Library / /8 of 48

, 2019

Geometry Difficulty 5.2 AIME, harder Prove it Greece

Let $\triangle AB\Gamma$ be a triangle with $AB > A\Gamma$ . Let point $\Delta$ be on the side $AB$ such that $B\Delta = A\Gamma$ . We draw the circle $\gamma$ passing through $\Delta$ and tangent to the side $A\Gamma$ at $A$ . The circumcircle $\omega$ of the triangle $AB\Gamma$ meets circle $\gamma$ at $A$ and $E$ . Prove that $E$ is the point of intersection of the perpendicular bisectors of the segments $B\Gamma$ and $A\Delta$ .

Solution

figure 3
It is enough to prove that $EB = E\Gamma$ and $EA = E\Delta$ . For that we compare the triangles $\triangle EB$ and $\triangle AE\Gamma$ which have: $B\Delta = A\Gamma$ , $\Delta\hat\{B\}E = A\hat\{\Gamma\}E$ (inscribed to the same arch $\omega$ ) and $B\hat\{\Delta\}E = 180^\circ - E\hat\{\Delta\}A = 180^\circ - E\hat\{A\}Z = E\hat\{A\}\Gamma$ , (we have used $E\hat\{\Delta\}A = E\hat\{A\}Z$ , inscribed angle - angle chord and tangent)
Therefore the triangles $\triangle EB$ and $\triangle AE\Gamma$ are equal and so $EB = E\Gamma$ and $E\Delta = EA$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.