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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12.213. A perpendicular, equal to $a$, is dropped from the base of the height of a regular triangular pyramid onto a lateral face. Find the volume of the pyramid if the angle of inclination of the lateral edge to the plane of the base is $\alpha$. | Solution.
Let $SO$ be the height of the regular pyramid $SABC$ (Fig. 12.80), $\angle SCO = \alpha$, and $D$ the midpoint of $BC$. Then $AD \perp BC$, $SD \perp BC$, and the edge $BC$ is perpendicular to the plane $ADS$. The planes $ADS$ and $BSC$ are perpendicular.
In the plane $ADS$, draw a perpendicular $OM$ from p... | \frac{\sqrt{3}^3\sqrt{(1+4\tan^2\alpha)^3}}{4\tan^2\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,128 |
12.214. The base of the pyramid is a rhombus with side $a$ and acute angle $\alpha$. Two lateral faces are perpendicular to the base, while the other two are inclined to it at an angle $\varphi$. Find the volume and lateral surface area of the pyramid. | Solution.
Let the rhombus $ABCD$ be the base of the pyramid $SABCD$ (Fig. 12.81), $AB=a$, and the acute angle of the rhombus $\alpha$.
Then the area of the rhombus $S=a^{2} \sin \alpha$, and its height $h=a \sin \alpha$.
The lateral faces $ABS$ and $CBS$ of the pyramid are perpendicular to the plane of the base, so ... | \frac{1}{3}^{3}\sin^{2}\alpha\operatorname{tg}\varphi;\frac{2^{2}\sin\alpha\cos^{2}(45-\frac{\varphi}{2})}{\cos\varphi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,129 |
12.216. The perpendicular dropped from the center of the base of a cone to its generatrix rotates around the axis of the cone. Find the angle between its generatrix and the axis, if the surface of revolution divides the volume of the cone in half. | Solution.
Let $\triangle A B C$ be the axial section of the given cone (Fig. 12.83), $B O$ be its height, and $O M$ be the perpendicular dropped from the center of the base of the cone to the generatrix $A B$. As a result of rotating $O M$ around the line $B O$, a cone is formed, the axial section of which is $\triang... | \arccos(\frac{1}{\sqrt[4]{2}}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,131 |
12.217. Find the angle between the generatrix and the base of a truncated cone, the total surface area of which is twice the surface area of a sphere inscribed in it. | Solution.
Let trapezoid $ABCD, AB=CD$ (Fig. 12.84) be the axial section of the given truncated cone, $O$ - the center of the inscribed sphere, $M, N, K$ - the points of tangency of the sphere with the diameters $BC$ and $AD$ of the bases of the cone and its generatrix $CD$.
If $r$ and $R$ are the radii of the upper a... | \arcsin\frac{2}{\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,132 |
12.218. The base of a right prism is a triangle with side $a$ and adjacent angles $\alpha$ and $\beta$. A plane is drawn through the side of the base at an angle $\varphi$ to it, intersecting the opposite lateral edge. Find the volume of the resulting triangular pyramid. | ## Solution.
Let $\triangle A B C$ be the base of the right prism $A B C A_{l} B_{1} C_{1}$ (Fig.12.85), $\angle A C B=\alpha, \angle A B C=\beta, B C=a, E$ be the point of intersection of the section of the prism passing through $B C$ with its lateral edge $A A_{1}$. Then $A E$ is the height
} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,133 |
12.219. When a circular sector rotates about one of its extreme radii, a body is formed whose spherical surface area is equal to the area of the conical surface. Find the sine of the central angle of the circular sector. | ## Solution.
As a result of the rotation, a body consisting of a cone and a spherical segment with a common base was obtained (Fig. 12.86): the center $O$ of the circular sector is the vertex of the cone, the radius $OB$, around which the rotation was performed, is the axis of the obtained body, the radius $OA$ is the... | \frac{4}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,134 |
12.220. The lateral edge of a regular quadrilateral pyramid forms an angle $\boldsymbol{\alpha}$ with the plane of the base. A plane is drawn through the vertex of the base and the midpoint of the opposite lateral edge, parallel to one of the diagonals of the base. Find the angle between this plane and the plane of the... | Solution.
Let SO be the height of the regular pyramid
Fig. 12.87
$S A B C D$ (Fig. 12.87), $L$ is the midpoint of $S C, \angle S C A=\alpha, M$ and $K$ are the points of intersection of ... | \operatorname{arctg}\frac{\operatorname{tg}\alpha}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,135 |
12.222. The base of the pyramid is a rhombus, one of the angles of which is $\alpha$. The lateral faces are equally inclined to the plane of the base. A plane is drawn through the midpoints of two adjacent sides of the base and the vertex of the pyramid, forming an angle $\beta$ with the plane of the base. The area of ... | ## Solution.
Let the rhombus $ABCD$ be the base of the pyramid $LABCD$ (Fig. 12.89), $\angle BAD=\alpha$, $LO$ - the height of the pyramid. Since the lateral faces are equally inclined to the base plane, $O$ is the center of the circle inscribed in the rhombus, the point of intersection of its diagonals. Let $E$ be th... | 2\sqrt{\frac{2S\cos\beta}{\sin\alpha}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,137 |
12.223. The base of the pyramid is a rhombus with an acute angle $\alpha$. All lateral faces form the same angle $\beta$ with the base plane.
The area of the section made through the larger diagonal of the base and the apex of the pyramid is $S$. Find the volume of the pyramid. | Solution.
Let the rhombus $ABCD$ be the base of the pyramid $LABCD$ (Fig. 12.90), $\angle BCD=\alpha, 0^{\circ}<\alpha<90^{\circ}, LO$-the height of the pyramid. All lateral faces form the same angle with the plane of the base, so $O$ is the center of the circle inscribed in the rhombus, which is the point of intersec... | \frac{2S\sqrt{S\sin\frac{\alpha}{2}\operatorname{ctg}\beta}}{3\cos\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,138 |
12.224. In a regular triangular pyramid, the dihedral angle at the base is equal to $\alpha$, the lateral surface area is $S$. Find the distance from the center of the base to the lateral face. | Solution.
Let $L O$ be the height of the regular pyramid $L A B C$ (Fig. 12.91). Draw a perpendicular $L D$ to $A C$ in the face $A L C$. Then $O D \perp A C, \angle L D O=\alpha$, side $A C$ is perpendicular to the plane $L D O$, and planes $L D O$ and $A L C$ are perpendicular.
In the plane $L D O$, draw a perpendi... | \frac{\sin\alpha}{3}\sqrt{S\sqrt{3}\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,139 |
12.225. The height of a regular triangular pyramid is $H$. The lateral face forms an angle $\alpha$ with the base plane. A plane is drawn through a side of the base and the midpoint of the opposite lateral edge. Find the area of the resulting section. | ## Solution.
Let $MO$ be the height of the regular pyramid $MABC$ (Fig. 12.92), $MO=H, F$ be the midpoint of $BC$. Then $AF \perp BC, MF \perp BC, \angle AFM=\alpha$.
In $\triangle MOF\left(\angle MOF=90^{\circ}\right): OF=$ $H \operatorname{ctg} \alpha . K$ is the midpoint of $AM$. In the plane $AMF$, draw $KL \para... | \frac{1}{2}H^{2}\sqrt{3}\operatorname{ctg}\alpha\sqrt{1+16\operatorname{ctg}^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,140 |
12.226. At the base of a triangular pyramid lies an isosceles triangle, which has an area of $S$ and the angle at the vertex is $\alpha$. Find the volume of the pyramid if the angle between each lateral edge and the height of the pyramid is $\beta$. | ## Solution.
Let $DO$ be the height of the pyramid $DABC$ (Fig. 12.93), $\angle ADO = \angle BDO = \angle CDO = \beta, AB = BC, \angle ABC = \alpha, S_{\triangle ABC} = S$.
$\triangle AOM = \triangle BOM = \triangle COM \Rightarrow OA = OB = OC, O$ is the center of the circumcircle of $\triangle ABC, OA$ is its radiu... | \frac{S\operatorname{ctg}\beta\sqrt{2S\sin\alpha}}{6\cos\frac{\alpha}{2}\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,141 |
12.227. The base of the pyramid is an isosceles trapezoid, where the lateral side is equal to $a$, and the acute angle is equal to $\alpha$. All lateral faces form the same angle $\beta$ with the base of the pyramid. Find the total surface area of the pyramid. | ## Solution.
Let trapezoid $ABCD$ be the base of the pyramid $FABCD$ (Fig. 12.94), $FO$ - the height, $AB=CD=a, \angle ABC=\alpha, 0^{\circ}<\alpha<90^{\circ}$.
