2013 AMC 12B Problems/Problem 19
- 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
, 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,
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,
is
and
is
. Using Pythagorean theorem, we now get
and can be found out noting that
is just 48/5 through area times height (12(9) = 15 *
, similar triangles gives AE =
), and that
is just
.
From there, is
Now, , and squaring and adding both sides and subtracting a 169 from both sides gives:
, so
x = .
Thus, the answer is
Solution 4
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 |
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