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

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==Solution==
 
==Solution==
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Let <math>C_1</math> be the reflection of <math>C</math> across <math>\overline{AB}</math>, and let <math>C_2</math> be the reflection of <math>C_1</math> across <math>\overline{AC}</math>.  Then it is well-known that the quantity <math>BE+DE+CD</math> is minimized when it is equal to <math>C_2B</math>.  (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see [http://hom.wikidot.com/heron Heron's Shortest Path Problem].)  As <math>A</math> lies on both <math>AB</math> and <math>AC</math>, we have <math>C_2A=C_1A=CA=6</math>.  Furthermore, <math>\angle CAC_1=2\angle CAB=80^\circ</math> by the nature of the reflection, so <math>\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ</math>.  Therefore by the Law of Cosines <cmath>BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{14\textbf{ (D)}}.</cmath>
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Let <math>C_1</math> be the reflection of <math>C</math> across <math>\overline{AB}</math>, and let <math>C_2</math> be the reflection of <math>C_1</math> across <math>\overline{AC}</math>.  Then it is well-known that the quantity <math>BE+DE+CD</math> is minimized when it is equal to <math>C_2B</math>.  (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.)  As <math>A</math> lies on both <math>AB</math> and <math>AC</math>, we have <math>C_2A=C_1A=CA=6</math>.  Furthermore, <math>\angle CAC_1=2\angle CAB=80^\circ</math> by the nature of the reflection, so <math>\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ</math>.  Therefore by the Law of Cosines <cmath>BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{14\textbf{ (D)}}.</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2014|ab=A|num-b=19|num-a=21}}
 
{{AMC12 box|year=2014|ab=A|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:20, 17 February 2016

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 \[BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{14\textbf{ (D)}}.\]

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AMC 12 Problems and Solutions

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