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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3.27. A plane is drawn through the side of the lower base of a cube, dividing the volume of the cube in the ratio $m: n$, counting from the lower base. Find the angle between this plane and the plane of the base, if $m \leq n$. | 3.27. From the condition $m \leq n$, it follows that the plane $A D N M$ intersects the face of the cube $B B_{1} C_{1} C$ (and not $A_{1} B_{1} C_{1} D_{1}$) along the line $M N \| A D$ and, therefore, $M N$ is perpendicular to the face $D D_{1} C_{1} C$ (Fig. 3.29). Let the edge of the cube be $a$, and $\angle N D C=... | \alpha=\operatorname{arctg}\frac{2}{+n} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,186 |
3.28. The base of the prism is an isosceles triangle with an angle \(\alpha\) at the vertex. The diagonal of the face opposite this angle is \(l\) and forms an angle \(\beta\) with the base plane. Find the volume of the prism. | 3.28. Given that $A B C A_{1} B_{1} C_{1}$ is a right prism, $A C=C B, \angle A C B=\alpha, A_{1} B=l$, $\angle A_{1} B A=\beta$ (Fig. 3.30), we need to find $V_{\text {pr }}=S_{\triangle A B C} \cdot A_{1} A$. From $\triangle A_{1} A B$ we find $A A_{1}=l \sin \beta$, $A B=l \cos \beta$, and from $\triangle A D C$ we ... | \frac{1}{8}^{3}\sin2\beta\cos\beta\operatorname{ctg}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,187 |
3.29. The side of the base of a regular quadrilateral prism is equal to $a$. The angle between the intersecting diagonals of two adjacent lateral faces is $\alpha$. Find the volume of the prism. | 3.29. According to the condition, in the regular quadrilateral prism \(ABCD A_1 B_1 C_1 D_1\), we have \(AB = a\), \(\angle A_1 DC_1 = \alpha\) (Fig. 3.31). Since \(A_1 C_1\) is the diagonal of the square, then \(A_1 C_1 = a \sqrt{2}\). Let \(A_1 D = x\); then \(DC_1 = x\). From \(\Delta A_1 DC_1\) by the cosine theore... | \frac{^3\sqrt{2\cos\alpha}}{2\sin\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,188 |
3.30. The lateral edge of a regular triangular prism is equal to the side of the base. Find the angle between the side of the base and the non-intersecting diagonal of the lateral face. | 3.30. According to the condition, in the prism $A B C A_{1} B_{1} C_{1}$ we have $A B=B C=A C=A A_{1}$, i.e., all lateral faces of the prism are equal squares (Fig. 3.32). Let $a$ be the side of these squares; we need to find the angle between the diagonal $A_{1} C$ and the side $A B$. Since $A_{1} B_{1} \| A B$ and $A... | \arccos\frac{\sqrt{2}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,189 |
3.31. The height of a regular triangular prism is $H$. A plane passing through the midline of the lower base and the parallel side of the upper base forms an acute dihedral angle $\alpha$ with the plane of the lower base. Find the area of the section formed by this plane. | 3.31. Given that $A B C A_{1} B_{1} C_{1}$ is a regular triangular prism, $A A_{1}=H, A N=N B, B M=M C$; $\left(\left(A_{1} C_{1} M\right) ;(A B C)\right)=\alpha$ (Fig. 3.33). Draw $B E \perp A C$ and $E F \| A A_{1} ; \angle F O E=\alpha$ as the linear angle of the dihedral angle. Let $B E \cap M N=O$; then $F O \perp... | \frac{H^{2}\sqrt{3}\operatorname{ctg}\alpha}{\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,190 |
3.32. At the base of a right prism lies a rhombus with an acute angle $\alpha$. The ratio of the height of the prism to the side of the base is $k$. A plane is drawn through the side of the base and the midpoint of the opposite lateral edge. Find the angle between this plane and the plane of the base. | 3.32. Given that $A B C D A_{1} B_{1} C_{1} D_{1}$ is a right prism, $A B C D$ is a rhombus, $\angle B A D=\alpha\left(\alpha<90^{\circ}\right), A A_{1}: A B=k, C_{1} E=E C, A D E F$ is a section of the prism (Fig. 3.34); it is required to find the angle between the section and the base $(A B C)$; (FAD). Let $A B=a$; t... | \operatorname{arctg}\frac{k}{2\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,191 |
3.33. The lateral edge of a regular triangular pyramid is equal to $l$ and forms an angle $\alpha$ with the base plane. Find the volume of the pyramid. | 3.33. By the condition, $S A B C-$ is a regular triangular pyramid, $S A=l, S O \perp(A B C), \angle S A O=\alpha$ (Fig. 3.35). From $\triangle S A O$ we find $S O=l \sin \alpha, A O=l \cos \alpha$, hence $A B=\sqrt{3} A O=l \sqrt{3} \cos \alpha$. Therefore,
$$
S_{\triangle A B C}=\frac{A B^{2} \sqrt{3}}{4}=\frac{3 \s... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,192 | |
3.34. The base of the pyramid is a right-angled triangle, the hypotenuse of which is equal to $c$, and one of the acute angles is equal to $\alpha$. Each lateral edge forms an angle $\beta$ with the plane of the base. Find the volume of the pyramid. | 3.34. Since the lateral edges of the pyramid $ABCD$ (Fig. 3.36) are equally inclined to the plane of the base, the vertex $D$ projects onto the center of the circle circumscribed around triangle $ABC$; for a right triangle, it projects onto the midpoint $E$ of the hypotenuse $AB$. The height $DE$ of triangle $ADB$ is t... | \frac{^3}{24}\sin2\alpha\operatorname{tg}\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,193 |
3.35. The base of a quadrilateral pyramid is a rhombus with side $a$ and acute angle $\alpha$. All lateral faces are inclined to the base plane at the same angle $\beta$. Find the total surface area of the pyramid. | 3.35. By condition, $S A B C D$ is a pyramid, $A B C D$ is a rhombus, $\angle B A D=$ $=\alpha\left(\alpha<90^{\circ}\right), A B=a$,
$((S A B) ;(A B C))=$
$=((S B C) ;(A B C))=$
$=((S C D) ;(A B C))=$
$=((S D \widehat{A}) ;(A B C))=\beta$,
$S O \perp(A B C)$ (Fig. 3.37). Let's draw the apothems of the pyramid $S ... | \frac{2^{2}\sin\alpha\cos^{2}\frac{\beta}{2}}{\cos\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,194 |
3.36. The base of the pyramid is an isosceles triangle with a lateral side equal to \(a\), and the angle at the vertex is \(\alpha\). All lateral edges are inclined to the base plane at an angle \(\beta\). Find the volume of the pyramid. | 3.36. Given that $S A B C-$ is a triangular pyramid, $A B=B C=$ $=a, \angle A B C=\alpha, S O \perp(A B C), \angle S A O=$ $=\angle S B O=\angle S C O=\beta$ (Fig. 3.38). Since $\triangle S O A=\triangle S O B=\triangle S O C$ (by the leg and adjacent angle), then $O A=O B=O C$, i.e., $O$ is the center of the circle ci... | \frac{1}{6}^{3}\sin\frac{\alpha}{2}\operatorname{tg}\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,195 |
3.37. The base of the pyramid is an equilateral triangle with side $a$. Two lateral faces of the pyramid are perpendicular to the base plane, and the equal lateral edges form an angle $\alpha$. Find the height of a right triangular prism equal in volume to the given pyramid and having a common base. | 3.37. By condition, $DABC$ is a pyramid, $ABC$ is an equilateral triangle, $(ABD) \perp (ABC)$, $(BCD) \perp (ABC)$, $\angle ADC = \alpha$, $AB = a$ (Fig. 3.39). It is required to find the height of the prism whose base is $\triangle ABC$, such that $V_{\text{pr}} = V_{\text{pir}}$. The height of the prism coincides wi... | H=\frac{}{6}\sqrt{\operatorname{ctg}^2\frac{\alpha}{2}-\operatorname{ctg}^2\frac{\pi}{6}}=\frac{\sqrt{\sin(\frac{\pi}{6}+\frac{\alpha}{2})\sin(\frac{\pi}{6}-\frac{\alpha}{2})}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,196 |
3.38. The base of the pyramid $ABCD$ is a right triangle $ABC (\angle C=90^{\circ})$. The lateral edge $AD$ is perpendicular to the base. Find the acute angles of triangle $ABC$ if $\angle DBA=\alpha$ and $\angle DBC=\beta (\alpha<\beta)$. | 3.38. By the condition, $DABC$ is a pyramid, $\angle ACB=90^{\circ}, AD \perp(ABC)$, $\angle DBA=\alpha, \angle DBC=\beta(\alpha<\beta)$ (Fig. 3.40). Let $AD=h$; then from $\triangle ADB$ we find $AB=h \operatorname{ctg} \alpha, DB=\frac{h}{\sin \alpha}$. Since $BC \perp AC$, then $DC \perp BC$ and from $\triangle DCB$... | \angleA=\arcsin\frac{\cos\beta}{\cos\alpha},\angleB=\frac{\pi}{2}-\arcsin\frac{\cos\beta}{\cos\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,197 |
3.39. The plane angle at the vertex of a regular hexagonal pyramid is equal to the angle between a lateral edge and the plane of the base. Find this angle. | 3.39. By condition, $O A B C D E F$ is a regular hexagonal pyramid, $O O_{1} \perp(A B C), \angle F O E=\angle O F O_{1}$ (Fig. 3.41). Let $O F=l$ and $\angle F O E=\alpha$; then $F E=\sqrt{2 l^{2}-2 l^{2} \cos \alpha}=2 l \sin \frac{\alpha}{2}$. Considering that $O_{1} F=F E$, in $\triangle O O_{1} F$ we have
$$
\beg... | \alpha=2\arcsin\frac{\sqrt{3-1}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,198 |
3.40. The ratio of one side of the base of a triangular pyramid to each of the other five edges of the pyramid is $k$. Find the dihedral angle between two equal lateral faces of the pyramid and the permissible values of $k$.
