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

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, $m\angle BAC = 40 * 3$).


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


~bjhhar

Would prime notation be clearer?

Solution 3

[asy] import graph; size(20.95cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -1.52, xmax = 19.43, ymin = -2.35, ymax = 10.68; /* image dimensions */   draw(arc((8.03,9.81),0.61,-124.95,-84.95)--(8.03,9.81)--cycle);  draw(arc((8.03,9.81),0.61,-164.95,-124.95)--(8.03,9.81)--cycle);  draw(arc((8.03,9.81),0.61,-84.95,-44.95)--(8.03,9.81)--cycle);  /* draw figures */ draw((8.03,9.81)--(2.3,1.61));  draw((8.03,9.81)--(8.56,3.83));  draw((2.3,1.61)--(8.56,3.83));  draw((8.03,9.81)--(2.24,8.25));  draw((8.03,9.81)--(15.11,2.74));  draw((2.24,8.25)--(15.11,2.74));  /* dots and labels */  label("$A$", (7.9,10.03), NE * labelscalefactor);   label("$B$", (1.91,1.75), NE * labelscalefactor);  label("$40^\circ$", (7.58,8.84), NE * labelscalefactor);   label("$C$", (8.82,3.42), NE * labelscalefactor);   label("$B'$", (15.47,2.6), NE * labelscalefactor);   label("$C'$", (1.85,8.42), NE * labelscalefactor);  label("$6$", (5.05,9.35), NE * labelscalefactor);  label("$10$", (11.45,6.7), NE * labelscalefactor);   label("$D$", (6.21,6.68), NE * labelscalefactor);   label("$E$", (7.86,6.11), NE * labelscalefactor);  label("$40^\circ$", (6.99,9.23), NE * labelscalefactor);  label("$40^\circ$", (8.23,8.86), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  [/asy]

(Diagram by dasobson) 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 $AB'C'$: $B'C' = \sqrt{6^2 + 10^2 - 2(6)(10)\cos 120} = 14$

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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

Invalid username
Login to AoPS