Library / /16 of 55

, 2019

Geometry Difficulty 8.5 Shortlist Prove it IMO

Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$ , meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$ . The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$ . Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$ .

Solution

Notice that $\angle TFB = \angle FDA$ because $FT$ is tangent to circle $BDF$ , and moreover $\angle FDA = \angle CGA$ because quadrilateral $ADFG$ is cyclic. Similarly, $\angle TGB = \angle GEC$ because $GT$ is tangent to circle $CEG$ , and $\angle GEC = \angle CFA$ . Hence,
$\begin{equation*} \angle TFB = \angle CGA \quad \text{and} \quad \angle TGB = \angle CFA. \tag{1} \end{equation*}$

Triangles $FGA$ and $GFT$ have a common side $FG$ , and by (1) their angles at $F, G$ are the same. So, these triangles are congruent. So, their altitudes starting from $A$ and $T$ , respectively, are equal and hence $AT$ is parallel to line $BFGC$ .

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.