| 3.40. Given that $S A B C$ is a triangular pyramid, $\frac{A C}{A B}=\frac{A C}{B C}=$ $=\frac{A C}{S A}=\frac{A C}{S B}=\frac{A C}{S C}=k$ (Fig. 3.42). Since
$A B=B C=S A=S B=S C=\frac{A C}{k}$,
then $\triangle A S B=\triangle B S C$ (by three sides). Draw $A D \perp S B$ and connect points $D$ and $C$. Then $\trian... | 2\arcsin\frac{k\sqrt{3}}{3},\quad0<k<\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,199 |
3.41. The side of the larger base of a regular quadrilateral truncated pyramid is equal to $a$. The lateral edge and the diagonal of the pyramid form angles with the base plane that are equal to $\alpha$ and $\beta$, respectively. Find the area of the smaller base of the pyramid. | 3.41. Given that $A B C D A_{1} B_{1} C_{1} D_{1}$ is a regular truncated quadrilateral pyramid, $A B=a, O O_{1} \perp(A B C), B_{1} E\left\|D_{1} F\right\| O O_{1}$, $\angle D_{1} D E=\alpha, \angle B_{1} D E=\beta$ (Fig. 3.43). Let $A_{1} B_{1}=x$; then $B_{1} D_{1}=x \sqrt{2}, B D=a \sqrt{2}$. We have
}{\sin^{2}(\alpha+\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,200 |
3.42. A plane passing through a generatrix of a cylinder forms an acute angle $\alpha$ with the plane of the axial section containing the same generatrix. The diagonal of the rectangle obtained by the intersection of the cylinder with this plane is equal to $l$ and forms an angle $\beta$ with the plane of the base. Fin... | 3.42. Given that $\angle B A C=\alpha, \angle B A B_{1}=\beta, A B_{1}=l$ (Fig. 3.44). From $\triangle B A B_{1}$ we find $A B=l \cos \beta, B B_{1}=l \sin \beta$. Since $O A=O B$ (radii of the base), $\triangle A O B$ is isosceles and $\alpha$ is the angle at its base $A B$. Therefore, $O B=\frac{1}{2} A B: \cos \alph... | \frac{\pi^{3}\sin2\beta\cos\beta}{8\cos^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,201 |
3.43. Find the angle at the vertex of the axial section of a cone if the central angle in the development of its lateral surface is $\alpha$ radians. | 3.43. Let $\triangle A B C$ be the axial section of a cone (Fig. 3.45). Then the lateral surface of the cone unfolds into a sector of a circle, with the radius $R$ equal to the slant height $B C$ of the cone; the length of the arc of the sector ($l$) is equal to the circumference of the base of the cone,
. Since $\cup B n K=\alpha^{\circ}$, then
Fig. 3.45
. According to the condition, $A O: O B_{1}=2: 1, \angle A O... | \frac{7\pi^{3}}{54}\sin\alpha\sin\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,204 |
3.46. Find the angle between the slant height and the height of a cone, the lateral surface area of which is the mean proportional between the area of the base and the total surface area. | 3.46. According to the condition, $S_{\text {osn }}: S_{\text {bok.pov }}=$ $=S_{\text {bokpov }}: S_{\text {poln }} ;$ we need to find $\angle C B O=$ $=x$ (Fig. 3.48). Let $O C=R$; then $S_{\text {osn }}=$ $=\pi R^{2}, S_{\text {bok.pov }}=\pi R l$, where $l=B C$. Therefore,
$$
\frac{\pi R^{2}}{\pi R l}=\frac{\pi R ... | \arcsin\frac{\sqrt{5}-1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,205 |
3.47. The bases of two cones with a common vertex lie in the same plane. The difference in their volumes is $V$. Find the volume of the smaller cone, if the tangents drawn from an arbitrary point on the circumference of the base of the larger cone to the circumference of the base of the smaller cone form an angle $\alp... | ### 3.47. Let $V_{1}$ be the volume of the larger cone, and $V_{2}$ be the volume of the smaller cone (i.e., the one we are looking for). According to the problem, $V_{1}-V_{2}=V, A B=A C$, where $B$ and $C$ are the points of tangency, $\angle B A C=\alpha$ (Fig. 3.49). Since $A B=A C$, then $\angle O A B = \angle O A ... | V\operatorname{tg}^{2}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,206 |
3.48. The height of the cone is $H$, the angle between the slant height and the height is $\alpha$. A smaller cone is inscribed in this cone such that the vertex of the smaller cone coincides with the center of the base of the larger cone, and the corresponding slant heights of both cones are perpendicular to each othe... | 3.48. According to the condition, $S O=H, O C \perp A S$, $O D \perp B S, \angle O S B=\alpha$ (Fig. 3.50); it is required to find the volume of the cone $O D C$. Obviously, $\angle O D O_{1}=\angle O S B$ as acute angles with mutually perpendicular sides. From $\triangle S O D$ we find $O D=H \sin \alpha$; further, fr... | \frac{1}{3}\piH^{3}\sin^{4}\alpha\cos^{2}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,207 |
3.49. The side of the rhombus is $a$, and its acute angle is $\alpha$. The rhombus rotates around a line passing through its vertex and parallel to the larger diagonal. Find the volume of the solid of revolution. | 3.49. By the condition, $ABCD$ is a rhombus, $AB=a, \angle BAD=\alpha\left(\alpha<90^{\circ}\right)$, $MN$ is the axis of rotation, $MN \| AC, B \in MN$ (Fig. 3.51); it is required to find $V_{\text {rot }}=2\left(V_{B E C D}-V_{B E C}\right)$. The trapezoid $B E C D$ forms a frustum of a cone upon rotation, and the tr... | 2\pi^{3}\sin\alpha\sin\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,208 |
3.50. The radius of the circle inscribed in a right trapezoid is $r$, and the acute angle of the trapezoid is $\alpha$. This trapezoid rotates around the smaller lateral side. Find the lateral surface area of the solid of revolution. | 3.50. By the condition, $ABCD$ is a trapezoid, $CD \perp BC, CD \perp AD, CD = 2r$ and $\angle BAD = \alpha$ (Fig. 3.52). When the trapezoid is rotated around the side $CD$, a truncated cone $ABB_1A_1$ is obtained, the lateral surface of which is equal to the difference between the lateral surfaces of the complete cone... | \frac{4\pir^2(1+\sin\alpha)}{\sin^2\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,209 |
3.51. Find the acute angle of a rhombus, knowing that the volumes of the bodies obtained by rotating the rhombus around its larger diagonal and around its side are in the ratio of $1: 2 \sqrt{5}$.
