# Consider an isosceles triangle

Consider an isosceles triangle. Let $r$ be the radius of its circumscribed circle and $p$ the radius of its inscribed circle. Prove that the distance $d$ between the centres of these two circles is $d =\sqrt {r(r-2p)}$.

I could not get any idea to solve. However I have tried to make a figure (partially).

• Your isosceles looks equilateral :P – Jaideep Khare Apr 1 '17 at 8:44
• @JaideepKhare, consider it as isosceles – pi-π Apr 1 '17 at 8:51
• Well... Making good diagram is a skill which save you from misunderstanding the problem... So yes... Consider making a good diagram. – Vidyanshu Mishra Apr 1 '17 at 9:27
• It's Euler's theorem. – StubbornAtom Apr 1 '17 at 12:01

This is another masterpiece of Euler. This is general result of what you have asked.

Source: H.S.M. Coxeter and S.L. Greitzer- Geometry Revisited.

• [+1] You have a good historical culture ! – Jean Marie Apr 1 '17 at 10:46
• Why is $R^2-d^2 = LI \times IA$? – rogerl Apr 1 '17 at 13:58
• @rogerl think about power of point, that's what you need to know to conclude what you are asking, I will include answer to that why of you don't understand it even after that hint. – Vidyanshu Mishra Apr 1 '17 at 16:14

Consider the following figure:

Using Euclid's theorem of sides in a right triangle one has $$2r(r+d+p)=b^2=4r^2-a^2=4r^2-\left({2r\over r+d}\>p\right)^2\ .$$ It follows that $$(r+d)^2(r+d+p)=2r\bigl((r+d)^2-p^2\bigr)=2r(r+d+p)(r+d-p)\ .$$ Removing the factor $r+d+p$ leads to $$r^2+2rd+d^2=2r(r+d-p)\ ,$$ from which the claim immediately follows.

• what does it mean 'Euclid's theorem of sides in a right triangle '? – pi-π Apr 2 '17 at 3:11
• It's the theorem that in a right triangle $b^2=cq$ where $c$ is the hypotenuse, $b$ a leg, and $q$ the projection of this leg onto the hypotenuse. The German word is "Kathetensatz". – Christian Blatter Apr 2 '17 at 6:36