problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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In triangle $A B C$, side $A B$ is equal to 21, the bisector $B D$ is $8 \sqrt{7}$, and $D C=8$. Find the perimeter of triangle $A B C$. | Apply the property of the bisector of a triangle and the sine and cosine theorems.
## Solution
Let $A D=x, B C=y, \angle A B D=\angle C B D=\alpha$.
By the property of the bisector of a triangle, $\frac{A B}{B C}=\frac{A D}{D C}$, or $\frac{21}{y}=\frac{x}{8}$, from which we find that $y=\frac{21 \cdot 8}{x}$. By th... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the village, there are 100 houses. What is the maximum number of closed, non-intersecting fences that can be built so that each fence encloses at least one house and no two fences enclose the same set of houses
# | In the maximum set of fences, there is a fence that limits exactly two houses. By combining these two houses into one, we reduce the number of houses by 1 and the number of fences by 2. In this process, both conditions of the problem are maintained. Continue this process. After 99 steps, one house and one fence will re... | 199 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Extremal properties (other) $)]$
In a city, there are 10 streets parallel to each other and 10 streets intersecting them at right angles. What is the smallest number of turns a closed bus route can have, passing through all intersections?
# | A closed route passing through all intersections can have 20 turns (see figure). It remains to prove that such a route cannot have fewer than 20 turns. After each turn, there is a transition from a horizontal street to a vertical one or vice versa. Therefore, the number of horizontal segments of the closed route is equ... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$[$ Extremal properties (other) $]$
What is the maximum number of cells on an $8 \times 8$ chessboard that can be cut by a single straight line? # | A line can intersect 15 cells (see the figure). Now let's prove that a line cannot intersect more than 15 cells. The number of cells a line intersects is one less than the number of intersection points with the segments that form the sides of the cells. Inside the square, there are 14 such segments. Therefore, inside t... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Relationships between the sides and angles of triangles (other).]
In a right triangle $ABC$ with a right angle at $A$, a circle is constructed on the altitude $AD$ as a diameter, intersecting side $AB$ at point $K$ and side $AC$ at point $M$. Segments $AD$ and $KM$ intersect at point $L$. Find the acute angles of tr... | Clearly, $A K D M$ is a rectangle and $L$ is the point of intersection of its diagonals. Since $A D \perp B C$ and $A M \perp B A$, we have $\angle D A M = \angle A B C$. Similarly, $\angle K A D = \angle A C B$. Drop a perpendicular $A P$ from point $A$ to line $K M$. Let's assume for definiteness that $\angle B < \an... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Extreme properties of a triangle (miscellaneous).]
In a triangle, two sides $a$ and $b$ are known. What should the third side be so that the smallest angle of the triangle has the greatest possible magnitude? | Let's assume for definiteness that $a \geq b$. Then the smallest angle of the triangle is angle $B$ or angle $C$. Consider a semicircle $S$ with radius $a$. Let $C$ be the center of this semicircle, and $B$ be a point on the extension of the diameter such that $CB = a$. Draw a tangent $BA_1$ from point $B$ to the semic... | 1699 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4+ [ Dirichlet's Principle (finite number of points, lines, etc.)
On a plane, 300 lines are drawn, and no two of them are parallel, and no three intersect at the same point. The plane is divided into pieces by these lines. Prove that among the pieces, there will be at least 100 triangles.
# | Let's take one of these lines and consider all the intersection points of the lines that do not lie on the chosen line, and select the nearest among them. Among the pieces into which the plane is cut, there is a triangle, one vertex of which is the chosen point, and the other two lie on the chosen line. Indeed, the tri... | 100 | Geometry | proof | Yes | Yes | olympiads | false |
[^1]
A spider web has the form of a grid of $100 \times 100$ nodes (in other words, it is a grid of $99 \times 99$ cells). In one of its corners sits a spider, and in some 100 nodes, flies are stuck to the web. In one move, the spider can move to any adjacent node. Can the spider guarantee to eat all the flies, spendi... | b) Divide the grid into 10 horizontal strips, each with 1000 nodes ($9 \times 99$ cells). Let the spider sit in the bottom-left corner. Initially, it moves from left to right along the bottom edge of the lower strip until it sees a fly stuck in the same strip above it. Then it crosses the strip vertically, eats the fly... | 1980 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Tangents touching circles $\quad$]
A circle of radius 2 touches another circle externally at point A. The common tangent to both circles, drawn through point $A$, intersects with another common tangent at point $B$. Find the radius of the second circle if $A B=4$. | If $r$ and $R$ are the radii of the circles, then $A B=\sqrt{r R}$.
## Solution
First method.
Let $r$ and $R$ be the radii of the circles ($r=2$), and $C$ and $D$ be the points of tangency of the circles with the second (external) tangent. Then
$$
B C=A B=B D=4
$$
Since $C D=2 \sqrt{r R}$, we have $\sqrt{2 R}=4$. ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In parallelogram $A B C D$, diagonal $B D$ is equal to 2, angle $C$ is $45^{\circ}$, and line $C D$ is tangent to the circumcircle of triangle $A B D$. Find the area of parallelogram $A B C D$. | Prove that $B D \perp B C$.
## Solution
It is clear that $\angle A=\angle C=45^{\circ}$. Angle $\angle B A D$ is inscribed in the circle, and $\angle B D C$ is the angle between the tangent and the chord $B D$, therefore,
$$
\angle B D C=\angle A=45^{\circ}
$$
Thus, $\angle C B D=90^{\circ}$. Therefore,
 }}\end{array}\right]}\end{array}\right]$
In a segment, the arc of which is $60^{\circ}$, a square is inscribed. Find the area of the square if the radius of the circle is $2 \sqrt... | Let $x$ be the side of the square and apply the Pythagorean theorem to the triangle with vertices: at the center of the circle, at the midpoint of the side of the square, and at the vertex of the square belonging to this side.
## Solution
Let $O$ be the center of the circle, $R$ its radius ($R=2 \sqrt{3}+\sqrt{17}$),... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two sides of a triangle are equal to 10 and 12, and the median drawn to the third side is equal to 5. Find the area of the triangle.
# | On the extension of the median $A M$ beyond point $M$, lay off the segment $M D$, equal to $A M$.
## Solution
Let $A M$ be the median of triangle $A B C$, where $A M=5, A B=10, A C=12$. On the extension of the median $A M$ beyond point $M$, we lay off the segment $M D$, equal to $A M$. Then $A B D C$ is a parallelogr... | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Pairing and grouping; bijections ]
What is the maximum number of pawns that can be placed on a chessboard (no more than one pawn per square), if:
1) a pawn cannot be placed on the e4 square;
2) no two pawns can stand on squares that are symmetric with respect to the e4 square?
