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532. Prove that the area of a triangle, whose vertices are a vertex of a parallelogram and the midpoints of the sides converging at the opposite vertex of the parallelogram, is $\frac{3}{8}$ of the area of this parallelogram. | 532. Let points $E$ and $F$ be the midpoints of sides $B C$ and $C D$. Triangles $A B E, A E C, A C F$ and $A F D$ are equal in area; hence, the area of $A E C F$ is half the area of parallelogram $A B C D$. By establishing that the area of triangle $C E F$ is $\frac{1}{8}$ of the area of the parallelogram, determine t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,108 |
533. A segment parallel to the bases $a$ and $c$ of a trapezoid and enclosed between its lateral sides is equal to the geometric mean of these bases. Prove that the ratio of the areas of the trapezoids into which this segment divides the given trapezoid is $a: c$
74 | 533. Use the fact that the ratio of the heights of the trapezoids formed is $\frac{a-\sqrt{a c}}{\sqrt{a c}-c}$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,109 |
534. Prove that if only one of the midlines of a quadrilateral divides its area in half, then the quadrilateral is a trapezoid. | 534. Prove that if in the given quadrilateral $A B C D$ the areas of $A M N B$ and $M N C D$ are equal, then by the equality $(M N C)=$ $=(M B N)$ the areas of triangles $M A B$ and $M C D$, having equal bases $A M$ and $M D$, are equal. From this follows the equality of the heights of the latter triangles, and thus th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,110 |
538. Through two opposite vertices of a convex quadrilateral, lines are drawn, each of which divides the area of the quadrilateral in half and meets the corresponding side of the quadrilateral at points $M$ and $N$. Prove that the line $M N$ is parallel to the diagonal of the quadrilateral connecting the specified vert... | 538. The lines $A M$ and $C N$, meeting the sides $C D$ and $A D$ at points $M$ and $N[(A B C)<(A C D)]$, divide the area of the quadrilateral in half. Therefore, triangles $A M D$ and $C N D$ are equal in area, from which it follows that triangles $A M C$ and $C N A$, which have the common base $A C$, are also equal i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,114 |
539. A quadrilateral is divided by its diagonals into four triangles. If the square of the area of one of these triangles is equal to the product of the areas of the adjacent triangles, then the quadrilateral is either a trapezoid or a parallelogram. | 539. Since $s_{1}{ }^{2}=s_{2} \cdot s_{4}$, then $s_{1}: s_{2}=s_{4}: s_{1}$ and $h_{1}: h_{2}=h_{4}: h_{3}$, where $h_{1}$ and $h_{2}, h_{3}$ and $h_{4}$ are the heights dropped from the vertices of the quadrilateral to its diagonals. From this, we conclude that the diagonals are divided into proportional segments at... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,115 |
540. The diagonals $A C$ and $B D$ of trapezoid $A B C D (A B \| C D)$ intersect at point $O$. Prove that
$$
\sqrt{(A B C D)}=\sqrt{(A B O)}+\sqrt{(C D O)}
$$ | 540. Through vertex $D$, draw a line parallel to diagonal $A C$ and intersecting the extension of $A B$ at point $E$. Establish the equality of area of triangle $E B D$ and the given trapezoid. Consider similar triangles $E B D, A B O$ and $C D O$. Use the theorem on the ratio of areas of similar figures. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,116 |
541. The product of the areas of the triangles into which a quadrilateral is divided by one of its diagonals is equal to the product of the areas of the triangles into which it is divided by the other diagonal. Prove that the quadrilateral is a trapezoid or, in particular, a parallelogram. | 541. F i r s t m e t h o d. Since $\left(s_{1}+s_{2}\right)\left(s_{3}+s_{4}\right)=\left(s_{1}+\right.$ $\left.+s_{4}\right)\left(s_{2}+s_{3}\right)$, after some transformations, we obtain that $\left(s_{4}-s_{2}\right) \cdot\left(s_{1}-s_{3}\right)=0$. From this, we conclude that if one of the factors is zero, the qu... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,117 |
543. Point $M$ is reflected with respect to the midpoints of the sides of a quadrilateral. Prove that the resulting points are the vertices of a quadrilateral, the area of which does not depend on the choice of point $M$. | 543. Show that the obtained quadrilateral is a parallelogram with sides equal to and parallel to the diagonals of the given quadrilateral. Therefore, the area of it does not depend on the choice of point $M$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,119 |
544. On the extensions of sides $A B, B C, C D, D A$ of quadrilateral $A B C D$, points $P, Q, R, S$ are taken respectively such that $A B = B P, B C = C Q, C D = D R, D A = A S$. Prove that the ratio of the areas of quadrilaterals $A B C D$ and $P Q R S$ is $1: 5$. | 544. Draw the diagonal $A C$ and establish that $(A D C)=$ $=(A D R)=(A R S),(A B C)=(B C P)=(C P Q)$, then draw the diagonal $B D$ and also find the equal areas. Then show that the desired ratio is $1: 5$. | 1:5 | Geometry | proof | Yes | Yes | olympiads | false | 44,120 |
545. On the sides $A B, B C, C D, D A$ (or their extensions) of quadrilateral $A B C D$, points $P, Q, R, S$ are taken such that $A P = m \cdot A B, B Q = m \cdot B C, C R = m \cdot C D, D S = m \cdot D A$. Prove that the ratio of the areas of quadrilaterals $P Q R S$ and $A B C D$ is $2 m^{2} - 2 m + 1$. | 545. Draw the diagonal $A C$ and establish that the areas of triangles $B P C$ and $Q C P$ are respectively $(m-1) \cdot(A B C)$ and $(m-1)^{2} \cdot(A B C)$, and the areas of triangles $A D R$ and $A R S$ are
$(m-1) \cdot(A D C)$ and $(m-1)^{2} \cdot(A D C)$. Similarly, by drawing the second diagonal, we find the sum... | 2^{2}-2+1 | Geometry | proof | Yes | Yes | olympiads | false | 44,121 |
546. On the sides $A B, B C, C D, D A$ of the parallelogram $A B C D$, points $M, N, P, Q$ are given respectively such that $\frac{A M}{M B}=k_{1}, \frac{B N}{N C}=k_{2}$, $\frac{C P}{P D}=k_{1}, \frac{D Q}{Q A}=k_{2}$. Prove that the lines $A P, B Q, C M, D N$, intersecting, form a second parallelogram and that the ra... | 546. Determine that
$$
\frac{(A M C P)}{(A B C D)}=\frac{k_{1}}{1+k_{1}} \text { and } \frac{\left(A_{1} B_{1} C_{1} D_{1}\right)}{(A M C P)}=\frac{k_{2}\left(1+k_{1}\right)}{1+\left(k_{1}+1\right)\left(k_{2}+1\right)}
$$
where $\left(A_{1} B_{1} C_{1} D_{1}\right)$ is the area of the second parallelogram. | \frac{k_{1}\cdotk_{2}}{(1+k_{1})(1+k_{2})+1} | Geometry | proof | Yes | Yes | olympiads | false | 44,122 |
548. Through the midpoint of each diagonal of quadrilateral $ABCD$, a line is drawn parallel to the other diagonal. Prove that if these lines intersect at point $M$, then the areas of quadrilaterals $M M_{1} A M_{4}, M M_{1} B M_{2}, M M_{2} C M_{3}, M M_{3} D M_{4}$ are equal to each other ( $M_{i}$ - midpoints of the... | 548. Connect point $M$ with the vertices of the quadrilateral and, by considering the pairwise equal-area triangles thus formed, establish that the sum of the areas of quadrilaterals $A M_{1} M M_{4}$ and $M M_{2} C M_{3}$ is equal to half the area of the given quadrilateral. Then, by determining the equal areas of qua... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,124 |
