problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
Described quadrilaterals Area of a trapezoid
The bases of an isosceles trapezoid circumscribed around a circle are 2 and 18. Find the area of the trapezoid.
# | Let $C H$ be the height of the given isosceles trapezoid $A B C D$ with bases $A D=18$ and $B C=2$. Then
$$
D H=\frac{1}{2}(A D-B C)=\frac{1}{2}(18-2)=8
$$
Since the trapezoid is circumscribed around a circle, the sums of its opposite sides are equal, meaning the sum of the lateral sides is equal to the sum of the ba... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Circle inscribed in an angle Area of a trapezoid
A circle with a radius of 3 is inscribed in a right trapezoid, the smaller base of which is 4. Find the larger base of the trapezoid. | Let $O$ be the center of a circle with radius 3, inscribed in trapezoid $ABCD$ with bases $AD$ and $BC=4$, $\angle C = \angle D = 90^\circ$, and the circle touching sides $CD$, $BC$, $AB$, and $AD$ at points $K$, $L$, $M$, and $N$ respectively. The quadrilaterals $ONDK$ and $OKCL$ are squares, so
$$
DN = OK = 3, \quad... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Projections of the bases, sides, or vertices of a trapezoid ] $[\quad$ The midline of a trapezoid $\quad]$
It is known that a circle can be inscribed in an isosceles trapezoid $A B C D$ with bases $A D>B C$; $C H-$ is the height of the trapezoid, $A H=7$. Find the lateral side of the trapezoid. | Isosceles trapezoid $A B C D$ is circumscribed around a circle, so $2 A B=A B+A B=A B+C D=A D+B C$, which means $A B=\frac{1}{2}(A D+B C)$. On the other hand, the projection of the diagonal of an isosceles trapezoid onto the larger base is equal to half the sum of the bases, i.e., $A H=\frac{1}{2}(A D+B C)$, and the ha... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Midline of the trapezoid ] $[\quad$ Area of the trapezoid $\quad]$
The diagonals of the trapezoid are 6 and 8, and the midline is 5. Find the area of the trapezoid. | Let $h$ be the height of trapezoid $ABCD$ with bases $AD$ and $BC$ and diagonals $AC=6$ and $BD=8$, and let $l$ be the midline of the trapezoid. Draw a line through vertex $C$ parallel to diagonal $BD$ until it intersects the extension of base $AD$ at point $M$. Then quadrilateral $BCMD$ is a parallelogram, so
$$
CM =... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Four points lying on one circle Auxiliary circle
In a convex quadrilateral $A B C D$, it is known that $\angle B C D=80^{\circ}, \angle A C B=50^{\circ}$ and $\angle A B D=30^{\circ}$. Find the angle $A D B$. | Quadrilateral $A B C D$ is convex, so ray $C A$ passes between the sides of angle $B C D$, hence,
$$
\angle A C D=\angle B C D-\angle A C B=80^{\circ}-50^{\circ}=30^{\circ}
$$
Points $B$ and $C$, lying on the same side of line $A D$, see segment $A D$ under the same angle, which means points $B, C, A$ and $D$ lie on ... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Four points lying on one circle Auxiliary circle $\quad]$
In a convex quadrilateral $A B C D$, it is known that $\angle A C B=25^{\circ}, \angle A C D=40^{\circ}$ and $\angle B A D=115^{\circ}$. Find the angle $A D B$ | Quadrilateral $A B C D$ is convex, so ray $C A$ passes between the sides of angle $B C D$, hence,
$$
\angle B C D=\angle A C B+\angle A C D=25^{\circ}+40^{\circ}=65^{\circ} .
$$
Then
$\angle B C D+\angle B A D=65^{\circ}+115^{\circ}=180^{\circ}$,
therefore quadrilateral $A B C D$ is cyclic, i.e., points $A, B, C$ a... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
} \underline{\text { Folklore }}
$$
In a right triangle $ABC$, angle $A$ is $60^{\circ}$, and $M$ is the midpoint of the hypotenuse $AB$.
Find the angle $IMA$, where $I$ is the center of the inscribed circle of the given triangle. | Let $r$ be the radius of the circle inscribed in triangle $ABC$, and let $D$, $E$, and $F$ be the points of tangency with sides $AB$, $BC$, and $AC$ respectively (see figure).
Obviously, qu... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Folklore
Find the maximum value of the expression $x^{2}+y^{2}$, if $|x-y| \leq 2$ and $|3 x+y| \leq 6$. | The set of points satisfying the system of inequalities $|x-y|<2$ and $|3x+y|<6$ forms the intersection of two strips, and thus represents the boundary and interior of a parallelogram $ABCD$, where $A(1,3), B(2, 0), C(-1,-3), D(-2,0)$ (see figure).
.
It is clear that $AM = MC = CN$ and $\angle MCB = \angle NCB$. But also $\angle CLN = \angle LCM$ (since $LN \parallel AC$),... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7,8,9,1 |
Avor: Kuyygin A.K:
Among any five nodes of a regular square grid, there will definitely be two nodes such that the midpoint of the segment between them is also a node of the grid. What is the minimum number of nodes of a regular hexagonal grid that must be taken so that among them there are definitely two n... | Lemma. Among any five nodes of a grid of equilateral triangles, there will be two such that the midpoint of the segment between them is also a grid node.
Proof of the lemma. Introduce the o... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Midline of a trapezoid ] $[$ Area of a trapezoid $]$
Find the area of trapezoid $A B C D$ with side $B C=5$, if the distances from vertices $A$ and $D$ to line $B C$ are 3 and 7, respectively. | Let $M$ be the midpoint of the side $A D$; $A_{1}, M_{1}$, and $D_{1}$ be the projections of points $A, M$, and $D$ onto the line $B C$. Since $M M_{1}$ is the midline of the trapezoid $A D D_{1} A_{1}$, then $M M_{1}=1 / 2\left(A A_{1}+D D_{1}\right)=5$. According to problem $\underline{54964} S_{A B C D}=2 S_{B M C}=... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9
On a circle, a point $A$ is taken, on the diameter $B C$ - points $D$ and $E$, and on its extension beyond point $B$ - point $F$. Find $B C$, if $\angle B A D=\angle A C D, \angle B A F=\angle C A E, B D=2, B E=5$ and $B F=4$. | Let $\angle B A D=\angle A C D=\alpha, \angle B A F=\angle C A E=\beta$. Since point A lies on the circle with diameter $B C$, then $\angle B A C=90^{\circ}$. Therefore,
$$
\begin{gathered}
\angle A D C=180^{\circ}-\angle A C D-\angle C A D=180^{\circ}-\alpha-\left(90^{\circ}-\alpha\right)=90^{\circ}, \\
\angle F A E=... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}{[\text { Midline of the trapezoid }]} \\ {[\quad \text { Area of the trapezoid }]}\end{array}\right]$
The product of the midline of the trapezoid and the segment connecting the midpoints of its diagonals is 25. Find the area of the trapezoid, if its height is three times the difference of the b... | It is known that the midline of a trapezoid is equal to the half-sum of the bases, and the segment connecting the midpoints of the diagonals is the half-difference of the bases. Let $a$ and $b$ be the bases of the trapezoid ($a > b$), $h$ be its height, and $S$ be its area. According to the problem,
$$
\frac{a+b}{2} \... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A circle with its center on the diagonal $A C$ of trapezoid $A B C D (B C \| A D)$ passes through vertices $A$ and $B$, is tangent to side $C D$ at point $C$, and intersects the base $A D$ at point $E$. Find the area of trapezoid $A B C D$, if $C D=6 \sqrt{13}$, $A E=8$. | Points $B$ and $E$ lie on the circle with diameter $A C$, so $\angle B A E=\angle A B C=\angle A E C=90^{\circ}$ and $A B C E-$
is a rectangle; $A C$ is the diameter of the circle tangent to the line $C D$ at point $C$, so $\angle A C D=90^{\circ}$. Therefore, $C E$ is the altitude of the right triangle $A C D$ drawn ... | 204 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[The volume of a body is equal to the sum of the volumes of its parts] Volume of a tetrahedron and a pyramid
A convex polyhedron $A B C D F E$ has five faces: $C D F, A B E, B C F E, A D F E$ and $A B C D$. Edge $A B$ is parallel to edge $C D$. Points $K$ and $L$ are located on edges $A D$ and $B C$ respectively, such... | Let $E 1, M 1$ and $F 1$ be the orthogonal projections of points $E, M$ and $F$ onto the plane of parallel lines $A B$ and $C D$. Then $E E 1$ is the height of the quadrilateral pyramid $E A K L B$ with vertex $E, M M 1$ is the height of the quadrilateral pyramid $M A B C D$ with vertex $M, F F 1$ is the height of the ... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Kaiibhanov A.K:
The audience has the shape of a regular hexagon with a side of 3 m. In each corner, a snoremeter is installed, determining the number of sleeping students at a distance not exceeding 3 m. How many sleeping students are there in total in the audience, if the sum of the snoremeter readings is 7?
