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[Trigonometric ratios in a right triangle] [Trigonometric equations
Isosceles triangles $ABC (AB = BC)$ and $A_{1}B_{1}C_{1} (A_{1}B_{1} = B_{1}C_{1})$ are similar, and $AC: A_{1}C_{1} = 5: \sqrt{3}$.
Vertices $A_{1}$ and $B_{1}$ are located on sides $AC$ and $BC$, respectively, and vertex $C_{1}$ is on the extension... | Prove that $C_{1} A_{1} \perp A C$ and formulate a trigonometric equation with respect to the angle at the base of the isosceles triangle $A B C$.
## Solution
Let $A_{1} B_{1}=B_{1} C_{1}=1, \angle C=\gamma$. Then $A B=\frac{5}{\sqrt{3}}, \angle C_{1} A_{1} C=\angle C_{1} A_{1} B_{1}+\angle B_{1} A_{1} C=\gamma+\left... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{cc}{\left[\begin{array}{l}\text { Law of Cosines } \\ {[\text { Similar triangles (other) }}\end{array}\right]}\end{array}\right]$
In triangle $ABC$, $AC=2\sqrt{3}$, $AB=\sqrt{7}$, $BC=1$. Outside the triangle, a point $K$ is taken such that segment $KC$ intersects segment $AB$ at a point differen... | Prove that $\angle A K C=\angle C$.
## Solution
$A C$ is the largest side of triangle $A B C$. Since angle $K A C$ is obtuse, then $\angle K A C=\angle B$. Angle $C$ is not equal to angle $K C A$, as it is larger. Therefore,
$\angle A K C=\angle C$.
By the cosine theorem, $\cos \angle C=\frac{A C^{2}+B C^{2}-A B^{2... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left.\frac{\text { Auxiliary similar triangles }}{[\quad \text { Law of Cosines }}\right]$
In triangle $ABC$, a point $D$ is taken on side $AC$, such that $AD=3$, $\cos \angle BDC=13/20$, and $\angle B+\angle ADB=180^{\circ}$. Find the perimeter of triangle $ABC$ if $BC=2$. | Triangles $A B C$ and $B D C$ are similar.
## Solution
Since $\angle B=180^{\circ}-\angle A D B=\angle B D C$, triangles $A B C$ and $B D C$ are similar by two angles. Therefore, $B C$ : $A C=D C: B C$, or $\frac{2}{3+D C}=1 / 2 D C$. From this, we get $D C=1, A C=4$. By the cosine rule, $A C^{2}=A B^{2}+B C^{2}-$
$... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\begin{aligned} & {\left[\begin{array}{l}\text { Sum of the angles of a triangle. Theorem about the exterior angle. } \\ \text { [ Criteria and properties of an isosceles triangle. }\end{array}\right]}\end{aligned}$
In triangle $ABC$, angle $B$ is $20^{\circ}$, angle $C$ is $40^{\circ}$. The bisector $AD$ is 2. Find ... | On side $B C$, lay off segment $B M$ equal to $A B$.
## Solution
On side $B C$, we lay off segment $B M$ equal to $A B$. In the isosceles triangle $A B M$, the angles at the base $A M$ are each $80^{\circ}$, so
$\angle C A M = \angle A M D - \angle A C B = 40^{\circ} = \angle A C M$.
Moreover, $\angle A D M = \angl... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$3+$ [Properties and characteristics of an isosceles triangle.]
On each side of a square, one point was taken. It turned out that these points are the vertices of a rectangle, the sides of which are parallel to the diagonals of the square. Find the perimeter of the rectangle if the diagonal of the square is 6.
# | 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 $K N \| B D$ and
$K L \| A C$, and the segments $K L$ and $M N$ intersect the diagonal $B D$ of the square at points $P$ and $Q$, respectively. Then $K P=B P$... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11 |
| :---: | :---: | :---: |
| | Coloring r | |
| | Regular tetrahedron | |
| | Volume helps solve the task | |
| | Pigeonhole Principle (other) | |
Author: Kanel-Beglov A.,
The faces of an icosahedron are colored in five colors (including red and blue) such that two faces of the same color do not share... | We will prove that up to rotations and color permutations, there is a unique coloring that satisfies the condition. Let's call the distance between two faces the minimum number of edge transitions required to move from one face to another. Then the distance from each face to its opposite is 5. Moreover, there are three... | 90 | Geometry | proof | Yes | Yes | olympiads | false |
8,9, |
In triangle $A B C \angle B=110^{\circ}, \angle C=50^{\circ}$. On side $A B$ a point $P$ is chosen such that $\angle P C B=30^{\circ}$, and on side $A C$ a point $Q$ is chosen such that
$\angle A B Q=40^{\circ}$. Find the angle $Q P C$. | From the condition, it follows that $\angle B P C=40^{\circ}, \angle Q B C=70^{\circ}$.
First method. Draw a ray symmetric to ray $C P$ with respect to line $A C$ (see the left figure). Let $M$ be the point of intersection of this ray with ray $B Q$. Since $\angle P C M=40^{\circ}=\angle P B M$, quadrilateral $P B C M... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.v.
Inside an isosceles triangle $\$ \mathrm{ABC} \$$, a point $\$ K \$$ is marked such that $\$ C K=\mathrm{AB}=\mathrm{BC} \$$ and $\angle K A C=30^{\circ}$. Find the angle $\$ A K B \$$.
# | 
line $BC$
 consecutive natural numbers. What is the greatest possible value of the area of tr... | Author: Shestakov S.A.
Let the areas of the triangles be $n, n+1, n+2$, and $n+3$. Then $S_{A B C D}=4 n+6$. $E F$ is the midline of triangle $B C D$, so $S_{B C D}=4 S_{E C F} \geq 4 n$. Therefore, $S_{A B D}=S_{A B C D}-S_{B C D} \leq 6$.
We will show that this area can equal 6. Consider an isosceles trapezoid $A B... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Sergeev I.N.
