Question

Equilateral $\triangle ABC$ is inscribed in a circle of radius $2.$ Extend $\overline{AB}$ through $B$ to point $D$ so that $AD = 13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $\ell_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $\ell_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $\ell_1$ and $\ell_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p,$ $q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime, find $p + q + r.$

Answer

Solution:

By construction $ADFE$ is a parallelogram with $AD = 13,$ $DF = AE = 11,$ and $\angle ADF = 180^\circ - \angle DAE = 120^\circ.$ Hence $[ADF] = \frac{1}{2} \cdot 13 \cdot 11 \sin 120^\circ = \frac{143\sqrt{3}}{4},$ and by the law of cosines, $AF^2 = 13^2 + 11^2 - 2 \cdot 13 \cdot 11 \cos 120^\circ = 169 + 121 + 143 = 433.$

Since $G$ lies on the circle, inscribed angles give $\angle GCB = \angle GAB = \angle FAD$ (both subtend arc $GB$ ) and $\angle CBG = \angle CAG$ (both subtend arc $CG$ ); and $\angle CAG = \angle AFD$ because $\overline{AE} \parallel \overline{DF}.$ So $\triangle CBG \sim \triangle AFD$ with ratio $\frac{CB}{AF}.$ The side of an equilateral triangle inscribed in a circle of radius $2$ is $BC = 2\sqrt{3}.$

Therefore $[CBG] = \left(\frac{2\sqrt{3}}{\sqrt{433}}\right)^2 \cdot \frac{143\sqrt{3}}{4} = \frac{12}{433} \cdot \frac{143\sqrt{3}}{4} = \frac{429\sqrt{3}}{433},$ and $p + q + r = 429 + 3 + 433 = 865.$

2006 AIME II Problem 12

Solution:

Problem 12 in Other Years