# | 
Each student is "seen" by 2, 3, or 6 snorometers (see fig). Therefore, 7 can be broken down into a sum of terms, each of which is equal to 2, 3, or 6. It is easy to see that 7 can be represe... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The area of the triangle is 16. Find the area of the trapezoid that is cut off from the triangle by its midline.
# | The median line of a triangle cuts off a similar triangle from it, with a similarity coefficient of $\frac{1}{2}$. Therefore, the area of the cut-off triangle is one-fourth of the area of the original triangle, i.e., $\frac{1}{\mathbf{4}} 16=4$. Consequently, the area of the remaining trapezoid is $16-4=12$.
## Answer... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a square, 20 points were marked and connected with non-intersecting segments to each other and to the vertices of the square, so that the square was divided into triangles. How many triangles were formed?
# | We will consider the marked points and the vertices of the square as vertices, and the segments connecting them and the sides of the square as edges of a planar graph. For each piece into which this graph divides the plane, we will count the number of edges bounding it, and sum all the obtained numbers. Since each edge... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[
[ Tangent Circles $]$
Two circles touch each other internally. A line passing through the center of the smaller circle intersects the larger circle at points $A$ and $D$, and the smaller circle at points $B$ and $C$. Find the ratio of the radii of the circles if $A B: B C: C D=2: 4: 3$.
# | Apply the theorem of the products of segments of intersecting chords.
## Solution
Let $M$ be the point of tangency of the circles, $r$ and $R (r < R)$ be their radii, and $O$ be the center of the smaller one. Denote: $AB = 2x, BC = 4x, CD = 3x$. Then
$$
r = OB = OC = 2x, DO \cdot OA = MO(2R - MO), \text{ or } 5x \cd... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
Construct a circle equidistant from four points on a plane. How many solutions does the problem have? | Answer: 7 solutions (in the non-degenerate case). Let $A, B, C, D$ be the given points, and $S$ be the sought circle. On one side of $S$, there are $k$ given points, and on the other side, there are 4 - $k$ given points. We will assume that the given points do not lie on the same circle (otherwise, $S$ can be any circl... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a convex 1950-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?
# | Answer: 1949. The same reasoning as in solving problem 1 for 7-8 grades shows that the resulting polygon has no more than 1950 sides, and if the number of its sides is 1950, then exactly two diagonals emanate from each vertex of the original polygon, bounding the resulting polygon. Let two diagonals $A_{1} A_{\mathrm{p... | 1949 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On side $AB$ of triangle $ABC$, point $K$ is marked, and on side $AC$, point $M$ is marked. Segments $BM$ and $CK$ intersect at point $P$. It turns out that angles $APB$, $BPC$, and $CPA$ are each $120^{\circ}$, and the area of quadrilateral $AKPM$ is equal to the area of triangle $BPC$. Find angle $BAC$. | To both sides of the equality $S_{A K P M}=S_{B P C}$, add the area of triangle $B P K$ (see the figure). We get that $S_{A B M} = S_{B C K}$. Therefore,
${ }^{A M} /_{A C} \cdot S_{A B C}={ }^{B K} /{ }_{A B} \cdot S_{A B C}$, that is, $B K: A B = A M: A C$. Thus, points $K$ and $M$ divide segments $B A$ and $A C$ in... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Given a cyclic quadrilateral $A B C D$. The rays $A B$ and $D C$ intersect at point $K$. It turns out that points $B, D$, as well as the midpoints $M$ and $N$ of segments $A C$ and $K C$, lie on the same circle. What values can the angle $A D C$ take? | $M N$ is the midline in triangle $A K C$, so $\angle B A C = \angle N M C$. Additionally, $\angle B A C = \angle B D C$ because quadrilateral $A B C D$ is cyclic.
Let points $M$ and $N$ lie on the same side of line $B D$. Then $M$ lies inside triangle $B C D$ and, therefore, inside triangle $B N D$, and thus inside it... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A regular hexagon $A B C D E F$ is inscribed in a circle. Points $P$ and $Q$ are chosen on the tangents to this circle at points $A$ and $D$ respectively, such that the line $P Q$ is tangent to the smaller arc $E F$ of this circle. Find the angle between the lines $P B$ and $Q C$. | Let $T$ be the point of tangency of $P Q$ with the circle, $M, N$ be the midpoints of segments $A T, D T$. Since $P B$ and $C Q$ are symmedians of triangles $A B T$ and $C D T$ respectively (see problem $\underline{\text { 56983) }}$), then $\angle A B P=\angle M B T, \angle D C Q=\angle N C T$. Since $M N$ is the midl... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Tokaeva I.
Let $F_{1}, F_{2}, F_{3}, \ldots$ be a sequence of convex quadrilaterals, where $F_{k+1}$ (for $k=1,2,3, \ldots$) is obtained by cutting $F_{k}$ along a diagonal, flipping one of the parts, and gluing it back along the cut line to the other part. What is the maximum number of different quadrilaterals that t... | Let $ABCD$ be the original quadrilateral $F_{1}$. We can assume that each time the half of the quadrilateral containing side $CD$ is flipped, while side $AB$ remains stationary. In this process, the sum of angle $A$ and the opposite angle does not change. Additionally, the set of side lengths does not change. However, ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Pythagorean Theorem (direct and inverse). ] [ Properties and characteristics of an isosceles triangle. ]
In an isosceles triangle \(ABC\), the side \(AB\) is 10, and the base \(AC\) is 12. The angle bisectors of angles \(A\) and \(C\) intersect at point \(D\). Find \(BD\). | If $M$ is the midpoint of $A C$, then $B D: D M=B C: C M$.