At the base $A_{1} A_{2} \ldots A_{n}$ of the pyramid $S A_{1} A_{2} \ldots A_{n}$ lies a point $O$, such that $S A_{1}=S A_{2}=\ldots=S A_{n}$ and $\angle S A_{1} O=\angle S A_{2} O=$ $\ldots=\angle S A_{n} O$.
For what least value of $n$ does it follow from this that $S O$ is the height of the pyramid? | By the Law of Sines for triangles $S A_{k} O (k=1,2, \ldots, n)$, we have $\sin \angle S O A_{k} = \frac{S A_{k}}{S O} \cdot \sin \angle S A_{k} O$.
Since the right-hand side of this equality does not depend on the choice of $k=1,2, \ldots, n$, the value of $\sin \angle S O A_{k}$ also does not depend on this choice. ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Parallelism of lines and planes $]$ [ Systems of segments, lines, and circles ]
In space, there are $n$ segments, no three of which are parallel to the same plane. For any two segments, the line connecting their midpoints is perpendicular to both segments. What is the largest $n$ for which this is possible? | Answer: $n=2$.
Suppose that $n \geq 3$. Take three segments $a, b$, and $c$ and draw three lines connecting the midpoints of each pair. Each of the segments $a, b, c$ is perpendicular to the two lines emanating from its midpoint, and therefore it is perpendicular to the plane passing through all three drawn lines. Thu... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
For what largest $n$ can one choose $n$ points on the surface of a cube so that not all of them lie in the same face of the cube and are vertices of a regular (planar) $n$-gon. | On one face of a cube, no more than two vertices of a polygon can lie (otherwise, the entire polygon would lie in this face). Therefore, the polygon cannot have more than 12 vertices.
To obtain a 12-sided polygon, recall that there exists a section of the cube by a plane that forms a regular hexagon. This section pass... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.G.
On the lateral sides $AB$ and $BC$ of an isosceles triangle $ABC$, points $K$ and $L$ are taken respectively, such that $AK + LC = KL$. From the midpoint $M$ of segment $KL$, a line parallel to $BC$ is drawn, and this line intersects side $AC$ at point $N$. Find the measure of angle $KNL$. | Draw a line through point $K$ parallel to $BC$, intersecting the base $AC$ at point $P$. Clearly, $KP = AK$. $MN$ is the midline of the trapezoid (or parallelogram) $KLCP$, so $MN = \frac{1}{2}(LC + KP) = \frac{1}{2}KL$. Therefore, angle $KNL$ subtends the diameter of the circle with center $M$, making it a right angle... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Inscribed Quadrilaterals (Miscellaneous) ]
Quadrilateral $ABCD$ is inscribed in a circle. Point $X$ lies on its side $AD$, such that $BX \parallel CD$ and $CX \parallel BA$. Find $BC$, if $AX = 3/2$ and $DX = 6$.
# | Triangles $A B C, B X C$ and $X C D$ are similar.
## Solution
Let the angles at vertices $A, B$, and $X$ of triangle $A B X$ be denoted as $\alpha, \beta$, and $\gamma$ respectively. Since $C X \| B A$ and $B X \| C D$, then
$\angle D C X = \angle B X C = \angle A B X = \beta, \quad \angle C D X = \angle B X A = \ga... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In trapezoid $K L M N$, the bases $K N$ and $L M$ are equal to 12 and 3, respectively. From point $Q$, lying on side $M N$, a perpendicular $Q P$ is dropped to side $K L$. It is known that $P$ is the midpoint of side $K L$, $P M=4$, and that the area of quadrilateral $P L M Q$ is four times smaller than the area of qua... | If $M A$ and $N B$ are the altitudes of triangles $L M P$ and $N K P$, then the right triangles $M A P$ and $N B P$ are similar with a ratio of $1 / 4$.
## Solution
Since $P$ is the midpoint of $K L$, the altitudes of triangles $P L M$ and $P K N$, dropped from vertex $P$, are equal. Therefore, $S_{P L M}: S_{P K N}=... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Diagonals of a convex quadrilateral $A B C D$ intersect at point $E, A B=A D, C A-$ is the bisector of angle $C$,
$\angle B A D=140^{\circ}, \angle B E A=110^{\circ}$.
Find the angle $C D B$.
# | Continue sides $B C$ and $A D$ until they intersect at point $F$ and prove that triangle $C D F$ is isosceles.
## Solution
The angles at the base $B D$ of the isosceles triangle $B A D$ are each $20^{\circ}$. Therefore, $\angle C A D = \angle A E B - \angle A D E = 90^{\circ}$.
Extend sides $B C$ and $A D$ until the... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Sum of angles in a triangle. Theorem about the exterior angle. ] [ Lengths of sides, heights, medians, and angle bisectors ]
In triangle $A B C \angle A=40^{\circ}, \angle B=20^{\circ}$, and $A B-B C=4$. Find the length of the angle bisector of angle $C$. | Let's set aside segment $BD$ on side $AB$, equal to $BC$. Then triangle $BCD$ is isosceles with the angle at the vertex $20^{\circ}$, so the angles at the base are $80^{\circ}$ (see the figure). Let $CE$ be the bisector of angle $C$. Then $\angle BCE = 60^{\circ}$, so $\angle AEC = 20^{\circ} + 60^{\circ} = 80^{\circ}$... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Examples and counterexamples. Constructions ]
Six identical parallelograms of area 1 were used to cover a cube with an edge of 1. Can we assert that all parallelograms are squares? Can we assert that all of them are rectangles? | Examples of such cube nets are shown in Figure 1.
Fig. 1
## [Elementary (basic) constructions with a compass and straightedge] Problem 107725 Topics: [ Construction of triangles by various e... | 3 | 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. Pythagorean Theorem (direct and inverse).
