problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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11.160. The base of the pyramid is a rhombus with diagonals $d_{1}$ and $d_{2}$. The height of the pyramid passes through the vertex of the acute angle of the rhombus. The area of the diagonal section, passing through the smaller diagonal, is $Q$. Calculate the volume of the pyramid given that $d_{1}>d_{2}$. | Solution.
The required volume $V=\frac{1}{3} S_{\text{base}} \cdot S A$, where $S_{\text{base}}=\frac{1}{2} d_{1} d_{2}$ (Fig. 11.55). In $\triangle S A O$, we have $S A=\sqrt{S O^{2}-A O^{2}}$, where $A O=\frac{d_{1}}{2}$, and $S O=\frac{2 Q}{d_{2}}$, since by the condition $\frac{1}{2} d_{2} \cdot S O=Q$. Therefore,... | \frac{2PQ}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,015 |
11.162. In a triangular pyramid, all four faces are equal isosceles triangles with base $a$ and lateral side $b$. Calculate the volume of the pyramid. Does the problem have a solution for any $a$ and $b$? | ## Solution.
In the pyramid $SABC$ (Fig. 11.57), $AB = AC = BS = CS = b, BC = AS = a$: $E$ is the midpoint of $BC$. Then $AE \perp BC, SE \perp BC$. Therefore, the line $BC$ is perpendicular to the plane $ASE$. Then the plane $ABC$ is also perpendicular to the plane $ASE$. In the plane $ASE$, drop the perpendicular $S... | \frac{^{2}\sqrt{4b^{2}-2a^{2}}}{12};0<<b\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,016 |
11.164. The side of the base of a regular triangular prism is less than the lateral edge and is equal to $a$. A plane is drawn through the side of the upper base, which forms an angle of $45^{\circ}$ with the base plane and divides the prism into two parts. Determine the volume and the total surface area of the upper p... | ## Solution.
In the right prism $A B C A_{1} B_{1} C_{1} \quad B_{1} C_{1}=a, B_{1} C_{1}<A A_{1}, E$ is the intersection point of the plane passing through the side $B_{1} C_{1}$ of the upper base and the line $A A_{1}$ (Fig. 11.58). Let $K$ be the midpoint of $B_{1} C_{1}$. Then $A_{1} K \perp B_{1} C_{1} ; A_{1} K=... | \frac{^{3}}{8};\frac{^{2}\sqrt{3}(3+\sqrt{2})}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,017 |
11.165. The diagonals of the faces of a rectangular parallelepiped are equal to $a, b$ and $c$. Determine its total surface area. | ## Solution.
In the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.59) $A C=a, A D_{1}=b, C D_{1}=c$. Introduce the unknowns $A D=x, C D=y$, $D D_{1}=z$. Consider the system of equations: $\left\{\begin{array}{l}a^{2}=x^{2}+y^{2} ; \\ b^{2}=x^{2}+z^{2} ; \\ c^{2}=y^{2}+z^{2} ;\end{array}\right.$... | \sqrt{^{4}-(b^{2}-^{2})^{2}}+\sqrt{b^{4}-(^{2}-^{2})^{2}}+\sqrt{^{4}-(^{2}-b^{2})^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,018 |
11.166. The lengths of the edges of the parallelepiped are $a, b$, and $c$. The edges of lengths $a$ and $b$ are mutually perpendicular, and the edge of length $c$ forms an angle of $60^{\circ}$ with each of them. Determine the volume of the parallelepiped. | Solution.
In the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.60) $A B=a, A D=b$, $\angle B A D=90^{\circ}, A A_{1}=c, \angle A_{1} A B=\angle A_{1} A D=60^{\circ}, A_{1} K$ is the height. Since $\angle A_{1} A B=\angle A_{1} A D$, then $A K$ is the bisector of $\angle B A D . A A_{1}$ is the oblique to t... | \frac{\sqrt{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,019 |
11.167. The base of a right parallelepiped is a parallelogram with an angle of $120^{\circ}$ and sides of 3 and 4 cm. The smaller diagonal of the parallelepiped is equal to the larger diagonal of the base. Find the volume of the parallelepiped. | Solution.
In the right parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.61) $A B=3$ cm, $A D=4$ cm, $\angle A B C=120^{\circ}$. Then $A C$ is the larger diagonal of the base, $B_{1} D$ is the smaller diagonal of the parallelepiped, and $B_{1} D=A C$. The area of the base $A B C D: S=A B \cdot B C \cdot \sin \... | 36\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,020 |
11.168. The base of the pyramid is a rectangle with an area of $S$. Two lateral faces are perpendicular to the base plane, while the other two are inclined to it at angles of 30 and $60^{\circ}$. Find the volume of the pyramid. | Solution.
According to the problem, the base of the pyramid $EABCD$ is a rectangle $ABCD$, and the lateral faces $ABE$ and $CBE$ are perpendicular to the plane of the base (Fig. 11.62). Therefore, their common edge $EB$ is perpendicular to the plane of the base and serves as the height of the pyramid. $BA$ is the proj... | \frac{S\sqrt{S}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,021 |
11.169. A plane is drawn through the vertex of the base and the midpoints of the two lateral edges of a regular triangular pyramid. Find the ratio of the lateral surface area of the pyramid to the area of its base, given that the intersecting plane is perpendicular to the lateral face. | Solution.
Let $K O$ be the height of the regular pyramid $K A B C$ (Fig. 11.63), $M$ be the midpoint of $A K$, and $N$ be the midpoint of $C K$. The plane $M B N$ is perpendicular to the plane $A K C$. $B M$ and $B N$ are medians of the equal isosceles $\triangle A K B$ and $\triangle C K B$, drawn to their lateral si... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,022 |
11.170. From the midpoint of the height of a regular triangular pyramid, perpendiculars are dropped to a lateral edge and to a lateral face. The lengths of these perpendiculars are $a$ and $b$, respectively. Find the volume of the pyramid. Does the problem have a solution for any $a$ and $b$? | ## Solution.
Let $O_{1}$ be the midpoint of the height $S \dot{O}$ of the regular pyramid $S A B C$ (Fig. 11.63), $S D$ be the apothem of the lateral face $A S B$. $O_{1} N$ and $O_{1} M$ are perpendiculars dropped from $O_{1}$ to $S C$ and $S D$, respectively, with $O_{1} N=a$.
Since the pyramid $S A B C$ is regular... | \frac{18^{3}b^{3}}{(^{2}-b^{2})\sqrt{4b^{2}-^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,023 |
11.171. A cube is inscribed in a hemisphere of radius $R$ such that four of its vertices lie on the base of the hemisphere, while the other four vertices are located on its spherical surface. Calculate the volume of the cube. | Solution.
Consider the section of the specified hemisphere and cube by a plane passing through the opposite lateral edges $A A_{1}$ and $C C_{1}$ of the cube (Fig. 11.65), $O$ - the center of the hemisphere and lies on the diagonal $A C$ of the base of the cube. Let the edge of the cube $A A_{1}=x$. Then $O A=\frac{x ... | \frac{2R^{3}\sqrt{6}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,024 |
11.172. The angle between the generatrix of the cone and the plane of the base is $30^{\circ}$. The lateral surface area of the cone is $3 \pi \sqrt{3}$ square units. Determine the volume of a regular hexagonal pyramid inscribed in the cone. | Solution.
A regular hexagon $ABCDEF$ is the base of a regular pyramid inscribed in a cone with height $MO$ (Fig. 11.66). Then $\angle MEO$ is the angle between the generatrix of the cone and the plane of the base, $\angle MEO = 30^\circ$. Let $EO = R, ME = l$.
Then from $\triangle MOE \left(\angle MOE = 90^\circ\righ... | \frac{27\sqrt{2}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,025 |
11.173. A regular hexagonal prism is circumscribed around a sphere of radius $R$. Determine its total surface area. | ## Solution.
A regular hexagon $A B C D E F$ is the base of a regular prism circumscribed around a sphere with center $O$ (Fig. 11.66). Point $O$ is the midpoint of the axis $L L_{1}$ of the prism, $L L_{1}=2 R$. $M$ is the point of tangency of the sphere with the face $C C_{1} D_{1} D$ of the prism, $K$ is the midpoi... | 12R^{2}\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,026 |
11.174. A regular truncated hexagonal pyramid is inscribed in a sphere of radius $R$, with the plane of the lower base passing through the center of the sphere, and the lateral edge forming an angle of $60^{\circ}$ with the base plane. Determine the volume of the pyramid. | Solution.
