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$\left[\begin{array}{l}{[\text { Problems on maximum and minimum (miscellaneous). }} \\ {[\quad \underline{\text { Rectangular parallelepipeds }}]}\end{array}\right]$
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 |
A sphere of radius $\sqrt{41}$ passes through the vertices $B, C, C_1$ and the midpoint of the edge $A_1 D_1$ of the cube $A B C D A_1 B_1 C_1 D_1$ ( $A A_1$ || $B B_1$ || $C C_1$ || $D D_1$ ). Find the surface area of the cube. | Let $M$ be the midpoint of edge $A_1D_1$. The center $O$ of the sphere is equidistant from the endpoints of segment $BC$, so point $O$ lies in the plane perpendicular to edge $BC$ and passing through its midpoint. Similarly, point $O$ lies in the plane perpendicular to edge $CC_1$ and passing through the midpoint of $C... | 384 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The generatrix of a cone forms an angle $\alpha$ with the plane of its base, $\cos \alpha=\frac{1}{4}$. A sphere is inscribed in the cone, and a plane is drawn through the circle of contact between the sphere and the lateral surface of the cone. The volume of the part of the cone enclosed between this plane and the bas... | Let $P$ be the vertex of the cone, $PH$ its height, and $O$ the center of the inscribed sphere. Consider the axial section of the cone - an isosceles triangle $PAB$ with angle $\alpha$ at the base $AB$. The section of the sphere is a circle with center $O$, inscribed in the triangle $PAB$ and touching the base $AB$ at ... | 27 | 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
$$
27=S_{\triangle A B C}=S_{\triangle A O C}+S_{\tria... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A circle with center $O$, inscribed in an isosceles triangle $A B C$, touches the lateral sides $A B$ and $B C$ at points $P$ and $Q$ respectively. Prove that a circle can be inscribed in the... | Quadrilateral $B P O Q$ is convex, $B P = B Q$ as tangent segments drawn from the same point to a circle, and $O P = O Q$ as radii of the same circle, so $B P + O Q = B Q + O P$. Therefore, a circle can be inscribed in quadrilateral $B P O Q$. Let $r$ be its radius. Then the radius of the inscribed circle of triangle $... | 90 | Geometry | proof | Yes | Yes | olympiads | false |
8,9
Segment $B D$ is the median of isosceles triangle $A B C (A B = B C)$. A circle with radius 4 passes through points $B, A, D$ and intersects side $B C$ at point $E$ such that $B E: B C = 7: 8$. Find the perimeter of triangle $A B C$. | Let $CE = x$. Then $BC = 8x$ and $BE = 7x$, and since $8x = BC = AB = 8$, we have $x = 1$, $BE = 7$. Point $E$ lies on the circle with diameter $AB$, so $\angle AEB = 90^\circ$. From the right triangles $ABE$ and $ACE$, we find that
$$
AE^2 = AB^2 - BE^2 = 64 - 49 = 15, \quad AC = \sqrt{AE^2 + CE^2} = \sqrt{15 + 1} = ... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Authors: $\underline{\text { Muraykin M.V., Berlov S.L., Bogdanov I.i. }}$
There are three commissions of bureaucrats. It is known that for each pair of bureaucrats from different commissions, among the members of the remaining commission, there are exactly 10 bureaucrats who are acquainted with both, and exactly 10 b... | Consider a graph where the vertices are bureaucrats, and two bureaucrats from different commissions are connected by a red edge if they are acquainted, and by a blue edge otherwise. Let the number of bureaucrats in three commissions be \(a\), \(b\), and \(c\). Consider any two bureaucrats \(A\) and \(B\) from the first... | 120 | Combinatorics | 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 |
In space, there is a regular dodecahedron. In how many ways can a plane be drawn so that it intersects the dodecahedron to form a regular hexagon?