untranslated part:
- Найти острый угол ромба, зная, что объемы тел, получен-
- ых от вращения ромба вокруг его большей ди... | ### 3.51. When the rhombus $ABCD$ is rotated around the diagonal $BD$, two cones with a common base $AC$ are formed (Fig. 3.53). The volume of this figure $V_{1}=\frac{1}{3} \pi O A^{2}(O B+O D)=$ $=\frac{1}{3} \pi O A^{2} \cdot B D$. When the same rhombus is rotated around the side $DC$, a cylinder $A B B_{1} A_{1}$ i... | \alpha=\arccos\frac{1}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,210 |
3.52. A triangular pyramid is inscribed in a cone, with its lateral edges being pairwise perpendicular to each other. Find the angle between the slant height of the cone and its height. | 3.52. Given that $DAB$ is a cone, $DABC$ is a pyramid inscribed in this cone, $AD \perp BD, \quad AD \perp CD, \quad DB \perp DC$, and $DO \perp (ABC)$ (Fig. 3.54). Since $DB = DC = DA$, $\triangle ADB = \triangle ADC = \triangle DBC$ as right triangles with two equal legs; therefore, $\triangle ABC$ is equilateral. Le... | \arcsin\frac{\sqrt{6}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,211 |
3.53. A sphere is inscribed in a cone. The radius of the circle where the cone and the sphere touch is $r$. Find the volume of the cone if the angle between the height and the slant height of the cone is $\alpha$. | 3.53. Let's depict the axial section of the figure. In the section, we get $\triangle SDE$, where $SD$ and $SE$ are the generators of the cone, $ED$ is the diameter of its base, $SB$ is the height of the cone, $O$ is the center of the largest circle inscribed in the cone by the sphere, $C$ is the point of tangency of t... | \frac{\pir^3\operatorname{tg}^3(\frac{\pi}{4}+\frac{\alpha}{2})}{3\sin\alpha\cos^2\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,212 |
3.54. A hemisphere is inscribed in a cone; the great circle of the hemisphere lies in the base plane of the cone, and the spherical surface touches the surface of the cone. Find the volume of the hemisphere if the slant height of the cone is $l$ and it makes an angle $\alpha$ with the base plane. | 3.54. The axial section of the figure is depicted as a triangle $S A B$, in which a semicircle $M n K$ is inscribed; $S A$ and $S B$ are the generators of the cone, $S O \perp A B$ is the height of the cone; $A B$ is the diameter of the base of the cone and $M K$ is the diameter of the semicircle; $O$ is the center of ... | \frac{1}{12}\pi^{3}\sin^{3}2\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,213 |
3.55. A truncated cone is described around a sphere, with the area of one base being 4 times larger than the area of the other. Find the angle between the slant height of the cone and the plane of its base.
## Group B | 3.55. The axial section of the figure is an isosceles trapezoid \( L M M_{1} L_{1} \), in which a circle with center \( O \) is inscribed
Fig. 3.57
(Fig. 3.57). By the problem statement, \(... | \arccos\frac{1}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,214 |
3.57. 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? | 3.57. By the condition, \( A B = B C, \angle A C B = \alpha, \angle C A M = \beta \) (Fig. 3.59). Let \( A B = a \); then \( A C = 2 a \cos \alpha \) and, therefore,
\[
\frac{S_{\triangle A M B}}{S_{\triangle A M C}} = \frac{0.5 A B \cdot A M \sin (\alpha - \beta)}{0.5 A C \cdot A M \sin \beta} = \frac{a \sin (\alpha ... | \frac{\sin(\alpha-\beta)}{2\cos\alpha\sin\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,216 |
3.58. 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 $\beta$. | 3.58. By the condition, $\angle B A C=\alpha, B C=a, \angle D A B=\angle D A C, A E \perp B C$, $\angle D A E=\beta$ (Fig. 3.60). We have $\angle B A D=\frac{\alpha}{2}, \angle A D E=90^{\circ}-\beta$; then $\angle A B D=90^{\circ}-\beta-\frac{\alpha}{2}$ (according to the property of the exterior angle of a triangle),... | \frac{\cos(\beta-\frac{\alpha}{2})\cos(\beta+\frac{\alpha}{2})}{\sin\alpha\cos\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,217 |
3.59. 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? | 3.59. By the condition, $\angle A C B=90^{\circ}$, $\angle C B A=\alpha$ (Fig. 3.61). Let $R$ and $r$ be the radii of the circumscribed and inscribed circles, respectively; then $A C=2 R \sin \alpha ; B C=2 R \cos \alpha$. Therefore, $S_{\triangle A B C}=\frac{1}{2} \cdot 4 R^{2} \sin \alpha \cos \alpha=2 R^{2} \sin \a... | \alpha=\frac{\pi}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,218 |
3.60. 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. | 3.60. According to the condition, in an equilateral triangle $ABC$ we have $AL: LC = n: m$ (Fig. 3.62); it is required to find the angle $ALB$. Let $AC = b$, $AL = nx$, $LC = mx$; then $nx + mx = b$, from which $x = \frac{b}{m+n}$, $AL = \frac{nb}{m+n}$, $LC = \frac{mb}{m+n}$. Draw $BB_1 \perp AC$;
\sqrt{3}}{-n} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,219 |
3.61. The tangent of the acute angle between the medians of a right triangle, drawn to its legs, is $k$. Find the angles of the triangle and the permissible values of $k$.