Since all lateral faces of the pyramid form the same angle $\beta$ with the base, point $O$ is equidistant from all sides of trapezoid $ABCD$ and is the cent... | \frac{2a^2\sin\alpha\cos^2\frac{\beta}{2}}{\cos\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,142 |
12.228. The dihedral angle at the base of a regular triangular pyramid is equal to $\alpha$, the lateral surface area of the pyramid is $S$. Find the distance from the center of the base to the midpoint of the slant height of a lateral face. | ## Solution.
Let $FO$ be the height of the regular pyramid $FABC$ (Fig. 12.95), $D$ be the midpoint of $AC$. Then $FD \perp AC$, $OD \perp AC$, $\angle FDO = \alpha$.
If $E$ is the midpoint of the apothem $FD$, then $OE$ is the median of the right $\triangle FOD$, drawn to the hypotenuse, $OE = \frac{1}{2} FD$.
, $\angle A_1CA_2 = \alpha, B$ be the midpoint of $CA_2$, and $OB = a$.
$OB$ is the median of the right $\angle COA_2$, drawn to the hypotenuse, hence $CA_2 = 2a, S_{\triangle A_1CA_2} = \frac{1}{2} A_1C^2 \sin \alpha = 2a^2 \sin ... | \frac{4na^2\sin\frac{\alpha}{2}\sin(\frac{\alpha}{2}+\frac{\pi}{n})}{\sin\frac{\pi}{n}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,144 |
12.230. Two cones have concentric bases and the same angle, equal to $\alpha$, between the height and the slant height. The radius of the base of the outer cone is $R$. The lateral surface area of the inner cone is half the total surface area of the outer cone. Find the volume of the inner cone. | Solution.
Let $\triangle A B C$ be the axial section of the outer cone (Fig. 12.97), $\triangle M L K$ be the axial section of the inner cone, $O$ be the common center of their bases, $O A=R, \angle A B O=\angle M L O=\alpha$.
\operatorname{ctg}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,145 |
12.231. A rectangular parallelepiped is inscribed in a cylinder, with its diagonal forming angles $\alpha$ and $\beta$ with the adjacent sides of the base. Find the ratio of the volume of the parallelepiped to the volume of the cylinder. | Solution.
Let in the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 12.98), inscribed in a cylinder, $\angle B_{1} D A=\alpha, \angle B_{1} D C=\beta, A D=a, C D=b$, the radius of the cylinder's base $R$, the common height of the cylinder and the parallelepiped $H$, $B_{1} D=d, A B$ - the projectio... | \frac{4\cos\alpha\cos\beta}{\pi(\cos^{2}\alpha+\cos^{2}\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,146 |
12.233. A cube is inscribed in a regular quadrilateral pyramid; the vertices of its upper base lie on the lateral edges, and the vertices of its lower base are in the plane of the pyramid's base. Find the ratio of the volume of the cube to the volume of the pyramid, if the lateral edge of the pyramid makes an angle $\a... | Solution.
Let $SO$ be the height of the regular pyramid $SABCD$ (Fig. 12.100), $\angle SAO=\alpha, MNKLM_1N_1K_1L_1$ be the cube from the problem's condition, $MNKL$ be its lower, and $M_1N_1K_1L_1$ be its upper base, $O_1$ be the intersection point of $SO$ and $M_1K_1$.
Let the edge of the cube be $a$.
$M_1K_1 \| A... | \frac{3\sqrt{2}\cot\alpha}{(\sqrt{2}\cot\alpha+1)^3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,148 |
12.234. The side of the base of a regular quadrilateral pyramid is equal to $a$; the lateral face forms an angle $\alpha$ with the base plane. Find the radius of the circumscribed sphere. | ## Solution.
Let $S O$ be the height of the regular pyramid $S A B C D$ (Fig. 12.101), $A B=a$, and $M$ be the midpoint of $C D$. Then $O M \perp C D, S M \perp C D, \angle S M O=\alpha$.
The radius $R$ of the sphere will be found as the radius of the circle circumscribed around $\triangle A S C$.
If $\angle A S C=x... | \frac{(3+\cos2\alpha)}{4\sin2\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,149 |
12.235. The angle between a lateral edge of a regular quadrilateral pyramid and the plane of the base is equal to the magnitude of the plane angle at the vertex of the pyramid. Find the angle between a lateral face and the plane of the base. | Solution.
Let $SO$ be the height of the regular pyramid $SABCD$ (Fig. 12.102), $\angle SDO = \angle CSD$, and $E$ be the midpoint of $CD$. Then $OE \perp CD$, $SE \perp CD$, and $\angle SEO$ is the angle between the lateral face and the base plane of the pyramid.
If $OE = 1$, $SO = H$, then $CD = 2$, $OC = \sqrt{2}$.... | \arctan\sqrt{1+\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,150 |
12.236. Find the ratio of the volume of a spherical segment to the volume of the entire sphere, if the arc in the axial section of the segment corresponds to a central angle equal to $\alpha$. | ## Solution.
Consider the axial section of the given sphere, $O$ - its center, $\angle A O C=\alpha$, $K$ - the vertex of the segment, $D$ - the point of intersection of $A C$ and the radius $O K, K D-$ the height of the segment (Fig. 12.103).
If $R$ is the radius of the given sphere, then its volume $V_{1}=\frac{4}{... | \sin^{4}\frac{\alpha}{4}(2+\cos\frac{\alpha}{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,151 |
12.237. The hypotenuse of a right triangle is equal to $c$, and its acute angle is equal to $\alpha$. The triangle rotates around the bisector of the external right angle. Find the volume of the solid of revolution. | ## Solution.
Let in $\triangle A C B \quad \angle A C B=\frac{\pi}{2}, \angle B A C=\alpha, A B=c$ (Fig. 12.104). Then $A C=c \cos \alpha, B C=c \sin \alpha$.
When $\triangle A C B$ is rotated around the bisector $O_{1} O_{2}$, a body is formed, the axial section of which is shown in Fig. 12.104.
The volume $V$ of t... | \frac{\pi^{3}}{6}\sin2\alpha\sin(\frac{\pi}{4}+\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,152 |
12.238. A sphere is inscribed in a truncated cone. The sum of the lengths of the diameters of the upper and lower bases of the cone is five times the length of the radius of the sphere. Find the angle between the generatrix of the cone and the plane of the base. | ## Solution.
Let the isosceles trapezoid $ABCD$ be the axial section of the given truncated cone, $E$ and $F$ be the centers of its lower and upper bases, $O$ be the center of the inscribed sphere, $EF$ be its diameter and the height of the trapezoid $ABCD$ (Fig. 12.105). If $R$ is the radius of the sphere, then $EF =... | \arcsin\frac{4}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,153 |
12.239. The ratio of the surface area of a sphere inscribed in a cone to the area of the base of the cone is $k$. Find the cosine of the angle between the slant height of the cone and the plane of its base and the permissible values of $k$. | ## Solution.
Let $\triangle A B C$ be the axial section of the cone (Fig. 12.106), $D$ be the center of its base, $B D$ be the height, $O$ be the center of the inscribed sphere, $\angle B A D$ be the angle between the generatrix of the cone and the plane of its base, $\angle B A D=\alpha$, $D A=R, O D=r$.
$\angle O A... | \frac{4-k}{4+k};0<k<4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,154 |
12.240. The ratio of the volume of a sphere inscribed in a cone to the volume of a circumscribed sphere is $k$. Find the angle between the slant height of the cone and the plane of its base and the permissible values of $k$. | ## Solution.
Let $\triangle A B C$ be the axial section of the cone, $D$ be the center of its base, $B D$ be the height, $O$ be the center of the inscribed sphere (Fig. 12.107), $\angle B A D$ be the angle we are looking for between the generatrix $B A$ of the cone and the plane of its base.
$\angle B A O=\alpha, r$ ... | \arccos\frac{1\\sqrt{1-2\sqrt[3]{k}}}{2},0<k\leq\frac{1}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,155 |
12.241. A cone is inscribed in a sphere with radius $R$; a cylinder with a square axial section is inscribed in this cone. Find the total surface area of the cylinder, if the angle between the generatrix of the cone and the plane of its base is $\alpha$. | ## Solution.
Consider the axial section of the given set of bodies. $BC$ is the generatrix of the cone, $FE$ is the generatrix of the cylinder, $D$ is the center of the base of the cone and the lower base of the cylinder, $K$ is the center of the upper base of the cylinder, $\angle BCD = \alpha$ (Fig. 12.108).