# | Answer: 30 ways. First of all, note that for each of the 10 large diagonals of the dodecahedron, there are exactly three different planes perpendicular to this diagonal and cutting out a regular hexagon. Indeed, let us move the plane perpendicular to the diagonal from one vertex to another. At first, a regular triangle... | 30 | 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, $C D=A D=4$.
## Answer
## [ Inscrib... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 | |
Segments $A M$ and $B H$ are the median and altitude of an acute triangle $A B C$, respectively. It is known that $A H=1$ and $2 \angle M A C=\angle M C A$. Find the side $B C$. | Let $\angle M A C=\alpha$. Then $\angle M C A=2 \alpha$. Segment $H M$ is the median of the right triangle $B H C$, drawn from the vertex of the right angle. Therefore, $H M=M C$. Triangle $C M H$ is isosceles, so $\angle C H M=\angle M C H=2 \alpha$.
By the external angle theorem of a triangle, $\angle A H M=\angle C... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Sum of angles in a triangle. Theorem of the exterior angle.] [Median line of a triangle $]
On the side $AC$ of triangle $ABC$, a point $D$ is chosen such that $2AD = DC$. $E$ is the foot of the perpendicular dropped from point $D$ to segment $BC$, and $F$ is the intersection point of segments $BD$ and $AE$. Find the ... | $\angle B D E=90^{\circ}-\angle D B E=90^{\circ}-60^{\circ}=30^{\circ}, \angle D E F=\angle B F E-\angle B D E=60^{\circ}-30^{\circ}=30^{\circ}$. Therefore, triangle $D F E$ is isosceles, $F D=F E=B F$. Hence, $F$ is the midpoint of segment $B D$. Let $K$ be the midpoint of $D C$. Then $K D=A D$, which means $D$ is the... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Vertices $A$ and $B$ of the prism $A B C A 1 B 1 C 1$ lie on the axis of the cylinder, while the other vertices lie on the lateral surface of the cylinder. Find the dihedral angle in this prism with edge $A B$. | When orthogonally projecting the data of a cylinder and a prism onto the plane of the cylinder's base, the lateral surface will transform into a circle equal to the base circle, vertices $A$ and $B$ will transform into the center $O$ of this circle, vertices $A 1$ and $B 1$ will transform into some point $M$ on this ci... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3 [ Constructions on a Projection Drawing ]
In a regular quadrilateral pyramid with a lateral edge of 20, the angle between the lateral edges lying in the same face is $\frac{\pi}{3}$. A line perpendicular to one of the lateral edges and passing through a point on this edge intersects the height of the pyramid. Find t... | ## Answer
17.00
## Problem | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10,11
Find the length of the shortest path on the surface of a unit regular tetrahedron between the midpoints of its opposite edges.
# | Let $M$ and $N$ be the midpoints of the opposite edges $A B$ and $C D$ of a regular tetrahedron $A B C D$ (Fig.1), 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 |
Given a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$, where $A B=4, A D=A A 1=14$. Point $M$ is the midpoint of edge $C C 1$. Find the area of the section of the parallelepiped by the plane passing through points $A 1, D$ and $M$. | Let the lines $D M$ and $D_1 C_1$ intersect at point $K$, and the lines $A_1 K$ and $B_1 C_1$ intersect at point $L$. Then the considered section is the quadrilateral $A_1 D M L$ (Fig.1). Its orthogonal projection onto the base plane $A_1 B_1 C_1 D_1$ is the quadrilateral $A_1 L C_1 D_1$. From the equality of triangles... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[
A circle inscribed in a right trapezoid divides its larger lateral side into segments of 1 and 4. Find the area of the trapezoid.
# | Let a circle with center $O$ and radius $r$, inscribed in a right trapezoid $A B C D$ with bases $A D$ and $B C$ and right angles at vertices $A$ and $B$, touch the bases $A D$ and $B C$ at points $K$ and $L$ respectively, and the larger lateral side $C D$ at point $M$. Given that $C M=1, D M=4$.
