problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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|---|---|---|---|---|---|---|---|---|
[ [Sphere touching the edges or sides of the pyramid]
The base of the pyramid $S A B C$ is an equilateral triangle $A B C$ with side $2 \sqrt{2}$. Edges $S B$ and $S C$ are equal. A sphere touches the sides of the base, the plane of the face $S B C$, and the edge $S A$. What is the radius of the sphere if $S A=\frac{3... | Let a sphere of radius $R$ touch the side $BC$ of the base at point $D$ (Fig.1). Then $D$ is the only common point of the sphere with the plane of the face $SBC$, i.e., the sphere touches this plane at point $D$. If the sphere touches the edges $AB$ and $AC$ at points $M$ and $N$ respectively, then the center $O$ of th... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
The base of the pyramid $S A B C D$ is a trapezoid $A B C D$ with bases $B C$ and $A D$, and $B C: A D=2: 5$.
The diagonals of the trapezoid intersect at point $E$, and the center $O$ of the sphere inscribed in the pyramid lies on the segment $S E$ and divides it in the ratio $S O: O E=7: 2$. Find the total sur... | Let
$$
\begin{gathered}
S_{\triangle B S C}=S 1, S_{\triangle A S B}=S 2, S_{\triangle A S D}=S 3, S_{\triangle C S D}=S 4, S_{\text {base }}=S_{A B C D}=S, \\
V_{S B E C}=V 1, V_{S A E B}=V 2, V_{S A E D}=V 3, V_{S C E D}=V 4, V_{S A B C D}=V .
\end{gathered}
$$
By the given condition, $S 1=8$. Note that
$$
\frac{S... | 126 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kovaldji A.K.
A football is sewn from 32 patches: white hexagons and black pentagons. Each black patch borders only white ones, and each white one borders three black and three white. How many white patches are there? | Calculate the number of white patches bordering with black ones in two ways.
## Solution
Let the desired number of white patches be denoted by $x$. Then the number of black patches will be 32 - $x$. To form an equation, we will count the number of "borders" between white and black patches in two ways.
Each white pat... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Find the perimeter of triangle $ABC$, if the coordinates of its vertices $A(-3,5), B(3,-3)$ and point $M(6$,
1), which is the midpoint of side $BC$.
# | Use the formulas for the coordinates of the midpoint of a segment and the formulas for the distance between two points.
## Solution
Let $(x, y)$ be the coordinates of vertex C. According to the condition, $1 / 2(x+3)=6, 1 / 2(y-3)=1$, from which $x=9, y=5$.
Using the distance formula between two points, we find the ... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In triangle $A B C$, it is known that $A B=10, B C=24$, and the median $B D$ is 13. The circles inscribed in triangles $A B D$ and $B D C$ touch the median $B D$ at points $M$ and $N$ respectively. Find $M N$. | Let's first prove the following statement. If a circle is inscribed in a triangle $X Y Z$, and $x$ is the distance from vertex $X$ to the point of tangency of the circle with side $X Y$, and $Y Z=a$, then $x=p-a$, where $p$ is the semiperimeter of the triangle.
Let the points of tangency of the inscribed circle with s... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The center of the circle circumscribed around the triangle coincides with the center of the inscribed circle. Find the angles of the triangle.
# | Let $O$ be the common center of the inscribed and circumscribed circles of triangle $ABC$. Then triangles $AOB$, $AOC$, and $BOC$ are isosceles, and $CO$, $AO$, and $BO$ are the angle bisectors of angles $C$, $A$, and $B$.
## Solution
Let $O$ be the common center of the inscribed and circumscribed circles of triangle... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 [ Angles subtended by equal arcs and equal chords]
In a convex quadrilateral $A B C D$, it is known that $\angle C B D=58^{\circ}, \angle A B D=44^{\circ}, \angle A D C=78^{\circ}$. Find the angle $C A D$. | Quadrilateral $A B C D-$ is inscribed.
## Solution
Since
$$
\angle A B C=\angle A B D+\angle D B C=44^{\circ}+58^{\circ}=102^{\circ},
$$
Therefore, quadrilateral $A B C D$ is inscribed. Hence,
$$
\angle C A D=\angle D B C=58^{\circ}
$$
 ]
Given two parallel lines at a distance of 15 from each other; between them is a point $M$ at a distance of 3 from one of them. A circle is drawn through point $M$, touching both lines. Find the distance between the projections ... | Drop a perpendicular from point $M$ to the radius of the circle, drawn to one of the points of tangency.
## Solution
Let $A$ and $B$ be the projections of point $M$ and the center $O$ of the circle onto one of the lines, with $M A=3$. Then $O B$ is the radius of the circle and $O B=\frac{15}{2}$.
Let $P$ be the proj... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
In a regular truncated quadrilateral pyramid, the height is 2, and the sides of the bases are 3 and 5. Find the diagonal of the truncated pyramid. | Let's make a section through the opposite lateral edges $A A_{1}$ and $C C_{1}$ of the given truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}$ with bases $A B C D$ and $A_{1} B_{1} C_{1} D_{1}\left(A B=5, A_{1} B_{1}=3\right)$. Let $O$ and $O_{1}$ be the centers of the bases $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$, r... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Power of a set. Bijective mappings] Triangle inequality (other).
We consider all possible triangles with integer sides and a perimeter of 2000, as well as all possible triangles with integer sides and a perimeter of 2001. Which set of triangles is larger? | Match each triangle with a perimeter of 2000 to some triangle with a perimeter of 2003.
## Solution
Let the sides of the triangle be integers $a, b, c$ and its perimeter be 2000. We will match this triangle to a triangle with sides $a+1, b+1, c+1$, whose perimeter is 2003 (it is easy to verify that the triangle inequ... | 2003 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.G.
Each side of an equilateral triangle is divided into $n$ equal segments, and lines parallel to the sides are drawn through all the division points. This triangle is divided into $n^{2}$ small triangular cells. Triangles located between two adjacent parallel lines form a strip.
a) What is the maximum nu... | a) Examples are shown in the figures.
Estimate. First method. Cut the triangle into four equal triangles using the midlines (left figure). Suppose we managed to mark a certain number of cells... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8,9,10,11 |
Avor: Teresein D.A.
