problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
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|---|---|---|---|---|---|---|---|---|
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,
$$
S=3 s, \text { or } \pi r l=3 \pi r^2 \text {, }
$$
$$
l=3 r,... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Khachaturyan A.V.
A rectangular sheet of paper was folded, aligning a vertex with the midpoint of the opposite shorter side (see figure). It turned out that triangles I and II are equal. Find the longer side of the rectangle if the shorter side is 8.
.
We see that the length of the smaller side is $a+b$. The... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On side $A B$ of triangle $A B C$, a point $K$ is marked. Segment $C K$ intersects the median $A M$ of the triangle at point $P$. It turns out that $A K = A P$.
Find the ratio $B K: P M$. | The first method. Draw a line through point $M$ parallel to $C K$, which intersects $A B$ at point $D$ (left figure). By Thales' theorem, $B D=K D$. By the theorem of proportional segments, $P M=K D=1 / 2 B K$.
, B C=\frac{4}{5} A D, \angle A S D=\angle$ $C D S=\frac{\pi}{2}$. All vertices of the pyramid lie on the circles of the bases of a cylinder, the height of which is 2, and the radius of the base is $\frac{\frac{5}{3}}{3}$. Find the vol... | If vertices $A$ and $D$ are located on different bases of the cylinder, then by the theorem of the intersection of two parallel planes by a third plane, $A B \| C D$, i.e., $A B C D$ is a parallelogram, not a trapezoid. Let vertices $A$ and $D$ be located on the circumference of one base of the cylinder. Then the media... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Angle Bisectors, Concurrency Problem $\underline{115562}$ topics: [Area of a Triangle (using semiperimeter and radius of inscribed or exscribed circle [ Inscribed, circumscribed, and exscribed circles; their radii
A circle with center $O$, inscribed in triangle $A B C$, touches its sides $A B$ and $A C$ at points $... | Let $r$ be the radius of the circle inscribed in triangle $ABC$, and $p$ be the semiperimeter of triangle $ABC$. Then,
$$
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 |
$\left[\begin{array}{l}{[\text { Problems on maximum and minimum (miscellaneous). }} \\ {[\quad \underline{\text { Rectangular parallelepipeds }}]}\end{array}\right]$
Consider all possible rectangular parallelepipeds, each with a volume of 4, and whose bases are squares. Find among them the parallelepiped with the sma... | Let $x$ be the side of the base of a rectangular parallelepiped. Then its lateral edge is $\frac{4}{x^{2}}$. If $P(x)$ is the perimeter of the lateral face of the parallelepiped, then
$$
P(x)=\frac{B}{x^{2}}+2 x
$$
Thus, the problem reduces to finding the minimum value of the function $P(x)=\frac{8}{x^{2}}+2 x$ on th... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11 Find the volume of a rectangular parallelepiped, the areas of the diagonal sections of which are equal to $\sqrt{13}, 2 \sqrt{10}$ and $3 \sqrt{5}$ | Let $A B C D A 1 B 1 C 1 D 1$ be a rectangular parallelepiped, and
$$
S_{B B} 1 D 1 D=\sqrt{13}, S_{C D A} 1 B 1=2 \sqrt{10}, S_{A D C} 1 B 1=3 \sqrt{5}
$$
Denote $A B=x, B C=y, A A 1=z$. Then
$$
B D 2=x 2+y 2, A 1 D 2=y 2+z 2, A B 1=x 2+z 2
$$
Since $B B 1 D 1 D, C D A 1 B 1$, and $A D C 1 B 1$ are rectangles with... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The center of a circle touching the legs $A C$ and $B C$ of a right triangle $A B C$ lies on the hypotenuse $A B$. Find the radius of the circle, if it is six times smaller than the sum of the legs, and the area of triangle $A B C$ is 27. | Let $O$ be the center of the circle, $M$ and $N$ be the points of tangency with the legs $A C$ and $B C$ respectively. Then $O M$ and $O N$ are the altitudes of triangles $A O C$ and $B O C$. Let $B C=a, A C=b, O M=O N=r$. According to the problem, $a+b=6 r$. Then
$$
27=S_{\triangle A B C}=S_{\triangle A O C}+S_{\tria... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
On the side $A D$ of the inscribed quadrilateral $A B C D$, there is the center of the circle that touches the other three sides of the quadrilateral. Find $A D$, if $A B=2$ and $C D=3$.
# | Prove that $A D=A B+C D$.
## Answer
5.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Sum of angles in a triangle. Theorem about the exterior angle. ] [ Criteria and properties of an isosceles triangle. ]
In triangle $A B C$, angle $C$ is three times angle $A$. A point $D$ is taken on side $A B$ such that $B D=B C$.
Find $C D$, if $A D=4$. | Let $\angle A=\alpha, \angle A C D=\varphi$, then $\angle C=3 \alpha, \angle B C D=\alpha+\varphi$. Therefore, $\varphi+(\alpha+\varphi)=3 \alpha$, that is, $\varphi=\alpha$. Thus, triangle $A D C$ is isosceles with base $A C$. Therefore, $C D=A D=4$.
, and the path between $M$ and $N$ intersects the edge $A C$. Consider the net of the tetrahedron, consisting of four equilateral triangles: $A B C, D_1 A B, D_2 A C$, and $D_3 B C$ (Fig.2). Then the shortes... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9
[ Ratio in which the bisector divides the side ]
In an isosceles triangle $A B C$, a rhombus $D E C F$ is inscribed such that vertex $E$ lies on side $B C$, vertex $F$ on side $A C$, and vertex $D$ on side $A B$. Find the length of the side of the rhombus if $A B=B C=12, A C=6$. | Let $x$ be the side of the rhombus. Since $D E \| A C$, triangles $D B E$ and $A B C$ are similar. Therefore, $B E: B C=D E: A C$, or $12-x / 12=x / 6$. From this, $x=4$.
## Answer
4. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In the square $A B C D$ with side 1, point $F$ is the midpoint of side $B C$, and $E$ is the foot of the perpendicular dropped from vertex $A$ to $D F$.
Find the length of $B E$.
# | Extend segment $D F$ to intersect line $A B$ at point $K$ (see figure). Right triangles $K B F$ and $D C F$ are equal (by leg and acute angle), so $K B = D C$. Therefore, $E B$ is the median of the right triangle $A E K$, drawn to the hypotenuse, hence $B E = 1 / 2 A K = 1$.
