D(a, b)$ the greatest common divisor of two numbers $a$ and $b$.
Prove the formula
$$
M(a, b) \cdot D(a, b)=a b
$$
For three numbers, prove the formula
$$
\frac{M(a, b, c) \cdot D(a, b) \cdot D(b, c) \cdot D(c, a)}{D(a, b... | Solution 3. First, we will prove the formula for two numbers. Let $a=p_{1}^{\alpha_{1}} \cdots p_{k}^{\alpha_{k}}$ and $b=p_{1}^{\beta_{1}} \cdots p_{k}^{\beta_{k}}$. Then
$$
\begin{aligned}
D(a, b) & =p_{1}^{\min \left\{\alpha_{1}, \beta_{1}\right\}} \cdots p_{k}^{\min \left\{\alpha_{k}, \beta_{k}\right\}} \\
M(a, b)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,165 |
3-ча 1. Solve the system of equations:
$$
\left\{\begin{aligned}
x+y & =a \\
x^{5}+y^{5} & =b^{5}
\end{aligned}\right.
$$ | Let $x y=t$. It is not hard to verify that
$$
x^{5}+y^{5}=(x+y)^{5}-5(x+y)^{3} x y+5(x+y) x^{2} y^{2}=a^{5}-5 a^{3} t+5 a t^{2}
$$
For $t$, we obtain the quadratic equation $t^{2}-a^{2} t+\frac{a^{5}-b^{5}}{5 a}=0$. Solving it, we find $t=\frac{1}{2}\left(a^{2} \pm \sqrt{\frac{a^{5}+4 b^{5}}{5 a}}\right)$. As a resul... | \frac{1}{2}(^{2}\\sqrt{\frac{^{5}+4b^{5}}{5}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,166 |
3-4. In how many different ways can 1000000 be represented as a product of three natural ${ }^{1}$ numbers? Products that differ only in the order of the factors are considered identical.
(This problem was not solved by any of the olympiad participants.) | Solution 4. Let the factors have the form $2^{a_{1}} 5^{b_{1}}, 2^{a_{2}} 5^{b_{2}}$ and $2^{a_{3}} 5^{b_{3}}$. Then $a_{1}+a_{2}+a_{3}=6$ and $b_{1}+b_{2}+b_{3}=6$. Here, the numbers $a_{i}$ and $b_{i}$ can be zero. If $a_{1}=k$, then for the decomposition $a_{2}+a_{3}=6-k$ we get $7-k$ options. Therefore, for the dec... | 139 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,169 |
3-5. In space, there are 3 planes and a sphere. In how many different ways can a second sphere be placed in space so that it touches the three given planes and the first sphere? (In this problem, it is actually about the tangency of spheres, i.e., it is not assumed that the spheres can only touch externally - ed. note.... | Solve 5. Answer: from 0 to 16 (depending on the arrangement of the planes and the sphere).
Let $O$ and $R$ be the center and radius of the given sphere $S$. Suppose that the sphere $S_{1}$ with center $O_{1}$ touches the given sphere and three given planes. We associate with the sphere $S_{1}$ the sphere $S_{1}^{\prim... | 0to16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,170 |
3-ча 1. Solve the system:
$$
\left\{\begin{aligned}
x+y+z & =a \\
x^{2}+y^{2}+z^{2} & =a^{2} \\
x^{3}+y^{3}+z^{3} & =a^{3}
\end{aligned}\right.
$$ | Solve 1. Answer: $(0,0, a),(0, a, 0)$ and $(a, 0,0)$.
The identity $(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)=2(x y+y z+x z)$ shows that
$$
x y+y z+x z=0
$$
The identity $(x+y+z)^{3}-\left(x^{3}+y^{3}+z^{3}\right)=3(x+y)(y+z)(z+x)$ shows that $(x+y)(y+z)(z+x)=0$. Considering equation (1), we get $3 x y z=0$. If $x=... | (0,0,),(0,,0),(,0,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,171 |
3-ча 1. In space, points $O_{1}, O_{2}, O_{3}$ and point $A$ are given. Point $A$ is symmetrically reflected with respect to point $O_{1}$, the resulting point $A_{1}$ - with respect to $O_{2}$, the resulting point $A_{2}$ - with respect to $O_{3}$.
