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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle with $BC = 2007$ , $CA = 2008$ , $AB = 2009$ . Let $\omega$ be an excircle of $ABC$ that touches the line segment $BC$ at $D$ , and touches extensions of lines $AC$ and $AB$ at $E$ and $F$ , respectively (so that $C$ lies on segment $AE$ and $B$ lies on segment $AF$ ). Let $O$ be the center of $\omega$ . Let $\ell$ be the line through $O$ perpendicular to $AD$ . Let $\ell$ meet line $EF$ at $G$ . Compute the length $DG$ .

Solution

Solution:

Answer: $2014024$

Let line $AD$ meet $\omega$ again at $H$ . Since $AF$ and $AE$ are tangents to $\omega$ and $ADH$ is a secant, we see that $DEHF$ is a harmonic quadrilateral. This implies that the pole of $AD$ with respect to $\omega$ lies on $EF$ . Since $\ell \perp AD$ , the pole of $AD$ lies on $\ell$ . It follows that the pole of $AD$ is $G$ .

Thus, $G$ must lie on the tangent to $\omega$ at $D$ , so $C, D, B, G$ are collinear. Furthermore, since the pencil of lines $(AE, AF ; AD, AG)$ is harmonic, by intersecting it with the line $BC$ , we see that $(C, B ; D, G)$ is harmonic as well. This means that
$\frac{BD}{DC} \cdot \frac{CG}{GB} = -1$
(where the lengths are directed.) The semiperimeter of $ABC$ is $s = \frac{1}{2}(2007 + 2008 + 2009) = 3012$ . So $BD = s - 2009 = 1003$ and $CD = s - 2008 = 1004$ . Let $x = DG$ , then the above equation gives
$\frac{1003}{1004} \cdot \frac{x + 1004}{x - 1003} = 1$
Solving gives $x = 2014024$ .

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