2014 AMC 12A Problem 20

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

Reflect

B

over line

AC

to

B'

and

C

over line

AB

to

C'

The path becomes

B'E+ED+DC'

, minimized as the straight segment

B'C'

; use the Law of Cosines with

\angle B'AC'=120^\circ

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.

Problem 20 in other years