# | Evaluation. All fields of the board except for the vertical a, the horizontal 8, and the field e4 itself can be divided into pairs symmetrical relative to e4. Such pairs amount to 24. According to the condition, no more than one pawn can be placed on the fields of each pair. Additionally, no more than one pawn can be p... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Counting in two ways ] [ Different tasks on cutting ]
Inside a square, 100 points are marked. The square is divided into triangles in such a way that the vertices of the triangles are only the 100 marked points and the vertices of the square, and for each triangle in the partition, each marked point either lies out... | Calculate the sum of the angles of all triangles.
## Solution
The sum of the angles of triangles with a vertex at some vertex of the square is $90^{\circ}$, each of the 100 marked points contributes $360^{\circ}$. Since there are no other vertices of the triangles, the sum of the angles of all triangles in the partit... | 202 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Properties and characteristics of an isosceles triangle. ]
In an isosceles triangle, the center of the inscribed circle divides the height in the ratio $12: 5$, and the lateral side is equal to 60. Find the base. | Let $CM$ be the altitude of the given triangle $ABC$, $AC=BC=60$, $O$ be the center of the inscribed circle. It is clear that $AM$ $<MC$. Since $AO$ is the bisector of angle $BAC$, then $AM=5/12 AC=25$, and $AB=2 AM=50$.
## Answer
50. | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Pythagorean Theorem (direct and inverse) ] [Auxiliary similar triangles]
In a right triangle $ABC$, $AC=16$, $BC=12$. A circle is described from the center $B$ with radius $BC$, and a tangent is drawn to it, parallel to the hypotenuse $AB$ (the tangent and the triangle lie on opposite sides of the hypotenuse). The l... | Consider similar triangles.
## Solution
Let $M$ be the point of tangency, $K$ be the point of intersection of the tangent with the extension of the leg $CB$. $AB^2 = CB^2 + CA^2 = 144 + 256 = 400$.
Triangles $BMK$ and $ACB$ are similar, so $BK : BM = AB : AC$. Therefore, $BK = 20 \cdot 16 : 12 = 15$.
 triangle ]
On the side $A B$ of the square $A B C D$, an equilateral triangle $A B M$ is constructed. Find the angle $D M C$.
# | Find the angles of the isosceles triangle $D A M$.
## Solution
Let point $M$ be located outside the square $A B C D$. Then the angle at vertex $A$ of the isosceles triangle $DAM$ is $90^{\circ}+60^{\circ}=150^{\circ}$, so $\angle A M D=\left(180^{\circ}-150^{\circ}\right): 2=15^{\circ}$.
Similarly, $\angle B M C=15^... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The centers of three pairwise externally tangent circles are located at points $A, B, C$,
$\angle A B C=90^{\circ}$. The points of tangency are $K, P$, and $M$; point $P$ lies on side $A C$. Find the angle $K P M$. | Express the desired angle in terms of the acute angles of triangle $ABC$.
## Solution
Let $\angle B A C=\alpha, \angle A C B=\gamma\left(\alpha+\gamma=90^{\circ}\right)$.
Suppose point K lies on segment $A B$. From the isosceles triangles KAP and MCP, we find that $\angle A P K=90^{\circ} - \alpha / 2, \angle M P C=... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Folkor
In a convex quadrilateral $A B C D: \angle B A C=20^{\circ}, \angle B C A=35^{\circ}, \angle B D C=40^{\circ}, \angle B D A=70^{\circ}$. Find the angle between the diagonals of the quadrilateral. | We will prove that point $D$ is the center of the circumcircle of triangle $ABC$. This can be done in various ways.
The first method. Describe a circle around triangle $ABC$ and extend segment $BD$ to intersect this circle at point $K$ (Fig. a). Since $\angle BKC = \angle BAC = 20^\circ$, then $\angle KCD = \angle BDC... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Tomong A.K.
On the plane, there is an open non-intersecting broken line with 31 segments (adjacent segments do not lie on the same straight line). Through each segment, a line containing this segment was drawn. As a result, 31 lines were obtained, some of which may have coincided. What is the smallest number of differ... | Evaluation. Except for the ends, the broken line has 30 vertices, and each is the intersection of two lines. If there are no more than eight lines, then there are no more than $7 \cdot 8: 2=28$ intersection points - a contradiction.
.]
The base of the pyramid is an isosceles triangle with a base of 6 and a height of 9. Each lateral edge is 13. Find the volume of the pyramid. | Let $ABCD$ be a triangular pyramid with base $ABC$, $DA=DB=DC=13$, $BC=6$, $AK=9$ ($K$ is the midpoint of $BC$). From the right triangle $AKC$, we find that
$$
AC=\sqrt{AK^{2}+CK^{2}}=\sqrt{81+9}=3\sqrt{10}, \sin \angle ACK=\frac{AK}{AC}=\frac{3}{\sqrt{10}}
$$
Since the lateral edges of the pyramid are equal, its hei... | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Regular pyramid ] [ Sections, unfoldings, and other frameworks. ]
On the extension of edge $S T$ beyond point $T$ of the regular quadrilateral pyramid $S P Q R T$ with vertex $S$, a point $B$ is taken such that the distance from it to the plane $S P Q$ is $\frac{9 \sqrt{7}}{2}$. Find the segment $B T$, if $Q R=12$, ... | Consider the section of the given pyramid by a plane passing through the vertex $S$ and the midpoints $A$ and $K$ of the sides $R T$ and $P Q$ of the square $P Q R T$, respectively. Let $S O$ be the height of the pyramid. Then $O$ is the center of the base $P Q R T$. From the right triangles $P K S$ and $K O S$, we fin... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The area of the base of a right triangular prism is 4, the areas of the lateral faces are 9, 10, and 17. Find the volume of the prism.
# | Let $V$ be the volume of the prism; $h$ be its height; $a, b$, and $c$ be the sides of the base of the prism; $p$ be the semi-perimeter of the base, and $S$ be the area of the base. According to the problem,
$$
a h=9, b h=10, c h=17
$$
from which
$$
a=\frac{9}{h}, b=\frac{19}{h}, c=\frac{17}{h} .
