Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it Bulgaria

Problem:
Let $k$ be the incircle of $\triangle ABC$ with $AC \neq BC$ , $I$ be the center of $k$ and let $D$ , $E$ and $F$ be the tangent points of $k$ to the sides $AB$ , $BC$ and $AC$ , respectively.

a) If $S = CI \cap EF$ , prove that $\triangle CDI \sim \triangle DSI$ .

b) Let $M$ be the second intersection point of $k$ and $CD$ . The tangent line to $k$ at $M$ intersects the line $AB$ at point $G$ . Prove that $GS \perp CI$ .

Solution

Solution:

a) From the right $\triangle CEI$ we have $EI^2 = SI \cdot CI = DI^2$ . Then we get $\frac{DI}{SI} = \frac{CI}{DI}$ and therefore $\triangle CDI \sim \triangle DSI$ .

b) The quadrilateral $DIM G$ is cyclic. Since a) implies that $\angle ISD = \angle IDC = \angle IMD$ we conclude that $S$ lies on the circumcircle of $DIM G$ . It is now obvious that $\angle GSI = \angle GMI = 90^\circ$ .

Remark. It is easy to see that b) implies that the points $E$ , $F$ and $G$ are colinear.

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