$, from point $E$ - the midpoint of $C D$, a perpendicular $E F$ to line $A B$ is drawn. Find the area of the trapezoid if $A B=5, E F=4$. | It is sufficient to prove that the area of the trapezoid is twice the area of triangle $ABE$ (which is 10). The first method. Draw a line through point $E$ parallel to the side $AB$, and mark the points $P$ and $Q$ where it intersects the lines $AD$ and $BC$ respectively (see figure).
]
Quadrilateral $ABCD$ is inscribed in a circle. The bisectors of angles $B$ and $C$ intersect at a point lying on the segment $AD$.
Find $AD$, if $AB=5, CD=3$. | Let $M$ be the point of intersection of the bisectors of angles $B$ and $C$.
First method. Denote: $\angle A B C=2 \alpha, \angle B C D=2 \beta$, then $\angle A D C=180^{\circ}-2 \alpha, \angle B A D=180^{\circ}-2 \beta$ (see fig.).
. It is known that through any point inside the circle, no more than two chords with endpoints at the marked points pass. We will call a matching a set of $N$ chords with endpoints at the marked points such that each marked point is the endpoint ... | Let the marked points be $A_{1}, A_{2}, \ldots, A_{2 N}$ in the order of clockwise traversal of the circle. We will prove by induction on $N$ that the number of even matchings is one more than the number of odd matchings. For $N=1$, the statement is obvious: there is only one matching, and it is even.
Inductive step. ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
The plan of the palace of the Shah is a $6 \times 6$ square, divided into rooms of size $1 \times 1$. In the middle of each wall between the rooms, there is a door. The Shah said to his architect: "Knock down some of the walls so that all rooms become $2 \times 1$, no new doors appear, and the path bet... | Consider an arbitrary path from the bottom-left corner of the palace to the top-right corner. Since one needs to "climb" 5 horizontal levels and "move right" 5 vertical levels, one has to pass through at least 10 doors, visiting at least 11 rooms (including the starting and ending rooms).
11 rooms of size $1 \times 1$... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Rubaev I.S.
There is a square grid paper with a size of $102 \times 102$ cells and a connected figure of unknown shape, consisting of 101 cells. What is the maximum number of such figures that can be guaranteed to be cut out from this square? A figure composed of cells is called connected if any two of its cells can b... | Lemma. Any connected figure composed of 101 cells can be enclosed in a rectangle with sides \(a\) and \(b\) such that \(a + b = 102\).
Take two cells of our figure that share a common side. They form a rectangle \(1 \times 2\), the sum of whose sides is 3. Due to the connectivity of the given figure, there will be a c... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Eveokkimov $M$.
A polyhedron is circumscribed around a sphere. We will call a face of the polyhedron large if the projection of the sphere onto the plane of the face is entirely contained within the face. Prove that there are no more than 6 large faces. | The first solution. If two large faces are not parallel, then the dihedral angle containing these faces cannot be obtuse (in Fig. 1, projections of the sphere and these two faces onto a plane perpendicular to their line of intersection are shown, with the projections of the large faces highlighted). Therefore, if we er... | 6 | Geometry | proof | Yes | Yes | olympiads | false |
Quadrilateral $ABCD$, whose diagonals are perpendicular to each other, is inscribed in a circle.
Perpendiculars dropped from vertices $B$ and $C$ to side $AD$ intersect diagonals $AC$ and $BD$ at points $E$ and $F$ respectively. Find $EF$, if $BC=1$. | Prove that BCFE is a parallelogram.
## Solution
First method.
Let $\angle C B D=\alpha$. Then
$$
\angle C A D=\alpha, \angle B E C=90^{\circ}-\alpha, \angle D B E=\alpha
$$
Therefore, $B E=B C$.
Let $\angle A C B=\beta$. Similarly, we can prove that $\angle A C F=\beta$. Therefore, $C F=B C$. Hence, $B E=C F$, an... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Segment $K L$ is the diameter of a certain circle. Through its endpoints $K$ and $L$, two lines are drawn, intersecting the circle at points $P$ and $Q$, respectively, lying on the same side of the line $K L$. Find the radius of the circle if $\angle P K L=60^{\circ}$ and the point of intersection of the lines $K P$ an... | Triangle $K M L-$ is equilateral ( $M$ - the intersection point of lines $K P$ and $L Q$).
## Solution
Let $M$ be the intersection point of lines $K P$ and $L Q$. Point $M$ cannot lie on the circle. If $M$ is inside the circle, then $K M \cdot M P = L M \cdot M Q$. Therefore, $K M = M L$, which is impossible.
If poi... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In trapezoid $A B C E$ base $A E$ is equal to $16, C E=8 \sqrt{3}$. The circle passing through points $A, B$ and $C$ intersects line $A E$ again at point $H ; \angle A H B=60^{\circ}$. Find $A C$. | Apply the Law of Cosines.
## Solution
Let $O_{1}$ be the center of the given circle, $N$ be its point of tangency with the line $A C$, and $K$ be with the side $B C$.
From the right triangle $A O_{1} N$, we find that
$$
A N=O_{1} N \operatorname{ctg} 30^{\circ}=\sqrt{3}
$$
On the other hand, $A N$ is equal to the ... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Based on the base $AB$ of isosceles triangle $ABC$, a point $D$ is taken such that $BD - AD = 4$. Find the distance between the points where the incircles of triangles $ACD$ and $BCD$ touch the segment $CD$. | If the circle inscribed in triangle $P Q R$ touches side $P Q$ at point $S$, then $P S=\frac{P Q+P R-R Q}{2}$.
## Solution
Let the circles inscribed in triangles $A C D$ and $B C D$ touch segment $C D$ at points $M$ and $N$ respectively. Since $A C=B C$, and
$$
C M=\frac{A C+C D-A D}{2}, C N=\frac{B C+C D-B D}{2},
$... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9
A regular octagon with side 1 is cut into parallelograms. Prove that among them there are at least two rectangles, and the sum of the areas of all rectangles is 2.
