problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
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Berlov S.L.
Petya is coloring 2006 points, located on a circle, in 17 colors. Then Kolya draws chords with endpoints at the marked points such that the endpoints of any chord are of the same color and the chords do not have any common points (including common endpoints). Kolya wants to draw as many chords as possible,... | Note that $2006=17 \cdot 118$; therefore, there will be two colors in which at least $2 \cdot 118=236$ points are painted in total.
We will prove by induction on $k$ that through $2 k-1$ points of two colors, it is always possible to draw $k-1$ non-intersecting chords with same-colored endpoints.
Base case $(k=1,2)$ ... | 117 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A thousand points are the vertices of a convex thousand-sided polygon, inside which there are another five hundred points such that no three of the five hundred lie on the same line. This thousand-sided polygon is cut into triangles in such a way that all the specified 1500 points are vertices of the triangles and thes... | Answer: 1998. The sum of all angles of the obtained triangles is equal to the sum of the angles of a 1000-gon and 500 angles of $360^{\circ}$, corresponding to 500 internal points. Therefore, the sum of the angles of the triangles is $998 \cdot 180^{\circ} + 500 \cdot 360^{\circ}$ $=(998 + 2 \cdot 500) 180^{\circ}$. He... | 1998 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Vasiliev N.B. }}$
How many maximum parts can the coordinate plane xOy be divided into by the graphs of 100 quadratic trinomials of the form
$y=a_{n} x^{2}+b_{n} x+c_{n}(n=1,2, \ldots, 100) ?$ | We will prove by induction that $n$ parabolas of the specified type can divide the plane into no more than $n^{2}+1$ parts.
Base case. One parabola divides the plane into $2=1^{2}+1$ parts.
Inductive step. The $n$-th parabola intersects each of the others at no more than two points. These intersection points, of whic... | 10001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Systems of points and segments. Examples and counterexamples ]
$$
\text { [ Combinatorial geometry (miscellaneous). ] }
$$
The grandfather called his grandson to visit him in the village:
- You'll see what an extraordinary garden I've planted! There are four pears growing there, and there are also apple trees, pla... | Various arrangements of apple and pear trees are possible, for example, as shown in the figure on the left. The maximum number of apple trees can be placed if the pears grow densely enough. For instance, if pears are planted in a row every 5 meters, there will be room for 12 apple trees (figure in the center).
. Since by the condition $\angle BXC = \angle AYB$, $\angle ABX = \angle CAY$, then $\angle BAX = \angle YCA$, which means that tri... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Place as many points as possible on a plane so that any three points do not lie on the same line and are vertices of an isosceles triangle.
# | The maximum number of points is six: a regular pentagon and its center (see figure).
 | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Tiling with domino bones and tiles]
Giuseppe has a sheet of plywood, measuring $22 \times 15$. Giuseppe wants to cut out as many rectangular blanks of size $3 \times 5$ as possible from it. How can he do this?
# | Note that more than 22 blanks cannot be obtained. Why?
## Solution
First of all, note that Giuseppe will not be able to get more blanks than \((22 \times 15) / (3 \times 5) = 22\) pieces. Now let's proceed to cutting.
Cut the sheet into three strips across the side of 22: \(5 \times 15\), \(5 \times 15\), and \(12 \... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
How to cut a square into unit squares a) $4 \times 4$ b) $5 \times 5$ with the fewest number of cuts. (Parts can be stacked on top of each other during cutting).
# | a) 4 cuts; b) 6 cuts. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Polygons (inequalities) $]$
A pentagon is inscribed in a circle of radius 1. Prove that the sum of the lengths of its sides and diagonals is less than 17.
# | It is possible to evaluate the sum of the sides and the sum of the diagonals of a pentagon separately.
## Solution
Since the length of each diagonal of the pentagon is no more than the diameter of the circle, the sum of the lengths of the diagonals is no more than $5 \cdot 2=10$. The sum of the lengths of the sides o... | 17 | Inequalities | proof | Yes | Yes | olympiads | false |
[ Geometry (miscellaneous).]
A sphere of radius $\sqrt{5}$ with center at point $O$ touches all sides of triangle $ABC$. The point of tangency $N$ bisects side $AB$. The point of tangency $M$ divides side $AC$ such that $AM=\frac{1}{2} MC$. Find the volume of the pyramid $OABC$, given that $AN=NB=1$. | The radius of the circle inscribed in a triangle is equal to the area of the triangle divided by its semi-perimeter.
## Solution
Let the given sphere touch the side $BC$ of triangle $ABC$ at point $K$. Then
$$
BK = BN = 1, AM = AN = 1, CM = 2 \cdot AM = 2, CK = CM = 2
$$
The section of the sphere by the plane of tr... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Geometry (other) ]
A sphere with radius $3 / 2$ has its center at point $N$. From point $K$, located at a distance of $3 \sqrt{5} / 2$ from the center of the sphere, two lines $K L$ and $K M$ are drawn, touching the sphere at points $L$ and $M$ respectively. Find the volume of the pyramid $K L M N$, given that $M L=... | From the right triangle $K L N$, we find that
$$
K L=\sqrt{K N^{2}-L N^{2}}=\sqrt{45 / 4-9 / 4}=3
$$
Therefore, $K M=K L=3$. In the isosceles triangles $L N M$ and $L K M$, the medians $N P$ and $K P$ are altitudes, so
$$
\begin{gathered}
N P^{2}=\sqrt{N L^{2}-L P^{2}}=\sqrt{9 / 4-1}=\sqrt{5} / 2 \\
K P^{2}=\sqrt{K ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Plane, divided by lines ]
How many maximum parts can 5 segments divide a plane into?
# | Each subsequent segment intersects with p already drawn segments in no more than p points.
## Solution
Let $\mathrm{n}$ segments have already been drawn, and they divide the plane into K(n) parts. Draw the ( $\mathrm{n}+1$ )-th segment. It intersects with the n already drawn segments in no more than n points, which c... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Law of Cosines ]
Find the sum of the squares of the distances from a point $M$, taken on the diameter of a certain circle, to the ends of any chord parallel to this diameter, if the radius of the circle is $R$, and the distance from the point $M$ to the center of the circle is $a$. | Apply the Law of Cosines.
## Solution
Let $O$ be the center of the circle. $AB$ is an arbitrary chord parallel to the given diameter. Denote $\angle BOM = \varphi$. Then $\angle AOM = 180^\circ - \varphi$.