If the... | \frac{6\piR^2\sin^22\alpha}{(1+2\operatorname{ctg}\alpha)^2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,156 |
12.243. The lateral face of a regular truncated triangular pyramid forms an angle $\alpha$ with the plane of the base. Find the ratio of the total surface area of the pyramid to the surface area of a sphere inscribed in it. | Solution.
Let $M$ be the center of the lower, $M_{1}$ the center of the upper bases of the regular truncated pyramid $A B C A_{1} B_{1} C_{1}$ (Fig. 12.110), $O$ the center of the inscribed sphere, $O$ lies on $M M_{1}$, $K$ is the midpoint of $B C$, $K_{1}$ is the midpoint of $B_{1} C_{1}$. $\angle K_{1} K M=\alpha$,... | \frac{3\sqrt{3}}{2\pi}(4\operatorname{ctg}^{2}\alpha+3) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,158 |
12.244. A sphere is inscribed in a cone. The radius of the circle of contact between the sphere and the lateral surface of the cone is $r$. A line passing through the center of the sphere and an arbitrary point on the circumference of the base of the cone makes an angle $\alpha$ with the height of the cone. Find the vo... | ## Solution.
Let $S$ be the vertex of the cone, $O$ the center of its base (the point of tangency of the inscribed sphere with the base plane), $C$ the center of the sphere, $D$ the center of the circle of tangency of the sphere's surface and the lateral surface of the cone, $AD$ the radius of this sphere, $AD = r$, $... | -\frac{\pir^3\tan2\alpha}{24\cos^6\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,159 |
12.245. The ratio of the volume of a cone to the volume of a sphere inscribed in it is $k$. Find the angle between the slant height and the base plane of the cone and the permissible values of $k$. | ## Solution.
Let $\triangle A B C$ be the axial section of the given cone (Fig. 12.112), $D$ the center of its base, $O$ the center of the inscribed sphere, $\angle B A D$ the angle we need to find between the generatrix and the base plane of the cone, $\angle B A D=\alpha$, and the radius of the base of the cone $O A... | 2\operatorname{arctg}\sqrt{\frac{k\\sqrt{k^{2}-2k}}{2k}};k\geq2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,160 |
12.246. Find the angle between the generatrix of the cone and the plane of the base, if the lateral surface of the cone is equal to the sum of the areas of the base and the axial section. | ## Solution.
Let $R$ be the radius of the base, $l$ the slant height, $H$ the height of the given cone, and $\alpha$ the angle we are looking for between the slant height and the base plane.
Then $H=R \operatorname{tg} \alpha, l=\frac{R}{\cos \alpha}$.
The lateral surface area of the cone $S_{1}=\pi R l=\frac{\pi R^... | 2\operatorname{arcctg}\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,161 |
12.247. The angle between the height and the slant height of a cone is $\alpha$. A regular triangular prism is inscribed in the cone; the lower base of the prism lies in the plane of the base of the cone. The lateral faces of the prism are squares. Find the ratio of the lateral surfaces of the prism and the cone. | ## Solution.
Let $S$ be the vertex, $O$ the center of the base of the given cone, $S A$ the generatrix of the cone passing through the vertex $C$ of the inscribed regular prism (Fig. 12.113), $\angle A S O=\alpha$. The lateral edge $C B$ of the prism is parallel to $S O$, so $\angle A C B=\angle A S O=\alpha$.
If the... | \frac{9\sin2\alpha\cos\alpha}{8\pi\sin^{2}(\alpha+\frac{\pi}{6})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,162 |
12.248. A right prism with a rhombus as its base is circumscribed around a sphere. The larger diagonal of the prism forms an angle $\alpha$ with the base plane. Find the acute angle of the rhombus. | Solution.
Let the rhombus $ABCD$ be the base of the given right prism $ABCD A_1 B_1 C_1 D_1$, circumscribed around a sphere (Fig. 12.114), and the radius of the sphere is $R$. Then the height of the prism $BB_1 = 2R$.
If $B_1 D$ is the major diagonal of the prism, then $\angle B_1 DB = \alpha$ and $BD$ is the major d... | 2\arcsin(\operatorname{tg}\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,163 |
12.249. The lateral edge of a regular truncated quadrilateral pyramid forms an angle $\alpha$ with the plane of the base. A rectangular parallelepiped is inscribed in the pyramid such that its upper base coincides with the upper base of the pyramid, and its lower base lies in the plane of the lower base of the pyramid.... | ## Solution.
Let in the right truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 12.115) $\angle D_{1} D B=\alpha . A_{1} B_{1} C_{1} D_{1} A_{2} B_{2} C_{2} A_{2}$ - the inscribed rectangular parallelepiped $\angle B_{1} D_{2} B_{2}=\beta$.
Assume the lateral edge of the parallelepiped is 1, then from $\Delta... | \frac{\sin(\alpha+\beta)\sqrt{2(1+\sin^{2}\alpha)}}{2\sin^{2}\alpha\cos\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,164 |
12.251. The center of the sphere inscribed in a regular quadrilateral pyramid divides the height of the pyramid in the ratio $m: n$, measured from the vertex of the pyramid. Find the angle between two adjacent lateral faces. | Solution.
The center of the sphere $M$, inscribed in the regular pyramid $SABCD$, is on the height $SO$ (Fig. 12.117) $SM: MO = m: n$.
$OC \perp BD, OC$ is the projection of $SC$ on the base of the pyramid. Then $SC \perp BD$.
In the plane $SDC$, draw $DK \perp SC$.
Since $DK \perp SC, BD \perp SC$, the plane $BKD$... | \pi-\arccos\frac{n^2}{^2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,166 |
12.253. A cylinder is inscribed in a cone; the lower base of the cylinder lies in the plane of the base of the cone. A line passing through the center of the upper base of the cylinder and a point on the circumference of the base of the cone makes an angle $\alpha$ with the plane of the base. Find the ratio of the volu... | ## Solution.
Let $O$ be the center of the bases of the given cone and the cylinder inscribed in it, $O_{1}$ be the center of the upper base of the cylinder, $SA$ be the generatrix of the cone, $CB$ be the generatrix of the cylinder, $\angle ASO=\beta, \angle O_{1}AO=\alpha$, (Fig. 12.119).
If the radius of the base o... | \frac{\cos^{3}\alpha\cos^{3}\beta}{3\sin\alpha\sin\beta\cos^{2}(\alpha+\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,168 |
12.254. The base of the pyramid is a rhombus with an acute angle $\alpha$. All lateral faces form the same angle $\beta$ with the base plane. Find the radius of the sphere inscribed in the pyramid if the volume of the pyramid is $V$. | ## Solution.
Let the rhombus $ABCD$ be the base of the pyramid $EABCD$, $\angle BCD = \alpha$, $0^{\circ} < \alpha < 90^{\circ}$, and $EO$ be its height (Fig. 12.120). Since all lateral faces of the pyramid form the same angle $\beta$ with the base plane, point $O$ is the center of the circle inscribed in the rhombus ... | \frac{1}{2}\tan\frac{\beta}{2}\sqrt[3]{6V\sin\alpha\cot\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,169 |
12.255. Two faces of a triangular pyramid are equal right-angled triangles with a common leg equal to $d$. The angle between these faces is $\alpha$. Two other faces of the pyramid form a dihedral angle $\beta$. Find the radius of the sphere circumscribed about the pyramid. | ## Solution.
Let the faces $SAB$ and $SAC$ of the pyramid $SABC$ (Fig. 12.121) be equal, $\angle SAB = \angle SAC = 90^\circ, SA = d$. Then the segment $SA$ is perpendicular to the plane of the base of the pyramid and is the height of the pyramid, $\triangle BAC$ is isosceles. If $D$ is the midpoint of $BC$, then $AD ... | \frac{}{2\cos^2\frac{\alpha}{2}}\sqrt{\cos^4\frac{\alpha}{2}+\operatorname{ctg}^2\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,170 |
12.256. The base of the pyramid is a rectangle, where the angle between the diagonals is $\alpha$. One of the lateral edges is perpendicular to the base plane, and the largest edge forms an angle $\beta$ with the base plane. The radius of the sphere circumscribed around the pyramid is $R$. Find the volume of the pyrami... | ## Solution.
Let the rectangle $ABCD$ be the base of the pyramid $EABCD$ (Fig. 12.122), the angle between its diagonals is $\alpha$, the lateral edge $EB$ is perpendicular to the plane of the base of the pyramid and is its height, $AB$, $CB$, $DB$ are the projections of the lateral edges $AE$, $EC$, $DE$ on the plane ... | \frac{2}{3}R^3\sin2\beta\cos\beta\sin\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,171 |
12.257. The base of the pyramid is a right-angled triangle inscribed in the base of the cone. The vertex of the pyramid coincides with the vertex of the cone. The lateral faces of the pyramid, containing the legs of the base, form angles $\alpha$ and $\beta$ with the plane of the base. Find the ratio of the volumes of ... | ## Solution.