## Solution
Since the angle bisectors of a triangle intersect at one point, $B D$ is the bisector of angle $B$. Extend $B D$ to intersect $A C$ at point $M$. Then $M$ is the midpoint of $A C, B M \perp A C$. Therefore, $B M^{2}=B C^{2}-M C^{2}=64$.
Since $C ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 | |
In a right triangle $A B C$, the hypotenuse $A B$ is equal to $c$ and $\angle B=\alpha$. Find all medians of this triangle. | As is known, the median of the right triangle is equal to $c / 2$.
Let $M$ be the midpoint of $B C$. Then $C M=1 / 2 B C=1 / 2 c \sin \alpha$.
By the Pythagorean theorem, $A M^{2}=A C^{2}+C M^{2}=1 / 4\left(4 \cos ^{2} \alpha+\sin ^{2} \alpha\right)=1 / 4\left(1+3 \cos ^{2} \alpha\right)$.
Similarly, we find the thi... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Properties and characteristics of the tangent [
A circle, the center of which lies on the hypotenuse $AB$ of a right triangle $ABC$, touches the two legs $AC$ and $BC$ at points $E$ and $D$, respectively.
Find the angle $ABC$, if it is known that $AE=1, BD=3$. | Let $O$ be the center of the circle. Consider similar triangles $B D O$ and $O E A$.
## Solution
Let $O$ be the center of the circle, $x$ be its radius. Then $O D=O E=x, O D \perp B C$ and $O E \perp A C$. Right triangles $B D O$ and $O E A$ are similar, so
$B D: D O=O E: A E$, or $3: x=x: 1$, from which $x=\sqrt{3}... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9
[ Rectangles and Squares. Properties and Characteristics]
The side of the square is 1. A line is drawn through its center. Calculate the sum of the squares of the distances from
the four vertices of the square to this line. | Let a line passing through the center $O$ of the square $A B C D$ intersect the side $A B$. Drop perpendiculars $A P$ and $B Q$ to this line. Triangles $A P O$ and $O Q B$ are equal by hypotenuse and acute angle. Therefore, $A P^{2}+B Q^{2}=$ $A P^{2}+O P^{2}=A O^{2}=1 / 2$
## Answer
1. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7,8,9 |
The height $A H$ of triangle $A B C$ is equal to its median $B M$. On the extension of side $A B$ beyond point $B$, point $D$ is marked such that $B D=A B$. Find the angle $B C D$. | From point $M$, drop a perpendicular $ME$ to side $BC$. Since $ME \parallel AH$, $ME$ is the midline of triangle $AHC$ (see figure). Therefore, $ME = \frac{1}{2} AH = \frac{1}{2} BM$.
In the right triangle $BME$, the leg $ME$ is half of the hypotenuse $BM$, so $\angle MBE = 30^{\circ}$. Moreover, $BM$ is the midline o... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow Metro from "Taganskaya" station to "Kievskaya" station, and in the evening - back (see diagram).
Upon... | Let $p$ be the probability that the Scientist boards a train going clockwise. Then the expected travel time from "Taganskaya" to "Kievskaya" is $11 p + 17(1-p) = 17 - 6p$.
On the return trip from "Kievskaya" to "Taganskaya," the expected travel time is $17 p + 11(1-p) = 11 + 6p$.
By the condition $(11 + 6p) - (17 - 6... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Segment visible from two points at the same angle] [Perpendicular bisector of the segment (LMT).]
Inside a non-isosceles triangle $A B C$, a point $O$ is taken such that $\angle O B C=\angle O C B=20^{\circ}$. Additionally, $\angle$ BAO + $\angle O C A=70^{\circ}$. Find the angle $A$. | Through point $C$, draw a line perpendicular to $AC$. Let this line intersect the extension of segment $AO$ at point $E$. Then
\[
\begin{gathered}
\angle B C E=90^{\circ}-\angle B C O-\angle O C A= \\
=90^{\circ}-20^{\circ}-\angle O C A=70^{\circ}-\angle O C A=\angle B A O=\angle B A E
\end{gathered}
\]
This means th... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Rectangular parallelepipeds ] [ Unfolding helps solve the task ]
In a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$, it is known that $A B = A A 1 = 12$ and $A D = 30$. Point $M$ is located in the face $A B B 1 A 1$ at a distance of 1 from the midpoint of $A B$ and at equal distances from vertices $A$ and $B$... | Suppose the path crosses edges A1B1 and C1D1 (or edges $A B$ and CD). In this case (Fig.2), the length of the shortest path is $11+30+1=42$. Suppose the path sequentially crosses edges BB1, B1C1, and C1D1. Consider the part of such an unfolding of the parallelepiped that contains rectangles A1B1C1D1 and BB1C1C with a c... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
Consider a rectangle $A B C D$ and a point $E$ not lying in its plane. Let the planes $A B E$ and $C D E$ intersect along the line $l$, and the planes $B C E$ and $A D E$ - along the line $p$. Find the angle between the lines $l$ and $p$. | Planes $A B E$ and $C D E$ pass through parallel lines $A B$ and $C D$ and intersect along line $l$, so $l \|$ $A B$. Similarly, line $p$ of the intersection of planes $B C E$ and $A D E$ is parallel to line $B C$. Since lines $A B$ and
BC are perpendicular, the lines $l$ and $p$ parallel to them are also perpendicula... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find the angle between the lines $A C$ and $B D$, if the distance between the midpoints of segments $A D$ and $B C$ is equal to the distance between the midpoints of segments $A B$ and $C D$.
# | Let $M, N, K$ and $L$ be the midpoints of segments $AD, BC, AB$ and $CD$ respectively. Since $ML$ and $KN$ are the midlines of triangles $ADC$ and $ABC$ with the common base $AC$, then $ML \| KN$. Therefore, points $M, N, K$ and $L$ lie in the same plane. Moreover, $MK \| LN$, so quadrilateral $KNLM$ is a parallelogram... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a triangular pyramid $A B C D$, find the angle between the lines $A D$ and $B C$, if $A B=A C$ and $\angle D A B=\angle D A C$. | Let $DO$ be the height of the pyramid, and points $M$ and $N$ be the bases of the perpendiculars dropped from point $O$ to lines $AB$ and $AC$, respectively. According to the theorem of three perpendiculars, $DM \perp AB$ and $DN \perp AC$. Right triangles $DAM$ and $DAN$ are equal by hypotenuse and acute angle, so $OM... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
In triangle $ABC$, it is known that $AC=12, AB=BC=3\sqrt{10}$. Two spheres touch the plane of triangle $ABC$ at points $A$ and $C$ and are located on opposite sides of this plane. The distance between the centers of these spheres is 15. The center of a third sphere is at point $B$, and this sphere externally tou... | Let $O_1$ and $O_2$ be the centers of the spheres that touch the plane of triangle $ABC$ at points $A$ and $C$ respectively, with radii $x$ and $y$. Let $z$ be the radius of the third sphere. Through the lines $O_1A$ and $O_2C$, which are perpendicular to the plane $ABC$, we draw a plane. In this plane, we drop a perpe... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8
How many sides can a convex polygon have if all its diagonals are of the same length? | Let's prove that the number of sides of such a polygon does not exceed 5. Suppose that all diagonals of the polygon $A_{1} \ldots A_{n}$ have the same length and $n \geq 6$. Then the segments $A_{1} A_{4}, A_{1} A_{5}, A_{2} A_{4}$, and $A_{2} A_{5}$ have the same length, as they are diagonals of this polygon. However,... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Extremal properties (miscellaneous).]
