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[The product of the lengths of the segments of chords and the lengths of the segments of secants] [Properties and characteristics of an isosceles triangle.] [Thales' theorem and the theorem of proportional segments]
On the extension of side $A D$ of rhombus $A B C D$ beyond point $D$, point $K$ is taken. Lines $A C$ a... | Let $O$ be the center of the circle. Use the similarity of triangles $K Q O$ and $K B A$.
## Solution
Let $O$ be the center of the circle, $F$ be the second intersection point of the circle with the line $A$. Since triangle $A O Q$ is isosceles $(O A=O Q=6)$, then
$\angle A Q O=\angle O A Q=\angle B A Q$. Therefore,... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ The inscribed angle is half the central angle ]
In an acute-angled triangle $A B C$, altitudes $C H$ and $A H_{1}$ are drawn. It is known that $A C=2$, and the area of the circle circumscribed around triangle $H B H_{1}$ is $\pi / 3$. Find the angle between the altitude $C H$ and the side $B C$.
# | Let $P$ be the point of intersection of the altitudes of triangle $ABC$. Points $H$ and $H_{1}$ lie on the circle with diameter $BP$.
Let the radius of this circle be $R$. Then
$\pi R^{2}=\pi / 3$. Hence, $R=\frac{1}{\sqrt{3}}$
First method. From the solution of problem $\underline{54814}$, it follows that $2 R=AC \... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In triangle $A B C$, the bisector $C D$ of the right angle $A C B$ is drawn; $D M$ and $D N$ are the altitudes of triangles $A D C$ and $B D C$, respectively.
Find $A C$, given that $A M=4, B N=9$.
# | Note that $M C=M D=D N$. From the similarity of triangles $A M D$ and $D N B$, it follows that $A M: M D=D N: N B$, that is, $M C^{2}=A M \cdot N B=36$. Therefore,
$A C=A M+M C=4+6=10$.
## Answer
10. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## [ Quadrilaterals (extreme properties).]
For what value of the height does a rectangular trapezoid with an acute angle of $30^{\circ}$ and a perimeter of 6 have the maximum area
# | Express the area of the given trapezoid through its height and apply the Cauchy inequality.
## Solution
Let $h$ be the height of the trapezoid, $3 x$ be the sum of the bases. Then the larger lateral side of the trapezoid is $2 h$, and the perimeter is $3 x + 3 h = 6$. The area of the trapezoid is $\frac{3 x h}{2} = \... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 |
On the sides $C B$ and $C D$ of the square $A B C D$, points $M$ and $K$ are taken such that the perimeter of triangle $C M K$ is equal to twice the side of the square.
Find the measure of angle $M A K$.
# | Let's rotate triangle $ABM$ around point $A$ by $90^{\circ}$ so that vertex $B$ moves to $D$. Let $M'$ be the image of point $M$ under this rotation. Since by the condition
$MK + MC + CK = (BM + MC) + (CK + KD)$, then $MK = BM + KD = KM'$. Moreover, $AM = AM'$, therefore triangles $AMK$ and $AM'K$ are congruent, which... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Consider $M . \mathbf{B}$.
On some cells of a $10 \times 10$ board, there is a flea. Once a minute, the fleas jump simultaneously, and each one jumps to an adjacent cell (along a side). A flea jumps strictly in one of the four directions parallel to the sides of the board, maintains its direction as long as possible, ... | Evaluation. On one vertical line, there can be no more than two fleas jumping along the vertical (otherwise, fleas in cells of the same color will meet). The same is true for the horizontals. In total, on 20 horizontals and verticals, there can be no more than 40 fleas.
Example. It is clear that fleas from cells of di... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the greatest possible number of rays in space emanating from a single point and forming obtuse angles with each other
# | Answer: 4. We will consider vectors instead of rays. We can assume that the first vector has coordinates (1, $0,0)$. Then the other vectors have coordinates ( $x_{i}, y_{i}, z_{i}$ ), where $x_{i}0$, and thus, $y_{\mathrm{i}}2$ the inequality $x_{\mathrm{i}} x_{\mathrm{j}}+y_{\mathrm{i}} y_{\mathrm{j}}+z_{\mathrm{i}} z... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
From a point located outside a circle, two mutually perpendicular tangents are drawn to the circle. The radius of the circle is 10. Find the length of each tangent.
# | ## Solution
The quadrilateral formed by the given tangents and radii drawn to the points of tangency is a square.
## Answer
10. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$C$ is a point on the extension of diameter $A B, C D$ is a tangent, angle $A D C$ is $110^{\circ}$. Find the angular measure of arc $B D$.
# | The angle between the tangent and the chord is equal to half the angular measure of the arc enclosed between them.
## Solution
$$
\begin{gathered}
\cup D B A=2 \angle A D C=220^{\circ}, \\
\cup D B=\cup D B A-\cup A B=220^{\circ}-180^{\circ}=40^{\circ} .
\end{gathered}
$$
## Answer
$40^{\circ}$. | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2+ [ Inscribed angle, based on the diameter ]
$A B$ is the diameter of the circle, $B C$ is the tangent. The secant $A C$ is divided by the circle at point $D$ in half. Find the angle $D A B$.
# | An inscribed angle that subtends a diameter is $90^{\circ}$.
## Solution
In triangle $A B C$, the height $B D$ is also a median. Therefore, this triangle is isosceles. Hence, $\angle D A B=45^{\circ}$.
## Answer
$45^{\circ}$ | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left.\frac{\text { [Inscribed and Circumscribed Circles ]}}{\text { Rhombuses Properties and Characteristics }}\right]$
The side of the rhombus is 8 cm, and the acute angle is $30^{\circ}$. Find the radius of the inscribed circle. | The diameter of the inscribed circle is equal to the height of the rhombus.
## Solution
The diameter of the inscribed circle is equal to the height of the rhombus, and the height, dropped from a vertex to the opposite side, is the leg of a right triangle opposite the angle of $30^{\circ}$.
Therefore, the height of t... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The Law of Cosines
[Pythagorean Theorem (direct and inverse).]
In a right triangle \(ABC\), the leg \(AC = 15\) and the leg \(BC = 20\). On the hypotenuse \(AB\), a segment \(AD\) equal to 4 is laid out, and point \(D\) is connected to \(C\). Find \(CD\). | Find $\cos \angle A$ from the right triangle $A B C$.
