Question

In $\triangle BAC,$ $\angle BAC=40^\circ,$ $AB=10,$ and $AC=6.$ Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC},$ respectively. What is the minimum possible value of $BE+DE+CD?$

$6\sqrt3+3$

$\dfrac{27}{2}$

$8\sqrt3$

$14$

$3\sqrt3+9$

Answer: D

Concepts:reflection (geometry)law of cosines optimization

Difficulty rating: 2110

Solution:

Reflect $B$ across line $AC$ to get $B',$ and reflect $C$ across line $AB$ to get $C'.$ Then $BE=B'E$ and $CD=C'D,$ so $BE+DE+CD$ $=B'E+ED+DC',$ a broken path from $B'$ to $C'.$

This is minimized when the path is the straight segment $B'C'.$ We have $AB'=AB=10,$ $AC'=AC=6,$ and $\angle B'AC'=3\cdot40^\circ=120^\circ.$

By the Law of Cosines, $\begin{gathered} B'C'^2=10^2+6^2\\ {}-2\cdot10\cdot6\cos120^\circ\\ =136+60\\ =196, \end{gathered}$ so $B'C'=14.$

Thus, the correct answer is D.

2014 AMC 12A Problem 20

Answer: D

Difficulty rating: 2110

Solution:

Problem 20 in Other Years