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A sphere of unit radius touches all edges of some triangular prism. What can be the volume of this prism?

The prism is regular because it is inscribed in the sphere. We express the radius by the Pythagorean theorem $1=\sqrt{(0,5 a)^{2}+\left(\frac{a}{\sqrt{3}}\right)^{2}}$. So, $a_{1}=-\frac{2 \sqrt{21}}{7}, a_{2}=\frac{2 \sqrt{21}}{7}$. And we have $\left(\frac{2 \sqrt{21}}{7}\right)^{2} \times 0,5 \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{7}$. Then the answer is $\frac{18 \sqrt{7}}{49}$. Am I right?

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1 Answer

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If the edge length is $2 a $, then the circumcircle of the the triangular face (which is an equilateral triangle) is

$ R = \sqrt{ 1^2 - a^2 }$

but $ R = \dfrac{2a}{\sqrt{3}} $

Hence, $1 - a^2 = \dfrac{4 a^2}{3} $

making $a^2 = \dfrac{3}{7} $

The volume is $(2 a) \dfrac{\sqrt{3}}{4} (4 a^2) = (6) (\dfrac{1}{\sqrt{7}}) \dfrac{3}{7} = \dfrac{ 18 }{7 \sqrt{7}} = \dfrac{ 18 \sqrt{7}}{49} $

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