By the Law of Cosines in triangles $BOM$ and $AOM$, we find that
\[
\begin{gathered}
BM^2 = a^2 + R^2 - 2aR \c... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ The product of the lengths of the chord segments and the lengths of the secant segments ]
On a line, points $A, B, C$, and $D$ are arranged in the given order. It is known that $B C = 3, A B = 2 \cdot C D$. A circle is drawn through points $A$ and $C$, and another circle is drawn through points $B$ and $D$. Their co... | Apply the theorem of intersecting chords twice.
## Solution
Let $C D=a, A B=2 a$. Denote $B K=x$. Then
$$
K C=3-x, K D=D C+K C=a+3-x, A K=A B+B K=2 a+x
$$
Let $M N$ be the common chord of the specified circles. Then
$$
A K \cdot K C=M K \cdot N K=B K \cdot K D
$$
from which
From this equation, we find that $x=2$... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Folklore
In an acute-angled triangle $A B C$, the bisector $A N$, the altitude $B H$, and the line perpendicular to side $A B$ and passing through its midpoint intersect at one point. Find the angle $BAC$.
# | Let $P$ be the intersection point of $A N$ and $B H$ (see the figure). Since the perpendicular bisector of side $A B$ passes through point $P$, we have $P A = P B$, which means triangle $A P B$ is isosceles. Therefore, $\angle A B P = \angle B A P = \angle P A H$, since $A P$ is the angle bisector of $\angle B A H$. Mo... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On sides $A B$ and $B C$ of an equilateral triangle $A B C$, points $D$ and $K$ are marked, respectively, and on side $A C$, points $E$ and $M$ are marked such that $D A + A E = K C + C M = A B$. Segments $D M$ and $K E$ intersect. Find the angle between them. | Consider triangles $A D M$ and $C E K$ (see the figure): $D A = A B - A E = A C - A E = C E$; $A M = A C - C M = A B - C M = K C$; $\angle D A M = 60^{\circ} = \angle E C K$. Therefore, these triangles are congruent (by two sides and the included angle). Then $\angle A M D = \angle C K E$.
?
# | Answer: 1304. Let $A$ be one of the selected points, $B$ and $C$ be the selected points, which are at distances 1 and 2 from it, respectively. The arrangement in the order $A B C$ is impossible, since in this case, for point $B$, there are two selected points at a distance of 1. Therefore, the points are arranged in th... | 1304 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Regular polygons ] [ Continuity considerations ]
A regular 100-gon is placed on a table, with the numbers $1,2, \ldots, 100$ written at its vertices. Then these numbers are rewritten in the order of their distance from the front edge of the table. If two vertices are at the same distance from the edge, the left numb... | Due to the symmetry of the regular 100-gon, each number appears in the sets at the 13th position the same number of times, which means the number of sets is a multiple of 100. If the regular 100-gon is continuously rotated counterclockwise around its center, then, first, all sets will appear, and second, the change of ... | 10100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
A circle of radius $1+\sqrt{2}$ is circumscribed around an isosceles right triangle. Find the radius of the circle that touches the legs of this triangle and internally touches the circle circumscribed around it. | Let $M$ and $N$ be the points of tangency of the desired circle with the legs $A C$ and $B C$ of triangle $A B C$, and $Q$ be the center of this circle. Then the quadrilateral $Q M C N$ is a square.
## Solution
Let $r$ be the radius of the desired circle, $Q$ its center, $M, N$ the points of tangency with the legs $A... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Common tangent to two circles [ Trigonometric ratios in a right triangle ]
On the plane, there are two circles with radii 12 and 7, centered at points $O_{1}$ and $O_{2}$, touching a certain line at points $M_{1}$ and $M_{2}$ and lying on the same side of this line. The ratio of the length of the segment $M_{1} M_{2... | Consider the right triangle $O_{1} P O_{2}$, where $P$ is the projection of point $O_{2}$ onto $O_{1} M_{1}$.
## Solution
Let $P$ be the projection of point $O_{2}$ onto the line $O_{1} M_{1}$. Denote $\angle O_{1} O_{2} P=\alpha$. Then
$$
\begin{gathered}
\cos \alpha=\frac{O_{2} P}{O_{1} O_{2}}=\frac{M_{1} M_{2}}{O... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}{[\text { Equilateral (regular) triangle }]} \\ {[\quad \text { Area of a circle, sector, and segment }}\end{array}\right]$
In an equilateral triangle $ABC$, a circle is drawn passing through the center of the triangle and touching side $BC$ at its midpoint $D$. A line is drawn from point $A$, t... | Find $\sin \angle D A E$ from the right triangle $E A Q(Q$ - the center of the given circle).
## Solution
Let $O$ be the center of triangle $A B C, Q$ be the center of the given circle. Denote the side of triangle $A B C$ by $a$. Then
$$
O D=\frac{1}{3} A D=\frac{a \sqrt{3}}{6}, O Q=Q D=\frac{a \sqrt{3}}{12}
$$
$$
... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## In a convex quadrilateral $A B C D$, the length of the segment connecting the midpoints of sides $A B$ and $C D$ is 1. Lines
$B C$ and $A D$ are perpendicular. Find the length of the segment connecting the midpoints of diagonals $A C$ and $B D$. | Prove that the midpoints of sides $AB$ and $CD$ and the midpoints of diagonals $AC$ and $BD$ are the vertices of a rectangle.
## Solution
Let $M$ and $N$ be the midpoints of sides $AB$ and $CD$ of quadrilateral $ABCD$, and $P$ and $Q$ be the midpoints of its diagonals $AC$ and $BD$, respectively. Then $MP$ is the mid... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In an isosceles trapezoid, the bases are 40 and 24, and its diagonals are perpendicular to each other. Find the area of the trapezoid.
# | Through the vertex of the trapezoid, draw a line parallel to the diagonal.
## Solution
Let the bases $AD$ and $BC$ of the isosceles trapezoid $ABCD$ be 40 and 24, respectively, and the diagonals $AC$ and $BD$ are perpendicular. Draw a line through vertex $C$ parallel to diagonal $BD$. Let this line intersect the exte... | 1024 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 | |
The height $B K$ of the rhombus $A B C D$, dropped to the side $A D$, intersects the diagonal $A C$ at point $M$. Find $M D$, if it is known that $B K=4, A K: K D=1: 2$. | 
## Solution
Since $AC$ is the perpendicular bisector of diagonal $BD$, then $MD = MB$. Since $AM$ is the angle bisector of triangle $BAK$, then
$$
\frac{BM}{MK} = \frac{AB}{AK} = \frac{AD}{AK... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Auxiliary area. The area helps to solve the problem_] Heron's formula
The sides of the triangle are 13, 14, and 15. Find the radius of the circle that has its center on the middle side and touches the other two sides. | Let the center $O$ of the given circle be located on the side $AB$ of triangle $ABC$. The area of triangle $ABC$ is equal to the sum of the areas of triangles $AOC$ and $BOC$.