## Solution
By the Pythagorean theorem
$$
A B=\sqrt{A C^{2}+B C^{2}}=\sqrt{15^{2}+20^{2}}=\sqrt{225+400}=\sqrt{625}=25
$$
From the right triangle $A B C$, we find that
$$
\cos \angle A=\frac{A C}{A B}=\frac{15}{25}=\frac{3}{5}
$$
By the cosine rule
$$
C D^{2}... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A circle is inscribed in a right angle. A chord connecting the points of tangency is equal to 2. Find the distance from the center of the circle to this chord.
# | The diagonals of a square are equal.
## Solution
The quadrilateral formed by the tangents and radii drawn to the points of tangency is a square.
The desired segment is half the diagonal of this square, i.e., 1.
. The six resulting triangles will be isosceles.
## Answer
All... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Rectangular triangles (other). ]
[Concurrency of heights. Angles between heights.]
A line perpendicular to the hypotenuse $A B$ of a right triangle $A B C$ intersects the lines $A C$ and $B C$ at points $E$ and $D$ respectively.
Find the angle between the lines $A D$ and $B E$. | Consider triangle $A B D$ (see figure). From the condition, it follows that lines $A C$ and $D E$ contain its altitudes, that is, $E$ is the orthocenter of this triangle. Therefore, line $B E$ contains the third altitude of the triangle, that is, $B E \perp A D$.
# Answer
$90^{\circ}$.
Send a comment | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$3+$ $[$ Convex Polygons $]$
In a convex 13-gon, all diagonals are drawn. They divide it into polygons. Among them, we take the polygon with the largest number of sides. What is the maximum number of sides it can have? | From each vertex of the original 13-gon, no more than two diagonals emerge, which are sides of the considered polygon. Each diagonal corresponds to two vertices, so the number of sides of the considered polygon does not exceed 13. An example of a regular 13-gon shows that the number of sides of the polygon obtained by ... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A circle is inscribed in a triangle, and the points of tangency with the sides of the triangle are connected to each other. A new circle is inscribed in the resulting triangle, and the points of tangency with the sides form the vertices of a third triangle, which has the same angles as the original triangle. Find these... | Let the inscribed circle touch the sides $A B, B C, C A$ at points $C_{1}, A_{1}, B_{1}$. Triangles $A_{1} B C_{1}$ and $A_{1} C B_{1}$ are isosceles; their angles at the bases are $1 / 2\left(180^{\circ}-\angle B\right)$ and $1 / 2\left(180^{\circ}-\angle C\right)$. Therefore, $\angle A_{1}=1 / 2$ $(\angle B+\angle C)... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Vassily H.S.
Consider all possible closed broken lines with six segments, all vertices of which lie on a circle.
a) Draw such a broken line that has the maximum possible number of self-intersections.
b) Prove that a broken line of this kind cannot have a greater number of self-intersections. | a) The figure shows a six-segment broken line with seven self-intersection points.
b) We will prove that such a broken line cannot have more than seven self-intersections. Let the total numb... | 7 | Geometry | proof | Yes | Yes | olympiads | false |
9
A point lying inside a circumscribed $n$-gon is connected by segments to all vertices and points of tangency. The triangles formed by this are alternately painted red and blue. Prove that the product of the areas of the red triangles is equal to the product of the areas of the blue triangles. | Let $h_{1}, \ldots, h_{n}$ be the distances from a given point to the corresponding sides, and $a_{1}, \ldots, a_{n}$ be the distances from the vertices of the polygon to the points of tangency. Then the product of the areas of both the red and blue triangles is $a_{1} \ldots a_{n} h_{1} \ldots h_{n} / 2^{n}$.
What is... | 3 | Geometry | proof | Yes | Yes | olympiads | false |
[Relationships between the sides and angles of triangles (other).]
Find the angle $B$ of triangle $A B C$, if the length of the height $C H$ is half the length of side $A B$, and $\angle B A C=75^{\circ}$. | Let $B^{\prime}$ be the point of intersection of the perpendicular bisector of segment $A C$ with line $A B$. Then $A B^{\prime}=C B^{\prime}$ and $\angle$ $A B^{\prime} C=180^{\circ}-2 \cdot 75^{\circ}=30^{\circ}$. Therefore, $A B^{\prime}=C B^{\prime}=2 C H=A B$, i.e., $B^{\prime}=B$ and $\angle B=30^{\circ}$. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[The triangle formed by the bases of two altitudes and a vertex]
The side of the triangle is $\sqrt{2}$, and the angles adjacent to it are $75^{\circ}$ and $60^{\circ}$.
Find the segment connecting the bases of the altitudes drawn from the vertices of these angles. | Let $B M$ and $C N$ be the altitudes of triangle $A B C$, $B C=\sqrt{2}$, $\angle B=75^{\circ}$, $\angle C=60^{\circ}$. Then $\angle A=45^{\circ}$.
Triangle $A M N$ is similar to triangle $A B C$ with a similarity coefficient of $\cos 45^{\circ}$. Therefore, $M N=$ $B C \cdot \cos 45^{\circ}=1$.
## Answer
1.
Probl... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A rectangle is divided by two vertical and two horizontal segments into nine rectangular parts. The areas of some of the resulting parts are indicated in the figure. Find the area of the upper right part.
| 30 | | $?$ |
| :---: | :---: | :---: |
| 21 | 35 | |
| | 10 | 8 | | It is not difficult to understand that if a rectangle is cut into 4 parts by a vertical and a horizontal segment, then the products of the areas of opposite parts are equal. Using this consideration, we sequentially find the areas of the upper middle part $(30 \cdot 35: 21=50)$, the right middle part $(8 \cdot 35: 10=2... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
What is the maximum number of colors that can be used to color the cells of an $8 \times 8$ chessboard so that each cell shares a side with at least two cells of the same color? | Let's divide the board into 16 squares of $2 \times 2$ and color each square with a different color. This coloring with 16 colors satisfies the condition.
It is impossible to achieve more colors. Indeed, if there are no more than three cells of a certain color, then only one of them can share a side with two cells of ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Algebraic problems on the triangle inequality ]
In a triangle, two sides are equal to 3.14 and 0.67. Find the third side, given that its length is an integer.
# | Apply the triangle inequality.
## Solution
Let the integer $n-$ be the length of the third side. Then
$$
3.14-0.67<n<3.14+0.67
$$
From this, we find that $n=3$.
## Answer
3.