Points $A_2, B_2$ and $C_2$ are the midpoints of the altitudes $A A_1, B B_1$ and $C C_1$ of an acute-angled triangle $A B C$. Find the sum of the angles $B_2 A_1 C_2, C_2 B_1 A_2$ and $A_2 C_1 B_2$. | Let $H$ be the point of intersection of the altitudes of triangle $ABC$, and $M$ be the midpoint of side $AB$. Prove that points $H, M, B_2, A_2$, and $C_1$ lie on the same circle.
## Solution
Let $H$ be the point of intersection of the altitudes of triangle $ABC$, and $M$ be the midpoint of side $AB$. Since $MB_2$ a... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Angles subtended by equal arcs and equal chords ] [ Properties and characteristics of the tangent $]$
In a circle, chords $A C$ and $B D$ intersect at point $E$, and the tangent to the circle passing through point $A$ is parallel to $B D$. It is known that $C D: E D=3: 2$ and $S_{\Delta \mathrm{ABE}}=8$. Find the ar... | Triangle $A C B$ is similar to triangle $A B E$, and triangle $A B E$ is similar to triangle $D C E$.
## Solution
The diameter of the circle passing through point $A$ is perpendicular to the given tangent, and therefore perpendicular to the chord $B D$, which means it bisects it. Thus, triangle $A B D$ is isosceles, ... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9}
The medians of a triangle are 3, 4, and 5. Find the area of the triangle.
# | Prove that the area of the triangle formed by the medians of a given triangle is $\frac{3}{4}$ of the area of the given triangle.
## Solution
Let $B_{1}$ be the midpoint of side $A C$ of triangle $A B C$, and $M$ be the point of intersection of its medians. On the extension of median $B B_{1}$ beyond point $B_{1}$, l... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Pлотников M.
Let $T_{1}, T_{2}$ be the points of tangency of the excircles of triangle $ABC$ with sides $BC$ and $AC$ respectively. It turns out that the point symmetric to the center of the inscribed circle of the triangle with respect to the midpoint of $AB$ lies on the circumcircle of triangle $C T_{1} T_{2}$. Find... | Let $D$ be the fourth vertex of the parallelogram $A C B D, J$ be the center of the inscribed circle of triangle $A B D, S_{1}$, $S_{2}$ be the points of tangency of this circle with $A D$ and $B D$. Then $D S_{1}=B T_{1}$ (see problem $\underline{55404}$), so $S_{1} T_{1} \| A C$.
Similarly, $S_{2} T_{2} \| B C$, whi... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Akopyan A.V.
On the plane, a point $M$ is marked, which does not lie on the coordinate axes. A point $Q$ moves along the y-axis, and a point $P$ moves along the x-axis such that the angle $P M Q$ is always right. Find the geometric locus of points $N$, which are symmetric to $M$ with respect to $P Q$. | Points $P, Q, M$ and the origin $O$ lie on a circle with diameter $P Q$. Therefore, point $N$ also lies on this circle and $\angle P O N = \angle P O M$ (see the figure). Thus, $N$ lies on the line symmetric to $O M$ with respect to the coordinate axes.
. Then
$S_{OBMC} - S_{ABMC} = S_{OBC} + 2S_{MBC} - S_{ABC}$. Therefore, the geometric locus of points for which $S_{OBMC} = S_{ABMC}$ is the perpendicular bisector of segment $OA$. Hence, $AM = ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kenooarov $\mathbf{P . 5}$.
In quadrilateral $A B C D$, side $A B$ is equal to diagonal $A C$ and is perpendicular to side $A D$, and diagonal $A C$ is perpendicular to side $C D$. A point $K$ is taken on side $A D$ such that $A C=A K$. The bisector of angle $A D C$ intersects $B K$ at point $M$. Find the angle $A C M... | Since triangle $B A K$ is a right isosceles triangle, $\angle A K B=45^{\circ}$. Let the bisector of angle $C A D$ intersect segment $B K$ at point $N$. Triangles $A N K$ and $A N C$ are congruent: $A N$ is a common side, $A C=A K, \angle C A N=$ $\angle K A N$. Therefore, $\angle N C A=\angle N K A=45^{\circ}$. Thus, ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.B.
On a line, 100 points are marked, and one more point is marked outside the line. Consider all triangles with vertices at these points.
What is the maximum number of them that can be isosceles? | Let $l$ be a given line, $O$ be a marked point outside it, and $O H$ be the perpendicular dropped to $l$.
Estimate. The base of an isosceles triangle can lie on $l$ or not. The triangles of the first type are symmetric with respect to $O H$, so there are no more than 50 - half of the marked points on $l$. Triangles of... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The distance from a fixed point $P$ on the plane to two vertices $A, B$ of an equilateral triangle $A B C$ are $A P=2 ; B P=3$. Determine the maximum value that the segment $P C$ can have.
# | Let $A, B, C$ and $P$ be points on a plane such that $AB = BC = CA$, $AP = 2$, and $BP = 3$. Draw a ray $BM$ from point $B$ such that $\angle CBM = \angle ABP$, and mark a segment $BP' = PB$ on this ray. From the equality of angles: $\angle CBM = \angle ABP$, it follows that $\angle PBP' = \angle ABC = 60^\circ$, and t... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
What is the smallest number of non-overlapping tetrahedra into which a cube can be divided?