From a point $A$ outside a circle, a tangent and a secant are drawn to the circle. The distance from point $A$ to the point of tangency is 16, and the distance from point... | Let a secant intersect a circle at points $B$ and $C$, and $M$ be the point of tangency. Then $A M=16, A C=32, B C=32$. By the tangent-secant theorem, $A M^{2}=A C \cdot A B$, or $16^{2}=32(32-B C)$. From this, $B C=24$.
Let $K$ be the projection of the center $O$ of the given circle onto the chord $B C$. The radius o... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Two tangents drawn from one point $]$
Given a circle of radius 1. From an external point $M$, two perpendicular tangents $MA$ and $MB$ are drawn to the circle. Between the points of tangency $A$ and $B$ on the smaller arc $AB$, a point $C$ is taken, and a third tangent $KL$ is drawn through it, forming a triangle $K... | Tangents drawn from a single point to a circle are equal to each other.
## Solution
Since $KA = KC$ and $BL = LC$, then
$$
\begin{gathered}
ML + LK + KM = ML + (LC + CK) + KM = \\
= (ML + LC) + (CK + KM) = (ML + LB) + (AK + KM) = \\
= MB + AM = 1 + 1 = 2.
\end{gathered}
$$
. [ Area of a triangle (using the semiperimeter and the radius of the inscribed or exscribed circle).
A circle is inscribed in a right triangle. One of the legs is divided by the point of tangency into segments of 6 and 10, measured from the vertex... | Given that the radius of the inscribed circle is 6, and one of the legs is 16. Let the second leg be $b$. Then the semiperimeter $p=b+16-6=b+10$. By calculating the area $S$ of the triangle in two ways, we get $8b = S = 6(b+10)$, from which $b=30$.
## Answer
240. | 240 | 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 [Pythagorean Theorem (direct and inverse).
A circle is tangent to side $B C$ of triangle $A B C$ at its midpoint $M$, passes through point $A$, and intersects segments $A B$ and $A C$ at points $D$ and $E$ respect... | Apply the tangent and secant theorem.
## Solution
By the tangent and secant theorem, \( B M^{2}=A B \cdot B D, C M^{2}=A C \cdot C E \), or \( 36=A B\left(A B-7 / 2\right), 36=\frac{9 A C}{\sqrt{5}} \), from which \( A B=8, A C=4 \sqrt{5} \).
Since \( A B^{2}+A C^{2}=64+80=144=B C^{2} \), triangle \( A B C \) is a r... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two circles with radii 5 and 3 touch each other internally. A chord of the larger circle touches the smaller circle and is divided by the point of tangency in the ratio $3: 1$. Find the length of this chord. | Let $O_{1}$ and $O_{2}$ be the centers of circles with radii 5 and 3, respectively. $AB$ is a given chord, $C$ is the point of tangency with the smaller circle ($AC: BC=1: 3$), and $P$ is the projection of point $O_{1}$ onto the radius $O_{2}C$ of the smaller circle.
If $AC = x$, then $BC = 3x$. Drop a perpendicular $... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two circles with radii 5 and 4 touch each other externally. A line tangent to the smaller circle at point $A$ intersects the larger circle at points $B$ and $C$, and
$A B = B C$. Find $A C$. | Let a circle of radius 4 with center $O_{1}$ and a circle of radius 5 with center $O_{2}$ touch each other externally at point $D$ (see the left figure). Then
$O_{1} O_{2}=O_{1} D+O_{2} D=9$
Drop perpendiculars $O_{2} M$ to the chord $B C$ and $O_{1} F$ to the line $O_{2} M$. Then $A O_{1} F M$ is a rectangle, so $M ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The height of a right triangle, dropped to the hypotenuse, is 1, and one of the acute angles is $15^{\circ}$.
Find the hypotenuse. | Draw the median from the vertex of the right angle.
## Solution
Let $C H$ be the height of the right triangle $A B C$, drawn from the vertex of the right angle $C, \angle A=15^{\circ}$. Draw the median $C M$. The angle $C M H$ as the external angle of the isosceles triangle $A M C$ is $30^{\circ}$. From the right tri... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{l}\text { Rhombuses. Properties and characteristics }\end{array}\right]$
The diagonals of a rhombus are 24 and 70. Find the side of the rhombus. | Use the properties of the diagonals of a rhombus.
## Solution
The diagonals of a rhombus are perpendicular to each other and bisect each other at their point of intersection. The side of the rhombus is the hypotenuse of a right triangle, with the legs being half the lengths of the diagonals. By the Pythagorean theore... | 37 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Diagonal $B D$ of parallelogram $A B C D$ forms angles of $45^{\circ}$ with side $B C$ and the height drawn from vertex $D$ to side $A B$.
Find angle $A C D$.
# | From the condition, it follows that the specified height is parallel to side $B C$, therefore, this height coincides with side $D A$, which means $A B C D$ is a rectangle (see figure). Since $\angle A B D = \angle C B D = 45^{\circ}$, $B D$ is the bisector of angle $B$ of this rectangle, hence $A B C D$ is a square. Co... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the sides of angle $A B C$, points $M$ and $K$ are marked such that angles $B M C$ and $B K A$ are equal, $B M=B K, A B=15, B K=8, C M$ $=9$.
Find the perimeter of triangle $С O K$, where $O$ is the intersection point of lines $А K$ and $С M$.
# | Triangles $A B K$ and $C B M$ are equal (by side and adjacent angles, see figure). Therefore, $\angle B C M = \angle B A K$ and $C B = A B = 15$, so $C K = A M = 7$.
Considering also that $\angle C K O = \angle A M O$ (they supplement equal angles to straight angles), we get that triangles СОК and АОМ are equal. There... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Central Angle. Arc Length and Circumference] [ $\quad$ Proof by Contradiction
Consider all moments in time when the hour and minute hands of a clock lie on the same straight line, forming a straight angle.
Will there be two such lines that are mutually perpendicular?
