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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 $d =\sqrt {r(r-2p)}$.

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

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

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    $\begingroup$ [+1] You have a good historical culture ! $\endgroup$ – Jean Marie Apr 1 '17 at 10:46
  • $\begingroup$ Why is $R^2-d^2 = LI \times IA$? $\endgroup$ – rogerl Apr 1 '17 at 13:58
  • $\begingroup$ @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. $\endgroup$ – Vidyanshu Mishra Apr 1 '17 at 16:14
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Consider the following figure:

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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.

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  • $\begingroup$ what does it mean 'Euclid's theorem of sides in a right triangle '? $\endgroup$ – pi-π Apr 2 '17 at 3:11
  • $\begingroup$ 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". $\endgroup$ – Christian Blatter Apr 2 '17 at 6:36

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