## Problem | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
[ Homothetic Circles]
Inside an angle, there are three circles $\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}$, each of which touches the two sides of the angle, and circle $S_{2}$ touches circles $S_{1}$ and $S_{3}$ externally. It is known that the radius of circle $S_{1}$ is 1, and the radius of circle $S_{3}$ is... | Homothety with the center at the vertex of the angle, translating circle $S_{1}$ to circle $S_{2}$, also translates circle $S_{2}$ to circle $S_{3}$.
## Solution
Consider the homothety $\mathrm{H}$ with the center at the vertex of the angle, translating circle $\mathrm{S}_{1}$ to circle $\mathrm{S}_{2}$. Circle $\mat... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ inscribed quadrilaterals]
Three consecutive sides of an inscribed quadrilateral are in the ratio 1:2:3. Find its sides if it is known that the perimeter is 24 m.
# | Since the sums of the opposite sides of a circumscribed quadrilateral are equal to each other, two parts fall on the fourth side. Therefore, the sides are $\frac{1}{8}, \frac{1}{4}, \frac{3}{8}$ and $\frac{1}{4}$ of the perimeter, i.e., 3, 6, 9, 6.
## Answer
3 cm, 6 cm, 9 cm, 6 cm. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9
Find the sum of the exterior angles of a convex $n$-gon, taken one at each vertex. | The sum of the interior angles of a convex $n$-gon is $180^{\circ}(n-2)$.
## Solution
The sum of the straight angles at all vertices of the $n$-gon is $180^{\circ} n$. The sum of the interior angles of a convex $n$-gon is $180^{\circ}(n-2)$. The sum of the exterior angles of this $n$-gon is the difference between the... | 360 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ The angle between two chords and two secants]
A circle with a radius equal to the height of an equilateral triangle rolls along one of its sides. Prove that the angular magnitude of the arc cut off on the circle by the sides of the triangle is always $60^{\circ}$. | Let the angular measure of the arc cut off by the sides of triangle $ABC$ on the circle be denoted by $\alpha$.
Consider the arc cut off by the extensions of the sides of the triangle on the circle, and denote its angular measure by $\alpha^{\prime}$. Then $\left(\alpha+\alpha^{\prime}\right) / 2=\angle B A C=60^{\cir... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
8,9 |
| :---: | :---: | :---: |
| | Inscribed and Circumscribed Circles | |
| | [ Properties and characteristics of isosceles triangles.] $]$ | |
In triangle $A B C$, $\angle B=36^{\circ}, \angle C=42^{\circ}$. On side $B C$, a point $M$ is taken such that $B M=R$, where $R$ is the radius of the circumscribed circ... | Prove that point $M$ lies on the radius connecting the center of the circumscribed circle with vertex $A$.
## Solution
Let $O$ be the center of the circumscribed circle. $\angle A O B = 2 \angle C = 84^{\circ}, \angle B O C = 2 \angle B + 2 \angle C = 156^{\circ}, \angle O B C = 90^{\circ} - 1 / 2$ $\angle B O C = 12... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In triangle $ABC$ with angle $A$ equal to $120^{\circ}$, the angle bisectors $AA_{1}, BB_{1}$, and $CC_{1}$ intersect at point $O$. Prove that $\angle A_{1} C_{1} O=30^{\circ}$. | Prove that $A_{1} C_{1}$ is the bisector of angle $A A_{1} B$ and apply the formula for the angle between the bisectors of a triangle.
## Solution
Notice that ray $A B$ is the bisector of the external angle $A$ of triangle $A A_{1} C$. Since $C C_{1}$ is the bisector of angle $C$, then $C_{1}$ is the center of the ex... | 30 | Geometry | proof | Yes | Yes | olympiads | false |
Shvecov D.V.
On the hypotenuse $AC$ of the right triangle $ABC$, a point $C_{1}$ is marked such that $BC = CC_{1}$. Then, on the leg $AB$, a point $C_{2}$ is marked such that
$AC_{2} = AC_{1}$; the point $A_{2}$ is defined similarly. Find the angle $AMC$, where $M$ is the midpoint of the segment $A_{2}C_{2}$. | Let $I$ be the center of the inscribed circle of triangle $ABC$. Since point $C_{1}$ is symmetric to $B$ with respect to $CI$, and $C_{2}$ is symmetric to $C_{1}$ with respect to $AI$, then
$BI = IC_{2}$ and $\angle BIC_{2} = 90^{\circ}$. Similarly, $BI = IA_{2}$ and $\angle BIA_{2} = 90^{\circ}$ (see figure). Therefo... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Yakubov A. }}$.
Parallelogram $A B C D$ is such that $\angle B<90^{\circ}$ and $A B<B C$. Points $E$ and $F$ are chosen on the circumcircle $\omega$ of triangle $A B C$ such that the tangents to $\omega$ at these points pass through point $D$. It turns out that $\angle E D A=$ $\angle F D C$. Find ... | Let $l$ be the bisector of angle $E D F$. Since $D E$ and $D F$ are tangents, the line $l$ passes through the center $O$ of the circle $\omega$.
Perform a reflection about $l$. Since $\angle... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A square piece of paper was punctured at 1965 points. From the puncture points and the vertices of the square, no three lie on the same straight line. Then, several straight, non-intersecting cuts were made, each starting and ending only at the punctured points or the vertices of the square. It turned out that the squa... | Answer: 5896 cuts, 3932 triangles. Let's solve the problem in the general case when the square is pierced at $n$ points. Let the number of resulting triangles be $x$. On the one hand, the sum of the angles of all these triangles is $x \cdot 180^{\circ}$. On the other hand, it is equal to $360^{\circ} + n \cdot 360^{\ci... | 5896 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Dirichlet's Principle (finite number of points, lines, etc.).] [ Central angle. Arc length and circumference length ]
On a circle of radius 1, a point $O$ is marked and a notch is made to the right with a compass of radius $l$. From the resulting point $O_{1}$, another notch is made in the same direction with the s... | We will prove by induction on $n$ that the number of different arcs after $n$ cuts does not exceed 3. For $n=2$, this is obvious. Let $A_{k}$ denote the cut with number $k$. Suppose $n$ cuts have been made and the point $A_{n}$ falls on the ARC $A_{k} A_{l}$. Then the point $A_{n-1} 1$ falls on the arc $A_{k}-1 A_{l-} ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Pechkovskiy A.n.