Let $DO$ be the common height of the given pyramid $DABC$ and the cone (Fig. 12.123). Since $O$ is the center of the circle circumscribed around the right triangle $ABC$, $O$ is the midpoint of the hypotenuse $AC$.
Draw perpendiculars $OE$ and $OF$ from point $O$ to the legs $AB$ and $BC$ in the plane of... | \frac{2\operatorname{ctg}\alpha\operatorname{ctg}\beta}{\pi(\operatorname{ctg}^{2}\alpha+\operatorname{ctg}^{2}\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,172 |
12.258. The side of the square lying at the base of a regular quadrilateral pyramid is equal to $a$. A regular quadrilateral prism is inscribed in the pyramid; the vertices of the upper base lie on the lateral edges, and the vertices of the lower base lie in the plane of the pyramid's base. The diagonal of the prism fo... | Solution.
Let $EO$ be the height of the regular pyramid $EABCD$, $AD=\alpha$, $\angle ECA=\alpha$, $MNKLM_1N_1K_1L_1$ be a regular prism, $\angle M_1KM=\varphi$ (Fig. 12.124). The vertices of the lower base of the prism lie on the diagonals of the square $ABCD$.
^3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,173 |
12.259. The side of the lower base of a regular truncated quadrilateral pyramid is equal to $a$, the side of the upper base is equal to $b$. The lateral face forms an angle $\alpha$ with the base plane. A plane is drawn through the side of the lower base and the midpoint of the segment connecting the centers of the bas... | Solution.
Let point $P$ be the midpoint of segment $O O_{1}$, connecting the centers of the bases of the regular truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}, A D=\mathrm{a}, A_{1} D_{1}=\mathrm{b}$ (Fig. 12.125), $M N$ - the line where the plane passing through side $C D$ of the lower base and point $P$ interse... | \frac{(-b)\operatorname{tg}\alpha}{3-b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,174 |
12.261. A plane is drawn through two generators of a cone, the angle between which is $\alpha$. The area of the section is to the total surface area of the cone as $2: \pi$. Find the angle between the generator and the height of the cone. | ## Solution.
Let $B O$ be the height of the cone, $\triangle A B C$ be the section of the cone, $\angle A B C=\alpha$, (Fig. 12.127), $A B=d$.
Then the area of the section is
$$
S_{1}=\frac{1}{2} A B^{2} \sin \angle A B C=\frac{1}{2} d^{2} \sin \alpha
$$
If the desired angle $A B C$ is $\beta$, then from
$\triangl... | \arcsin(\sqrt{2}\sin\frac{\alpha}{4}\sin(45-\frac{\alpha}{4})) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,176 |
12.263. The height of a regular truncated triangular pyramid is $H$ and is the mean proportional between the sides of the bases. The lateral edge makes an angle $\alpha$ with the base. Find the volume of the pyramid. | Solution.
Let $a$ and $\beta$ be the sides of the bases of the given truncated pyramid, with $a > \beta$.
Then $H = \left(\frac{a \sqrt{3}}{3} - \frac{b \sqrt{3}}{3}\right) \tan \alpha = \frac{(a - b) \sqrt{3}}{3} \tan \alpha, a - b = H \sqrt{3} \cot \alpha$, the areas of the bases $S_{1} = \frac{a^{2} \sqrt{3}}{4}, ... | \frac{H^{3}\sqrt{3}}{4\sin^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,178 |
12.266. The base of the pyramid is an equilateral triangle. Two lateral faces are perpendicular to the plane of the base. The sum of two unequal plane angles at the vertex is $\frac{\pi}{2}$. Find these angles. | Solution.
Let the faces $SBA$ and $SBC$ of the pyramid $SABC$ be perpendicular to the plane $ABC$ of the base (Fig. 12.131), and $\triangle ABC$ is equilateral. Then $\triangle SBA = \triangle SBC$ by two legs, and $\angle ASB = \angle CSB, AS = SC$.
If $\angle ASC = \alpha$, then $\angle ASB = \frac{\pi}{2} - \alpha... | \arccos(\sqrt{3}-1);\frac{\pi}{2}-\arccos(\sqrt{3}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,181 |
12.267. The ratio of the total surface area of a cone to the area of its axial section is $k$. Find the angle between the height and the slant height of the cone and the permissible values of $k$. | ## Solution.
Let $\alpha$ be the angle between the height and the slant height of the cone, $l$ be the slant height, and $r$ be the radius of the base of the cone.
Then $r = l \sin \alpha$ and the area of the total surface of the cone is
$$
S_{1} = \pi r^{2} + \pi r l = \pi l^{2} \sin ^{2} \alpha + \pi l^{2} \sin \a... | \frac{\pi}{2}-2\operatorname{arctg}\frac{\pi}{k},k>\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,182 |
12.268. One of the faces of a triangular prism inscribed in a cylinder passes through the axis of the cylinder. The diagonal of this face forms angles $\alpha$ and $\beta$ with the adjacent sides of the base of the prism. Find the volume of the prism if the height of the cylinder is $H$. | Solution.
Let the face $A A_{1} B_{1} B$ of the prism $A B C A_{1} B_{1} C_{1}$ pass through the axis $O O_{1}$ of the cylinder in which it is inscribed (Fig. 12.132), $A A_{1}=H, \angle A B A_{1}=\alpha$, $\angle C B A_{1}=\beta$, the center $O$ of the base of the cylinder is the midpoint of the side $A B$ of the tri... | \frac{H^{3}\cos\beta}{2\sin^{2}\alpha}\sqrt{\sin(\beta+\alpha)\sin(\beta-\alpha)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,183 |
12.269. Two vertices of an equilateral triangle with side $a$ lie on the circumference of the upper base of a cylinder, while the third vertex lies on the circumference of the lower base. The plane of the triangle forms an angle $\alpha$ with the generatrix of the cylinder. Find the lateral surface area of the cylinder... | ## Solution.
Let the side $BC$ of the equilateral triangle $ABC$ (Fig. 12.133) lie on the upper base of the cylinder, and the vertex $A$ on the circumference of the lower base, $BC=a, AM$ is the generator of the cylinder, $K$ is the midpoint of $BC$. Then $BM=CM, MK \perp BC, AK \perp BC$. Therefore, the line $BC$ is ... | \frac{1}{4}\pi^{2}\operatorname{ctg}\alpha(3\sin^{2}\alpha+1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,184 |
12.270. Find the dihedral angle at the vertex of a regular quadrilateral pyramid if it is equal to the angle between a lateral edge and the plane of the pyramid's base. | Solution.
Let $S O$ be the height of the regular pyramid $S A B C D$ (Fig. 12.134), $\angle S D O = \angle C S D, \angle S D O = \angle C S D = \alpha$.
Then $\angle S D C = 90^{\circ} - \frac{\alpha}{2}$:
$S O$ is the oblique to the plane of the base of the pyramid, $O D$ is its projection onto this plane, $C D$ is... | \arccos\frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,185 |
12.273. Point $A$ lies on the circumference of the upper base of the cylinder, point $B$ - on the circumference of the lower base. The line $A B$ forms an angle $\alpha$ with the base plane, and an angle $\beta$ with the plane of the axial section passing through point $B$. Find the volume of the cylinder if the length... | ## Solution.
Let $O$ and $O_{1}$ be the centers of the upper and lower bases of the cylinder, and $CD$ be its generatrix (Fig. 12.137), then $BD$ is the projection of $BC$ on the plane of the lower base and $\angle CBD = \alpha$.
The plane of the axial section $BMNA$ is perpendicular to the planes of the bases. Draw ... | \frac{\pi^3\sin2\alpha\cos^3\alpha}{8\cos(\alpha+\beta)\cos(\alpha-\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,188 |
12.274. A cube is inscribed in a cone (one of the faces of the cube lies in the plane of the base of the cone). The ratio of the height of the cone to the edge of the cube is $k$. Find the angle between the slant height and the height of the cone. | ## Solution.
Let the edge of the cube be $a$ (Fig. 12.138).
Then $O_{1} C_{1}=\frac{a \sqrt{2}}{2}, S O=k a, S O_{1}=S O-O O_{1}=a(k-1)$.