$$
Then
$$
p=\fra... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
The base of a right prism is an isosceles trapezoid $ABCD$, where $AB=CD=13, BC=11, AD=$ 21. The area of the diagonal section of the prism is 180. Find the area of the complete surface of the prism. | Let $A B C D A 1 B 1 C 1 D 1$ be the given prism, $S$ - the area of its total surface. The diagonal section of the prism is a rectangle $A C C 1 A 1$. Dropping a perpendicular $C K$ from vertex $C$ of the smaller base of the trapezoid to the larger base $A D$. Then
$$
D K=\frac{1}{2}(A D-B C)=\frac{1}{2}(21-11)=5
$$
... | 906 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Quadrilateral pyramid ] [ Volume of a tetrahedron and pyramid $]$
The base of the pyramid is a parallelogram, with adjacent sides equal to 9 and 10, and one of the diagonals equal to 11. The opposite lateral edges are equal, and each of the larger edges is equal to $10^{\frac{1}{2}}$. Find the volume of the pyramid. | Let $P A B C D$ be the given pyramid with vertex $P, P O$ - the height of the pyramid. Since $A P = C P$, point $O$ is equidistant from points $A$ and $C$. Therefore, point $O$ lies on the perpendicular bisector of segment $A C$. Similarly, point $O$ lies on the perpendicular bisector of segment $B D$, and since the pe... | 200 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 $[\quad$ Volume of a parallelepiped $\quad]$
The base of a right parallelepiped is a parallelogram, one of the angles of which is $30^{\circ}$. The area of the base is 4. The areas of two lateral faces of the parallelepiped are 6 and 12. Find the volume of the parallelepiped. | Let $a$ and $b$ be the sides of the parallelogram, $c$ be the height of the parallelepiped, and $V$ be its volume. From the problem statement, we have:
$$
a b \sin 30^{\circ}=\frac{1}{2} a b=4, a c=6, b c=12
$$
Therefore,
$$
a b c 2=6 \cdot 12=72, c 2=\frac{\frac{T 2}{\alpha b}}{\alpha i}=\frac{72}{8}=9
$$
from whi... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$[\quad$ Volume of a parallelepiped $\quad]$
In the base of a rectangular parallelepiped lies a square with side $2 \sqrt{3}$. The diagonal of one lateral face forms an angle of $30^{\circ}$ with the plane of the adjacent lateral face. Find the volume of the parallelepiped. | Let the diagonal $A B 1$ of the lateral face $A A 1 B 1 B$ of the rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$ form an angle of $30^{\circ}$ with the plane of the lateral face $B B 1 C 1 C$. The line $A B$ is perpendicular to the plane of the face $B B 1 C 1 C$, since it is perpendicular to two intersecting lin... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The lateral edge of a regular triangular pyramid is $\sqrt{5}$, and the height of the pyramid is 1. Find the dihedral angle at the base.
# | Let $M$ be the center of the base $ABC$ of a regular triangular pyramid $ABCD$, and $K$ be the midpoint of $BC$. Then $DM$ is the height of the pyramid, $AK$ is the height of the equilateral triangle $ABC$, and $AK$ passes through point $M$. Therefore, $MKD$ is the linear angle of the dihedral angle at the base of the ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ The sum of the angles of a triangle. The theorem about the exterior angle.] [ Criteria and properties of an isosceles triangle. ]
On the side $B C$ of the isosceles triangle $A B C(A B=B C)$, points $N$ and $M$ ( $N$ is closer to $B$ than $M$) are taken such that $N M=A M$ and $\angle M A C=\angle B A N$.
Find $\an... | Let $\angle M A C=\angle B A N=\alpha, \angle N A M=\angle A N M=\varphi$. Since $A M C$ is the exterior angle of triangle $A M N$, then $\angle A M C=2 \varphi . \angle A C M=\angle C A B=2 \alpha+\varphi$. Therefore, the sum of the angles of triangle $A M C$ is $(2 \alpha+\varphi)+2 \varphi+\alpha=3(\alpha+$\varphi).... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
It is known that the point symmetric to the center of the inscribed circle of triangle $ABC$ with respect to side $BC$ lies on the circumcircle of this triangle. Find angle $A$.
# | If $O$ is the center of the inscribed circle of triangle $ABC$, then $\angle BOC = 90^{\circ} + \frac{1}{2} \angle BAC$.
## Solution
Let $\angle BAC = \alpha$. Suppose $O$ is the center of the inscribed circle of triangle $ABC$. Then $BO$ and $CO$ are the angle bisectors of $\angle ABC$ and $\angle ACB$. Therefore,
... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Properties and characteristics of an isosceles triangle.] [Sum of the angles of a triangle. Theorem of the exterior angle.]
Triangle $A B C$ is isosceles $(A B=B C)$. Segment $A M$ divides it into two isosceles triangles with bases $A B$ and $M C$. Find angle $B$. | Use the theorems about the exterior angle of a triangle and the sum of the angles of a triangle.
## Solution
Let $\angle B=\beta$. Then $\angle B A M=\beta, \angle A M C=2 \beta, \angle A=\angle C=2 \beta$. Therefore, $\beta+2 \beta+2 \beta=180^{\circ}$, which means $\beta=36^{\circ}$.
## Answer
$36^{\circ}$. | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In triangle $A B C$, the altitudes $A A_{1}$ and $C C_{1}$ intersect at point $H$, which lies inside the triangle. It is known that $H$ is the midpoint of $A A_{1}$, and $C H: H C_{1}=2: 1$. Find the measure of angle $B$. | Mark the midpoint of segment $C H$.
## Solution
Let $M$ be the midpoint of segment $C H$. Then $A C_{1} A_{1} M$ is a parallelogram. Therefore, $\angle A_{1} M H=\angle A C_{1} H=90^{\circ}$. Segment $A_{1} M$ is the height and median of the right triangle $H A_{1} C$, which means this triangle is isosceles. Therefor... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The Law of Cosines
The bases of a 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 |
3 [Mutual arrangement of altitudes, medians, bisectors, etc.]
In triangle $ABC$, the bisector drawn from vertex $A$, the altitude drawn from vertex $B$, and the perpendicular bisector of side $AB$ intersect at one point. Find the angle at vertex $A$. | Let $\angle A=2 \alpha, M$ - the intersection point of the three lines specified in the condition. Since point $M$ lies on the perpendicular bisector of segment $A B$, then
$\angle A B M=\angle B A M=\alpha$.
From the right triangle $A B H$ we get $\alpha+2 \alpha=90^{\circ}$, that is $\alpha=30^{\circ}$. Therefore, ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a convex quadrilateral circumscribed around a circle, the products of opposite sides are equal. The angle between a side and one of the diagonals is $20^{\circ}$. Find the angle between this side and the other diagonal. | Let $A B C D$ be the given quadrilateral, $\angle B A C=20^{\circ}$. Denote $A B=a, B C=b, C D=c, A D=d$. From the condition of the problem, it follows that $a c=b d$ and $a+c=b+d$ (property of a cyclic quadrilateral). Then, by the theorem converse to Vieta's theorem, either $a=b$ and $c=d$, or $a=d$ and $c=b$. Suppose... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a convex quadrilateral $A B C D A B=B C$. The rays $B A$ and $C D$ intersect at point $E$, and the rays $A D$ and $B C$ intersect at point $F$. It is also known that $B E=B F$ and
$\angle D E F=25^{\circ}$. Find the angle $E F D$. | Triangles $B C E$ and $B A F$ are equal by two sides and the included angle, so $\angle B E C=\angle B F A$. Triangle $E B F$ is isosceles, so
$\angle B E F=\angle B F E$. Therefore, $\angle E F D=\angle E F A=\angle B F E-\angle B F A=\angle B E F-\angle B E C=\angle C E F=\angle D E F=25^{\circ}$.