What is the largest number of points that can be placed on a segment of length 1 so that on any segment of length $d$, contained in this segment, there are no more than $1+1000 d^{2}$ points? | First, let's prove that it is impossible to place 33 points in such a way. Indeed, if 33 points are placed on a segment of length 1, then the distance between some two of them does not exceed $1 / 32$. The segment with endpoints at these points contains two points, but it should contain no more than $1 + 1000 / 32^{2}$... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Markelov S.v.
In parallelogram $A B C D$, angle $A C D$ is equal to $30^{\circ}$. It is known that the centers of the circles circumscribed around triangles $A B D$ and $B C D$ are located on diagonal $A C$. Find angle $A B D$. | Consider two cases: the centers of the circles are different points; the centers of the circles coincide.
## Solution
If the centers of the given circles do not coincide, then the line of centers (i.e., the diagonal $AC$ of the parallelogram) is perpendicular to the common chord (diagonal $BD$). Therefore, the given ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Perpendicular bisector of a segment_([MT)]
[ Inscribed angle subtended by a diameter ]
A circle is circumscribed around triangle $A M B$, and its center is 10 units away from side $A M$.
The extension of side $A M$ beyond vertex $M$ intercepts a segment $C B$ from the tangent to the circle at vertex $B$, which is 2... | Let $O$ be the center of the given circle, $K$ the midpoint of side $AM$, $P$ the projection of point $B$ onto line $AC$, and $F$ the intersection point of ray $BO$ with line $AC$.
Assume that point $O$ lies between points $B$ and $F$. Then
$$
BP = CB \sin \angle C = 29 \cdot \frac{20}{29} = 20
$$
Since $OK = 10$, p... | 210 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9
In trapezoid $ABCD$, points $K$ and $M$ are the midpoints of the bases $AB=5$ and $CD=3$. Find the area of the trapezoid if triangle $AMB$ is a right triangle and $DK$ is the height of the trapezoid. | Draw the height of the trapezoid from point $M$.
## Solution
Notice that in triangle $A M B \angle M=90^{\circ}$ (otherwise, $M A$ or $M B$ would be perpendicular to $A B$, which is impossible).
Let $M P$ be another height of the trapezoid. Then
\[
\begin{gathered}
K P=D M=\frac{1}{2} D C=\frac{3}{2}, B P=K B-K P=\... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The Law of Sines
[The ratio in which the bisector divides the side.]
Two circles intersect at points $A$ and $B$. A line passing through point $A$ intersects these circles again at points $C$ and $D$, with point $A$ lying between $C$ and $D$, and the chords $A C$ and $A D$ are proportional to the radii of their respec... | Prove that $BA-$ is the bisector of triangle $BCD$.
## Solution
Let point $C$ lie on a circle of radius $r$, and point $D$ - on a circle of radius $R$. Then
$$
\begin{gathered}
BC = 2r \sin \angle BAC \\
BD = 2R \sin \angle BAD = 2R \sin (180^\circ - \angle BAC) = 2R \sin \angle BAC
\end{gathered}
$$
Therefore,
$$... | 13 | Geometry | proof | Yes | Yes | olympiads | false |
Two circles, the radii of which are in the ratio $9-4 \sqrt{3}$, touch each other internally. Two chords of the larger circle, equal in length and tangent to the smaller circle, are drawn. One of these chords is perpendicular to the segment connecting the centers of the circles, while the other is not. Find the angle b... | Let circles with radii $R>r$ and centers $O_1$ and $O_2$ respectively touch internally at point $K$. Chord $AB$ of the larger circle is perpendicular to $O_1 O_2$ and touches the smaller circle at point $P$, and an equal chord $CD$ of the larger circle touches the smaller circle at point $Q$ and intersects chord $AB$ a... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Iveliev B.M.
Let $ABC$ be an acute-angled triangle, $CC_1$ its bisector, $O$ the center of the circumscribed circle. The intersection point of the line $OC_1$ with the perpendicular dropped from vertex $C$ to side $AB$ lies on the circumscribed circle $\Omega$ of triangle $AOB$. Find the angle $C$. | Let $D$ be the point of intersection of $OC_{1}$ with the given perpendicular (see figure). Since $D$ lies on $\Omega$ and $AO=OB$, then $\angle ADC_{1}=\angle BDC_{1}$. Therefore,
$AD:BD = AC_{1}:BC_{1} = AC:BC$. On the other hand, since $CD \perp AB$, then $AC^2 + BD^2 = AD^2 + BC^2$. From these equalities, it follo... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Shnol D....
Given a triangle $ABC$ and an excircle with center $O$, touching side $BC$ and the extensions of sides $AB$ and $AC$. Point $O_{1}$ is symmetric to point $O$ with respect to line $BC$. Find the measure of angle $A$, if it is known that point $O_{1}$ lies on the circumcircle of triangle $ABC$. | From the condition, it follows that $\angle B O C=\angle B O_{1} C=\angle A$. On the other hand, $\angle B O C=180^{\circ}-1 / 2\left(180^{\circ}-\angle B\right)-1 / 2\left(180^{\circ}-\right.$ $\angle C)=1 / 2\left(180^{\circ}-\angle A\right)$. From this, we obtain that $\angle A=60^{\circ}$. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Properties of an isosceles triangle. ] [ Right triangle with an angle of $\$ 30 \wedge$ ^circ\$ ]
[ Angle bisector (GMT) ]
Inside triangle $A B C$, a point $M$ is chosen on the bisector of its angle $B$ such that $A M=A C$ and $\angle B C M=30^{\circ}$.
Prove that $\angle A M B=150^{\circ}$. | Let $K$ and $L$ be the feet of the perpendiculars dropped from point $M$ to sides $AB$ and $BC$, and let $N$ be the midpoint of the hypotenuse $CM$ of the right triangle $CML$. A point lying on the angle bisector is equidistant from the sides of the angle, so $MK = ML$. The leg opposite the $30^\circ$ angle is half the... | 150 | Geometry | proof | Yes | Yes | olympiads | false |
8,9
[ Angles subtended by equal arcs and equal chords]
Trapezoid $A E F G(E F \| A G)$ is placed in square $A B C D$ with side length 14 such that points $E, F$, and $G$ lie on sides $A B, B C$, and $C D$ respectively. Diagonals $A F$ and $E G$ are perpendicular, and $E G=10 \sqrt{2}$. Find the perimeter of the trapez... | Prove that the given trapezoid is isosceles, and the diagonal of the square passes through the point of intersection of its diagonals.
## Solution
Let \( M \) be the point of intersection of the diagonals of the trapezoid, and \( P \) be the projection of point \( E \) onto side \( CD \) of the square. Angles \( \ang... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9,10,11 |
Authors: Bverlov S.L., $\underline{\text { Yakubov A. }}$.