In $\triangle S O_{1} C_{1}\left(\angle S O_{1} C_{1}=90^{\circ}\right):$
\[
\begin{aligned}
& \operatorname{ctg} \angle O_{1} S C_{1}=\frac{S O_{1}}{O_{1} C_{1}}=\frac{a(k-1)}{\... | \operatorname{arcctg}(\sqrt{2}(k-1)) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,189 |
12.275. The base of the pyramid is a rectangle. Two lateral faces are perpendicular to the base plane, and the other two form angles $\alpha$ and $\beta$ with it. Find the lateral surface area of the pyramid if the height of the pyramid is $H$. | Solution.
Let the rectangle $ABCD$ be the base of the pyramid $KACBD$ (Fig. 12.139).
The lateral faces $ABK$ and $CBK$ are perpendicular to the base plane of the pyramid, their common edge $KB$ is the height of the pyramid, $KB=H$.
\sin(45+\frac{\beta}{2})}{\sin\alpha\sin\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,190 |
12.276. One of the sides of the base of a right triangular prism is equal to $a$, and the angles adjacent to it are $\alpha$ and $\beta$. Find the lateral surface area of the prism if its volume is $V$. | ## Solution.
If $\mathrm{b}$ and $c$ are the sides of the base of the prism, opposite to its angles $\beta$ and $\alpha$, and $\gamma$ is the angle of the base opposite to side $a$, then
$$
\begin{aligned}
& b=\frac{a \sin \beta}{\sin \gamma}=\frac{a \sin \beta}{\sin \left(180^{\circ}-(\alpha+\beta)\right)}=\frac{a \... | \frac{2V\sin\frac{\alpha+\beta}{2}}{\sin\frac{\alpha}{2}\sin\frac{\beta}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,191 |
12.277. One lateral edge of a triangular pyramid is perpendicular to the base plane and equals $l$, the other two form an angle $\alpha$ between them, and with the base plane - the same angle $\beta$. Find the volume of the pyramid. | ## Solution.
Let the lateral edge $DA = l$ of the pyramid $DABC$ (Fig. 12.140), perpendicular to the plane of the base, be its height, $AB$ and $AC$ - the projections of the lateral edges $DB$ and $DC$ of the pyramid onto the plane of the base. Then $\angle ABD = \angle ACD = \beta, \angle BDC = \alpha$.
$\triangle DA... | \frac{^3\sin\frac{\alpha}{2}\sqrt{\cos(\frac{\alpha}{2}+\beta)\cos(\frac{\alpha}{2}-\beta)}}{3\sin^2\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,192 |
12.278. The base of the pyramid is an isosceles trapezoid, where the acute angle is $\alpha$, and the area is $S$. All lateral faces form the same angle $\beta$ with the base plane. Find the volume of the pyramid. | ## Solution.
Let trapezoid $ABCD$ be the base of the pyramid $EABCD$ (Fig. 12.141), $AB=CD, \angle ABC=\alpha, 0^{\circ}<\alpha<90^{\circ}, EO$ - the height of the pyramid.
Fig. 12.142
Si... | \frac{S\operatorname{tg}\beta\sqrt{S\sin\alpha}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,193 |
12.279. The cosine of the angle between two adjacent lateral faces of a regular quadrilateral pyramid is $k$. Find the cosine of the angle between a lateral face and the plane of the base and the permissible values of $k$. | ## Solution.
Let $S O$ be the height of the regular pyramid $S A B C D$ (Fig. 12.142), $\angle B E D$ be the angle between the lateral faces $B S C$ and $D S C$, $F$ be the midpoint of $C D$, $\angle S F O$ be the angle between the lateral face and the base plane, $\angle B E D=\alpha, \angle S C D=\beta, \angle S F O... | \sqrt{-k};-1<k<0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,194 |
12.280. The base of the pyramid is a rectangle $A B C D$ ($A B \| C D$). The lateral edge $O A$ is perpendicular to the base. The edges $O B$ and $O C$ form angles with the base, respectively equal to $\alpha$ and $\beta$. Find the angle between the edge $O D$ and the base. | Solution.
Edge $OA$ is the height of the pyramid $OABCD$ (Fig. 12.143), $\angle OBA=\alpha$, $\angle OCA=\beta$, $\angle ODA$ is the angle between edge $OD$ and the base to be found.
Let $OA=1, \angle ODA=\gamma$, then from $\triangle OAB\left(\angle OAB=90^{\circ}\right): AB=OA$ $\operatorname{ctg} \angle OBA=\opera... | \operatorname{arctg}\frac{\sin\alpha\sin\beta}{\sqrt{\sin(\alpha-\beta)\sin(\alpha+\beta)}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,195 |
12.281. A plane is drawn through the diagonal of the base and the height of a regular quadrilateral pyramid. The ratio of the area of the section to the lateral surface area of the pyramid is $k$. Find the cosine of the angle between the apothems of opposite lateral faces and the permissible values of $k$. | ## Solution.
Let $SO$ be the height, $SE$ and $SF$ be the apothems of the opposite lateral faces of the regular pyramid $SABCD$ (Fig. 12.144), $AD=a, \angle SFO=\alpha$.
Then $AC=a \sqrt{2}$, and from $\triangle \operatorname{SOF} \left(\angle SOF=90^{\circ}\right)$:
$$
SO = OF \operatorname{tg} \angle SFO = \frac{a... | 16k^{2}-1;0<k<0.25\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,196 |
12.283. The side of the base of a regular quadrilateral pyramid is equal to $a$. The angle between adjacent lateral faces is $\alpha$. Find the lateral surface area of the pyramid. | Solution.
Let $SO$ be the height of the regular pyramid $SABCD$ (Fig. 12.146), $AD = \alpha$.
In the plane of the face $SDC$, draw the perpendicular $DE$ to $SC$.
$SO \perp BD, AC \perp BD$, therefore the line $BD$ is perpendicular to the plane $ASC$. Hence, $BD \perp SC$.
$SC \perp BD, SC \perp DE$. We have that t... | \frac{^2}{\sqrt{-\cos\alpha}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,198 |
12.284. In a regular triangular pyramid, a plane is drawn through a lateral edge and the height. The ratio of the area of the section to the total surface area of the pyramid is $k$. Find the dihedral angle at the base and the permissible values of $k$.
In a regular triangular pyramid, a plane is drawn through a later... | ## Solution.
Let $\triangle D E C$ be the section of the regular pyramid $D A B C$ (Fig. 12.147) by a plane passing through the lateral edge $D C$ and the height of the pyramid $D O$. Then $E$ is the midpoint of $A B, \angle D E O$ is the linear angle of the dihedral angle at the base.
If $\angle D E O=\alpha, A B=1$... | 2\operatorname{arctg}(2k\sqrt{3});0<k<\frac{\sqrt{3}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,199 |
12.285. The angle between the height and the slant height of a cone is $\alpha$. A plane passing through the vertex of the cone forms an angle $\beta$ with the height ( $\beta<\alpha$ ). In what ratio does this plane divide the circumference of the base? | ## Solution.
Let SO be the height of the given cone, $\triangle A S B$ be its section by a plane (Fig. 12.148), $\angle O S B=\alpha$.
If $C$ is the midpoint of $A B$, then $S C \perp A B$, $\dot{O} C \perp A B, \angle B O C=\frac{1}{2} \angle A O B$.
Draw a perpendicular $O D$ from point $O$ to the plane $A S B$. T... | \frac{\arccos\frac{\operatorname{tg}\beta}{\operatorname{tg}\alpha}}{\pi-\arccos\frac{\operatorname{tg}\beta}{\operatorname{tg}\alpha}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,200 |
12.287. The edges of a rectangular parallelepiped are in the ratio $3: 4: 12$. A diagonal section is made through the larger edge. Find the sine of the angle between the plane of this section and a diagonal of the parallelepiped not lying in it. | ## Solution.
Let $O$ and $O_{1}$ be the points of intersection of the diagonals of the bases of the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 12.150), $A_{1} B_{1}=3 ; B_{1} C_{1}=$ $=4 ; B B_{1}=12$.
Then $A_{1} C_{1}=\sqrt{A_{1} B_{1}^{2}+B_{1} C_{1}^{2}}=5$;
$$
B_{1} D=\sqrt{A_{1} B_{1}^{... | \frac{24}{65} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,202 |
12.288. The lateral face of a regular triangular pyramid forms an angle with the base plane, the tangent of which is equal to $k$. Find the tangent of the angle between the lateral edge and the apothem of the opposite face. | ## Solution.
Let $S O$ be the height of the regular pyramid $S A B C$ (Fig. 12.151), and $F$ be the midpoint of $B C$.