We obtain some point $A_{3}$, which we also sequentially reflect with... | Solve 1. It is easy to check that $\overrightarrow{A A_{2}}=2 \overrightarrow{O_{1} O_{2}}, \overrightarrow{A_{2} \overrightarrow{A_{4}}}=2 \overrightarrow{O_{3} O_{1}}$ and $\overrightarrow{A A_{6}}=2 \overrightarrow{O_{2} O_{3}}$. Therefore, $\overrightarrow{A A_{6}}=\overrightarrow{0}$, i.e., $A=A_{6}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,177 |
3-4. How many natural ${ }^{2}$ numbers less than a thousand are there that are not divisible by 5 or 7? | Solution 4. Answer: 686 numbers. First, let's strike out from the set of numbers $1,2, \ldots, 999$ the numbers that are multiples of 5; their quantity is $\left[\frac{999}{5}\right]=199$. Then, from the same set of numbers $1,2, \ldots, 999$, let's strike out the numbers that are multiples of 7; their quantity is $\le... | 686 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,180 |
3-ча 1. Solve the system of equations:
$$
\left\{\begin{aligned}
3 x y z - x^{3} - y^{3} - z^{3} & = b^{3} \\
x + y + z & = 2 b \\
x^{2} + y^{2} - z^{2} & = b^{2}
\end{aligned}\right.
$$ | Consider first the case when $b=0$. In this case, the last two equations can be written as $z=-x-y$ and $z^{2}=x^{2}+y^{2}$. Squaring the first of these, we get $x y=0$. Therefore, $x=0, z=-y$ or $y=0, z=-x$. The first equation of the original system is satisfied in these cases.
Now consider the case when $b \neq 0$. ... | (1\\sqrt{\frac{-1}{2}})b,(1\\sqrt{\frac{-1}{2}})b,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,181 |
3-ча 2. Prove that $\cos \frac{2 \pi}{5}+\cos \frac{4 \pi}{5}=-\frac{1}{2}$. | Solution 2. Let $A B C D E$ be a regular pentagon inscribed in a circle of radius 1 with center $O$. The sum of the vectors from point $O$ to the vertices of this pentagon is zero, since upon rotation by an angle $\frac{2 \pi}{5}$, this sum transforms into itself. Therefore, the sum of the projections of the vectors $\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,182 |
3-4. Solve the equation $\sqrt{a-\sqrt{a+x}}=x$.
---
(Translation provided as requested, maintaining the original formatting and line breaks.) | Solve 4. By eliminating the radicals, we arrive at the equation
$$
x^{4}-2 a x^{2}-x+a^{2}-a=0
$$
With respect to $a$, this is a quadratic equation. Solving it, we get two solutions:
$$
\begin{aligned}
& a=x^{2}+x+1 \\
& a=x^{2}-x
\end{aligned}
$$
Solving these quadratic equations with respect to $x$, we obtain fou... | \begin{aligned}&x_{1,2}=-\frac{1}{2}\\sqrt{-\frac{3}{4}}\\&x_{3,4}=\frac{1}{2}\\sqrt{+\frac{1}{4}}\end{aligned} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,184 |
3-5. Prove that in any triangle, the bisector lies between the median and the altitude drawn from the same vertex. | Problem 5. Let $A D, A M$ and $A H$ be the bisector, median, and altitude of triangle $A B C$. Let $D^{\prime}$ be the point of intersection of line $A D$ with the circumcircle of this triangle. Then $D^{\prime}$ is the midpoint of arc $B C$, so $M D^{\prime} \| A H$. From this, it follows that point $D$ lies on segmen... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,185 |
3-ча 1. Factor the expression into integral rational factors
$$
a^{10}+a^{5}+1
$$ | Solve 1. Answer: $\left(a^{2}+a+1\right)\left(a^{8}-a^{7}+a^{5}-a^{4}+a^{3}-a+1\right)$. | (^{2}++1)(^{8}-^{7}+^{5}-^{4}+^{3}-+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,186 |
3-5. Given a regular pyramid. From an arbitrary point $P$ of its base, a perpendicular to the plane of the base is erected. Prove that the sum of the segments from point $P$ to the points of intersection of the perpendicular with the planes of the pyramid's faces does not depend on the choice of point $P$ on the base. | Solution 5. Let $P$ be the plane of the base of the pyramid, $Q$ be the point of intersection of the perpendicular to the plane $P$ drawn from point $P$ with the plane of the face of the pyramid, and $R$ be the base of the perpendicular dropped onto the edge of this face lying in the plane $P$. Then $P Q = P R \operato... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,189 |
3-4. Draw a circle with the given radius that touches the given line and the given circle. How many solutions does this problem have? | Solve 4. Let $r$ be the given radius, $R$ and $O$ be the radius and center of the given circle. The center of the desired circle lies on the circle $S$ of radius $|R \pm r|$ with center $O$. On the other hand, its center lies on the line $l$, parallel to the given line and at a distance $r$ from it; there are two such ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,194 |