According to the condition, $\angle O A A_{1}=60^{\circ}$ (Fig. 11.68); hence, $\angle O_{1} O A_{1}=30^{\circ}$ and $A_{1} O_{1}=\frac{1}{2} A_{1} O=\frac{R}{2}, O O_{1}=\frac{R \sqrt{3}}{2}$. We find $S_{\text {lower.base }}=6 \frac{R^{2} \sqrt{3}}{4}=\frac{3 R^{2} \sqrt{3}}{2}$, $S_{\text {upper.base }}=\... | \frac{21R^{3}}{16} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,027 |
11.175. A right parallelepiped is described around a sphere, with the diagonals of the base being $a$ and $b$. Determine the total surface area of the parallelepiped. | Solution.
Let the radius of the sphere be $R$. In the section of the sphere by a plane passing through its center and parallel to the base of the parallelepiped, we obtain a parallelogram circumscribed about a circle of radius $R$. Since the sums of the opposite sides of such a circumscribed parallelogram are equal, i... | 3ab | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,028 |
11.176. A regular quadrilateral pyramid is inscribed in a sphere of radius $R$. Determine the volume of this pyramid if the radius of the circle circumscribed around its base is $r$. | ## Solution.
Let $EO$ be the height of the regular pyramid $EABCD$ inscribed in a sphere of radius $R$ (Fig. 11.69), $OC=r$. Points $A, E, C$ belong to the surface of the sphere. Therefore, $R$ is the radius of the circle circumscribed around $\triangle AEC$. Then $R=\frac{AE \cdot EC \cdot AC}{4 S_{\triangle AEC}}=\f... | \frac{2r^{2}(R\\sqrt{R^{2}-r^{2}})}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,029 |
11.177. A cone is formed by rotating a right triangle with area $S$ around one of its legs. Find the volume of the cone if the length of the circumference described by the point of intersection of the triangle's medians during its rotation is $L$.
| Solution.
The required volume $V=\frac{1}{3} \pi r^{2} h$. Let the cone be formed by rotating $\triangle A B C$ around the leg $B C$ (Fig. 11.70); then $A C=r, B C=h$. By the condition $\frac{1}{2} r h=S$; then $V=\frac{2}{3} \pi r S$. Further, by the condition, $2 \pi \cdot D N=L$, where $D-$
, $A_{1} B_{1}=c, A A_{1} \perp A_{1} B_{1}, B B_{1} \perp A_{1} B_{1}, A A_{1}=a, B \dot{B}_{1}=b$. In the plane $A A_{1} B_{1}$, draw the perpendicular $B_{1} C$ to $A_{1} B_{1}, B_{1} C=A_{1} A$. Then $\angle B B_{1} C$ is t... | \sqrt{^{2}+b^{2}+^{2}+} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,033 |
11.181. The total surface area of the cone is $\pi S$ sq. units. The lateral surface of the cone, when unfolded onto a plane, forms a sector with an angle of $60^{\circ}$. Determine the volume of the cone. | Solution.
Let $R$ be the radius of the base of the cone, $l$ its slant height, and $H$ its height. The lateral surface of the cone unfolds into a sector of a circle with radius $l$ and a central angle of $60^{\circ}$, the length of the arc of which is $2 \pi R$. Then $2 \pi R = \frac{1}{6} \cdot 2 \pi l ; l = 6 R . H ... | \frac{\piS\sqrt{5S}}{21} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,034 |
11.182. The radius of the base of the cone is $R$, and the lateral surface area is equal to the sum of the areas of the base and the axial section. Determine the volume of the cone. | ## Solution.
Let $H$ be the height, and $l$ the slant height of the cone. Then $l=\sqrt{H^{2}+R^{2}}$, the area of the axial section $S_{1}=R H$, the area of the base $S_{2}=\pi R^{2}$, and the lateral surface area $S_{3}=\pi R l=\pi R \sqrt{H^{2}+R^{2}}$. According to the condition, $S_{3}=S_{1}+S_{2}$.
$$
\text { T... | \frac{2\pi^{2}R^{3}}{3(\pi^{2}-1)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,035 |
11.183. A regular triangular prism is described around a sphere, and a sphere is described around it. Find the ratio of the surfaces of these spheres. | Solution.
Let $r$ and $R$ be the radii of the inscribed and circumscribed spheres (Fig. 11.73); then $B D=3 r, A D^{2}+B D^{2}=A B^{2}=4 A D^{2}$, $B D^{2}=3 A D^{2}, A D^{2}=3 r^{2}$.
From $\triangle A K D$ we find that
$$
K A^{2}=K D^{2}+A D^{2}=r^{2}+3 r^{2}=4 r^{2}, \mathrm{a}
$$
. Find the ratio of their volumes. | Solution.
Let $R$ be the radius of the base of the cylinder and the radius of the sphere, $H$ be the height, $S_{1}$ be the total surface area, $V_{1}$ be the volume of the cylinder, $S_{2}$ be the surface area, and $V_{2}$ be the volume of the sphere. Then $\frac{S_{1}}{S_{2}}=\frac{2 \pi R^{2}+2 \pi R H}{4 \pi R^{2}... | \frac{6-3n}{4n} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,037 |
11.185. Find the surface area of a sphere inscribed in a pyramid, whose base is a triangle with sides 13, 14, and $15 \mathrm{~cm}$, if the vertex of the pyramid is 5 cm away from each side of the base. | Solution.
Let $MO$ be the height of the pyramid $MABC$ (Fig. 11.74), $AC=13$ cm, $BC=14$ cm, $AB=15$ cm, $MK \perp AC$, $MP \perp AB$, $MF \perp BC$, $MK=MP=MF=5$ cm.
Then $OK \perp AC$, $OP \perp AB$, $OF \perp BC$, $OK=OP=OF$, and therefore, point $O$ is the center of the circle inscribed in $\triangle ABC$, and $O... | \frac{64\pi}{9}^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,038 |
11.186. The height of the cone is $h$. The lateral surface of this cone unfolds into a sector with a central angle of $120^{\circ}$. Calculate the volume of the cone. | ## Solution.
Let $R$ be the radius of the base of the cone, and $l$ be its slant height. The length of the arc of the lateral surface development is $2 \pi R$ or $\frac{2 \pi l}{3}$. Therefore, $2 \pi R = \frac{2}{3} \pi l$, which gives $l = 3R$. Since $l^2 = R^2 + h^2$, we have $9R^2 = R^2 + h^2$, leading to $R^2 = \... | \frac{\pi^3}{24} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,039 |
11.187. Calculate the surface area of a sphere inscribed in a triangular pyramid, all edges of which are equal to $a$. | Solution.
Consider the plane passing through the height of the pyramid and the apothem (Fig. 11.75). The radius of the circle in the obtained section is equal to the radius of the sphere. Since all edges of the pyramid are equal to $a$, then $S D=\frac{a \sqrt{3}}{2}$, $O D=\frac{a \sqrt{3}}{6}$. From $\triangle S O D... | \frac{\pi^{2}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,040 |
11.188. Determine the lateral surface area and volume of a truncated cone with a generatrix of length $l$, circumscribed around a sphere of radius $r$. | Solution.
To find the lateral surface area of a truncated cone, we use the formula \( S_{\text {bok }}=\pi\left(r_{1}+r_{2}\right) l \), where \( r_{1} \) and \( r_{2} \) are the radii of the bases of the truncated cone. We draw a plane through the height of the cone. In the cross-section, we get an isosceles trapezoi... | \pi^{2};\frac{2\pir(^{2}-r^{2})}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,041 |
11.190. The radius of the base of the cone is $R$. Two mutually perpendicular generators divide the lateral surface area of the cone into parts in the ratio $1: 2$. Find the volume of the cone. | Solution.
VO is the height of the given cone, $BA$ and $BC$ are the generators mentioned in the problem (Fig. 11.77). Let $BA = l$. Since $BA = BC$ and $\angle ABC = 90^{\circ}$, then $AC = l \sqrt{2}$. Since the generators $BA$ and $BC$ divide the lateral surface area of the cone in the ratio $1:2$, the length of the... | \frac{\piR^3\sqrt{2}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,043 |
11.192. A plane passing through the vertex of a cone intersects the base along a chord, the length of which is equal to the radius of this base. Determine the ratio of the volumes of the resulting parts of the cone. | Solution.