# | Let's highlight two mutually perpendicular pairs of opposite sides in a regular octagon and consider, as in problem 25.1, chains of parallelograms connecting opposite sides. At the intersections of these chains, there are rectangles. By considering two other pairs of opposite sides, we will obtain at least one more rec... | 2 | Geometry | proof | Yes | Yes | olympiads | false |
Find the distance between the points of tangency of the circles inscribed in triangles $A B C$ and $C D A$ with side $AC$, if
a) $A B=5, B C=7, C D=D A$;
b) $A B=7, B C=C D, D A=9$. | The distance from the vertex of a triangle to the nearest point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side of the triangle ($x = p - a$).
## Solution
a) Let the inscribed circle of triangle $ABC$ touch side $AC$ at point $K$, and the inscribed circ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underline{53619}$ topics:
In an isosceles triangle $ABC$ with base $AB$, the bisector $BD$ is drawn. A point $E$ is taken on the line $AB$ such that $\angle EDB=90^{\circ}$.
Find $BE$, if $AD=1$. | Connect point $D$ with the midpoint of segment $C E$.
## Solution
Let $M$ be the midpoint of $B E$. Then $D M$ is the median of the right triangle $E D C$, drawn to the hypotenuse $E B$, so $D M=1 / 2 B E=B M$. Therefore,
$\angle D M A=2 \angle M B D=\angle B=\angle A=\angle D A M$. Hence, $D M=D A=1, B E=2 D M=2$.
... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ $\left[\begin{array}{l}\text { The ratio in which the bisector divides the side } \\ {[\text { Law of Cosines }}\end{array}\right]$
In triangle $ABC$, the bisectors $BL$ and $AE$ of angles $ABC$ and $BAC$ respectively are drawn, intersecting at point $O$. It is known that $AB=BL$, the perimeter of triangle $ABC$ is ... | Use the property of the bisector of a triangle and the cosine theorem.
## Solution
Let $O L=2 a, B O=4 a$. Then $A B=B L=6 a$. Since $A O$ is the bisector of triangle $A B L$, then $\frac{A B}{A L}=\frac{B O}{O L}=2$, so $A L=\frac{1}{2} A B=3 a$.
Let $C L=b$. Since $B L$ is the bisector of triangle $A B C$, then $\... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Radii of the inscribed, circumscribed, and exscribed circles (other) [ Area of a triangle (through the semiperimeter and the radius of the inscribed or exscribed circle).
Through the center $O$ of the inscribed circle $\omega$ of triangle $A B C$, a line parallel to side $B C$ is drawn, intersecting sides $A B$ and ... | Let $r$ be the radius of $\omega, p$ be the semiperimeter of triangle $ABC, P$ be the point of tangency of $\omega$ with side $AB$. Then $AP^2 = AO^2 - OP^2 = 15r^2$, $AP = p - BC = AP = r\sqrt{15} + 2$,
$\sqrt{15} = S_{ABC} = pr = (r\sqrt{15} + 2)r$, from which $AP = r\sqrt{15} = 3, p = 5$.
The perimeter of triangle... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## angle between the tangent and the chord [Angles subtended by equal arcs and equal chords]
A circle touches the sides $A C$ and $B C$ of triangle $A B C$ at points $A$ and $B$ respectively. On the arc of this circle, lying inside the triangle, there is a point $K$ such that the distances from it to the sides $A C$ a... | Let $M, H$ and $N$ be the feet of the perpendiculars dropped from point $K$ to $AC, AB$ and $BC$ respectively. Points $M$ and $H$ lie on the circle with diameter $AK$. From the theorem about the angle between a tangent and a chord, $\angle K M H = \angle K A H = \angle K A B = \angle K B N$.
Points $N$ and $H$ lie on ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 [ Properties and characteristics of an isosceles triangle.]
Quadrilateral \(ABCD\) is inscribed in a circle, \(M\) is the intersection point of its diagonals, \(O_1\) and \(O_2\) are the centers of the inscribed circles of triangles \(ABM\) and \(CMD\) respectively, \(K\) is the midpoint of the arc \(AD\) not con... | Point $K$ is the midpoint of arc $AD$, so $BK$ is the bisector of the inscribed angle $ABD$, which means point $O_{1}$ lies on segment $BK$. Similarly, point $O_{2}$ lies on segment $CK$. The rays $MO_{1}$ and $MO_{2}$ are the bisectors of the vertical angles
$AMB$ and $CMD$, so point $M$ lies on segment $O_{1}O_{2}$ ... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The radii of two circles are 2 and 4. Their common internal tangents are perpendicular to each other. Find the length of each of them. | For each of the given circles, the quadrilateral formed by the indicated tangents and radii drawn to the points of tangency is a square.
## Solution
For each of the given circles, the quadrilateral formed by the tangents and radii drawn to the points of tangency is a square. Therefore, the length of the desired tange... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Pythagorean Theorem (direct and inverse).] [Tangent properties and criteria]
Circles with radii 8 and 3 touch each other internally. A tangent is drawn from the center of the larger circle to the smaller circle. Find the length of this tangent. | The line connecting the centers of two tangent circles passes through their point of tangency.
## Solution
Let $O$ and $O_{1}$ be the centers of circles with radii 8 and 3, respectively, and $A$ be the point of tangency, $O M-$ be the desired tangent. Then
$$
O O_{1}=O A-O O_{1}=8-3=5
$$
Therefore,
$$
O M^{2}=O O^... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Properties and signs of tangents ] [ Rectangles and squares. Properties and signs ]
Two circles are given. Their common internal tangents are perpendicular to each other. The chords connecting the points of tangency are 3 and 5. Find the distance between the centers of the circles. | For each of the circles, the quadrilateral formed by the tangents and radii drawn to the points of tangency is a square.
## Solution
For each of the circles, the quadrilateral formed by the tangents and radii drawn to the points of tangency is a square. The segment connecting the point of intersection of the tangents... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The perimeter of triangle $ABC$ is 8. An incircle is inscribed in the triangle, and a tangent is drawn to it, parallel to side $AB$. The segment of this tangent, enclosed between sides $AC$ and $CB$, is equal to 1. Find the side $AB$.
