Oh, that angle!

Geometry Level 4

A regular Icosahedron initially has two of its opposing vertices aligned with the z-axis (vertical). You want to rotate it, such it sits on one of its equilateral triangular faces. Find the angle of rotation $\theta$ . What will be $\cos \theta$ ?

\dfrac{ \tan 30^{\circ} }{\tan 72^{\circ} }

\dfrac{ \sin 60^{\circ} }{\tan 36^{\circ} }

\dfrac{ \cos 60^{\circ} }{\tan 72^{\circ} }

\dfrac{ \tan 30^{\circ} }{\tan 36^{\circ} }

1 solution

Otto Bretscher
Oct 19, 2018

Let's give the sides a length of 1, for simplicity. Let $P$ be the lowest vertex (in the initial position, before the rotation), let $Q$ be the center of the lower horizontal pentagon, and let $R$ be the midpoint of a side of that pentagon. We have $PR=\frac{\sqrt{3}}{2}$ and $QR=\frac{1}{2\tan(\pi/5)}$ , the radius of the incircle of the pentagon. Now $\cos(\theta)=\frac{QR}{PR}=\frac{1}{\sqrt{3}\tan(\pi/5)}=\boxed{\frac{\tan(\pi/6)}{\tan(\pi/5)}}$ . A simple but very endearing problem! Thank you for sharing, Hosam!

Oh, that angle!

1 solution

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