# | Answer: 5. If a tetrahedron $A^{\prime} B C^{\prime} D$ is cut out from the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, the remaining part of the cube splits into 4 tetrahedra, i.e., the cube can be cut into 5 tetrahedra. We will prove that it is impossible to cut the cube into fewer than 5 tetrahedra. ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$[\underline{\text { Properties of Sections }}]$
The edge of the cube $A B C D A 1 B 1 C 1 D 1$ is 12. Point $K$ lies on the extension of edge $B C$ at a distance of 9 from vertex $C$. Point $L$ on edge $A B$ is 5 units away from $A$. Point $M$ divides the segment $A 1 C 1$ in the ratio $1: 3$, counting from $A 1$. Fi... | Let the line $L K$ intersect the edge $C D$ at point $F$ (Fig.1). According to the theorem on the intersection of two parallel planes by a third intersecting plane, the plane intersects the base $A 1 B 1 C 1 D 1$ along a line $a$, passing through point $M$ parallel to $L F$. Let the line $a$ intersect the lines $A 1 B ... | 156 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9}
In a triangular pyramid $A B C D$, it is known that $D C=9, D B=A D$, and the edge $A C$ is perpendicular to the face $A B D$. A sphere of radius 2 touches the face $A B C$, the edge $D C$, and the face $D A B$ at the point of intersection of its medians. Find the volume of the pyramid. | Let $C$ be the vertex, and $A B D$ be the base of the triangular pyramid $C A B D$ (Fig.1); the given sphere with center $O$ touches the lateral edge $C D$ at point $P$, the base $A B D$ at point $M$ (the intersection of the medians of the isosceles triangle $A B C$), and the lateral face $A B C$ at point $Q$. The medi... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9
The volume of the pyramid $A B C D$ is 5. A plane is drawn through the midpoints of the edges $A D$ and $B C$, intersecting the edge $C D$ at point $M$. In this case, $D M: M C=2: 3$. Find the area of the section of the pyramid by the specified plane, if the distance from it to the vertex $A$ is 1. | Since the secant plane passes through the midpoints of opposite edges of the tetrahedron, it divides its volume in half. Let the secant plane intersect edge $AB$ at point $K$, and let $P$ and $Q$ be the midpoints of edges $AD$ and $BC$. Then the volume of the polyhedron $PMQKAC$ is $\frac{5}{2}$. On the other hand, it ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$[$ [extension of a tetrahedron to a parallelepiped]
Segment $A B(A B=1)$, being a chord of a sphere with radius 1, is positioned at an angle of $60^{\circ}$ to the diameter $C D$ of this sphere. The distance from the end $C$ of the diameter to the nearest end $A$ of the chord $A B$ is $\sqrt{2}$. Find $B D$.
# | Complete the tetrahedron $ABCD$ to a parallelepiped $AKBLNDMC (AN\|KD\|BM\|LC)$ by drawing pairs of parallel planes through its opposite edges. Let $O$ and $Q$ be the centers of the faces $NDMC$ and $AKBL$ respectively. Then $O$ is the center of a sphere, $KL=CD=2, OQ \perp AB, OA=OB=1$ as radii of the sphere, $KL \| C... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4 $[\quad$ Equifacetal Tetrahedron $\quad]$
In the triangular pyramid $A B C D$, the sums of the three dihedral angles at each of the vertices $B$ and $C$ are $180^{\circ}$ and $A D=B C$. Find the volume of the pyramid if the area of the face $B C D$ is 100, and the distance from the center of the circumscribed sphere... | Consider the unfolding $D 1 A D 2 B D 3 C$ of the tetrahedron $A B C D$ onto the plane of triangle $A B C$, where points $D 1, D 2$, and $D 3$ are the vertices of triangles with bases $A C, A B$, and $B C$ respectively (Fig.2). Since the sums of the three plane angles at each of the vertices $B$ and $C$ of the tetrahed... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ $\underline{\text { Cone }}$
Through the edge $B C$ of the triangular pyramid $P A B C$ and the point $M$, the midpoint of the edge $P A$, a section $B C M$ is drawn. The vertex of the cone coincides with the vertex $P$ of the pyramid, and the circle of the base is inscribed in the triangle $B C M$, touching the sid... | Let $E$ and $D$ be the points of intersection of the medians of triangles $A P B$ and $A P C$, $K$ be the midpoint of $B C$, $O$ be the center of the circle inscribed in triangle $B M C$, and $r$ be the radius of the base of the cone, with $2 r$ being its height. Then
$$
P D=P E=P K=\sqrt{r^{2}+4 r^{2}}=r \sqrt{5}
$$
... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two cones have a common vertex, and the generatrix of the first cone is the height of the second. The angle at the vertex of the axial section of the first cone is $\arccos \frac{1}{3}$, and that of the second is $-120^{\circ}$. Find the angle between the generatrices at which the lateral surfaces of the cones intersec... | Since $120^{\circ}>\arccos \frac{1}{\frac{1}{3}}$, the height of the first cone is inside the second (Fig.1). Let $A$ be the common vertex of the cones, $O$ be the center of the base of the second cone, $A O=1, A M$ and $A N$ be the common generators of the cones. Through the line $M N$, we draw a plane perpendicular t... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Distance between skew lines]
On the line $l$ in space, points $A, B$ and $C$ are sequentially located, with $A B=18$ and $B C=14$. Find the distance between the lines $l$ and $m$, if the distances from points $A, B$ and $C$ to the line $m$ are 12, 15 and 20, respectively.
# | Let $A 1, B 1$ and $C 1$ be the feet of the perpendiculars dropped from points $A, B$ and $C$ to the line $m$ respectively (Fig.1). According to the problem,
$$
A A 1=12, B B 1=15, C C 1=20 .
$$
Assume that lines $l$ and $m$ lie in the same plane. It is clear that they cannot be parallel. If the intersection point of... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Consider $M$.
$ABC$ is an isosceles right triangle. On the extension of the hypotenuse $AB$ beyond point $A$, a point $D$ is taken such that $AB = 2AD$. Points $M$ and $N$ on side $AC$ are such that $AM = NC$. On the extension of side $CB$ beyond point $B$, a point $K$ is taken such that $CN = BK$. Find the angle betw... | Let $L$ be the projection of $M$ onto $AB$. Note that $\frac{ML}{CN} = \frac{AL}{BK} = \frac{AD}{BC} = \frac{1}{\sqrt{2}}$; therefore, $\frac{LD}{CK} = \frac{AL + AD}{BK + BC} = \frac{1}{\sqrt{2}}$. Hence, the right triangles $MLD$ and $NCK$ are similar, and $\angle MLD = \angle NKC$ (see the figure). Therefore, the an... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
S. Rukhin
A closed eight-segment broken line is drawn on the surface of a cube, with its vertices coinciding with the vertices of the cube.
What is the minimum number of segments of this broken line that can coincide with the edges of the cube?
# | The segments of the broken line are either edges of the cube or diagonals of its faces. To do this, let's color the vertices of the cube in two colors in a checkerboard pattern (see fig.). Note that an edge of the cube connects vertices of different colors, while a diagonal connects vertices of the same color. For the ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$3+$ [ The transfer helps solve the task_ ]
On the side AB of the square ABCD, an equilateral triangle AKB was constructed (outside). Find the radius of the circle circumscribed around triangle CKD, if $\mathrm{AB}=1$.