untranslated text:
3.61. Тангенс острого угла между медианами прямоугольного греугольника, проведенными к его катетам, равен $k$. Найти утлы треуг... | 3.61. According to the condition, $\angle A B C = 90^{\circ}, A B_{1} = B_{1} C ; C A_{1} = A_{1} B$, $\operatorname{tg} \angle A O B_{1} = k$ (Fig. 3.63); it is required to find $\angle A$ and $\angle B$. Let $\angle C A A_{1} = \psi, \angle C B_{1} B = \varphi, \angle A O B_{1} = \alpha$; then $\dot{\alpha} = \varphi... | \angleA=\frac{1}{2}\arcsin\frac{4k}{3},\quad0<k\leq\frac{3}{4},\quad\angleB=\frac{\pi}{2}-\frac{1}{2}\arcsin\frac{4k}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,220 |
3.62. Find the sine of the angle at the vertex of an isosceles triangle, given that the perimeter of any rectangle inscribed in it, with two vertices lying on the base, is a constant value. | 3.62. Given that $A B=A C$, $D E F L$ is a rectangle inscribed in $\triangle A B C$, the perimeter of which does not depend on the choice of point $E$ on $A B$ (Fig. 3.64); we need to find $\sin A$. Draw $A A_{1} \perp B C$ and let $A A_{1}=h, B C=a$. Construct another rectangle inscribed in the same way, one of whose ... | \frac{4}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,221 |
3.63. 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 $ABC$. | 3.63. By the condition, $\angle B A C=\alpha, \angle B C A=\gamma(\alpha>\gamma), A D=D C, B E$ is the bisector (Fig. 3.65); it is required to find $S_{\triangle B D E}: S_{\triangle A B C}$. Draw $B F \perp A C$; then $A B=\frac{B F}{\sin \alpha} ; B C=\frac{B F}{\sin \gamma} \cdot$ We have
; it is required to find $S_{\triangle K L M}: S_{\triangle A B C}$. Let $B M=h$; then $A M=h \operatorn... | \frac{1}{4}\sin2\alpha\sin2\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,223 |
3.65. 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 $k$. | 3.65. According to the condition, $B D \perp A C, D C: A D=k, D C>A D, \angle D B C > \angle D B A, \angle D B C: \angle D B A=2: 1$ (Fig. 3.67). Let $\angle D B A=\alpha$; then $\angle D B C=2 \alpha$ and $\angle C=90^{\circ}-2 \alpha2$. Therefore,
$$
\sin C=\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\oper... | \frac{1}{k-1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,224 |
3.66. 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. | 3.66. By the condition, $A B \| D C$; $A D=B C ; \angle B O C=\alpha ; S_{A B C D}=S$ (Fig.3.68). Since $\angle B O C$ is the exterior angle of $\triangle D O C$, then $\angle B O C=$
$=2 \angle O D C$, i.e., $\angle O D C=\frac{\alpha}{2}$. Draw $B N \perp C D, E F \| B N$; from $\triangle O F D$ and $\triangle O E B... | \sqrt{S\operatorname{tg}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,225 |
3.67. 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. | 3.67. By the condition, $A D \| B C, B C=2, \angle A B C=\angle B C D=135^{\circ}$, $\angle A O D=150^{\circ}$ (Fig. 3.69). Since the angles at the base of the trapezoid are equal, the trapezoid is isosceles, i.e., $A B=C D$. We find
$, the larger base is equal to $a$. The angles at the larger base are in the ratio $2: 1$. Find the smaller base. | 3.68. I n d i c a t i o n. D e n o t i n g t h e a n g l e s a t t h e l a r g e r b a s e o f t h e t r a p e z o i d a s $\alpha$ a n d $2 \alpha$, d r o p t h e h e i g h t s o f t h e t r a p e z o i d o n t o t h i s b a s e a n d u s e t h e i r e q u a l i t y.
A n s w e r: $\frac{p^{2}+a p-q^{2}}{p}$.
, and the angles adjacent to it contain \(45^{\circ}\) and \(15^{\circ}\). A circle with a radius equal to the height dropped to this base is drawn from the vertex opposite the base. Find the area of the part of the corresponding circle that is enclosed within the triangl... | 3.69. By the condition, $\angle A C B=$ $=45^{\circ}, \angle A B C=15^{\circ}, B C=a$, $A M \perp B C, A M$ - radius of the circle, $E$ and $F$ - points of intersection of the circle with sides $A B$ and $A C$; it is required to find $S_{A F M E}$ (Fig. 3.70). From $\triangle A M C$ and $\triangle A M B$ we find $M C=A... | \frac{\pi^{2}(2-\sqrt{3})}{18} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,228 |
3.70. 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. | 3.70. By the condition, in $\triangle A B C$ we have: $B C=a, \angle C=\alpha$ radians, $\angle B=\beta$ radians, $A D \perp B C, A D-$ is the radius of the circle intersecting $A C$ and $A B$ at points $K$ and $L$ (Fig. 3.71); we need to find $l_{\cup K L}$. From $\triangle A D C$ and $\triangle A D B$ we find $C D=$ ... | \frac{(\pi-\alpha-\beta)\sin\alpha\sin\beta}{\sin(\alpha+\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,229 |
3.71. Find the ratio of the area of a sector with a given central angle $\alpha$ radians to the area of the circle inscribed in it. | 3.71. By the condition, $O A n B$ is a sector, $\angle A O B=\alpha$ radians, $O_{1}$ is the center of the circle inscribed in the sector $O A n B$ (Fig. 3.72). Let $R$ be the radius of the sector, $r$ be the radius of the inscribed circle; then $S_{O A n B}=\frac{1}{2} R^{2} \alpha, S_{O_{1}}=\pi r^{2}$. In $\Delta O_... | \frac{2\alpha\cos^{2}(\frac{\pi}{4}-\frac{\alpha}{4})}{\pi\sin^{2}\frac{\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,230 |
3.72. 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 $\alpha$. | 3.72. By the condition, $AB$ is a chord, $\cup ANB=\alpha^{\circ}; AC=BC, DE$ is a chord, $DC:CE=1:3$ (Fig. 3.73). Let $CD=x$; then $EC=3x, ED=4x$. Draw $OF \perp ED$ and connect points $O$ and $C, O$ and $A$. Since $OC \perp AB$ (diameter $MN$ is perpendicular to $AB$), we have $\angle FOC=\angle ACD$ (acute angles wi... | \alpha\leq120,\angleACD=\arcsin\frac{\operatorname{tg}\frac{\alpha}{2}}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,231 |
3.73. The height of an isosceles triangle is equal to $h$ and forms an angle $\alpha\left(\alpha \leq \frac{n}{6}\right)$ with the lateral side. Find the distance between the centers of the inscribed and circumscribed circles of the triangle. | 3.73. By the condition, $A B=B C$, $B B_{1} \perp A C, B B_{1}=h, \angle A B B_{1}=\alpha$ ( $\alpha \leq \frac{\pi}{6}$ ), $O$ - the center of the circumcircle of $\triangle A B C, O_{1}$ the center of the incircle of $\triangle A B C$ (Fig. 3.74). From $\triangle A B_{1} B$ we find $A B_{1}=h \operatorname{tg} \alpha... | \frac{2\sin(\frac{\pi}{12}-\frac{\alpha}{2})\cos(\frac{\pi}{12}+\frac{\alpha}{2})}{\cos^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,232 |
3.74. 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. | 3.74. Given that $ABCD$ is a rhombus, $AD \in \beta$, $\angle A$ is acute; the diagonals of the rhombus form angles $\alpha$ and $2\alpha$ with $\beta$ (Fig. 3.75). Let $O$ be the point of intersection of the diagonals. Draw $OO_1 \perp \beta$; thus, $\sin \angle OAO_1 = \frac{OO_1}{OA}$ and $\sin \angle ODO_1 = \frac{... | 2\operatorname{arcctg}(2\cos\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,233 |
