Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it

One, (40 points) As shown in Figure $1, \triangle ABC (AB > AC)$ has an incircle $\odot I$ that touches sides $BC$ , $CA$ , and $AB$ at points $D$ , $E$ , and $F$ respectively. Through a point $P$ on the extension of $BC$ , draw another tangent to $\odot I$ , which touches $\odot I$ at point $G$ and intersects $AB$ and $AC$ at points $M$ and $N$ respectively. Let $MD$ intersect $BG$ at point $Q$ , and $ND$ intersect $CG$ at point $R$ . Prove that $P$ , $R$ , and $Q$ are collinear if and only if $A$ , $G$ , and $D$ are collinear.

Solution

First, prove that $P, Q, F$ and $P, E, R$ are collinear respectively.
By the tangent length theorem, we have
$\begin{array}{l} P G=P D, M G=M F, B D=B F . \\ \text { Then } \frac{P G}{G M} \cdot \frac{M F}{F B} \cdot \frac{B D}{D P}=1 . \end{array}$

By the converse of Ceva's theorem, we know that $B G, M D, P F$ are concurrent, i.e., $P, Q, F$ are collinear.
Similarly, $P, E, R$ are collinear.
Thus, $P, R, Q$ are collinear
$\Leftrightarrow P, E, F$ are collinear.
Next, prove:
$P, E, F$ are collinear
$\Leftrightarrow A, G, D$ are collinear.
In fact,
$\begin{array}{l} P, E, F \text { are collinear } \Leftrightarrow I A \perp P F \\ \Leftrightarrow A P^{2}-A F^{2}=I P^{2}-I F^{2}, \\ A, G, D \text { are collinear } \Leftrightarrow I P \perp A D \\ \Leftrightarrow P A^{2}-P D^{2}=I A^{2}-I D^{2} . \end{array}$

Combining $I D=I F$ and
$I P^{2}-P D^{2}=I D^{2}=I F^{2}=I A^{2}-A F^{2} \text {, }$

we know that equation (1) $\Leftrightarrow$ equation (2).
In summary, the necessary and sufficient condition for $P, R, Q$ to be collinear is that $A, G, D$ are collinear.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.