Since $C O$ and $D O$... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$3-$ Area of a trapezoid $\quad$ K [ Mean proportionals in a right triangle ]
A circle inscribed in an isosceles trapezoid divides its lateral side into segments equal to 4 and 9. Find the area of the trapezoid. | Let a circle with center $O$ and radius $r$, inscribed in an isosceles trapezoid $A B C D$, touch the bases $A D$ and $B C$ at points $K$ and $L$ respectively, and the lateral side $C D$ at point $M$. Given that $C M=4, D M=9$.
Since $C O$ and $D O$ are the angle bisectors of angles $B C D$ and $A D C$, the sum of whi... | 156 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$:$ Folkiore
Find the midline of an isosceles trapezoid if its diagonal is 25 and its height is 15.
# | Let $ABCD$ be a given trapezoid ($AD \| BC$), $EF$ be its midline, and $CG$ be its height (see figure). We will prove that $EF = AG$.
The first method. Construct parallelogram $ABCH$.
Trian... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9
[ Two tangents drawn from one point ]
In an isosceles triangle with a lateral side of 100 and a base of 60, a circle is inscribed. Find the distance between the points of tangency located on the lateral sides.
# | The segment connecting the points of tangency of a circle with the lateral sides of an isosceles triangle cuts off a triangle similar to the given one.
## Solution
Let \( M \) and \( N \) be the points of tangency of the circle with the lateral sides \( AB \) and \( AC \) of the isosceles triangle \( ABC \), and \( K... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
| Two tangents drawn from one point | |
| :---: | :---: | :---: |
| | Properties and characteristics of tangents | Slo |
| | Auxiliary similar triangles Pythagoras' theorem (direct and inverse) | |
Two tangents are drawn from one point to a circle. The length of the tangent is 156, and the distance between the po... | Let $O$ be the center of the circle, $M A$ and $M B$ be the tangents, $A$ and $B$ be the points of tangency, and $K$ be the midpoint of segment $A B$. Then $M K^{2} = A M^{2} - A K^{2} = 156^{2} - 60^{2} = 96 \cdot 216 = 144^{2}$.
From the similarity of triangles $M A O$ and $M K A$, it follows that $O A : A M = A K :... | 65 | 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 $B E = 1 / 2 A K = 1$.
. As a result, Petrov's apartment becam... | From the condition, it follows that the number of entrances to the left and right of Petrov's entrance differ by 0.2 or 4, that is, by an even number. At the same time, the difference between the apartment numbers of Petrov under the two numbering options shows how many more apartments are in the entrances to the left ... | 985 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$:$ Folkpor
In trapezoid $A B C D(A D \| B C)$, 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).
.
Assume that there exists a square $n \times n$, where $n>2000$. This leads to a contradiction.
## Ans... | 2000 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Completion of a Tetrahedron to a Parallelepiped]
On the faces of a regular tetrahedron with edge $a$, equal regular pyramids are constructed. The plane angles at the vertices of these pyramids, opposite to the faces of the tetrahedron, are right angles. Consider the polyhedron formed by the tetrahedron and the constr... | Consider the cube $A B C D A 1 B 1 C 1 D 1$. All edges of the triangular pyramid $A C B 1 D 1$ are equal. Therefore, this pyramid is a regular tetrahedron. On the lateral faces of the regular tetrahedron $A C B 1 D 1$ as bases, regular triangular pyramids $D A C D 1$ with vertex $D$, $A 1 A B 1 D 1$ with vertex $A 1$, ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Inscribed Quadrilaterals (Miscellaneous)]
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 $A B$ is the diameter of a circle, and point $C$ lies outside this circle. Segments $A C$ and $B C$ intersect the circle at points $D$ and $M$ respectively. Find the angle $C B D$, if the areas of triangles $D C M$ and $A C B$ are in the ratio $1: 4$. | Triangle DCM is similar to triangle BCA with a coefficient of $\frac{1}{2}$.