# | The truncated triangle is similar to the given one with a coefficient equal to the ratio of their perimeters.
## Solution
Let the points of intersection of the tangent with sides $A C$ and $C B$ be $M$ and $N$, and the points of tangency of these sides with the inscribed circle be $P$ and $Q$. Then the semiperimeter ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Right triangle $A B C$ is divided by the height $C D$, drawn to the hypotenuse, into two triangles: $B C D$ and $A C D$. The radii of the circles inscribed in these triangles are 4 and 3, respectively. Find the radius of the circle inscribed in triangle $A B C$. | Let $r$ be the desired radius, $r_{1}$ and $r_{2}$ be the radii of the given circles. From the similarity of triangles $A C D$, $C B D$, and $A B C$, it follows that $r_{1}: r_{2}: r = A C: B C: A B$. Therefore, $r^{2} = r_{1}^{2} + r_{2}^{2} = 25$.
=30^{\circ}
$$
Let $H$ be the foot of the perpendicular dropped from vertex $C$ to $A D$. Then $A H=\frac{1}{2}(A D+B C)$, i.e., segment $A H$ is equal to the ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The circle constructed on the larger lateral side $AB$ of the right trapezoid $ABCD$ as its diameter intersects the base $AD$ at its midpoint. It is known that $AB=10, CD=6$. Find the midline of the trapezoid. | Let $M$ be the midpoint of $A D$. Then $\angle A M B=90^{\circ}$, so the quadrilateral $M B C D$ is a rectangle, $B M=C D=6$. From the right triangle $A B M$ we find that
$$
A M=\sqrt{A B^{2}-B M^{2}}=\sqrt{100-36}=8
$$
Then $B C=M D=A M=8$. Therefore, the midline of the trapezoid $A B C D$ is
$$
\frac{1}{2}(A D+B C... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Isosceles, inscribed, and circumscribed trapezoids ] [ Projections of the bases, sides, or vertices of a trapezoid ]
The height of an isosceles trapezoid, dropped from the vertex of the smaller base to the larger base, divides the larger base into segments that are in the ratio 2:3. How does the larger base relate t... | Let $CH$ be the height of the isosceles trapezoid $ABCD$, dropped from the vertex $C$ of the smaller base $BC$ to the larger base $AD$. Let $DH = 2t$, $AH = 3t$. Drop a perpendicular $BQ$ from vertex $B$ to the base $AD$. From the equality of the right triangles $AQB$ and $DHC$, it follows that $AQ = DH = 2t$, and sinc... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Isosceles, inscribed and circumscribed trapezoids]
The perimeter of a trapezoid circumscribed about a circle is 40. Find the midline of the trapezoid. | Since the trapezoid is circumscribed around a circle, the sums of its opposite sides are equal, meaning the sum of the bases is equal to the semiperimeter of the trapezoid, i.e., 20, and the midline of the trapezoid is equal to the half-sum of the bases, i.e., 10.
## Answer
10.00 | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\underline{115625}$ topics: [ Tangent circles [Mean proportionals in a right triangle]
Two circles touch each other externally at point $C$. A line is tangent to the first circle at point $A$ and to the second circle at point $B$. The line $A C$ intersects the second circle at point $D$, different from $C$. Find $B ... | Let the common tangent to the circles passing through point $C$ intersect segment $A B$ at point $M$. Then $M A = M C = M B$, i.e., the median $C M$ of triangle $A B C$ is equal to half of side $A B$, which means $\angle A C B = 90^{\circ}$.
The adjacent angle to $\angle A C B$, $\angle B C D$, is also $90^{\circ}$, s... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Folklore
Points $K$ and $L$ are the midpoints of sides $A B$ and $B C$ of a regular hexagon $A B C D E F$. Segments $K D$ and $L E$ intersect at point $M$. The area of triangle $D E M$ is 12. Find the area of quadrilateral KBLM. | Since quadrilateral $K B C D$ is the image of quadrilateral $L C D E$ under a rotation around the center of $A B C D E F$ by an angle of $60^{\circ}$, these quadrilaterals are equal, and therefore, have the same area (see figure). By subtracting the area of quadrilateral $L C D M$ from each of them, we get that $S_{K B... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $A$ be the following.
A numerical sequence is defined by the conditions: $a_{1}=1, a_{n+1}=a_{n}+\left[\sqrt{a_{n}}\right]$.
How many perfect squares occur among the first terms of this sequence, not exceeding
1000000? | According to the solution of problem $\underline{9152}$, all perfect squares in this sequence have the form $4^{m}$. There are exactly 10 numbers of this form within the specified limits
$\left(4^{10}=1024^{2}>10^{6}\right)$. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Boodanov I.I.
On the board, nine quadratic trinomials are written: $x^{2}+a_{1} x+b_{1}, x^{2}+a_{2} x+b_{2}, \ldots, x^{2}+a_{9} x+b_{9}$. It is known that the sequences $a_{1}, a_{2}, \ldots, a_{9}$ and $b_{1}, b_{2}, \ldots, b_{9}$ are arithmetic progressions. It turned out that the sum of all nine trinomials has a... | Let $P_{i}(x)=x^{2}+a_{i} x+b_{i}, P(x)=P_{1}(x)+\ldots+P_{9}(x)$. Note that $P_{i}(x)+P_{10-i}(x)=2 P_{5}(x)$. Therefore, $P(x)=$ $9 P_{5}(x)$, and the condition is equivalent to $P_{5}(x)$ having at least one root.
Let $x_{0}$ be one of its roots. Then $P_{i}\left(x_{0}\right)+P_{10-i}\left(x_{0}\right)=2 P_{5}\left... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For the numbers $1000^{2}, 1001^{2}, 1002^{2}, \ldots$, the last two digits are discarded. How many of the first terms of the resulting sequence form an arithmetic progression? | The general term of the original sequence is: $a_{n}=\left(10^{3}+n\right)^{2}=10^{6}+2 n \cdot 10^{3}+n^{2}$ ( $a_{0}-$ the first term).
Let the general term of the resulting sequence be denoted by $b_{n}$, then
$b_{n}=\left[\frac{a_{n}}{100}\right]=\left[\frac{10^{6}+2 \cdot n \cdot 10^{3}+n^{2}}{100}\right]=10^{4}... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$2+$ $[$ higher degree equations (miscellaneous). ]
Solve the equation $\left[x^{3}\right]+\left[x^{2}\right]+[x]=\{x\}-1$.
