# A question on an isosceles obtuse angled triangle.

Given an isosceles triangle with equal sides of length $b$ and base angle $A$ $(<π/4)$. $O$ and $I$ are the circumcenter and incenter of the triangle respectively. Then true or false $$OI= \left|\frac{b \cos(3A/2)}{2\sin(A) \cos (A/2)}\right|.$$

I have found out the values of the inradius and circumradius to be $b \sin(2 A)/2(1+\cos A)$ and $b/2 \sin(A)$ respectively and also that the isosceles triangle is obtuse angled. I also know that in an isosceles triangle the circumcenter and incenter lie on the same straight line. But I have no idea how to find $OI$. Please help.

I think you are right.

Let in $\Delta ABC$ we have: $AC=CB=b$, $\measuredangle A=\alpha$,

$D$ be midpoint of $AB$ and $E$ be a midpoint of $BC$.

Thus, $CD=b\sin\alpha$, $CO=\frac{b}{2\sin\alpha}$, $ID=b\cos\alpha\tan\frac{\alpha}{2}$ and $$OI=|CD-CO-ID|=\left|b\sin\alpha-\frac{b}{2\sin\alpha}-b\cos\alpha\tan\frac{\alpha}{2}\right|=$$ $$=\frac{b}{2\sin\alpha\cos\frac{\alpha}{2}}\left|2\sin^2\alpha\cos\frac{\alpha}{2}-\cos\frac{\alpha}{2}-2\sin\alpha\cos\alpha\sin\frac{\alpha}{2}\right|=$$ $$=\frac{b}{2\sin\alpha\cos\frac{\alpha}{2}}\left|\cos2\alpha\cos\frac{\alpha}{2}+\sin2\alpha\sin\frac{\alpha}{2}\right|=$$ $$=\frac{b\left|\cos\frac{3\alpha}{2}\right|}{2\sin\alpha\cos\frac{\alpha}{2}},$$ which you got.

• By ID you mean inradius right? Sep 12, 2017 at 17:20
• @Lokesh Sangewar Yes, of course. Sep 12, 2017 at 17:22
• hahah....ok thanks a lot for your help. Sep 12, 2017 at 17:23
• @Lokesh Sangewar You are welcome! Sep 12, 2017 at 17:23

this distance is given by $$d=R^2-2Rr$$ this is the formula of Euler

• Any short and sweet proof for this? Sep 12, 2017 at 16:38