[Math] Area of the triangle formed by circumcenter, incenter and orthocenter

Question

Lets say we have $\triangle$$ABC$ having $O,I,H$ as its circumcenter, incenter and orthocenter. How can I go on finding the area of the $\triangle$$HOI$.

I thought of doing the question using the distance (length) between $HO$,$HI$ and $OI$ and then using the Heron's formula, but that has made the calculation very much complicated. Is there any simple way to crack the problem?

Answer 1

Hint:

$AH=2R\cos A$, $AI=4R\sin\frac{B}{2}\sin\frac{C}{2}$, $OA=R$, $\angle OAH=2\angle OAI=B-C$

$\triangle HOI=\triangle AHO-\triangle AHI-\triangle AIO$

Final answer $$2R^2\sin\dfrac{B-C}{2}\sin\dfrac{C-A}{2}\sin\dfrac{A-B}{2}$$