Given a parallelogram $A B C D$, where $A B<A C<B C$. 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 $D$; moreover, segments $A D$ and $C E$ inters... | Since $D$ lies outside $\omega$, angle $A B C$ is acute. Let $A^{\prime}$ be the second intersection point of $D C$ and $\omega$. Since $B C > A C$, we have
$\angle D C A = \angle C A B > \angle C B A = \angle D A ' A$; hence, $A^{\prime}$ lies on the extension of segment $D C$ beyond point $C$. Notice that $-E C A^{\... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
What is the minimum length a piece of wire must have so that a cube frame with an edge of 10 cm can be bent from it?
(The wire can pass along the same edge twice, can be bent at $90^{\circ}$ and $180^{\circ}$, but it cannot be broken.) | How to bend a frame from a wire of length 150 cm, as shown in the figure.
Let's assume for a moment that we are only allowed to bend the wire at the vertices of the cube, and that some frame... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Volchkovich M.A.
Point $O$ lies inside rhombus $A B C D$. Angle $D A B$ is $110^{\circ}$. Angles $A O D$ and $B O C$ are $80^{\circ}$ and $100^{\circ}$, respectively. What can the angle $A O B$ be? | Note that the geometric locus of points $O$, from which the segment $A D$ is seen at an angle of $80^{\circ}$, and lying on the same side of $A D$ as point $B,$ is an arc of a circle passing through points $A$ and $D$, and the geometric locus of points $O$, from which the segment $B C$ is seen at an angle of $100^{\cir... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Proyasov B.Y.
Two circles with radii 1 and 2 have a common center at point $O$. Vertex $A$ of an equilateral triangle $ABC$ lies on the larger circle, and the midpoint of side $BC$ lies on the smaller circle. What can the angle $BOC$ be equal to? | ## Author: Oschekina M
Consider the case when points $O$ and $A$ lie in the same half-plane relative to the line $B C$ (see the left figure). Let $K$ be the midpoint of $B C$, and $G$ be the point of intersection of the medians of triangle $A B C$. Extend the segment $O K$ until it intersects the larger circle at poin... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Khachaturyan A.V.
Bus stop B is located on a straight highway between stops A and C. After some time following departure from A, the bus found itself at a point on the highway such that the distance from this point to one of the three stops is equal to the sum of the distances to the other two. After the same amount o... | At both moments in time mentioned in the problem, the sum will obviously be the distance from the bus to the farthest stop from it. This cannot be $B$, as it is closer than $C$. Therefore, these were $C$ (before the bus had traveled halfway from $A$ to $C$) and $A$ (after this moment).
 $\mathrm{AN}=\mathrm{NM}=\mathrm{MK}=1$. Then $\mathrm{NB}=1, \mathrm{MB}=2$, so $\mathrm{MN}+\mathrm{NB}=\mathrm{MB}$. 2) $\mathrm{KA}=\mathrm{AN}=\mathrm{NM}=1$. Then $\mathrm{AC}... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { The ratio in which the bisector divides the side. } \\ {[\underline{\text { Properties of bisectors, concurrency }}]}\end{array}\right]$
In triangle $ABC$, points $M$ and $N$ are marked on sides $AB$ and $BC$ respectively, such that $BM = BN$. A line is drawn through point $M$ perpendicu... | Prove that $B P$ is the bisector of triangle $A B C$, apply the bisector theorem of a triangle and the formula for the square of the bisector of a triangle: $B P^{2}=A B \cdot B C-A P \cdot P C$.
## Solution
In the isosceles triangle $B M N$, point $O$ is the intersection of the altitudes. Therefore, $B P$ is the bis... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$[\underline{C}$ Tangent Circles
Two circles touch each other externally at point $A$. Their common tangent touches the first circle at point $B$ and the second circle at point $C$. A line passing through points $A$ and $B$ intersects the second circle at point $D$. It is known that $A B=5$ cm, $A D=4$ cm. Find $C D$... | Prove that $C A$ is the altitude of the right triangle $B C D$, drawn from the vertex $C$ of the right angle.
## Solution
Let the common tangent, drawn to the given circles at point $A$, intersect the segment $B C$ at point $K$. Then $B K=K A=K C$. Therefore, $\angle B A C=90^{\circ}$. This means that the chord $C D$... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In triangle $ABC$, the lines containing the altitudes $AP$, $CR$, and $BQ$ (points $P$, $R$, and $Q$ lie on the lines containing the corresponding sides of triangle $ABC$) intersect at point $O$. Find the areas of triangles $ABC$ and $POR$, given that $RP$ is parallel to $AC$, $AC=4$, and $\sin \angle ABC=\frac{24}{25}... | Prove that the given triangle is isosceles. Consider two cases: $\angle A B C = 90^{\circ}$.
## Solution
Since from points $P$ and $R$ the segment $A C$ is seen at a right angle, these points lie on a circle with diameter $AC$. Therefore, the trapezoid ACPR (Fig.1) (or ACRP (Fig.2)) is isosceles. Hence, triangle $ABC... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The area of the right triangle ABC ( $\angle C=90^{\circ}$ ) is 6, and the radius of the circumscribed circle around it is $\frac{5}{2}$. Find the radius of the circle inscribed in this triangle.
# | The radius of the circle inscribed in a right-angled triangle with legs $a, b$ and hypotenuse $c$ is $\frac{a+b-c}{2}$.
## Solution
It is known that the radius $r$ of the circle inscribed in a right-angled triangle with legs $a, b$ and hypotenuse $c$ is $\frac{a+b-c}{2}$, and the area is $\frac{ab}{2}$.
Since the ce... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Angles subtended by equal arcs and equal chords]
## Law of Sines
The lengths of three sides of a cyclic quadrilateral inscribed in a circle of radius $2 \sqrt{2}$ are equal and each is 2. Find the fourth side. | Prove that the given quadrilateral is an isosceles trapezoid. Then apply the Law of Sines.
## Solution
Let the quadrilateral $A B C D$ be inscribed in a circle with radius $R=2 \sqrt{2}$, and $A B=B C=C D=2$. Let $\angle C A D=\alpha$. Since equal chords subtend equal arcs, and inscribed angles subtending equal chord... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Auto: Witness $M$.
In quadrilateral $A B C D$, sides $A B, B C$ and $C D$ are equal, $M$ is the midpoint of side $A D$. It is known that $\angle B M C=90^{\circ}$. Find the angle between the diagonals of quadrilateral $A B C D$. | Let $O, K, L$ be the midpoints of segments $B C, A C$ and $B D$ respectively, $P$ be the intersection point of lines $A C$ and $B D$. Points $K, L$ are distinct (otherwise $A B C D$ is a rhombus and
$\angle B M C<\angle B P C=90^{\circ}$). Since angles $B K C, B M C$ and $B L C$ are right (medians in isosceles triangl... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Through the opposite vertices $A$ and $C$ of the quadrilateral $A B C D$, a circle is drawn, intersecting
sides $A B, B C, C D$ and $A D$ at points $M, N, P$ and $Q$ respectively. It is known that $B M=B N=D P=D Q=R$, where $R$ is the radius of the given circle. Prove that in this case the sum of angles $B$ and $D$ of... | Let $O$ be the center of the given circle. From the condition, it follows that $O M B N$ and $O P D Q$ are rhombuses. Therefore, $\angle A M O = \angle B$ and $\angle$ $A Q O = \angle D$. Therefore, $\angle A = \angle O A M + \angle O A Q = \angle O M A + \angle O Q A = \angle B + \angle D$. Similarly, $\angle C = \ang... | 120 | Geometry | proof | Yes | Yes | olympiads | false |
[ Isosceles, Inscribed, and Circumscribed Trapezoids ]
[Properties and characteristics of isosceles triangles. ]
Let $M$ be the point of intersection of the diagonals of a convex quadrilateral $ABCD$, in which sides $AB$, $AD$, and $BC$ are equal to each other.