# | Construct an equilateral triangle on side $\mathrm{CD}$ inside the square $\mathrm{ABCD}$.
## Solution
Construct an equilateral triangle CDM on side CD inside the square ABCD. Then CM=1 and DM=1. Moreover, triangle ABK is obtained from triangle CDM by a parallel translation by vector CB, the length of which is 1. The... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[
Diameter, main properties
On the leg $B C$ of the right triangle $A B C$ as a diameter, a circle is constructed, intersecting the hypotenuse at point $D$ such that $A D: B D=1: 3$. The height dropped from the vertex $C$ of the right angle to the hypotenuse is 3. Find the leg $B C$.
# | $C D$ - the height of triangle $A C D$.
## Solution
Since angle $B D C$ is inscribed in the given circle and rests on its diameter $B C$, then $\angle B D C=90^{\circ}$. Therefore, $C D$ is the height of triangle $A B C$.
Let $A D=x, B D=3 x$. Since $C D^{2}=A D \cdot D B$, then $3 x^{2}=9$. From this, we find that ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}{[\text { Inscribed quadrilaterals }]} \\ \text { Area of a trapezoid }]\end{array}\right.$
The midline of an isosceles trapezoid is 5. It is known that a circle can be inscribed in the trapezoid.
The midline of the trapezoid divides it into two parts, the ratio of the areas of which is $\frac{... | Solve the system of equations with two unknowns - the lengths of the bases of the trapezoid.
## Solution
Let $x$ and $y$ be the bases of the trapezoid. Then
$$
\left\{\begin{array}{l}
x+y=10 \\
\frac{x+5}{y+5}=\frac{7}{13}
\end{array}\right.
$$
From this, we find that $x=2$ and $y=8$. Therefore, the lateral side of... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
In the pyramid $ABCD$, the edges are given: $AB=7, BC=8, CD=4$. Find the edge $DA$, if it is known that the lines $AC$ and $BD$ are perpendicular. | Let's draw the height $B K$ of triangle $A B C$. The line $A C$ is perpendicular to two intersecting lines $B K$ and $B D$ in the plane $B D K$. Therefore, the line $A C$ is perpendicular to each line in this plane, particularly to the line $D K$. Hence, $D K$ is the height of triangle $A D C$. We can apply the Pythago... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Sum of interior and exterior angles of a polygon]
Find the number $n$ of sides of a convex $n$-gon if each of its interior angles is not less than $143^{\circ}$ and not more than $146^{\circ}$.
# | The sum of the interior angles of a convex $n$-gon is $180^{\circ}(n-2)$, therefore
from which we find that $9 \frac{27}{37} \leqslant n \leqslant 10 \frac{20}{27}$. This inequality is satisfied by the unique natural number 10.
## Answer
10.00 | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Given points $A, B, C$, and $D$ such that segments $A C$ and $B D$ intersect at point $E$. Segment $A E$ is 1 cm shorter than segment $A B$, $A E = D C$, $A D = B E$,
$\angle A D C = \angle D E C$. Find the length of $E C$. | $\angle B E A=\angle D E C=\angle A D C$, therefore triangles $A D C$ and $B E A$ are equal by two sides and the included angle. Thus, $A B=A C$. Therefore,
$E C=A C-A E=A B-A E=1$.
## Answer
1 cm. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Equilateral Triangle]
Two equilateral triangles with perimeters of 12 and 15 are positioned such that their sides are respectively parallel (see Fig.1). Find the perimeter of the resulting hexagon.
# | Let $a, b, c, d, e$ and $f$ be the sides of the formed equilateral triangles (Fig.2). Then $b+a+f=15$ and $c+d+e=12$. Therefore, the perimeter of the hexagon is
$$
a+b+c+d+e+f=(a+b+f)+(c+d+e)=15+12=27
$$
## Answer
27.00 | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Perimeter of a Triangle ]
A large triangle is divided by three bold segments into four triangles and three quadrilaterals. The sum of the perimeters of the quadrilaterals is 25 cm. The sum of the perimeters of the four triangles is 20 cm. The perimeter of the original large triangle is 19 cm. Find the sum of the len... | The sum of the perimeters of all triangles and quadrilaterals is equal to the perimeter of the large triangle plus twice the sum of the lengths of the bold segments. Therefore, this sum is equal to $(25 + 20 - 19) : 2 = 13$.
## Answer
13 cm. | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Blinkov A.D:
On the side $AB$ of rectangle $ABCD$, a point $M$ is chosen. Through this point, a perpendicular to the line $CM$ is drawn, which intersects the side $AD$ at point $E$. Point $P$ is the foot of the perpendicular dropped from point $M$ to the line $CE$. Find the angle $APB$. | Since
$$
\angle M A E=\angle M P E=\angle C P M=\angle M B C=90^{\circ}
$$
(see figure), quadrilaterals $A E P M$ and $B M P C$ are cyclic. Therefore, $\angle B P A=\angle B P M+\angle M P A=\angle B C M+\angle M E A=$ $=180^{\circ}-\angle E M A-\angle C M B=\angle E M C=90^{\circ}$.
, then $\alpha = 30^{\circ}$.
## Answer
... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Ratio of areas of triangles with a common angle ]
On the sides $A B, B C$ and $A C$ of triangle $A B C$, whose area is 75, points $M, N$ and $K$ are located respectively. It is known that $M$ is the midpoint of $A B$, the area of triangle $B M N$ is 15, and the area of triangle $A M K$ is 25. Find the area of triang... | Let $S_{\triangle A B C}=S$. Then
$$
\frac{S_{\Delta \mathrm{H} M K}}{S}=\frac{A M}{A B} \cdot \frac{A K}{A C}=\frac{1}{2} \cdot \frac{A K}{A C}
$$
Since according to the problem, $\frac{S_{\Delta+\Lambda M K}}{S}=\frac{25}{75}=\frac{1}{3}$, we have $\frac{1}{2} \cdot \frac{A K}{A C}=\frac{1}{3}$, from which we find ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[
Find the height of a right triangle dropped to the hypotenuse, given that the base of this height divides the hypotenuse into segments of 1 and 4.
# | Let $A B C$ be a right triangle with hypotenuse $A B$, and $C H$ be its altitude, $B H=1, A H=4$.
According to the altitude theorem of a right triangle drawn from the vertex of the right angle,
$$
C H^2=B H \cdot A H=1 \cdot 4=4
$$
Therefore, $C H=2$.