# | The number on the left is an integer, and, consequently, the number on the right is also an integer, that is, $\{x\}=0$. Therefore, $x$ is an integer, so the equation can be rewritten as $x^{3}+x^{2}+x=-1$, or $(x+1)\left(x^{2}+1\right)=0$.
## Answer
$x=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Folklore
The function $f(x)$ is defined for all $x$, except 1, and satisfies the equation: $(x-1) f\left(\frac{x+1}{x-1}\right)=x+f(x)$. Find $f(-1)$. | Substitute the values $x=0$ and $x=-1$ into the given equation. We get: $\left\{\begin{array}{c}-f(-1)=f(0), \\ -2 f(0)=-1+f(-1)\end{array}\right.$. Therefore, $2 f(-1)=-1+f(-1)$, which means $f(-1)=-1$.
## Answer
-1. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[
For all real $x$ and $y$, the equality $f\left(x^{2}+y\right)=f(x)+f\left(y^{2}\right)$ holds. Find $f(-1)$.
# | Substituting $x=0, y=0$, we get $f(0)=f(0)+f(0)$, that is, $f(0)=0$.
Substituting $x=0, y=-1$, we get $f(-1)=f(0)+f(1)$, that is, $f(-1)=f(1)$.
Substituting $x=-1, y=-1$, we get $f(0)=f(-1)+f(1)$. Therefore, $2 f(-1)=0$.
## Answer
0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Monotonicity, boundedness $\quad$ ]
Find the maximum value of the expression $a+b+c+d-ab-bc-cd-da$, if each of the numbers $a, b, c$ and $d$ belongs to the interval $[0,1]$.
# | Notice that $a+b+c+d-a b-b c-c d-d a=(a+c)+(b+d)-(a+c)(b+d)$. Let $a+c=x, b+d=y, 0$ $\leq x \leq 2$ and $0 \leq y \leq 2$.
$x+y-x y=(x-1)(1-y)+1$, where $|x-1| \leq 1$ and $|1-y| \leq 1$. Therefore, $(x-1)(1-y) \leq 1$, and $x+y-x y \leq 2$.
The value 2 is achieved, for example, if $a=c=1, b=d=0$.
## Answer
2
Solv... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Does there exist a natural number $n$ such that $\sqrt[n]{17 \sqrt{5}+38}+\sqrt[n]{17 \sqrt{5}-38}=2 \sqrt{5}$? | Note that $(\sqrt{5} \pm 2)^{3}=5 \sqrt{5} \pm 30+12 \sqrt{5} \pm 8=17 \sqrt{5} \pm 38$. Therefore, the given equality is true for $n=3$.
## Answer
It exists. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Rybnikov I.g.
A store received 20 kg of cheese, and a queue formed. After selling cheese to the next customer, the saleswoman accurately calculates the average weight of the purchase for all the cheese sold and informs how many people the remaining cheese will last if everyone buys exactly this average weight. Could t... | Let $s_{k}$ be the average weight of cheese sold to the first $k$ customers. According to the condition, $20 - k s_{k} = 10 s_{k}$, hence $s_{k} = \frac{20}{k+10}$ kg, and after the $k$-th customer, there remains
200
$k+10$
message after each customer (not limited to the first ten). After the 10th customer, exactly ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$4-$ [Increasing and Decreasing. Function Analysis]
Solve the equation $2 \sin \pi x / 2 - 2 \cos \pi x = x^{5} + 10 x - 54$.
# | Transferring all terms to one side, we get the equation $x^{5}+10 x-54-2 \sin \pi x / 2+2 \cos \pi x=0$. Consider the function
$f(x)=x^{5}+10 x-54-2 \sin ^{\pi x} / 2+2 \cos \pi x$. Note that $f^{\prime}(x)=5 x^{4}+10-\pi \cos \pi x / 2-2 \sin \pi x>5 x^{4}+10-3 \pi>0$. Therefore, $f(x)$ is an increasing function, so ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Ilya Muromets meets the three-headed Zmei Gorynych. And the battle begins. Every minute Ilya cuts off one of Zmei's heads. With a probability of $1 / 4$, two new heads grow in place of the severed one, with a probability of $1 / 3$ only one new head grows, and with a probability of $5 / 12$ - no heads grow. The Zmei is... | Blows by Ilya Muromets, in which the number of heads changes, are called successful.
Let's find the probability that at some point there will be a last successful blow. This means that starting from this point, there will be no more successful blows, that is, all blows will be unsuccessful. The probability of this is ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Sequence $A$.
The sequence of numbers $x_{1}, x_{2}, \ldots$ is such that $x_{1}=1 / 2$ and $x_{k+1}=x_{k}^{2}+x_{k}$ for any natural number $k$.
Find the integer part of the sum $\frac{1}{x_{1}+1}+\frac{1}{x_{1}+1}+\ldots+\frac{1}{x_{100}+1}$. | Note that $\frac{1}{x_{k+1}}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1}$, which means $\frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}}$. Therefore,
$\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\ldots+\frac{1}{x_{100}+1}=\frac{1}{x_{1}}-\frac{1}{x_{2}}+\frac{1}{x_{2}}-\frac{1}{x_{3}}+\ldots+\frac{1}{x_{100}}-\frac{1}{x_{101}}=\frac... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Given natural numbers $M$ and $N$, both greater than ten, consisting of the same number of digits, and such that $M = 3N$. To obtain the number $M$, one of the digits in $N$ must be increased by 2, and each of the other digits must be increased by an odd digit. What digit could the number $N$ end with? | The number $A=M-N=2 N$ is even. However, by the condition, the number $A$ is composed of odd digits and the digit 2. Therefore, $A$ ends in 2. Thus, the number $N$, which is half of $A$, ends in either 1 or 6.
If $N$ ends in 1, then when it is doubled, there is no carry from the last digit to the second last digit. Th... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
A bag of seeds was passed around the table. The first person took 1 seed, the second took 2, the third took 3, and so on: each subsequent person took one more seed than the previous one. It is known that in the second round, the total number of seeds taken was 100 more than in the first round. How many... | Let there be $n$ people sitting at the table. Then on the second round, each took $n$ more seeds than on the first, and all together took $n \cdot n=n^{2}$ more seeds than on the first. Since $n^{2}=100$, then $n=10$.