Find the angle $CMD$, given that $DM = MC$, and $\angle... | Let $\angle A B D=\angle A D B=\alpha, \angle B A C=\angle A C B=\beta$. By the external angle theorem of a triangle, $\angle B M C=\alpha+\beta$. Draw a line through point $A$ parallel to side $C D$. Let this line intersect line $D B$ at point $K$. Triangle $AMK$ is isosceles because it is similar to the isosceles tri... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Sharygin I.F.
The angle at the vertex $A$ of the isosceles triangle $ABC (AB = AC)$ is $20^{\circ}$. On the side $AB$, we lay off the segment $AD$, equal to $BC$. Find the angle $BCD$. | We will sequentially construct points $E, F$, and $G$ on sides $AB$ and $AC$ such that $BC = CE = EF = FG$ (see the left figure).
This will result in isosceles triangles $BCE$ (with an angle of $20^{\circ}$ at vertex $E$), $CEF$ (with an angle of $60^{\circ}$ at the base $CF$), $EFG$ (with an angle of $100^{\circ}$ at... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Galperin G.A.
Point $P$ lies inside isosceles triangle $ABC (AB = BC)$, and $\angle ABC = 80^{\circ}, \angle PAC = 40^{\circ}$, $\angle ACP = 30^{\circ}$. Find the angle $BPC$. | Let $Q$ be the point of intersection of segment $CP$ with the height $BH$ of the given isosceles triangle, dropped to the base.
Then $\angle QAC = \angle ACQ = 30^{\circ}$, $\angle BAQ = 20... | 100 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Inscribed Circles in a Segment ]
On the diameter $AB$ of circle $S$, a point $K$ is taken, and a perpendicular is erected from it, intersecting $S$ at point $L$. Circles $S_{\mathrm{A}}$ and $S_{\mathrm{B}}$ are tangent to circle $S$, segment $LK$, and diameter $AB$, specifically, $S_{\mathrm{A}}$ is tangent to segm... | Let $\angle L A B=\alpha$ and $\angle L B A=\beta\left(\alpha+\beta=90^{\circ}\right)$. According to problem Z.42, b) $A B_{1}=A L$, so $\angle A B_{1} L=90^{\circ}-\alpha / 2$. Similarly, $\angle B A_{1} L=90^{\circ}-\boldsymbol{\beta} / 2$. Therefore, $\angle A_{1} L B_{1}=(\boldsymbol{\alpha}+\boldsymbol{\beta}) / 2... | 45 | Geometry | proof | Yes | Yes | olympiads | false |
Quadrilateral $ABCD$ is inscribed in a circle. It is known that $AC \perp BD$. Find the length of $BC$, if the distance from the center of the circle to side $AD$ is 2. | Draw the diameter $B B_{1}$. Then $C B_{1}=A D$.
## Solution
Let $B_{1}$ be the point diametrically opposite to point $B$. Since $B_{1} D \perp B D\left(B B_{1}\right.$ - the diameter of the circle) and $A C \perp B D$, then $B_{1} D \| A C$. Therefore, $C B_{1}=A D$, and since equal chords are equidistant from the c... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Two pairs of similar triangles ] [ Quadratic equations. Vieta's theorem ]
In the angles $B$ and $C$ of triangle $ABC$, two circles with radii 2 and 3 are inscribed, touching the angle bisector of angle $A$ of the triangle.
Find this bisector, if the distance between the points where the circles touch $BC$ is 7. | Let a circle of radius 2 with center $O_{1}$ touch the bisector $A D$ and side $B C$ of triangle $A B C$ at points $M$ and $K$ respectively, and a circle of radius 3 with center $O_{2}$ touch $A D$ and $B C$ at points $N$ and $L$ respectively. Denote $\angle O_{1} D K=\alpha$. Since $D O_{1}$ and $D O_{2}$ are bisector... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In an acute-angled triangle $A B C$, altitudes $A A_{1}$ and $B B_{1}$ are drawn. The bisector of the exterior angle at vertex $C$ intersects the lines $A B$ and $A_{1} B_{1}$ at points $L$ and $K$ respectively. It turns out that $C L=2 C K$. Find the angle $C$. | As is known, triangle $C A_{1} B_{1}$ is similar to triangle $C A B$ with a similarity coefficient of $\cos \angle C$.
Let's assume that point $L$ lies on the extension of side $A B$ beyond point $B$. We will prove that point $C$ lies between $K$ and $L$. Suppose this is not the case. Then, if $P$ is a point on the ex... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
Find the lateral surface area of a regular triangular pyramid if its height is 4 and the apothem is 8. | Let $M$ be the center of the base $ABC$ of a regular triangular pyramid $ABCD$, and $K$ be the midpoint of $BC$. Then $DM$ is the height of the pyramid, $DK$ is the apothem of the pyramid, and $AK$ is the height of the equilateral triangle $ABC$, with $AK$ passing through point $M$. Denote $AB = BC = AC = a$. Then
$$
... | 288 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The height of the cone is 20, and the radius of the base is 25. Find the area of the section passing through the vertex if its distance from the center of the base of the cone is 12.
# | Let $A$ be the vertex of the cone, $O$ the center of the base, $BC$ a chord where the secant plane intersects the base of the cone, $M$ the midpoint of $BC$, and $OK$ the height of triangle $AOM$. Since $BC \perp OM$ and $BC \perp AO$, $BC$ is perpendicular to the plane $AOM$. Therefore, $AM \perp BC$ and $OK \perp BC$... | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 [ [sphere, circumscribed around a pyramid]
A regular quadrilateral pyramid is inscribed in a sphere. The radius of the sphere is 1. The plane angle at the vertex of the pyramid is $45^{\circ}$. Find the lateral surface area of the pyramid.
# | The center $O$ of the sphere is equidistant from the vertices of the base $A B C D$ of the regular pyramid $P A B C D$, so the point $O$ lies on the height $P Q$ of the pyramid (Fig.1). Let
$$
A B=B C=C D=A D=a, P A=P B=P C=P D=l
$$
By the cosine theorem,
$$
A B 2=P A 2+P B 2-2 P A \cdot P B \cos \angle A P B
$$
or... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
The edge of a regular tetrahedron is $4 \sqrt{6}$. Find the radius of the sphere that touches the lateral faces of the tetrahedron at points lying on the sides of the base. | Let the sphere with center $O$ touch the planes of the faces $A B D, B C D$, and $A C D$ of the given regular tetrahedron $A B C D$ at points $N, M$, and $K$, respectively, which lie on the sides $A B, B C$, and $A C$ of the base $A B C$. Then the section of the sphere by the plane of the base $A B C$ is a circle inscr... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 [ Perpendicularity of a line and a plane (other).]