549. A line passing through the midpoints of the diagonals of a quadrilateral meets its opposite sides $A B$ and $C D$ at points $M$ and $N$. Prove that $A M: M B=C N: N D$ and that triangles $A B N$ and $C D M$ are equal in area. | 549. 1) Use Menelaus' theorem. 2) Drop perpendiculars from the vertices of the quadrilateral to the line $M N$ and establish that $A A_{1}=C C_{1}$ and $B B_{1}=D D_{1}$. From this, it follows that triangles $A B N$ and $C D M$ are equal in area as triangles composed of equal-area triangles with the common base $M N$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,125 |
550. The midpoint of each side of a parallelogram is connected to the vertices belonging to the opposite side. Prove that the ratio of the area of the resulting octagon to the area of the parallelogram is $\frac{1}{6}$. | 550. For the convenience of the proof and without loss of generality, we can assume that the given parallelogram is a square, and in this case, the octagon is semiregular. Consider one-eighth of the square and the one-eighth of the octagon that belongs to it, representing two triangles with a common angle and with the ... | \frac{1}{6} | Geometry | proof | Yes | Yes | olympiads | false | 44,126 |
551. Through each vertex of a triangle, two lines are drawn, dividing the opposite sides into three equal parts. Prove that the ratio of the area of the hexagon formed by these lines to the area of the given triangle is $\frac{1}{10}$. | 551. If $B A_{1}=A_{1} A_{2}=A_{2} C, \quad C B_{1}=B_{1} B_{2}=B_{2} A, \quad A C_{1}=$ $=C_{1} C_{2}=C_{2} B$ and $A A_{1} \times B B_{2} \equiv M, \quad C M \times A B \equiv M_{3}, \quad$ then $G M \times$ $\times G M_{3}=2: 5$; if $A A_{1} \times B G \equiv N$, then $G N: G B=1: 4$ (fig. 76). Therefore, $(G M N):\... | \frac{1}{10} | Geometry | proof | Yes | Yes | olympiads | false | 44,127 |
554. On the sides of triangle $A B C$, points $A_{1}, B_{1}, C_{1}$ are taken respectively. Prove that the area of triangle $A_{1} B_{1} C_{1}$ is not less than the smallest area of triangles $A B_{1} C_{1}, A_{1} B C_{1}$, $A_{1} B_{1} C$. | 554. Let $A B_{1}: B_{1} C=\lambda, C A_{1}: A_{1} B=\mu, B C_{1}: C_{1} A=\nu$ (Fig. 77). Let 1) $\lambda \geqslant \nu \geqslant \mu \geqslant 1$. In this case $\frac{\left(C A_{1} B_{1}\right)}{\left(C_{1} A_{1} B_{1}\right)}=$ $=\frac{\mu(1+\nu)}{1+\lambda \mu \nu}$. From this it follows that
$1+\lambda \mu \nu-\m... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,130 |
556. On the sides $AC$ and $BC$ of triangle $ABC$, there are two points $M_{1}$ and $M_{2}$, $N_{1}$ and $N_{2}$ respectively, such that
$$
A M_{1}: M_{1} C = C N_{1}: N_{1} B = k_{1}, A M_{2}: M_{2} C = C N_{2}: N_{2} B = k_{2}
$$
The segments $M_{1} N_{1}$ and $M_{2} N_{2}$, intersecting at point $L$, are divided b... | 556. Establish that points $P_{1}$ and $L$ divide the segment $M_{1} N_{1}$ (as well as points $L$ and $P_{2}$ divide the segment $M_{2} N_{2}$) in the same way that points $M_{1}$ and $M_{2}$ divide the side $A C$. Consequently, the sets of points on the lines $A C, M_{1} N_{1}, M_{2} N_{2}$ are similar, and $A M_{1}:... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,132 |
557. Through a point $M$ taken on the side $AB$ of a quadrilateral $ABCD$, a line is drawn parallel to the diagonal $AC$ and intersects the side $BC$ at point $N$; through $N$ a line is drawn parallel to $BD$ and intersects $CD$ at point $P$; through $P$ a line is drawn parallel to $AC$ and intersects $AD$ at point $Q$... | 557. 1) Use the properties of proportional segments cut by parallel lines on the sides of an angle and the converse theorem; 2) express the areas of triangles $M B N$, $N C P$, $P D Q$ and $Q A M$ through $k$. | \frac{(1+k)^{2}}{2k} | Geometry | proof | Yes | Yes | olympiads | false | 44,133 |
558. Corresponding vertices of two $n$-gons $A_{1} A_{2} A_{3} \ldots A_{n}$ and $B_{1} B_{2} B_{3} \ldots B_{n}$ are connected by lines. Prove that if these lines are parallel and the segments $A_{i} B_{i}$ are intersected by some line in a given ratio $k$, then $\left(A_{1} A_{2} A_{3} \ldots A_{n}\right):\left(B_{1}... | 558. Use the result of problem No. 526, first dividing both n-gons into triangles by corresponding diagonals emanating from one vertex. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,134 |
559. In the plane of triangle $ABC$, a point $M$ is given, through which three rays are drawn, perpendicular to the sides of the triangle and not lying in the same half-plane. On these rays from point $M$, segments $M A_{1}, M B_{1}, M C_{1}$ are laid off, respectively equal to the sides of the triangle. Prove that: 1)... | 559. Establish that each of the angles formed at point M, when added to the corresponding angle of the triangle, sums up to $180^{\circ}$, and using the fact that the area of a triangle is equal to half the product of two of its sides and the sine of the angle between them, show that the area of each of the triangles $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,135 |
560. On the sides $A C$ and $C B$ of an equilateral triangle $A B C$, equal segments are laid off: $A M=B N=\frac{1}{4} A B$. Prove that the area of triangle $A B C$ is bisected by the broken line $M O N$ (point $O$ is the center of the triangle). | 560. Using the fact that the area of a triangle is equal to half the product of two of its sides by the sine of the angle between them, establish that the area of the quadrilateral $M O N C$ is $\frac{1}{4} a h$, where $a$ is a side of the triangle $A B C$, and $h$ is its height. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,136 |
562. Perpendiculars $A A_{1}, B B_{1}, C C_{1}$ are dropped from the tangents to the circle inscribed in triangle $A B C$ and intersecting its sides $C A$ and $C B$. Prove that
$$
C C_{1} \cdot A B - A A_{1} \cdot B C - B B_{1} \cdot A C = 2(A B C)
$$ | 562. For the secant passing through the incenter of triangle $ABC$ (Fig. 78), we have:
$C C_{0} \cdot A B=A A_{0} \cdot B C+B B_{0} \cdot C A$
(the secant meets the extension of $AB$). But $A A_{1}=A_{0} A+r$, $B B_{1}=B B_{0}+r, C C_{1}=C C_{0}-r$ which leads to the equality:
$$
C C_{1} \cdot A B=A A_{1} \cdot B C+... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,138 |
563. Prove that among all triangles having a common base and equal perimeters, the isosceles triangle has the largest area. | 563. First solution. Consider an isosceles triangle $A B C$ with a given base $A B$ and a given perimeter, and another triangle $A B D$ with the same perimeter. Establish that point $D$ lies outside triangle $A B C$ and point $C$ lies outside triangle $A B D$. If $A D$ intersects $B C$ at point $E$, then: 1) $B E C E$,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,139 |
566. A right-angled triangle is inscribed in a circle, and its vertices are reflected relative to the center of the circle. Prove that the area of the resulting hexagon is twice the area of the given triangle. | 566. If triangle $A_{1} B_{1} C_{1}$ is symmetric to triangle $A B C$, then $2(O A B)=\left(O B A_{1}\right)+\left(O B_{1} A\right) ; 2(O B C)=\left(O C B_{1}\right)+\left(O C_{1} B\right) ; 2(O C A)=$ $=\left(O A C_{1}\right)+\left(O A_{1} C\right)$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,142 |
