2014 AMC 12A Problems/Problem 20

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 1

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

Solution 2

In $\triangle BAC$ , the three lines look like the Chinese character 又. Let $\triangle DEA$ , $\triangle CDA$ , and $\triangle BEA$ have bases $DE$ , $CD$ , and $BE$ respectively. Then, $\triangle DEA$ has the same side $DA$ as $\triangle CDA$ and the same side $EA$ as $\triangle BEA$ . Connect all three triangles with $\triangle DEA$ in the center and the two triangles sharing one of its sides. Then, $\pentagon BACDE$ is formed with $BE+DE+CD$ forming the base.

Intuitively, the pentagon's base is minimized when all three bottom sides are collinear. This is simply the original $\triangle BAC$ except that $\angle BAC =120^\circ$ . (In $\triangle DEA$ , $\triangle CDA$ , and $\triangle BEA$ , $\angle A = 40^\circ$ , and the three triangles connect at $A$ to form the pentagon). Thus, $m\angle BAC = 40 \cdot 3$ ).

$BC$ in this new triangle is then the minimum of $BE+DE+CD$ . Applying law of cosines, $BC=\sqrt{6^2+10^2-2(6)(10)\cos (120^\circ)}=\sqrt{196}=14 \implies \boxed{\textbf{(D) }14}$

~bjhhar

Would prime notation be clearer?

Solution 3

$[asy] size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,Ep,Bp,Cp; A = (0,0); B = 10*dir(-110);C = 6*dir(-70); Bp = 10*dir(-30);Cp = 6*dir(-150); D = IP(Cp--Bp, A--B); Ep = IP(Cp--Bp, A--C); draw(A--B--C--A--Cp--Bp--A); draw(Cp--B); draw(C--Bp); draw(C--D); draw(B--Ep); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$B'$", Bp, E); dot("$C'$", Cp, W); dot("$D$", D, dir(-70)); dot("$E$", Ep, dir(60)); MA("40^\circ",Cp,A,D, 1); MA("40^\circ",D,A,Ep, 1); MA("40^\circ",Ep,A,Bp, 1); label("$6$", A--Cp); label("$10$", Bp--A); [/asy]$

(Diagram by shihan) Reflect $C$ across $AB$ to $C'$ . Similarly, reflect $B$ across $AC$ to $B'$ . Clearly, $BE = B'E$ and $CD = C'D$ . Thus, the sum $BE + DE + CD = B'E + DE + C'D$ . This value is minimized when $B'$ , $C'$ , $D$ and $E$ are collinear. To finish, we use the law of cosines on the triangle $AB'C'$ : $B'C' = \sqrt{6^2 + 10^2 - 2(6)(10)\cos 120^{\circ}} = \boxed{\textbf{(D) } 14}.$

Note (note by JiYang): $\quad \because \angle BAC \textless \frac{\pi}{3}$ $\qquad \therefore \angle C'AB' \textless \pi$ $\qquad \therefore$ the line C'B' would pass through $\Delta ABC$

Remark

This problem is similar to Fagnano's Problem.

~isabelchen

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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