Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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==Solution== | ==Solution== | ||
Since <math>\angle{AFB}=\angle{ADB}=90</math>, quadrilateral <math>ABDF</math> is cyclic. It follows that <math>\angle{ADE}=\angle{ABF}</math>. In addition, since <math>\angle{AFB}=\angle{AED}=90</math>, triangles <math>ABF</math> and <math>ADE</math> are similar. It follows that <math>AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})</math>. By [[Ptolemy's Theorem|Ptolemy]], we have <math>13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})</math>. Cancelling <math>13</math>, the rest is easy. We obtain <math>DF=\frac{16}{5}\implies{16+5=21}\implies{\boxed{(B)}}</math> | Since <math>\angle{AFB}=\angle{ADB}=90</math>, quadrilateral <math>ABDF</math> is cyclic. It follows that <math>\angle{ADE}=\angle{ABF}</math>. In addition, since <math>\angle{AFB}=\angle{AED}=90</math>, triangles <math>ABF</math> and <math>ADE</math> are similar. It follows that <math>AF=(13)(\frac{4}{5}), BF=(13)(\frac{3}{5})</math>. By [[Ptolemy's Theorem|Ptolemy]], we have <math>13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})</math>. Cancelling <math>13</math>, the rest is easy. We obtain <math>DF=\frac{16}{5}\implies{16+5=21}\implies{\boxed{(B)}}</math> | ||
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== See also == | == See also == |
Revision as of 20:18, 2 February 2014
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
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
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
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.