Fig. 12.151
Then $\angle S F O$ is the angle between a lateral fac... | \frac{3k}{k^{2}-2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,203 |
12.289. All lateral faces of the pyramid form the same angle with the base plane. Find this angle if the ratio of the total surface area of the pyramid to the area of the base is $k$. For what values of $k$ does the problem have a solution | Solution.
Let $S$ be the total surface area, $S_{1}$ be the lateral surface area of the pyramid, and $S_{2}$ be the area of its base.
Since all lateral faces form the same angle $\alpha$ with the plane of the base, then $S_{1}=\frac{S_{2}}{\cos \alpha}$.
Further, $S=S_{1}+S_{2}=\frac{S_{2}}{\cos \alpha}+S_{2}=\frac{... | \arccos\frac{1}{k-1},k>2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,204 |
12.291. The cosine of the angle between the lateral edges of a regular quadrilateral pyramid, not lying in the same face, is $k$. Find the cosine of the dihedral angle at the vertex of the pyramid. | ## Solution.
Let $S O$ be the height of the regular pyramid $S A B C D$ (Fig. 12.153), $\cos \angle A S C=k$.
If $\angle A S C=\alpha$, then $\angle D S O=\angle C S O=\frac{\alpha}{2}$.
Consider the plane $S O C$. $S D$ is inclined to this plane, $S O$ is the projection of $S D$ onto this plane, and $S C$ is a line... | \frac{1+k}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,206 |
12.292. A plane is drawn through a side of a rhombus, forming angles $\alpha$ and $2 \alpha$ with the diagonals. Find the acute angle of the rhombus. | Solution.
Let $ABCD$ be a rhombus, through the side of which a plane $\pi$ is drawn (Fig. 12.154), $O$ - the point of intersection of the diagonals of the rhombus.
Draw perpendiculars $BB_1$ and $CC_1$ to the plane $\pi$, $\angle CAC_1 = \alpha$, $\angle BDB = 2\alpha$.
$BC \| AD$, and $AD$ belongs to the plane $\pi... | 2\operatorname{arcctg}(2\cos\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,207 |
12.293. The base of the oblique prism $A B C A_{1} B_{1} C_{1}\left(A A_{1}\left\|B B_{1}\right\| C C_{1}\right)$ is an isosceles triangle, where $A B=A C=a$ and $\angle C A B=\alpha$. The vertex $B_{1}$ of the upper base is equidistant from all sides of the lower base, and the edge $B_{1} B$ forms an angle $\beta$ wit... | ## Solution.
Draw a perpendicular $B_{1} O$ from point $B_{1}$ to the plane $A B C$ and perpendiculars $B_{1} F, B_{1} K, B_{1} E$ to the sides $A B, B C, A C$ of triangle $A B C$ (Fig. 12.155), $\angle B{ }_{1} B O=\beta$.
$O F, O K, O E$ are the projections of $B_{1} F, B_{1} K, B_{1} E$ on the plane $A B C, O F \pe... | \frac{^{3}\sin\alpha\sin\frac{\alpha}{2}\operatorname{tg}\beta}{2\cos\frac{\pi-\alpha}{4}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,208 |
12.295. The base of a right prism described around a sphere of radius $r$ is a right triangle with an acute angle $\alpha$. Find the volume of the prism. | ## Solution.
Let $\triangle A B C$ be the lower base of the given right prism, $\angle A C B=90^{\circ}, \angle B A C=\alpha$ (Fig. 12.157), the height of the prism $H=2 r$, the projection of the sphere onto the plane $A B C$ is a circle inscribed in $\triangle A B C$, with radius $r$, $O$ is the center of this circle... | 2r^{3}\operatorname{ctg}\frac{\alpha}{2}\operatorname{ctg}(\frac{\pi}{4}-\frac{\alpha}{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,209 |
12.296. The diagonals $A B_{1}$ and $C B_{1}$ of two adjacent lateral faces of a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ form angles with the diagonal $A C$ of the base $A B C D$ that are equal to $\alpha$ and $\beta$, respectively. Find the angle between the plane of triangle $A B_{1} C$ and the p... | ## Solution.
In the plane $A B_{1} C$, draw the perpendicular $B_{1} E \perp A C$ (Fig. 12.158) $\angle B_{1} A E=\alpha, \angle B_{1} C E=\beta$.
$B E$ is the projection of $B_{1} E$ on the base plane $A B C D$. Then $B E \perp A C$ and $\angle B_{1} E B$ is the angle between the plane $\triangle \mathrm{A} B_{1} C$... | \arccos\sqrt{\operatorname{ctg}\alpha\operatorname{ctg}\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,210 |
12.297. In a regular triangular prism, the side of the base is equal to $a$, and the angle between the non-intersecting diagonals of two lateral faces is $\alpha$. Find the height of the prism. | ## Solution.
Let's complete the regular prism $A B C A_{1} B_{1} C_{1}$ to a parallelepiped $A D B C A_{1} D_{1} B_{1} C_{1}$ (Fig. 12.159). Since $B D_{1} \| C A_{1}$, the angle between the skew lines $A_{1} C$ and $B C_{1}$ is $\angle C_{1} B D_{1}=\alpha$.
Since $\Delta A_{1} B_{1} C_{1}$ is equilateral, $A_{1} D_... | \frac{\sqrt{\sin(\frac{\pi}{3}-\frac{\alpha}{2})\sin(\frac{\pi}{3}+\frac{\alpha}{2})}}{\sin\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,211 |
12.298. In a right-angled triangle, a plane is drawn through its hypotenuse, forming an angle $\alpha$ with the plane of the triangle, and an angle $\beta$ with one of the legs. Find the angle between this plane and the second leg. | Solution.
Through the hypotenuse $AB$ of the triangle $ABC$, a plane $\pi$ is drawn (Fig. 12.160).
Fig. 12.160
$CD \perp \pi$. Then $\angle CAD$ and $\angle CBD$ are the angles between t... | \arcsin\sqrt{\sin(\alpha+\beta)\sin(\alpha-\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,212 |
12.299. In a right-angled triangle with an acute angle $\alpha$, a plane is drawn through the smallest median, forming an angle $\beta$ with the plane of the triangle. Find the angles between this plane and the legs of the triangle. | ## Solution.
Let the median $D E$, drawn to the larger side of the triangle
Fig. 12.162
$A D F$ - its hypotenuse $A F$ - the smaller median of the given triangle (Fig. 12.161), through wh... | \arcsin(\sin\beta\cos\alpha)\arcsin(\sin\beta\sin\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,213 |
12.301. The base of a right prism is an isosceles triangle with a lateral side $a$ and an angle $\alpha$ between the lateral sides. The diagonal of the lateral face opposite the given angle forms an angle $\varphi$ with the adjacent lateral face. Find the volume of the prism. | ## Solution.
In $\triangle ABC$, which is the base of the right prism $ABC A_1 B_1 C_1$ (Fig. 12.163), $AC = BC = a, \angle ACB = \alpha$. Then $S_{\triangle ABC} = \frac{1}{2} a^2 \sin \alpha$,
$AB = 2a \sin \frac{\alpha}{2}$. In the plane $ABC$, draw the perpendicular $BD$ to $AC$. Since the planes of the base and ... | \frac{^3\sin\alpha\sin\frac{\alpha}{2}}{\sin\varphi}\sqrt{\cos(\varphi+\frac{\alpha}{2})\cos(\varphi-\frac{\alpha}{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,214 |
12.303. The base of an oblique prism is a rectangle with sides $a$ and $b$. Two adjacent lateral faces form angles $\alpha$ and $\beta$ with the base plane. Find the volume of the prism if the lateral edge is equal to $c$. | Solution.
Let the rectangle $ABCD$ be the base of the oblique prism $ABCD A_1B_1C_1D_1$, where $AD = a$, $CD = b$, $DD_1 = c$, and $D_1O$ is the height of the prism (Fig. 12.165).
Draw perpendiculars $D_1E$ to $AD$ and
.
Fig. 12.167
Since the edges $D A$ and $D C$ are perpendicular to the faces $A A... | ^{3}\sin\alpha\sin\beta\sqrt{\cos(\alpha+\beta)\cos(\alpha-\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,216 |
12.305. In a regular triangular prism, a plane passing through the center of the base and the centers of symmetry of two lateral faces forms an acute angle $\alpha$ with the base plane. Find the area of the section formed by this plane if the side of the base is equal to $a$. | ## Solution.