## Solution
Let $AC=13$, $AB=14$, $BC=15$, $O$ be the center of the given circle (with $O$ on side $AB$), $R$ be its radius, and $P$ and $Q$ b... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In parallelogram $A B C D$, the longer side $A D$ is equal to 5. The angle bisectors of angles $A$ and $B$ intersect at point $M$.
Find the area of the parallelogram if $B M=2$, and $\cos \angle B A M=\frac{4}{5}$. | Prove that triangle $A M B$ is a right triangle.
## Solution
Since $\angle B A D+\angle A B C=180^{\circ}$, triangle $A M B$ is a right triangle. Therefore,
$$
A B=\frac{B M}{\sin \angle B A M}=2 \cdot \frac{5}{3}=\frac{10}{3}
$$
Using the formula $\sin 2 \boldsymbol{\alpha}=2 \sin \boldsymbol{\alpha} \cos \boldsym... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Theorem of the tangent and secant; the product of the entire secant and its external part [Pythagorean Theorem (direct and inverse).
One of the two lines passing through point $M$ touches the circle at point $C$, while the second intersects this circle at points $A$ and $B$, with $A$ being the midpoint of segment $B ... | Prove that triangle $B M C$ is a right triangle.
## Solution
Let $A M=A B=x$. By the tangent-secant theorem, $B M \cdot A M=M C^{2}$, or $2 x^{2}=4$, from which $x=\sqrt{2}$.
In triangle $B M C$, the sides $M C=2, B M=2 x=2 \sqrt{2}$ and the angle between them $\angle B M C=45^{\circ}$ are known. Consider a right tr... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[
doubling the median ] [Theorem of the sum of the squares of the diagonals ]
In a triangle, two sides are equal to 11 and 23, and the median drawn to the third side is 10. Find the third side.
# | Use the theorem about the sum of the squares of the diagonals of a parallelogram or the formula for the median of a triangle.
## Solution
First method.
Let $C M$ be the median of triangle $A B C$, where $A C=11, B C=23, C M=10$. Then
$$
C M^{2}=\frac{1}{4}\left(2 A C^{2}+2 B C^{2}-A B^{2}\right) \text {, or } 100=\... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kovaldji A.K.
Vladimir is running on a circular track at a constant speed. There are two photographers standing at two points on the track. After the start, Vladimir was closer to the first photographer for 2 minutes, then closer to the second photographer for 3 minutes, and then closer to the first photographer again... | Let's mark the first and second photographers on the circle with points $A$ and $B$ respectively, and points $C$ and $D$ will denote the midpoints of the arcs connecting $A$ and $B$. Then on the semicircle $C A D$, Vasya is closer to the first photographer, and on the semicircle $C B D$ (marked in bold on the diagram) ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Shaovalov A.B.
Inside a rectangular grid with a perimeter of 50 cells, a rectangular hole with a perimeter of 32 cells is cut out along the cell boundaries (the hole does not contain any boundary cells). If this figure is cut along all horizontal grid lines, 20 strips 1 cell wide will be obtained. How many strips will... | The first solution: Let the rectangle occupy $a$ cells vertically and $b$ cells horizontally, with $a + b = 50 / 2 = 25$. Similarly, let the dimensions of the hole be $x$ cells vertically and $y$ cells horizontally, with $x + y = 32 / 2 = 16$.
Fig. 1
 is $\sqrt{3}$ times the length of the base edge. Point $E$ is the midpoint of the apothem lying in the face $A S B$. Find the angle between the line $D E$ and the plane $A S C$.
# | Let $S H$ be the height of the pyramid, $M$ and $N$ be the midpoints of edges $A B$ and $S A$ respectively (Fig.1). Let $A B=a, S H=a \sqrt{3}$. On the extension of edge $C D$ beyond point $D$, lay off the segment $D P=\frac{1}{4} C D=\frac{1}{4} a$. Since $E N$ is the midline of triangle $A S M$, then $N E\|A M\| D P$... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The base of the right prism $A B C D A_{1} B_{1} C_{1} D_{1}$ is an isosceles trapezoid $A B C D$, where $B C \| A D, B C=5$, $A D=10, \angle B A D=\operatorname{arctg} 2$. A plane, perpendicular to the line $A_{1} D$, intersects the edges $A D$ and $A_{1} D_{1}$ at points $M$ and $N$ respectively, and $M D=A_{1} N=1$.... | 31; cross-section - hexagon.
Send comment | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9
In a sphere of radius 9, three equal chords $A A 1, B B 1$, and $C C 1$ are drawn through a point $S$ such that $A S=4, A 1 S=8, B S < B 1 S, C S < C 1 S$. Find the radius of the sphere circumscribed about the pyramid $S A B C$. | The products of the segments of intersecting chords are equal, so $B S \cdot S B 1=A S \cdot S A 1=4 \cdot 8=32$. In addition, $B B 1=A A 1=A S+S A 1=4+8=12$. From the system
$$
\left\{\begin{array}{l}
B S \cdot S B_{1}=32 \\
B S+S B_{1}=12
\end{array}\right.
$$
and the condition $B S<B 1 S$, we find that $B S=4$ and... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the pyramid $A B C D$, the dihedral angle at edge $A C$ is $90^{\circ}, A B=B C=C D, B D=A C$. Find the dihedral angle at edge $A D$.
# | Since the planes $A B C$ and $A C D$ are perpendicular (the dihedral angle between them is $90^{\circ}$), the height $B M$ of triangle $A B C$ is perpendicular to the plane $A C D$, meaning that triangle $A M D$ is the orthogonal projection of triangle $A B D$ onto the plane $A C D$. Since triangle $A B C$ is isosceles... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 [Radii of the inscribed, circumscribed, and exscribed circles (other).]