On a chessboard of size $8 \times 8$, 64 points are marked - the centers of all cells. Is it possible to separate all the points from each other by drawing 13 lines that do not pass through these points? | Answer: No. Indeed, consider a square passing through the centers of all 28 boundary cells. It is clear that 13 lines intersect it in no more than 26 points, and therefore divide it into no more than 26 parts, meaning that two "boundary" centers will end up in the same part. Therefore, to separate the 28 boundary, and ... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Cube [ Properties of parts obtained by cutting ]
The surface of a 3 x 3 x 3 Rubik's cube consists of 54 cells. What is the maximum number of cells that can be marked so that the marked cells do not share any vertices? | In Fig. 1, it is shown how to mark 7 cells on three adjacent faces of a cube. On the three "invisible" faces, seven cells symmetrical to these need to be marked.
Now let's prove that it is impossible to mark more than 14 cells in the required manner. We will do this in two ways.
The first way. Let's count the total n... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In the arena of the circus, which is a circle with a radius of 10 m, a lion is running. Moving along a broken line, he ran 30 km. Prove that the sum of all his turning angles is not less than 2998 radians. | Suppose the lion ran along the broken line $A_{1} A_{2} \ldots A_{n}$. Let's straighten the lion's trajectory as follows. Rotate the circus arena and the subsequent trajectory around point $A_{2}$ so that point $A_{3}$ falls on the ray $A_{1} A_{2}$. Then rotate the circus arena and the subsequent trajectory around poi... | 2998 | Geometry | proof | Yes | Yes | olympiads | false |
How long will the strip be if a cubic kilometer is cut into cubic meters and laid out in a single line?
# | In one cubic kilometer - a billion cubic meters.
## Solution
In one cubic kilometer, there is a billion cubic meters (1000 in length, 1000 in width, and 1000 in height). If all of them are laid out in a row, it would form a strip one billion meters long, i.e., one million kilometers.
## Answer
1000000 km. | 1000000 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.] [ Volume of a parallelepiped ]
A cube with a side of 1 m was sawn into cubes with a side of 1 cm and laid in a row (in a straight line). What length did the row turn out to be
# | We get $100 \times 100 \times 100=1000000$ (cm) or 10000 m = 10 km. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9}
On a ruler, three marks are made: 0, 2, and 5. How can you measure a segment equal to 6 using it?
# | $5-2=3$, so you can lay down a segment of length 3. Two such segments give a segment equal to 6.
Send a comment | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Find the sum of all dihedral angles of a triangular pyramid.
# | This sum is equal to the sum of the angles of four triangles (faces of the pyramid), that is, $180^{\circ} \cdot 4=720^{\circ}$.
## Answer
$720^{\circ}$. | 720 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Sharygin I.F.
A rectangle is composed of six squares (see the right figure). Find the side of the largest square if the side of the smallest one is 1.
 | The side of the largest square is equal to the sum of the sides of two squares: the next one in the clockwise direction and the smallest one.
## Solution
Note that the side of the largest square is equal to the sum of the sides of two squares: the next one in the clockwise direction and the smallest one. Denoting the... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Auto: Zastassky A. A.
Two convex quadrilaterals are such that the sides of each lie on the perpendicular bisectors of the sides of the other. Find their angles. | Let the side $C^{\prime} D^{\prime}$ of the quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ lie on the perpendicular bisector of the side $A B$ of the quadrilateral $A B C D$, and the side $D^{\prime} A^{\prime}$ on the perpendicular bisector of $B C$. Then $D^{\prime}$ is the center of the circumcircle of ... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Prokopenko D:
Around triangle $A B C$, a circle $\Omega$ is described. Let $L$ and $W$ be the points of intersection of the angle bisector of angle $A$ with side $B C$ and circle $\Omega$, respectively. Point $O$ is the center of the circumcircle of triangle $A C L$. Restore triangle $A B C$ if the circle $\Omega$ and... | Let $O^{\prime}$ be the center of $\Omega$. Then the lines $O^{\prime} O$ and $O^{\prime} W$ are perpendicular to the sides $A C$ and $B C$, so the directions of these sides are known. Moreover,
$\angle C O L=2 \angle C A L=2 \angle L C W$ and, therefore, $\angle O C W=90^{\circ}$ (see the figure). Consequently, $C$ i... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 |
In triangle $A B C$, it is known that $\angle B A C=75^{\circ}, A B=1, A C=\sqrt{6}$. A point $M$ is chosen on side $B C$, such that $\angle$ ВАМ $=30^{\circ}$. Line $A M$ intersects the circumcircle of triangle $A B C$ at point $N$, different from $A$. Find $A N$. | Denote $A N$ as $x$, express $B N$ and $C N$ through the circumradius of triangle $A B C$, and apply the Law of Cosines to triangles $A B N$ and $A C N$ (or use Ptolemy's theorem).
## Solution
## First Method.
Let $A N=x, R$ be the circumradius of the triangle. Then
$$
B N=2 R \sin 30^{\circ}=R, C N=2 R \sin 45^{\c... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4 [ Proof by contradiction $]$
Inside an isosceles right triangle $ABC$ with hypotenuse $AB$, a point $M$ is taken such that angle $MAB$ is $15^{\circ}$ greater than angle $MAC$, and angle $MCB$ is $15^{\circ}$ greater than angle $MBC$. Find angle $BMC$. | Let the point $X$ of intersection of $A M$ and the altitude $C H$ of triangle $A B C$ lie on the segment $A M$ (see the figure; at the end of the solution, we will show that the other case is impossible).
. $\angle BAC = \angle ACB$ (as the base angles of an isosceles triangle), and $\angle BCA = \angle CAD$ (as a... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Trapezoids Auxiliary similar triangles $]$
Point $M$ is located on the lateral side $A B$ of trapezoid $A B C D$, such that $A M: B M=2: 1$. A line passing through point $M$ parallel to the bases $A D$ and $B C$ intersects the lateral side $C D$ at point $N$. Find $M N$, if $A D=18, B C=6$. | Let's draw the diagonal $B D$. Suppose it intersects the segment $M N$ at point $P$. According to the theorem of proportional segments,
$B P: P D = C N: N D = B M: A M = 1: 2$. Triangle $B M P$ is similar to triangle $B A D$ with a similarity ratio of $B M / B A = 1 / 3$, and triangle $D N P$ is similar to triangle $D... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 On the hypotenuse $BC$ of the right triangle $ABC$, points $D$ and $E$ are marked such that $AD \perp BC$ and $AD = DE$. On the side $AC$, a point $F$ is marked such that $EF \perp BC$. Find the angle $ABF$.} | Points $A$ and $E$ lie on a circle with diameter $B F$. The inscribed angles $A B F$ and $A E F$ are equal. In the isosceles right triangle $A D E$, the angles at the base are $45^{\circ}$ each, so $\angle A B F=\angle A E F$ $=90^{\circ}-\angle A E D=45^{\circ}$.
