Regular pyramid, centre of circumscribed ball=centre of inscribed ball

Question

Prove that if the sum of plane angles in the vertex of a regular pyramid equals $180 ^{\circ}$, then in this pyramid the centers of the inscribed and circumscribed balls are equal.

Could you help me a bit?

Thank you!

Answer 1

To prove it we should easy calculate all.

Let the pyramid's lateral edge is equals to one.

Calculate a length $a$ of a side of the regular polygon M with n sides which is the base of the pyramid: $$a=2\sin \frac{\phi}{2} = 2\sin \frac{\pi}{2n}$$

Here $\phi=\pi \,/\, n$ is a plane angle at the vertex of the pyramid.

Let R is a radius of the circumscribed circle of the polygon M. Then $a=2R\sin \frac{\pi}{n}$. Hence $$R=\frac{\sin \frac{\pi}{2n}}{\sin \frac{\pi}{n}} = \frac{1}{2\cos\frac{\pi}{2n}}$$ $$\cos\frac{\pi}{2n} = \frac{1}{2R}$$

Calculate a radius r of the inscribed circle of the polygon M:

$$r=R\cos\frac{\pi}{n}=R(2\cos^2\frac{\pi}{2n}-1)=R(\frac{1}{2R^2}-1)=\frac{1-2R^2}{2R}$$

Calculate a height of the pyramid by the Pythagorean theorem: $$H=\sqrt{ 1 - R^2}$$

Calculate a radius $\rho$ of the inscribed ball of the pyramid by using the formula that a radius of an inscribed circle for a triangle is equals to its area divided by semiperimeter:

$$\rho = \frac{rH}{r+\cos\frac{\pi}{2n}}=H\frac{1}{1+\frac{\cos\frac{\pi}{2n}}{r}} = H\frac{1}{1+\frac{1\,/\,2R}{(1-2R^2)\,/\,2R}}=H\frac{1-2R^2}{2(1-R^2)}$$

Here $\cos\frac{\pi}{2n}$ is a height of a lateral side of the pyramid.

By the formula that a radius of a circumscribed circle of a triangle is equals to a product of lengths of its sides divided by four times the area of a triangle we get a radius of the circumscribed ball of the pyramid:

$$\Omega = \frac{2R \times 1 \times 1}{4HR}=\frac{1}{2H}$$

Finally, to prove our theorem we need to check that $\rho + \Omega = H$:

$$\rho + \Omega= \frac{1}{2H}(2H^2\frac{1-2R^2}{2(1-R^2)} + 1) = \frac{1 - 2R^2 + 1}{2H} = \frac{1-R^2}{H}=H$$