3-4. In a plane, two lines are given. Find the geometric locus of points, the difference of whose distances from these lines is equal to a given segment. | Solution 4. Let $a$ be a given segment. Further, let the points of intersection of the given lines $l_{1}$ and $l_{2}$ with lines parallel to $l_{1}$ and $l_{2}$ and at a distance $a$ from them form a rectangle $M_{1} M_{2} M_{3} M_{4}$. We will show that the desired locus of points is the extensions of the sides of th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,197 |
3-5. The factorial of a number $n$ is defined as the product of all integers from 1 to $n$ inclusive. Find all three-digit numbers that are equal to the sum of the factorials of their digits. | Solve 5. Answer: 145. Let $N=100x+10y+z$ be the desired number, for which $N=x!+y!+z!$. The number $7!=5040$ is four-digit, so no digit of the number $N$ exceeds 6. Therefore, the number $N$ is less than 700. But then no digit of the number $N$ exceeds 5, since $6!=720$. The inequality $3 \cdot 4!=72<100$ shows that at... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,198 |
3-ча 1. Find a four-digit number that is a perfect square, and such that the first two digits are the same as each other and the last two digits are also the same. | Solution 1. Let $a$ be the first and second digits, $b$ be the third and fourth. Then the given number is equal to $11(b+100a)$, so $b+100a=11x^2$ for some natural number $x$. Moreover, $100 \leqslant b+100a \leqslant 908$, which means $3 \leqslant x \leqslant 9$. By calculating the squares of the numbers $33, 44, \ldo... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,199 |
3-4. How many pairs of integers $x, y$, lying between 1 and 1000, are there such that $x^{2}+y^{2}$ is divisible by 7. | Solve 4. Answer: $142^{2}=20164$. The number $x^{2}+y^{2}$ is divisible by 7 if and only if both numbers $x$ and $y$ are divisible by 7. Indeed, the square of an integer when divided by 7 gives remainders of 0, 2, and 4. The number of integers between 1 and 1000 that are divisible by 7 is 142. Therefore, the desired nu... | 20164 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,201 |
3-3. The center $O$ of the circumcircle of triangle $A B C$ is reflected symmetrically with respect to each of the sides. Restore triangle $A B C$ from the three resulting points $O_{1}, O_{2}, O_{3}$ if everything else is erased. | Solution 3. Let points $O_{1}, O_{2}$, and $O_{3}$ be symmetric to point $O$ with respect to sides $BC, CA$, and $AB$ respectively. Let $A_{1}, B_{1}$, and $C_{1}$ be the midpoints of sides $BC, CA$, and $AB$. Then $BC \parallel B_{1}C_{1} \parallel O_{2}O_{3}$ and $OA_{1} \perp BC$. Therefore, $OO_{1} \perp O_{2}O_{3}... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,204 |
3-5. How many positive integers $x$, less than 10000, are there for which $2^{x}-x^{2}$ is divisible by 7? | Solve 5. Answer: 2857. The remainders of the division by 7 of the numbers $2^{x}$ and $x^{2}$ repeat with periods of 3 and 7, respectively, so the remainders of the division by 7 of the numbers $2^{x} - x^{2}$ repeat with a period of 21. Among the numbers $x$ from 1 to 21, the numbers that give equal remainders from th... | 2857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,205 |
3-4. Through a point $P$ lying outside a circle, all possible lines intersecting this circle are drawn. Find the set of midpoints of the chords cut off by the circle on these lines. | Solution 4. Let $O$ be the center of the given circle, $M$ be the midpoint of the chord cut off from the circle by a line passing through point $P$. Then $\angle P M O=90^{\circ}$. Therefore, the desired set is the part of the circle with diameter $O P$, lying inside the given circle. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,209 |
3-2. On the sides of a parallelogram, squares are constructed outside it. Prove that their centers lie at the vertices of some square. | Solution 2. Let $P, Q$ and $R$ be the centers of the squares constructed on the sides $DA, AB$ and $BC$ of a parallelogram with an acute angle $\alpha$ at vertex $A$. It is easy to verify that $\angle PAQ=90^{\circ}+\alpha=\angle RBQ$, and thus, $\triangle PAQ = \triangle RBQ$. The sides $AQ$ and $BQ$ of these triangle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,211 |
3-ча 3. Prove that a polynomial with integer coefficients
$$
a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}
