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Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:

Equilateral triangles $\triangle ABC$ and $\triangle DEF$ are drawn such that points $B$ , $E$ , $F$ , and $C$ lie on a line in this order, and point $D$ lies inside triangle $\triangle ABC$ . If $BE = 14$ , $EF = 15$ , and $FC = 16$ , compute $AD$ .

Solution

Solution:

Extend $DE$ to meet $AC$ at $X$ . Observe that $ABEX$ and $DFCX$ are isosceles trapezoids (both with base angles of $60^\circ$ ), so we have

$AX = BE = 14$

$DX = FC = 16$

and $\angle AXD = 120^\circ$

By Law of Cosines on $\triangle ADX$ , the answer is

$AD = \sqrt{AX^2 + DX^2 - 2 \cos(120^\circ) \cdot AX \cdot DX}$

$= \sqrt{14^2 + 16^2 + 14 \cdot 16} = \boxed{26}.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.