Question

Triangle $ABC$ has side lengths $AB = 11,$ $BC = 24,$ and $CA = 20.$ The bisector of $\angle BAC$ intersects $\overline{BC}$ in point $D,$ and intersects the circumcircle of $\triangle ABC$ in point $E \neq A.$ The circumcircle of $\triangle BED$ intersects the line $AB$ in points $B$ and $F \neq B.$ What is $CF?$

$28$

$20\sqrt{2}$

$30$

$32$

$20\sqrt{3}$

Solution:

Points $A,$ $D,$ $E$ are collinear on the bisector, and $A,$ $B,$ $F$ are collinear on line $AB.$ The power of $A$ with respect to the circle through $B,$ $E,$ $D$ gives $AB \cdot AF = AD \cdot AE.$

Since $\angle BAE = \angle DAC$ and $\angle AEB = \angle ACB$ (subtending $AB$ ), triangles $ABE$ and $ADC$ are similar, so $AD \cdot AE = AB \cdot AC.$ Therefore $AF = AC = 20.$

Place $A = (0, 0),$ $B = (11, 0).$ From $CA = 20,$ $CB = 24,$ point $C = \left(-\tfrac52, \tfrac{\sqrt{1575}}{2}\right).$ Point $F$ lies on ray $AB$ with $AF = 20,$ so $F = (20, 0).$

Then $CF^2 = \left(20 + \tfrac52\right)^2 + \tfrac{1575}{4} = \tfrac{2025}{4} + \tfrac{1575}{4} = 900,$ so $CF = 30.$

Thus, the correct answer is C.

2021 AMC 12B Fall Problem 24

Solution:

Problem 24 in Other Years