Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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==Solution 1== | ==Solution 1== | ||
− | Since <math>\angle{AFB}=\angle{ADB}=90^{\circ}</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's Theorem]], we have <math>13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})</math>. Cancelling <math>13</math>, | + | Since <math>\angle{AFB}=\angle{ADB}=90^{\circ}</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's Theorem]], we have <math>13DF+(5)(13)(\frac{4}{5})=(12)(13)(\frac{3}{5})</math>. Cancelling <math>13</math>, we obtain <math>DF=\frac{16}{5}</math>, so our answer is <math>16+5=\boxed{21\,\textbf{(B)}}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 15:41, 4 February 2019
- 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
?
Solution 1
Since , quadrilateral
is cyclic. It follows that
. In addition, since
, triangles
and
are similar. It follows that
. By Ptolemy's Theorem, we have
. Cancelling
, we obtain
, so our answer is
.
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,
and our answer is
.
Solution 3
If we draw a diagram as given, but then add as an altitude to use the Pythagorean theorem, we end up with similar triangles
and
. Thus,
and
. Using Pythagorean theorem, we now get
and
can be found out noting that
is just
through area times height (since
, similar triangles gives
), and that
is just
. From there,
Now,
, and squaring and adding both sides and subtracting a 169 from both sides gives
, so
. Thus, the answer is
.
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 |
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.