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Sphere inscribed in pyramid

I need help:

Sphere is inscribed in right pyramid with equilateral triangle in base.We know :

angle of inclination = $\displaystyle \alpha$.

side of the triangle = a

Is it possible express the radius of sphere using a and \alfa? Or do we have to know the $\displaystyle \beta$? Use trygonometric functions.

Attachment 24873

Re: Sphere inscribed in pyramid

Hi, aleschio.

I don't think we need to know $\displaystyle \beta.$ I will use $\displaystyle m$ to denote the measure of a length. My idea is something like this:

1) Since the base is equilateral we know $\displaystyle m(AD)=\frac{\sqrt{3}}{2}a$ (see Equilateral triangle - Wikipedia, the free encyclopedia). There are many ways to show that $\displaystyle m(DE)=\frac{1}{3}m(AD)$ (for example see Viviani's Theorem Viviani's theorem - Wikipedia, the free encyclopedia). Hence

$\displaystyle m(DE)=\frac{\sqrt{3}}{6}a.$

2) Since the tangents to the circle DF and DA are drawn from the same point (i.e. D), we know $\displaystyle DG\cong DE.$ Therefore, $\displaystyle m(DG)=\frac{\sqrt{3}}{6}a$ and $\displaystyle \Delta DGO\cong \Delta DEO.$ Since $\displaystyle \Delta DGO\cong \Delta DEO$ we also know that $\displaystyle m(\angle GDO)=\alpha/2.$

3) We now use the right triangle $\displaystyle \Delta DGO$ and the tangent function to obtain the radius.

Does this answer your question? Let me know if anything is unclear/incorrect.

Good luck!

Re: Sphere inscribed in pyramid

Thanks GJA :)

I have a problem that I have to count the ratio of this pyramid to sphere inscribed in it when $\displaystyle \alpha$ and $\displaystyle \beta$ are given. I wondered if it's possible to count it when only $\displaystyle \alpha$ is known. Your solution says that it is possible and fairly easy. But I have one more question, Is obvious that sphere is tangent to the base in point which is the end of the height of this pyramid or do we have to proof it?