567. Prove that if the area of a triangle is equal to the product of the segments into which its side is divided by the point of tangency of the inscribed circle, then the triangle is a right triangle. | 567. Let the side $AC$ of triangle $ABC$ be divided by the point of tangency $K$ into segments $KA$ and $KC$, the product of which is equal to the area of the triangle. Then $AK = p - a, \quad KC = p - c$ and $\sqrt{p(p-a)(p-b)(p-c)} = (p-a)(p-c)$, from which it follows after some transformations that $b^2 = a^2 + c^2$... | b^2=^2+^2 | Geometry | proof | Yes | Yes | olympiads | false | 44,143 |
571. On the extensions of sides $A B, B C, C A$ of an equilateral triangle $A B C$, points $C_{1}$ and $C_{2}, A_{1}$ and $A_{2}$, $B_{1}$ and $B_{2}$ are given such that $A C_{1}=B C_{2}, B A_{1}=C A_{2}, C B_{1}=A B_{2}$. Prove that triangles $A_{1} B_{1} C_{1}$ and $A_{2} B_{2} C_{2}$ are equal in area and that they... | 571. Let $A C_{1}=x, B A_{1}=y, C B_{1}=z$ (Fig. 79) and compute the sum of the areas of triangles $A B_{1} C_{1}, B C_{1} A_{1}, C A_{1} B_{1}$, as well as the sum of the squares of the sides $B_{1} C_{1}, C_{1} A_{1}, A_{1} B_{1}$ of these triangles. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,147 |
572. Prove that for any triangle, the inequality holds: $S>2 R^{\frac{1}{2}} r^{\frac{3}{2}}$. | 572. In triangle $A B C$, $h_{i}>2 r$, or $2 R h_{i}>4 R r$. But $2 h_{1} R=$ $=b c, 2 R h_{2}=c a, 2 R h_{3}=a b$; therefore, $b c>4 R r, c a>4 R r$, $a b>4 R r$, from which we have: $a b c>8 R^{\frac{3}{2}} r^{\frac{3}{2}}$, or $4 R S>8 R^{\frac{3}{2}} r^{\frac{3}{2}}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 44,148 |
573. Prove that among all quadrilaterals with given sides, the one inscribed in a circle has the largest area. | 573. Using the formulas: $S=\frac{1}{2}$ ef $\sin \varphi, \quad \cos \varphi=$ $=\frac{a^{2}+c^{2}-b^{2}-d^{2}}{2 e f}$, prove that $16 S^{2}=(b+d-a+c)(b+d+a-$ $-c)(a+c-b+d)(a+b+c-d)-4(a c+b d-e f)(a c+$ $+b d+e f)$. The area of the quadrilateral $S$ will be maximal if the second term is zero, i.e., if $e f=a c+b d$. ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,149 |
574. From the midpoints $M_{1}$ and $M_{3}$ of the sides $A B$ and $C D$ of the quadrilateral $A B C D$, perpendiculars $M_{1} N_{1}$ and $M_{3} N_{3}$ are dropped to its opposite sides $C D$ and $A B$. Prove that the area of the given quadrilateral can be calculated by the formula:
$$
(A B C D)=\frac{1}{2}\left(A B \... | 574. Establish that the area of quadrilateral $A M_{1} C M_{3}$ is equal to $\frac{1}{4}\left(A B \cdot M_{3} N_{3}+C D \cdot M_{1} N_{1}\right)$, considering its area as the sum of the areas of two triangles. Then verify that the area of this quadrilateral is half the area of the given quadrilateral (draw the diagonal... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,150 |
575. The bases of a trapezoid are equal to $a$ and $c$. A segment, whose ends lie on the lateral sides of the trapezoid and is parallel to its bases, divides the area of the trapezoid in half. Prove that the length of this segment is
$$
\sqrt{\frac{a^{2}+c^{2}}{2}}
$$ | 575. Establish that the ratio of the heights of the given trapezoid $ABCD$ and the formed trapezoid $EBCF$ is equal to the ratio $\frac{AD-BC}{EF-BC}$, and, using the condition of the problem, find that $EF^{2}=\frac{AD^{2}+BC^{2}}{2}$. | \sqrt{\frac{^{2}+^{2}}{2}} | Geometry | proof | Yes | Yes | olympiads | false | 44,151 |
576. The extensions of the lateral sides $A B$ and $C D$ of trapezoid $A B C D$ intersect at point $E$. At the midpoint $M_{1}$ of side $A B$, a perpendicular $M_{1} N_{1}$ is erected, equal to $\frac{1}{2} A B$. On the ray $E A$, a segment $E N$ is laid off, equal to $E N_{1}$. Prove that the line drawn through point ... | 576. Denoting the length of segment $A B$ by $b$, express the lengths of segments $B E$ and $N E$ in terms of $a, b, c$ and establish that
$$
A N: N B=\left(a-\sqrt{\frac{a^{2}+c^{2}}{2}}\right):\left(\sqrt{\frac{a^{2}+c^{2}}{2}}-c\right) .
$$
Verify that the length of the line segment located inside the trapezoid an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,152 |
578. A quadrilateral with perpendicular diagonals is inscribed in a circle 0. Perpendiculars are dropped from the point of intersection of the diagonals to the sides, the bases of which are the vertices of a new quadrilateral. Prove that a circle $O^{\prime}$ can be inscribed in the latter quadrilateral and that the ar... | 578. Express the sums of the opposite sides of the second quadrilateral in terms of the diagonals of the given quadrilateral and the corresponding sines of its angles, and establish the equality of these sums. Use the known formulas for calculating the area of a quadrilateral circumscribed about a circle and the area o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,154 |
579. Inside a circle of radius $R$, in which an equilateral triangle $ABC$ is inscribed, a point $P$ is given. Prove that between the area $S_{1}$ of the triangle constructed on the segments $PA, PB, PC$, and the distance $d$ from the point $P$ to the center of the circle, the following relationship holds: $d^{2}+\frac... | 579. Taking into account that $P A^{2}+P B^{2}+P C^{2}=a^{2}+3 d^{2}$, where $a=AB$ (see Leibniz's theorem). Using the solution of problem № 325, obtain: $9 d^{4}-6 a^{2} d^{2}-48 S_{1}^{2}+a^{4}=0$, from which $d^{2}=R^{2} \pm \frac{4 S_{1}}{\sqrt{3}}$. | ^{2}=R^{2}\\frac{4S_{1}}{\sqrt{3}} | Geometry | proof | Yes | Yes | olympiads | false | 44,155 |
581. On the plane, two segments $A B$ and $C D$ are given, not lying on the same line. Prove that the geometric locus of points for which the areas of triangles $A B M$ and $C D N$ are equal is a pair of lines. | 581. If the lines $A B$ and $C D$ intersect, then the required geometric locus of points represents a pair of lines passing through their point of intersection. If the lines $A B$ and $C D$ are parallel, then the geometric locus represents a pair of lines parallel to the given ones. [If the lines $A B$ and $C D$ coinci... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,157 |
583. Given an equilateral triangle $A B C$. Prove that the geometric locus of points $P$, for which the areas of triangles with sides equal to $P A, P B, P C$ are equal to each other and less than $\frac{1}{3}(A B C)$, is a pair of circles with a common center at the center of the given triangle, and in the opposite ca... | 583. Between the segments $x=P A, y=P B, z=P C$ and the side $a$ of the given triangle, there is the following relationship: $a^{4}+x^{4}+y^{4}+z^{4}=$ $=a^{2}\left(x^{2}+y^{2}+z^{2}\right)+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}$ (see problem No. 325). From this, it follows that $\left(x^{2}+y^{2}+z^{2}\right)^{2}-4 a^{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,159 |
585. Given a quadrilateral $A B C D$, different from a parallelogram. Prove that the geometric locus of points $M$, for which the areas of the oriented triangles satisfy the relation $(A M B)+(C D M)=(B C M)+(D A M)$, is a line passing through the midpoints of the diagonals of the quadrilateral.
