Christmas Streak 75/88: Geometry Rush #5

Geometry Level 4

For a triangle $\rm ABC$ and some point $\rm D,$ $\overline{\rm AD}$ bisects the exterior angle of $\angle \rm A.$

$\rm \overline{AD}//\overline{BE}//\overline{FG}$ and $\rm \overline{AB}//\overline{EF}$ where points $\rm E,~G$ are on $\overline{\rm AC}$ and point $\rm F$ is on $\overline{\rm CD}.$

Also, $\rm \overline{BD}=12,~\overline{CB}=5,~\overline{AB}=4,~\overline{CD}=17.$

The length of $\overline{\rm CG}$ is $\dfrac{p}{q}$ for some coprime positive integers $p,~q.$

Find the value of $p+q.$

This problem is a part of <Christmas Streak 2017> series .

The answer is 76.

1 solution

Miles Koumouris
Dec 22, 2017

Since $BD+CB=CD=12+5=17$ , it is evident that $B$ lies on $CD$ . Note that $\triangle ADC\sim \triangle EBC\sim \triangle GFC,$ and since $AB$ and $EF$ are common, the side lengths of these triangles follow a geometric progression with common ratio $\tfrac{5}{17}$ . Therefore, $CG=\dfrac{25}{289}CA.$ By the Exterior Angle Bisector theorem, we have $\dfrac{AC}{AB}=\dfrac{CD}{BD},$ and thus $AC=4\times \dfrac{17}{12}=\dfrac{17}{3}.$ Therefore, $CG=\dfrac{25}{289}\times \dfrac{17}{3}=\dfrac{25}{51}.$ Hence, $p+q=25+51=\boxed{76}$ .

Christmas Streak 75/88: Geometry Rush #5

The answer is 76.

1 solution

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