Difference between revisions of "2014 AMC 12A Problems/Problem 20"
(Created page with "==Problem== In <math>\triangle BAC</math>, <math>\angle BAC=40^\circ</math>, <math>AB=10</math>, and <math>AC=6</math>. Points <math>D</math> and <math>E</math> lie on <math>\ov...") |
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==Solution== | ==Solution== | ||
− | 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 <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> | + | 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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==See Also== | ==See Also== | ||
{{AMC12 box|year=2014|ab=A|num-b=17|num-a=19}} | {{AMC12 box|year=2014|ab=A|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:51, 7 February 2014
Problem
In , , , and . Points and lie on and respectively. What is the minimum possible value of ?
Solution
Let be the reflection of across , and let be the reflection of across . Then it is well-known that the quantity is minimized when it is equal to . (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem. As lies on both and , we have . Furthermore, by the nature of the reflection, so . Therefore by the Law of Cosines
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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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