My Math Forum Diameter of inscribed sphere

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 June 4th, 2012, 06:19 AM #1 Newbie   Joined: Jun 2012 Posts: 1 Thanks: 0 Diameter of inscribed sphere The right four edges pyramid has all edges square root of 5. What is the diameter of inscribed sphere? Can someone help me in solving the task? Thanks. P.S. sorry for my English, I hope that you understand the task.
 June 4th, 2012, 07:44 PM #2 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: Diameter of inscribed sphere Can you be more specific as to which pyramid this is? For instance, there are several different types of pyramids, all depending upon the base of the pyramid. I assume this is a square pyramid. However, I do not think this is enough information. My thoughts show that the diameter of the inscribed sphere is equal to $\sqrt{2}s$ where s is the length of any side of the square base. I cannot calculate s in terms of $\sqrt{5}$, the length of the edges. There is not an obvious relationship between these two that allows one to calculate the diameter of the inscribed sphere. I am not saying this is impossible. I am saying I do not see how to do it. You may try to figure out a method via Mathworld or Wikipedia. It may very well be possible, who knows.
June 4th, 2012, 08:45 PM   #3
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Re: Diameter of inscribed sphere

If we assume a pyramid having a square base and unknown height h, the sides of which are $\sqrt{5}$ units in length, then if we cut the pyramid and inscribed sphere with a vertical plane passing through the center of the sphere and perpendicular to two opposing sides of the base, the cross-section becomes the problem of a circle inscribed in an isosceles triangle:

[attachment=0:30166kt0]sphereinpyramid.jpg[/attachment:30166kt0]

By similarity and Pythagoras, we may state:

$\frac{r}{h-r}=\frac{\frac{\sqrt{5}}{2}}{\sqrt{\frac{5}{4}+h^2 }}$

$r\sqrt{\frac{5}{4}+h^2}=(h-r)\frac{\sqrt{5}}{2}$

$$$\sqrt{\frac{5}{4}+h^2}+\frac{\sqrt{5}}{2}$$r=\fr ac{\sqrt{5}}{2}h$

$r=\frac{\frac{\sqrt{5}}{2}h}{\sqrt{\frac{5}{4}+h^2 }+\frac{\sqrt{5}}{2}}=\frac{h}{\sqrt{\frac{4}{5}h^ 2+1}+1}$

Hence, the diameter d of the sphere as a function of the unknown height of the pyramid is:

$d=2r=\frac{2h}{\sqrt{\frac{4}{5}h^2+1}+1}$
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a sphere is inscribed inside a pyramid with a square as a base

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