## Answer
2. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Ratio of areas of triangles with a common angle]
On the sides $AB$ and $AC$ of triangle $ABC$, whose area is 50, points $M$ and $K$ are taken such that $AM: MB=1: 5$, and $AK: KC=3: 2$. Find the area of triangle $AMK$.
# | $$
S_{\triangle A M K}=\frac{A M}{A B} \cdot \frac{A K}{A C} S_{\triangle A B C}=\frac{1}{6} \cdot \frac{3}{5} \cdot 50=5
$$
## Answer
5. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$[$ Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part
A point $M$ is connected to a circle by two lines. One of them touches the circle at point $A$, while the other intersects the circle at points $B$ and $C$, with $B C=7$ and $B M=9$. Find $A M$.
# | Use the Tangent-Secant Theorem. Consider two cases.
## Solution
Let point $B$ lie between points $M$ and $C$. According to the Tangent-Secant Theorem,
$$
A M^{2}=M C \cdot M B=(9+7) 9=16 \cdot 9=12^{2}
$$
Therefore, $A M=12$.
If point $C$ lies between points $B$ and $M$, then similarly we get $A M=3 \sqrt{2}$.
##... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The diagonal of a rectangular parallelepiped is 13, and the diagonals of the side faces are $4 \sqrt{10}$ and $3 \sqrt{17}$.
Find its volume. | Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a rectangular parallelepiped, in which $B D_{1}=13, A D_{1}=3 \sqrt{17}, C D_{1}=4 \sqrt{10}$.
Apply the Pythagorean theorem to the right triangles $A B D_{1}, B C D_{1}$, and $C D D_{1}$.
## Solution
Let $A B C D A_{1} B_{1} C_{1} D_{1}$ be a rectangular parallelepiped, in w... | 144 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Rectangles and Squares. Properties and Characteristics ]
In a plane, there is a square with vertices $A, B, C, D$ in sequence and a point $O$ outside the square. It is known that $A O=O B=5$ and $O D=\sqrt{13}$. Find the area of the square. | Compose an equation with respect to the side of the square. One of the solutions of this equation contradicts the condition of the problem.
## Solution
Let $x$ be the side of the square. Let $P$ be the projection of point $O$ onto the line $A D$. From the condition of the problem, it follows that points $O$ and $A$ l... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Sum of angles in a triangle. Theorem about the exterior angle.]
$A B C$ is an isosceles triangle with base $A C, C D$ is the bisector of angle $C, \angle A D C=150^{\circ}$. Find $\angle B$.
# | Apply the theorem about the sum of the angles of a triangle.
## Solution
Let $\angle A=\angle C=2 \alpha$. Then $\angle D C A=\angle B C D=\alpha$. By the condition $2 \alpha+\alpha=180^{\circ}-150^{\circ}$, from which $\alpha=10^{\circ}$. Therefore, $\angle B=180^{\circ}-4 \alpha=140^{\circ}$.
.]
One of the legs of a right triangle is 10 more than the other and 10 less than the hypotenuse. Find the hypotenuse of this triangle. | Let the first leg be $x$. Then the second leg is $x-10$, and the hypotenuse is $x+10$. By the Pythagorean theorem, $(x+10)^{2}=x^{2}+(x-10)^{2}$, from which $x=40$. Therefore, $x+10=50$.
## Answer
50. | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$[$ Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part
From a point $M$, located outside the circle at a distance of $\sqrt{7}$ from the center, a secant is drawn, the internal part of which is half the external part and equal to the radius of the circle.
Find the... | Apply the tangent-secant theorem.
## Solution
Let $r$ be the radius of the circle. According to the tangent-secant theorem, $2 r \cdot 3 r = (\sqrt{7} - r)(\sqrt{7} + r) = 7 - r^2$, from which we get $r = 1$.
## Answer
1. Send a comment
Point $B$ lies on segment $AC$, which is equal to 5. Find the distance between... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Measurement of segment lengths and angle measures. Adjacent angles.]
On a straight line, points $A, B, C$ and $D$ are marked sequentially, and $A B=B C=C D=6$.
Find the distance between the midpoints of segments $A B$ and $C D$.
# | Let $M$ and $N$ be the midpoints of segments $AB$ and $CD$ respectively. Then $MN = MB + BC + CN = 3 + 6 + 3 = 12$.
## Answer
12. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Intuitive Geometry in Space ]
On three faces of a cube, diagonals were drawn such that a triangle was formed. Find the angles of this triangle.
# | Got an equilateral triangle, therefore, the angles of this triangle are 60 degrees each. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Measuring lengths of segments and measures of angles. Adjacent angles.] [Arithmetic operations. Numerical identities]
On a ruler 9 cm long there are no divisions.
Mark three intermediate divisions on it so that it can measure distances from 1 to 9 cm with an accuracy of 1 cm
# | It is sufficient to make intermediate divisions at the points 1 cm, 4 cm, 7 cm. Then we have 4 segments: 1 cm, 3 cm, 3 cm, and 2 cm. It is not difficult to verify that with such a ruler, one can measure any whole distance from 1 to 9 cm.
Send a comment | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Inscribed and Circumscribed Circles ]
The hypotenuse of a right triangle is 4 m. Find the radius of the circumscribed circle.
# | The center of the circle circumscribed around a right triangle coincides with the midpoint of the hypotenuse.
## Solution
The center of the circle circumscribed around a right triangle coincides with the midpoint of the hypotenuse. Therefore, the radius of the circle is half the hypotenuse, i.e., 2 m.
 })]\end{array}\right]$
The inner chambers of Sultan Ibrahim ibn-Said's palace consist of 100 identical square rooms, arranged in a $10 \times 10$ square. If two rooms share a wall, there is exactly one door in ... | Note that the sultan's palace has 4 outer walls and 18 internal partitions.
## Solution
Since the sultan's palace has 4 outer walls, each 10 rooms long, and 18 internal partitions (9 longitudinal and 9 transverse), each also 10 rooms long, we can determine the number of windows $(10 \times 4=40)$ and doors $(10 \time... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Cutting (other).]
Two people had two square cakes. Each made 2 straight cuts from edge to edge on their cake. As a result, one ended up with three pieces, and the other with four. How could this happen? | Note: Cuts can intersect.
