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A sphere with a radius of 4 has a regular hexagonal pyramid with a height of 6 inscribed in it, and a second sphere is inscribed in the pyramid. Find the radius of the second sphere.

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Let $P A B C D E F$ be a regular hexagonal pyramid with vertex $P$ (Fig.1), $P H$ - its height, $M$ and $N$ - the midpoints of sides $A B$ and $D E$, respectively, $O$ - the center of the inscribed sphere. Denote $A B = A H = a, R$ - the radius of the circumscribed sphere. According to the problem, $P H = 6, R = 4$. Consider the section of the pyramid by a plane passing through the lateral edges $P A$ and $P D$ (Fig.2). Then the height $P H$, and therefore the center of the circumscribed sphere, lies in this plane. The section of the sphere circumscribed around the pyramid is a circle of radius $R$, circumscribed around the isosceles triangle $A P D$. Extend the height $P H$ to intersect this circle at point $Q$. Then $D H = a$ is the height of the right triangle $P D Q$, drawn from the vertex of the right angle $P D Q$. Therefore, $D H^2 = P H \cdot H Q$, or

$$
a^2 = P H (P Q - P H) = 6 (2 R - 6) = 6 (8 - 6) = 12
$$

from which $a = 2 \sqrt{3}$. Then

$$
M H = \frac{a \sqrt{3}}{2} = \frac{2 \sqrt{3} \sqrt{3}}{2} = 3
$$

The center $O$ of the sphere inscribed in the regular pyramid also lies on its height, and the points of tangency with the lateral faces lie on the apothems of the pyramid. Let the radius of the sphere be $r$. Consider the section of the pyramid by a plane passing through points $P, M$, and $N$ (Fig.3). We obtain an isosceles triangle $P M N$ and a circle inscribed in it with radius $r$ and center $Q$. Denote $\angle P M H = \beta$. From the right triangle $P M H$ we find that

$$
\tan \beta = \frac{P H}{M H} = \frac{6}{3} = 2
$$

The center of the circle inscribed in an angle lies on the bisector of this angle, so $\angle O_1 M H = \frac{\beta}{2}$. Substituting $\tan \beta = 2$ into the left side of the formula $\tan \beta = \frac{2 \tan \frac{\beta}{2}}{1 - \tan^2 \frac{\beta}{2}}$, we find that $\tan \frac{\beta}{2} = \frac{\sqrt{5} - 1}{2}$. Therefore,

$$
r = O_1 H = M H \tan \frac{\beta}{2} = 3 \cdot \frac{\sqrt{5} - 1}{2} = \frac{3(\sqrt{5} - 1)}{2}
$$

## Answer

$\frac{3(\sqrt{5} - 1)}{2} = \frac{6}{\sqrt{5} + 1}$

## Regular Pyramid $\quad]$  Problem 110448 Topics: [ Inscribed Sphere in Pyramid ]  Difficulty: 3  [  Classes: 10,11

A regular hexagonal pyramid is circumscribed around a sphere of radius 2 with a height of 5, and a second sphere is circumscribed around it. Find the radius of the second sphere.

## Answer

$\frac{31}{6}$

## [Theorem of Three Perpendiculars ]

![](https://cdn.mathpix.com/cropped/2024_05_06_dc8ed1d6cdac2bf77445g-01.jpg?height=116&width=644&top_left_y=2678&top_left_x=472)
Difficulty: 3 Classes: 10,11 From point $M$ to plane $\alpha$ a perpendicular $M H$ of length $\sqrt{3}$ is dropped, and two oblique lines are drawn, making angles of $60^{\circ}$ with the perpendicular. The angle between the oblique lines is $120^{\circ}$. a) Find the distance between the bases

$A$ and $B$ of the oblique lines. b) On segment $A B$ as the leg in plane $\alpha$ a right triangle $A B C$ (angle $A$ is right) is constructed. Find the volume of the pyramid $M A B C$, knowing that $\cos \angle B M C = \frac{1}{-3}$.

## Solution

From the right triangles $A H M$ and $B H M$ (Fig. 1) we find that

$$
A M = B M = \frac{M H}{\cos 60^{\circ}} = \frac{\sqrt{3}}{\frac{1}{2}} = 2 \sqrt{3}, A H = B H = 3
$$

By the cosine rule

$$
A B = \sqrt{A M^2 + B M^2 - 2 A M \cdot B M \cos 120^{\circ}} = \sqrt{12 + 12 + 2 \cdot 2 \sqrt{3} \cdot 2 \sqrt{3} \cdot \frac{1}{2}} = \sqrt{36} = 6
$$

Since $A H + B H = 3 + 3 = 6 = A B$, points $A, H$, and $B$ lie on the same line, with $H$ being the midpoint of segment $A B$, and points $A, H, B$, and $M$ lie in the same plane (Fig.2). Since $A H$ is the orthogonal projection of the oblique line $A M$ onto plane $\alpha$ and $A H \perp A C$, by the theorem of three perpendiculars $A M \perp A C$. Therefore, triangle $M A C$ is also a right triangle. Let $A C = x$. By the Pythagorean theorem

$$
M C = \sqrt{A C^2 + A M^2} = \sqrt{x^2 + 12}, B C = \sqrt{A C^2 + A B^2} = \sqrt{x^2 + 36}
$$

By the cosine rule

$$
B C^2 = M C^2 + M B^2 - 2 M C \cdot M B \cos \angle B M C
$$

or

$$
x^2 + 36 = (x^2 + 12) + 12 - 2 \sqrt{x^2 + 12} \cdot 2 \sqrt{3} \cdot \left(-\frac{1}{3}\right).
$$

From this equation, we find that $x = \sqrt{15}$. Therefore,

$$
V_{M A B C} = \frac{1}{3} S_{\triangle A B C} \cdot M H = \frac{1}{3} \cdot \frac{1}{2} A B \cdot A C \cdot M H = \frac{1}{6} \cdot 6 \cdot \sqrt{15} \cdot \sqrt{3} = 3 \sqrt{5}
$$

## Answer

a) 6 ; b) $3 \sqrt{5}$.