## Answer
$25^{\... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
What is the maximum area that a triangle with sides \(a\), \(b\), and \(c\) can have, given the following constraints:
\[
0 < a \leq 1 \leq b \leq 2 \leq c \leq 3 \text{ ? }
\] | First, assume that the length of the third side c is not subject to any conditions. The area of a triangle with sides a and b and the angle $\alpha$ between them is determined by the formula $1 / 2$ ab sin $\alpha$. If a and b are subject to the conditions 0 , then the product ab $\sin \alpha$ will be maximal when each... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find the volume of a triangular pyramid, five edges of which are equal to 2, and the sixth is equal to $\sqrt{6}$.
# | Let $A B C D$ be a triangular pyramid, in which
$$
A D=B D=C D=A B=A C=2, B C=\sqrt{6} \text {. }
$$
If $D O$ is the height of the pyramid, then $O$ is the center of the circumcircle of triangle $A B C$ (since $A D=B D=C D$). Let $R$ be the radius of this circle, $M$ be the midpoint of $B C$, and $\angle A B C=\alpha... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find the height of a regular tetrahedron with edge $a$.
# | Let $A B C D$ be a regular tetrahedron, $D O$ its height, and $M$ the midpoint of $B C$. Since $A M$ is the height of the equilateral triangle $A B C$, and $O$ is the center of this triangle, we have
$$
A M = a^{\frac{\sqrt{3}}{2}}, \quad O A = \frac{2}{3}, \quad A M = \frac{a \sqrt{3}}{3}
$$
From the right triangle ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ [ Cross-sectional area $]$
The lateral edge of the pyramid is divided into 100 equal parts, and planes parallel to the base are drawn through the points of division. Find the ratio of the areas of the largest and smallest of the resulting cross-sections.
# | Let $S$ be the area of the base of the pyramid, $S_1$ and $S_2$ be the areas of the largest and smallest sections, respectively. The largest and smallest sections are polygons similar to the polygon of the base of the pyramid with coefficients $\frac{99}{100}$ and $\frac{1}{100}$, respectively. Therefore,
$$
S_1 = \le... | 9801 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11 |
The lateral surface development of a cylinder is a square with a side length of $2 \sqrt[3]{\pi}$. Find the volume of the cylinder. | Let the generating line of the cylinder be $h$, and the radius of the base be $r$, the volume of the cylinder be $V$. Since the development of the lateral surface of the cylinder is a square with side $2 \sqrt[3]{\pi}$, the length of the circumference of the base and the generating line of the cylinder are also equal t... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Points $E$ and $F$ lie on the sides $A B$ and $B C$ of the rhombus $A B C D$, respectively, such that $A E=5 B E, B F=5 C F$. It is known that triangle $D E F$ is equilateral. Find the angle $B A D$. | On side $AB$, we lay off the segment $AK = CF = BE$. From the equality of triangles $AKD$ and $CFD$ (by two sides and the included angle), it follows that
$DK = DF = ED$. The angles at the base $KE$ of the isosceles triangle $DKE$ are equal, so the adjacent angles $AKD$ and $BED$ are also equal, and triangles $AKD$ an... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the diagonal $B D$ of the rectangular trapezoid $A B C D\left(\angle D=90^{\circ}, B C \| A D\right)$, a point $Q$ is taken such that $B Q: Q D=1: 3$. A circle with center at point $Q$ touches the line $A D$ and intersects the line $B C$ at points $P$ and $M$. Find the length of the side $A B$, if $B C=9, A D=8, P M... | Let $E$ be the point of tangency of the circle with the line $A D$, and $F$ be the projection of the center $Q$ of the circle onto the line $B C$. Then the points $E, Q$, and $F$ lie on the same line, and $F$ is the midpoint of $P M$. Therefore, $F P = F M = 2$. Let $Q E = Q P = Q M = R$. Suppose that the point $P$ lie... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ $\quad \underline{\text { Tangents to Spheres }}$]
In a regular hexagonal pyramid $\operatorname{SABCDEF}$ ( $S$ - apex) the side of the base is $2 \sqrt{3}$, and the height of the pyramid $S H$ is 6. A plane passes through point $E$ perpendicular to line $A S$ and intersects segment $S H$ at point $O$. Points $P$ a... | Let the diagonals $A D$ and $C E$ of the regular hexagon $A B C D E F$ intersect at point $N$. From the properties of a regular hexagon, it follows that $N$ is the midpoint of segments $D H$ and $C E$, and $A D \perp C E$. Then, by the theorem of three perpendiculars, $C E \perp S A$, which means that the line $C E$ li... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{c}\text { Sphere inscribed in a pyramid } \\ \text { Regular pyramid }\end{array}\right]$
In a regular quadrilateral pyramid with a height not less than $h$, a hemisphere of radius 1 is placed such that it touches all the lateral faces of the pyramid, and the center of the hemisphere lies on the b... | Let $O$ be the center of the base $ABCD$ of a regular quadrilateral pyramid $PABCD$. The center of the hemisphere coincides with point $O$, and the hemisphere touches the lateral faces at points lying on the apothems. Let $M$ be the midpoint of side $BC$ of the square $ABCD$, and $K$ be the point of tangency of the hem... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 [
In triangle $ABC$, angle $B$ is equal to $\arccos \frac{15}{17}$. On side $AC$, a point $K$ is taken such that $AK=12$, $KC=4$. Find the radius of the circle passing through vertex $B$, touching side $AC$ at point $K$, and touching the circumcircle of triangle $ABC$.
# | 
## Answer
12.00 | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9}
In triangle $A B C$, angle $A$ is equal to $\arccos \frac{5}{13}$, side $B C$ is equal to 12. On the extension of $C B$ beyond point $C$, a point $M$ is taken such that $C M=6$. Find the radius of the circle passing through vertex $A$, tangent to line $B C$ at point $M$, and tangent to the circumcircle of triangl... | 
## Answer
6.00
Send a comment | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
In triangle $A B C A B=B C, \angle B=20^{\circ}$. Point $M$ on the base $A C$ is such that $A M: M C=1: 2$, point $H$ is the projection of $C$ on $B M$. Find the angle $A H B$. | Let's complete the triangle to a rhombus $ABCD$. Let $O$ be the center of the rhombus. Then the line $BM$ divides the median $AO$ of triangle $ABD$ in the ratio $2:1$. Therefore, this line is also a median, meaning it passes through the midpoint $K$ of segment $AD$.
Notice that points $O$ and $H$ lie on the circle wit... | 100 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Trapezoids $A B C D$ and $A C D E$ with equal larger bases $A D$ and $A C$ respectively are inscribed in a circle. What is the radius of this circle if the area of triangle $A D E$ is $1+\sqrt{3}$, and the angle $C O D$ is $60^{\circ}$, where $O$ is the point of intersection of the diagonals of trapezoid $A B C D$? | $\angle C O D=\frac{U C D+U A B}{2}$.
## Solution
Arcs enclosed between parallel chords are equal. Therefore, $\cup A B=\cup C D=\cup A E$. Since chords $A C$ and $A D$ are equal, then
$$
\cup A B+\cup B C=\cup A E+\cup E D
$$
Thus, $\cup B C=\cup E D$.