Let $O$ be the center of the circle circumscribed around triangle $A B C, \angle A O C=60^{\circ}$. Find the angle $A M C$, where $M-$ is the center of the circle inscribed in triangle $A B C$. | If points $O$ and $B$ lie on opposite sides of line $A C$ (Fig.1), then the degree measure of arc $A C$, not containing point $B$, is $360^{\circ}-60^{\circ}=300^{\circ}$, therefore
$$
\angle A B C=\frac{1}{2} \cdot 300^{\circ}=150^{\circ}
$$
The sum of the angles at vertices $A$ and $C$ of triangle $A B C$ is $180^{... | 165 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Right triangle with an angle in ]
The hypotenuse $AB$ of the right triangle $ABC$ is 2 and is a chord of a certain circle. The leg $AC$ is 1 and lies inside the circle, and its extension intersects the circle at point $D$, with $CD=3$. Find the radius of the circle. | The leg $A C$ is equal to half the hypotenuse $A B$, so $\angle A=60^{\circ}$. In addition, $A D=2 A B$, so the angle $A B D-$ is a right angle. Therefore, $A D$ is the diameter of the circle.
## Answer
2. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
\[
\begin{aligned} & {\left[\begin{array}{l}\text { Angle between the tangent and the chord }\end{array}\right]} \\ & \text { [Inscribed angle is half of the central angle]}\end{aligned}
\]
Secant $A B C$ intercepts arc $B C$, which contains $112^{\circ}$; tangent $A D$ at point of tangency $D$ divides this arc in the... | Consider the exterior angle of triangle $A D C$, adjacent to angle $A D C$.
## Solution
Let $K$ be a point lying on the extension of the tangent $A D$ beyond point $D$. The angle $K D C$ is an exterior angle for triangle DAC. Therefore,
$\angle A=\angle K D C-\angle D C B=1 / 2-C D-1 / 2-D B=1 / 2\left(9 / 16-{ }^{7... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Inscribed Quadrilaterals [ Angles subtending equal arcs and equal chords $]$
In a convex quadrilateral $A B C D$, it is given that $\angle A B C=116^{\circ}, \angle A D C=64^{\circ}, \angle C A B=35^{\circ}$ and $\angle C A D=52^{\circ}$. Find the angle between the diagonals subtending side $A B$.
# | Let the diagonals of the given quadrilateral intersect at point $M$. Since
$$
\angle A B C+\angle A D C=116^{\circ}+64^{\circ}=180^{\circ},
$$
a circle can be circumscribed around the given quadrilateral. The desired angle $A M B$ is the exterior angle of triangle $A M D$. Therefore, it is equal to
$$
\begin{gathere... | 81 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 [Pythagorean Theorem (direct and inverse)]
Three circles of different radii touch each other externally. The segments connecting their centers form a right triangle. Find the radius of the smallest circle if the radii of the largest and medium circles are 6 and 4.
# | Let $x$ be the radius of the smaller circle. The sides of the resulting triangle are $10, 6+x$, and $4+x$. Since 10 is the largest side, it is the hypotenuse. Therefore, $(x+6)^{2}+(x+4)^{2}=100$, from which $x=2$.
 is 9, and the area is $9 \cdot 8=72$
## Answer | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
\left.\begin{array}{l}{[\text { Angles subtending equal arcs and equal chords }]} \\ {[\text { Inscribed angle subtending the diameter }]\end{array}\right]
A circle passes through vertices $A$ and $C$ of triangle $ABC$, intersecting side $AB$ at point $E$ and side $BC$ at point $F$. Angle $AEC$ is five times the angle... | Let $\angle B A F=\alpha$. Then $\angle A E C=5 \alpha$. Inscribed angles $E C F$ and $E A F$ subtend the same arc, so $\angle B C E=\angle E C F=\angle E A F=\angle B A F=\alpha$.
By the exterior angle theorem of a triangle, $\angle A B C+\angle B C E=\angle A E C$, or $72^{\circ}+\alpha=5 \alpha$, from which we find... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two parallel lines are intersected by a third. Find the angle between the bisectors of the internal same-side angles.
# | Use the theorem about the sum of the angles of a triangle.
## Solution
Let lines $l$ and $m$ be parallel, and a third line intersects them at points $A$ and $B$ respectively. Take point $C$ on line $l$ and point $D$ on line $m$ such that these points lie on the same side of line $A B$. Then angles $B A C$ and $A B D$... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 | |
Find the radius of the smallest circle in which a triangle with sides 7, 9, and 12 can be placed. | The given triangle is obtuse.
## Solution
$7^{2}+9^{2}<127^{2}$, therefore the given triangle is obtuse. Consequently, the circle constructed with the largest side of the triangle as its diameter contains the triangle.
## Answer
6. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Linear element relationships of similar triangles] [Rectangles and squares. Properties and characteristics]
In a triangle with a base of 30 and a height of 10, an isosceles right triangle is inscribed such that its hypotenuse is parallel to the base of the given triangle, and the vertex of the right angle lies on thi... | Use the equality of the ratios of corresponding heights to the bases in similar triangles.
## Solution
Let the vertices $M$ and $N$ of the hypotenuse lie on the sides $A C$ and $B C$ of triangle $A B C$, respectively, and the vertex of the right angle $K$ lies on the base $A B$. Denote by $x$ the height of triangle $... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the plane, there are 16 points (see the figure).
a) Show that it is possible to erase no more than eight of them so that no four of the remaining points lie at the vertices of a square.
b) Show that it is possible to manage with erasing six points.
c) Find the minimum number of points that need to be erased for t... | b) All these points form 20 squares: 9 with side 1, 4 with side 2, 1 with side 3, 4 with side $\sqrt{2}$, and 2 with side $\sqrt{5}$. Let's denote the points with letters (Fig. 1) and count how many squares each point is a vertex of (Fig. 2).
| $A$ | $B$ | $C$ | $D$ |
| :--- | :--- | :--- | :--- |
| $E$ | $F$ | $G$ | ... | 6 | Combinatorics | proof | Yes | Yes | olympiads | false |
Chekanov Yu.V.
A rectangle of size $1 \times k$ for any natural $k$ will be called a strip. For which natural $n$ can a rectangle of size $1995 \times n$ be cut into pairwise distinct strips? | Idea of the solution: we take the maximum strip (equal to the maximum side of the rectangle). The remaining strips will be combined in pairs, giving the sum of the maximum strip. If we have filled the rectangle, the problem is solved; otherwise, reasoning with areas shows that the rectangle cannot be cut into different... | 3989 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[The ratio of the areas of triangles with a common base or common height] Complex
A quadrilateral is divided into four triangles by its diagonals. The areas of three of them are 10, 20, and 30, and each is less than the area of the fourth triangle. Find the area of the given quadrilateral.