## Answer
$45^{\circ}$. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Triangle Inequality (other).]
The two altitudes of a triangle are 12 and 20. Prove that the third altitude is less than 30. | Since $c>|b-a|$ and $a=2 S / h_{\mathrm{a}}, b=2 S / h_{\mathrm{b}}, c=2 S / h_{\mathrm{c}}$, then $\frac{1}{h_{c}}>\left|\frac{1}{h_{a}}-\frac{1}{h_{b}}\right|$. Therefore, in our case $h_{\mathrm{c}}<20 \cdot 12 / 8=$ 30. | 30 | Inequalities | proof | Yes | Yes | olympiads | false |
[ Quadrilaterals (extreme properties) ]
The diagonals of a convex quadrilateral $A B C D$ intersect at point $O$. What is the smallest area that this quadrilateral can have if the area of triangle $A O B$ is 4 and the area of triangle $C O D$ is 9? | Since $S_{\mathrm{AOB}}: S_{\mathrm{BOC}}=A O: O C=S_{\mathrm{AOD}}: S_{\mathrm{DOC}}$, then $S_{\mathrm{BOC}} \cdot S_{\mathrm{AOD}}=S_{\mathrm{AOB}} \cdot S_{\mathrm{DOC}}=36$. Therefore, $S_{\mathrm{BOC}}+S_{\mathrm{AOD}}$ $\geq 2 \sqrt{S_{B O C} \cdot S_{A O D}}=12$, and equality is achieved if $S_{\mathrm{BOC}}=S_... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Extreme properties (continued).]
If five points are given on a plane, then by considering all possible triples of these points, one can form 30 angles. Let the smallest of these angles be $\alpha$. Find the maximum value of $\alpha$.
# | First, assume that the points are the vertices of a convex pentagon. The sum of the angles of a pentagon is $540^{\circ}$, so one of its angles does not exceed $540^{\circ} / 5=108^{\circ}$. The diagonals divide this angle into three angles, so one of them does not exceed $108^{\circ} / 3=36^{\circ}$. In this case, $\a... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Mutual relations between sides and angles of triangles (other).]
In triangle $ABC$, the height $AH$ is equal to the median $BM$. Find the angle $MBC$.
# | Drop a perpendicular $M D$ from point $M$ to line $B C$. Then $M D=A H / 2=B M / 2$. In the right triangle $B D M$, the leg $M D$ is equal to half of the hypotenuse $B M$. Therefore, $\angle M B C=\angle M B D=30^{\circ}$.
Send a comment | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Mutual relations between sides and angles of triangles (other).]
In triangle $ABC$, the bisectors $AD$ and $BE$ are drawn. Find the measure of angle $C$, given that $AD \cdot BC = BE \cdot AC$ and $AC \neq BC$. | The quantities $A D \cdot B C \sin A D B$ and $B E \cdot A C \sin A E B$ are equal, as they are equal to twice the area of triangle $A B C$. Therefore, $\sin A D B=\sin A E B$. There are two possible cases.
1. $\angle A D B=\angle A E B ;$ in this case, points $A, E, D, B$ lie on the same circle, so $\angle E A D=\ang... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ [Assist
In triangle $ABC$, the height $BD$ is 11.2 and the height $AE$ is 12. Point $E$ lies on side $BC$ and $BE$ : $EC=5: 9$. Find the side $AC$.
# | The product of the base and height for the given triangle is constant.
## Solution
Let $B E=5 x, C E=9 x$. By the Pythagorean theorem from the right triangle $A E C$, we find that
$$
\begin{aligned}
A C= & \sqrt{E C^{2}+A E^{2}}= \\
& \sqrt{81 x^{2}+144}
\end{aligned}
$$
From the formula for the area of a triangle,... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ [ Relationship,
In triangle $ABC$, the three sides are given: $AB=26$, $BC=30$, and $AC=28$. Find the part of the area of this triangle enclosed between the height and the bisector drawn from vertex $B$.
# | Apply Heron's formula and the property of the angle bisector of a triangle.
## Solution
Let $B P$ and $B Q$ be the height and the angle bisector of the given triangle $A B C$. By Heron's formula,
On the other hand, $S=1 / 2$ AC$\cdot$BP.
Therefore, $B P=\frac{2 S_{\triangle A B C}}{A C}=$
$\frac{2 \cdot 336}{28}=2... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The lateral side, the smaller base, and the diagonal of an isosceles trapezoid are 10, 6, and 14, respectively. Find the larger base.
# | The projection of the lateral side of an isosceles trapezoid onto the larger base is equal to half the difference of the bases, and the projection of the diagonal is half the sum of the bases.
## Solution
Let $H$ be the projection of vertex $C$ of the smaller base of trapezoid $ABCD$ onto the larger base $AD$.
Let $... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Circles with centers $O_{1}$ and $O_{2}$ have a common chord $A B, \angle A O_{1} B=60^{\circ}$. The ratio of the length of the first circle to the length of the second is $\sqrt{2}$. Find the angle $A O_{2} B$. | Express the sides of triangle $A O_{2} B$ in terms of the radius of the smaller circle.
## Solution
Let $R$ and $r$ be the radii of the circles with centers $O_{1}$ and $O_{2}$, respectively. According to the problem,
$$
\frac{2 \pi R}{2 \pi r}=\frac{R}{r}=\sqrt{2} \text {. }
$$
Therefore, $R=r \sqrt{2}$. Triangle ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## [ Auxiliary area. The area helps to solve the problem ]
Pythagorean Theorem (direct and inverse)In triangle $A B C$, heights $A E$ and $C D$ are drawn. Find the side $A B$, if $B D=18, B C=30, A E=20$. | $A B \cdot C D=B C \cdot A E$
## Solution
By the Pythagorean theorem from the right triangle $B D C$, we find that
$$
C D^{2}=B C^{2}-B D^{2}=30^{2}-18^{2}=(30-18)(30+18)=12 \cdot 48=12^{2} \cdot 2^{2}
$$
thus, $C D=24$.