# | Suppose such lines do exist. Then at these moments, the hour hands are perpendicular. This means that a whole number of hours has passed (specifically $6n+3$, where $n$ is an integer). Then the positions of the minute hands at these moments are the same, and they cannot form an angle of $90^{\circ}$, which contradicts ... | 90 | Geometry | proof | Yes | Yes | olympiads | false |
5,6,7 |
Giuseppe has a sheet of plywood measuring $22 \times 15$. Giuseppe wants to cut as many rectangular blanks of size $3 \times 5$ as possible from it. How can he do this? | Note: more than 22 blanks cannot be obtained. Why?
## Solution
First, note that Giuseppe cannot obtain more blanks than $(22 \times 15) /(3 \times 5)=22$ pieces. Now let's proceed with the cutting.
Author: Folkpor
In a hexagon, five angles are $90^{\circ}$, and one angle is $-270^{\circ}... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Folklore
What is the smallest value that the perimeter of a scalene triangle with integer side lengths can take?
# | Let $a, b$ and $c$ be the integer lengths of the sides of a triangle and $a>b>c$. According to the triangle inequality, $c>a-b$. Since $a$ and $b$ are different natural numbers, $c \geq 2$, thus $b \geq 3$ and $a \geq 4$. Therefore, $a+b+c \geq 9$. Equality is achieved for a triangle with sides 2, 3, and 4.
## Otвет
... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
One side of the rectangle was increased by 3 times, and the other side was reduced by 2 times, resulting in a square.
What is the side of the square if the area of the rectangle is 54 m²
# | Let's reduce one side of the given rectangle by half. Then the area of the resulting rectangle will be $27 \mathrm{~m}^{2}$. Next, we will triple the other side. The area of the resulting figure will become $27 \cdot 3=81$ ( $\mathrm{m}^{2}$ ). Since, by the condition, a square is formed, its side is 9 m.
## Answer
9... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Through vertices $A$ and $B$ of an acute-angled triangle $ABC$, a circle is drawn, intersecting side $AC$ at point $X$ and side $BC$ at point $Y$. It turns out that this circle passes through the center of the circumcircle of triangle $XCY$. Segments $AY$ and $BX$ intersect at point $P$. It is known that $\angle ACB = ... | Let $O$ be the center of the circumcircle $s$ of triangle $X C Y$, and $s 1$ be the circle passing through points $A, B$, $X$ and $Y$. Denote $\angle A P X = a$. Then,
$$
\angle X C Y = \angle A C B = 2a, \angle X O Y = 2 \angle X C Y = 4a
$$
(the central angle $X O Y$ of circle $s$ is twice the inscribed angle $X C ... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
What is the minimum number of three-cell corners that can be placed in an $8 \times 8$ square so that no more such corners can be placed in this square? | In each $2 \times 2$ square, at least two cells must be covered by corners (otherwise, another corner can fit into such a square).
A $8 \times 8$ square can be divided into 16 squares of size $2 \times 2$ each, meaning that at least 32 cells must be covered by corners, which requires no fewer than 11 corners.
 = a, \quad C N = 2 B C \cos 20^{\circ} = 2 b \cos 20^{\circ... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4 [ Angles subtended by equal arcs and equal chords]
On side $A B$ of rectangle $A B C D$, a point $M$ is chosen. Through this point, a perpendicular to line $C M$ is drawn, which intersects side $A D$ at point $E$. Point $P$ is the foot of the perpendicular dropped from point $M$ to line $C E$. Find the angle $A P B$... | Let $\angle B C M=\alpha$. Then
$$
\angle B M C=90^{\circ}-\alpha, \angle A M E=90^{\circ}-\left(90^{\circ}-\alpha\right)=\alpha, \angle A E M=90^{\circ}-\alpha .
$$
Points $B$ and $P$ view the segment $C M$ at a right angle, so these points lie on the circle with diameter $C M$. The inscribed angles $B P M$ and $B C... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[The area of a triangle does not exceed half the product of two sides] Complex
What is the maximum area that a quadrilateral with side lengths of 1, 4, 7, and 8 can have? | Consider a quadrilateral, two adjacent sides of which are equal to 1 and 8. Then the other two adjacent sides are equal to 4 and 7. If the sides equal to 1 and 8 are opposite, we draw a diagonal and reflect one of the resulting triangles relative to its perpendicular bisector. Then the sides equal to 1 and 8 will be ad... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 [ Investigation of a quadratic trinomial $\quad$ ]
Authors: Kosukhin O.N., Bednov B.B.
Three athletes started simultaneously from point $A$ and ran in a straight line to point $B$ each at their own constant speed. Upon reaching point $B$, each of them instantly turned around and ran back to the finish line at poi... | Let's assign numbers to the athletes in descending order of their starting speeds. We will draw graphs of their movements, plotting time on the x-axis and the distance to point $A$ on the y-axis. Let $O$ be the origin, $S$ the point on the y-axis corresponding to point $B \quad (O S=60 \mathrm{~m}), K, L, M$ - points o... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8,9 Two circles touch each other externally at point $A$, and a third circle at points $B$ and $C$. The extension of chord $A B$ of the first circle intersects the second circle at point $D$, the extension of chord $A C$ intersects the first circle at point $E$, and the extensions of chords $B E$ and $C D$ intersect t... | Let $S 1, S 2$ and $S 3$ be the first, second, and third circles, respectively. Draw the common tangents $l_{a}, l_{b}, l_{c}$ through points $A, B$ and $C$ to the circles $S 1$ and $S 2, S 1$ and $S 3, S 2$ and $S 3$ respectively. Then the tangents $l_{a}$ and $l_{b}$ form equal angles with the chord $A B$. Denote the... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Touching Circles $\quad$ [
Two circles touch each other externally at point $A$, and a third circle at points $B$ and $C$. The extension of chord $A B$ of the first circle intersects the second circle at point $D$, the extension of chord $A C$ intersects the first circle at point $E$, and the extensions of chords $B... | Let $S 1, S 2$ and $S 3$ be the first, second, and third circles, respectively. Draw the common tangents $l_{a}, l_{b}, l_{c}$ through points $A, B$ and $C$ to the circles $S 1$ and $S 2, S 1$ and $S 3, S 2$ and $S 3$ respectively. Then the tangents $l_{a}$ and $l_{b}$ form equal angles with the chord $A B$. Denote the... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4 [ Sphere inscribed in a dihedral angle ]
Point $O$ is located in the section $A A^{\prime} C^{\prime} C$ of a rectangular parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with dimensions $2 \times 6 \times 9$ such that $\angle O A B + \angle O A D + \angle O A A^{\prime} = 180^{\circ}$. A sphere ... | Let $\angle O A B=\alpha, \angle O A D=\beta, \angle O A A^{\prime}=\gamma$. According to the problem, $\alpha+\beta+\gamma=180^{\circ}$.