## Answer
10 people. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$3+$
$p$ and $8 p^{2}+1$ are prime numbers. Find $p$. | If $p \neq 3$, then $8 p^{2}+1$ is divisible by 3.
Answer
$p=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Decimal numeral system ] [ Divisibility rules for 3 and 9 ]
For the number $2^{100}$, the sum of its digits was found, then the sum of the digits of the result, and so on. Eventually, a single-digit number was obtained. Find it. | $2^{100}=\left(2^{6}\right)^{16} \cdot 2^{4} \equiv 1^{16} \cdot 16 \equiv 7(\bmod 9)$.
## Answer
7. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the greatest common divisor of the numbers $9 m+7 n$ and $3 m+2 n$, if the numbers $m$ and $n$ have no common divisors other than one?
# | Any common divisor of the numbers $9 m+7 n$ and $3 m+2 n$ must also be a divisor of the numbers $(9 m+7 n)-3(3 m+2 n) = n$ and $7(3 m+2 n)-2(9 m+7 n) = 3 m$. Since the numbers $m$ and $n$ are coprime, any common divisor of the numbers $n$ and $3 m$ must be a divisor of the number 3, that is, it cannot be greater than 3... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Tokoroiev $C$. $\mathbf{L}$.
Numbers $a, b$ and $c$ are such that $(a+b)(b+c)(c+a)=a b c,\left(a^{3}+b^{3}\right)\left(b^{3}+c^{3}\right)\left(c^{3}+a^{3}\right)=a^{3} b^{3} c^{3}$. Prove that $a b c=0$. | First, note that $x^{2}-x y+y^{2}>|x y|$ for any distinct numbers $x$ and $y$.
Assume that $a b c \neq 0$. Then, dividing the second equality by the first, we get $\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right)=|a b| \cdot|b c| \cdot|a c|$
All the brackets on the left and all the... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
Berdnikov A.
Let's call a natural number good if all its digits are non-zero. A good number is called special if it has at least $k$ digits and the digits are in strictly increasing order (from left to right).
Suppose we have some good number. In one move, it is allowed to append a special number to either end or ins... | Obviously, a special number does not have more than nine digits. If $k=9$, then with each operation, the number of digits changes by exactly 9, meaning the remainder of the number of its digits divided by 9 does not change, and a single-digit number cannot be turned into a two-digit number.
Let $k=8$. Since all operat... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Sindarov V.A.
Find all such natural $k$ that the product of the first $k$ prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first power).
# | Let $n \geq 2$, and $2=p_{1}1$; then $k>1$. The number $a$ is odd, so it has an odd prime divisor $q$. Then $q>p_{k}$, otherwise the left side of the equation (*) would be divisible by $q$, which is not the case. Therefore, $a>p_{k}$.
Without loss of generality, we can assume that $n$ is a prime number (if $n=s t$, th... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Three square paths with a common center are spaced 1 m apart from each other (see figure). Three ants start simultaneously from the lower left corners of the paths and run at the same speed: Mu and Ra counterclockwise, and Wei clockwise. When Mu reaches the lower right corner of the largest path, the o... | The lengths of the sides of two adjacent paths differ by 2 meters. Therefore, at the moment when Mu reached the corner, Ra had run 2 meters along the right side of the path and was at a distance of $2+1=3$ meters from the "lower" side of the outer path. Since Ra is halfway between Mu and Vey, Vey is at twice the distan... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Дубанов И.С.
In five pots standing in a row, Rabbit poured three kilograms of honey (not necessarily into each and not necessarily equally). Winnie-the-Pooh can take any two adjacent pots. What is the maximum amount of honey that Winnie-the-Pooh can guarantee to eat?
# | Evaluation. Let's assume Winnie-the-Pooh cannot take at least a kilogram of honey. This means that in any pair of adjacent pots, there is less than a kilogram of honey. This is true for both the two rightmost pots and the two leftmost pots. However, then in the middle pot, there must be more than a kilogram of honey (o... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Xichaturyan A.V.
Bus stop B is located on a straight highway between stops A and C. At some point after leaving A, the bus found itself at a point on the highway such that the distance from it to one of the three stops is equal to the sum of the distances to the other two. After the same amount of time, the bus was ag... | 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).
, and then all odd numbers not exceeding 15 - for any of them (let's say for the number $n$) the even number $2n$ is already m... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Work problem. $]$
The bathtub fills up in 23 minutes from the hot water tap, and in 17 minutes from the cold water tap. Masha first turned on the hot water tap. After how many minutes should she turn on the cold water tap so that by the time the bathtub is full, 1.5 times more hot water is used than cold water? | To have 1.5 times more hot water than cold water in the bathtub, the cold water tap should fill 2/5 of the bathtub, and the cold water $-3 / 5$. Then the hot water tap should be open for $3 / 5 \cdot 23=69 / 5$ minutes, and the cold water tap for $2 / 5 \cdot 17=34 / 5$ minutes. Therefore, the cold water tap should be ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On a straight line, ten points were placed at equal intervals, and they occupied a segment of length $a$. On another straight line, one hundred points were placed at the same intervals, and they occupied a segment of length $b$. How many times is $b$ greater than $a$? | On a segment of length $a$, 9 intervals fit, and on a segment of length $b-99$. Therefore, $b=11a$.
C
11 times. | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
It is known that $\operatorname{tg} A+\operatorname{tg} B=2$ and $\operatorname{ctg} A+\operatorname{ctg} B=3$. Find $\operatorname{tg}(A+B)$. | $\frac{2}{\operatorname{tg} A \operatorname{tg} B}=\frac{\operatorname{tg} A+\operatorname{tg} B}{\operatorname{tg} A \operatorname{tg} B}=\operatorname{ctg} A+\operatorname{ctg} B=3$, hence $\operatorname{tg} A \operatorname{tg} B=2 / 3$. Therefore,
$\operatorname{tg}(A+B)=\frac{\operatorname{tg} A+\operatorname{tg} ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Arithmetic. Mental calculation, etc.]
It is known that 9 cups of tea cost less than 10 rubles, and 10 cups of tea cost more than 11 rubles. How much does a cup of tea cost
# | Write the condition in the form of inequalities and use the fact that the cost of a glass of tea is expressed as an integer number of kopecks.
## Solution
Let the cost of a glass of tea, expressed in kopecks, be \( x \). Then, according to the condition, \( 9x \leq 1100 \).