The base of the pyramid is an equilateral triangle with a side length of 6. One of the lateral edges is perpendicular to the base plane and equals 4. Find the radius of the sphere circumscribed around the pyramid.
# | Let $Q$ be the center of the base $ABC$ of the triangular pyramid $ABCD$, where the lateral edge $AD$ is perpendicular to the plane of the base, and $AB = BC = AC = a = 6$, $AD = h = 4$. The center $O$ of the sphere with radius $R$, circumscribed around the pyramid $ABCD$, lies on the line perpendicular to the plane of... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A sphere with a radius of 1 is inscribed in a cone, the height of which is twice the diameter of the sphere. Find the ratio of the total surface area of the cone to the surface area of the sphere. | Consider the axial section of a cone with vertex $A$. We obtain an isosceles triangle $A B C$ with vertex $A$, in which a circle is inscribed, the center $O$ of which coincides with the center of the sphere inscribed in the cone. Let $M$ and $K$ be the points of tangency of the circle with the sides $B C$ and $A B$, re... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A metallic sphere with a radius of $\sqrt[z]{16}$ is recast into a cone, the lateral surface area of which is three times the area of the base. Find the height of the cone. | Let $R=\sqrt[3]{16}$ be the radius of the sphere, $V$ its volume, $r$ the radius of the base of the cone, $l$ the slant height, $h$ the height of the cone, $S$ the lateral surface area of the cone, and $s$ the area of its base. According to the problem,
$$
S=3 s, \text { or } \pi r l=3 \pi r^2 \text {, }
$$
$$
l=3 r,... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The base of an oblique parallelepiped is a rhombus with a side length of 60. The plane of the diagonal section passing through the larger diagonal of the base is perpendicular to the plane of the base. The area of this section is 7200. Find the smaller diagonal of the base, if the lateral edge is 80 and forms an angle ... | Let the plane of the diagonal section $A A1 C1 C$, passing through the larger diagonal $AC$ of the base $ABCD$ of the parallelepiped $ABCD A1 B1 C1 D1$, be perpendicular to the base $ABCD$. Then the perpendicular $C1K$, dropped from the vertex $C1$ to the plane of the base $ABCD$, is the height of the parallelepiped $A... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the plane $\alpha$, two perpendicular lines are drawn. The line $l$ forms angles of $45^{\circ}$ and $60^{\circ}$ with them. Find the angle between the line $l$ and the plane $\alpha$. | Let the line $l$ form angles of $60^{\circ}$ and $45^{\circ}$ with the lines $a$ and $b$ in the plane $\alpha$, respectively, and intersects this plane at point $P$. Take a point $M$ on the line $l$ such that $P M=1$. Let $O$ be the orthogonal projection of point $M$ onto the plane $\alpha$, and let $A$ and $B$ be the ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Ivan the Tsarevich went to look for Vasilisa the Beautiful, who had been kidnapped by Koschei. On his way, he met the Forest Spirit.
- I know, - he said, - the way to Koschei's Kingdom. I have been there before. I walked for four days and four nights. In the first day and night, I covered a third of the way - a straig... | Try to draw the path of the Leshy.
## Solution
The path of the Leshy is shown in the figure.
Ivan the Tsarevich can only travel $1 / 3$ of the Leshy's path - $1 / 6$ to the north and $1 /... | 300 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Folklore
Petya wants to make an unusual die, which, as usual, should have the shape of a cube, with dots drawn on its faces (different numbers of dots on different faces), but at the same time, on any two adjacent faces, the number of dots must differ by at least two (at the same time, it is allowed for some faces to ... | Arrange six numbers (according to the number of dots on the faces of a cube) in ascending order. We will prove that there are no three consecutive numbers among them. Suppose such three numbers exist: $a, a+1$, and $a+2$. Then the numbers $a$ and $a+1$ must be on opposite faces of the cube. No matter where we place the... | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[ Measurement of lengths of segments and measures of angles. Adjacent angles. ] [ Central angle. Length of an arc and circumference. ]
Determine the angle between the hour and minute hands of a clock showing 1 hour and 10 minutes, given that both hands move at constant speeds. | At 1:00, the minute hand was "behind" the hour hand by $30^{\circ}$. In the 10 minutes that passed after this moment, the hour hand will "travel" $5^{\circ}$, and the minute hand $-60^{\circ}$ (see problem $\underline{54776}$ ), so the angle between them is $60^{\circ}-30^{\circ}-5^{\circ}=25^{\circ}$.
## Answer
Prob... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ [Measuring lengths of segments and measures of angles. Adjacent angles. $]$ Central angle. Length of an arc and circumference ]
Determine the angle between the hour and minute hands of a clock showing 4 hours and 10 minutes, given that both hands move at constant speeds.
# | At 4:00, the minute hand was "behind" the hour hand by $120^{\circ}$. In the 10 minutes that passed after this moment, the hour hand will "travel" $5^{\circ}$, and the minute hand will travel $60^{\circ}$ (see problem $\underline{54776}$), so the angle between them is $120^{\circ}+5^{\circ}-60^{\circ}=65^{\circ}$.
## ... | 65 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The bases of an isosceles trapezoid are 10 and 24, and the lateral side is 25. Find the height of the trapezoid.
# | Drop perpendiculars from the vertices of the smaller base to the larger base.
## Solution
Drop perpendiculars $B M$ and $C N$ from the vertices $B$ and $C$ of the smaller base $B C$ of trapezoid $A B C D$ to the larger base $A D$. Then $A M = D N = \frac{1}{2}(A D - B C) = 7$.
By the Pythagorean theorem, $B M^{2} = ... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Khachaturyan A.V.
A rectangular sheet of paper was folded, aligning a vertex with the midpoint of the opposite shorter side (see figure). It turned out that triangles I and II are equal. Find the longer side of the rectangle if the shorter side is 8.
.
We see that the length of the smaller side is $a+b$. The... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
| $\left[\begin{array}{l}\text { [ Right triangle with an angle of \$30^{\circ}\$ } \\ \text { [ Parallel lines, properties and criteria. Transversals ] }\end{array}\right.$ |
| :---: | :---: |
| | $\left[\begin{array}{ll}\text { Chords and secants (other) }\end{array}\right]$ |
In a circle, a diameter $A B$ and a ch... | Let $O$ be the center of the given circle, and $K$ be the foot of the perpendicular dropped from $O$ to $CD$ (see the figure). Then, in the right triangle $OSK$, the leg $OK$ is half the hypotenuse $OC$, so $\angle OCK = 30^{\circ}$. Since $CD \parallel AB$, then $\angle AOC = \angle OCK = 30^{\circ}$. Triangle $AOC$ i... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Biyankov Yu.A.
In an isosceles triangle \(ABC\), a point \(M\) is marked on the lateral side \(BC\) such that the segment \(MC\) is equal to the height of the triangle dropped to this side, and a point \(K\) is marked on the lateral side \(AB\) such that the angle \(KMC\) is a right angle. Find the angle \(ACK\). | Given that $CL = CM$ (see the figure). Therefore, right triangles $CLK$ and $CMK$ are congruent by hypotenuse and leg, and $CK$ is the angle bisector of $\angle LCM$. Let $\angle BAC = \angle BCA = \alpha$, then $\angle LCA = 90^\circ - \alpha$, and $\angle LCB = \alpha - (90^\circ - \alpha) = 2\alpha - 90^\circ$. Thus... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Pythagorean Theorem (direct and inverse) $]$ [ Diameter, chords, and secants $]$
The tangent and the secant, drawn from the same point to the circle, are perpendicular to each other. The tangent is 12, and the inner part of the secant is 10. Find the radius of the circle. | The distance from the center of the circle to the secant is equal to the given tangent segment.