We mark a segment $A E=1$ on the ray $A O$ (Fig.1). Then the projections of point $E$ onto the lines $A B, A D$, and $A A^{\prime}$ are $\cos \alpha, \cos \beta$, and $\cos \gamma$ ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Rectangular parallelepipeds ] [ Sphere inscribed in a dihedral angle ]
Point $O$ is located in the section $A C C^{\prime} A^{\prime}$ of a rectangular parallelepiped $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with dimensions $2 \times 3 \times 6$ such that $\angle O C B + \angle O C D + \angle O C C^{\pr... | Let $\angle O C B=\alpha, \angle O C D=\beta, \angle O C C^{\prime}=\gamma$. According to the problem, $\alpha+\beta+\gamma=180^{\circ}$.
We mark a segment $C E=1$ on the ray $C O$. Then the projections of point $E$ on the lines $C B, C D$, and $C C^{\prime}$ are $\cos \alpha, \cos \beta$, and $\cos \gamma$ respective... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Mumchov D:
On a circle of length 2013, 2013 points are marked, dividing it into equal arcs. A chip is placed at each marked point. We define the distance between two points as the length of the shorter arc between them. For what largest $n$ can the chips be rearranged so that there is again one chip at each marked poi... | Evaluation. Let's number the points and the chips on them in a clockwise direction with consecutive non-negative integers from 0 to 2012. Consider an arbitrary permutation and the chips with numbers 0, 671, and 1342, initially located at the vertices of an equilateral triangle. The pairwise distances between them are 6... | 670 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shveiov D.V.
The circle inscribed in the right triangle $ABC\left(\angle ABC=90^{\circ}\right)$ touches the sides $AB, BC, AC$ at points $C_{1}, A_{1}, B_{1}$ respectively. The excircle opposite to $A$ touches the side $BC$ at point $A_{2}$. $A_{0}$ is the center of the circumcircle of triangle $A_{1} A_{2} B_{1}$; th... | Since points $A_{1}$ and $A_{2}$ are symmetric with respect to the midpoint of segment $B C$ (see problem $\underline{55404}$), then $A_{0} B=A_{0} C$. On the other hand, $A_{0}$ lies on the perpendicular bisector of segment $A_{1} B_{1}$, which coincides with the bisector of angle $C$. Therefore, $\angle C B A_{0}=\an... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kalinin A.
Anya and Borya, whose speeds are constant but not necessarily the same, simultaneously set out from villages A and B towards each other. If Anya had set out 30 minutes earlier, they would have met 2 km closer to village B. If Borya had set out 30 minutes earlier, the meeting would have taken place closer to... | Let Kolya (half an hour after Borya) and Tolya (half an hour before Borya) also leave from B at the same speed, and from A - Tanya (half an hour before Anya).
First method. Anya meets Tolya, Borya, and Kolya sequentially at equal time intervals. Therefore, the distances between the adjacent meeting points are also equ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
|
At a familiar factory, they cut out metal disks with a diameter of 1 m. It is known that a disk with a diameter of exactly 1 m weighs exactly 100 kg. During manufacturing, there is a measurement error, and therefore the standard deviation of the radius is 10 mm. Engineer Sidorov believes that a stack of 100 disks wi... | Given $\mathrm{E} R=0.5 \mathrm{~m}, \mathrm{D} R=10^{-4} \mathrm{M}^{2}$. Let's find the expected value of the area of one disk:
$\mathrm{E} S=\mathrm{E}\left(\pi R^{2}\right)=\pi \mathrm{E} R^{2}=\pi\left(\mathrm{D} R+\mathrm{E}^{2} R\right)=\pi\left(10^{-4}+0.25\right)=0.2501 \pi$.
Therefore, the expected value of... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The altitudes of an unequal-sided acute triangle $A B C$ intersect at point $H$. $O$ is the center of the circumscribed circle of triangle $B H C$. The center $I$ of the inscribed circle of triangle $A B C$ lies on the segment $O A$.
Find the angle $A$.
# | From the condition, it follows that point $O$ lies at the intersection of the angle bisector of $\angle A$ and the perpendicular bisector of side $BC$. Since these lines intersect on the circumcircle of triangle $ABC$, $O$ lies on this circle and is the midpoint of arc $BC$. Moreover, $\angle BHC = 180^{\circ} - \angle... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11 | |
A right circular cone with base radius $R$ and height $H=3 R \sqrt{7}$ is laid on its side on a plane and rolled so that its vertex remains stationary. How many revolutions will the base make before the cone returns to its original position? | The base circle rolls along the circumference of a circle of radius $L$, where $L$ is the slant height of the cone. Therefore, the number of revolutions is $\frac{2 \pi L}{2 \pi R}=\frac{\sqrt{H^{2}+R^{2}}}{R}=8$.
## Answer
8 revolutions. | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Properties of bisectors, concurrency $]$
On the sides $B C$ and $C D$ of the square $A B C D$, points $M$ and $K$ are marked such that $\angle B A M=\angle C K M=30^{\circ}$. Find $\angle A K D$.