## Solution
Triangles $D C M$ and $B C A$ are similar with a coefficient of $\frac{1}{2}$. Therefore, $B C=2 C D$. Consequently,
$$
\sin \angle C B D=\frac{C D}{B C}=\frac{1}{2}
$$
since angle $C B D$ is acute, $\angle C B D=30^{\circ}$.
... | 30 | 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 |
$\left[\begin{array}{l}\text { Symmetry helps solve the problem } \\ \text { [The inscribed angle is half the central angle]}\end{array}\right]$
$A B$ is the diameter of the circle; $C, D, E$ are points on the same semicircle $A C D E B$. On the diameter $A B$, points $F$ and $G$ are taken such that $\angle C F A = \a... | Let $C_{1}$ and $E_{1}$ be the points of intersection of the rays $D F$ and $D G$ with the given circle. Then $\cup A C_{1}=\cup A C$ and $\cup$ $B E_{1}=\cup B E$.
## Solution
Let $C_{1}$ and $E_{1}$ be the points of intersection of the rays $D F$ and $D G$ with the given circle. Since the circle is symmetric with r... | 50 | 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 |
[ Angles between angle bisectors $]$
The sides of a triangle are 1 and 2, and the angle between them is $60^{\circ}$. A circle is drawn through the incenter of this triangle and the endpoints of the third side. Find its radius. | Apply the formula $a=2 R \sin \alpha$.
## Solution
Let $O$ be the center of the inscribed circle of triangle $ABC$ with sides $AC=1, AB=2$ and angle $CAB$ equal to $60^{\circ}$. By the cosine theorem, we find that $BC=\sqrt{3}$. Therefore, triangle $ABC$ is a right triangle, $\angle ACB=90^{\circ}, \angle ABC=30^{\ci... | 340 | 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
$$
C M=\frac{A C+C D-A D}{2}, C N=\frac{B C+C D-B D}{2},
$... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}{[\text { Transfer of sides, diagonals, etc. ] }} \\ {[\text { Area of a trapezoid }}\end{array}\right]$
Find the area of a trapezoid with bases 11 and 4 and diagonals 9 and 12. | Through the vertex of the trapezoid, draw a line parallel to the diagonal.
## Solution
Through vertex $C$ of the smaller base $BC$ of trapezoid $ABCD (BC=4, AD=11, AC=9, BD=12)$, draw a line parallel to diagonal $BD$, intersecting line $AD$ at point $K$. In triangle $ACK$,
$$
AC=9, CK=BD=12, AK=AD+DK=AD+BC=11+4=15.
... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
## Area of a Trapezoid
Find the area of a trapezoid where the parallel sides are 16 and 44, and the non-parallel sides are 17 and 25. | Through the vertex of the trapezoid, draw a line parallel to the lateral side.
## Solution
Through vertex $C$ of trapezoid $A B C D (B C=16, A D=44, A B=17, C D=25)$, draw a line parallel to side $A B$ until it intersects the base $A D$ at point $K$.
In triangle $C K D$
$$
C K=17, C D=25, K D=A D-B C=28 .
$$
By He... | 450 | 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 |
[ Various cutting problems ]
Can a regular triangle be cut into 1000000 convex polygons such that any line intersects no more than 40 of them?
# | If cuts are made close to the vertices of a convex $n$-gon, then one can cut off $n$ triangles from it and obtain a convex $2n$-gon. It is easy to check that in this case, any line intersects no more than two of the cut-off triangles.
Cut off 3 triangles from an equilateral triangle, then 6 triangles from the resultin... | 1699 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\left.\begin{array}{l}{[\text { Auxiliary equal triangles }} \\ \text { [ Congruent triangles. Criteria for congruence }]\end{array}\right]$
Given triangle $A B C$, where $A B=A C$ and $\angle A=80^{\circ}$. Inside triangle $A B C$, a point $M$ is taken such that $\angle M B C=30^{\circ}$, and $\angle M C B=10^{\circ... | Let $N$ be the point of intersection of the line $B M$ with the bisector of angle $B A C$. Prove that triangles $A C N$ and $M C N$ are equal.