Let $AO = r, SO = h$ (Fig. 11.79), $V$ be the volume of the cone, and $V_{1}$ and $V_{2}$ be the volumes of its parts. We will find $V_{1}$ as the difference between the volumes of the part of the cone whose base is the sector $AOB$ and the pyramid whose base is $\triangle AOB$. According to the condition, $... | \frac{2\pi-3\sqrt{3}}{10\pi+3\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,044 |
11.193. The base of the pyramid is a right-angled triangle. The lateral edges of the pyramid are equal, and the lateral faces passing through the legs form angles of 30 and $60^{\circ}$ with the base plane. Find the volume of the cone circumscribed around the pyramid if the height of the pyramid is $h$. | ## Solution.
Let $D O$ be the height of the pyramid $D A B C$ (Fig. 11.80), $\angle A C B=90^{\circ}$, $D O=h$. Since the lateral edges of the given pyramid are equal, $O$ is the center of the circle circumscribed around $\triangle A B C$, and since $\triangle A B C$ is a right triangle, $O$ is the midpoint of the hyp... | \frac{10\pi^{3}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,045 |
11.194. A parallelogram with a perimeter of $2 p$ rotates around an axis perpendicular to a diagonal of length $d$ and passing through its end. Find the surface area of the solid of revolution. | ## Solution.
The surface $S$ of the body of revolution consists of the lateral surfaces of two truncated cones obtained by rotating segments $B C$ and $C D$ (Fig. 11.81), and two cones obtained by rotating segments $A B$ and $A D$.
, from which $l = 4R$.
Next, we find the height o... | \frac{\piR^3\sqrt{15}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,047 |
12.131. In an acute-angled triangle $A B C$, the altitude $A D=a$, the altitude $C E=b$, and the acute angle between $A D$ and $C E$ is $\alpha$. Find $A C$. | Solution.
Since $\triangle A B C$ is an acute-angled triangle, the point $O$ of intersection of its altitudes lies inside the triangle, $\angle A O E=\alpha$ (Fig. 12.4).
$\triangle A E O$ and $\triangle A D B$ are right triangles with a common angle $\angle D A B$. Therefore, $\angle B=\angle A O E=\alpha$.
From $\... | \frac{\sqrt{^{2}+b^{2}-2\cos\alpha}}{\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,048 |
12.132. The acute angle of a right triangle is $\alpha$. Find the ratio of the radius of the inscribed circle to the radius of the circumscribed circle. For what value of $\alpha$ is this ratio the greatest? | Solution.
In $\triangle ABC$ (Fig. 12.5) $\angle ABC=90^{\circ}, \angle ABC=\alpha, O$ is the center of the inscribed circle, $D$ is the point of tangency with the hypotenuse $AB$.
Let the radius of the inscribed circle $OD=r$.
Then from $\triangle BDO\left(\angle BDO=90^{\circ}\right): BD=OD \operatorname{ctg} \ang... | \alpha=45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,049 |
12.133. The arc $AB$ of sector $AOB$ contains $\alpha$ radians. A line is drawn through point $B$ and the midpoint $C$ of radius $OA$. In what ratio does this line divide the area of the sector? | ## Solution.
Let the radius of the given sector be $\mathrm{R}$.
The area of sector $A O B$ (Fig. 12.6) $S=\frac{R^{2} \alpha}{2}$.
$$
S_{\triangle B C O}=\frac{1}{2} O C \cdot O B \sin \angle B O C=\frac{R^{2} \sin \alpha}{4}
$$
The area of figure $A B C$
$$
S_{1}=S-S_{\triangle B C O}=\frac{R^{2} \alpha}{2}-\fra... | \frac{\sin\alpha}{2\alpha-\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,050 |
12.134. The bases of an isosceles trapezoid are equal to $a$ and $b(a>b)$, and the angle at the larger base is $\alpha$. Find the radius of the circle circumscribed around the trapezoid. | Solution.
Let $BL$ be the height of the given trapezoid $ABCD$ (Fig. 12.7), $BC=b, AD=a$, $\angle A=\alpha, 0^{\circ}<\alpha<90^{\circ}$.
Then $AL=\frac{a-b}{2}$ and from $\triangle ALB\left(\angle ALB=90^{\circ}\right)$:
$$
AB=\frac{a-b}{2 \cos \alpha}
$$
, and the center $O_{1}$ of the circle inscribed in it lies on the bisector $O K$ of the angle $A O B$.
Further, let $R$ be the radius of the sector, and $r$ be the radius of the circle.
Then the area of the sector $S_{1}=\frac{\alpha R^{2}}{2... | \frac{2\alpha\cos^{4}(\frac{\pi}{4}-\frac{\alpha}{4})}{\pi\sin^{2}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,052 |
12.136. The lateral sides of the trapezoid are equal to $p$ and $q(p<q)$, the larger base is equal to $a$. The angles at the larger base are in the ratio 2:1.
Find the smaller base. | Solution.
Let $B M$ and $C N$ be the heights of trapezoid $A B C D$ (Fig. 12.9a), $A B=p, C D=q, A D=a$.
Then $\angle A: \angle D=2: 1, M N=B C=x$. Let $\angle D=\alpha$. Then $\angle A=2 \alpha$.
In $\triangle A M B\left(\angle A M B=90^{\circ}\right): B M=p \sin 2 \alpha ; A M=p \cos 2 \alpha$.
In $\triangle D N ... | \frac{p^{2}+-q^{2}}{p} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,053 |
12.137. The area of an isosceles trapezoid is $S$, the angle between its diagonals, opposite the lateral side, is $\alpha$. Find the height of the trapezoid. | ## Solution.
Let $O$ be the point of intersection of the diagonals of the given trapezoid $ABCD, AB=CD$.
Fig. 12.10 (Fig. 12.10), $\angle AOB = \alpha$.
Since $\angle AOB$ is the extern... | \sqrt{S\operatorname{tg}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,054 |
12.138. The larger base of a trapezoid inscribed in a circle is equal to the diameter of the circle, and the angle at the base is $\alpha$. In what ratio does the point of intersection of the diagonals of the trapezoid divide its height? | Solution.
Let the base $AD$ of the isosceles trapezoid $ABCD$ (Fig. 12.11) be the diameter of the circle circumscribed around the trapezoid, then the center $O$ of the circle is the midpoint of $AD$.
The height $KO$ of the trapezoid passes through the point $L$ of intersection of the diagonals, $\triangle BLC \sim \t... | -\cos2\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,055 |
12.139. An equilateral triangle $A_{1} B_{1} C_{1}$ is inscribed in an equilateral triangle $A B C$. Point $A_{1}$ lies on side $B C$, point $B_{1}$ - on side $A C$, and point $C_{1}$ - on side $A B$. The angle $A_{1} B_{1} C$ is equal to $\alpha$. Find the ratio $A B : A_{1} B_{1}$. | Solution.
In $\triangle A_{l} B_{l} C$ (Fig. 12.12) $\angle C=60^{\circ}, \angle B_{l} A_{l} C=120^{\circ}-\alpha$.
\[
\begin{aligned}
& \frac{A B}{A_{1} B_{1}}=\frac{A_{1} C+B_{1} C}{A_{1} B_{1}}=\frac{A_{1} C}{A_{1} B_{1}}+\frac{B_{1} C}{A_{1} B_{1}}=\frac{\sin \angle A_{1} B_{1} C}{\sin \angle C}+\frac{\sin \angle... | 2\sin(30+\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,056 |
12.140. In what ratio does the point $O$, from which all three sides of the isosceles triangle $ABC$ are seen at the same angle $(\angle AOB = \angle BOC = \angle COA)$, divide the height of the isosceles triangle $ABC$ if the angle at the base of the triangle is $\alpha\left(\alpha>\frac{\pi}{6}\right)$? | ## Solution.
Let $B D$ be the height, bisector, and median of $\triangle A B C$ (Fig.12.13).
$\angle A O B=\angle B O C=\angle A O C=\frac{2 \pi}{3}$.
$\angle A O K=\frac{1}{2} \angle A O C=\frac{\pi}{3}$.