# | Let $M$ be the point of intersection of the diagonals $AC$ and $BD$ of quadrilateral $ABCD$. If
$$
S_{DAMD}=30, S_{DAMB}=10, S_{DCMD}=20,
$$
then
$$
\begin{gathered}
\frac{BM}{MD}=\frac{S_{\triangle AMB}}{S_{\triangle AMD}}=\frac{1}{3} \\
S_{DBMC}=\frac{1}{3} S_{DCMD}=\frac{20}{3}<10 .
\end{gathered}
$$
By examinin... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{l}\text { [ Homothety helps solve the problem. } \\ \text { [Two tangents drawn from one point] }\end{array}\right]$
On the sides $A C$ and $B C$ of triangle $A B C$, points $P$ and $Q$ are marked, respectively. It turned out that $A B=A P=B Q=1$, and the point of intersection of segments $A Q$ an... | Let the incircle of triangle $ABC$ touch its sides $AB, BC$, and $AC$ at points $S, R$, and $T$; $M$ is the intersection point of segments $AQ$ and $BP$. We can observe that point $P$ lies on segment $CT$, and point $Q$ lies on segment $CR$. Then segments $AM$ and $BM$ intersect the circle again. Let the points of inte... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { Angles between lines and planes } \\ {[\underline{\text { Linear dependence of vectors }}]}\end{array}\right]$
The lateral face of a regular quadrilateral pyramid forms an angle of $45^{\circ}$ with the plane of the base. Find the angle between the apothem of the pyramid and the plane of... | Let $A B C D P$ be a given regular quadrilateral pyramid with vertex $P ; A B=B C=C D=A D=a ; M-$ the center of the square $A B C D ; K, L, N$ and $E$ - the midpoints of segments $A B, B C, C D$ and $A D$ respectively. Since $P K \perp A B$ and $M K \perp A B$, the angle $P K M$ is the linear angle of the dihedral angl... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Coordinate method in space ] [Area and orthogonal projection]
The areas of the projections of a certain triangle onto the coordinate planes Oxy and Oyz are $\sqrt{6}$ and $\sqrt{7}$, respectively, and the area of the projection onto the plane $O x z$ is an integer. Find the area of the triangle itself, given that it... | Let the vector perpendicular to the plane of the original triangle form angles $\alpha, \beta$, and $\gamma$ with the coordinate axes $O x, O y$, and $O z$ respectively. Then
$$
\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=1
$$
Let the area of the original triangle be denoted by $S$, and the areas of the projections on t... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a triangle, the angle bisectors $A L$ and $B M$ are drawn. It is known that one of the intersection points of the circumcircles of triangles $A C L$ and $B C M$ lies on the segment $A B$. Prove that $\angle A C B=60^{\circ}$. | Let $\angle C A B=2 \alpha, \angle A B C=2 \beta, \angle A C B=2 \gamma$. Let $K$ be the intersection point of the circles mentioned in the condition, lying on side $A B$. Since quadrilaterals $A K L C$ and $B K M C$ are cyclic,
$$
\angle A K M=\angle B C M=2 \gamma, \angle B K L=\angle A C L=\gamma .
$$
By the inscr... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
In a triangular pyramid PABC, the sums of the three plane angles at each of the vertices A and B are both $180^{\circ}$ and $PC = AB$. Inside the pyramid, a point $D$ is taken, the sum of the distances from which to the three lateral faces PAB, PAC, and PBC is 7. Find the distance from the center of the circumscribed s... | Consider the unfolding P1AP2CP3BP1 of the tetrahedron ABCD onto the plane of triangle ABC (Fig.1), where points P1, P2, and P3 are the vertices of triangles with bases AB, AC, and BC, respectively. Since the sums of the three plane angles at each of the vertices A and B of the tetrahedron PABC are $180^{\circ}$, point ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Through each face of the cube, a plane was drawn. Into how many parts will these planes divide the space? | The planes divide the space into 27 parts.
Indeed, let's first draw two planes through opposite faces. They will divide the space into three "layers". Now, let's draw the remaining four planes. They will cut out nine sections in each layer. (It might be helpful here to look at what happens if you draw lines through th... | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Coverings ]
It is known that a set $M$ of points on a line can be covered by three segments of length 1.
What is the smallest number of segments of length 1 that can certainly cover the set of midpoints of segments with endpoints in the points of the set $M$?
# | If point $A$ is covered by the segment $[a, a+1]$, and point $B$ by the segment $[b, b+1]$, then the midpoint of segment $A B$ is covered by the segment $\left[a+b / 2, a+b / 2+1\right]$.
## Solution
Let two points $A$ and $B$ be covered by one unit segment. Then the midpoint of segment $A B$ is also covered by this ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Two pairs of similar triangles]
Through a point taken inside an arbitrary triangle, segments are drawn parallel to its sides with endpoints on the sides of the triangle.
Prove that the sum of the three ratios of these segments to the sides of the triangle parallel to them is equal to 2. | Consider similar triangles.
## Solution
Let $O$ be an arbitrary point taken inside triangle $A B C ; A_{1} A_{2} \| B C, B_{1} B_{2} \| A C, C_{1} C_{2} \| A B$ (see figure).
Denote $A_{2} C = O B_{2} = x, A_{2} C_{1} = y$,
$A C_{1} = O B_{1} = z$. Then $\frac{C_{1} C_{2}}{A B} = \frac{x + y}{A C}, \frac{A_{1} A_{2... | 2 | Geometry | proof | Yes | Yes | olympiads | false |
[ Measurement of segment lengths and angles. Adjacent angles.]
How many times in a day do the hour and minute hands coincide? Form a straight angle? Form a right angle? | In a day, the minute hand makes 24 revolutions, while the hour hand makes 2. Therefore, the minute hand overtakes the hour hand 22 times. At the moment of overtaking, they coincide. (In this case, one midnight is counted, but the second one is not.)
The moments of coincidence of the hands divide the day into 22 equal ... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Product of the lengths of the chord segments and the lengths of the secant segments ]
A line passing through point $A$, located outside the circle at a distance of 7 from its center, intersects the circle at points $B$ and $C$. Find the radius of the circle, given that $AB=3$, $BC=5$.
# | The product of the entire secant and its external part for a given point and a given circle is constant.
## Solution
Since $A B<B C$, point $B$ lies between points $A$ and $C$. Let $O$ be the center of the circle, $R$ its radius, and the line $A O$ intersects the circle at points $D$ and $E$ (with $D$ between $A$ and... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Systems of points and segments (other) ] [ Classical combinatorics (other) ]
On a circle, 20 points are marked. How many triples of chords with endpoints at these points exist such that each chord intersects the other two (possibly at the endpoints)? | The ends of the chords in a triplet can be 3, 4, 5, or 6 points. Let's consider these cases. 1) The ends of the chords are 3 points (see fig.). They can be chosen in $C_{20}^{3}$ ways. Each triplet of points can be connected by chords in a unique way.