Let $O$ be the center of the base $ABC$ of the regular prism $ABC A_1 B_1 C_1$. Points $P$ and $Q$ are the centers of symmetry of the diagonals of the faces $AA_1 C_1 C$ and $AA_1 B_1 B$ (Fig. 12.167). The intersecting plane passing through points $O$, $P$, and $Q$ intersects the parallel faces $ABC$ and ... | \frac{^2\sqrt{3}}{12\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,217 |
12.306. In 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)$, the sides of the base $A B$ and $B C$ are equal to $a$ and $b$ respectively, and the angle between them is $\alpha$. A plane is drawn through the bisector of this angle and vertex $A_{1}$, forming an acute angle $\beta... | Solution.
Let $B D$ be the bisector of $\triangle A B C, \triangle A_{1} B D$ - the section of the given
prism (Fig. 12.168) $\frac{A C}{A D}=$
$$
=\frac{A D+D C}{A D}=1+\frac{D C}{A D}=... | \frac{^{2}b\sin\alpha}{2(+b)\cos\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,218 |
12.308. The base of the prism is an equilateral triangle with side $a$. The lateral edge is equal to $b$ and forms angles, each equal to $\alpha$, with the intersecting sides of the base. Find the volume of the prism and the permissible values of $\alpha$. | ## Solution.
Let the equilateral triangle $ABC$ be the base of the prism $ABC A_1 B_1 C_1$ (Fig. 12.170), $AB = a, BB_1 = b, B_1 E$ be the height of the prism.
Since $\angle B_1 B C = \angle B_1 B A = \alpha$, the point $E$ lies on the bisector $BD$ of angle $ABC$.
$\angle B_1 B E$ is the angle of inclination of $B ... | \frac{^2b}{2}\sqrt{\sin(\alpha+30)\sin(\alpha-30)},30<\alpha<150 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,220 |
12.309. The base of the prism is a rectangle. The lateral edge forms equal angles with the sides of the base and is inclined to the plane of the base at an angle $\alpha$. Find the angle between the lateral edge and the side of the base. | Solution.
Let the rectangle $ABCD$ be the base of the prism $ABCD A_1 B_1 C_1 D_1$, and $A_1 E$ be the height of the prism (12.171). Then $\angle A_1 A E$ is the angle of inclination of the edge
. Then $O_{1} O_{2}$ is the required distance from the center of ... | \frac{\piH^{2}\cos^{4}\alpha}{4\cos^{4}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,222 |
12.312. The side of the base of a regular triangular pyramid is equal to $a$, and the dihedral angle at the base is $\alpha$. The pyramid is intersected by a plane parallel to the base. The area of the section is equal to the lateral surface area of the resulting frustum of the pyramid. Find the distance from the cutti... | Solution.
Let $DO$ be the height of the regular pyramid $DABC$ (Fig. 12.174), $A_1B_1C_1$ be the section of the pyramid, $E$ be the midpoint of $AB$, $E_1$ be the midpoint of $A_1B_1$, and $O_1$ be the center of $A_1B_1C_1$.
Then $EE_1$ is the apothem of the regular truncated pyramid $ABCA_1B_1C_1$, $OO_1$ is the dis... | \frac{\sqrt{3}\tan\alpha}{3\cos\frac{\alpha}{2}}\sin(\frac{\pi}{8}-\frac{\alpha}{4})\sin(\frac{\pi}{8}+\frac{\alpha}{4}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,223 |
12.313. The height of the cone is $H$, and the angle between the generatrix and the base plane is $\alpha$. The total surface area of this cone is divided in half by a plane perpendicular to its height. Find the distance from this plane to the base of the cone. | Solution.
Let $R$ be the radius of the base of the given cone, $L$ its slant height, $r, l, h$ - respectively the radius of the base, slant height, and height of the truncated cone.
Then $R=H \operatorname{ctg} \alpha, L=\frac{H}{\sin \alpha}, r=h \operatorname{ctg} \alpha, l=\frac{h}{\sin \alpha}$, the total surface... | 2H\sin^{2}\frac{\alpha}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,224 |
12.317. The generatrix of a truncated cone circumscribed around a sphere is equal to $a$, the angle between the generatrix and the plane of the base is $\alpha$. Find the volume of the cone whose base is the circle of contact between the spherical surface and the lateral surface of the truncated cone, and whose vertex ... | Solution.
Consider the axial section of the given set of bodies (Fig. 12.178): an isosceles trapezoid \(A A_{1} B_{1} B \left(A A_{1}=a, \angle A_{1} A B=\alpha\right)\), a circle inscribed in it with center \(E\), touching \(A A_{1}, B B_{1}, A B\) and \(A_{1} B_{1}\) at points \(C, D, O\) and \(O_{1}\) respectively,... | \frac{\pi^{3}}{12}\sin^{5}\alpha\cos^{2}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,228 |
12.318. Two cones with a common base are inscribed in a sphere of radius $R$; the vertices of the cones coincide with the opposite ends of the sphere's diameter. The spherical segment containing the smaller cone has an arc in the axial section equal to $\alpha^{\circ}$. Find the distance between the centers of the sphe... | ## Solution.
The given set of bodies has an axis of symmetry; let's consider its axial section (Fig. 12.179): $O$ - the center of the given sphere, $\triangle A B C$ - the axial section of the smaller cone, $\triangle A D C$ - the axial section of the larger cone, $E$ - the center of their common base, $O_{1}$ and $O_... | \frac{R\sqrt{2}\sin\frac{\alpha}{2}}{2\cos\frac{\alpha}{8}\cos(45-\frac{\alpha}{8})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,229 |
12.319. Find the ratio of the volume of a regular $n$-sided pyramid to the volume of the circumscribed sphere, if the angle between the lateral edge and the plane of the base of the pyramid is $\alpha$. | Solution.
Let $\mathrm{CO}_{2}$ be the height of a regular $n$-sided pyramid, and $A_{1} A_{2}$ be the side of its base. Then $\angle C A_{1} O_{2}=\alpha, \angle A O_{1} B=\frac{2 \pi}{\mathrm{n}}$.
The center $\mathrm{O}_{1}$ of the circumscribed sphere is the intersection of the segment $\mathrm{CO}_{1}$ (Fig. 12.... | \frac{n\sin^{2}\alpha\sin^{2}2\alpha\sin\frac{2\pi}{n}}{4\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,230 |
12.320. The lateral faces of a regular triangular prism are squares. Find the angle between the diagonal of a lateral face and a non-intersecting side of the base of the prism. | ## Solution.
The angle between the lateral diagonal $B_{1} C$ of the lateral face of a regular prism $A B C A_{1} B_{1} C_{1}$ (Fig. 12.181) and the side $A B$ of the base is the angle between $B_{1} C$ and $A_{1} B_{1}$.
If the length of the edge of the prism is 1, then the diagonal of the square $B_{1} C = \sqrt{2}... | \arccos\frac{\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,231 |
12.323. The base of a right prism is an equilateral triangle. A plane is drawn through one of its sides, cutting off a pyramid from the prism, the volume of which is equal to $V$. Find the area of the section, if the angle between the cutting plane and the base plane is $\alpha$. | Solution.
Let $\triangle B E C$ be the section of the regular prism $A B C A_{1} B_{1} C_{1}$ (Fig. 12.184),
Fig. 12.185
$D$ is the midpoint of $B C$. Then $\angle E D A=\alpha, E A$ is t... | \sqrt[3]{\frac{3\sqrt{3}V^{2}}{\sin^{2}\alpha\cos\alpha}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,234 |
12.324. In a regular quadrilateral pyramid, a section parallel to the base is made. A line passing through the vertex of the base and the opposite (i.e., not belonging to the same face) vertex of the section forms an angle $\alpha$ with the plane of the base. Find the area of the section if the lateral edge of the pyra... | Solution.
Let $EO$ be the height of the regular pyramid $EABCD$, and let the square $A'B'C'D'$ be a section of the pyramid, with $\angle BDB'=\alpha$ (Fig. 12.185).
Since $BE=DE=BD$, then $\angle EBD=\frac{\pi}{3}, \angle BB'D=\frac{2\pi}{3}-\alpha$.
In $\triangle BB'D: \frac{DB'}{\sin \angle EBD}=\frac{BD}{\sin \an... | \frac{^2\sin^2(\frac{\pi}{3}-\alpha)}{2\sin^2(\frac{\pi}{3}+\alpha)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,235 |
12.325. The base of the pyramid is an acute isosceles triangle with a side length of $b$ and an angle at the base of $\alpha$. All lateral edges of the pyramid form the same angle $\beta$ with the base plane. Find the area of the section of the pyramid by a plane passing through the vertex of the given angle $\alpha$ a... | Solution.