## Solution
Let $N$ be the point of intersection of the line $B M$ with the bisector of angle $B A C$. Since $\angle B N C=120^{\circ}$, then $\angle A N C=$ $\angle A N B=120... | 70 | Geometry | math-word-problem | 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 |
Two circles intersect at points $A$ and $B$. A line through point $B$ intersects the circles at points $C$ and $D$, which lie on opposite sides of line $A B$. The tangents to these circles at points $C$ and $D$ intersect at point $E$. Find $A E$, if $A B=10, A C=16, A D=15$. | Prove that a circle can be circumscribed around quadrilateral $ABCD$, and also that triangles $ABC$ and $ADE$ are similar.
## Solution
From the theorem about the angle between a tangent and a chord, it follows that $\angle BAC = \angle ECD$ and $\angle BAD = \angle EDC$. Since ray $AB$ lies between the sides of angle... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Berroov S.L.
Given a $15 \times 15$ board. Some pairs of centers of adjacent cells by side are connected by segments such that a closed non-self-intersecting broken line is formed, which is symmetric with respect to one of the diagonals of the board. Prove that the length of the broken line does not exceed 200. | Clearly, the broken line intersects the diagonal. Let $A$ be one of the vertices of the broken line lying on the diagonal. We will move along the broken line until we first reach another vertex $B$ lying on the diagonal. By symmetry, if we move along the broken line from $A$ in the other direction, $B$ will also be the... | 200 | Geometry | proof | 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 |
In triangle $ABC$, the angles are known: $\angle A=45^{\circ}, \angle B=15^{\circ}$. On the extension of side $AC$ beyond point $C$, point $M$ is taken such that $CM=2AC$. Find $\angle AMB$. | Mark off on segment $C B$ a segment $C K$ equal to segment $A C$.
## Solution
Let $P$ be the midpoint of segment $C M$. Then $\quad A C=C P=P M$. Mark a point $K$ on side $C B$ such that $C K=C A$. By the external angle theorem of a triangle,
$\angle P C K=\angle C A B+\angle A B C=45^{\circ}+15^{\circ}=60^{\circ}$.... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Bikin A.d:
In triangle $ABC$, the midpoints of sides $AC$ and $BC$ are marked as points $M$ and $N$ respectively. Angle $MAN$ is $15^{\circ}$, and angle $BAN$ is $45^{\circ}$.
Find angle $ABM$. | Extend segment $M N$ by its length in both directions and obtain points $K$ and $L$ (see the left figure). Since $M$ is the common midpoint of segments $A C$ and $K N$, $A K C N$ is a parallelogram. Therefore, $\angle C K M=45^{\circ}, \angle K C M=15^{\circ}$. Mark a point $P$ on segment $C M$ such that $\angle C K P=... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Obukhov B B.
Given a convex pentagon $A B C D E$, all sides of which are equal. It is known that angle $A$ is $120^{\circ}$, angle $C$ is $135^{\circ}$, and angle $D$ is $n^{\circ}$.
Find all possible integer values of $n$.
# | Let's start by showing that the answer is unique. Consider two equilateral pentagons $A B C D E$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}$, in which $\angle A=\angle A^{\prime}=120^{\circ}$,
$\angle C=\angle C^{\prime}=135^{\circ}$. We can assume that the side lengths of the pentagons are equal to 1... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
From four photographs, three different rectangles can be formed (see figure). The perimeter of one of them is 56 cm. Find the perimeters of the other two rectangles, if the perimeter of the photograph is 20 cm.
, we get: $A K=$ $A P, C K=C H=M H=M P$. Therefore, $A C=A M$. Moreover, $C M=0.5 A B=A M$.
Thus, triangle $C A M$ is equilateral, which means $\angle B A C=60^{\circ}$.
