Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 27\qquad\textbf{(E)}\ 30</math> | <math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 27\qquad\textbf{(E)}\ 30</math> | ||
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==Solution 1== | ==Solution 1== |
Revision as of 10:51, 13 August 2014
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
Contents
[hide]Problem
In triangle ,
,
, and
. Distinct points
,
, and
lie on segments
,
, and
, respectively, such that
,
, and
. The length of segment
can be written as
, where
and
are relatively prime positive integers. What is
?
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 27\qquad\textbf{(E)}\ 30$ (Error compiling LaTeX. Unknown error_msg)
Solution 1
Since , quadrilateral
is cyclic. It follows that
. In addition, since
, triangles
and
are similar. It follows that
. By Ptolemy, we have
. Cancelling
, the rest is easy. We obtain
Solution 2
Using the similar triangles in triangle gives
and
. Quadrilateral
is cyclic, implying that
= 180°. Therefore,
, and triangles
and
are similar. Solving the resulting proportion gives
. Therefore,
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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.