$$
taking odd values at $x=0$ and $x=1$, does not have integer roots. | Solution 3. Let $P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}$. By the condition, the numbers $a_{n}=P(0)$ and $a_{0}+a_{1}+\cdots+a_{n}=P(1)$ are odd. If $x$ is an even number, then $P(x) \equiv a_{n}(\bmod 2)$. If $x$ is an odd number, then $P(x) \equiv a_{0}+a_{1}+\cdots+a_{n}(\bmod 2)$. In both cases, we g... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,212 |
3-4. Construct triangle $ABC$ given points $M$ and $N$ - the feet of the altitudes $AM$ and $BN$ - and the line on which side $AB$ lies. | Sol 4. Points $M$ and $N$ lie on a circle with diameter $AB$. The center $O$ of this circle is the intersection of line $AB$ and the perpendicular bisector of segment $NM$, so we can construct it. Points $A$ and $B$ are the points of intersection of line $AB$ and the circle centered at $O$ passing through point $M$. Th... | 63 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,213 |
3-3. Given a triangle $ABC$. A point $M$, located inside it, moves parallel to side $BC$ until it intersects side $CA$, then parallel to side $AB$ until it intersects side $BC$, then parallel to side $CA$, and so on. Prove that after a certain number of such steps, the point will return to its original position, and fi... | Solution 3. Let $A_{1}, B_{1}, B_{2}, C_{2}, C_{3}, A_{3}, A_{4}, B_{4}, \ldots$ be the consecutive points of the trajectory on the sides of triangle $A B C$. Since $A_{1} B_{1} \| A B_{2}, B_{1} B_{2} \| C A_{1}$ and $B_{1} C \| B_{2} C_{2}$, triangle $A B_{2} C_{2}$ is obtained from triangle $A_{1} B_{1} C$ by a para... | 7 | Geometry | proof | Yes | Yes | olympiads | false | 26,215 |
3-4. Find the integer $a$, for which
$$
(x-a)(x-10)+1
$$
can be factored into the product $(x+b)(x+c)$ of two factors with integer $b$ and $c$. | Solution 4. Let $(x-a)(x-10)+1=(x+b)(x+c)$. By setting $x=-b$, we get $(b+a)(b+10)=-1$. Since $a$ and $b$ are integers, $b+a$ and $b+10$ are also integers. The number -1 can be represented as the product of two factors in two ways. Accordingly, we get two cases: 1) $b+10=1$ and $b+a=-1$; 2) $b+10=-1$ and $b+a=1$. There... | =8or=12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,216 |
3-5. Prove that the square of any prime number $p>3$ when divided by 12 leaves a remainder of 1. | Solution 5. Let's see what remainders a prime number $p>3$ can give when divided by 6. It cannot give a remainder of 2 or 4, since otherwise it would be even. It cannot give a remainder of 3, since otherwise it would be divisible by 3. Therefore, a prime number $p>3$ gives a remainder of 1 or 5 when divided by 6, i.e.,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 26,217 |
3-6. Construct triangle $ABC$ given three points $H_{1}, H_{2}$, and $H_{3}$, which are the reflections of the orthocenter of the desired triangle across its sides. | Solution 6. Let points $H_{1}, H_{2}$, and $H_{3}$ be symmetric to the orthocenter $H$ with respect to the sides $BC$, $CA$, and $AB$. The construction easily follows from the following fact: if triangle $ABC$ is acute-angled, then its vertices are the points of intersection of the circumcircle of triangle $H_{1} H_{2}... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,218 |
3-2. Some number of points are located on a plane such that any 3 of them can be enclosed in a circle of radius $r=1$. Prove that then all the points can be enclosed in a circle of radius 1. | Solution 2. Consider a circle containing all the given points. We will decrease the radius of such a circle as much as possible. Let $R$ be the radius of the resulting circle. At least two of the given points lie on the boundary of this circle. First, consider the case when exactly two points $A$ and $B$ lie on the bou... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,220 |
3-3. Find such non-zero, unequal integers $a, b, c$ that the expression
$$
x(x-a)(x-b)(x-c)+1
$$
can be factored into a product of two polynomials with integer coefficients. | Solution 3. Let $x(x-a)(x-b)(x-c)+1=P(x) Q(x)$, where $P(x)$ and $Q(x)$ are polynomials with integer coefficients. It is clear that $P(x)$ and $Q(x)$ are polynomials with leading coefficient 1. For $x=0, x=a$, $x=b$ and $x=c$, we have $P(x) Q(x)=1$, i.e., either $P(x)=1$ and $Q(x)=1$, or $P(x)=-1$ and $Q(x)=-1$. In bot... | (1,2,3),(-1,-2,-3),(1,-2,-1),(2,-1,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,221 |
3-6. Construct a right triangle given two medians drawn to the legs.