3.75. 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. | 3.75. Given that $A B C D A_{1} B_{1} C_{1} D_{1}$ is a right parallelepiped, $A B C D$ is a rhombus, $A A_{1}=h, \angle A C A_{1}=\alpha, \angle D B D_{1}=\beta$ (Fig. 3.76). From $\triangle A_{1} A C$ and $\triangle D_{1} D B$ we find $A C=h \operatorname{ctg} \alpha, B D=\boldsymbol{h} \operatorname{ctg} \beta$; the... | S_{\text{side}}=2^{2}\sqrt{\operatorname{ctg}^{2}\alpha+\operatorname{ctg}^{2}\beta},\quadV=\frac{1}{2}^{3}\operatorname{ctg}\alpha\operatorname{ctg}\beta | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,234 |
3.76. 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 $V$. | 3.76. According to the condition, $A B C A B_{1} C_{1}$ is a straight prism, $A B=a, \angle B A C=\alpha$, $\angle A B C=\beta, V_{\text {mp }}=V$ (Fig. 3.77). Further, considering that $V=S_{\triangle A B C} H$, where the area of the base is expressed by the equality
$ 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 $\beta$ with the base p... | 3.77. Given that $A B C A_{1} B_{1} C_{1}$ is a right prism, $A B=A C, A B+A C+$ $+B C=2 p, \angle B A C=\alpha, A_{1} B C$ is a section of the prism, $\angle A_{1} B C A=\beta$ (Fig. 3.78). Draw $A_{1} K \perp B C$ and let $A A_{1}=h$; then from $\triangle A_{1} A K$ we find $A K=h \operatorname{ctg} \beta$, and from ... | p^{3}\operatorname{tg}^{3}\frac{\pi-\alpha}{4}\operatorname{tg}\beta\operatorname{tg}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,236 |
3.78. 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 $\alpha$. | 3.78. By the condition, $A B C A_{1} B_{1} C_{1}$ is a regular triangular prism, $S A B C$ is a pyramid, $V_{S A B C}=V,((S A B);(A B C))=$ $=\alpha$ (Fig. 3.79). Let $A B=a, S C=h$; then $V=\frac{1}{3} \frac{a^{2} \sqrt{3}}{4} h, S_{\triangle S A B}=\frac{a^{2} \sqrt{3}}{4 \cos \alpha}$ and $h=\frac{a \sqrt{3}}{2} \op... | S_{\triangleSAB}=\sqrt[3]{\frac{3\sqrt{3}V^{2}}{\sin^{2}\alpha\cos\alpha}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,237 |
3.79. Find the cosine of the angle between the non-intersecting diagonals of two adjacent lateral faces of a regular triangular prism, where the lateral edge is equal to the side of the base. | 3.79. I m e t h o d. By the condition, $A B C A_{1} B_{1} C_{1}$ is a regular triangular prism, $A A_{1}=A B, A B_{1}$ and $B C_{1}$ are non-intersecting diagonals (Fig. 3.80); it is required to find $\angle (AB, BC_{1})$. Let $A B=A A_{1}=a$. Draw $B_{1} D \| B C_{1}, D \in (B C)$; then $\angle A B_{1} D = \varphi$ is... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,238 |
3.80. The base of the pyramid is an isosceles trapezoid, with the lateral side equal to \(a\), and the acute angle 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. | 3.80. By the condition, $S A B C D$ is a pyramid, $A B \| D C, A D=B C, A D=a$, $\angle A D C=\alpha, \angle S A B C=\angle S B C D=$ $=\angle S D C A=\angle S A D B=\beta$ (Fig. 3.81). Construct the linear angles of the dihedral angles at the base. Draw $S O \perp(A B C)$ and $O E, O F, O G, O H$ - perpendiculars to t... | \frac{2^{2}\sin\alpha\cos^{2}\frac{\beta}{2}}{\cos\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,239 |
3.81. 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 $\alpha$. | 3.81. By the condition, $S A B C$ is a regular triangular pyramid, $O$ is the center of the base, $O K \perp S B, O K=p, \angle A S B C=\alpha$ (Fig. 3.82). We construct the linear angle of the dihedral angle $A S B C$. We have $A C \perp S B$, since $S B$ is a slant to ( $A B C$ ), $O B$ is its projection onto this pl... | \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 | 50,240 |
3.82. The ratio of the area of the diagonal section of a regular quadrilateral pyramid to the area of its base is $k$. Find the cosine of the plane angle at the vertex of the pyramid. | 3.82. Given that $P A B C D$ is a regular quadrilateral pyramid, $S_{\triangle P B D}: S_{A B C D}=k$ (Fig. 3.83); we need to find $\cos \angle D P C$. Let $D C=a$ and $O P \doteq h$; then $\frac{0.5 a \sqrt{2} h}{a^{2}}=k, h=a k \sqrt{2}$. From $\triangle P O D$ we find $D P^{2}=2 a^{2} k^{2}+\frac{a^{2}}{2}=\frac{a^{... | \frac{4k^{2}}{4k^{2}+1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,241 |
3.83. 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. | 3.83. Given that $S A B C$ is a regular triangular pyramid, $S O \perp (A B C)$, $S O=H$, $\angle S A C B=\alpha$, $S M=M B$, and $A M C$ is a section (Fig. 3.84). Draw $S K \perp A C$; then $O K \perp A C$ and $\angle S K O=\alpha$ as the linear angle of the dihedral angle $A C$. From $\triangle S O K$ we find $O K=H$... | \frac{H^{2}\sqrt{3}\operatorname{ctg}\alpha}{2}\sqrt{1+16\operatorname{ctg}^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,242 |
3.84. 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? | 3.84. Instruction. Use the formula $S_{\text {lat }}=\frac{S_{\text {base }}}{\cos \alpha}$, where $\alpha$ is the angle between the base of the pyramid and its lateral faces.
Answer: $\alpha=\arccos \frac{1}{k-1}$, where $k>2$. | \alpha=\arccos\frac{1}{k-1},wherek>2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,243 |
3.85. 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 $k$. | 3.85. By the condition, $S A B C D$ is a regular pyramid, $S O \perp(A B C)$, $S_{\triangle S A C}: \mathrm{S}_{\text {bok }}=k, S F$ and $S E$ are the apothems of the pyramid (Fig. 3.85). Let $A B=$
$=a, \angle E S F=\alpha ;$ we have $S F=\frac{a}{2 \sin \frac{\alpha}{2}}$
(from $\triangle S O F), \quad S O=\frac{a... | \cos\alpha=16k^2-1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,244 |
3.86. In a regular triangular pyramid with an angle $\alpha$ between a lateral edge and a side of the base, a section is made through the midpoint of the lateral edge parallel to a lateral face. Given the area $S$ of this section, find the volume of the pyramid. What are the possible values of $\alpha$? | 3.86. By the condition, $PABC$ is a regular triangular pyramid, $\angle PBC=\alpha, PM=MC, (KMN) \|(APB)$, $S_{\triangle KMN}=S$ (Fig. 3.86). Since $PC=2MC$, we have $S_{\triangle APB}=4S_{\triangle KMN}=4S$. Further, we have $S_{\triangle APB}=\frac{1}{2} AP^2 \sin (180^\circ - 2\alpha)=$
\sin(\alpha-\frac{\pi}{6})}{3\sin2\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,245 |
3.87. 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. | 3.87. Given $A B C A_{1} B_{1} C_{1}$ is a regular truncated triangular pyramid, $O O_{1}=H, \quad O O_{1}=\sqrt{B C \cdot B_{1} C_{1}}$, $\angle A_{1} A O=\alpha$ (Fig. 3.87). Let $B C=a$, $B_{1} C_{1}=b$; then $O A=\frac{a \sqrt{3}}{3}, O_{1} A_{1}=\frac{b \sqrt{3}}{3}$.