## Solution
This could happen if in the first case the cuts did not intersect each other, while in the second case they did. For example, if in the first case the cuts were parallel to each other, and in the second case they were perpendicular.
## Answer
In the first case, the cuts were pa... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[The angle between two intersecting chords and two secants]
A circle is divided by points $A, B, C, D$ such that $\sim A B:-B C:-C D:-D A=3: 2: 13: 7$. Chords $A D$ and $B C$ are extended to intersect at point $M$.
Find the angle $AMB$ | $\angle A M B=1 / 2(-D C--A B)=1 / 2 \cdot \cdot 10 / 25 \cdot 360^{\circ}=72^{\circ}$
Answer
$72^{\circ}$. | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a right parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ with bases $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$, it is known that $A B=29, A D=36$, $B D=25, A A_{1}=48$. Find the area of the section $A B_{1} C_{1} D$. | Drop a perpendicular $B K$ from vertex $B$ to line $A D$. Since the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ is a right parallelepiped, $B_{1} B$ is perpendicular to the base plane $A B C D$. Therefore, $B K$ is the orthogonal projection of the slant $B_{1} K$ onto the base plane $A B C D$. By the theorem of th... | 1872 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Angles between angle bisectors $\quad]$
In triangle $ABC$, the angle bisectors of the angles at vertices $A$ and $C$ intersect at point $D$. Find the radius of the circumcircle of triangle $ABC$, if the radius of the circumcircle of triangle $ADC$ with center at point $O$ is $R=6$, and $\angle ACO=30^{\circ}$. | Prove that $\angle A D C=90^{\circ}+\frac{1}{2} \angle B$.
## Solution
Since $\angle B A C+\angle A C B=180^{\circ}-\angle B$, then $\angle D A C+\angle A C D=90^{\circ}-\frac{1}{2} \angle B$. Therefore,
$$
\angle A D C=180^{\circ}-\left(90^{\circ}-\frac{1}{2} \angle B=90^{\circ}+\frac{1}{2} \angle B>90^{\circ}-\ang... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Triangle Inequality
Point $C$ divides the chord $A B$ of a circle with radius 6 into segments $A C=4$ and $C B=5$. Find the minimum distance from point $C$ to the points on the circle.
# | Let $O$ be the center of the circle. Extend the segment $OC$ beyond point $C$ to intersect the circle at point $M$, and using the triangle inequality, prove that the length of the segment $CM$ is the smallest of the distances from point $C$ to the points on the circle.
## Solution
Let $O$ be the center of the circle.... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ The Law of Cosines [Isosceles, Inscribed, and Circumscribed Trapezoids]
The bases of the trapezoid are 3 cm and 5 cm. One of the diagonals of the trapezoid is 8 cm, and the angle between the diagonals is $60^{\circ}$. Find the perimeter of the trapezoid.
# | Through the vertex of the smaller base of the trapezoid, draw a line parallel to one of the diagonals.
## Solution
Let $A B C D$ be the given trapezoid, with $B C=3$ and $A D=5$ as its bases, and $A C=8$ as the given diagonal. Let $O$ be the point of intersection of the diagonals. Draw a line through vertex $C$ paral... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The number of edges of a convex polyhedron is 99. What is the maximum number of edges that a plane, not passing through its vertices, can intersect?
# | The plane does not intersect at least one edge in each face.
## Solution
The answer is 66. We need to prove that this number is possible, and that a larger number is impossible. First, let's prove the latter. Each face of the polyhedron is intersected by the plane in no more than two edges, meaning that at least one ... | 66 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
A sphere inscribed in the pyramid $SABC$ touches the faces $SAB, SBC, SCA$ at points $D, E, F$ respectively.
Find all possible values of the sum of the angles $SDA, SEB$, and $SFC$. | Since triangles SCE and $S C F$ are equal, $\angle S F C=\angle S E C$. Similarly, $\angle S E B=\angle S D B$ and $\angle S D A=\angle S F A$. Therefore,
$\angle S D A+\angle S E B+\angle S F C=\angle S F A+\angle S D B+\angle S E C=1 / 2\left(3 \cdot 360^{\circ}-(\angle A D B+\angle B E C+\angle C F A)\right)$. But ... | 360 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A rook has traversed the chessboard, visiting each square at least once. What is the minimum number of turns it could have made
# | Either the rook made at least one move in each row, or it made at least one move in each column.
## Solution
An example with 14 turns is shown in the picture. Let's prove that there could not have been fewer than 14 turns. We will show that either the rook made at least one move in each row, or it made at least one m... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Mutual Position of Two Circles]
What is the mutual position of two circles if:
a) the distance between the centers is 10, and the radii are 8 and 2;
b) the distance between the centers is 4, and the radii are 11 and 17;
c) the distance between the centers is 12, and the radii are 5 and 3? | Let $O_{1}$ and $O_{2}$ be the centers of the circles, $R$ and $r$ be their radii, and $R \geqslant r$.
If $R+r=O_{1} O_{2}$, then the circles touch.
If $R+r<O_{1} O_{2}$, then one circle is located outside the other.
If $O_{1} O_{2}<R-r$, then one circle is located inside the other.
## Answer
a) touch; b) one ins... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Krekov D:
In an acute scalene triangle $ABC$, the altitudes $AA'$ and $BB'$ intersect at point $H$, and the medians of triangle $AHB$ intersect at point $M$. The line $CM$ bisects the segment $A'B'$. Find the angle $C$. | Let $C_{0}$ be the midpoint of $AB$, and $H^{\prime}$ be the point symmetric to $H$ with respect to $C_{0}$ (it is known that $H^{\prime}$ is a point on the circumcircle of triangle $ABC$, diametrically opposite to $C$). The medians $CC_{0}$ and $CM$ of similar triangles $ABC$ and $A^{\prime}B^{\prime}C$ are symmetric ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Emelyanov L.A.
The median $A A_{0}$ of triangle $A B C$ is laid off from point $A_{0}$ perpendicular to side $B C$ to the exterior of the triangle. Denote the second end of the constructed segment by $A_{1}$. Similarly, points $B_{1}$ and $C_{1}$ are constructed. Find the angles of triangle $A_{1} B_{1} C_{1}$, if the... | 
Since triangle $ABC$ is isosceles, $BB_0$ is the perpendicular bisector of the base $AC$. Therefore, $B_1$ lies on this perpendicular and $CB_0 \perp BB_1$. Thus, $CB_0$ is the height and me... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Bogov and I.I.