Construct a right triangle given two medians drawn to the legs. | Problem 6. Let $a$ and $b$ be the lengths of the medians, and $x$ and $y$ be the lengths of the legs. Then $\frac{x^{2}}{4}+y^{2}=a^{2}$ and $x^{2}+\frac{y^{2}}{4}=b^{2}$. Therefore, $x^{2}=\frac{16 b^{2}-4 a^{2}}{15}$ and $y^{2}=\frac{16 a^{2}-4 b^{2}}{15}$.
## VIII Olympiad (1945)
## First Round
## 7th - 8th Grade... | x^{2}=\frac{16b^{2}-4^{2}}{15},\quady^{2}=\frac{16^{2}-4b^{2}}{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,223 |
3-ча 1. Divide $a^{128}-b^{128}$ by $(a+b)\left(a^{2}+b^{2}\right)\left(a^{4}+b^{4}\right)\left(a^{8}+b^{8}\right)\left(a^{16}+b^{16}\right)\left(a^{32}+b^{32}\right)\left(a^{64}+b^{64}\right)$. | Solve 1. Answer: $a-b$. It is easy to check that if we multiply $(a+b)\left(a^{2}+b^{2}\right) \ldots\left(a^{64}+b^{64}\right)$ by $(a-b)$, we get $a^{128}-b^{128}$. | -b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,224 |
3-ча 2. Prove that for any positive integer $n$ the sum
$$
\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}
$$
is greater than $\frac{1}{2}$. | Solution 2. Add $n-1$ inequalities $\frac{1}{n+1}>\frac{1}{2 n}, \frac{1}{n+2}>\frac{1}{2 n}, \ldots, \frac{1}{2 n-1}>\frac{1}{2 n}$ and the equality $\frac{1}{2 n}=\frac{1}{2 n}$. As a result, we obtain the required | proof | Inequalities | proof | Yes | Yes | olympiads | false | 26,225 |
3-3. A two-digit number, when added to the number written with the same digits but in reverse order, gives a perfect square. Find all such numbers. | Solve 3. Answer: $29,38,47,56,65,74,83,92$.
Let $10a+b$ be the desired number. According to the condition, the number $(10a+b)+(10b+a)=11(a+b)$ is a square of some number $k$. Then $k$ is divisible by 11, which means $a+b$ is also divisible by 11. But $a+b \leqslant 18$, so $a+b=11$. | 29,38,47,56,65,74,83,92 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,226 |
3-4. Prove that a scalene triangle cannot be cut into two equal triangles. | Solution 4. Suppose that segment $C D$ cuts the scalene triangle $A B C$ into two equal triangles $A C D$ and $B C D$. These triangles have, in particular, equal areas, so $A D=B D$. Moreover, side $C D$ is common. Therefore, the remaining sides are equal, i.e., $A C=B C$. This leads to a contradiction. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,227 |
3-5. To two externally tangent circles, common external tangents are drawn and the points of tangency are connected. Prove that in the resulting quadrilateral, the sums of the opposite sides are equal. | Solution 5. Let $O$ be the point of tangency of the circles, $A$ and $D$ be the points of tangency with one tangent, and $B$ and $C$ be the points of tangency with another tangent (points $A$ and $B$ lie on one circle, $C$ and $D$ on the other). Draw a common tangent through point $O$ to the circles. Let it intersect l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,228 |
3-2. Find a three-digit number, every integer power of which ends in the same three digits as the original number (in the same order). | Let $N$ be the desired number. Then $N^{2}-N=N(N-1)$ is divisible by 1000. The numbers $N$ and $N-1$ are coprime, so one of them is divisible by 8, and the other by 125. Let's first assume $N=125 k$. Then $k \leqslant 8$. Among the numbers $125 k-1, k=1, \ldots, 8$, only the number 624 is divisible by 8. Now let $N-1=1... | 376 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,230 |
3-ча 3. A system of second-order equations
$$
\left\{\begin{aligned}
x^{2}-y^{2} & =0 \\
(x-a)^{2}+y^{2} & =1
\end{aligned}\right.