. ]
On a plane, 10 equal segments were drawn, and all their points of intersection were marked. It turned out that each point of intersection divides any segment passing through it in the ratio $3: 4$. What is the... | On each segment, there are no more than two points. On the other hand, each intersection point belongs to at least two segments. Therefore, there are no more than $10 \cdot 2: 10=10$ points. An example with 10 points is shown in the figure.
. When the spy is at point $A$, he counts the poles on one side of $AB$, and when he is at point $B$, he counts the poles on the other side of... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
**Zaslavsky A.**.
A sphere is circumscribed around a regular tetrahedron $ABCD$. On its faces, as bases, regular pyramids $ABCD', ABD'C, ACD'B, BCD'A$ are constructed outwardly, with their vertices lying on this sphere. Find the angle between the planes $ABC_1$ and $ACD'$. | Consider a cube, vertices $A, B, C$ and $D$ of which coincide with the vertices of a tetrahedron (see figure). Since the sphere circumscribed around the cube is also circumscribed around the tetrahedron, $A^{\prime}, B^{\prime}, C^{\prime}$ and $D^{\prime}$ are the remaining four vertices of this cube, and the specifie... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kanel-Belov A.Y.
A cube with a side of 20 is divided into 8000 unit cubes, and a number is written in each cube. It is known that in each column of 20 cubes, parallel to the edge of the cube, the sum of the numbers is 1 (columns in all three directions are considered). In some cube, the number 10 is written. Through t... | Through the given cube K, one horizontal layer G and two vertical layers pass. The sum of all numbers in 361 vertical columns, not included in the last two layers, is 361. From this sum, we need to subtract the sum $S$ of the numbers lying in the cubes at the intersection of these columns with G (there are 361 such cub... | 333 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shaovalov A.V.
Among the angles of each lateral face of a pentagonal prism, there is an angle $\varphi$. Find all possible values of $\varphi$. | In a right prism, each lateral face is a rectangle, so $\varphi=90^{\circ}$ is suitable.
Suppose $\varphi \neq 90^{\circ}$. We can assume that $\varphi<90^{\circ}$ (if the obtuse angles in the parallelograms are equal, then the acute angles are also equal). Draw in the plane of the base through one of the vertices $V$... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Shaovalov A.v.
Two players take turns coloring the sides of an $n$-gon. The first player can color a side that borders with zero or two colored sides, the second player - a side that borders with one colored side. The player who cannot make a move loses. For which $n$ can the second player win, regardless of how the f... | For $n=3$, it is obvious that the first player wins, while for $n=4$, the second player wins. Let's show how the first player wins when $n>4$. After the first move of the second player, two adjacent sides are painted. The first player can paint a side "one apart" from them, creating an unpainted "hole" of one side. Thi... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Similar auxiliary triangles $]$ [ Right triangle with an angle at ]
On the cathetus $A C$ of the right triangle $A B C$, a circle is constructed with $A C$ as its diameter. It intersects side $A B$ at point $E$. A point $G$ is taken on side $B C$ such that segment $A G$ intersects the circle at point $F$, and segment... | Triangles $A B C$ and $G A C$ are similar.
## Solution
Let $C G=t, \angle C A G=\alpha$. Then $B C=3 t$, and since $C E$ is the height, $\angle B=\angle A C E$. Since trapezoid $A E F C$ is inscribed in a circle, it is isosceles, so $\angle A C E=\alpha$. Therefore, $\angle B=\alpha$. Consequently, right triangles $A... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Doubling the Median ] [ Law of Cosines ]
Determine the angle $A$ between the sides of lengths 2 and 4, if the median drawn from vertex $A$ is $\sqrt{7}$. | On the extension of the median $A M$ of the given triangle, lay off the segment $M D$, equal to the segment $A M$.
## Solution
On the extension of the median $A M$ of the given triangle $A B C$ with sides $A B=2$ and $A C=4$, lay off the segment $M D$, equal to the segment $A M$. Then the quadrilateral $A B D C$ is a... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Method of Coordinates on the Plane [ Ratio of Areas of Triangles with a Common Base or Common Height]
On the coordinate plane, points $A(1 ; 3)$, $B(1 ; 9)$, $C(6 ; 8)$, and $E(5 ; 1)$ are given. Find the area of the pentagon $A B C D E$, where $D$ is the point of intersection of the lines $A C$ and $B E$. | If $y_{1} \neq y_{2}$ and $x_{1} \neq x_{2}$, then the equation of the line passing through the points ( $x_{1} ; y_{1}$ ) and ( $x_{2} ; y_{2}$ ) is
$$
\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}
$$
## Answer
21.00 | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[The ratio of the areas of triangles with a common base or common height $]$ Complex Trapezoids (other)
The area of trapezoid $A B C D$ with bases $A D$ and $B C (A D > B C)$ is 48, and the area of triangle $A O B$, where $O-$ is the point of intersection of the diagonals of the trapezoid, is 9. Find the ratio of the ... | $S_{\triangle \mathrm{AOD}}=\frac{D O}{O B} \cdot S_{\triangle \mathrm{AOB}}$
## Solution
Notice that triangles $A B D$ and $A C D$ are equal in area, as they share the same base and have equal heights. Therefore, triangles $C O D$ and $A O B$ are also equal in area.
Let $\frac{A D}{B C}=x>1$. From the similarity of... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Proizvolov V.V.
On the sides $AB$ and $BC$ of an equilateral triangle $ABC$, points $D$ and $K$ are taken, and on the side $AC$ - points $E$ and $M$, such that $DA + AE = KC + CM = AB$.
Prove that the angle between the lines $DM$ and $KE$ is $60^{\circ}$. | Prove the equality of triangles $M A D$ and $E C K$.
## Solution
Let the lines $D M$ and $K E$ intersect at point $P$. From the condition, it follows that $C E = A C - A E = A D$. Similarly, $C K = A M$. Therefore, triangles $M A D$ and $E C K$ are equal by two sides and the angle between them. Thus, $\angle M P E = ... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
What is the greatest number of acute angles that can occur in a convex polygon?