# | From the condition, it follows that $\angle B M A=\angle C M K=60^{\circ}$, and then $\angle A M K=60^{\circ}$ (see the left figure). We can reason in different ways.
First method. Let $A H$ be the perpendicular dropped from vertex $A$ to $M K$. Then the right triangles $A M B$ and $A M H$ are equal by hypotenuse and ... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The altitudes $A D$ and $B E$ of an acute-angled triangle $A B C$ intersect at point $H$. The circumcircle of triangle $A B H$ intersects sides $A C$ and $B C$ at points $F$ and $G$ respectively. Find $F G$, if $D E=5$ cm. | Let $\angle F A H = \angle H B F = \alpha$. Right triangles $A D C$ and $E C B$ share the common angle $C$, so $\angle E B C = \alpha$.
Thus, $B E$ is both the altitude and the angle bisector of triangle $F B C$, which means this triangle is isosceles and $B E$ is its median, i.e.,
$F E = E C$. Similarly, $C D = D G$... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Equilateral (Equiangular) Triangle]
[Properties and Characteristics of a Parallelogram]
[Auxiliary Congruent Triangles]
[Rotation Helps Solve the Problem.
On the sides $AB, BC$, and $CA$ of an equilateral triangle $ABC$, points $D, E$, and $F$ are chosen respectively such that $DE \parallel AC$ and $DF \parallel B... | Let $A E$ and $B F$ intersect at point $N$. Since $C E D F$ is a parallelogram and triangle $B D E$ is equilateral, then $B E = C F$ (see the left figure). We can reason in different ways.
F... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$ABCD$ - a convex quadrilateral. It is known that $\angle CAD = \angle DBA = 40^{\circ}, \angle CAB = 60^{\circ}, \angle CBD = 20^{\circ}$. Find the angle $\angle CDB$. | Since $\angle C A B=60^{\circ}, \angle A B C=\angle A B D+\angle D B C=60^{\circ}$, the triangle $A B C$ is equilateral (see the left figure). We can reason in different ways.
The first method. In triangle $A B D$, $\angle A B D=40^{\circ}, \angle B A D=\angle B A C+\angle C A D=100^{\circ}$, so $\angle B D A=180^{\ci... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
Given a square $ABCD$. On the extension of the diagonal $AC$ beyond point $C$, a point $K$ is marked such that $BK=AC$. Find the angle $BKC$.
# | Since the picture is symmetrical with respect to the line $A C$, then $D K=B K=A C$. And since the diagonals in a square are equal, $A C=B D$. Thus, triangle $B K D$ is equilateral, and angle $B K D$ is $60^{\circ}$. Again, due to symmetry with respect to the line $A C$, $K$ is the bisector of this angle.
. Since the angles at the common vertex $C$ of triangles $C D E$ and $C B E$ are equal, and $\angle C D E > \angle C B E$ (one is obtuse, the other is acute), then $\angle C E D < \angle C E B$. Therefore, if from the r... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Construct a circle of a given radius passing through a given point and tangent to a given line.
# | The center of the desired circle is at a given distance from a given line and from a given point.
## Solution
The center of the desired circle is at the intersection of a line parallel to the given one and at a distance equal to the given radius, and a circle with the given radius, whose center is the given point.
!... | 91 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}{[\text { The inscribed angle is half the central angle }} \\ {[\quad \text { Properties and characteristics of tangents }}\end{array}\right]$
$A B$ and $A C$ are two chords forming angle $B A C$, which is equal to $70^{\circ}$. Tangents are drawn through points $B$ and $C$ until they intersect ... | $$
\cup B C=2 \angle B A C=140^{\circ}, \angle B M C=180^{\circ}-2 \angle B A C=180^{\circ}-140^{\circ}=40^{\circ}
$$
?C
The angle between the tangents drawn from a point to a circle complements the corresponding central angle to $180^{\circ}$.
$. | Prove that $ABCD$ is a cyclic quadrilateral with mutually perpendicular diagonals.
## Solution
Let $C_{1}$ be the point symmetric to vertex $C$ with respect to the perpendicular bisector of diagonal $BD$. Then
$$
\begin{gathered}
S_{\triangle \mathrm{ABCD}}=S_{\Delta \mathrm{ABC}_{1} \mathrm{D}}=S_{\Delta \mathrm{AB... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Given a parallelogram $ABCD$, where $AB=3$, $AD=\sqrt{3}+1$, and $\angle BAD=60^{\circ}$. On side $AB$, a point $K$ is taken such that $AK:KB=2:1$. A line parallel to $AD$ is drawn through point $K$. On this line, inside the parallelogram, a point $L$ is chosen, and on side $AD$, a point $M$ is chosen such that $AM=KL$... | Prove that the lines $K D, B M$, and $C L$ intersect at one point. For this, use the following statement. Through a point $X$ inside a parallelogram, lines parallel to its sides are drawn. Then the two parallelograms formed with the only common vertex $X$ are equal in area if and only if the point $X$ lies on the diago... | 105 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Given a parallelogram $ABCD$, where $AB=5$, $AD=2\sqrt{3}+2$, and $\angle BAD=30^{\circ}$. On side $AB$, a point $K$ is taken such that $AK:KB=4:1$. A line parallel to $AD$ is drawn through point $K$. On this line, inside the parallelogram, a point $L$ is chosen, and on side $AD$, a point $M$ is chosen such that $AM=KL... | Prove that the lines $K D, B M$ and $C L$ intersect at one point. For this, use the following statement. Through a point $X$ lying inside a parallelogram, lines parallel to its sides are drawn. Then the two parallelograms formed in this way with the only common vertex $X$ are equal in area if and only if the point $X$ ... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$4+$Given a parallelogram $K L M N$, where $K L=8, K N=3 \sqrt{2}+\sqrt{6}$ and $\angle K N L=45^{\circ}$. On side $K L$, a point $A$ is taken such that $K A: A L=3: 1$. Through point $A$, a line parallel to $L M$ is drawn, on which a point $B$ is chosen inside the parallelogram, and a point $C$ is chosen on side $K N$... | Prove that the lines $A N, B M$, and $C L$ intersect at one point. For this, use the following statement. Through a point $X$ lying inside a parallelogram, lines parallel to its sides are drawn. Then the two parallelograms formed in this way, with the only common vertex $X$, are equal in area if and only if the point $... | 105 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find the angles of a triangle if it is known that all squares inscribed in it are equal (each of the squares is inscribed such that two of its vertices lie on one side of the triangle, and the other vertices on the other two sides of the triangle).
