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

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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>
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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{\textbf{(D) }14}.</cmath>
  
 
==Solution 2==
 
==Solution 2==
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<asy>
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In <math>\triangle BAC</math>, the three lines look like the Chinese character 又. Let <math>\triangle DEA</math>, <math>\triangle CDA</math>, and <math>\triangle BEA</math> have bases <math>DE</math>, <math>CD</math>, and <math>BE</math> respectively. Then, <math>\triangle DEA</math> has the same side <math>DA</math> as <math>\triangle CDA</math> and the same side <math>EA</math> as <math>\triangle BEA</math>. Connect all three triangles with <math>\triangle DEA</math> in the center and the two triangles sharing one of its sides. Then, <math>\pentagon BACDE</math> is formed with <math>BE+DE+CD</math> forming the base.
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import graph; size(20.95cm);
 
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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 */  
 
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real xmin = -1.52, xmax = 19.43, ymin = -2.35, ymax = 10.68; /* image dimensions */
 
  
 +
Intuitively, the pentagon's base is minimized when all three bottom sides are collinear. This is simply the original <math>\triangle BAC</math> except that <math>\angle BAC =120^\circ</math>. (In <math>\triangle DEA</math>, <math>\triangle CDA</math>, and <math>\triangle BEA</math>, <math>\angle A = 40^\circ</math>, and the three triangles connect at <math>A</math> to form the pentagon). Thus, <math>m\angle BAC = 40 \cdot 3</math>).
  
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draw(arc((8.03,9.81),0.61,-124.95,-84.95)--(8.03,9.81)--cycle);
 
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draw(arc((8.03,9.81),0.61,-164.95,-124.95)--(8.03,9.81)--cycle);
 
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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));
 
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draw((8.03,9.81)--(8.56,3.83));
 
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draw((2.3,1.61)--(8.56,3.83));
 
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draw((8.03,9.81)--(2.24,8.25));
 
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draw((8.03,9.81)--(15.11,2.74));
 
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draw((2.24,8.25)--(15.11,2.74));
 
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/* dots and labels */
 
  
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label("$A$", (7.9,10.03), NE * labelscalefactor);
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<math>BC</math> in this new triangle is then the minimum of <math>BE+DE+CD</math>. Applying law of cosines, <math>BC=\sqrt{6^2+10^2-2(6)(10)\cos (120^\circ)}=\sqrt{196}=14 \implies \boxed{\textbf{(D) }14}</math>
  
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label("$B$", (1.91,1.75), NE * labelscalefactor);
 
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label("$40^\circ$", (7.58,8.84), NE * labelscalefactor);
 
  
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label("$C$", (8.82,3.42), NE * labelscalefactor);
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~bjhhar
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Would prime notation be clearer?
  
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label("$B'$", (15.47,2.6), NE * labelscalefactor);  
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==Solution 3==
 +
<asy>
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size(300);
 +
defaultpen(linewidth(0.4)+fontsize(10));
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pen s = linewidth(0.8)+fontsize(8);
  
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label("$C'$", (1.85,8.42), NE * labelscalefactor);  
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pair A,B,C,D,Ep,Bp,Cp;
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label("$6$", (5.05,9.35), NE * labelscalefactor);  
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A = (0,0);
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label("$10$", (11.45,6.7), NE * labelscalefactor);  
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B = 10*dir(-110);C = 6*dir(-70);
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Bp = 10*dir(-30);Cp = 6*dir(-150);
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D = IP(Cp--Bp, A--B); Ep = IP(Cp--Bp, A--C);
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draw(A--B--C--A--Cp--Bp--A);
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draw(Cp--B);
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draw(C--Bp);
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draw(C--D);
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draw(B--Ep);
  
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label("$D$", (6.21,6.68), NE * labelscalefactor);  
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dot("$A$", A, N);
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dot("$B$", B, SW);
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dot("$C$", C, SE);
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dot("$B'$", Bp, E);
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dot("$C'$", Cp, W);
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dot("$D$", D, dir(-70));
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dot("$E$", Ep, dir(60));
  
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label("$E$", (7.86,6.11), NE * labelscalefactor);  
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MA("40^\circ",Cp,A,D, 1);
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label("$40^\circ$", (6.99,9.23), NE * labelscalefactor);  
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MA("40^\circ",D,A,Ep, 1);
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label("$40^\circ$", (8.23,8.86), NE * labelscalefactor);  
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MA("40^\circ",Ep,A,Bp, 1);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  
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label("$6$", A--Cp);
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label("$10$", Bp--A);
 
</asy>
 
</asy>
  
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(Diagram by dasobson)
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(Diagram by shihan)
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Reflect <math>C</math> across <math>AB</math> to <math>C'</math>. Similarly, reflect <math>B</math> across <math>AC</math> to <math>B'</math>. Clearly, <math>BE = B'E</math> and <math>CD = C'D</math>. Thus, the sum <math>BE + DE + CD = B'E + DE + C'D</math>. This value is minimized when <math>B'</math>, <math>C'</math>, <math>D</math> and <math>E</math> are collinear. To finish, we use the law of cosines on the triangle <math>AB'C'</math>: <math>B'C' = \sqrt{6^2 + 10^2 - 2(6)(10)\cos 120} = 14</math>
+
Reflect <math>C</math> across <math>AB</math> to <math>C'</math>. Similarly, reflect <math>B</math> across <math>AC</math> to <math>B'</math>. Clearly, <math>BE = B'E</math> and <math>CD = C'D</math>. Thus, the sum <math>BE + DE + CD = B'E + DE + C'D</math>. This value is minimized when <math>B'</math>, <math>C'</math>, <math>D</math> and <math>E</math> are collinear. To finish, we use the law of cosines on the triangle <math>AB'C'</math>: <math>B'C' = \sqrt{6^2 + 10^2 - 2(6)(10)\cos 120^{\circ}} = \boxed{\textbf{(D) } 14}.</math>
 +
 
 +
==Remark==
 +
 
 +
This problem is similar to [https://en.wikipedia.org/wiki/Fagnano%27s_problem Fagnano's Problem].
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==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}}

Latest revision as of 03:35, 22 January 2024

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 $AB'C'$: $B'C' = \sqrt{6^2 + 10^2 - 2(6)(10)\cos 120^{\circ}} = \boxed{\textbf{(D) } 14}.$

Remark

This problem is similar to Fagnano's Problem.

~isabelchen

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
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

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