Let $O K=1$, then from $\triangle A D O\left(\angle A D O=\frac{\pi}{2}\right)$ :
$$
A D=O D \operatorname{tg... | \frac{2\sin(\alpha-\frac{\pi}{6})}{\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,057 |
12.141. The height of an isosceles triangle is $h$ and forms an angle $\alpha\left(\alpha \leq \frac{\pi}{6}\right)$ with the lateral side. Find the distance between the centers of the inscribed and circumscribed circles of the triangle. | ## Solution.
Let $B D$ be the height of $\triangle A B C, A B=B C, B D=h, \angle A B D=\alpha, O_{1}$ and $O_{2}$ be the centers of the circumcircle and incircle of triangle $A B C$, respectively.
Since $\angle A B C=2 \alpha \leq \frac{\pi}{3}$, the points $O_{1}$ and $O_{2}$ are located on the segment $B D$ such th... | \frac{\cos(\frac{\pi}{4}+\frac{3\alpha}{2})}{2\cos^{2}\alpha\cos(\frac{\pi}{4}-\frac{\alpha}{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,058 |
12.142. A triangle is inscribed in a circle of radius $R$, with its vertices dividing the circle into three parts in the ratio $2: 5: 17$. Find the area of the triangle. | ## Solution.
Let $\cup A B=2 x$ (Fig. 12.15), $\cup B C=5 x$, $\cup A m C=17 x$ and $2 x+5 x+17 x=360^{\circ}, x=15^{\circ}$ $\Rightarrow \angle A O B=\cup A B=30^{\circ}, \angle B O C=\cup B C=75^{\circ}$, $\angle A O C=\angle A O B+\angle B O C=105^{\circ}$.
$S_{\triangle A B C}=S_{\triangle A O B}+S_{\triangle B O... | \frac{R^{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,059 |
12.143. The tangent of the angle at the base of an isosceles triangle is 3/4. Find the tangent of the angle between the median and the bisector drawn to the lateral side. | ## Solution.
In $\triangle ABC$, $AB=BC$, $AM$ is the angle bisector, $AN$ is the median, $BD$ is the altitude, (Fig.12.16). $\angle C=\alpha, \operatorname{tg} \alpha=\frac{3}{4}$.
Fig. 1... | \frac{1}{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,060 |
12.145. A line is drawn through the vertex of angle $\alpha$ at the base of an isosceles triangle, intersecting the opposite lateral side and forming an angle $\beta$ with the base. In what ratio does this line divide the area of the triangle? | Solution.
In $\triangle ABC$, $AB = BC$, $\angle BAC = \alpha$, $\angle DAC = \beta$ (Fig. 12.18)
Then $\angle BAD = \alpha - \beta$.
Fig. 12.19
}{2\cos\alpha\sin\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,062 |
12.146. Through the vertices of an equilateral triangle $A B C$, parallel lines $A D, B E$ and $C F$ are drawn. The line $B E$ lies between the lines $A D$ and $C F$ and divides the distance between them in the ratio $m: n$, measured from the line $A D$. Find the angle $B C F$. | ## Solution.
Let $A K$ be the common perpendicular of the given parallel lines, $L$ be the intersection point of $B E$ and $A K, A L: L K=m: n, B F \perp F K$ (Fig. 12.19).
Let $B C=a, \angle B C F=\alpha$.
In $\triangle B F C\left(\angle B F C=90^{\circ}\right): B F=a \sin \alpha, K L=B F=a \sin \alpha$,
$$
\begin... | \operatorname{arctg}\frac{n\sqrt{3}}{2+n} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,063 |
12.147. Find the cosine of the acute angle of the rhombus if a line drawn through its vertex divides the angle in the ratio $1: 3$, and the opposite side - in the ratio $3: 5$. | ## Solution.
Let $L$ be the point of intersection of the given line $BL$ with side $AD$ of the rhombus $ABCD, \angle ABL: \angle CBL = 1: 3, AL: LD = 3: 5$ (Fig. 12.20).
$$
\angle ABL = \alpha, AL = 3x
$$
Then $\angle CBL = 3\alpha, LD = 5x, AB = 8x, \angle ABC = 4\alpha, \angle ABO = 2\alpha$.
If $BL$ is the bisec... | \frac{7}{18} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,064 |
12.148. The ratio of the area of rectangle $ABCD (BC \| AD)$ to the square of its diagonal is $k$. Find $\angle EAF$, where $E$ and $\boldsymbol{F}$ are the midpoints of sides $BC$ and $CD$, respectively. | Solution.
Let $AB = y, AD = x, \angle BAE = \alpha, \angle DAF = \beta, \angle EAF = \gamma$ (Fig. 12.21). The diagonal of the rectangle is $\sqrt{x^2 + y^2}$, the area is $xy$, and $\frac{xy}{x^2 + y^2} = k \Rightarrow$
, then $B L=2 r$. In $\triangle A L B\left(\angle A L B=90^{\circ}\right): A B=\frac{B L}{\sin \angle B A L}=\frac{2 r}{\sin \alpha}$. Since a circle is inscribed in the given trapezoid, then $A B+B C=A B+C D=2 A B$, $A B=C D \Rig... | \frac{r\sqrt{1+\sin^{2}\alpha}}{\sin^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,066 |
12.150. The height of a triangle divides the angle of the triangle in the ratio $2: 1$, and the base - into segments, the ratio of which (larger to smaller) is $k$. Find the sine of the smaller angle at the base and the permissible values of $k$. | Solution.
Let $A D$ be the height of $\triangle A B C$
Fig. 12.23
(Fig. 12.23), $\angle C A D: \angle B A D=2: 1 \Rightarrow$ $\Rightarrow A C>A B, C D>B D, C D: B D=k$.
If $\angle B A D... | \frac{1}{k-1},k>2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,067 |
12.151. The hypotenuse of a right-angled triangle is divided by the point of tangency of the inscribed circle into segments, the ratio of which is $k$. Find the angles of the triangle. | ## Solution.
Let $O$ be the center of the circle inscribed in triangle $ABC$, $\angle BAC = \pi / 2$, and $D$ be the point of tangency of this circle with the hypotenuse $BC$ (Fig. 12.24), $BD: DC = k$, $\angle BAC = \alpha$. Then $\angle OBD = \frac{\alpha}{2}$, $\angle OCD = \frac{\pi}{4} - \frac{\alpha}{2}$.
$$
\t... | \frac{\pi}{4}\\arcsin\frac{\sqrt{2}(k-1)}{2(k+1)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,068 |
12.152. The ratio of the lateral sides of the trapezoid is equal to the ratio of its perimeter to the length of the inscribed circle and is equal to $k$. Find the angles of the trapezoid and the permissible values of $k$. | ## Solution.
Let in trapezoid $ABCD$ (Fig. 12.25) $BC \| AD$,
$BC < AD, BE$ and $CK$ - the heights of the trapezoid, $R$ - the radius of the circle inscribed in the trapezoid, $\angle A = \alpha, \angle D = \beta$.
Then $BE = CK = 2R, AB = \frac{2R}{\sin \alpha}, CD = \frac{2R}{\sin \beta}$.
Since $AB + CD = BC + A... | \arcsin\frac{2(k+1)}{\pik^2},\pi-\arcsin\frac{2(k+1)}{\pik^2},\arcsin\frac{2(k+1)}{\pik},\pi-\arcsin\frac{2(k+1)}{\pik},k\geq\frac{2}{\pi-2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,069 |
12.153. A circle of radius $r$ is inscribed in a sector of radius $R$. Find the perimeter of the sector. | ## Solution.
Let $D$ be the point of tangency of the circle inscribed in the sector $A O B$ (Fig. 12.26), with the center $O_{1}$ located on the bisector $O K$ of the angle $A O B$.
Let $\angle A O B=\alpha$, then $\angle D O K=\frac{\alpha}{2}$ and from $\triangle O D O_{1}\left(\angle O D O_{1}=\frac{\pi}{2}\right)... | 2R(1+\arcsin\frac{r}{R-r}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,070 |
12.154. In an acute isosceles triangle, the radius of the inscribed circle is 4 times smaller than the radius of the circumscribed circle. Find the angles of the triangle. | Solution.