# | Let the sides of the triangle be \(a, b\), and \(c\), and the heights dropped to them be \(h_{a}, h_{b}\), and \(h_{c}\). According to problem 53756, the sides of the inscribed squares are \(\frac{a h_{a}}{a + h_{a}}, \frac{b h_{b}}{b + h_{b}}, \frac{c h_{c}}{c + h_{c}}\). By the condition, these numbers are equal.
Th... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Point $M$ is the midpoint of chord $A B$. Chord $C D$ intersects $A B$ at point $M$. A semicircle is constructed on segment $C D$ as its diameter. Point $E$ lies on this semicircle, and $M E$ is perpendicular to $C D$. Find the angle $A E B$. | $M E$ - the height of the right triangle $C E D$, drawn from the vertex of the right angle, so $E M^{2}=C M \cdot M D$ $=A M \cdot M B=A M^{2}$, that is, the median $E M$ of triangle $A E B$ is equal to half of the side $A B$. Therefore, $\angle A E B=$ $90^{\circ}$.
## Answer
$90^{\circ}$. | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $M$ and $N$ be the midpoints of sides $C D$ and $D E$ of a regular hexagon $A B C D E F$. Find the angle between the lines $A M$ and $B N$.
# | When rotating by $60^{\circ}$ around the center $X$ of the regular hexagon $A B C D E F$, which maps vertex $A$ to vertex $B$, vertex $C$ transitions to vertex $D$, and vertex $D$ transitions to vertex $E$. Therefore, the midpoint $M$ of segment $C D$ transitions to the midpoint $N$ of segment $E D$, and the line $A M$... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
On the side $B C$ of an equilateral triangle $A B C$, points $K$ and $L$ are marked such that $B K=K L=L C$, and on the side $A C$ a point $M$ is marked such that $A M=1 / 3 A C$. Find the sum of the angles $A K M$ and $A L M$. | Note that triangle $M K C$ is also equilateral, since $C M=\frac{2}{3} C A=2 / 3 C B=C K$ and $\angle M C K=60^{\circ}$ (see the left figure). Therefore, $M K \| A B$, so $\angle A K M=\angle K A B$.
The median $M L$ of triangle $M K C$ is its altitude and, therefore, parallel to the altitude $A D$ of triangle $A B C$... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Sum of angles in a triangle. Theorem about the exterior angle.]
$[\quad$ Auxiliary equal triangles Angles between bisectors
The bisectors of triangle $ABC$ intersect at point $I, \angle AB... | Note that $\angle A B Q=\angle C B P=\angle A B I=\angle C B I=60^{\circ}$.
Let $\angle B A C=2 x$, and $\angle B C A=2 y$, then (from triangle $A B C$) $2 x+2 y+120^{\circ}=180^{\circ}$, w... | 90 | Geometry | proof | Yes | Yes | olympiads | false |
[ Trigonometric ratios in a right triangle [ Theorem of the length of a tangent and a secant; the product of the entire secant and its external part [ Two tangents drawn from the same point
In parallelogram $K L M N$, side $K L$ is equal to 8. A circle, tangent to sides $N K$ and $N M$, passes through point $L$ and in... | Apply the tangent-secant theorem.
## Solution
Let $A B$ be the hypotenuse of the right triangle $A B C$. Then $A B=2 R$. If $\angle C A B=\boldsymbol{\alpha}$, then
$$
B C=A B \sin \alpha=2 R \sin \alpha .
$$
If $C D$ is the height of the triangle $A B C$, then
$$
C D=B C \cos \angle D C B=B C \cos \alpha=2 R \sin... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Malkin M.i.
On the board, 101 numbers are written: $1^{2}, 2^{2}, \ldots, 101^{2}$. In one operation, it is allowed to erase any two numbers and write down the absolute value of their difference instead.
What is the smallest number that can result from 100 operations? | From four consecutive squares (in three operations), the number 4 can be obtained: $(n+3)^{2}-(n+2)^{2}-((n+$ $\left.1)^{2}-n^{2}\right)=(2 n+5)-(2 n+1)=4$.
We can get 24 such fours from the numbers $6^{2}, 7^{2}, \ldots, 101^{2}$. 20 fours can be turned into zeros by pairwise subtraction. From the numbers $4,9,16,25$... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Examples and counterexamples. Constructions ] [ Symmetry helps solve the task ]
What is the minimum number of cells that need to be marked on a chessboard so that each cell of the board (marked or unmarked) shares a side with at least one marked cell? | Let's highlight 20 white cells (on the diagram, they are marked with the sign "x").
Any black cell is adjacent to no more than two highlighted white cells. Therefore, to ensure that these whit... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.i.
