Question

What is the ratio of the inscribed sphere's radius to the regular octahedron's edge length?


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Answer 1

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Triangle ABC, where A is a vertex, B is a mid-point of a side, and CC the center of the octahedron, has a right angle at C and sides CB

$= > \frac12CB$

and

$AB=\frac{\sqrt3}2$

Point of contact between the sphere and the octahedron, D is in the center of the face, therefore the median AB is divided in the ration of 2:1 making

$DB= \frac{1}{3} \\ \\ AB=\frac1{2\sqrt3}$

Triangle CBD has a right angle at D, so

$CD=\sqrt{CB^2-DB^2} \\ \\ =\sqrt{\frac14-\frac1{12}} \\ \\ =\boxed{\frac1{\sqrt6}}$

So your final answer is

1/√6.