Let $S$ be the area of the triangle. Then
$$
S=\frac{1}{2} A B \cdot C D \text { and } S=\fra... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$3+$
Find the area of trapezoid $ABCD (AD \| BC)$, if its bases are in the ratio $5:3$, and the area of triangle $ADM$ is 50, where $M$ is the point of intersection of lines $AB$ and $CD$. | The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.
## Solution
Let ${ }^{B C} /{ }_{A D}=3 / 5$. The similarity coefficient of triangles $M B C$ and $M A D$ is $3 / 5$. Therefore, $S_{M B C}=9 / 25 S_{M A D}=18$. Hence, $S_{A B C D}=S_{M A D}-S_{M B C}=50-18=32$.
 measure the distance in centimeters between two given points; b) compare two given numbers. What is the minimum number of operations this device needs to perform to definitely det... | To determine whether $A B C D$ is a rectangle, it is sufficient to check the equalities $A B=C D$, $B C=A D$, and $A C=B D$ - a total of 9 operations (3 operations for each equality: two measurements and one comparison).
A rectangle $ABCD$ will be a square if $A B=B C$ - for this, one more, the 10th, operation of comp... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
On the table, there are 9 apples, forming 10 rows with 3 apples in each (see the figure).
It is known that the weights of nine rows are the same, while the weight of the te... | Before spending money, let's think
Let $l_{1}, l_{2}, l_{3}$ be the weights of the three highlighted (see fig. a) diagonal rows, and $v_{1}, v_{2}, v_{3}$ be the weights of the three vertical rows. Then $l_{1}+l_{2}+l_{3}=v_{1}+v_{2}+v_{3}:$ both are simply the sum of the weights of all nine apples. But at least five ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Emelyanov L.A.
A square is cut into $n$ rectangles of size $a_{i} \times b_{i}, i=1, \ldots, n$.
For what smallest $n$ can all the numbers in the set $\left\{a_{1}, b_{1}, \ldots, a_{n}, b_{n}\right\}$ be distinct? | First, let's show that no rectangle (in particular, a square) can be cut into two, three, or four rectangles with different sides.
Obviously, if a rectangle is cut into two rectangles, they share a common side. Suppose the rectangle is cut into three rectangles. Then one of them contains two vertices of the original r... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8,9
In a convex quadrilateral $A B C D$, diagonal $A C$ is drawn, $A D=7, B C=3, \angle A C D=60^{\circ}$. It is known that points $A, B, C, D$ lie on the same circle, and the perpendicular from point $A$ to side $C D$ bisects angle $\angle B A D$. Find the diagonal $A C$. | Find the sine of angle $A D C$.
## Solution
Let $A M$ be the perpendicular dropped from point $A$ to side $C D$. The ray $A C$ passes between the sides of angle $B A M$, so
$$
\angle B A M=\angle B A C+\angle C A M .
$$
Let $R$ be the radius of the circumscribed circle of quadrilateral $A B C D$. Then
$$
\begin{al... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Rectangles and Squares. Properties and Characteristics]
Inside the square $A B C D$, a point $M$ is taken such that $\angle M A B=60^{\circ}, \angle M C D=15^{\circ}$. Find $\angle M B C$. | Construct an equilateral triangle on $A B$ inside the square.
## Solution
Construct an equilateral triangle $A B K$ on $A B$ inside the square. Then the point $M$ lies on the ray $A K$. Moreover,
$$
\begin{gathered}
\angle K B A=60^{\circ}, \angle K B C=30^{\circ}, B C=B K, \\
\angle B C K=\frac{1}{2}\left(180^{\cir... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Orthocenter and orthic triangle ] [ [ median line of a triangle ]
Segments connecting the feet of the altitudes of an acute-angled triangle are 8, 15, and 17. Find the radius of the circumscribed circle around the triangle.
# | Continue the heights $A D, B E$ and $C F$ of the given triangle $A B C$ until they intersect the circumscribed circle at points $A^{\prime}, B^{\prime}$ and $C^{\prime}$ respectively. Then the triangle $A^{\prime} B^{\prime} C^{\prime}$ is similar to triangle $D E F$ with a similarity coefficient of 2.
## Solution
Le... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Rhombi. Properties and characteristics ] [Area of a triangle (using two sides and the angle between them).]
In an acute-angled triangle $A B C$, point $D$ is chosen on side $A B$ such that $\angle D C A=45^{\circ}$. Point $D_{1}$ is symmetric to point $D$ with respect to line $B C$, and point $D_{2}$ is symmetric to... | $\angle B C D=30^{\circ}$, quadrilateral $D B D_{1} C-$ is a rhombus.
## Solution
Let $Q$ and $P$ be the points of intersection of segments $D D_{1}$ and $D_{1} D_{2}$ with lines $B C$ and $A C$. Denote $\angle A C B=\gamma$. Then
$$
\begin{gathered}
\angle D_{1} C P=\angle D_{2} C P=\angle A C B=\gamma \\
\angle Q ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 | |
A circle can be inscribed in quadrilateral $A B C D$. Let $K$ be the intersection point of its diagonals. It is known that $A B>B C>K C, B K=4+\sqrt{2}$, and the perimeter and area of triangle $B K C$ are 14 and 7, respectively.
Find $D C$. | Applying Heron's formula, find the sides of triangle $B K C$. Prove that $A C \perp B D$. Also prove that if the diagonals of a convex quadrilateral are perpendicular to each other, then the sums of the squares of the opposite sides are equal.
## Solution
Let $p=7$ be the semi-perimeter, $S=7$ be the area of triangle... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ $\left.\begin{array}{l}\text { Inscribed quadrilaterals } \\ \text { [Quadrilateral: calculations, metric relations.] }\end{array}\right]$
In the quadrilateral $A B C D$, a circle can be inscribed. Let $K$ be the point of intersection of its diagonals. It is known that $B C>A B>B K, K C=\sqrt{7}-1$, the cosine of an... | Let $\angle K B C=\alpha, B K=x, B C=y$. Let $P=2 \sqrt{7}+4$ be the perimeter of triangle $B K C$. Then
$$
x+y=P-K C=2 \sqrt{7}+4-(\sqrt{7}-1)=5+\sqrt{7}
$$
By the cosine rule,
$$
K C^{2}=B K^{2}+B C^{2}-2 \cdot B K \cdot B C \cdot \cos \alpha, \text { or }(\sqrt{7}-1)^{2}=x^{2}+y^{2}-2 x y \cdot \frac{\sqrt{7}+1}{... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In triangle $K L M$, the ratio of the radii of the circumscribed and inscribed circles is 3. The inscribed circle touches the sides of triangle $K L M$ at points $A, B$, and $C$. Find the ratio of the area of triangle $K L M$ to the area of triangle $A B C$.