Let $EO$ be the height of the pyramid $EABC$, $AC=AB=b$, $\angle ACB=\alpha$ (Fig. 12.186). Since all lateral edges form the same angle with the base plane, point $O$ is the center of the circle circumscribed around $\triangle ABC$, and since $\triangle ABC$ is isosceles and acute-angled, point $O$ lies on i... | -\frac{b^{2}\cos\alpha\operatorname{tg}\beta}{2\cos3\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,236 |
12.326. A plane broken line consists of $n$ equal segments, connected in a zigzag pattern at an angle $\alpha$ to each other. The length of each segment of the broken line is $a$. This line rotates around a straight line passing through one of its ends and parallel to the bisector of the angle $\alpha$. Find the surfac... | ## Solution.
The first segment, having a common point with the axis of rotation, describes the lateral surface of a complete cone $S_{1}$, and all
Fig. 12.187 others - the lateral surface... | \pi^{2}n^{2}\sin\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,237 |
12.327. Two cones have a common height; their vertices lie at opposite ends of this height. The generatrix of one cone is equal to $l$ and forms an angle $\alpha$ with the height. The generatrix of the other cone forms an angle $\beta$ with the height. Find the volume of the common part of both cones. | ## Solution.
Consider the axial section of the given set of cones (Fig. 12.188): $\triangle A B C$ - the axial section of the cone with slant height $l, \triangle M L K-$ the axial section of the second cone, $B L-$ their common height, $\angle A B L=\alpha$, $\angle M L B=\beta$.
The common part of these cones is tw... | \frac{\pi^{3}\sin^{2}2\alpha\cos\alpha\sin^{2}\beta}{12\sin^{2}(\alpha+\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,238 |
12.329. The lateral side of an isosceles triangle is equal to $a$, and the angle at the base is $\alpha$. This triangle rotates around a line passing through the vertex opposite the base, parallel to the bisector of angle $\alpha$. Find the surface area of the solid of revolution. | Solution.
In $\triangle BAC, BA = AC = a, \angle ACB = \alpha, CD$ is the bisector of $\angle ACB$. Consider the axial section of the resulting solid of revolution (Fig. 12.190). Its surface area $S$ is the sum of the lateral surface area $S_{1}$ of the cone with axial section
\cos(\frac{\pi}{6}-\frac{\alpha}{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,240 |
12.330. The base of the pyramid is a right-angled triangle, with the radius of the inscribed circle being $r$, and one of the acute angles being $\alpha$. All lateral edges of the pyramid form the same angle $\beta$ with the base plane. Find the volume of the pyramid. | ## Solution.
Let $EO$ be the height of the pyramid $EABC$, $\angle ACB = \frac{\pi}{2}$, $\angle BAC = \alpha$, $O_1$ be the center of the circle inscribed in $\triangle ABC$, $L$ and $K$ be the points of tangency of this circle with sides $AC$ and $BC$ respectively (Fig. 12.191).
Then $O_1L \perp AC$, $O_1K \perp BC... | \frac{r^3\sqrt{2}\operatorname{tg}\beta\operatorname{ctg}(\frac{\pi}{4}-\frac{\alpha}{2})\operatorname{ctg}\frac{\alpha}{2}}{12\sin(\frac{\pi}{4}-\frac{\alpha}{2})\sin\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,241 |
12.331. A sphere is inscribed in a cone, and a tangent plane is drawn to the sphere, parallel to the base plane of the cone. In what ratio does this plane divide the lateral surface of the cone, if the angle between the generatrix and the base plane is $\alpha$? | ## Solution.
Consider the axial section of the given set of bodies (Fig. 12.192): $\triangle A E C$ - the axial section of the given cone with lateral surface $S_{1}$, $O$ - the center of the given sphere - the center of the circle inscribed in the trapezoid $A B C D$, which is the axial section of the frustum of the ... | \frac{\cos\alpha}{\sin^{4}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,242 |
12.333. The base of the right prism $A B C A_{1} B_{1} C_{1}\left(A A_{1}\left\|B B_{1}\right\| \mathrm{CC}_{1}\right)$ is an isosceles triangle $A B C(A B=A C)$, with a perimeter of $2 p$ and the angle at vertex $A$ equal to $\alpha$. A plane is drawn through side $B C$ and vertex $A_{1}$, forming an angle $\beta$ wit... | ## Solution.
Let $A D$ be the height of $\triangle A B C$ (Fig. 12.194). Then $\angle A_{1} D A$ is the angle of inclination of the section $B A_{1} C$ to the base plane, $\angle A_{1} D A=\beta$.
. Then $\angle M E O=\alpha$, the center $F$ of the inscribed sphere lies on the segment $L O$, $\angle O E F=\angle L E F=\frac{\alpha}{2} \cdot \angle L F E$ is the external angle of triangle $F O E$, ... | 8^{2}\cos\alpha\cos^{2}\frac{\alpha}{2}\operatorname{ctg}^{2}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,245 |
12.335. The base of the pyramid is a rectangle, where the angle between the diagonals is $\alpha$. A sphere of radius $R$ is circumscribed around this pyramid. Find the volume of the pyramid if all its lateral edges form an angle $\beta$ with the base. | Solution.
Let $EO$ be the height of the pyramid $EABCD$, and the rectangle $ABCD$ be its base (Fig. 12.196). Since all lateral edges form the same angle with the plane of the base, point $O$ is the center of the circle circumscribed around the rectangle $ABCD$, the point of intersection of its diagonals, and the cente... | \frac{4}{3}R^3\sin^22\beta\sin^2\beta\sin\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,246 |
12.336. The slant height of the cone is equal to $l$ and forms an angle $\alpha$ with the height. A plane is drawn through two slant heights of the cone, the angle between which is $\beta$. Find the distance from this plane to the center of the sphere inscribed in the cone. | Solution.
Let $SO$ be the height of the given cone, $\triangle ASB$ be the section of the cone, $\angle ASO = \alpha, AS = l, \angle ASB = \beta$, the center $O_1$ of the inscribed sphere belongs to the segment $SO$ (Fig. 12.197).
$C$ is the midpoint of $AB$. Then $OC \perp AB$, $SC \perp AB$. Therefore, the line $AB... | \frac{\cot(\frac{\pi}{4}+\frac{\alpha}{2})\sqrt{\sin(\alpha+\frac{\beta}{2})\sin(\alpha-\frac{\beta}{2})}}{\cos\frac{\beta}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,247 |
12.337. The base of the pyramid is an isosceles triangle, with an area of $S$, and the angle between the lateral sides is $\alpha$. All lateral edges of the pyramid form the same angle with the base plane. Find this angle if the volume of the pyramid is $V$. | ## Solution.
Let $DO$ be the height of the pyramid $DABC$, with $AB = AC$ and $\angle BAC = \alpha$ (Fig. 12.198). Since $\angle DAO = \angle DBO = \angle DCO$, $O$ is the center of the circumcircle of $\triangle ABC$.
If the radius of this circle is $R$, then $AB = 2R \sin \angle ACB = 2R \sin \left(90^\circ - \frac... | \operatorname{arctg}\frac{3V\cos\frac{\alpha}{2}}{S}\sqrt{\frac{2\sin\alpha}{S}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,248 |
12.338. The side of the base of a regular quadrilateral prism is equal to $a$, its volume is equal to $V$. Find the cosine of the angle between the diagonals of two adjacent lateral faces. | ## Solution.
The lateral edge of the prism (Fig. 12.199) $B B_{1}=\frac{V}{S_{\text {base }}}=\frac{V}{a^{2}} \Rightarrow A B_{1}^{2}=B_{1} C^{2}=$ $=\frac{V^{2}}{a^{4}}+a^{2}=\frac{V^{2}+a^{6}}{a^{4}}$.
$\mathrm{B} \triangle A B_{1} C$ :
$A C^{2}=A B_{1}^{2}+B_{1} C^{2}-2 A B_{1} \cdot B_{1} C \cos \angle A B_{1} C... | \frac{V^{2}}{V^{2}+^{6}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,249 |
12.339. The acute angle of the rhombus at the base of a quadrilateral pyramid is $\alpha$. The ratio of the total surface area of the pyramid to the square of the side of the base is $k$. Find the sine of the angle between the apothem and the height of the pyramid, given that all its lateral faces are equally inclined ... | Solution.
If $EO$ is the height of the pyramid $EABCD$ (Fig. 12.200), and $EF$ is the height of its lateral face $CED$, then $\angle EFO$ is the angle of inclination of this lateral face to the plane of the base, and since all lateral faces are equally inclined to it, $O$ is the center of the circle inscribed in the b... | \frac{\sin\alpha}{k-\sin\alpha};k>2\sin\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,250 |
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