## Answer
$60^{\circ}$. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
A rectangle was divided by two lines, parallel to its sides, into four rectangles. One of them turned out to be a square, and the perimeters of the rectangles adjacent to it are 20 cm and 16 cm. Find the area of the original rectangle. | Let's introduce the notations as shown in the figure. Express the half-perimeters of the rectangles specified in the condition: \(a + x = 8\) and \(b + x = 10\). Then the area of the original rectangle is \((a + x)(b + x) = 80\) (sq units).
.] [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 |
[ Dirichlet's Principle (area and volume).] $[$ Geometry on graph paper $]$
A corner square has been cut out of a chessboard. What is the minimum number of equal-area triangles the resulting figure can be cut into? | Evaluation. Let's take the area of one cell as a unit. The given figure is a non-convex hexagon $A B C D E F$ with an area of 63 and an angle of $270^{\circ}$ at vertex $D$ (see the figure on the right). If the figure is divided into triangles, it is clear that point $D$ must belong to at least two triangles, one of wh... | 18 | 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$.
. Area of a trapezoid
A circle passes through vertices $A$ and $B$ of rectangle $A B C D$ and is tangent to side $C D$ at its midpoint. A line is drawn through vertex $D... | Apply the Tangent-Secant Theorem.
## Solution
Let $M$ be the midpoint of $CD$. Then $DE = DM = 5$. Let $AK = 2x$, then $KE = 3x$. By the Tangent-Secant Theorem, $KE^2 = BK \cdot AK$, or $9x^2 = (10 + 2x) \cdot 2x$, from which $x = 4$. Therefore, $AK = 8$, $BK = 18$, $KE = 12$, $KD = KE + ED = 12 + 5 = 17$.
By the Py... | 210 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Tokarev S.i.
The path from platform A to platform B was traveled by the electric train in $X$ minutes ($0<X<60$). Find $X$, given that at the moment of departure from $A$ and upon arrival at $B$, the angle between the hour and minute hands was $X$ degrees. | Since more than an hour passes between two consecutive overtakes of the hour hand by the minute hand, no more than one overtake occurred during the specified time of the electric train's movement. Let O be the center of the clock face, and let $T_{A}$ and $T_{B}$ be the points where the tip of the hour hand was located... | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
A convex polyhedron with vertices at the midpoints of the edges of a certain cube is called a cuboctahedron. When a cuboctahedron is intersected by a plane, a regular polygon is obtained. What is the maximum number of sides it can have? | Let the edge of the original cube, from which the cuboctahedron is obtained, be 1. Consider the sections of the cuboctahedron by a plane parallel to the base of the cube at a distance of $08$. Then the vertices of this $n$-gon must lie on the edges of the cuboctahedron, and no edge can have more than two vertices of th... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In an acute-angled triangle, two altitudes are equal to 3 and $2 \sqrt{2}$, and their point of intersection divides the third altitude in the ratio 5:1, counting from the vertex of the triangle. Find the area of the triangle. | Let $\alpha$ and $\beta$ be the acute angles of a triangle, from the vertices of which the given altitudes are drawn. Formulate a system of trigonometric equations in terms of $\alpha$ and $\beta$.
## Solution
Let $A A_{1}, B B_{1}$, and $C C_{1}$ be the altitudes of triangle $ABC$, and $H$ be their point of intersec... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The height of the isosceles trapezoid $ABCD$ with bases $AD$ and $BC$ is $4 \sqrt{3}$, the diagonals of the trapezoid intersect at point $O, \angle AOD=120^{\circ}$. Find the midline of the trapezoid. | The diagonals of an isosceles trapezoid form equal angles with the base, therefore
$$
\angle O A D=\angle O D A=\frac{1}{2}\left(180^{\circ}-120^{\circ}\right)=30^{\circ}
$$
Let $H$ be the foot of the perpendicular dropped from vertex $C$ to $A D$. Then $A H=\frac{1}{2}(A D+B C)$, i.e., segment $A H$ is equal to the ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$[$ Transfer of side, diagonal, etc.]
