Difference between revisions of "2014 AMC 12A Problems/Problem 20"
m (→Solution 2) |
Happyman25 (talk | contribs) m (→Solution 2) |
||
Line 57: | Line 57: | ||
(Diagram by dasobson) | (Diagram by dasobson) | ||
− | 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' | + | 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 maximized 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> |
==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}} |
Revision as of 05:39, 1 August 2018
Contents
[hide]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
(Diagram by dasobson)
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.