Let $\cup A B=\alpha, \cup B C=\beta$. Then
$$
\angle C O D=... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The distance from the orthocenter of triangle $ABC$ to vertex $C$ is equal to the radius of the circumscribed circle. Find the angle $ACB$.
# | Let $H$ be the orthocenter of triangle $ABC$, $C_{1}$ the midpoint of $AB$, $O$ the center of the circumscribed circle, and $R$ its radius. Since $CH=2OC_{1}$, then
$$
OC_{1}=\frac{1}{2}CH=\frac{1}{2}R=\frac{1}{2}OA
$$
Therefore,
$$
\angle AOC_{1}=60^{\circ}, \angle AOB=2\angle AOC_{1}=120^{\circ}.
$$
If points $C$... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Bogganov I.I.
A $100 \times 100$ grid is given, with cells colored black and white. In all columns, there are an equal number of black cells, while in all rows, there are different numbers of black cells. What is the maximum possible number of pairs of adjacent cells of different colors?
# | Number the rows from top to bottom and the columns from left to right with numbers from 1 to 100.
In each row, there can be from 0 to 100 black cells. Since the number of black cells in all rows are different, these numbers are all from 0 to 100, except for one (say, except for \( k \)). Therefore, the total number of... | 14751 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.v.
In a white $2016 \times 2016$ table, some cells are painted black. We will call a natural number $k$ successful if $k \leq 2016$, and in each of the grid squares with side $k$ located in the table, exactly $k$ cells are painted black. (For example, if all cells are black, then the only successful number i... | Evaluation. Consider an arbitrary coloring of the table. Suppose there are at least two successful numbers, and let $a$ be the smallest of them, and $b$ be the largest.
Divide $b$ by $a$ with a remainder: $b = q a + r$, where $0 \leq r < a$. If $r \neq 0$, then $r$ is a successful number, and $a < r < b$, which contra... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Tiling with domino bones and tiles ] ..... ]
[ Examples and counterexamples. Constructions ]
On a grid sheet of size 100×100, several pairwise non-overlapping cardboard isosceles right triangles with a leg of 1 were placed; each triangle occupies exactly half of one of the cells. It turned out that each unit segment... | Let $n=50$. We will call a triangle upper if it is located above the line containing its horizontal leg, and lower otherwise. We will number the horizontal lines of the grid from bottom to top with numbers from 0 to $2n$.
Estimate. Let $u_{k}$ (respectively $d_{k}$) be the number of segments on the $k$-th line that ar... | 2450 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let $\Gamma$.
The teacher is going to give the children a problem of the following type. He will inform them that he has thought of a polynomial $P(x)$
of degree 2017 with integer coefficients, the leading coefficient of which is 1. Then he will inform them of $k$ integers $n_{1}, n_{2}, \ldots, n_{k}$ and separately... | Evaluation. Let the teacher use some $k \leq 2016$, thinking of a polynomial $P(x)$.
Consider the polynomial $P(x)=P(x)+\left(x-n_{1}\right)\left(x-n_{2}\right) \ldots\left(x-n_{k}\right)$. Notice that the degree of the polynomial $Q(x)$ is also 2017, and its leading coefficient is also 1. In this case, $P\left(n_{1}\... | 2017 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kozhevnikov P.A
On each of two parallel lines $a$ and $b$, 50 points were marked.
What is the maximum possible number of acute-angled triangles with vertices at these points? | Obviously, the maximum is achieved when the points on both lines are sufficiently close, so that each segment with marked ends on one of the lines is seen from any point on the other line at an acute angle. Let's paint the points on one of the lines blue, and the projections of the points of the second line onto this l... | 41650 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Didin $M$.
There are 100 piles, each with 400 stones. In one move, Petya selects two piles, removes one stone from each, and earns as many points as the absolute difference in the number of stones in these two piles. Petya must remove all the stones. What is the maximum total number of points he can earn? | Evaluation. Let's assume that the stones in the piles are stacked on top of each other, and Petya takes the top (at the moment) stones from the selected piles. We will number the stones in each pile from bottom to top with numbers from 1 to 400. Then the number of points Petya gets on each move is the difference betwee... | 3920000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8,9 | |
Two identical polygons were cut out of cardboard, aligned, and pierced with a pin at some point. When one of the polygons is rotated around this "axis" by $25^{\circ} 30^{\prime}$, it aligns again with the second polygon. What is the smallest possible number of sides of such polygons? | Answer: 240. First, note that $\frac{1}{360} \cdot 25 \frac{1}{2}=\frac{17}{240}$, and the numbers 17 and 240 are coprime. Consider a ray emanating from the center to a vertex of the first polygon. Rotations of this ray around the center by angles $k \cdot 25^{\circ} 30^{\prime}$, where $k=1,2, \ldots, 240$, are distin... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Rearrangement of areas]
a) Four vertices of a regular dodecagon are located at the midpoints of the sides of a square (see figure). Prove that the area of the shaded part is 12 times smaller than the area of the dodecagon. b) Prove that the area of a dodecagon inscribed in a circle of radius 1 is 3.
 Let's cut the original square into four squares and consider one of them (see figure). Let point $B^{\prime}$ be symmetric to point $B$ with respect to the line $P Q$. We will prove that $\triangle A P B = \triangle O B^{\prime} P$. Triangle $A P B$ is isosceles, and the angle at its base is $15^{\circ}$, so triangl... | 3 | Geometry | proof | Yes | Yes | olympiads | false |
In an equilateral (irregular) pentagon $A B C D E$, angle $A B C$ is twice the angle $D B E$. Find the measure of angle $A B C$.
# | Since $\angle E B D=\angle A B E+\angle C B D$, we can take a point $P$ on side $E D$ such that $\angle E B P=\angle A B E=\angle A E B$, i.e., $B P \| A E$. Then $\angle P B D=\angle E B D-\angle E B P=\angle C B D=\angle B D C$, i.e., $B P \| C D$. Therefore, $A E \| C D$, and since $A E = C D$, $C D E A$ is a parall... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Bikin Yu.A.
In trapezoid $ABCD$, sides $AD$ and $BC$ are parallel, and $AB = BC = BD$. The height $BK$ intersects diagonal $AC$ at point $M$. Find $\angle CDM$.
# | From the equality of sides $AB$ and $BC$ and the parallelism of $BC$ and $AD$, it follows that $AC$ is the bisector of angle $BAD$ (see figure).
Since triangle $ABD$ is isosceles, $DM$ is t... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
a) Six different colors are chosen; it is required to paint 6 faces of a cube, each in a special color from the chosen ones. In how many geometrically distinct ways can this be done? Geometrically distinct are two such colorings that cannot be made to coincide with each other by rotating the cube around its cente... | a) The first method. Suppose the procedure of painting a cube goes as follows: an unpainted cube is placed in a machine in a certain fixed position, and then its faces are painted in a certain order. There are as many such ways as there are permutations of 6 colors, that is, 6!. However, the cube can be placed in a fix... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
In triangle $ABC$, the bisectors $AD$ and $BE$ are drawn. It is known that $DE$ is the bisector of angle $ADC$. Find the measure of angle $A$. | Prove that $E$ is the center of the excircle of triangle $ADB$.