## § 20. Mixed Problem... | 585. Take a point $M$ on the given line and establish that $(D A M)+(B C M)=(D A F)+(B C F)$ and $(A B M)+(C D M)=(A B F)+$ $+(C D F)$, where $F$ is the midpoint of diagonal $B D$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,161 |
586. Prove that the distance $m$ between the bases of the perpendiculars dropped from the base of the altitude of triangle $ABC$ to its two other sides is given by the formula:
$$
m=\frac{(ABC)}{R}
$$ | 586. Notice that $m=h_{3} \sin C, h_{3}=\frac{2 S}{c}, c=2 R \sin C$. | \frac{(ABC)}{R} | Geometry | proof | Yes | Yes | olympiads | false | 44,162 |
587. Prove that if the height of a triangle is equal to $R \sqrt{2}$, then the segment of the line passing through the bases of the perpendiculars dropped from the base of this height to the other two sides divides the area of the triangle in half. | 587. Notice that the segment $m$ cuts off a triangle from the given triangle with sides $R \sqrt{2} \sin A, R \sqrt{2} \sin B$ and angle $C$ between them. Use the formula: $a b c=4 R S$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,163 |
588. From the foot of the height of a triangle, perpendiculars are dropped to its other two sides. Prove that if the segment connecting the bases of these perpendiculars passes through the center of the circle circumscribed around the triangle, then this segment divides the area of the triangle into two equal parts. | 588. Determine that the segment $m$, connecting the bases of the dropped perpendiculars, is antiparallel to the side of the triangle and therefore it is parallel to the tangent to the circumcircle of the triangle at the opposite vertex of the side. Hence, the height of the triangle cut off by the segment $m$ is equal t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,164 |
589. In a triangle with no obtuse angles, two squares are inscribed such that the vertices of the squares lie on the sides of the triangle. Prove that: 1) if the squares are equal, then the triangle is isosceles or right-angled, 2) if the squares are not equal, then two vertices of the larger square lie on the side of ... | 589. 1) Calculate the sides of the squares: $x=\frac{c h_{3}}{c+h_{3}}$ and $y=$ $=\frac{b h_{2}}{b+h_{2}}$. From the condition $x=y$ it follows: $c+h_{3}=b+h_{2}$. In addition, $h_{3} c=h_{2} b$, therefore, $b=c$, or $b=h_{3}$. 2) If $x>y$ then $c+h_{3}>b+h_{2}$, hence $c<b$. If $2S$ is considered, then $c<b$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,165 |
590. Prove that the ratio of the area of a quadrilateral to the area of a triangle, whose vertices are the midpoints of the diagonals and the point of intersection of the extensions of a pair of opposite sides of the quadrilateral, is 4. | 590. Let $K$ and $L$ be the midpoints of the diagonals $AC$ and $BD$ of a quadrilateral, and let $E$ be the point of intersection of $AD$ and $BC$ (Fig. 80). Then $(BCDK) = \frac{1}{2}(ABCD), \quad(CDKL) = \frac{1}{2}(BCDK)$. Therefore, $(CDKL) = \frac{1}{4}(ABCD)$. Connect points $K$ and $L$ with the midpoint $M$ of s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,166 |
592. Prove that any line passing through the center of a polygon circumscribed about a circle divides its area and perimeter in equal ratios. | 592. Connect the center of the circle with the vertices of the polygon and consider that in the obtained triangles one of the heights is equal to the radius of the circle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,168 |
593. On a semicircle with diameter $A B$, points $C$ and $D$ are given. Prove that among all quadrilaterals $A C D B$, the one with the maximum area is the one where points $C$ and $D$ divide the semicircle into three equal parts. | 593. Let's draw the radii $O C$ and $O D$. The area of the quadrilateral $A C D B$ is $\frac{1}{2} R^{2}(\sin \alpha+\sin \beta+\sin \gamma)$, where $\alpha+\beta+\gamma=180^{\circ} ; \quad \sin \alpha+$ $+\sin \beta+\sin \gamma=4 \cos _{\sin } \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \times$ $\times \cos \frac{\gam... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,169 |
595. Prove that the segment $d$ of a line parallel to the bases of a trapezoid and dividing its area in the ratio $p: q$, can be determined by the formula:
$$
d=\sqrt{\frac{q a^{2}+p c^{2}}{p+q}}
$$
where $a$ and $c$ are the bases of the trapezoid. | 595. Let the desired segment be denoted by $x$. Then
$$
\frac{(a+x) \cdot u}{(c+x) \cdot v}=\frac{p}{q}
$$
where the ratio $u: v$ is the ratio of the segments into which the secant divides the lateral side of the trapezoid. On the other hand,
$$
\frac{u}{v}=\frac{a-x}{x-c}
$$
After eliminating $\frac{u}{v}$, we fin... | \sqrt{\frac{^{2}+p^{2}}{p+q}} | Geometry | proof | Yes | Yes | olympiads | false | 44,171 |
596. Through the vertices of the trapezoid $A B C D$, parallel lines are drawn which intersect an arbitrary line at points $A_{1}, B_{1}, C_{1}, D_{1}$. The segments $A A_{1}, B B_{1}, C C_{1}$, $D D_{1}$ are divided by points $A_{0}, B_{0}, C_{0}, D_{0}$ in equal ratios, counting from the vertices of the trapezoid. Pr... | 596. Using the result of problem № 526. From the equality $\frac{(A B C)}{\left(A_{0} B_{0} C_{0}\right)}=\frac{(A B D)}{\left(A_{0} B_{0} D_{0}\right)}$ it follows that $\left(A_{0} B_{0} C_{0}\right)=\left(A_{0} B_{0} D_{0}\right)$, since $(A B C)=(A B D)$. Therefore, the quadrilateral $A_{0} B_{0} C_{0} D_{0}$ is a ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,172 |
597. Prove that 1) $(A B C) \leqslant \frac{1}{2}\left(a^{2}-a b+b^{2}\right)$;
2) $(A B C) \leqslant\left(\frac{a+b}{2 \sqrt{2}}\right)^{2}$. | 597. 598) $(A B C)=\frac{a b \sin C}{2} \leqslant \frac{(a-b)^{2}+a b}{2}=\frac{1}{2}\left(a^{2}-a b+b^{2}\right)$.
2) $(A B C)=\frac{a b \sin C}{2}=\frac{4 a b \sin C}{8} \leqslant \frac{(a-b)^{2}+4 a b}{8}=\left(\frac{a+b}{2 \sqrt{2}}\right)^{2}$.
Equality holds for an isosceles right triangle. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 44,173 |
600. Inside triangle $ABC$, a point $M$ is taken, through which lines $AM, BM, CM$ are drawn, intersecting the corresponding sides of the triangle at points $A_1, B_1, C_1$. Prove that the sum
$$
\frac{AM}{MA_1} + \frac{BM}{MB_1} + \frac{CM}{MC_1}
$$
and the product
$$
\frac{AM}{MA_1} \cdot \frac{BM}{MB_1} \cdot \fr... | 600. Notice that
$$
\frac{A M}{M A_{1}}=\frac{S_{2}+S_{3}}{S_{1}}, \frac{B M}{M B_{1}}=\frac{S_{3}+S_{1}}{S_{2}}, \frac{C M}{M C_{1}}=\frac{S_{1}+S_{2}}{S_{3}}
$$
where $(M B C)=S_{1}, \quad(M C A)=S_{2}, \quad(M A B)=S_{3}$.
Further,
$$
\frac{A M}{M A_{1}}+\frac{B M}{M B_{1}}+\frac{C M}{M C_{1}}=\left(\frac{S_{1}}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,176 |
601. On a plane, five lines are given; for each set of four of them, a Gauss line is constructed. Prove that the five Gauss lines either intersect at one point or are parallel to each other. | 601. Use the result of problem No. 585. Assume that two Gaussian lines intersect at some point, and prove that this point belongs to the third Gaussian line. If the two Gaussian lines are parallel, then all five Gaussian lines are parallel to each other. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,177 |
603. In the plane of quadrilateral $A B C D$, a point $O$ is given. Prove that
$$
[(A O B)+(C O D)]-[(B O C)-(D O A)]=4(K O L)
$$
where $K$ is the midpoint of diagonal $A C$, and $L$ is the midpoint of diagonal $B D$ (the areas of the triangles given in the formula are to be considered positive or negative depending ... | 603. Use the equality $(A B C)=(O A B)+(O B C)+$ $+(O C A)$, where $O$ is any point in the plane, as well as the fact that the median of a triangle divides its area in half. | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,179 |
604. Two equal parallelograms $A B C D$ and $A B_{1} C_{1} D_{1}$ are positioned such that angle $A$ is common to both. Prove that the line $C C_{1}$ intersects the extensions of sides $A B$ and $A D$ of parallelogram $A B C D$ at points $M$ and $M_{1}$, respectively, for which the equality $C M = C M_{1}$ holds. | 604. To establish the parallelism of the lines $B_{1} D, B D_{1}$, and $C C_{1}$.
