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Problem 1838

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.5 Prove it IMO 2019 Shortlisted Problems · IMO · 2019

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 $BC$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official 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 $BFGC$ .

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