Difference between revisions of "2014 AMC 12A Problems/Problem 20"

Revision as of 17:52, 7 February 2014

Problem

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 $BE+DE+CD$ ?

$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$

Solution

Let $C_1$ be the reflection of $C$ across $\overline{AB}$ , and let $C_2$ be the reflection of $C_1$ across $\overline{AC}$ . Then it is well-known that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$ . (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.) As $A$ lies on both $AB$ and $AC$ , we have $C_2A=C_1A=CA=6$ . Furthermore, $\angle CAC_1=2\angle CAB=80^\circ$ by the nature of the reflection, so $\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ$ . Therefore by the Law of Cosines $BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{14\textbf{ (D)}}.$

2014 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
-{{AMC12 box|year=2014|ab=A|num-b=17|num-a=19}}
+{{AMC12 box|year=2014|ab=A|num-b=19|num-a=21}}
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Difference between revisions of "2014 AMC 12A Problems/Problem 20"

Revision as of 17:52, 7 February 2014

Problem

Solution

See Also