Therefore, \( 9x \) does not exceed 999, a... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
[ Absolute value ]
Solve the equation $|x-2|+|x-1|+|x|+|x+1|+|x+2|=6$.
# | The number |x-a| is equal to the distance from the point on the number line with coordinate x to the point with coordinate a.
## Solution
Consider a point on the number line with coordinate x. The sum $|\mathrm{x}-2|+|\mathrm{x}-1|+|\mathrm{x}|+|\mathrm{x}+1|+|\mathrm{x}+2|$ is equal to the sum of the distances from ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[Text problems] [Case enumeration]
In a box, there are blue, red, and green pencils. In total, there are 20 pieces. The number of blue pencils is 6 times the number of green pencils, and the number of red pencils is less than the number of blue pencils. How many red pencils are in the box? | Think about how many blue pencils there can be.
## Solution
Since there are 20 pencils in total, and blue and green pencils together make up 7 parts. This means there can be 6 or 12 blue pencils, and green and red pencils would then be 1 and 13 or 2 and 6, respectively. Since there are fewer red pencils than blue one... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Fedorov P.M.
It is known that $a+\frac{b^{2}}{a}=b+\frac{a^{2}}{b}$. Is it true that $a=b$? | Let's bring the left side to a common denominator: $a+\frac{b^{2}}{a}=\frac{a^{2}+b^{2}}{a}$. Similarly, we will do the same for the right side.
We obtain the equality:
$$
\frac{a^{2}+b^{2}}{a}=\frac{a^{2}+b^{2}}{b}
$$
From the condition, it is clear that $a \neq 0, b \neq 0$. Therefore, $a^{2}+b^{2}>0$ (see comment... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9,10,11 |
Solve the equation $x^{3}+x-2=0$ by trial and error and using Cardano's formula. | The equation has an obvious root $x=1$. Dividing by $x-1$, we obtain the quadratic equation $x^{2}+x+2=0$, which has no real roots.
Cardano's formula gives $x=\sqrt[3]{-1+\sqrt{\frac{28}{27}}}+\sqrt[3]{-1-\sqrt{\frac{28}{27}}}$.
From this, it follows that $\sqrt[3]{-1+\sqrt{\frac{28}{27}}}+\sqrt[3]{-1-\sqrt{\frac{28}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}{[\text { Identical Transformations }]} \\ {[\text { Factorization }]}\end{array}\right]$
For different numbers $a$ and $b$, it is known that $\frac{a}{b}+a=\frac{b}{a}+b$. Find $\frac{1}{a}+\frac{1}{b}$. | The given equality can be written as $\frac{a}{b}-\frac{b}{a}=b-a$, from which $\frac{a^{2}-b^{2}}{a b}=b-a$ or $\frac{(a-b)(a+b)}{a b}=b-a$. Since the numbers $a$ and $b$ are different, we can divide both sides of the equation by $a-b$, after which we get: $\frac{a+b}{a b}=-1$. This is the desired value, since $\frac{... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { [Rectangles and squares. Properties and characteristics } \\ \text { Word problems (miscellaneous). }\end{array}\right]$
A frame for three square photographs has the same width everywhere (see figure). The perimeter of one opening is 60 cm, and the perimeter of the entire frame is 180 cm... | From the condition, it follows that the length of the side of the opening is 15 cm. Let $d$ cm be the width of the frame, then the perimeter of the rectangle is
$8 \cdot 15 + 12 d = 120 + 12 d = 180$. Therefore, $d=5$.
## Answer
5 cm. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Find the nearest integer to the number $x$, if $x=\frac{1}{\sqrt[4]{\frac{5}{4}+1}-\sqrt[4]{\frac{5}{4}-1}}$. | 
If the inequalities are verified by squaring, then \(1.9 < \frac{\sqrt{6} + \sqrt{2}}{2} < 2\).
## Answer
2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Gukhanov $H . X$.
Given quadratic trinomials $f_{1}(x), f_{2}(x), \ldots, f_{100}(x)$ with the same coefficients for $x^{2}$, the same coefficients for $x$, but different constant terms; each of them has two roots. From each trinomial $f_{i}(x)$, one root was chosen and denoted as $x_{i}$. What values can the sum $f_{... | Let the $i$-th quadratic polynomial have the form $f_{i}(x)=a x^{2}+b x+c_{i}$. Then $f_{2}\left(x_{1}\right)=f_{1}\left(x_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1}$, since $f_{1}\left(x_{1}\right)=$ 0. Similarly, we obtain the equalities
$f_{3}\left(x_{2}\right)=c_{3}-c_{2}, \ldots, f_{100}\left(x_{99}\right)=c... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Kalinin D.A.
In the room, there are 20 chairs of two colors: blue and red. On each chair sits either a knight or a liar. Knights always tell the truth, and liars always lie. Each of the seated individuals claimed that they were sitting on a blue chair. Then they somehow rearranged themselves, after which half of those... | Initially, all knights sit on blue chairs, and all liars on red ones. Therefore, the number of knights who moved to red chairs is equal to the number of liars who moved to blue chairs. Both groups claimed they were sitting on red chairs. In total, 10 people said they were sitting on red chairs. Therefore, the number of... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Shapovalov A.V.
Several natural numbers are written in a row with a sum of 20. No number and no sum of several consecutive numbers equals 3. Could there be more than 10 numbers written? | Example with 11 numbers: $1,1,4,1,1,4,1,1,4,1,1$.
## Answer
It could. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Shnop D.
In the park, there were lindens and maples. Maples made up $60 \%$ of them. In the spring, lindens were planted in the park, after which maples accounted for $20 \%$. In the fall, maples were planted, and maples again made up $60 \%$. By what factor did the number of trees in the park increase over the year? | At first, there were 1.5 times fewer lindens than maples, and in the summer, there became 4 times more. At the same time, the number of maples did not change. Therefore, the number of lindens became $1.5 \cdot 4=6$ times more.
By the end of the year, the ratio of the number of lindens to the total number of trees beca... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$3-$ $[$ Algorithm Theory (miscellaneous). $\quad]$
Author: Berdikov A.