# | Let $O$ be the center of the circle inscribed in triangle $KLM$. Then point $O$ lies inside triangle $ABC$ and
$$
S_{\triangle \mathrm{ABC}}=S_{\triangle \mathrm{BOC}}+S_{\triangle \mathrm{AOC}}+S_{\triangle \mathrm{AOB}}.
$$
Next, apply the Law of Sines and formulas for the area of a triangle.
## Solution
Let the ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Blinkov Yu.A.
On the sides $AB$ and $CD$ of the square $ABCD$, points $K$ and $M$ are taken respectively, and on the diagonal $AC$ - point $L$ such that $ML = KL$. Let $P$ be the point of intersection of segments $MK$ and $BD$. Find the angle $KPL$. | The first method. Let $O$ be the intersection point of $AC$ and $BD$; $N$ be the midpoint of segment $KM$ (see Fig. a). Since the midpoint of a segment with endpoints on parallel lines lies on a line equidistant from them, $ON \parallel AK$ and $\angle AON = 45^{\circ}$. On the other hand, since $LN \perp MK$, the poin... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Dorichenko S.A.
Which are there more of: rectangles with integer sides and a perimeter of 1996, or rectangles with integer sides and a perimeter of $1998$?
(Rectangles $a \times b$ and $b \times a$ are considered the same.) | If the perimeter of a rectangle is 1996, then the sum of the lengths of its adjacent sides is 998.
## Solution
If the perimeter of a rectangle is 1996, then the sum of the lengths of its adjacent sides is 998. This means the length of the shorter side can take values from 1 to 499. If the perimeter of the rectangle i... | 499 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Linear dependence of vectors $]$ [ Angles between lines and planes $]$
The height of a regular hexagonal pyramid is equal to the side of the base. Find the angle between a lateral edge and the plane of the base. | Let ABCDEFP be a given regular hexagonal pyramid with vertex $P$; $M$ - the center of the regular hexagon $A B C D E F$. Denote $A B=B C=C D=D E=E F=A F=a$. Since the pyramid is regular, $P M$ is its height. Therefore, the angle of the lateral edge with the base plane is the angle $\angle PAM$. In the right triangle $A... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Chords and secants (etc.). $]$
A chord subtends an arc of $90^{\circ}$ and is equal to 16. Find its distance from the center. | The foot of the perpendicular dropped from the center of the circle to the given chord bisects it.
## Solution
Let $M$ be the foot of the perpendicular dropped from the center $O$ of the circle to the chord $A B$. Then $M$ is the midpoint of $A B$ and the right triangle $O M A$ is isosceles. Therefore, $O M = M A = 8... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The radius of the circle is 13, the chord is 10. Find its distance from the center.
# | A diameter perpendicular to a chord bisects it. Therefore, the square of the desired distance is $13^{2}-5^{2}=$ $12^{2}$.
Answer
12. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ [tangents drawn from one point]
Two tangents are drawn from a common point to a circle. The radius of the circle is 11, and the sum of the tangents is 120.
Find the distance from the center to the common point of tangents.
# | The length of each tangent is 60. The square of the desired distance is $60^{2}+11^{2}=61^{2}$.
Answer
61. | 61 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\begin{aligned} & \text { [ Congruent triangles. Criteria for congruence ] } \\ & \text { [Central symmetry helps solve the problem].] }\end{aligned}$
Segments $A B$ and $C D$ intersect at point $O$, which is the midpoint of each of them. What is the length of segment $B D$ if segment $A C=10$? | Triangle $B O D$ is equal to triangle $A O C$ by two sides and the included angle. Therefore, $B D=A C=10$.
## Answer
10. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
\left.\begin{array}{l}\text { [Sum of angles in a triangle. Theorem about the exterior angle.] } \\ {[\text { Concurrency of altitudes. Angles between altitudes. ] }}\end{array}\right]
The altitudes of triangle $ABC$, drawn from vertices $A$ and $C$, intersect at point $M$. Find $\angle AMC$, if $\angle A=$ $70^{\circ... | Find the angle between side $B C$ and the height drawn from vertex $C$.
## Solution
Let $A A_{1}$ and $C C_{1}$ be the heights.
$\angle A C C_{1}=90^{\circ}-70^{\circ}=20^{\circ}, \angle C A A_{1}=90^{\circ}-80^{\circ}=10^{\circ}, \angle A M C=180^{\circ}-20^{\circ}-10^{\circ}=150^{\circ}$.
. Since the diameter perpendicular to a chord bisects it, points $M$ and $N$ are the midpoints of segments $A... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The hypotenuse $AB$ of the right triangle $ABC$ is a chord of a circle with radius 10. The vertex $C$ lies on the diameter of the circle, which is parallel to the hypotenuse, $\angle A=75^{\circ}$. Find the area of triangle $ABC$. | From the center $O$ of the given circle, drop a perpendicular $OM$ to the hypotenuse $AB$. Then $MC = MA = MB$.
Therefore, $\angle MCB = \angle B = 15^{\circ}$, $\angle BCO = \angle B = 15^{\circ}$. Consequently, $\angle MCO = 30^{\circ}$.
Let $OM = x$. From the right triangle $MCO$, we find that $MC = 2x$. By the Py... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Pythagorean Theorem (direct and inverse).]
## [ Perimeter of a triangle $\quad$]
The perimeter of a right triangle $ABC (\angle C=90^{\circ})$ is 72, and the difference between the median $CK$ and the altitude $CM$ is 7.
Find the area of triangle $ABC$. | Let $A C=b, B C=a, C K=x$. Then $A B=2 x, C M=x-7$. According to the condition $a+b=72-2 x$.
Since $2 S_{A B C}=a b=2 x(x-7),(a+b)^{2}-\left(a^{2}+b^{2}\right)=2 a b$, then $4(36-x)^{2}-4 x^{2}=4 x(x-7)$. From this, $(36-x)^{2}-x^{2}$ $=x^{2}-7 x, x^{2}+65 x-1296=0, \quad x=\frac{-65+\sqrt{4225+5184}}{2}=\frac{-65+97}... | 144 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Properties and characteristics of tangents ] [ Pythagorean Theorem (direct and inverse) ]
Find the geometric locus of points from which tangents are drawn to a given circle, equal to a given segment.
