Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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==Solution 3== | ==Solution 3== | ||
− | If we draw a diagram as given, but then add <math>DG</math> as an altitude to use the Pythagorean theorem, we end up with similar triangles <math>\triangle{DFG}</math> and <math>\triangle{DCE}</math>. Thus, <math>FG</math> is <math>\ | + | If we draw a diagram as given, but then add <math>DG</math> as an altitude to use the Pythagorean theorem, we end up with similar triangles <math>\triangle{DFG}</math> and <math>\triangle{DCE}</math>. Thus, <math>FG</math> is <math>\3/5 * x</math> and <math>DG</math> is <math>4/5 *x</math>. Using Pythagorean theorem, we now get |
− | <math>BF = \sqrt{(4/ | + | <math>BF = \sqrt{(4/5 *x + 5)^2 + (3/5 * x)^2}</math> |
− | and <math>AF</math> can be found out noting that <math>AE</math> is just 48/5 through area times height (12(9) = 15<math>36/5</math>, similar triangles gives AE = <math> | + | and <math>AF</math> can be found out noting that <math>AE</math> is just 48/5 through area times height (12(9) = 15 * <math>36/5</math>, similar triangles gives AE = <math>48/5</math>), and that <math>EF</math> is just <math>36/5 - x</math>. |
From there, <math>AF</math> is <math>\sqrt{((36/5 - x)^2 + (48/5)^2)}</math> | From there, <math>AF</math> is <math>\sqrt{((36/5 - x)^2 + (48/5)^2)}</math> |
Revision as of 20:01, 3 February 2017
- 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, 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 $\3/5 * x$ (Error compiling LaTeX. Unknown error_msg) 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
Using algebra, we get that , so
x = .
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.