The bases of the trapezoid are 1 and 7, and the diagonals are 6 and 10. Find the area of the trapezoid. | Through the vertex $C$ of the smaller base $BC$ of the trapezoid $ABCD (BC=1, AD=7, AC=6, BD=10)$, we draw a line parallel to the diagonal $BD$ until it intersects the line $AD$ at point $K$. In triangle $ACK$, it is known that
$$
AC=9, CK=BD=10, AK=AD+DK=AD+BC=7+1=8.
$$
Since $CK^2 = AC^2 + AD^2$, triangle $ACK$ is ... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The circle constructed on the larger lateral side $AB$ of the right trapezoid $ABCD$ as its diameter intersects the base $AD$ at its midpoint. It is known that $AB=10, CD=6$. Find the midline of the trapezoid. | Let $M$ be the midpoint of $A D$. Then $\angle A M B=90^{\circ}$, so the quadrilateral $M B C D$ is a rectangle, $B M=C D=6$. From the right triangle $A B M$ we find that
$$
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 |
[The ratio of the areas of triangles with a common base or common height] Class
The diagonals $AC$ and $BD$ of trapezoid $ABCD$ with bases $AD$ and $BC$ intersect at point $O$. It is known that $AD=2BC$ and the area of triangle $AOB$ is 4. Find the area of the trapezoid. | Triangle BOC is similar to triangle DOA with a coefficient of $\frac{1}{2}$, so $O C=\frac{1}{2} A O$, therefore
$$
\begin{aligned}
& S_{\triangle B O C}=\frac{1}{2} S_{\triangle A O B}=\frac{1}{2} \cdot 4=2 \\
& S_{\triangle D O A}=2 S_{\triangle A O B}=2 \cdot 4=8
\end{aligned}
$$
since triangles $ABD$ and $ACD$ ar... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8,9 |
The center of a circle with a radius of 5, circumscribed around an isosceles trapezoid, lies on the larger base, and the smaller base is equal to 6. Find the area of the trapezoid. | Let the center of the circle circumscribed around the isosceles trapezoid $ABCD$ with bases $AB$ and $CD$ lie on the larger base $AD$. Then $AD$ is the diameter of the circle, so $AD=10$ and $\angle ACD=90^{\circ}$. Let $CH$ be the height of the trapezoid. Then
$$
AH=\frac{1}{2}(AD+BC)=\frac{1}{2}(10+6)=8, DH=\frac{1}... | 32 | 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 |
In triangle $A B C$, the median $B M$ is twice as small as side $A B$ and forms an angle of $40^{\circ}$ with it. Find the angle $A B C$. | Extend median $B M$ beyond point $M$ to its length and obtain point $D$ (see figure). Since $A B=2 B M=B D$, triangle $A B D$ is isosceles. Therefore, $\angle B A D=\angle B D A=\left(180^{\circ}-40^{\circ}\right): 2=70^{\circ}$.
. 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 |
Trazyrin A.
In coordinate space, all planes with equations $x \pm y \pm z=n$ (for all integers $n$) were drawn. They divided the space into tetrahedra and octahedra. Let the point ( $x_{0}, y_{0}, z_{0}$ ) with rational coordinates not lie on any of the drawn planes. Prove that there exists a natural number $k$ such t... | Lemma. Let the rational numbers $a, b, c$ and $a+b+c$ be non-integers. Then there exists a natural number $k$ such that the numbers $k a, k b$, and $k c$ are non-integers, and
$12$. This means there is a smallest natural number $k$ for which $f(k)>1$ (then $f(k-1) \leq 1$). We will show that this $k$ satisfies all the... | 1414 | Geometry | proof | 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 |
[
The numerical function $f$ is such that for any $x$ and $y$, the equality $f(x+y)=f(x)+f(y)+80xy$ holds. Find $f(1)$, if $f(0.25)=2$.