## Solution
Point $E$ is equidistant from the lines $AD$, $BC$, and $AB$, since it lies on the angle bisectors $DE$ and $BE$ of angles $ADC$ and $ABC$. Therefore, $E$ is the center of the excircle of triangle $ADB$. Hence, point $E$ lies on the bisector ... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Sonkin M.
Let $O$ be the center of the circumcircle of an acute-angled triangle $ABC$, and $S_{A}, S_{B}, S_{C}$ be circles centered at $O$ that are tangent to the sides $BC, CA$, and $AB$ respectively. Prove that the sum of the three angles: between the tangents to $S_{A}$ from point $A$, to $S_{B}$ from point $B$, an... | Let the circle with center $O$ touch the side $AC$ at point $B_{1}$, and the tangents to this circle drawn from point $B$ touch it at points $B_{2}$ and $B_{3}$. Then the right triangles $B O B_{2}$, $B O B_{3}$, $A O B_{1}$, and $C O_{1}$ are equal by the leg (the radius of this circle) and the hypotenuse (the radius ... | 180 | Geometry | proof | Yes | Yes | olympiads | false |
Berolov s.l.
The incircle of triangle $ABC$ touches sides $AB$ and $AC$ at points $X$ and $Y$ respectively. Point $K$ is the midpoint of the arc $AB$ of the circumcircle of triangle $ABC$ (not containing point $C$). It turns out that line $XY$ bisects segment $AK$. What can the angle $BAC$ be? | Let segments $X Y$ and $A I$ intersect at point $S$. As known, $K I = K A$ (see problem $\underline{53119}$), which means the height $K T$ of triangle $A K I$ is also its median. Since $X Y \perp A I$, then $X Y \| K T$, and because $X Y$ bisects side $A K$, $X Y$ is the midline of triangle $A K T$. Therefore, $A S: S ... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Given a triangle $A B C$. On its sides $A B$ and $B C$, points $C_{1}$ and $A_{1}$ are fixed, respectively. Find a point $P$ on the circumcircle of triangle $A B C$ such that the distance between the centers of the circumcircles of triangles $A P C_{1}$ and $C P A_{1}$ is minimized. | Let $K$ be the second intersection point of the circumcircles of triangles $A P C_{1}$ and $C P A_{1}$ (see the left figure). We will prove that $K$ lies on the line $A_{1} C_{1}$.
Let point $P$ lie on the arc $A C$. Quadrilaterals $A C_{1} K P$ and $K A_{1} C P$ are cyclic, hence $\angle C_{1} K P = 180^{\circ} - \an... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$$
\begin{aligned}
& \text { [ Convex Polygons ]} \\
& \text { [ Union, Intersection, and Difference of Sets ]} \\
& \text { [ Pigeonhole Principle (finite number of points, lines, etc.) ]} \\
\end{aligned}
$$
[ Estimation + Example $\quad$]
Find the smallest $n$ such that any convex 100-gon can be obtained as the in... | Answer: $n=50$. First, note that 50 triangles are sufficient. Indeed, let $\Delta_{k}$ be the triangle whose sides lie on the rays $A_{\mathrm{k}} A_{\mathrm{k}-1}$ and $A_{\mathrm{k}} A_{\mathrm{k}+1}$ and which contains the convex polygon $A_{1} \ldots A_{100}$. Then this polygon is the intersection of the triangles ... | 50 | Geometry | proof | Yes | Yes | olympiads | false |
$[$ Area of a triangle (through semiperimeter and radius of the inscribed or exscribed circle)
In the triangular pyramid $S A B C$, the height $S O$ passes through the point $O$ - the center of the circle inscribed in the base $A B C$ of the pyramid. It is known that $\angle S A C=60^{\circ}, \angle S C A=45^{\circ}$,... | Let the circle inscribed in triangle $ABC$ touch the sides $AB, AC$, and $BC$ at points $K, L$, and $M$ respectively. Right triangles $OKS, OLS$, and $OMC$ are equal by two legs, so $SK = SL = SM$. Since $OK \perp AB$, by the theorem of three perpendiculars, $SK \perp AB$. Similarly, $SL \perp AC$ and $SM \perp BC$. Le... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The base of the pyramid is an equilateral triangle with a side length of 6. One of the lateral edges is perpendicular to
the base plane and equals 4. Find the radius of the sphere circumscribed around the pyramid.
# | Let $A B C D$ be a given triangular pyramid, the base of which is an equilateral triangle $A B C$, and the lateral edge $C D$ is perpendicular to the plane of the base. $O$ is the center of the sphere circumscribed around the pyramid $A B C D$, and $R$ is the radius of the sphere. Since point $O$ is equidistant from po... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the cube $A B C D A 1 B 1 C 1 D 1$, where $A A 1, B B 1, C C 1$ and $D D 1$ are parallel edges, the plane $P$ passes through point $D$ and the midpoints of edges $A 1 D 1$ and $C 1 D 1$. Find the distance from the midpoint of edge $A A 1$ to the plane $P$, if the edge of the cube is 2. | Let $M, N, K$ and $L$ be the midpoints of edges $A1D1, C1D1$, $AA1$ and $CC1$ respectively (Fig.1). The line $KL$ is parallel to the line $MN$, so the line $KL$ is parallel to the plane $P$. Moreover, the line $KL$ passes through the center $O$ of the cube $ABCD A1B1C1D1$. Therefore, the distance from the midpoint $K$ ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Vassily H.b.
In each cell of an $8 \times 8$ square, one of the diagonals is drawn. Consider the union of these 64 diagonals. It consists of several connected parts (a part includes points that can be reached from each other by moving along one or more diagonals). Can the number of these parts be
a) more than 15?
b)... | b) In the figure, an example is given where the number of parts is 21. One of them is a multiply connected "window frame," in which single diagonals are placed in the cells.
## Answer
It can... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Intersecting lines, angle between them]
Let $A B C$ be an equilateral triangle, and $B C K M$ be a parallelogram. Find the angle between the lines $A B$ and $K M$.
# | Since $V S K M$ is a parallelogram, line $K M$ is parallel to line $B C$. Therefore, the angle between lines $A B$ and $K M$ is equal to the angle between lines $A B$ and $B C$, i.e., $60^{\circ}$.
## Answer
$60^{\circ}$. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Perpendicularity of a line and a plane (other).]