Let in $\triangle A B C(A B=B C) \quad A C=a, \angle B A C=\alpha, B D$ - altitude, $O-$ center of the inscribed circle, $R$ and $r$ - radii of the circumscribed and inscribed circles (Fig. 12.27)
In $\triangle A D O\left(\angle A D O=90^{\circ}\right):$
$r=O D=A D \operatorname{tg} \angle O A D=\frac{a}{2}... | \arccos\frac{2-\sqrt{2}}{4};\arccos\frac{2\sqrt{2}+1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,071 |
12.155. In triangle $ABC$, the acute angles $\alpha$ and $\gamma (\alpha > \gamma)$ adjacent to side $AC$ are given. From vertex $B$, the median $BD$ and the angle bisector $BE$ are drawn. Find the ratio of the area of triangle $BDE$ to the area of triangle $ABC$. | Solution.
In $\triangle ABC$, $\angle A = \alpha$, $\angle C = \gamma$ (Fig. 12.28). Since $\alpha > \gamma$, point $D$ is located between points $A$ and $E$.
Fig. 12.28
$\triangle ABC$... | \frac{\tan\frac{\alpha-\gamma}{2}}{2\tan\frac{\alpha+\gamma}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,072 |
12.156. The angle at vertex $A$ of trapezoid $A B C D$ is $\alpha$. The lateral side $A B$ is twice the length of the shorter base $B C$. Find the angle $B A C$. | Solution. Let $\angle B A C=\beta$ (Fig. 12.29).
Then $\angle B C A=\angle C A D=\alpha-\beta$. From $\triangle A B C: \frac{\sin \beta}{\sin (\alpha-\beta)}=\frac{B C}{A B}=\frac{1}{2} \Leftrightarrow$
$\Leftrightarrow 2 \sin \beta=\sin \alpha \cos \beta-\sin \beta \cos \alpha \Leftrightarrow 2=\sin \alpha \operator... | \operatorname{arctg}\frac{\sin\alpha}{2+\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,073 |
12.157. In a right-angled triangle, find the angle between the median and the bisector drawn from the vertex of the acute angle equal to $\alpha$. | Solution.
Let in $\triangle A B C$ (Fig. 12.30) $\angle A=90^{\circ}, \angle A B C=\alpha$,
$B D$ - bisector, $B E$ - median. Since $\angle A>\angle C$, point $D$ is located between points $A$ and $E$. If $A B=x$, then $A C=x \operatorname{tg} \alpha$,
$$
\begin{aligned}
& A E=\frac{1}{2} x \operatorname{tg} \alpha ... | \operatorname{arctg}(\frac{\operatorname{tg}\alpha}{2})-\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,074 |
12.158. Find the cosines of the acute angles of a right triangle, given that the product of the tangents of half these angles is $1 / 6$. | Solution.
Let one of the angles be $\alpha$. Then the second angle will be $90^{\circ}-\alpha$ and by the condition $\operatorname{tg} \frac{\alpha}{2} \operatorname{tg}\left(45^{\circ}-\frac{\alpha}{2}\right)=\frac{1}{6} \Rightarrow \operatorname{tg} \frac{\alpha}{2} \cdot \frac{1-\operatorname{tg} \frac{\alpha}{2}}{... | \frac{3}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,075 |
12.159. The sides of the parallelogram are in the ratio $p: q$, and the diagonals are in the ratio m:n. Find the angles of the parallelogram. | ## Solution.
Let in parallelogram $A B C D$ (Fig. 12.31) $A D=p y, B D=m x$. Then $A B=q y, A C=n x$. From $\triangle B A D$ and $\triangle A B C$ by the cosine theorem: $m^{2} x^{2}=p^{2} y^{2}+q^{2} y^{2}-2 p q y^{2} \cos \angle A ; m^{2} x^{2}=p^{2} y^{2}+q^{2} y^{2}+2 p q y^{2} \cos \angle A$. $\Rightarrow\left(n^... | \arccos\frac{(p^{2}+q^{2})(n^{2}-^{2})}{2pq(^{2}+n^{2})};\pi-\arccos\frac{(p^{2}+q^{2})(n^{2}-^{2})}{2pq(^{2}+n^{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,076 |
12.160. The ratio of the perimeter of a rhombus to the sum of its diagonals is $k$. Find the angles of the rhombus and the permissible values of $k$. | ## Solution.
Let the side of the rhombus $ABCD$ be $a$, and the acute angle $\alpha$ (Fig. 12.32). Then the diagonals of the rhombus are $2 a \sin \frac{\alpha}{2}$ and $2 a \cos \frac{\alpha}{2}$.
## According to the condition
$$
\begin{aligned}
& k=\frac{4 a}{2 a \sin \frac{\alpha}{2}+2 a \cos \frac{\alpha}{2}}=\f... | \arcsin\frac{4-k^{2}}{k^{2}},\pi-\arcsin\frac{4-k^{2}}{k^{2}};\sqrt{2}\leqk<2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,077 |
12.162. The perimeter of a sector is equal to $l$. Find the distance from the vertex of the central angle of the sector to the center of the circle inscribed in this sector, if the radius of the arc of the sector is $R$. | ## Solution.
Let $E$ be the point of tangency of the inscribed circle and the radius $O_{1} A$ of the sector, the center $O$ of the circle lies on the bisector $O_{1} C$ of the angle $A O_{1} B$ (Fig. 12.34). If $\angle A O_{1} B=\alpha$, then the length of the arc $A C B$ is $\alpha R$ and
$$
\begin{aligned}
& l=2 R... | \frac{R}{2\cos^{2}\frac{(\pi+2)R-}{4R}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,079 |
12.163. Show that if in a triangle the ratio of the sum of the sines of two angles to the sum of their cosines is equal to the sine of the third angle, then the triangle is a right triangle. | ## Solution.
Let $\alpha, \beta, \gamma$ be the angles of the given triangle and $\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \sin \gamma$.
$$
\begin{aligned}
& \text{Then } \frac{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}}{2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}} = \s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 49,080 |
12.167. The radius of the sector's arc is $R$, the central angle $A O B$ is $\alpha$. Through the midpoint $C$ of the radius $O A$, a line is drawn parallel to the radius $O B$ and intersects the arc $A B$ at point $D$. Find the area of triangle $O C D$. | Solution.
Let $\angle B O D=\beta$ (Fig. 12.38). Then $\angle C D O=\beta, \angle D O C=\alpha-\beta$, $\angle D C O=180^{\circ}-\alpha$.
In $\triangle C O D: \frac{D O}{\sin \angle D C O}=\frac{C O}{\sin \angle C D O} \Leftrightarrow \frac{R}{\sin \left(180^{\circ}-\alpha\right)}=\frac{R}{2 \sin \beta} ; \sin \beta=... | \frac{R^{2}\sin\alpha}{8}(\sqrt{4-\sin^{2}\alpha}-\cos\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,084 |
12.168. In a triangle, the side $a$, the angle $\alpha$ opposite to it, and the height $h$ drawn to the given side are given. Find the sum of the other two sides. | ## Solution.
Let $b$ and $c$ be the unknown sides of the triangle. Then the area of the triangle $S=\frac{1}{2} b c \sin \alpha=\frac{1}{2} a h, b c=\frac{a h}{\sin \alpha}$.
$$
\begin{aligned}
& a^{2}=b^{2}+c^{2}-2 b c \cos \alpha \Leftrightarrow b^{2}+c^{2}=a^{2}+2 b c \cos \alpha \Leftrightarrow(b+c)^{2}= \\
& =b^... | \sqrt{^{2}+2\operatorname{ctg}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,085 |
12.169. In the square \(ABCD\), an isosceles triangle \(AEF\) is inscribed; point \(E\) lies on side \(BC\), point \(F\) lies on side \(CD\), and \(AE = EF\). The tangent of angle \(AEF\) is 2. Find the tangent of angle \(FEC\).
.
Then $\angle B E A=180^{\circ}-(\alpha+x)$.
In $\triangle A B E\left(\angle A B E=90^{\circ}\right)$ :
$$
\begin{aligned}
& A E=\frac{A B}{\sin \angle B E A}=\frac{a}{\sin (\alpha+x)} \\
& B E=A B \operatorname{ctg} \angle B E A=-a \operato... | 3-\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,086 |
12.170. In triangle $A B C$, the acute angles $\alpha$ and $\gamma (\alpha>\gamma)$ at the base $A C$ are given. From vertex $B$, the altitude $B D$ and the median $B E$ are drawn.