The distance between two cells on an infinite chessboard is defined as the minimum number of moves in the path of a king between these cells. On the board, three cells are marked, the pairwise distances between which are 100. How many cells exist such that the distances from them to all three marked cel... | Consider two arbitrary cells $A$ and $B$. Let the difference in the abscissas of their centers be $x \geq 0$, and the difference in the ordinates be $y \geq 0$. Then the distance $\rho(A, B)$ between these cells is $\max \{x, y\}$.
Let cells $A, B, C$ be marked. Then for each pair of cells, there exists a coordinate i... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.
Segments connecting an inner point of a convex non-equilateral $n$-gon with its vertices divide the $n$-gon into $n$ equal triangles.
For what smallest $n$ is this possible? | Let's prove that the specified situation is impossible for $n=3,4$.
The first method. For $n=3$, the angles of the triangles in the partition that meet at an internal point are equal, since the sum of any two different angles in them is less than $180^{\circ}$. But then the sides opposite to these angles, which are si... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The diagonals of the trapezoid are 3 and 5, and the segment connecting the midpoints of the bases is 2. Find the area of the trapezoid.
# | Through the vertex of the smaller base of the trapezoid, draw a line parallel to the diagonal.
## Solution
Let $M$ and $K$ be the midpoints of the bases $BC$ and $AD$ of trapezoid $ABCD$. Through vertex $C$ of the smaller base $BC (AC=3, BD=5)$, draw a line parallel to diagonal $BD$, intersecting line $AD$ at point $... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A circle with radius $\frac{2}{\sqrt{3}}$ is inscribed in an isosceles trapezoid. The angle between the diagonals of the trapezoid, subtending the larger base, is $2 \operatorname{arctg} \frac{2}{\sqrt{3}}$. Find the segment connecting the points of tangency of the circle with the larger base of the trapezoid and one o... | Denote the halves of the bases of the trapezoid as $x$ and $y$; find $x, y$ and the angle between the lateral side and the larger base.
## Solution
Let $Q$ be the point of intersection of the diagonals $A C$ and $B D$ of trapezoid $A B C D$; $N, M$, and $K$ be the points of tangency of the inscribed circle with the s... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ [tangents drawn from one point] [ Midline of a triangle $\quad$]
A circle of radius 1 is inscribed in triangle $ABC$, where $\cos \angle B=0.8$. This circle touches the midline of triangle $ABC$, parallel to side $AC$. Find the side $AC$. | The distance from the vertex of a triangle to the nearest point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side.
## Solution
Let $O$ be the center of the inscribed circle, and $K$ be its point of tangency with side $AB$. Then
$$
\operatorname{tg} \frac... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The altitudes of the acute-angled triangle $ABC$, drawn from points $B$ and $C$, were extended to intersect the circumscribed circle at points $B_1$ and $C_1$. It turned out that the segment $B_1 C_1$ passes through the center of the circumscribed circle. Find the angle $BAC$.
# | Let $\angle B A C=\alpha$. Inscribed angles $B A C$ and $B B 1 C$ subtend the same arc, so $\angle$ $B B 1 C=\angle B A C=\alpha$. Point $C$ lies on the circle with diameter $B 1 C 1$, therefore $\angle B 1 C C 1=90^{\circ}$. Lines $B 1 C$ and $A B$ are parallel, as they are perpendicular to the same line $C C 1$, henc... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## [ The larger side is opposite the larger angle ]
[Sum of the angles of a triangle. Theorem about the exterior angle.] In triangle $ABC$, it is known that $AB < BC < AC$, and one of the angles is twice as small as another and three times as small as the third. Find the angle at vertex $A$. | Against the larger side of a triangle lies the larger angle.
## Solution
Since in a triangle, the larger angle lies opposite the larger side, the smallest angle of triangle $A B C$ lies opposite side $A B$, that is, it is angle $A C B$. Let $\angle C=\gamma$. Then $\angle A=2 \gamma, \angle B=3 \gamma$. By the theore... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Varignon Parallelogram ] [ Area of a Quadrilateral ]
The segments connecting the midpoints of opposite sides of a convex quadrilateral are equal to each other. Find the area of the quadrilateral if its diagonals are 8 and 12. | Let $K, L, M$ and $N$ be the midpoints of the sides $A B, B C, C D$ and $A D$ of the given convex quadrilateral $A B C D$. Since $K L$ and $M N$ are the midlines of triangles $A B C$ and $A D C$, we have $K L \| M N$ and $K L = M N$, thus quadrilateral $K L M N$ is a parallelogram. Since its diagonals $K M$ and $L N$ a... | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Pythagorean Theorem (direct and inverse). ] [ The larger angle lies opposite the larger side ]
In a triangle, the larger angle at the base is equal to $45^{\circ}$, and the height divides the base into segments of 20 and 21. Find the larger lateral side. | In a triangle, the side opposite the larger angle is the longer side.
## Solution
Let $\angle A=45^{\circ}$. Since $\angle A>\angle C$, then $B C>A B$, which means $B C$ is the longer lateral side. Therefore, $B D=A D=20, B C^{2}=B D^{2}+D C^{2}=841=29^{2}$.
. $\quad]$
On a wooden ruler, three marks are made: 0, 7, and 11 centimeters. How can you measure with it a segment equal to: a) 8 cm $;$ b) 5 cm
# | a) Using divisions of 7 and 11, you can measure 4 cm. Doing this twice, you will get a segment of 8 cm.
b) First method. Knowing how to measure 8 and 7, you can measure 1 cm. Doing this 5 times, you will get 5 cm.
Second method. $5=3 \cdot 11-4 \cdot 7$, so it is enough to measure 3 times 11 cm and then 4 times 7 in ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Measuring lengths of segments and measures of angles. Adjacent angles.] Extreme properties (miscellaneous).
In a village, along a straight road, there are two houses $A$ and $B$ located 50 meters apart from each other.
At what point on the road should a well be dug so that the sum of the distances from the well to t... | Any point $X$ on the road segment between $A$ and $B$ is equally good, since the segments $A X$ and $B X$ together total 50 m. If a point is taken outside this segment, then the sum of the distances from it to points $A$ and $B$ will be greater than $50 \mathrm{M}$.