On the plane, the curves $y=\cos x$ and $x=100 \cos (100 y)$ were drawn, and all points of their intersection with positive coordinates were marked. Let $a$ be the sum of the abscissas, and $b$ be the sum of the ordinates of these points. Find $a / b$. | After replacing $x=100 u$, the equations will take the form: $y=\cos (100 u), u=\cos (100 y)$. The ordinates of the corresponding intersection points of the new curves will be the same, while the abscissas will be reduced by a factor of 100. Let $c$ be the sum of the abscissas of the new intersection points (with posit... | 100 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folklore }}$
In triangle $ABC$, the median drawn from vertex $A$ to side $BC$ is four times smaller than side $AB$ and forms an angle of $60^{\circ}$ with it. Find angle $A$. | Extending median $A M$ to its length, we obtain parallelogram $A B D C$ (see figure). In triangle $A B D$, we draw median $D E$, then $A E=1 / 2 A B=A D$. Thus, triangle $A D E$ is isosceles with an angle of $60^{\circ}$, meaning it is equilateral. Therefore, the median $D E$ of triangle $A B D$ is equal to half the si... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Convex Polygons ]
Among all such numbers $n$, that any convex 100-gon can be represented as the intersection (i.e., common part) of $n$ triangles, find the smallest. | 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 $\Delta_{2}, \De... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
Dominoes $1 \times 2$ are placed without overlapping on a chessboard $8 \times 8$. In this case, dominoes can extend beyond the board's edges, but the center of each domino must lie strictly inside the board (not on the edge). Place
a) at least 40 dominoes;
b) at least 41 dominoes;
c) more than 41 do... | a) First method. Take a square with vertices $(4, \pm 4)$ and rotate it slightly around its center so that its vertical sides do not intersect the lines with half-integer abscissas. The original square intersected eight such lines. After the rotation, the square cuts off segments longer than 8 on them. Therefore, insid... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bakayev E.V.
Inside an isosceles triangle $ABC$, a point $K$ is marked such that $AB = BC = CK$ and $\angle KAC = 30^{\circ}$.
Find the angle $AKB$. | The first solution. Construct an equilateral triangle $ACL$ on $AC$ such that points $L$ and $B$ lie on the same side of $AC$ (see figure).
Draw the altitude $BM$ in triangle $ABC$, which is ... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[
The ratio in which the bisector divides the side. [The ratio of the areas of triangles with a common base or common height]
In triangle $ABC$, the median $AD$ and the bisector $BE$ are perpendicular and intersect at point $F$. It is known that $S_{DEF}=5$. Find $S_{ABC}$.
# | Triangle $A B D$ is isosceles.
## Solution
Triangle $A B D$ is isosceles because its bisector $B F$ is also an altitude. Therefore, $A F=F D \Rightarrow S_{A F E}$
$=S_{D F E}=5$.
Moreover, $B C=2 B D=2 A B$. By the property of the bisector of a triangle, $E C: A E=2: 1$. Therefore, $S_{A B C}=2 S_{A D C}=6 S_{A D ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
From an external point $A$, a tangent $A B$ and a secant $A C D$ are drawn to a circle. Find the area of triangle $C B D$, if $A C: A B=2: 3$ and the area of triangle $A B C$ is 20. | The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.
## Solution
By the theorem on the angle between a tangent and a chord, $\angle A B C=\angle A D B$. Therefore, triangle $A D B$ is similar to triangle $A B C$ (by two angles). The similarity coefficient is the ratio of t... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[
The Perimeter of a Triangle
The bisector drawn from vertex $N$ of triangle $M N P$ divides side $M P$ into segments of 28 and 12.
Find the perimeter of triangle $M N P$, given that $M N - N P = 18$.
# | Apply the property of the bisector of a triangle.
## Solution
By the property of the bisector of a triangle, the sides are proportional to 28 and 12. Since $28-12=16$, the coefficient of proportionality is $9 / 8$. Therefore, the perimeter is $(9 / 8+1)(28+12)=85$.
## Answer
85. | 85 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Helper area. The area helps solve the task] Pythagorean Theorem (direct and inverse) ]
Find the area of an isosceles triangle if the height dropped to the base is 10, and the height dropped to the side is 12. | Compose an equation with respect to half the base of the triangle.
## Solution
Let $B K$ and $C M$ be the altitudes of the isosceles triangle $A B C (A B=B C, B K=10, C M=12)$. Denote $A K=$ КС $=x$. Then
$$
\begin{gathered}
A B^{2}=B K^{2}+A K^{2}=100+x^{2} \\
S_{\Delta \mathrm{ABC}}=\frac{1}{2} A C \cdot B K=\frac... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 | |
In parallelogram $A B C D$, the diagonals $A C=15, B D=9$ are known. The radius of the circle circumscribed around triangle $A D C$ is 10. Find the radius of the circle circumscribed around triangle $A B D$. | Apply the formula $a=2 R \sin \boldsymbol{\alpha}$.
## Solution
Let $R=10$ be the radius of the circumcircle of triangle $A D C$. Then
$$
\sin \angle A D C=\frac{A C}{2 R}=\frac{15}{20}=\frac{3}{4}
$$
Since $\angle B A D=180^{\circ}-\angle A D C$, then
$$
\sin \angle B A D=\sin \angle A D C=\frac{3}{4}
$$
If $R_{... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Given a triangle with sides $A B=2, B C=3, A C=4$. A circle is inscribed in it, and the point $M$ where the circle touches side $B C$ is connected to point $A$. Circles are inscribed in triangles $A M B$ and $A M C$. Find the distance between the points where these circles touch the line $A M$. | Since $K$ is the point of tangency of the incircle inscribed in triangle $AMB$ with side $AM$, $AK = (AB + AM - BM) / 2$.
Similarly, expressing $AL$ and subtracting the obtained expressions from each other, we get
$KL = (AB - AC - BM + MC) / 2$.
| 3 Topics: | $\left[\begin{array}{lc}\text { [ Cube } & \text { [ } \\... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## [ Rectangles and Squares. Properties and Characteristics ] [ Inscribed Quadrilaterals (other). ]
Point $F$ is the midpoint of side $B C$ of square $A B C D$. A perpendicular $A E$ is drawn to segment $D F$. Find the angle $C E F$.
# | Let the line $A E$ intersect the side $C D$ of the square at point $M$ (see the figure). Then triangles $A D M$ and $D C F$ are equal (by the leg and acute angle). Therefore, point $M$ is the midpoint of side $C D$. Hence, triangle $C F M$ is a right isosceles triangle, so $\angle C M F=45^{\circ}$.
$, which means $\angle A = 6... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
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