## Solution
Let $O$ be the center of the circle, $M$ be the common point of the tangent and the secant, $A$ be the point of tangency, $B C$ be the inner part of the secant $M C$, and $K$ be the midpoint of $B C$. Then,
$O... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The altitudes of an acute-angled triangle $ABC$, drawn from vertices $A$ and $B$, intersect at point $H$, and $\angle AHB = 120^{\circ}$. The angle bisectors drawn from vertices $B$ and $C$ intersect at point $K$, and $\angle BKC = 130^{\circ}$. Find the angle $ABC$. | Since $\angle A H B=120^{\circ}$, and angle $A C B-$ is acute, then $\angle C=180^{\circ}-120^{\circ}=60^{\circ}$.
Since $\angle B K C=90^{\circ}+1 / 2 \angle A$, then $\angle A=80^{\circ}$.
Therefore, $\angle B=180^{\circ}-60^{\circ}-80^{\circ}=40^{\circ}$.
## Answer
$40^{\circ}$. | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In an isosceles trapezoid $A B C D$ the bases $A D=12, B C=6$, the height is 4. The diagonal $A C$ divides the angle $B A D$ of the trapezoid into two parts. Which one is larger? | Consider triangle $A B C$.
## Solution
Let $P$ and $Q$ be the projections of vertices $B$ and $C$ onto the base $A D$ of trapezoid $A B C D$. Then $A P=D Q=1 / 2(A D-B C)=3$. By the Pythagorean theorem, $A B^{2}=A P^{2}+B P^{2}=25$.
In triangle $A B C$, side $B C$ is greater than side $A B$ (6>5). Therefore, $\angle... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the hypotenuse $AB$ of the right triangle $ABC$, points $M$ and $N$ are taken such that $BC = BM$ and $AC = AN$.
Prove that $\angle MCN = 45^{\circ}$. | $\angle C M N=90^{\circ}-1 / 2 \angle B, \angle C N M=90^{\circ}-1 / 2 \angle A$, therefore, $\angle M C N=1 / 2(\angle A+\angle B)=45^{\circ}$.
Send a comment | 45 | Geometry | proof | Yes | Yes | olympiads | false |
On side $A B$ of triangle $A B C$, a point $K$ is marked. Segment $C K$ intersects the median $A M$ of the triangle at point $P$. It turns out that $A K = A P$.
Find the ratio $B K: P M$. | The first method. Draw a line through point $M$ parallel to $C K$, which intersects $A B$ at point $D$ (left figure). By Thales' theorem, $B D=K D$. By the theorem of proportional segments, $P M=K D=1 / 2 B K$.
. Then $O O_{1}$ is the bisector of angle $A O B$,
$$
A O_{1}=1, O O_{1}=2, \angle O A O_{1}=90^{\cir... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a quadrilateral pyramid $S A B C D$, the base is a trapezoid $A B C D (B C \| A D), B C=\frac{4}{5} A D, \angle A S D=\angle$ $C D S=\frac{\pi}{2}$. All vertices of the pyramid lie on the circles of the bases of a cylinder, the height of which is 2, and the radius of the base is $\frac{\frac{5}{3}}{3}$. Find the vol... | If vertices $A$ and $D$ are located on different bases of the cylinder, then by the theorem of the intersection of two parallel planes by a third plane, $A B \| C D$, i.e., $A B C D$ is a parallelogram, not a trapezoid. Let vertices $A$ and $D$ be located on the circumference of one base of the cylinder. Then the media... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Sum of angles in a triangle. Theorem of the exterior angle.] Pentagons
In an equilateral (irregular) pentagon $A B C D E$, angle $A B C$ is twice the angle $D B E$. Find the measure of angle $A B C$.
# | Let the angles at the base $BE$ of the isosceles triangle $ABE$ be $\alpha$, and the angles at the base $BD$ of the isosceles triangle $BCD$ be $\beta$. From the condition, it follows that $\alpha + \beta = \angle DBE$. Therefore, $\angle AED + \angle CDE = (\alpha + \angle BED) + (\beta + \angle BDE) = (\alpha + \beta... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In triangle $A B C$, the lengths of two sides are given: $A B=6, B C=16$. Additionally, it is known that the center of the circle passing through vertex $B$ and the midpoints of sides $A B$ and $A C$ lies on the bisector of angle $C$. Find $A C$. | Let $N$ and $M$ be the midpoints of sides $AB$ and $AC$ of triangle $ABC$, $O$ be the center of the circle passing through points $B$, $N$, and $M$, and $K$ be the point of intersection of this circle with side $BC$ other than $B$.
Since $MN$ is the midline of triangle $ABC$, chords $MN$ and $BK$ are parallel, therefo... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Angle Bisectors, Concurrency Problem $\underline{115562}$ topics: [Area of a Triangle (using semiperimeter and radius of inscribed or exscribed circle [ Inscribed, circumscribed, and exscribed circles; their radii
A circle with center $O$, inscribed in triangle $A B C$, touches its sides $A B$ and $A C$ at points $... | Let $r$ be the radius of the circle inscribed in triangle $ABC$, and $p$ be the semiperimeter of triangle $ABC$. Then,
$$
p=\frac{AB+BC+AC}{2}=\frac{13+14+15}{2}=21
$$
By Heron's formula,
$$
S_{\triangle ABC}=\sqrt{p(p-AB)(p-BC)(p-AC)}=\sqrt{21 \cdot 8 \cdot 6 \cdot \bar{i}}=84
$$
Thus,
$$
r=\frac{S_{\triangle A}+... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$[$ Properties of Sections $]$ $[$ Tetrahedron and Pyramid $]$
The base of a quadrilateral pyramid $S A B C D$ is a parallelogram $A B C D$. 1) Construct the section of the pyramid by a plane passing through the midpoint of edge $A B$ and parallel to the plane $S A D$. 2) Find the area of the resulting section if the ... | 1) Let $M$ be the midpoint of edge $A B$. According to the theorem of the intersection of two parallel planes by a third intersecting plane, the intersecting plane intersects the plane of face $A S B$ along a line parallel to $A S$, i.e., along the midline $M N$ of triangle $A S B$. Similarly, the intersecting plane in... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Area and Volume (Extremum Problems).]
Consider all possible rectangular parallelepipeds whose bases are squares and each of whose lateral faces has a perimeter of 6. Find among them the parallelepiped with the greatest volume and calculate this volume. | Let's denote the side of the base of the rectangular parallelepiped by $x$. Then its lateral edge is $\frac{1}{2}(6-2 x) = 3 - x$. If $V(x)$ is the volume of the parallelepiped, then
$$
V(x) = x^2 (3 - x),
$$
so the problem reduces to finding the maximum value of the function $V(x) = x^2 (3 - x)$ on the interval $(0 ... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.