# | The sum of all exterior angles of a convex polygon is $360^{\circ}$. Therefore, a convex polygon cannot have more than three obtuse exterior angles, i.e., it cannot have more than three acute interior angles. Three acute angles can be found, for example, in a triangle. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The sides of the blue and green equilateral triangles are respectively parallel. The perimeter of the blue triangle is 4, and the perimeter of the green triangle is 5. Find the perimeter of the hexagon formed by the intersection of these triangles.
# | The sum of the perimeters of the blue and green triangles is equal to the sum of the perimeters of the six small triangles surrounding the hexagon.
## Solution
Notice that the sum of the perimeters of the blue and green triangles is equal to the sum of the perimeters of the six small equilateral triangles surrounding... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Law of Cosines]
In a triangle, the base is 12; one of the angles at the base is $120^{\circ}$; the side opposite this angle is 28.
Find the third side. | Apply the Law of Cosines.
## Solution
Let in triangle $ABC$ it is known that $\angle C=120^{\circ}, AC=12, AB=28$. Denote $BC=x$. By the Law of Cosines,
$$
AB^{2}=AC^{2}+BC^{2}-2 AC \cdot BC \cos \angle C
$$
or
$$
784=x^{2}+144+12 x, x^{2}+12 x-640=0
$$
The positive root of this equation is $x=20$
. Squares of $3 \times 3$ can be drawn 36, and so on. The total number of ... | 204 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ [geometry on graph paper ]
a) In how many ways can a rectangle $8 \times 2$ be divided into rectangles $1 \times 2$? b) Invent and describe a figure that can be cut into rectangles $1 \times 2$ in exactly 555 ways.
# | a) The number of ways is $f_{9}=34$, where $f_{n}$ is the $n$-th Fibonacci number. This statement follows from the following easily provable by induction lemma: a rectangle $\mathrm{n} \times 2$ can be divided into rectangles $1 \times 2$ in exactly $\mathrm{f}_{\mathrm{n}}$ ways.
## Answer
a) 34 ways. | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On graph paper, a square with a side of 5 cells is drawn. It needs to be divided into 5 parts of equal area by drawing segments inside the square only along the grid lines. Can it be such that the total length of the drawn segments does not exceed 16 cells? | ## Answer
Yes, it can. One possible example is shown in Figure 7.5 (the total length of the drawn segments is 16).
Problem | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A figure on a plane has exactly two axes of symmetry. Find the angle between these axes.
# | Sequential execution of two symmetries with respect to axes, the angle between which is $\varphi$, is a rotation by an angle of $2 \varphi$.
## Solution
Let the angle between the axes of symmetry be $\varphi$. By performing the symmetries sequentially with respect to these axes, we obtain a rotation by an angle of $2... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Two equal circles touch the inside of a third circle and touch each other. By connecting the three centers, a triangle with a perimeter of 18 is obtained. Find the radius of the larger circle. | The line connecting the centers of two tangent circles passes through their point of tangency.
## Solution
Let the radii of the given circles be $r, r$ and $R (r < R)$. Then the sides of the specified triangle are $R - r, R - r$, and $2r$. Therefore,
Consequently, $R = 9$.
 = 90^{\circ}$. Therefore, $\angle K M C = 90^{\circ} - \angle M K C = 90^{\ci... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Pythagorean Theorem (direct and inverse).]
Perpendiculars and two obliques are drawn from one point to a given straight line.
Find the length of the perpendicular, if the obliques are 41 and 50, and their projections on the given straight line are in the ratio $3: 10$ | Express the desired perpendicular using the Pythagorean theorem from two right triangles and solve the resulting equation.
## Solution
Let $AB=50$ and $AC=41$ be the obliques, $AD$ be the perpendicular, $BD=10x, CD=3x$. Then $50^{2}-(10x)^{2}=AD^{2}=$ $41^{2}-9x^{2}$, or $91x^{2}=91 \cdot 9$, from which $x=3$, $BD=30... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a right-angled triangle, the medians drawn from the vertices of the acute angles are equal to $\sqrt{52}$ and $\sqrt{73}$. Find the hypotenuse of the triangle. | Let the legs be denoted by $x$ and $y$. Then, by the Pythagorean theorem, $x^{2} + y^{2} / 4 = 52$, and $y^{2} + x^{2} / 4 = 73$.
Adding these equations, we get $5 / 4\left(x^{2} + y^{2}\right) = 125 \Rightarrow x^{2} + y^{2} = 100$.
 \cdot S_{A B C}=1,02 S_{A B C}$.
Answer
$102 \%$. | 102 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kustarev A.A.
The faces of a cube are numbered from 1 to 6. The cube was rolled twice. The first time, the sum of the numbers on the four side faces was 12, the second time it was 15. What number is written on the face opposite the one where the digit 3 is written? | The sum of the numbers on all faces of the cube is 21.
## Solution
Notice that the sum of all numbers written on the cube is 21. The sum of the numbers on the top and bottom faces in the first and second cases is 9 and 6, respectively.
After the first roll, it is clear that either 3 is opposite 6, or 4 is opposite 5... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.I.
Can the natural numbers from 1 to 81 be written in the cells of a $9 \times 9$ table so that the sum of the numbers in each $3 \times 3$ square is the same? | Table $T$, depicted in the left figure, contains each of the numbers $1,2, \ldots, 9$ nine times and has the property that the sum of the numbers in each $3 \times 3$ square is 36.
| 1 | 4 | 7 | 1 | 4 | 7 | 1 | 4 | 7 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 2 | 5 | 8 | 2 | 5 | 8 | 2 | 5 | 8... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Groovitz $\mathbf{B . M}$.
A regular polygon is drawn on the board. Vova wants to mark $k$ points on its perimeter so that there is no other regular polygon (not necessarily with the same number of sides) that also contains the marked points on its perimeter.
Find the smallest $k$ sufficient for any initial polygon.... | Example. Let's prove that five points are sufficient. Let $A, B, C, D$ be four consecutive vertices of the polygon (possibly $A=D$). Mark points $A, B$, an arbitrary point $X$ on side $AB$, point $Y$ on side $BC$, sufficiently close to $B$, and point $Z$ on side $CD$, sufficiently close to $C$.
Let $P$ be the original... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Through the vertices $A$ and $C$ of triangle $ABC$, a circle $K$ is drawn, the center of which lies on the circumcircle of triangle $ABC$. Circle $K$ intersects the extension of side $BA$ beyond point $A$ at point $M$.
Find the angle $C$, if $MA: AB=2: 5$, and $\angle B=\arcsin 3 / 5$.