Draw $A_{1} K \| O O_{1};$ from $\triangle A_... | \frac{H^{3}\sqrt{3}}{4\sin^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,246 |
3.88. A line segment connecting a point on the circumference of the upper base of a cylinder with a point on the circumference of the lower base is equal to $l$ and forms an angle $\alpha$ with the base plane. Find the distance from this line to the axis of the cylinder, if the axial section of the cylinder is a square... | 3.88. Given that $A K L M$ is a cylinder, $O O_{1}$ is its axis, $A$ belongs to the circumference of the upper base, $B$ belongs to the circumference of the lower base, $A B=l, \angle A B K=\alpha$, and the axial section $A M L K$ is a square (Fig. 3.88); we need to find the distance from $O O_{1}$ to $A B$. Draw $B S ... | \frac{1}{2}\sqrt{-\cos2\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,247 |
3.89. 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. | 3.89. By the condition, $S C D$ and $S_{1} C_{1} D_{1}$ are cones with a common center $O, \angle C S O=\angle C_{1} S_{1} O=\alpha, O C=R$,
Fig. 3.89
Solving the equation $\frac{2 \pi R^{2... | \frac{1}{3}\piR^{3}\cos^{3}(\frac{\pi}{4}-\frac{\alpha}{2})\operatorname{ctg}\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,248 |
3.90. In a truncated cone, the diagonals of the axial section are mutually perpendicular and the length of each is a. The angle between the generatrix and the base plane is $\alpha$. Find the total surface area of the truncated cone. | 3.90. Let's depict the axial section of a truncated cone. According to the problem, \( A B_{1} \perp A_{1} B \), \( O O_{1} \) is the axis of the truncated cone, \( B B_{1} \) is the generatrix of the truncated cone, \( A B_{1} = a \). \(\angle B_{1} B O = \alpha\) (Fig. 3.90). Let \( O B = R_{1} \), \( O_{1} B_{1} = R... | \frac{\pi^{2}}{\sin^{2}\alpha}\sin(\frac{\alpha}{2}+\frac{\pi}{12})\cos(\frac{\alpha}{2}-\frac{\pi}{12}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,249 |
3.91. An obtuse isosceles triangle rotates around a line passing through the point of intersection of its altitudes parallel to the larger side. Find the volume of the solid of revolution if the obtuse angle is $\alpha$, and the side opposite to it is $a$. | 3.91. By the condition, $\angle B A C=\alpha-$ is an obtuse angle, $A B=A C, B M \perp A C$, $C N \perp A B, A D \perp B C, O$ - the orthocenter (i.e., the point of intersection of the altitudes) of $\triangle A B C$, $O_{1} O_{2} \| B C, O \in O_{1} O_{2}, B C=a$ (Fig. 3.91). The volume of the solid of revolution of $... | \frac{\pi^{3}}{12}(3-\operatorname{ctg}^{2}\frac{\alpha}{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,250 |
3.92. On the segment $A B$, equal to $2 R$, a semicircle is constructed as if on a diameter and a chord $C D$ is drawn parallel to $A B$. Find the volume of the body formed by the rotation of triangle $A C D$ around the diameter $A B$, if the inscribed angle subtending the arc $A C$ is $\alpha$ $(A C<A D)$. | 3.92. By the condition, $A B=2 R, C D \| A B, O$ - the center of the semicircle, $\angle A D C=\alpha, A C<A D$ (Fig. 3.92). The volume of the body of revolution of $\triangle A C D$ around $A B$ will be found using the formula
$$
\begin{aligned}
& V_{\text {T. } \mathrm{pp}}=V_{A O_{2} C}+V_{O_{2} C D O_{1}}-V_{A O_{... | \frac{2}{3}\piR^{3}\sin2\alpha\sin4\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,251 |
3.93. The larger base of an isosceles trapezoid is equal to $a$, the acute angle is equal to $\alpha$. The diagonal of the trapezoid is perpendicular to its lateral side. The trapezoid rotates around its larger base. Find the volume of the solid of revolution. | 3.93. By the condition, \( A B \| D C, A D = B C, A B = a, \angle D A B = \alpha, B D \perp A D \) (Fig. 3.93). The volume of the solid of revolution of trapezoid \( A B C D \) around \( A B \) will be
\sin(\alpha+\frac{\pi}{6}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,252 |
3.94. 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 $\alp... | 3.94. Given that $P A B C D$ is a regular quadrilateral pyramid, $A_{1} B_{1} C_{1} D_{1} A_{2} B_{2} C_{2} D_{2}$ is a cube inscribed in it, points $A_{1}, B_{1}, C_{1}, D_{1}$ lie on the lateral edges, and $A_{2}, B_{2}, C_{2}, D_{2}$ lie on the base of the pyramid, $P O \perp(A B C), \angle P A O=\alpha$ (Fig. 3.94)... | \frac{3\sqrt{2}\operatorname{ctg}\alpha}{(1+\sqrt{2}\operatorname{ctg}\alpha)^{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,253 |
3.95. The base of the pyramid is an isosceles triangle with an angle $\alpha$ between the lateral sides. The pyramid is placed in a certain cylinder such that its base is inscribed in the base of this cylinder, and the vertex coincides with the midpoint of one of the cylinder's generators. The volume of the cylinder is... | 3.95. By condition, $S A B C$ is a pyramid, $C A=C B, \angle A C B=a$, $\triangle A B C$ is inscribed in the base of the cylinder, $S \in K M, K M$ is the generatrix of the cylinder, $K S=S M, V_{\text {cyl }}=V$ (Fig. 3.95). Let the radius of the base of the cylinder be $R$, and its height be $H$; then $V=\pi R^{2} H$... | \frac{V}{3\pi}\sin\alpha\cos^{2}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,254 |
3.96. A cube is inscribed in a cone (one of the cube's faces lies in the base plane 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. | 3.96. By the condition, $SAB$ is a cone, $SO$ is its height, $\operatorname{CDEFC}_{1}D_{1}E_{1}F_{1}$ is a cube inscribed in the cone, $C_{1}, D_{1}, E_{1}, F_{1}$ are points lying on the lateral surface of the cone, $SO: CC_{1}=k$ (Fig. 3.96). In $\Delta \mathrm{SO}_{1}C_{1}$, we have $\operatorname{ctg} \angle ASO =... | \operatorname{arcctg}(\sqrt{2}(k-1)) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,255 |
3.97. 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 volume... | 3.97. Let $PMN$ be the axial section of a cone, $KLN_1M_1$ be the axial section of an inscribed cylinder, $\angle O_1MK=\alpha$, $PO$ be the height of the cone, and $\angle MPO=\beta$ (Fig. 3.97); we need to find $V_{\text{cone}}: V_{\text{cyl}}$. Let the radius of the base of the cone be $R$; then $OP=R \operatorname{... | \frac{\cos^{3}\alpha\cos^{3}\beta}{3\sin\alpha\sin\beta\cos^{2}(\alpha+\beta)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,256 |
3.98. The ratio of the volume of a right parallelepiped to the volume of a sphere inscribed in it is $k$. Find the angles at the base of the parallelepiped and the permissible values of $k$.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 3.98. Let's represent the axial section of the figure by a plane passing through the center of the inscribed sphere and parallel to the base. In the section, we obtain a parallelogram \(ABCD\), equal to the base of the right parallelepiped, and the great circle of the inscribed sphere inscribed in it with the center at... | \alpha=\arcsin\frac{6}{\pik},\angleADC=\pi-\arcsin\frac{6}{\pik},k\geq\frac{6}{\pi} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 50,257 |
3.100. 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. | 3.100. By the condition, $S A B C D-$ is a pyramid, $A B C D$ is a rectangle, $S E \perp(A B C), \angle S A E=\angle S B E=$ $=\angle S C E=\angle S D E=\beta$, from which it follows that $E$ is the point of intersection of the diagonals $A C$ and $B D$ (Fig. 3.100), $\angle A E B=\alpha$ and $O$ is the center of the s... | \frac{4}{3}R^{3}\sin^{2}2\beta\sin^{2}\beta\sin\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,259 |
3.101. 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. | 3.101. Given that $A B C A_{1} B_{1} C_{1}$ is a regular truncated triangular pyramid, $\angle B_{1} B C A=\alpha$, and $L$ is the center of the inscribed sphere (Fig. 3.101); we need to find the ratio $S_{\text {full trunc. pyr }}: S_{\text {sphere }}$. Let $A K$ and $A_{1} K_{1}$ be the heights of the bases; then $K_... | \frac{3\sqrt{3}}{2\pi}(3+4\operatorname{ctg}^{2}\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,260 |
3.102. 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 $k$. | 3.102. Let's depict the axial section of the figure; $S A=S B$ are the generators of the cone, $S O$ is the height of the cone, $O_{1}$ and $O_{2}$ are the centers of the circumscribed and inscribed spheres (Fig. 3.102), the ratio of their volumes is $k$; we need to find $\angle S A O$. Let $R$ and $r$ be the radii of ... | \angleSAO=\alpha=\arccos\frac{1\\sqrt{1-2\sqrt[3]{k}}}{2},0<k\leq\frac{1}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,261 |
3.103. 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 $k$. | 3.103. Let's depict the axial section of the figure: $S B=S C$ - the generatrix of the cone, $S P$ - its height, $O$ the center of the inscribed sphere (Fig. 3.103), $V_{\text {cone }}: V_{\text {sphere }}=k$; we need to find $\angle S B P$. Let $A$ be the point of tangency of the sphere with the lateral surface of the... | \alpha=2\operatorname{arctg}\sqrt{\frac{k\\sqrt{k^{2}-2k}}{2k}},k\geq2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,262 |
3.104. 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. | 3.104. Given, $A B C D-$
Fig. 3.104 is a truncated cone, $S$ - the center of the inscribed sphere (Fig. 3.104); $S_{\text {full trunc. cone }}: S_{\text {sphere }}=2 ;$ we need to find $\ang... | \alpha=\arcsin\frac{2\sqrt{5}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,263 |
3.105. Find the ratio of the volume of a spherical segment to the volume of the whole sphere, if the arc in the axial section of the segment corresponds to a central angle equal to $\alpha$.