Find the area of triangle $B D E$, if the area of triangle $A C B$ is $S$. | Solution.
In $\triangle ABC, \angle A=\alpha, \angle C=\gamma$ (Fig. 12.40).
Since $\alpha>\gamma$, point $D$ is located between points $A$ and $E$.
$A E=A D+D E, C D=C E+D E$. Since $C E=A E$, then $C D=(A D+D E)+D E$,
$D E=\frac{C D-A D}{2}$.
Let $B D=H$. Then from $\triangle A B D$ we get $A D=H \operatorname{c... | \frac{S\sin(\alpha-\gamma)}{2\sin(\alpha+\gamma)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,087 |
12.171. In a right triangle $ABC$, the acute angle at vertex $A$ is $\alpha$. A line through the midpoint $D$ of the hypotenuse $AB$ intersects the leg $AC$ at point $E$. In what ratio does this line divide the area of triangle $ABC$, if $\angle DEA=\beta, AE>0.5 AC?$ | Solution.
Let $S_{1}$ be the area of triangle $A D E$, $S_{2}$ be the area of quadrilateral $B C E D$, and $S$ be the area of triangle $A B C$ (Fig. 12.41).
Then
$$
\begin{aligned}
& S_{1}=\frac{1}{2} A E \cdot A D \sin \angle A=\frac{1}{4} A E \cdot A B \sin \angle A, S=\frac{1}{2} A C \cdot A B \sin \angle A \\
& ... | \frac{3\operatorname{tg}\beta-\operatorname{tg}\alpha}{\operatorname{tg}\alpha+\operatorname{tg}\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,088 |
12.172. A trapezoid is inscribed in a circle. The larger base of the trapezoid forms an angle $\alpha$ with a lateral side, and an angle $\beta$ with a diagonal. Find the ratio of the area of the circle to the area of the trapezoid. | ## Solution.
Let $AD$ be the larger base of the given trapezoid $ABCD$, $\angle BAD = \alpha$, $\angle BDA = \beta$ (Fig. 12.42), $BN$ be the height of the trapezoid, and $BD = 1$.
Then, from $\triangle BMD \left(\angle BMD = 90^\circ\right)$:
$BM = BD \sin \angle BDM = \sin \beta$
$DM = BD \cos \angle BDM = \cos \... | \frac{\pi}{2\sin^2\alpha\sin2\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,089 |
12.173. In triangle $ABC$, angle $A$ is equal to $\alpha$ and side $BC=a$. Find the length of the bisector $AD$, if the angle between the bisector $AD$ and the altitude $AE$ is $\beta$. | ## Solution.
If $A D=x$ (Fig. 12.43), then from $\triangle A E D\left(\angle A E D=90^{\circ}\right)$:
$A E=A D \cos \angle E A D=x \cos \beta$, and from $\triangle A E B\left(\angle A E B=90^{\circ}\right)$:
$$
B E=A E \operatorname{tg} \angle B A E=x \cos \beta \operatorname{tg}\left(\frac{\alpha}{2}-\beta\right)
... | \frac{\cos(\frac{\alpha}{2}-\beta)\cos(\frac{\alpha}{2}+\beta)}{\cos\beta\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,090 |
12.175. The radius of the arc of sector $A O B$ is $R$, the central angle $A O B$ is $\alpha$. A regular triangle is inscribed in this sector such that one of its vertices coincides with the midpoint of the arc $A B$, and the other two vertices lie on the radii $O A$ and $O B$ respectively. Find the sides of the triang... | Solution.
Let $M$ be the midpoint of the arc $AB$, and $\Delta LMK$ be the equilateral triangle mentioned in the problem (Fig. 12.45).
Then $\angle LOM = \frac{\alpha}{2}, \angle MLK = 60^{\circ}, \angle OLK = 90^{\circ} - \frac{\alpha}{2}, \angle OLM = $
$$
= \angle OLK + \angle MLK = 150^{\circ} - \frac{\alpha}{2}... | \frac{R\sin\frac{\alpha}{2}}{\sin(30+\frac{\alpha}{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,092 |
12.176. A circle is inscribed in an isosceles triangle with base $a$ and angle $\alpha$ at the base. Find the radius of the circle that is tangent to the inscribed circle and the lateral sides of the triangle. | Solution.
In $\triangle ABC$ (Fig. 12.46) $AB = BC, AC = a, \angle BAC = \alpha, O$ is the center of the inscribed circle, $R$ is its radius, $O_1$ is the center of the circle that is tangent to the inscribed circle and the lateral sides of the triangle, $r$ is its radius. Points $O$ and $O_1$ lie on the height $BM$ o... | \frac{}{2}\tan^3\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,093 |
12.177. Inside the given angle $\alpha$ there is a point at a distance $a$ from the vertex and at a distance $b$ from one side. Find the distance of this point from the other side. | Solution.
Let $OA$ be the distance from point $O$ to vertex $A$ of the given angle, $OB$ be the distance to one of the sides, $OC$ be the distance to the other side, $OA=a, OB=b, OC$, and $\angle BAC=\alpha$ (Fig. 12.47), $\angle OAB=x$.
In $\triangle ABO\left(\angle ABO=90^{\circ}\right): \sin x=\frac{OB}{OA}=\frac{... | \sin\alpha\sqrt{^{2}-b^{2}}-b\cos\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,094 |
12.178. In a right triangle $ABC$, the bisector $AD$ of the acute angle $A$, equal to $\alpha$, is drawn. Find the ratio of the radii of the circles inscribed in triangles $ABD$ and $ADC$. | Solution.
Let $O$ be the center of the circle inscribed in $\triangle A B D$, $F$ be the point of tangency of this circle with $A D$, $O_{1}$ be the center of the circle inscribed in $\triangle A D C$, and $E$ be the point of tangency of this circle with $A D$ (Fig. 12.48).
\[
\begin{aligned}
& \angle B A D=\frac{\al... | \frac{\sqrt{2}\operatorname{tg}(45+\frac{\alpha}{4})}{2\sin(45+\frac{\alpha}{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,095 |
12.179. Find the sine of the angle at the vertex of an isosceles triangle, given that the perimeter of any inscribed rectangle, two vertices of which lie on the base, has a constant value. | Solution.
Let $K L M N$ and $D E F G$ be two arbitrary rectangles inscribed in $\triangle A B C, A B=B C$ (Fig. 12.49), $L K=x_{1}, L M=y_{1}, E D=x_{2}$, $D G=y_{2}, \angle A B C=\alpha$. By the condition $x_{1}+y_{1}=x_{2}+y_{2}$.
$$
\text { Then } x_{2}-x_{1}=y_{1}-y_{2}
$$
If $B S$ is the altitude of $\triangle ... | \frac{4}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,096 |
12.180. A side of the triangle is equal to 15, the sum of the other two sides is 27. Find the cosine of the angle opposite the given side, if the radius of the inscribed circle in the triangle is 4. | ## Solution.
Let $a$ be the given side of the triangle, $a=15$, $b$ and $c$ be the other two sides, $b+c=27$, $c=27-b$. For definiteness, we will assume that $b>c$. The semiperimeter of the triangle $p=\frac{a+b+c}{2}=21$, its area $S=p r=84$.
By Heron's formula $S=\sqrt{p(p-a)(p-b)(p-c)}$, thus
$$
\begin{aligned}
&... | \frac{5}{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,097 |
12.181. The smaller arc of a circle subtended by the chord $A B$ contains $\alpha^{\circ}$. Through the midpoint $C$ of the chord $A B$, a chord $D E$ is drawn such that $D C$ : $C E=1: 3$. Find the acute angle $A C D$ and the permissible values of $\alpha$. | ## Solution.
Let $A C=B C=1, D C=x$ (Fig. 12.50).
Then $C E=3 x$.
If $L$ is the midpoint of $D E$, then $O C \perp A B, O L \perp D E, \angle C O L=\angle A C D$ as acute angles with mutually perpendicular sides, $D L=E L=2 x$, $C L=x$. Since $D C \cdot C E=A C \cdot B C$, then $x \cdot 3 x=1 \cdot 1 ; x=\frac{1}{\s... | \arcsin\frac{\operatorname{tg}\frac{\alpha}{2}}{\sqrt{3}},\alpha\leq120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,098 |
12.183. The area of an isosceles obtuse triangle is 8, and the median drawn to its lateral side is $\sqrt{37}$. Find the cosine of the angle at the vertex. | ## Solution.
