AoPS - Problem

Source: Spanish Mathematical Olympiad. 2015. Problem 2

Tags: geometry, incenter, circumcircle

tchebytchev

30.04.2015 18:00

In triangle $ABC$, let $A'$ is the symmetrical of $A$ with respect to the circumcenter $O$ of $ABC$. Prove that: a) The sum of the squares of the tangents segments drawn from $A$ and $A'$ to the incircle of $ABC$ equals $$4R^2-4Rr-2r^2$$ where $R$ and $r$ are the radii of the circumscribed and inscribed circles of $ABC$ respectively. b) The circle with center $A'$ and radius $A'I$ intersects the circumcircle of $ABC$ in a point $L$ such that $$AL=\sqrt{ AB.AC}$$ where $I$ is the centre of the inscribed circle of $ABC$.

My solution: Let $ \mathcal{P} (P, \odot) $ be the power of a point $ P $ WRT circle $ \odot $ . Let $ M=AI \cap \odot (ABC) $ and $ I_a $ be the A-excenter of $ \triangle ABC $ . (a) From Stewart theorem we get $ \tfrac{1}{2} {AI}^2 + \tfrac{1}{2} {A'I}^2 -R^2={OI}^2=R^2-2Rr $ , so $ \mathcal{P} (A, \odot(I) ) + \mathcal{P} (A', \odot(I) ) =( {AI}^2-r^2)+({A'I}^2-r^2)=4R^2-4Rr-2r^2 $ . (b) Notice that $ AI \cdot IM=-\mathcal{P}(I, \odot (O))=2Rr $ . From the proof of (a) we get $ {AI}^2+{A'I}^2=4R^2-4Rr $ , so $ {AL}^2={AA'}^2-{A'I}^2=4R^2-{A'I}^2=4R^2-{A'I}^2-{AI}^2+{AI}^2 $ $ =4Rr+{AI}^2=2AI \cdot IM+{AI}^2=AI \cdot II_a+AI^2=AI \cdot AI_a=AB \cdot AC $ . Q.E.D