$$
generally has four solutions. For what values of $a$ does the number of solutions of the system decrease to three or to two? | Solve 3. Answer: the number of solutions decreases to three when $a= \pm 1$, and the number of solutions decreases to two when $a= \pm \sqrt{2}$.
From the first equation, we get $y= \pm x$. Substituting this expression into the second equation, we obtain
$$
(x-a)^{2}+x^{2}=1
$$
The number of solutions of the system ... | \1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,231 |
3-4. A right triangle $ABC$ moves on a plane so that its vertices $B$ and $C$ slide along the sides of a given right angle. Prove that the set of points $A$ is a segment and find its length. | Solution 4. Let $O$ be the vertex of the given right angle. Points $O$ and $A$ lie on the circle with diameter $BC$, so $\angle AOB = \angle ACB = \angle C$. From this, it follows that point $A$ moves along a line that forms an angle with the side of the given right angle, equal to $\angle C$. In the extreme positions,... | BC-BA | Geometry | proof | Yes | Yes | olympiads | false | 26,232 |
3-ча 1. Given 6 digits: $0,1,2,3,4,5$. Find the sum of all four-digit even numbers that can be written using these digits (the same digit can be repeated in a number). | Solution 1. Answer: 1769580.
We will separately calculate the sum of thousands, hundreds, tens, and units for the considered numbers. The first digit can be any of the five digits $1,2,3,4,5$. The number of all numbers with a fixed first digit is $6 \cdot 6 \cdot 3=108$, since the second and third places can be any of... | 1769580 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,233 |
3-2. Two identical polygons were cut out of cardboard, aligned, and pierced with a pin at some point. When one of the polygons is rotated around this "axis" by $25^{\circ} 30^{\prime}$, it aligns again with the second polygon. What is the smallest possible number of sides of such polygons? | Solution 2. Answer: 240. First, note that $\frac{1}{360} \cdot 25 \frac{1}{2}=\frac{17}{240}$, and the numbers 17 and 240 are coprime. Consider a ray emanating from the "axis" to the vertex of the first polygon. Rotations of this ray around the "axis" by angles $k \cdot 25^{\circ} 30^{\prime}$, where $k=1,2, \ldots, 24... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,234 |
3-3. Side $AD$ of parallelogram $ABCD$ is divided into $n$ equal parts. The first division point $P$ is connected to vertex $B$. Prove that line $BP$ cuts off a segment $AQ$ on diagonal $AC$ which is $\frac{1}{n+1}$ of the diagonal $\left(AQ=\frac{AC}{n+1}\right)$. | Solution 3. Divide side $B C$ into $n$ equal parts and draw lines through the division points parallel to line $B P$. They will divide diagonal $A C$ into $n+1$ equal parts. | \frac{AC}{n+1} | Geometry | proof | Yes | Yes | olympiads | false | 26,235 |
3-ча 1. Solve the equation in integers
$$
x y+3 x-5 y=-3
$$ | Sol 1. The equation under consideration can be rewritten in the form $(x-5)(y+3)=-18$. Its solutions in integers correspond to the representations of the number -18 as the product of two integers.