For this, consider the equal triangles $A D_{1} P$ and $A B P$, where $P$ is the point of intersection of the sides $C D$ and $B_{1} C_{1}$. Then verify that $A P$ bisects $B D_{1}$ and therefore $B D_{1} \| B_{1} D$. Similarly, discover... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,180 |
1. Given three points $A, B, C$, not lying on the same straight line. Prove that any straight line not passing through points $A, B$ and $C$ either does not intersect any of the segments $B C, C A$ and $A B$, or intersects exactly two of them.
Note that in some geometry courses, the statement of this problem is accept... | 1. Three points not lying on the same straight line can assume, with respect to a given straight line, only two possible positions: 1) all three belong to one of the half-planes determined by the given straight line. In this case, no segment is intersected by the straight line; 2) two points belong to one of the mentio... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,181 |
2. A plane is divided by a finite number of lines into several regions. Prove that it is sufficient to have only two colors to color the resulting "map" so that regions sharing a common boundary (segment) are colored with different colors. | 2. The problem is solved by the method of mathematical induction.
1) The statement of the problem is true for one line: one half-plane can be painted with one color, the other with another.
2) Suppose that the required coloring can be carried out for any partition of the plane by $n$ lines.
Consider a partition of the... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 44,182 |
4. Points $A, B, C$ do not lie on the same line, point $P$ does not lie on the lines $B C, C A$ and $A B$. Prove that at least one of the lines $P A, P B$ or $P C$ intersects the respective segment $B C, C A$ or $A B$. | 4. According to the previous task, three lines divide the plane into 7 regions: one - the inner region of triangle $ABC$, three - the inner points of the three angles vertical to the angles of the triangle, and three - the inner points of the regions adjacent to the sides of the triangle. If point $P$ lies inside trian... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,184 |
5. Given the axis of symmetry $l$ and points $A$ and $A^{\prime}$, symmetric with respect to the axis $l$. Construct the point $B^{\prime}$, symmetric to an arbitrary point $B$ with respect to the axis $l$, using only a ruler. | 5. Let points $A$ and $B$ be in the same half-plane relative to $l$. We have: $(B A) \cap l=M,\left(B A^{\prime}\right) \cap l=N$. The lines $M A$ and $M A^{\prime}$, as well as ($A N$) and ($A^{\prime} N$), are symmetric with respect to $l$. But $(M A) \cap\left(N A^{\prime}\right)=B$, therefore $\left(M A^{\prime}\ri... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,185 |
6. Construct a triangle $ABC$ of the smallest perimeter, if:
a) the vertices $A$ and $B$ are given, and the third vertex $C$ must lie on a given line $l$;
b)* the given vertex $A$ is inside a given acute angle, and the vertices $B$ and $C$ must lie on the two sides of this angle;
c)* the vertices $A, B$, and $C$ of ... | 6. a) Suppose points $A$ and $B$ are in the same half-plane relative to line $l$. Let point $B'$ be symmetric to point $B$ with respect to $l$, and $(A B') \cap l = C$. The triangle $A B C$ has the smallest perimeter because $|A C| + |C B| = |A C| + |C B'| = |A B'|$, and for any other point $C'$ on line $l$, we get $|A... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,186 |
7.* Using the result of the previous problem, prove that the three altitudes of a triangle intersect at one point.
${ }^{1}$ Moritz Pasch (1843-1930) - German geometer. | 7. In the previous problem, we inscribed a triangle $ABC$ of minimal perimeter in the given acute triangle $MNP$. We will consider the following as known: the three bisectors of the interior angles of a triangle intersect at one point; any two bisectors of the exterior angles intersect at the same point as one bisector... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,187 |
10*. A certain transformation maps three points $A, B, C$, not lying on the same line, to points $A^{\prime}, B^{\prime}$, $C^{\prime}$, respectively. Prove that the image $P^{\prime}$ of any point $P$ can be constructed using only a ruler and a "sharp-legged" compass, i.e., using the compass only for transferring segm... | 10. According to the solution of problem 4, point $P$ can end up in one of the seven regions into which the plane is divided by the lines $A B, A C, B C$. In all these cases, one of the lines $P A$, $P B$, $P C$ will intersect the line $B C, C A$, or $A B$ respectively. Let, for example, $(P A)$ intersect $(B C)$ at po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,189 |
11. Prove the following properties of the composition of axial symmetries:
a) the composition of two axial symmetries with intersecting axes is a rotation by an angle twice the angle between the axes, with the center at the point of intersection of the axes;
b) the composition of two axial symmetries with parallel ax... | 11. a) Let the axes of symmetry $l_{1}$ and $l_{2}$ intersect at point $O$ (Fig. 9). Symmetry with respect to axis $l_{1}$ maps point $A$ to point $A_{1}$, and symmetry with respect to axis $l_{2}$ maps point $A_{1}$ to point $A_{2}$. Consider the acute angle $B_{1} O B_{2}$ between the axes $l_{1}$ and $l_{2}$ and ass... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,190 |
12. Prove the following properties of rotations and parallel translations:
a) every rotation can be represented as the composition of two axial symmetries with axes passing through the center of rotation and forming an angle that is half the angle of rotation, and one of the axes can be drawn through the center of rot... | 12. a) Let us have a rotation by an angle $\alpha$ with center $O$, where $\alpha \leqslant 180^{\circ}$. Draw through the point $O$ two axes $l_{1}$ and $l_{2}$: one arbitrarily, and the other such that the angle between the axes is equal to $\frac{\alpha}{2}$. Then, according to what was proven in problem 11 a), the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,191 |
14*. Three squares are constructed outside a triangle on its three sides. How can you reconstruct (construct) the triangle if only the centers of these squares are given? | 14. Let the center of the square built on side $[B C]$ of the triangle be point $O_{1}$, on side $[C A]$ - point $O_{2}$, and on side $[A B]$ - point $O_{3}$. When rotating by $90^{\circ}$ around $O_{1}$, point $B$ will move to point $C$, when rotating by $90^{\circ}$ around $O_{2}$, point $C$ will move to point $A$, a... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,192 |
15. Prove that the composition of two central symmetries is a parallel translation - a vector. How to find this vector? Conversely: any translation can be represented as the composition of two central symmetries, and one of the centers can be chosen arbitrarily. | 15. Let the symmetry with center $O_{1}$ map point $A$ to point $A_{1}$; then $\overrightarrow{A O_{1}}=O_{1} \vec{A}_{1}$. The symmetry with center $O_{2}$ maps $A_{1}$ to $A_{2}$. Therefore, $\overrightarrow{A_{1}} O_{2}=\overrightarrow{O_{2} A_{2}}$. Adding all these vectors, we get:
$$
\overrightarrow{A O}_{1}+\ov... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,193 |
16. What kind of movement is the composition of an even number of central symmetries? of an odd number? | 16. If the number of central symmetries is even and equal to \(2n\) (\(n\) is a natural number), then our composition is the composition of \(n\) pairwise compositions, i.e., according to problem 15, the composition of \(n\) translations. Since the composition of translations is a translation, the composition of an eve... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,194 |
17*. Given 5 points - the midpoints of the sides of some pentagon. Construct this pentagon.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 17. Let $O_{1}, O_{2}, O_{3}, O_{4}, O_{5}$ be the midpoints of the sides. Consider the central symmetries with respect to these points. The first one maps the first vertex of the pentagon to the second, the second one maps the second vertex to the third, and so on. Finally, the fifth symmetry maps the fifth vertex to ... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,195 | |
18*. Prove that any two congruent and identically oriented ${ }^{1}$ triangles can be mapped onto each other either by a rotation or a parallel translation, and in both cases the transformation is uniquely determined. | 18. Let $A B C$ and $A_{1} B_{1} C_{1}$ be similarly oriented triangles; let us indicate how to find a rotation or a parallel translation that maps $A$ to $A_{1}$ and $B$ to $B_{1}$.