Let $\triangle A B C A B=B C, A D-$ median, $A D=\sqrt{37}$ (Fig. 12.52), $\angle B=\alpha, B D=x$. Then $A B=2 x$.
From $\triangle A B D:$
$A D^{2}=A B^{2}+B D^{2}-2 A B \cdot B D \cos \angle B$
$37=4 x^{2}+x^{2}-4 x^{2} \cos \alpha$
$37=x^{2}(5-4 \cos \alpha)$.
$S_{\triangle A B C}=\frac{1}{2} A B^... | -\frac{3}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,099 |
12.184. In an isosceles triangle, the angle at the base is $\alpha$. The height dropped to the base is greater than the radius of the inscribed circle by $m$. Find the radius of the circumscribed circle. | ## Solution.
In $\triangle A B C$ (Fig. 12.53) $A B=B C, \angle B A C=\alpha, B D$ is the altitude, $O$ is the center of the inscribed circle.
Then, by the condition, $B O=B D-O D=m . \angle B A O=\angle D A O=\frac{\alpha}{2}, \angle A O B=$ $=\angle O A D+\angle A D O=\frac{\alpha}{2}+90^{\circ}$ as the external an... | \frac{}{4\sin^{2}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,100 |
12.185. In a triangle, the area $S$, side $a$, and the angle $\alpha$ opposite to it are known. Find the sum of the other two sides. | Solution.
Fig. 12.53
Fig. 12.54
Let \( x \) and \( y \) b... | \sqrt{^{2}+4S\operatorname{ctg}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,101 |
12.186. Let $O A$ be a fixed radius of a circle with center at point $O ; B$ be the midpoint of radius $O A ; M$ be an arbitrary point on the circle. Find the maximum value of angle $O M B$. | Solution.
Let $\angle O M B=\alpha, \angle O B M=\beta$ (Fig. 12.54).
From $\triangle M O B: \frac{\sin \beta}{\sin \alpha}=\frac{O M}{O B}=2 ; \sin \alpha=\frac{1}{2} \sin \beta$.
The maximum value of $\sin \beta$ is 1, therefore the maximum value of $\sin \alpha$ is $1 / 2$.
Since $O B<O M$, then $\alpha<\beta$, ... | \pi/6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,102 |
12.187. In an acute isosceles triangle, the angle at the base is $\alpha$, and the area is $S$. Find the area of the triangle whose vertices are the feet of the altitudes of the given triangle. | Solution.
In $\triangle ABC$ (Fig. 12.55) $AB=BC, \angle BAC=\alpha; BD, AF, CE$ are altitudes.
Let $AB=a$, then from $\triangle ADB\left(\angle ADB=90^{\circ}\right): AD=a \cos \alpha, BD=a \sin \alpha$. $AC=2AD=2a \cos \alpha$.
In $\triangle AEC\left(\angle AEC=90^{\circ}\right)$:
$AE=AC \cos \angle BAC=2a \cos^2... | -\frac{1}{2}S\sin4\alpha\operatorname{ctg}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,103 |
12.189. A ray drawn from the vertex of an equilateral triangle divides its base in the ratio $m: n$. Find the obtuse angle between the ray and the base. | Solution.
Ray $B D$ divides side $A C$ of an equilateral triangle $A B C$ in the ratio $m: n$ (Fig. 12.57).
Let $\angle B D C=\alpha$ be the acute angle between the ray and the base, and assume that $n>m$. Then $\frac{A D}{D C}=\frac{m}{n}$.
If $A D=m x$, then $D C=n x, A B=$ $=A C=(m+n) x$.
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,105 |
12.190. A line is drawn through the vertex of an equilateral triangle, dividing the base in the ratio $2: 1$. At what angles is it inclined to the lateral sides of the triangle? | Solution.
The line $B D$ divides the side $A C$ of the equilateral triangle $A B C$ in the ratio $2: 1, C D: A D=2: 1$.
Let $A D=x, \angle A B D=\alpha$, then $C D=2 x, A B=3 x, \angle A D B=120^{\circ}-\alpha$. From $\triangle A D B$ :
$$
\begin{aligned}
& \frac{\sin \angle A D B}{\sin \angle A B D}=\frac{A B}{A D}... | \operatorname{arctg}\frac{\sqrt{3}}{5},60-\operatorname{arctg}\frac{\sqrt{3}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,106 |
12.191. The base of the triangle is equal to $a$, and the angles at the base are $\alpha$ and $\beta$ radians. From the opposite vertex of the triangle, a circle is drawn with a radius equal to the height of the triangle. Find the length of the arc of this circle that is enclosed within the triangle. | ## Solution.
Let in $\triangle A B C$ (Fig.12.58) $B C=a, \angle C=\alpha, \angle B=\beta, A K$-height, $A K=H$.
In $\triangle A K C\left(\angle A K C=\frac{\pi}{2}\right): C K=H \operatorname{ctg} \alpha$.
From $\triangle B K A\left(\angle B K A=\frac{\pi}{2}\right): B K=H \operatorname{ctg} \beta$.
Since $C K+B K... | (\pi-\alpha-\beta)\cdot\frac{\sin\alpha\sin\beta}{\sin(\alpha+\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,107 |
12.192. Given two sides $a$ and $b$ of a triangle and the bisector $l$ of the angle between them. Find this angle. | Solution.
Let $B D$ be the bisector of $\triangle A B C, B D=l, B C=a, A B=b$ (Fig. 12.59), $\angle A B C=\alpha$.
Fig. 12.59
Then $\angle A B D=\angle C B D=\frac{\alpha}{2}$.
$$
\begin... | 2\arccos\frac{(+b)}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,108 |
12.193. The base of the triangle is 4, and its median is $\sqrt{6}-\sqrt{2}$. One of the angles at the base is $15^{\circ}$. Show that the acute angle between the base of the triangle and its median is $45^{\circ}$. | Solution.
Let $BD$ be the median of $\triangle ABC, BD=\sqrt{6}-\sqrt{2}, AC=4, \angle A=15^{\circ}$ (Fig. 12.60), $\angle ADB=x$.
Then $\angle ABD=180^{\circ}-\left(x+15^{\circ}\right)$.
By the Law of Sines
from $\triangle ABD:$
$\frac{AD}{\sin \angle ABD}=\frac{BD}{\sin \angle A} \Rightarrow \frac{2}{\sin \left(... | 45 | Geometry | proof | Yes | Yes | olympiads | false | 49,109 |
12.194. In a trapezoid, the smaller base is equal to 2, the adjacent angles are $135^{\circ}$ each. The angle between the diagonals, facing the base, is $150^{\circ}$. Find the area of the trapezoid.
.
In $\triangle B O C: \angle A C B=\angle D B C=15^{\circ}$, then $\angle B A C=180^{\circ}-(\angle A B C+$ $+\angle A C B)=30^{\circ}$.
By ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,110 |
12.195. Prove that if the bisector of one of the angles of a triangle is equal to the product of the sides enclosing it, divided by their sum, then this angle is $120^{\circ}$.
Fig. 12.62
... | ## Solution.
Let $B D$ be the bisector of $\triangle A B C$ (Fig.12.62), $B C=a, A B=b, B D=l$, $\angle A B C=\alpha$.
$S_{\triangle A B C}=S_{\triangle A B D}+S_{\triangle C B D} \Rightarrow \frac{a b}{2} \sin \alpha=\frac{a l}{2} \sin \frac{\alpha}{2}+\frac{b l}{2} \sin \frac{\alpha}{2} \Leftrightarrow$
$\Leftrigh... | 120 | Geometry | proof | Yes | Yes | olympiads | false | 49,111 |
12.197. In triangle $ABC$, the altitude $BM$ is drawn, and a circle is constructed on it as a diameter, intersecting side $AB$ at point $K$ and side $BC$ at point $Z$. Find the ratio of the area of triangle $K Z M$ to the area of triangle $A B C$, if $\angle A=\alpha$ and $\angle C=\beta$. | Solution.
$\angle A B C=180^{\circ}-(\alpha+\beta)($ Fig. 12.64)
$D A M B$ and $\triangle M K B, \triangle C M B$ and $\triangle M Z B$ are right triangles with common acute angles at vertex $B$.