In the pyramid $A B C D$, the edges $A D, B D$ and $C D$ are equal to 5, the distance from point $D$ to the plane $A B C$ is 4. Find the radius of the circle circumscribed around triangle $A B C$. | Let $R$ be the radius of the circle circumscribed around triangle $ABC$. Since the lateral edges of the pyramid $ABCD$ with vertex $D$ are equal, the height $DO$ of the pyramid passes through the center $O$ of the circle circumscribed around the base $ABC$. The line $DO$ is perpendicular to the plane of the base $ABC$,... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
What is the greatest number of axes of symmetry that a spatial figure consisting of three lines, none of which are parallel or coincident, can have?
# | A spatial figure consisting of two non-parallel and non-coincident lines $l_{1}$ and $l_{2}$ has exactly three axes of symmetry. Indeed, consider the plane $\Pi$, parallel to the lines $l_{1}$ and $l_{2}$ and equidistant from them (in the case of intersecting lines, this will be the plane containing them). The axes of ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$4-$ [ Volume of a Tetrahedron and Pyramid $\quad$]
In the triangular pyramid $A B C D$, it is known that $A B=8, C D=12$, the distance between the lines $A B$ and $C D$ is 6, and the volume of the pyramid is 48. Find the angle between the lines $A B$ and $C D$. | Let's complete the pyramid $ABCD$ to a parallelepiped by drawing three pairs of parallel planes through the opposite edges. Then the volume of the pyramid is one-third of the volume of the parallelepiped. As the planes of the bases of the parallelepiped, we take the parallel planes passing through the lines $AB$ and $C... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
Five identical balls are moving in one direction along a straight line, some distance apart from each other, while five other identical balls are moving towards them. The speeds of all the balls are the same. When any two balls collide, they fly apart in opposite directions with the sa... | After the collision, the balls scatter at the same speed, so the situation will not change if we allow the balls to pass through each other upon collision, maintaining their speed. Then each ball rolling "to the right" will meet each of the balls rolling "to the left" once, meaning there will be 25 encounters.
## Answ... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Authors: Feldman G. Baranov D.V.
An angle is drawn, and only a compass is available.
a) What is the minimum number of circles that need to be drawn to definitely determine whether the given angle is acute?
b) How can one determine if the given angle is $31^{\circ}$ (any number of circles can be drawn)? | a) Let's draw a circle with its center on one side of the angle, passing through its vertex $O$. If it intersects the other side of the angle, the angle was acute.
b) Let's draw a circle with center $O$. Suppose it intersects the sides of the angle at points $A$ and $B$. Starting from point $A$, we will lay off chords... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Sorrelkova N.P.
Under one of the cells of an $8 \times 8$ board, a treasure is buried. Under each of the other cells, there is a sign indicating the minimum number of steps required to reach the treasure from that cell (one step allows moving from a cell to an adjacent cell by side). What is the minimum number of cell... | Let's dig up the corner cell $U$. Suppose there is a sign there. All cells at the specified distance from $U$ form a diagonal perpendicular to the main diagonal drawn from $U$. Let's dig up the corner cell $W$ on the same side as $U$. If there is also a sign there, then another diagonal perpendicular to the first one i... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Yashchenko I.V.
On a grid sheet of paper, several cells were colored in such a way that the resulting figure had no axes of symmetry. Vanya colored one more cell. Could the resulting figure have four axes of symmetry? | To construct an example, one needs to take a figure with four axes of symmetry and remove a cell that does not lie on any of these axes. For example, a square has four axes of symmetry (see the left image). This results in the example in the center.
There are other examples as well (for example, on the right image).
!... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
In triangle $A B C$, a point $K$ is chosen on side $A B$, and the bisector $K E$ of triangle $A K C$ and the altitude $K H$ of triangle $B K C$ are drawn. It turns out that angle $E K H$ is a right angle. Find $B C$, if $H C=5$. | Since angles $A K C$ and $B K C$ are adjacent, their bisectors are perpendicular. Therefore, the bisector of angle $B K C$ coincides with $K H$.
Thus, in triangle $B K C$, segment $K H$ is bo... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 [Inscribed, circumscribed, and exscribed circles; their radii]
Triangle $A B C$ is isosceles. The radius $O A$ of the circumscribed circle forms an angle $O A C$ with the base $A C$, which is $20^{\circ}$. Find the angle $B A C$. | The inscribed angle is equal to half of the corresponding central angle. Consider two cases of the positions of points $O$ and $B$ relative to the line $A C$.
## Solution
If points $O$ and $B$ lie on opposite sides of the line $A C$, then
$$
\cup A B C=\angle A O C=180^{\circ}-2 \cdot 20^{\circ}=140^{\circ} \text {.... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 [
From one point, two tangents are drawn to a circle. The length of each tangent is 12, and the distance between the points of tangency is 14.4. Find the radius of the circle.
# | The angle between the given chord and the radius drawn to the point of tangency is equal to half the angle between the tangents.
## Solution
Let $A$ be the given point, $B$ and $C$ be the points of tangency, and $O$ be the center of the circle. Since the line $O A$ is perpendicular to the segment $B C$ and passes thr... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { [Mutual position of two circles] } \\ \text { Triangle inequality }\end{array}\right]$
One circle is inside the other. Their radii are 28 and 12, and the shortest distance between points on these circles is 10. Find the distance between the centers. | Determine the location of the center of the larger circle relative to the smaller circle.
## Solution
It is known that the shortest distance between points of two such circles is the smallest of the two segments of the line of centers enclosed between the circles. Let $O$ and $O_{1}$ be the centers of the larger and ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Inscribed, circumscribed, and exscribed circles; their radii $]$
[ Auxiliary area. The area helps to solve the task_ ]
In an isosceles triangle, the base is 30, and the lateral side is 39. Find the radius of the inscribed circle. | The radius of the circle inscribed in a triangle is equal to the area of the triangle divided by its semi-perimeter.
## Solution
First method.
Let's find the height of the triangle dropped to the base:
$$
h^{2}=39^{2}-15^{2}=24 \cdot 54=36^{2}, h=36
$$
The desired radius is equal to the area of the triangle divide... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9
In an isosceles trapezoid, the height is 10, and the diagonals are perpendicular to each other. Find the midline of the trapezoid. | Through the vertex of the smaller base of the trapezoid, draw a line parallel to the diagonal.
## Solution
First method.
Through the vertex $C$ of the smaller base $BC$ of trapezoid $ABCD$, draw a line parallel to diagonal $BD$. Let $M$ be the point of intersection of this line with the extension of base $AD$, and $... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The base of an isosceles triangle is $4 \sqrt{2}$, and the median drawn to the lateral side is 5.
Find the lateral sides. | Complete the given triangle to a parallelogram.
## Solution
Let $x$ be the side $A B$ of the isosceles triangle $A B C (B C=4 \sqrt{2})$. On the extension of the median $B M$ beyond point $M$, we lay off the segment $D M$ equal to $B M$. Then $B A D C$ is a parallelogram. Therefore,
$$
A C^{2}+B D^{2}=2\left(A B^{2}... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
\begin{aligned} & {[\text { Law of Cosines }]} \\ & {[\text { Law of Sines }]}\end{aligned}
Find the radius of the circle if an inscribed angle with sides of lengths 1 and 2 subtends an arc of $120^{\circ}$. | Apply the Law of Cosines.