Let the symmetry with axis $l_{1}$ map $A$ to $A_{1}$ and $B$ to some point $B^{\prime}$ (how to construct the axis $l_{1}$?). Now perfo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,196 |
19*. Prove that the composition of three axial symmetries with axes passing through the same point is an axial symmetry.
20". Prove that the composition of three axial symmetries with axes not passing through the same point and not parallel to one line can be represented as the composition of an axial symmetry and a p... | 19. Let the axes of three symmetries $l_{1}, l_{2}, l_{3}$ pass through the point $O$. The composition of the first two symmetries is a rotation, which can be replaced by the composition of other symmetries with axes $l_{1}^{\text {' }}$ and $l_{2}^{\prime}$, passing through the same point $O$ and intersecting at the s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,197 |
21*. Given two congruent but oppositely oriented triangles (see note to problem 18): $A_{1} B_{1} C_{1}$ and $A_{2} B_{2} C_{2}$. Points that map to each other are denoted by the same letters. Prove that the midpoints of segments $A_{1} A_{2}$, $B_{1} B_{2}, C_{1} C_{2}$ lie on the same line. | 21. Apply an axial symmetry to triangle $A_{1} B_{1} C_{1}$ such that point $A_{1}$ is mapped to point $A_{2}$. The resulting triangle $A_{2} B_{1}^{\prime} C_{1}^{\prime}$ will have the same orientation as triangle $A_{2} B_{2} C_{2}$, and they can be transformed into each other by a rotation around point $A_{2}$. Sin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,198 |
22. Prove that a given scalene triangle cannot be divided into two congruent triangles by any straight line.
$23 *$. A sheet of paper has a face and a back. On this paper, a scalene triangle is drawn. How can you cut this triangle into the smallest number of pieces so that these pieces can be reassembled into a triang... | 22. Let in the given scalene triangle $ABC$, $|AB| > |AC|$. Draw a line through vertex $A$ that intersects $[BC]$ at point $D$. Can triangles $ABD$ and $ACD$ be congruent? We will consider the notation $\triangle ABD \cong \triangle ACD$ as indicating that $[AB] \cong [AC]$, $[BD] \cong [CD]$, $[AD] \cong [AD]$, $\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,199 |
24. A strip is called a set of points in the plane lying between two parallel lines, i.e., belonging to segments connecting points of one of the parallel lines with points of the other; alternatively, a strip is defined as the non-empty intersection of two half-planes with different parallel boundaries.
a) Prove that ... | 24. a) If $O$ is the center of symmetry of a strip, and $M$ is a point on one of the boundary lines, then the symmetric point $M^{\prime}$ lies on the other boundary line (prove this!). Since $O$ is the midpoint of $[M M^{\prime}]$, point $O$ is equidistant from the boundary lines of the strip. Conversely, it is easy t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,200 |
27. How to construct the bisector of an angle when the vertex is inaccessible? Can this construction be performed using only a two-sided ruler? | 27. Let $A$ denote the inaccessible point of intersection of two given lines. Intersect
62
these lines with an arbitrary third line to obtain points $B$ and $C$. The desired bisector of angle $A$ passes through the points of intersection of the bisectors of the two interior and two exterior angles of triangle $A B C$ ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,202 |
28*. Given a point $P$ and two lines that do not pass through it, the intersection point of which is inaccessible. Using only a drafting triangle, draw a line through point $P$ and the intersection point of the given lines.
29". A triangle is inscribed in a circle. Prove the following statements:
a) the three points ... | 28. Let $A$ be the inaccessible point of intersection of the given lines. Draw a perpendicular from $P$ to one of these lines and denote by $B$ the point of intersection of this perpendicular with the other line. Then draw a perpendicular from $P$ to the second line and denote by $C$ the point of its intersection with ... | notfound | Geometry | proof | Yes | Yes | olympiads | false | 44,203 |
31. Four lines intersecting each other form four triangles. Prove:
a)* the four circumcircles of these triangles intersect at one point;
b)* the four centers of these circumcircles lie on one circle passing through the same point;
c)** the four orthocenters of these triangles lie on one line. | 31. a) When four lines intersect, four triangles $A B C, A E F, B F D$ and $C D E$ are formed (Fig. 19). Applying the conclusions from the previous problem to triangle $A B C$ and the three points on its sides: $D \in(B C), E \in(A C)$ and $F \in(A B)$, it immediately follows that the circumcircles of triangles $A E F,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,204 |
32**. A hexagon $A B C C^{\prime} B^{\prime} A^{\prime}$ is inscribed in a circle. Let $\left(B C^{\prime}\right) \cap\left(B^{\prime} C\right)=A_{1},\left(C A^{\prime}\right) \cap\left(C^{\prime} A\right)=B_{1}, \quad\left(A B^{\prime}\right) \cap\left(A^{\prime} B\right)=C_{1}$. Prove that the points $A_{1}, B_{1}, C... | 32. We will prove that the circumcircles of triangles $BCA_1$, $CAB_1$, and $ABC_1$ intersect at one point $D$ (see Fig. 20). Let the circumcircles of $BCA_1$ and $ABC_1$ intersect at point $D$ (they already have the common point $B$). By the property of inscribed angles, we have:
$, then $\overrightarrow{A C}=z \overrightarrow{A B}$, i.e., $\overrightarrow{O C}-\overrightarrow{O A}=z(\overrightarrow{O B}-\overrightarrow{O A})$, from which we obtain the expression for $\overrightarrow{O C}$ in terms of $\overrightarrow{O A}$ and $\overrightarrow{O B}: \overrightarrow{O C}=(1-z) ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,207 |
3. Using the result of the previous task, prove that the three medians of a triangle intersect at one point. | 3. Let $A B C$ be a given triangle, $M$ the midpoint of $[B C]$, $N$ the midpoint of $[A C]$, and $P$ the point of intersection of the medians $A M$ and $B N$. We introduce the following notations: $\overrightarrow{A B}=\vec{b}, \overrightarrow{A P}=\vec{p}, \overrightarrow{A N}=\vec{n}$. Then $\overrightarrow{A C}=\ov... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,208 |
4. Let's fix a point $O$ on the plane. Any other point $A$ is uniquely determined by its radius vector $\overrightarrow{O A}$, which we will simply denote as $\vec{A}$. We will use this concept and notation repeatedly in the following. The fixed point $O$ is conventionally called the initial point. We will also introdu... | 4. Since $(A B C)=k$, then $\frac{\overrightarrow{C A}}{\overrightarrow{C B}}=k$, or $\left(\frac{\overrightarrow{O A}-\overrightarrow{O C}}{\overrightarrow{O B}-\overrightarrow{O C}}=k\right) \Rightarrow(\overrightarrow{O A}-$ $-\overrightarrow{O C}=k \overrightarrow{O B}-k \overrightarrow{O C}) \Rightarrow((k-1) \ove... | \vec{C}=\frac{k\vec{B}-\vec{A}}{k-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,209 |
5. Let $[A N]$ be the bisector of the angle at vertex $A$ in triangle $A B C$. Derive the formula expressing the vector $\overrightarrow{A N}=\vec{n}$ in terms of the vectors $\overrightarrow{A B}=\vec{c}$ and $\overrightarrow{A C}=\vec{b}$. Prove that
$$
\frac{\overrightarrow{N B}}{\overrightarrow{N C}}=-\frac{|\vec{... | 5. Consider the unit vectors $\vec{b}$ and $\frac{\vec{c}}{c}$ (here $b=|\vec{b}|, c=|\vec{c}|$). Since their lengths are equal, the vector $\frac{\vec{b}}{b}+\frac{\vec{c}}{c}$, their sum, is directed along the bisector of angle $A$, i.e., along the vector $\vec{n}$. Thus, we have the equality:
$$
\vec{n} = x\left(\f... | \frac{\overrightarrow{NB}}{\overrightarrow{NC}}=-\frac{|\vec{}|}{|\vec{b}|} | Geometry | proof | Yes | Yes | olympiads | false | 44,210 |
6. In triangle $ABC$, $|AB| \neq |AC|$. Keeping the notation from the previous exercise, derive the formula expressing the vector $\overrightarrow{A N^{\prime}}=\overrightarrow{n^{\prime}}$, where $[A N^{\prime}]$ is the external angle bisector of triangle $ABC$, and $N^{\prime} \in (BC)$, in terms of the vectors $\ove... | 6. The task is solved in the same way as the previous one, only the vector $\frac{\vec{b}}{b}$ needs to be replaced with the opposite - $-\frac{\vec{b}}{b}$. Then we get: $\overrightarrow{n^{\prime}}=\frac{\overrightarrow{b c}-c \vec{b}}{c-b},\left(B C N^{\prime}\right)=\frac{c}{b}$. | (BCN^{\})=\frac{}{b} | Geometry | proof | Yes | Yes | olympiads | false | 44,211 |
