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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> | 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> |
Revision as of 02:07, 27 February 2018
Contents
Problem
In ,
,
, and
. Points
and
lie on
and
respectively. What is the minimum possible value of
?
Solution 1
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
Solution 2
Reflect across
to
. Similarly, reflect
across
to
. Clearly,
and
. Thus, the sum
. This value is maximized when
,
,
and
are collinear. To finish, we use the law of cosines on the triangle
:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.