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

Revision as of 12:17, 29 August 2021

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

$[asy] size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,Ep,Bp,Cp; A = (0,0); B = 10*dir(-110);C = 6*dir(-70); Bp = 10*dir(-30);Cp = 6*dir(-150); D = IP(Cp--Bp, A--B); Ep = IP(Cp--Bp, A--C); draw(A--B--C--A--Cp--Bp--A); dot("$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(-130)); 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 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 $AB'C'$ : $B'C' = \sqrt{6^2 + 10^2 - 2(6)(10)\cos 120} = 14$

@@ Line 28: / Line 28: @@
 ==Solution 3==
 <asy>
-import graph; size(10cm);
+size(300);
-real labelscalefactor = 0.75; /* changes label-to-point distance */
+defaultpen(linewidth(0.4)+fontsize(10));
-pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */
+pen s = linewidth(0.8)+fontsize(8);
-pen dotstyle = black; /* point style */
-real xmin = -1.52, xmax = 19.43, ymin = -2.35, ymax = 10.68; /* image dimensions */
+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(arc((8.03,9.81),0.61,-124.95,-84.95)--(8.03,9.81)--cycle);
+dot("$A$", A, N);
-draw(arc((8.03,9.81),0.61,-164.95,-124.95)--(8.03,9.81)--cycle);
+dot("$B$", B, SW);
-draw(arc((8.03,9.81),0.61,-84.95,-44.95)--(8.03,9.81)--cycle);
+dot("$C$", C, SE);
-/* draw figures */
+dot("$B'$", Bp, E);
-draw((8.03,9.81)--(2.3,1.61));
+dot("$C'$", Cp, W);
-draw((8.03,9.81)--(8.56,3.83));
+dot("$D$", D, dir(-70));
-draw((2.3,1.61)--(8.56,3.83));
+dot("$E$", Ep, dir(-130));
-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);
+MA("40^\circ",Cp,A,D, 1);
+MA("40^\circ",D,A,Ep, 1);
-label("$B$", (1.91,1.75), NE * labelscalefactor);
+MA("40^\circ",Ep,A,Bp, 1);
-label("$40^\circ$", (7.58,8.84), NE * labelscalefactor, dps);
+label("$6$", A--Cp);
+label("$10$", Bp--A);
-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, dps);
-label("$40^\circ$", (8.23,8.86), NE * labelscalefactor, dps);
-clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 </asy>

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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