To find the ratio of the volume of a spherical segment to the volume of the whole sphere, where the arc in the axial section of ... | 3.105. By the condition, \(A \cap B\) is a spherical segment, \(O\) is the center of the sphere, \(\cup A \cap B = \angle AOB = \alpha\) (Fig. 3.105). Let \(R\) be the radius of the sphere, \(OC\) be perpendicular to the base of the segment, \(O_1\) be the center of the base of the segment, \(O_1C = H\); then \(H = R -... | \sin^4\frac{\alpha}{4}(2+\cos\frac{\alpha}{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,264 |
4.1. Find the hypotenuse of a right triangle if the point of tangency of the inscribed circle divides one of the legs into segments of length $m$ and $n(m<n)$.
untranslated text:
4.1. Найти гипотенузу прямоугольного треугольника, если точка касания вписанной в него окружности детит один из катетов на отрезки длиной $... | 4.1. Let $P, N$ and $K$ be the points of tangency of the incircle with the sides of $\triangle ABC$ (Fig. 4.1). By the given condition, $KB = n, CK = m$, where $n > m$. Let $AB = x$; then $AN = AB - NB = x - n$. According to the Pythagorean theorem, we have $x^2 = (x - n + m)^2 + (m + n)^2; x^2 = x^2 + n^2 + m^2 -$
$$... | \frac{^2+n^2}{n-} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,265 |
4.2. The sum of the lengths of the legs of a right triangle is $8 \mathrm{~cm}$. Can the length of the hypotenuse be 5 cm? | 4.2. Let $a$ and $b$ be the legs, and $c$ be the hypotenuse of a right triangle. According to the condition, $a+b=8$, i.e., $b=8-a$. Suppose $c=5$; then we get the equation $a^{2}+(8-a)^{2}=5^{2}$, or $a^{2}+$ $+64-16a+a^{2}-25=0$, or $2a^{2}-16a+39=0$. But $\frac{D}{4}=64-2 \cdot 39<$ $<0$ and, therefore, this equatio... | no,itcannot | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,266 |
4.3. Prove that in a right triangle, the magnitude of the angle between the median and the altitude drawn to the hypotenuse is equal to the absolute value of the difference in the magnitudes of the acute angles of the triangle. | 4.3. Let $\angle C=\frac{\pi}{2}, CD$ be the altitude, and $CE$ be the median (Fig. 4.2). We need to prove that $\angle DCE=|\angle B-\angle A|$. Let $\angle DCE=\angle x$; then $\angle DCA=\angle B$ (since both angles complement angle $A$ to $\frac{\pi}{2}$). In a right triangle, the length of the median to the hypote... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,267 |
4.4. Prove that for any point $M$, belonging to an arbitrary triangle $A B C$ with sides $a, b$ and $c$ and heights $h_{a}, h_{b}$ and $h_{c}$, the equality $\frac{x}{h_{a}}+\frac{y}{h_{b}}+\frac{z}{h_{c}}=1$ holds, where $x, y$ and $z$ are the distances from point $M$ to the sides $B C, A C$ and $A B$, respectively. F... | ## 4.4. By connecting point $M$ with vertices $A, B$, and $C$, we obtain three triangles $BMC, AMC$, and $AMB$, the heights of which are respectively $x, y$, and $z$ (Fig. 4.3). Let $S$ be the area of triangle $ABC$; then $S=\frac{1}{2}(a x+b y+c z)$. On the other hand, $S=\frac{1}{2} a h_{a}, S=\frac{1}{2} b h_{b}$, $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,268 |
4.5. Prove that in any triangle, the ratio of the sum of all pairwise products of the lengths of the sides of the triangle to the sum of the lengths of its three altitudes is equal to the diameter of the circumscribed circle. | 4.5. We will prove that $\frac{a b+b c+a c}{h_{a}+h_{b}+h_{c}}=2 R$
(Fig. 4.4). We will use the Law of Sines: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$, from which we get
Fig. 4.4
$$
\frac{a c}{c \sin A}=\frac{a b}{a \sin B}=\frac{b c}{b \sin C}=2 R
$$
But $S_{\triangle A B C}=\frac{1}{2} a b \sin C=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,269 |
4.6. The lengths of the sides of an acute-angled triangle form an arithmetic progression with a common difference of 5 cm. Find the greatest number with the following property: the length of the largest side of any triangle of the specified type is greater than this number. | 4.6. Let in $\triangle A B C$ (Fig. 4.5) $A B=x, B C=$ $=x+5$ and $A C=x+10$, where $x>5$. According to the Law of Cosines, we have
$$
(x+10)^{2}=x^{2}+(x+5)^{2}-2 x(x+5) \cos B
$$
from which
$$
\cos B=\frac{x^{2}-10 x+75}{2 x(x+5)}=\frac{(x+5)(x-15)}{2 x(x+5)}=\frac{x-15}{2 x}
$$
; it is required to prove that $\triangle A B C$ is isosceles. In $\triangle A B D$ and $\triangle A C D$ by the cosine rule we have
$$
\begin{gathered}
a^{2}=c^{2}+m^{2}-2 m c \cos \alpha \\
a^{2}=b^{2}+m^{2}-2 m b \cos \alpha
\end{gathe... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,271 |
4.8. Find the angle at the vertex of an isosceles triangle if the medians drawn to the lateral sides are perpendicular to each other. | 4.8. Let $A C=x$ (Fig. 4.7); since $\angle O A F=\angle A O F=45^{\circ}$, we have $O F=A F=\frac{x}{2}$. But $B O=2 O F=x$, hence $B F=\frac{3 x}{2}$. Assuming $\angle A B F=\beta$, we find $\operatorname{tg} \beta=\frac{A F}{B F}=\frac{\frac{x}{2}}{\frac{3 x}{2}}=\frac{1}{3}$.