## Solution
Let $A$ be the vertex of the given angle, $AC$ and $AB$ be the sides of the angle, and points $B$ and $C$ lie on the circle, with $AC=1$ and $AB=2$. By the inscribed angle theorem, $\angle BAC = \frac{1}{2} \cdot 120^{\circ} = 60^{\circ}$. By the Law of Cosines,
$$
BC = \sqrt{AC... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Cutting into parts with special properties]
What is the minimum number of equal-area triangles that an $8 * 8$ square with a corner cell cut out can be divided into?
# | Consider the triangles adjacent to the cut-out cell and evaluate the area of one of them.
## Solution
Let the square $8 * 8$ with a corner cell cut out (its area is 63) be cut into n triangles, each of which has an area of 63/n. Let A, B, C be the vertices of the cut-out cell. Consider the triangles containing point ... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a right triangle $A B C$, the leg $A B$ is 21, and the leg $B C$ is 28. A circle with center $O$ lying on the hypotenuse $A C$ touches both legs.
Find the radius of the circle. | Let $P$ be the point of tangency of the circle with the leg $B C$. Consider similar triangles $C P O$ and $C B A$.
## Solution
Let $R$ be the radius of the circle.
First method. Let $P$ be the point of tangency of the circle with the leg $B C$. From the similarity of triangles $C P O$ and $C B A$, it follows that ${... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Triangle $A B C$ has no obtuse angles. On the side $A C$ of this triangle, a point $D$ is taken such that $A D=3 / 4 A C$. Find the angle $A$, given that the line $B D$ divides triangle $A B C$ into two similar triangles. | Prove that $B D \perp A C$.
## Solution
If angle $A D B$ is obtuse, then in triangle $C B D$ one of the angles is obtuse. Then in triangle $A B C$ one of the angles is obtuse, which contradicts the condition. Similarly, angle $A D B$ cannot be acute. Therefore, $\angle A D B=90^{\circ}$. If $\angle B C D=\angle B A D... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A circle touches two parallel lines and their secant. The segment of the secant, enclosed between the parallel lines, is divided by the point of tangency in the ratio $1: 3$. At what angle does the secant intersect each of the parallel lines? | A segment of the secant, enclosed between parallel lines, is seen from the center of the circle at a right angle.
## Solution
Let the line $l$ intersect the given parallel lines $a$ and $b$ at points $A$ and $B$, respectively, and let the circle with center $O$ touch the lines $a$, $b$, and $l$ at points $C$, $D$, an... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The altitudes of an acute-angled triangle $ABC$, drawn from vertices $B$ and $C$, are 7 and 9, respectively, and the median $AM$ is 8. Points $P$ and $Q$ are symmetric to point $M$ with respect to sides $AC$ and $AB$, respectively. Find the perimeter of quadrilateral $APMQ$.
# | Segments $M Q$ and $M P$ are equal to the given heights.
## Solution
Let $B B_{1}$ and $C C_{1}$ be the altitudes of triangle $A B C$, with $B B_{1}=7$ and $C C_{1}=9$. Let $K$ and $N$ be the midpoints of segments $M P$ and $M Q$. Then $M K$ and $M N$ are the midlines of triangles $C B B_{1}$ and $C B C_{1}$, respect... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The diagonals of a trapezoid are perpendicular to each other. One of them is equal to 6, and the second one forms an angle of $30^{\circ}$ with the base. Find the midline of the trapezoid. | Through the vertex of the trapezoid, draw a line parallel to the first diagonal.
## Solution
Let the diagonal $AC$ of trapezoid $ABCD$ be 6, and the diagonal $BD$ forms an angle of $30^{\circ}$ with the larger base $AD$. Let the line drawn through vertex $B$ parallel to $AC$ intersect the extension of base $AD$ at po... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$[\quad$ Area of a Trapezoid
Zad [Isosceles, Inscribed, and Circumscribed Trapezoids]
The diagonal of an isosceles trapezoid bisects its obtuse angle. The smaller base of the trapezoid is 3, and the perimeter is 42.
Find the area of the trapezoid.
# | Prove that the larger base of the trapezoid is equal to the lateral sides.
## Solution
Let $BC$ and $AD$ be the bases of trapezoid $ABCD$, with $BC=3$, and $CA$ being the bisector of angle $BCD$. Since $\angle CAD = \angle BCA = \angle DCA$, triangle $ACD$ is isosceles. Therefore, $AD = CD = AB = (42 - 3) : 3 = 13$. ... | 96 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kolmogorov A.N.
a) In the left figure, six points are shown, with three points lying on each of four lines. Prove that there are 24 different ways to map this set of six points onto itself such that any three points lying on one line are mapped to three points lying on one line.
 If for each of our four lines it is specified which line it is mapped to, then the image of each point will also be determined (each point of our configuration is the intersection point of certain two lines included in the configuration). The set of four lines can be mapped to itself in $4!=24$ ways.
b) Let us mark... | 108 | Combinatorics | proof | Yes | Yes | olympiads | false |
In the park, there are 10,000 trees planted in a square grid (100 rows of 100 trees each). What is the maximum number of trees that can be cut down so that the following condition is met: if you stand on any stump, you will not see any other stump? (The trees can be considered thin enough.)
# | Let's divide the trees into 2500 quadruples, as shown in the figure. In each such quadruple, no more than one tree can be cut down. On the other hand, all the trees growing in the upper left corners of the squares formed by our quadruples of trees can be cut down. Therefore, the maximum number of trees that can be cut ... | 2500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ The area of a figure is equal to the sum of the areas of the figures into which it is divided ]
A circle with its center on side $AC$ of isosceles triangle $ABC$ ($AB = BC$) touches sides $... | Let a circle of radius $r$ with center $O$ on the base $A C$ of an isosceles triangle $A B C$ touch the lateral sides $A B$ and $A C$ at points $M$ and $N$ respectively and intersect the base $A C$ at points $P$ and $Q$, such that $A P=P Q=Q C$. Then $O M$ and $O N$ are the altitudes of triangles $A O B$ and $C O B$, $... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Properties and characteristics of an isosceles triangle. ]
[ Pythagorean Theorem (direct and inverse). ]
The lateral sides of a right trapezoid are 10 and 8. The diagonal of the trapezoid, drawn from the vertex of the acute angle, bisects this angle.
Find the area of the trapezoid. | Let the acute angle at vertex $D$ of the rectangular trapezoid $A B C D$ with bases $A D$ and $B C$ be bisected by the diagonal $B D$, and $A B=8$ and
$C D=10$. Triangle $B C D$ is isosceles because $\angle C B D=\angle B D A=\angle C D B$, so $B C=C D=10$.
Drop the height $C H$ to the base $A D$. Since $A B C H$ is ... | 104 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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