## Answer
At any point on the segment $A B$. | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Unusual Constructions (Miscellaneous). ] [ Measuring the lengths of segments and measures of angles. Adjacent angles. $]$
There is an angle ruler with an angle of $70^{\circ}$. How can you construct an angle of $40^{\circ}$ with it?
# | First method. $40^{\circ}=180^{\circ}-2 \cdot 70^{\circ}$.
Second method. If one angle of the triangle is $70^{\circ}$, then the second one is $20^{\circ} .40^{\circ}=2 \cdot 20^{\circ}$. | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a square with an area of 18, a rectangle is inscribed such that one vertex of the rectangle lies on each side of the square. The sides of the rectangle are in the ratio $1: 2$.
Find the area of the rectangle. | Let the vertices $K, L, M$ and $N$ of the rectangle $K L M N$ be located on the sides $A B, B C, C D$ and $A D$ of the square $A B C D$, respectively, such that $K N=2 K L$. Let $K L=x, \angle A K N=\alpha$. Then
$M N=x, L M=K N=2 x, \angle C M L=\angle B L K=\angle A K N=\alpha, A K=2 x \cos \alpha, B K=x \sin \alpha... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 ( A square with an area of 24 has a rectangle inscribed in it such that one vertex of the rectangle lies on each side of the square. The sides of the rectangle are in the ratio $1: 3$.
Find the area of the rectangle.
# | ## Answer
9.
## Problem
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 [ Algebraic problems on the triangle inequality ]
How many (non-degenerate) triangles with a perimeter of 100 and integer side lengths exist?
# | We need to find the number of triples $(a, b, c)$ of natural numbers where $a \leq b \leq c, a+b+c=100, a+b>c$.
It is clear that $c$ can take values from 34 to 49. For each of these values, $a+b=100-c$, so $b$ can take values from
$\frac{1}{2}(100-c)$ to $c$, more precisely, from $50-\frac{c}{2}$ to $c$ for even $c$ ... | 208 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Moskvitin N.A.
In triangle $A B C$, $A B = B C$. A perpendicular $E D$ is dropped from point $E$ on side $A B$ to $B C$. It turns out that $A E = E D$. Find the angle $D A C$. | According to the exterior angle theorem, $\angle A E D=90^{\circ}+\angle B=270^{\circ}-2 \angle A$ (see figure). Therefore, $\angle E A D=1 / 2\left(180^{\circ}-\right.$ $\angle A E D)=\angle A-45^{\circ}$.
 line if during its move it moves from one square of this horizontal (vertical) to another. Suppose that the rook has made a tour of the chessboard, visiting each square exactly once. Let's show that during the tour, the rook must either move along each vert... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Ruubanov I.S.
A square is divided by lines into 25 smaller square cells. In some cells, one of the diagonals is drawn such that no two diagonals share a common point (even a common endpoint). What is the maximum possible number of drawn diagonals?
# | Example with 16 diagonals, see the figure.
Estimate. Suppose it was possible to draw 17 diagonals. We will present two ways to arrive at a contradiction.
First method. Each diagonal has tw... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Frame }}$ D:
On a circle, 2012 points are marked, dividing it into equal arcs. From these, $k$ points are chosen and a convex $k$-gon is constructed with vertices at the chosen points. What is the largest $k$ for which it could turn out that the polygon has no parallel sides? | Let $A_{1}, A_{2}, A_{2012}$ be the marked points in the order of traversal (we will assume that $A_{2013}=A_{1}, A_{2014}=A_{2}$). We will divide them into quadruples
$\left(A_{1}, A_{2}, A_{1007}, A_{1008}\right),\left(A_{3}, A_{4}, A_{1009}, A_{1010}\right), \ldots,\left(A_{1005}, A_{1006}, A_{2011}, A_{2012}\right... | 1509 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Each face of a cube $6 \times 6 \times 6$ is divided into $1 \times 1$ cells. The cube is covered with $2 \times 2$ squares such that each square covers exactly four cells, no squares overlap, and each cell is covered by the same number of squares. What is the maximum value that this identical number c... | Evaluation. A cell in the corner of a face can be covered in three ways (entirely within the face, with a fold over one edge of the corner, with a fold over the other edge of the corner). This means that each cell is covered by no more than three squares.
Example. Consider the usual covering of a cube with squares, wh... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Triangle Inequality (other).]
At vertex $A$ of a unit square $A B C D$, there is an ant. It needs to reach point $C$, where the entrance to the ant hill is located. Points $A$ and $C$ are separated by a vertical wall, which has the shape of an isosceles right triangle with hypotenuse $B D$. Find the length of the sh... | Let $K$ be the vertex of the right angle of triangle $B K D$ (a vertical wall). The path of the ant consists of four segments: $A M, M P, P M$, and $M C$, where point $M$ lies on line $B D$, and point $P$ lies on segment $D K$ or $B K$. Clearly, $A M = M C$, so it is sufficient to specify points $M$ and $P$ such that t... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
C
[Median line of a triangle
[ Trigonometric ratios in a right-angled triangle ]
In triangle $ABC$, the median $BM$ and the altitude $AH$ are drawn. It is known that $BM = AH$. Find the angle $MBC$.
# | Drop a perpendicular from point $M$ to line $B C$.
## Solution
Drop the perpendicular $M K$ to line $B C$. Then $M K$ is the midline of triangle $A H C$. Therefore, $M K = 1/2 A H = 1/2 B M$, which means $\angle M B K = 30^{\circ}$. Consequently, $\angle M B C = 30^{\circ}$ or $150^{\circ}$.
![](https://cdn.mathpix.c... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Given an isosceles triangle with a base of 12 and a side length of 18. What segments need to be marked off from the vertex of the triangle on its sides so that connecting their ends forms a trapezoid with a perimeter of 40?
# | The smaller base of the trapezoid cuts off a similar triangle from the given triangle.
## Solution
Let $ABC$ be the given triangle, $AB=BC=18, AC=12$. Denote the sought segments $BM$ and $BN$ by $x$. Then $AM=CN=18-x$.
From the similarity of triangles $MBN$ and $ABC$, we find that $MN=\frac{BM \cdot AC}{AB}=2x / 3$.... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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