7. Prove that for any choice of the initial point $O$, the centroid $Q$ (the point of intersection of the medians) of triangle $A B C$ is determined by the formula:
$$
\vec{Q}=\frac{\vec{A}+\vec{B}+\vec{C}}{3}
$$ | 7. If $Q$ is the point of intersection of the medians $[A M],[B N],[C K]$ of triangle $A B C, \quad O$ is an arbitrary origin, then, since $\overrightarrow{A Q}=$ $=\frac{2}{3} \overrightarrow{A M}$ (see, for example, problem 3), and $\overrightarrow{A M}=-\overrightarrow{A B}+\overrightarrow{B M}=\overrightarrow{A B}+... | \vec{Q}=\frac{1}{3}(\vec{A}+\vec{B}+\vec{C}) | Geometry | proof | Yes | Yes | olympiads | false | 44,212 |
8*. Prove that the vector $\overrightarrow{O P}=\frac{m_{1} \overrightarrow{O A}_{1}+m_{2} \overrightarrow{O A}_{2}+\cdots+m_{n} \overrightarrow{O A}_{n}}{m_{1}+m_{2}+\cdots+m_{n}}$ defines the corresponding point $P=\vec{P}(O)$ uniquely, independently of the choice of the origin $O$ of all radius vectors. The point $P... | 8. According to the formula given in the problem, we have:
$$
\left(m_{1}+\ldots+m_{n}\right) \overrightarrow{O P}=m_{1} \overrightarrow{O A}_{1}+\ldots+m_{n} \overrightarrow{O A}_{n}
$$
Assume that replacing point $O$ with $O^{\prime}$ will also replace $P$ with $P^{\prime}$, then we get: $\left(m_{1}+\ldots+m_{n}\r... | proof | Other | proof | Yes | Yes | olympiads | false | 44,213 |
9*. Find the position of the center of gravity of a wire triangle (assume that the entire wire is of the same diameter and consists of the same metal).
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly. | 9. Given the homogeneity of the material, we can assume that the masses of the sides of the triangle are concentrated at their midpoints and are proportional to their lengths: \(a = |BC|, b = |CA|\), and \(c = |AB|\). The midpoints of the sides \([BC]\), \([CA]\), and \([AB]\) of triangle \(ABC\) are denoted as \(A'\),... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,214 |
10*. Equal masses are placed at the vertices of an arbitrary quadrilateral. Find the position of their centroid. | 10. If $A B C D$ is the given quadrilateral, then the general formula for the position vector of the centroid is
$$
\vec{P}=\frac{\vec{A}+\vec{B}+\vec{C}+\vec{D}}{4}
$$
At the same time, denoting by $M_{1}, M_{2}, M_{3}, M_{4}$ the centroids of triangles $B C D, A C D, A B D$ and $A B C$, we obtain that $P$ is the po... | \vec{P}=\frac{\vec{A}+\vec{B}+\vec{C}+\vec{D}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,215 |
11*. Accepting without proof that the center of gravity of a homogeneous triangular plate coincides with the centroid of the triangle, i.e., the point of intersection of its medians, prove that the center of gravity of a homogeneous quadrilateral plate (the quadrilateral is convex) coincides with the center of the para... | 11. For the given quadrilateral \(ABCD\) (Fig. 21), we retain the same notations for centroids as in the previous problem: \(M_{2}\) and \(M_{4}\) are the centroids of triangles \(ACD\) and \(ABC\) respectively. Let \(A_{1}\) and \(A_{2}\) be the points of division on sides \(AB\) and \(AD\), closest to vertex \(A\); \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,216 |
13. A homothety is defined by its center $S$ and coefficient $k$. Write the formula connecting $\vec{A}$ and $\vec{A}^{\prime}$, where $A^{\prime}$ is the image of point $A$ under this homothety. (Note that the answer does not depend on the choice of origin O.) | 13. If a homothety with center $S$ and coefficient $k$ maps point $A$ to $A^{\prime}$, then $\left(\vec{A}^{\prime}-\vec{S}\right)=k(\vec{A}-\vec{S})$, from which:
$$
\vec{A}^{\prime}=\overrightarrow{k A}+(1-k) \vec{S}
$$ | \vec{A}^{\}=\overrightarrow{kA}+(1-k)\vec{S} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,218 |
14*. Prove that the composition of two homotheties is either a homothety or a parallel translation. How is the center and the coefficient of the resulting homothety determined in the first case, and how is the corresponding vector determined in the second case? | 14. Let points $A_{1}$ and $B_{1}$ be mapped to points $A_{2}$ and $B_{2}$, respectively, by a homothety with center $S_{1}$ and coefficient $k_{1}$. Then we have: $\left(A_{1} A_{2}\right) \cap\left(B_{1} B_{2}\right)=S_{1}$ and $\overrightarrow{A_{2} B_{2}}=k_{1} \overrightarrow{A_{1} B_{1}}$. A second homothety with... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,219 |
15**. Given three circles with centers not lying on the same line and with different radii. By establishing that for each pair of these circles there exist two homotheties transforming one into the other, prove that the six centers of the obtained homotheties are located three on four lines. (This statement is known as... | 15. Let us have circles with non-coincident centers $O_{1}$ and $O_{2}$ and different radii $r_{1}>r_{2}$. We draw the radius vector $\overrightarrow{O_{1} A_{1}}$, as well as the radius vector $\overrightarrow{O_{2} A_{2}}$, which is collinear with the first. The homothety with center $S_{1}=\left(O_{1} O_{2}\right) \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,220 |
16*. Given a triangle $ABC$. On the lines $BC$, $CA$, and $AB$ are located points $X$, $Y$, and $Z$ respectively. Prove that the necessary and sufficient condition for points $X$, $Y$, and $Z$ to lie on one straight line is the equality $(BCX) \cdot (CAY) \cdot (ABZ) = 1$, or $\frac{\overrightarrow{XB}}{\overrightarrow... | 16. Necessity of the condition. Let a line intersect the lines $B C, C A$, and $A B$ at points $X, Y, Z$ respectively (Fig. 24). Draw a line $l$ perpendicular to the line $X Z$. It will intersect the latter at point $O$. Denote by $A^{\prime}, B^{\prime}, C^{\prime}$ the projections of points $A, B$, and $C$ onto the l... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,221 |
17*. Given a triangle $ABC$, in which the bisectors of the external angles at vertices $A, B$, and $C$ intersect the opposite sides or their extensions at points $M, N$, and $P$ respectively. Prove that these three points lie on the same line. | 17. The external bisector divides the opposite side into segments proportional to the adjacent sides (see problem 5). By substituting the corresponding values of the segment lengths into the expression $\frac{|M B|}{|M C|} \cdot \frac{|N C|}{|N A|} \cdot \frac{|P A|}{|P B|}$, we can verify that these ratios satisfy the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,222 |
18*. One side of the angle intersects a circle. Construct a circle that touches both sides of the angle and the given circle. | 18. Let $ACB$ be the given angle, the circle with center $O'$ be the given one, and the circle with center $O$ be the one to be found (Fig. 25). Let the point of tangency $S$ of the circles be the center of homothety that transforms these circles into each other. This same homothety transforms the angle $ACB$ into the ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,223 |
19*. Prove that two similar and identically oriented triangles, which are neither congruent nor homothetic to each other, can always be mapped onto one another by a composition of homothety and rotation with a common, uniquely determined center. | 19. Let $ABC$ and $A'B'C'$ be given similar triangles, and some similarity transformation maps point $A$ to point $A'$ and point $B$ to point $B'$, with $(AB)$ not parallel to $(A'B')$. (If parallelism were the case, the transformation would reduce to a homothety, which has already been considered in problem 14.) Note ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,224 |
20**. Prove that two similar and oppositely oriented triangles can always be transformed into each other by the composition of a reflection about any (any) of two mutually perpendicular axes and a homothety with the center at the point of intersection of these axes. These axes are uniquely determined. | 20. As in the previous problem, the similarity transformation is defined by the correspondence of two pairs of points: $A$ maps to $A^{\prime}$ and $B$ to $B^{\prime}$ (Fig. 27), only in this case the orientation of the triangle changes to the opposite. Therefore, if axial symmetry is applied to the second triangle, th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,225 |
21. Prove that a line passing through the point of intersection of the extensions of the non-parallel sides of a trapezoid and the point of intersection of its diagonals passes through the midpoints of the bases of the trapezoid.