# | Let $A N$ be the height of triangle $A B C$. Prove that triangle $A N C$ is isosceles.
## Solution
Let $Q$ be the center of circle $K, A B=5 x, A M=2 x, A N$ be the height of triangle $A B C$. Then $A N=A B \sin \angle B=3 x$, $B N=4 x$,
$\angle A M C=1 / 2 \angle A Q C=1 / 2\left(180^{\circ}-\angle B\right)=90^{\ci... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4 [
On a sphere of radius 11, points $A, A 1, B, B 1, C$ and $C 1$ are located. Lines $A A 1, B B 1$ and $C C 1$ are pairwise perpendicular and intersect at point $M$, which is at a distance of $\sqrt{59}$ from the center of the sphere. Find $A A 1$, given that $B B 1=18$, and point $M$ divides the segment $C C 1$ in ... | Let $C M=(8+\sqrt{2}) x, C 1 M=(8-\sqrt{2}) x$. Consider the section of the sphere by a plane passing through its center $O$ and the line $C C 1$. We obtain a circle of radius 11 with center $O$ and chord $C C 1$. Extend the segment $O M$ to intersect the circle at points $D$ and $E$. By the intersecting chords theorem... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ [sphere touching the edges of a tetrahedron]
A sphere touches the edges $A S, B S, B C$ and $A C$ of the triangular pyramid $S A B C$ at points $K, L, M$ and $N$ respectively.
Find the segment $K L$, if $M N=7, N K=5, L N=2 \sqrt{29}$ and $K L=L M$.
# | Let's prove that points $K, L, M$, and $N$ lie in the same plane (Fig.1). Denote
$$
A K=A N=a, B L=B M=b, C M=C N=c, S K=S L=d .
$$
If $c=d$, then lines $K N$ and $L M$ are parallel to line $S C$, so $K N \| L M$. Therefore, points $K, L, M$, and $N$ lie in the same plane. Let $c \neq d$. Therefore, points $K, L, M$,... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the quadrilateral pyramid $S A B C D$, the base $A B C D$ has its axis of symmetry as the diagonal $A C$, which is equal to 9, and the point $E$ of intersection of the diagonals of the quadrilateral $A B C D$ divides the segment $A C$ such that the segment $A E$ is smaller than the segment $E C$. A plane is drawn th... | From the condition, it follows that the diagonals $A C$ and $B D$ of the quadrilateral $A B C D$ are perpendicular, and the triangles $A B D$ and $C B D$ are isosceles. Since the truncated pyramid $A B C D A 1 B 1 C 1 D 1$ has 6 faces, the plane $\alpha$ intersects each face along a segment, the ends of which do not co... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
At the base of the pyramid lies an isosceles triangle $ABC$ with base $AC=2$ and lateral side $\sqrt{7}$. The face $ACD$ is perpendicular to the base plane and is an equilateral triangle. Find the edge $BD$, as well as the areas of all sections of the pyramid that are squares. | Let $DM$ be the height of the equilateral triangle $ADC$. Then $M$ is the midpoint of $AC$ and
$$
DM = AD \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
$$
Since the plane of the face $ADC$ is perpendicular to the plane of the base $ABC$, the line $DM$ is perpendicular to the plane of the base. Therefore, $DM$ is the height of ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4 [ Excircle ]
Let $A B C$ be an equilateral triangle. Three planes through the lines $A B, B C$, and $A C$ form an angle $\phi$ with the plane $A B C$ and intersect at point $D 1$. In addition, three planes through the same lines form an angle $2 \phi$ with the plane $A B C$ and intersect at point $D 2$. Find $\phi$,... | Let's denote the side of the equilateral triangle $ABC$ by $a$. Then the radius of the inscribed circle of the triangle is $\frac{\sqrt{2}}{6}$, and the radius of the exscribed circle is $-\frac{\sqrt{2} \sqrt{3}}{2}$. The lateral faces of the triangular pyramid $ABCD_1$ form equal angles with the plane of the base $AB... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
Through a point in space, four planes are drawn, no three of which have a common line. Into how many parts do these planes divide the space? What are the names of the resulting parts of space? | 
A plane passing through the vertex of a trihedral angle and a point inside the trihedral angle divides the trihedral angle into a trihedral and a tetrahedral angle. Consider planes $\alpha$ and... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The area of the triangle is $4 \sqrt{21}$, the perimeter is 24, and the segment of the bisector from one of the vertices to the center of the inscribed circle is $\frac{\sqrt{30}}{3}$. Find the largest side of the triangle. | If the inscribed circle touches the side $A C$ of triangle $A B C$ at point $M$, and $p$ is the semiperimeter of the triangle, then $A M=p-B C$.
## Solution
Let $O$ be the center of the circle inscribed in the given triangle $A B C, r$ be its radius, $S=4 \sqrt{21}$ be the area, $2 p$
$=24$ be the perimeter, and $M$... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Concerning the Homothety of Tangent Circles helps to solve the problem, $\quad]$
In a right triangle $ABC$, angle $C$ is a right angle, and side $CA=4$. A point $D$ is taken on the leg $BC$, such that $CD=1$. A circle with radius $\frac{\sqrt{5}}{2}$ passes through points $C$ and $D$ and is tangent at point $C$ to the... | Since point $D$ lies inside the circumcircle of triangle $A B C$, the given circles touch internally. If $M$ is the point of intersection of the first circle with the leg $A C$ other than $C$, then $M D$ is the diameter of this circle,
$$
M C=\sqrt{M C D^{2}-C D^{2}}=\sqrt{5-1}=2
$$
Let $Q$ and $O$ be the midpoints o... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Zassovstiy A.A.
Given a square sheet of paper with a side length of 2016. Is it possible, by folding it no more than ten times, to construct a segment of length 1? | Note that we can bisect any segment by overlapping its ends. In addition, we can fold the paper along a line passing through a given point, perpendicular to a given line. In other words, we can construct the midpoints of segments and a perpendicular to a given line passing through a given point.
We will use these cons... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{l}\text { Tangent circles }\end{array}\right]$
Two circles touch each other internally at point $A$. From the center $O$ of the larger circle, a radius $O B$ is drawn, touching the smaller circle at point $C$. Find $\angle B A C$. | Express the desired angle in terms of angle $A O B$.
## Solution
Let $\angle A O B=\alpha$. Let $Q-$ be the center of the smaller circle. Since triangle $A O B$ is isosceles, then
$$
\angle O A B=\angle O B A=90^{\circ}-\frac{\alpha}{2}
$$
Since triangle $O Q C$ is a right triangle, then
$$
\angle O Q C=90^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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