# | Consider a right triangle with vertices at the considered point, the center of the given circle, and the point of tangency.
## Solution
Let the tangent line drawn from point $M$ to the given circle be $d$ (i.e., the segment with endpoints at point $M$ and the point of tangency is equal to the given length), and the r... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 |
Points $A$ and $B$ are connected by two arcs of circles, convex in opposite directions: $\cup A C B=$ $117^{\circ} 23^{\prime}$ and $\cup A D B=42^{\circ} 37^{\prime}$. The midpoints $C$ and $D$ of these arcs are connected to point $A$. Find the angle $C A D$. | Ray $A B$ passes between the sides of angle $C A D$. Therefore, $\angle C A D=\angle C A B+\angle B A D$.
## Solution
Since ray $A B$ passes between the sides of angle $C A D$, then
$$
\begin{gathered}
\angle C A D=\angle C A B+\angle B A D=\frac{1}{2} \cup B C+\frac{1}{2} \cup B D= \\
=\frac{\frac{1}{2} \cup A C B+... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Diameter $A B$ and chord $C D$ intersect at point $M, \angle C M B=73^{\circ}$, the angular magnitude of arc $B C$ is $110^{\circ}$. Find the magnitude of arc $B D$. | $C M B$ - the external angle of triangle $M D B$.
## Solution
Since $C M B$ is the external angle of triangle $M D B$, then
$$
\angle A B D=\angle C M B-\angle C D B=73^{\circ}-55^{\circ}=18^{\circ} .
$$
Therefore,
$$
\cup A D=36^{\circ}, \cup B D=180^{\circ}-36^{\circ}=144^{\circ}
$$
 Triangle ]
In a given circle with a radius of 3, six equal circles are inscribed, each touching the given circle; moreover, each of these six circles touches two adjacent ones. Find the radii of the circles.
# | Consider a triangle with vertices at the centers of the external and two adjacent internal circles.
## Solution
Let $r$ be the radius of one of the six internal circles. The triangle with vertices at the centers of the external and two adjacent internal circles is equilateral, so $3-r=2r$. Therefore, $r=1$.
 base $AD$ of trapezoid $ABCD$, according to the midline theorem, is 12. Through the vertex $C$ of the smaller base $BC$, draw a line parallel to the lateral side $AB$, until it intersects... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$C D$ is the median of triangle $A B C$. The circles inscribed in triangles $A C D$ and $B C D$ touch the segment $C D$ at points $M$ and $N$. Find $M N$, if $A C - B C = 2$. | If the circle inscribed in triangle $P Q R$ touches side $P Q$ at point $S$, then $P S=1 / 2(P Q+P R-R Q)$.
## Solution
Since $A D=D B$, and
$$
C M=1 / 2(A C+C D-A D) \text{ and } C N=1 / 2(B C+C D-B D)
$$
then
$$
\begin{aligned}
M N=|C M-C N| & =|1 / 2(A C+C D-A D)-1 / 2(B C+C D-B D)|= \\
& =1 / 2|A C-B C|=1 / 2 ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Acute angle $A$ of rhombus $A B C D$ is $45^{\circ}$, the projection of side $A B$ on side $A D$ is 12. Find the distance from the center of the rhombus to side $C D$. | Use the theorem of the midline of a triangle.
## Solution
Let $M$ be the projection of vertex $B$ of the rhombus $A B C D$ onto side $A D$. According to the problem, $A M=12$. The acute angle at vertex $A$ of the right triangle $A M B$ is $45^{\circ}$, so $B M=A M$.
Let $P$ and $Q$ be the projections of the center $... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 | |
The perimeter of the rhombus is 48, and the sum of the diagonals is 26. Find the area of the rhombus. | Let \( x \) and \( y \) be the halves of the diagonals of the rhombus and form a system of equations with respect to \( x \) and \( y \).
## Solution
Let \( x \) and \( y \) be the halves of the diagonals of the rhombus. Since the side of the rhombus is \( \frac{48}{4}=12 \), then
\[
\left\{\begin{array}{l}
x+y=13 \... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Zaslavsky A.A.
In triangle $A B C$, the bisectors $B B_{1}$ and $C C_{1}$ are drawn. It is known that the center of the circumcircle of triangle $B B_{1} C_{1}$ lies on the line $A C$. Find the angle $C$ of the triangle. | Continuing the ray $B C$ to intersect the circumcircle of triangle $B B_{1} C_{1}$, we obtain point $K$ (see the figure). The inscribed angles $\angle C_{1} B B_{1}$ and $\angle K B B_{1}$ are equal (since $B B_{1}$ is the angle bisector), which means the arcs they subtend, $B_{1} C_{1}$ and $B_{1} K$, are equal. Point... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Inside triangle $A B C$, there are points $P$ and $Q$ such that point $P$ is at distances 6, 7, and 12 from the lines $A B, B C, C A$ respectively, and point $Q$ is at distances 10, 9, and 4 from the lines $A B, B C, C A$ respectively. Find the radius of the inscribed circle of triangle $A B C$. | Prove that the midpoint of segment $P Q$ is the center of the inscribed circle of triangle $A B C$.
## Solution
Let $D, E, F$ be the feet of the perpendiculars dropped from points $P, Q, O$ to line $A B$ respectively. By the given condition, $P D=6, Q E=10$. $O F$ is the midline of trapezoid $D P Q E$, so the length ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Construction on a projection drawing ]
The base of a right prism $A B C A 1 B 1 C 1$ is a right triangle $A B C\left(\angle B=90^{\circ}, A B=B C=10\right)$; $A A 1=B B 1=C C 1=12$. Point $M$ is the midpoint of the lateral edge $A A 1$. A plane is drawn through points $M$ and $B 1$, forming an angle of $45^{\circ}$ ... | Let the lines $M E$ and $A 1 C 1$ intersect at point $P$. Then the plane of the section intersects the plane of the base $A 1 B 1 C 1$ along the line $B 1 P$ (Fig.1). Suppose that point $P$ lies on the extension of the edge $A 1 C 1$ beyond point $A 1$. Drop the perpendicular $A 1 K$ from point $A 1$ to the line $B 1 P... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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