Answer: $\operatorname{arctg} \frac{1}{... | \operatorname{arctg}\frac{1}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,272 |
4.9. The circumference of each of two equal circles of radius $R$ passes through the center of the other. Find the area of the common part of these circles. | 4.9. Since in $\triangle O A O_{1}$ all sides are equal to $R$, then $\angle A O B=120^{\circ}$ (Fig. 4.8). We find
$$
S_{\triangle A O B}=\frac{1}{2} R^{2} \sin 120^{\circ}=\frac{1}{2} R^{2} \sin 60^{\circ}=\frac{R^{2} \sqrt{3}}{4}, S_{\mathrm{ceKT} A O B}=\frac{1}{3} \pi R^{2}
$$
from which $S_{\text {segm }}=\frac... | \frac{R^{2}(4\pi-3\sqrt{3})}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,273 |
4.10. From point $A$, two rays are drawn intersecting a given circle: one - at points $B$ and $C$, the other - at points $D$ and $E$. It is known that $A B=7, B C=7, A D=10$. Determine $D E$. | 4.10. Since $\triangle A D C \sim \triangle A B E$ (Fig. 4.9), we have $\frac{A E}{A B}=\frac{A C}{A D}$, where $A B=B C=7, A D=10$. Let $D E=x$; then $\frac{10+x}{7}=\frac{14}{10}$, from which $100+10 x=98$, i.e., $x=-0.2$. Such a value of $x$ indicates that point $E$ is located between $A$ and $D$, i.e., $\frac{A D}{... | 0.2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,274 |
4.12. Prove that in any trapezoid $A B C D(B C \| A D)$ triangles $A O B$ and $C O D$ are equal in area ($O$ - the point of intersection of the diagonals). | 4.12. Let's find the areas of $\triangle A B C$ and $\triangle B C D$ (Fig. 4.10): $S_{\triangle A B C}=\frac{1}{2} B C \cdot h, S_{\triangle B C D}=\frac{1}{2} B C \cdot h$, where $h=B K$, i.e., $S_{\triangle A B C}=S_{\triangle B C D}$. But $S_{\triangle A O B}=S_{\triangle A B C}-S_{\triangle B O C}, S_{\triangle C ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,276 |
4.13. The lateral edges of a triangular pyramid are pairwise perpendicular and have lengths $a, b$ and $c$. Find the volume of the pyramid. | 4.13. Let's consider the base of the pyramid to be the right triangle \( A D B \) (Fig. 4.11), and the vertex to be point \( C \). Since \( C D \perp A D \) and \( C D \perp B D \), by the theorem of perpendicularity of a line and a plane, \( C D \perp (A D B) \) and, therefore, the edge \( C D \) is the height of the ... | \frac{}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,277 |
4.15. Find the angle between the intersecting diagonals of adjacent faces of a cube. | 4.15. Let the edge of the cube be $a$; we need to find the angle between the diagonals $A_{1} B$ and $B_{1} C$ (Fig. 4.13). This angle is equal to the angle between the lines $D_{1} C$ and $B_{1} C$, since $D_{1} C \| A_{1} B$. In $\Delta B_{1} D_{1} C$, we have $B_{1} D_{1}=D_{1} C=B_{1} C=a \sqrt{2}$, and therefore, ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,279 |
4.16. The cube $A B C D A_{1} B_{1} C_{1} D_{1}\left(A A_{1}\left\|B B_{1}\right\| C C_{1} \| D D_{1}\right)$ is intersected by a plane passing through the vertices $A, C$ and the midpoint $E$ of the edge $D D_{1}$. Show that the volume of the pyramid $A C D E$ is equal to $\frac{1}{12}$ of the volume of the cube. | 4.16. We have (Fig. 4.14) $D E=\frac{1}{2} a$ (a- edge of the cube), $S_{\triangle A D C}=\frac{1}{2} a^{2}$, $V_{A C D E}=\frac{1}{3} \cdot \frac{1}{2} a \cdot \frac{1}{2} a^{2}=\frac{1}{12} a^{3}, V_{\text {cube }}=a^{3}$ Thus, $\frac{V_{A C D E}}{V_{\text {cube }}}=\frac{1}{12}$. | \frac{1}{12} | Geometry | proof | Yes | Yes | olympiads | false | 50,280 |
4.17. Given a regular tetrahedron \( S A B C \). At what angle is the edge \( A B \) seen from the midpoint of the edge \( S C \)? | 4.17. By the condition, $S M=M C$ (Fig. 4.15); we need to find $\angle A M B$. Let $a$ be the edge of a regular tetrahedron; then in $\triangle A M B$ we have $A M=M B=\frac{a \sqrt{3}}{2}$, from which, by the cosine theorem, we get
$A B^{2}=A M^{2}+M B^{2}-2 A M \cdot M B \cos \angle A M B$, or
 is a variable.
It is known that $\frac{S_{\triangle A_{1} B_{1} C_{1}}}{S_{\triangle A B C}}=\left(\frac{h}{H}\right)^{2}$, i.e., $S_{\triangle A_{1} B_{1} C_{1}}=h^{2} \frac{S_{\triangle A B C}}{H^{2}}$, where $S_{\triangle A B C}$ and $H$ are constant values... | f(x)=kx^{2}, | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,282 |
4.19. Find the area of the total surface of a cone if its lateral surface can be unfolded into a sector with a radius of 1 and a right central angle.
## Group 5 | 4.19. The sector of the cone's development has an angle of $90^{\circ}$ and radius $R=1$; therefore, its area $S_{1}=\pi R \cdot \frac{1}{4}=\frac{\pi}{4}$. The lateral surface area of the cone $S_{2}=\pi r l=\pi r$ (since $l=R=1$). But $S_{1}=S_{2}$, so $\frac{\pi}{4}=\pi r$, from which $r=\frac{1}{4}$. Now we find th... | \frac{5\pi}{16} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,283 |
4.20. The length of the hypotenuse of an isosceles right triangle is 40. A circle with a radius of 9 touches the hypotenuse at its midpoint. Find the length of the segment cut off by this circle on one of the legs. | 4.20. First, let's determine whether the problem has a solution given the values of the hypotenuse length and the radius of the circle. From geometric considerations (Fig. 4.17), it is clear that for the circle with center \( O \), lying on the height \( B K \) of triangle \( A B C \), to intersect the leg \( B C \) at... | \sqrt{82} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,284 |
4.21. The side, bisector, and height of a triangle, all emanating from the same vertex, are equal to 5, 5, and \(2 \sqrt{6}\) cm, respectively. Find the other two sides of the triangle. | 4.21. We find (Fig. 4.18) $A H=H D=\sqrt{25-24}=1$ (cm), i.e., $A D=2$ cm. Further, we have $\frac{B C}{D C}=\frac{A B}{A D}=\frac{5}{2}$. It is known that the square of the length of the angle bisector is expressed by the formula $B D^{2}=A B \cdot B C - A D \cdot D C$ (see formula (1.37)). Thus, $25=5 \cdot 5 x - 2 \... | 4\frac{8}{21} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,285 |
4.22. In a right-angled triangle, find the bisector of the right angle, if the hypotenuse of the triangle is equal to $c$, and one of the acute angles is equal to $\alpha$. | 4.22. In $\triangle A B C$ (Fig. 4.19), we have $A C=c \sin \alpha$, $B C=c \cos \alpha, B D=x, A D=c-x, l$ - the angle bisector of angle $C$. Since $\frac{x}{c-x}=\frac{c \cos \alpha}{c \sin \alpha}=\operatorname{ctg} \alpha$, then $x=$ $=c \operatorname{ctg} \alpha-x \operatorname{ctg} \alpha$, from which
} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,286 |
4.25. The centers of the circumcircle and the incircle of a triangle are symmetric with respect to one of the sides of the triangle. Find the angles of the triangle. | 4.25. Let $\angle B D C=x$; then $\angle B O C=$ $=2 x$ (Fig. 4.22). Therefore, $\angle O B C=$ $=90^{\circ}-x$, i.e., $\angle O_{1} B C=90^{\circ}-x$ (due to the symmetry of points $O$ and $O_{1}$ relative to side $B C$). Let's write the sum of the angles of $\triangle A B C$: $180^{\circ}-x+180^{\circ}-2 x+180^{\circ... | 108,36,36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,287 |
4.26. In square $A B C D$, points $M$ and $N$ are the midpoints of sides $D C$ and $B C$. Find $\angle M A N$. | 4.26. Let's introduce a coordinate system with the origin at point $A(0 ; 0)$ and coordinate axes directed along the sides $A D$ and $A B$ of the square (Fig. 4.23). In this system, the vector $\overline{A M}$ has coordinates $(0.5 ; 1)$, and the vector $\overline{A N}$ has coordinates $(1 ; 0.5)$. We will use the form... | \arccos0.8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,288 |
4.27. Each side of a convex quadrilateral is less than $a$. Prove that its area is less than $a^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 4.27. Let $x<a, y<a, z<a$, $t<a$ be the sides of a convex quadrilateral $A B C D$ (Fig. 4.24). The diagonal $A C$ divides the quadrilateral into two triangles $A B C$ and $A C D$, with areas $S_{\triangle A B C}=\frac{x y}{2} \sin B$ and $S_{\triangle A D C}=\frac{z t}{2} \sin D$. Therefore,
$$
S_{A B C D}=\frac{x y}{... | Logic and Puzzles | other | Yes | Yes | olympiads | false | 50,289 |
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