Thus, $\angle K M B=\angle A=\alpha, \angle Z M B=\angle C=\beta, \angle K M Z=\alpha+\beta$. Let $B M=1$, then from $\tr... | \frac{1}{4}\sin2\alpha\sin2\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,113 |
12.198. A circle is inscribed in a rhombus. In the resulting curvilinear triangle (with an acute angle), another circle is inscribed. Find its radius if the height of the rhombus is \( h \), and the acute angle is \( \alpha \).
 $\angle B A D=\alpha, 0^{\circ}<\alpha<90^{\circ}, O$ - the point of intersection of the diagonals - the center of the inscribed circle, $K$ - the center of the circle inscribed in the curvilinear triangle.
$N$ and $M$ are the points of tangency of the constructed ci... | \frac{}{2}\operatorname{tg}^{2}(45-\frac{\alpha}{4}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,114 |
12.200. The side of the base of a regular quadrilateral pyramid is equal to $a$, the lateral edge forms an angle $\alpha$ with the base plane. A cube is inscribed in this pyramid such that four of its vertices lie on the apothems of the pyramid, and four lie on the base of the pyramid. Find the edge of the cube. | ## Solution.
Let in the regular pyramid $S A B C D$ (Fig. 12.67) $A B=a, S O-$ be the height, $\angle S C O=\alpha, S E$ be the apothem, and $F$ and $F_{1}$ be the vertices of the cube mentioned in the problem. Then the edge of the cube $F F_{1}$ is perpendicular to $O E$. $C O=\frac{a \sqrt{2}}{2}, O E=\frac{a}{2} \c... | \frac{\sin\alpha}{2\sin(\frac{\pi}{4}+\alpha)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,116 |
12.201. The area of a lateral face of a regular dodecagonal pyramid is $S$. The planar angle at the vertex is $\alpha$. Find the volume of the pyramid. | Solution.
Let $C$ be the vertex of a regular dodecagonal pyramid (Fig. 12.68), $AB$ be a side of its base, $\angle ACB = \alpha$, $S_{\triangle ACB} = S$, $CO$ be the height of the pyramid, $CA = d$, $\angle CAO = x$.
Then $\frac{1}{2} d^{2} \sin \alpha = S$, $d = \sqrt{\frac{2 S}{\sin \alpha}}$,
$\angle CAB = 90^{\... | \frac{S}{\cos\frac{\alpha}{2}\sin^{3}15}\sqrt{S\tan\frac{\alpha}{2}\sin(15+\frac{\alpha}{2})\sin(15-\frac{\alpha}{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,117 |
12.205. In the base of a right prism $A B C A_{1} B_{1} C_{1}\left(A A_{1}\left\|B B_{1}\right\| C C_{1}\right)$ lies a right triangle $A B C$, where the larger leg $A B$ is equal to $a$, and the angle opposite to it, $C$, is $\alpha$. The hypotenuse $B C$ is the diameter of the base of a cone, whose vertex lies on the... | Solution.
Let $M$ be the vertex, $L$ the center of the base, and $M B$ and $M C$ the generators of the cone (Fig. 12.72) from the problem statement. The cone is positioned such that its height $M L$ is perpendicular to $B C$. Drop a perpendicular $M N$ from point $M$ to the plane $A B C$. Then $M N \parallel A A_{1}, ... | \frac{\sqrt{\operatorname{ctg}^{2}\alpha+\sin^{2}\alpha}}{2\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,120 |
12.206. A section is made through the vertex of a regular triangular pyramid and the midpoints of two sides of the base. Find the area of the section and the volume of the pyramid, given the side $a$ of the base and the angle $\alpha$ between the section and the base. | Solution.
Let $D$ and $E$ be the midpoints of sides $AB$ and $AC$ of the base of a regular pyramid $SABC$ (Fig. 12.73), $\triangle DSE$ - a section of this pyramid, $SO$ the height of the pyramid, $K$ the point of intersection of $AO$ and $DE$. Then $OK \perp DE$, $DE=\frac{a}{2}$, $AK=\frac{a \sqrt{3}}{4}$, $OA=\frac... | \frac{^{2}\sqrt{3}}{48\cos\alpha};\frac{^{3}\operatorname{tg}\alpha}{48} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,121 |
12.207. A perpendicular, equal to $p$, is dropped from the base of the height of a regular triangular pyramid onto a lateral edge. Find the volume of the pyramid if the dihedral angle between its lateral faces is $\alpha$. | Solution.
Let SO be the height of the regular pyramid $SABC$ (Fig. 12.74), $OM \perp SA, OM=p, AL$ - the height of $\triangle ABC$.
In the plane $ABC$, through point $O$, draw $FK$ parallel to $BC$.
The line $AL$ is the projection of $SA$ onto the plane $ABC, AL \perp BC$. Thus, $SA \perp BC. SA \perp BC, FK \| BC$,... | \frac{9p^{3}\operatorname{tg}^{3}\frac{\alpha}{2}}{4\sqrt{3\operatorname{tg}^{2}\frac{\alpha}{2}-1}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,122 |
12.208. A perpendicular, equal to p, is dropped from the base of the height of a regular triangular pyramid onto a lateral edge. Find the volume of the pyramid if the dihedral angle between a lateral face and the base of the pyramid is $\alpha$.
untranslated text:
12.208. Из основания высоты правильной треугольной пи... | ## Solution.
Let $S O$ be the height of the regular pyramid $S A B C$ (Fig. 12.75), $O E \perp B S$, $O E=p$, and $D$ be the midpoint of $A C$. Then $\angle O D S=\alpha$.
Let $A C=a$. Then $O D=\frac{a \sqrt{3}}{6}, O B=\frac{a \sqrt{3}}{3}, S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}$.
In $\triangle D O S\left(\a... | \frac{\sqrt{3}p^{3}\sqrt{(4+\operatorname{tg}^{2}\alpha)^{3}}}{8\operatorname{tg}^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,123 |
12.209. Find the lateral surface area and volume of a right parallelepiped if its height is $h$, the diagonals form angles $\alpha$ and $\beta$ with the base, and the base is a rhombus. | Solution.
Let the rhombus $ABCD$ be the base of a right prism
Fig. 12.76
$ABCD A_{1} B_{1} C_{1} D_{1}$ (Fig. 12.76), $AA_{1}=BB_{1}=h, O$ - the point of intersection of the diagonals $A... | 2^{2}\sqrt{\operatorname{ctg}^{2}\alpha+\operatorname{ctg}^{2}\beta};\frac{1}{2}^{3}\operatorname{ctg}\alpha\operatorname{ctg}\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,124 |
12.210. The base of a right prism is an isosceles triangle, the base of which is equal to $a$, and the angle at the base is equal to $\alpha$. Find the volume of the prism if its lateral surface area is equal to the sum of the areas of the bases. | ## Solution.
Let $\triangle A B C$ be the base of the right prism $A B C A_{1} B_{1} C_{\mathrm{i}}$ (Fig. 12.77). $A B=B C, A C=a$, $\angle B A C=\alpha, B D$ - height.
$$
\text { In } \triangle A D B\left(\angle A D B=90^{\circ}\right): B D=\frac{a}{2} \operatorname{tg} \alpha, A B=\frac{a}{2 \cos \alpha}
$$
If $H... | \frac{1}{8}^{3}\operatorname{tg}\alpha\operatorname{tg}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,125 |
12.211. The base of the pyramid is a rhombus with an acute angle $\boldsymbol{\alpha}$. Find the volume of the pyramid if its lateral faces form the same dihedral angle $\beta$ with the base and the radius of the inscribed sphere is $r$. | Solution.
Let the rhombus $ABCD$ be the base of the pyramid $SABCD$ (Fig. 12.78), $SO$ - its height, $\angle BCD=\alpha, 0^{\circ}<\alpha<90^{\circ}$.
From the vertex $S$ of the pyramid, drop a perpendicular $SE$ to the side $CD$ of the base. Then $OE \perp CD$ and $\angle SEO-$
 be perpendicular to the plane $ABC$ of its base. Then their common edge $SA$ is the height of the pyramid. In $\triangle ABC$, $AB = AC = b$, $\angle BAC = \alpha$.
If $E$ is the midpoint of $BC$, then $\angle BAE = \frac{\alpha}{2}... | \frac{b\sin\alpha}{4\cos^2\frac{\alpha}{4}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,127 |
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