Plane Geometry problem I came across this problem in a mathematics-related facebook group. Could anyone advise on the solution to it(i.e. hints only)? Thank you.



Since triangle ABC is equilateral, we know that angle BAC is 60º.  Therefore, angle PAQ is also 60º; we have a circle theorem that tells us that angle POQ is thus twice that measure or 120º.  The diameter AD bisects that angle, so angle POD is 60º.
Diameter AD is also an altitude of equilateral triangle ABC:  thus, if  $ \ s \ $ is the length of one side of that triangle, the diameter of the circle is $ \ \frac{\sqrt{3}}{2}s \ $ . The radius OD therefore has length $ \ \frac{\sqrt{3}}{4}s \ $ .  Since the altitude AD bisects side BC, then the segment BD has length $ \ \frac{1}{2}s \ $ .  Hence, $ \ \tan(\angle BOD ) \ = \ \frac{2}{\sqrt{3}} \  $ .
Now, the measure of angle POB, designated as $ \ \phi \ , $ is the difference given by the measure of angle POD (call it $ \ \alpha \ $ ) minus the measure of angle BOD (call that $ \ \beta \ $ ) .  We can then use the formula for the tangent of the difference of two angles, with $ \ \tan(\angle POD) \ = \ \tan(\alpha) \ = \ \tan \ 60º \ = \ \sqrt{3} \ $ , to obtain
$$ \tan \phi \ = \ \tan( \alpha - \beta ) \ = \ \frac{\tan \alpha \ - \ \tan \beta}{1 \ + \ \tan \alpha \cdot \tan \beta} \ = \ \frac{\sqrt{3} \ - \ \frac{2}{\sqrt{3}}}{1 \ + \ \sqrt{3} \cdot \frac{2}{\sqrt{3}}} \ = \ \frac{ \frac{3 \ - \ 2}{\sqrt{3}}}{1 \ + \ 2 } \ = \ \frac{1}{3 \ \sqrt{3}} \ .  $$
We thereby conclude that  $ \ \cot \phi \ = \ 3 \ \sqrt{3} \ . $