# | $f(0,5)=f(0,25+0,25)=f(0,25)+f(0,25)+80 \cdot 0,25 \cdot 0,25=2+2+5=9$. Similarly $f(1)=f(0,5+0,5)=f(0,5)+$ $f(0,5)+80 \cdot 0,5 \cdot 0,5=9+9+20=38$.
## Answer
38.
## [ Rational and irrational numbers ] Problem 116563 Topics: [Identities (trigonometry).] [ Proof by contradiction
Author: Senderov V...
Does there ... | 38 | Algebra | 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 |
$\left[\begin{array}{l}{[\text { Integer and fractional parts. Archimedes' principle ] }} \\ {[\quad \underline{\text { equations in integers }} \underline{\text { ] }}}\end{array}\right]$
How many solutions in natural numbers does the equation $\left[{ }^{x} / 10\right]=\left[{ }^{x} / 11\right]+1$ have? | Let $x=11 n+r$, where $n \geq 0,0 \leq r \leq 10$. Then $\left[{ }^{x} / 11\right]=n, n+1=\left[{ }^{x} / 10\right]=n+\left[{ }^{n+r} / 10\right]$, that is, $10 \leq n+r<20,10$ $-r \leq n \leq 19-r$. For each $r$ from 0 to 10, we get 10 solutions.
## Answer
110 solutions. | 110 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Folklore
Find the number of solutions in natural numbers of the equation $\left[{ }^{x} /{ }_{10}\right]=\left[{ }^{x} /{ }_{11}\right]+1$.
# | Let $x=11 n+r$, where $n \geq 0,0 \leq r \leq 10$. Then $\left[{ }^{x} / 11\right]=n, n+1=\left[{ }^{x} / 10\right]=n+\left[{ }^{n+r} / 10\right]$, that is, $10 \leq n+r<20,10$ $-r \leq n \leq 19-r$. For each $r$ from 0 to 10, we get 10 solutions.
## Answer
110 solutions. | 110 | Number Theory | 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 |
[Limitedness, monotonicity]
For what value of $K$ is the quantity $A_{\mathrm{k}}=\frac{19^{k}+66^{k}}{k!}$ maximal? | Answer: For $k=65$. Let $B_{\mathrm{k}}=\frac{19^{k}}{k!}, C_{\mathrm{k}}=\frac{66^{k}}{k!}$. Then $A_{\mathrm{k}}=B_{\mathrm{k}}+C_{\mathrm{k}}, \frac{B_{k+1}}{B_{k}}=\frac{19}{k+1}, \frac{C_{k+1}}{C_{k}}=\frac{66}{k+1}$.
Therefore, for $k \leq 19$ both sequences are non-decreasing, and for $k \geq 65$ both sequences... | 65 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
acting as follows. First, he arbitrarily chose two numbers from the list, added them, discarded any digits after the decimal point (if there were any), and recorded the result in place of the two chosen numbers. With the resulting list of 99 numbers, he did the same, and so on, until only one integer remained in the li... | Evaluation. Accountants each time calculated the integer part of the sum of some two numbers, which is equal to the sum of their integer parts plus, possibly, one. Let's call this addition of one an incident. It could only happen when both addends were non-integers. Each accountant ended up with the sum of the integer ... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Galererin G.A. Each pair of numbers $x$ and $y$ is associated with some number $x^{*} y$. Find 1993*1935, given that for any three numbers $x, y, z$ the identities $x^{*} x=0$ and $x^{*}\left(y^{*} z\right)=\left(x^{*} y\right)+z$ are satisfied. | $x=x^{*} x+x=x^{*}\left(x^{*} x\right)=x^{*} 0=x^{*}\left(y^{*} y\right)=x^{*} y+y$. Therefore, $x^{*} y=x-y$. Hence $1993 * 1935=1993-1935=58$.
## Answer
58. | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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