Part 2. Some of the numbers $a_{1}, a_{2}, \ldots a_{n}$ are equal to +1, the others are equal to -1. Prove that
$$
\begi... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 26,237 | |
3-3. A circle with a radius equal to the height of a certain equilateral triangle rolls along the side of this triangle. Prove that the arc intercepted by the sides of the triangle on the circle is always $60^{\circ}$. | Solution 3. Let the angular measure of the arc cut off by the sides of the given equilateral triangle $ABC$ be denoted by $\alpha$. We will assume that the circle is tangent to the side $BC$. Consider the arc cut off by the extensions of the sides of the triangle $ABC$ on the circle, and denote its angular measure by $... | 60 | Geometry | proof | Yes | Yes | olympiads | false | 26,238 |
3-2. On a line, there are 3 points $A, B, C$. On the segment $A B$, an equilateral triangle $A B C_{1}$ is constructed, and on the segment $B C$, an equilateral triangle $B C A_{1}$ is constructed. Point $M$ is the midpoint of segment $A A_{1}$, and point $N$ is the midpoint of segment $C C_{1}$. Prove that triangle $B... | Solve 2. When rotated by an angle of $60^{\circ}$ around point $B$, segment $C C_{1}$ transforms into segment $A_{1} A$, so point $N$ transforms into point $M$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,240 |
3-ча 3. Find a four-digit number that, when divided by 131, gives a remainder of 112, and when divided by 132, gives a remainder of 98. | Solve 3. Answer: 1946. Let $N$ be the desired number. By the condition, $N=131 k+112=132 l+98$, where $k$ and $l$ are natural numbers. Moreover, $N<10000$, so $l=\frac{N-98}{132}<\frac{10000-98}{132} \leqslant 75$. Further, $131 k+112=$ $132 l+98$, so $131(k-l)=l-14$. Therefore, if $k \neq l$, then $|l-14| \geqslant 13... | 1946 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 26,241 |
3-ча 1. In space, two intersecting planes $\alpha$ and $\beta$ are given. On the line of their intersection, a point A is given. Prove that among all lines lying in the plane $\alpha$ and passing through point $A$, the one that forms the greatest angle with the plane $\beta$ is the one that is perpendicular to the line... | Let $l$ be a line lying in the plane $\alpha$ and passing through the point $A$. On the line $l$, we lay off the segment $A B$ of length 1. Let $B^{\prime}$ be the projection of the point $B$ onto the plane $\beta$, and $O$ be the projection of the point $B$ onto the line of intersection of the planes $\alpha$ and $\be... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,243 |
3-2. Through a point $A$ lying inside an angle, a line is drawn, cutting off from this angle the smallest possible area triangle. Prove that the segment of this line, enclosed between the sides of the angle, is bisected at point $A$. | Solve 2. Let $P Q$ be a segment that is bisected by point $A$, and $P^{\prime} Q^{\prime}$ be another segment passing through point $A$. We will show that the segment $P^{\prime} Q^{\prime}$ cuts off a triangle of greater area than the segment $P Q$. Let for definiteness $P^{\prime} A \geqslant Q^{\prime} A$. Lay off o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 26,244 |
3-rd 1. In a chess tournament ${ }^{3}$, two students from grade VII and a certain number of students from grade VIII participated. The two seventh-graders scored 8 points, and each of the eighth-graders scored the same number of points. How many eighth-graders participated in the tournament? Find all solutions. | Let $x$ - the number of eighth graders, $y$ - the number of points scored by each eighth grader. By calculating the total points scored by all participants in the tournament in two ways, we arrive at the equation
$$
x y+8=\frac{(x+2)(x+1)}{2}
$$
i.e.
$$
2 y=\frac{(x+2)(x+1)-16}{x}=x+3-\frac{14}{x}
$$
Therefore, $x$... | 714 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26,245 |
3-ча 2. Prove that the expression
$$
x^{5}+3 x^{4} y-5 x^{3} y^{2}-15 x^{2} y^{3}+4 x y^{4}+12 y^{5}
$$
is never equal to 33 for any integer values of $x$ and $y$.[^2] | Solve 2. Let's represent the given expression as
$$
(x+2 y)(x-y)(x+y)(x-2 y)(x+3 y)
$$
For $y \neq 0$, all five factors of this product are pairwise distinct, and the number 33 cannot be represented as the product of five pairwise distinct integers (although it can be represented as the product of four pairwise disti... | proof | Algebra | proof | Yes | Yes | olympiads | false | 26,246 |
3-4. From thirty points $A_{1}, A_{2}, \ldots, A_{30}$, located on a straight line $M N$ at equal distances from each other, thirty straight roads extend. These roads are located on one side of the line $M N$ and form the following angles with $M N$:
| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15... | Solve 4. Answer: cars with numbers 14, 23, and 24 will not be detained anywhere.
Let $a_{n}$ be the road leading out of point $A_{n}, \alpha_{n}$ be the angle that road $a_{n}$ forms with the line $M N$, and $P_{m n}$ be the intersection of roads $a_{n}$ and $a_{m}$.
(1) If car $a_{m}$ is detained at intersection $P_... | 14,23,24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 26,248 |
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