Sasha writes a sequence of natural numbers on the board. The first number $N>1$ is written in advance. New natural numbers he gets by subtracting from or adding to the last written number any of its divisors greater than 1. Can Sasha, for any nat... | The first method. By adding $N$ each time, we get 2011N. By subtracting 2011 each time, we get 2011.
The second method. If $N$ is odd, add $N$ and get an even number. By adding or subtracting twos from it, we get 4022. By subtracting 2011, we get 2011.
## Answer
In any case.
Author: $\underline{\text { Folklore }}$... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Folkopr }}$
Three companies $A, B$, and $C$ decided to jointly build a road 16 km long, agreeing to finance this project equally. In the end, $A$ built 6 km of the road, $B$ built 10 km, and $C$ contributed her share in money - 16 million rubles. How should companies $A$ and $B$ divide this money b... | Each company was supposed to build $51 / 3$ km of road. Instead of company $C$, company $A$ built $6-5 \frac{1}{3}=2 / 3$ km of road, and company $B$ built
$10-5 \frac{1}{3}=14 / 3$ km. Therefore, 16 million rubles should be divided between them in the ratio $2: 14$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Folklore
In the class, there are 17 people. It is known that among any ten, there is at least one girl, and there are more boys than girls. How many girls are there in this class? | Since there are more boys than girls in the class, there are no fewer than nine boys. If there are more than nine boys, the condition that among any ten people there is at least one girl will not be met. Therefore, there are exactly 9 boys and 8 girls.
## Answer
8. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[ Processes and operations [ Examples and counterexamples. Constructions ]
Several boxes together weigh 10 tons, and each of them weighs no more than one ton. How many three-ton trucks are definitely enough to haul this cargo?
# | We will load the trucks with boxes in any order, only ensuring that the "overloading" of the truck does not occur. As a result, more than 2 tons will be loaded onto each truck. Therefore, five trucks will definitely be enough. Four trucks may not be enough. For example, let's say we have 13 boxes, each weighing ${ }^{1... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Covering with percentages and ratios ] [ Inclusion-exclusion principle ]
Three crazy painters started painting the floor each in their own color. One managed to paint $75 \%$ of the floor red, another $70 \%$ green, and the third $65 \%$ blue. What part of the floor is definitely painted with all three colors? | Evaluation. 25% of the floor is not painted red, 30% of the floor is not painted green, and 35% of the floor is not painted blue.
25+30+35=90. From this, it follows that at least 10% of the floor is painted with all three colors.
An example where exactly 10% of the floor is painted with all three colors is clear from... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Prove that $M . \mathbf{B}$.
The sequence of numbers $a_{1}, a_{2}, \ldots$ is defined by the conditions $a_{1}=1, a_{2}=143$ and $a_{n+1}=5 \cdot \frac{a_{1}+a_{2}+\ldots+a_{n}}{n}$ for all $n$ $\geq 2$.
Prove that all members of the sequence are integers. | The number $a_{3}=5 \cdot 72=360$ is an integer. For $n \geq 4$
$$
a_{n}=5 \cdot \frac{a_{1}+\ldots+a_{n-1}}{n-1} \text { and } a_{n-1}=5 \cdot \frac{a_{1}+\ldots+a_{n-2}}{n-2} \text {, hence } a_{n}=\frac{5}{n-1}\left(\frac{n-2}{5} a_{n-1}+a_{n-1}\right)=\frac{n+3}{n-1} a_{n-1} \text {. Therefore, if } n \geq
$$
$$
... | 4 | Number Theory | proof | Yes | Yes | olympiads | false |
$\underline{\text { Martynova } H}$.
To play the game of Hat, Nadya wants to cut a sheet of paper into 48 identical rectangles. What is the minimum number of cuts she will have to make if she can rearrange any pieces of paper but cannot fold them, and Nadya can cut as many layers of paper at once as she likes? (Each c... | With each cut, the number of paper pieces can increase by no more than twice. Therefore, after five cuts, you can get no more than 32 pieces, which is not enough.
Six cuts will suffice. For example, you can cut the sheet in half, align the two rectangular pieces and cut them in half again (resulting in four equal rect... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Senderov V.A.
At the vertices of a cube, numbers from 1 to 8 were written, and on each edge - the absolute difference of the numbers at its ends. What is the smallest number of different numbers that can be written on the edges?
# | Three numbers must be necessarily present: for this, it is sufficient to consider the three edges of the cube coming out of the vertex where the number 1 (or 8) is written. Let's prove that there will be an arrangement of numbers for which exactly three numbers will be required. Consider 2 squares. In the vertices of t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
[Examples and Counterexamples. Constructions] [ Proof by Contradiction ]
## Auto: Zassavsiy A.A.
In a volleyball tournament, each team met every other team once. Each match consisted of several sets - until one of the teams won three times. If a match ended with a score of $3: 0$ or $3: 1$, the winning team received ... | Evaluation. If there are no more than three teams, then the "Simpletons" won all the matches, which means they have the most points. Contradiction.
If there are four or five teams, then each team will play three or four matches. This means that the "Cunning" won no more than one match and scored a maximum of $5=3+1+1$... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$\underline{\text { Glibov A. }}$.
Let $n$ be a natural number. We call a sequence $a_{1}, a_{2}, \ldots, a_{n}$ interesting if for each $i=1,2$, $\ldots, n$ one of the equalities $a_{i}=i$ or $a_{i}=i+1$ holds. We call an interesting sequence even if the sum of its terms is even, and odd otherwise. For each odd inter... | Denoting the sum containing the term $2 \cdot 3 \cdot \ldots \cdot n(n+1)$ by $A_{n}$, and the other by $B_{n}$, we will prove the equality $A_{n}-B_{n}=1$ by induction.
Base case. $A_{1}-B_{1}=2-1=1$.
Inductive step. Represent the sum $A_{n}$ as $A^{\prime}+A^{\prime \prime}$, where $A^{\prime}$ contains all terms o... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
\left.\begin{array}{l}{[\text { Systems of exponential equations and inequalities] }} \\ {[\quad \text { Monotonicity and boundedness }}\end{array}\right]
Solve the system in positive numbers:
$$
\begin{cases}x^{y} & =z \\ y^{z} & =x \\ z^{x} & =y\end{cases}
$$ | First, note that if one of the unknowns is equal to one, then the others are also equal to one. Indeed,
Answer
$x=y=z=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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