Is the converse theorem true: if a line passing through the point of intersection of the ... | 21. Let in the given trapezoid $A B C D(A D) \|(B C),(A B) \cap(C D)=$ $=S,|A D|>|B C|,(B D) \cap(A C)=P$. Take point $S$ as the center of homothety, mapping $B$ to $A$ and $C$ to $D$. In this case, the line $B D$ is mapped to a line $a$ parallel to it, passing through point $A$, and the line $A C$ is mapped to a line ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,226 |
22. Use the propositions proved in the previous problem for the following constructions performed only with a ruler:
a) on one of two parallel lines, a segment is given. Construct its midpoint;[^2]b) on a given line, a segment with its midpoint marked is given. Through a given point outside the line, draw a line paral... | 22. Constructions a) and b) are based on the application of two theorems from the previous problem.
c) Draw the diameter $[A B]$. The center of the circle $O$ is its midpoint. Through an arbitrary point $M$ on the circle, we draw the chord $[M N] \|(A B)$ using the previous construction. Let $(A M) \cap(B N)=S$. Due t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,227 |
23. To introduce rectangular (Cartesian) coordinates on a plane, a reference (repère - marker) is constructed, consisting of the origin - point $O$ and two unit vectors (orths) $\vec{i}$ and $\vec{j}$, perpendicular to each other. The vector $\vec{i}$ defines the direction of the $O x$ axis, and the vector $\vec{j}$ de... | 23. a) Let $\vec{r}=x \vec{i}+y \vec{j}$. Taking the origin $O$ as the center of homothety with coefficient $k$, we get: $k \vec{r}=k x \vec{i}+k y \vec{j}$.
b) Consider the sum $\overrightarrow{O M}+$ $+\overrightarrow{M N}=\overrightarrow{O N}$. Let $\overrightarrow{O M}=x_{1} \vec{i}+$ $+y_{1} \vec{j}, \overrightar... | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,228 |
28. The scalar product of non-zero vectors is the product of the lengths of these vectors by the cosine of the angle between them: $\overrightarrow{a b}=a b \cos \varphi$. Prove the main properties of the scalar product:
a) the product is zero when $a \neq 0, b \neq 0$ and $\varphi=$ $=90^{\circ}$
b) commutativity: $... | 28. Properties a), b), and c) are proved based on the definition of the scalar product of vectors, the fact that $\cos (-\varphi)=\cos \varphi$, and $\cos 90^{\circ}=0$.
d) Let $\vec{a}+\vec{b}=\vec{c}$ (Fig. 29). Multiply all vectors by the vector $\vec{m}$ and we get: $\vec{m} \vec{a}=m a^{\prime}, \vec{m} \vec{b}=m... | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,231 |
29. Apply the properties of the scalar product to derive the main metric relations of planimetry: a) the cosine and sine theorems for a triangle; b) the relationships between the sides and diagonals of a parallelogram. | 29. a) To prove the cosine and sine theorems, we square both sides of the equation $\vec{a}=\vec{b}-\vec{c}$, where $\vec{a}=\overrightarrow{B C}, \vec{b}=\overrightarrow{C A}, \vec{c}=\overrightarrow{A B}$ in triangle $A B C$. We obtain: $a^{2}=$ $=b^{2}+c^{2}-2 b c \cos \alpha$. From this, applying the formula $\sin ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,232 |
30. a) Prove the identity $\vec{a}(\vec{b}-\vec{c})+\vec{b}(\vec{c}-\vec{a})+\vec{c}(\vec{a}-\vec{b})=0$; b)* Apply this identity to prove that the altitudes of a triangle intersect at one point. | 30. a) To prove the identity, it is sufficient to expand all the brackets and perform the addition.
b) Let in triangle $ABC$ the altitudes drawn from vertices $B$ and $C$ intersect at point $H$. Denote: $\overrightarrow{A A}=\vec{a}, \overrightarrow{H B}=\vec{b}$, $\overrightarrow{H C}=\vec{c}$. Then, due to the perpe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 44,233 |
31. Prove that the sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of its diagonals plus four times the square of the distance between the midpoints of the diagonals. | 31. Let $a, b, c, d$ denote the lengths of the sides of the given quadrilateral $P Q R S$ (Fig. 30) and $A, B, C, D$ the midpoints of the sides; let $m$ and $n$ be the lengths of the diagonals, $M$ and $N$ their midpoints. Also, let $|A C|=p,|B D|=q$, $|M N|=t$. Using the property of the midline of a triangle, we can s... | ^{2}+b^{2}+^{2}+^{2}=^{2}+n^{2}+4^{2} | Geometry | proof | Yes | Yes | olympiads | false | 44,234 |
32. Express the length of a) the median and b)* the bisector of a triangle in terms of the lengths of its sides. | 32. a) Let $m_{a}$ denote the length of the median [ $A M$ ] of triangle $A B C$. By drawing lines through vertices $B$ and $C$ parallel to the opposite sides of the triangle, we obtain a parallelogram with sides of lengths $b$ and $c$, and diagonals of lengths $2 m_{a}$ and $a$, where $a, b, c$ are the lengths of the ... | m_{}^{2}=\frac{b^{2}+^{2}}{2}-\frac{^{2}}{4},\quadn^{2}=\frac{(b+)^{2}-^{2}}{(b+)^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,235 |
33. Write the scalar product of vectors $\overrightarrow{r_{1}}\left(x_{1} ; y_{1}\right)$ and $\overrightarrow{r_{2}}\left(x_{2} ; y_{2}\right)$ in coordinates. | 33. Let $\vec{r}_{1}=x_{1} \vec{i}+y_{1} \vec{j} ; \overrightarrow{r_{2}}=x_{2} \vec{i}+y_{2} \vec{j}$. Since $\vec{i}^{2}=\overrightarrow{j^{2}}=1$, $\vec{i}:=0$, multiplying $\vec{r}_{1}$ and $\vec{r}_{2}$, we get:
$$
\vec{r}_{1} \vec{r}_{2}=x_{1} x_{2}+y_{1} y_{2}
$$
78 | x_{1}x_{2}+y_{1}y_{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,236 |
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