Difference between revisions of "2016 AIME II Problems/Problem 10"
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By Ptolemy's Theorem applied to quadrilateral <math>ASTB</math>, we find | By Ptolemy's Theorem applied to quadrilateral <math>ASTB</math>, we find | ||
<cmath>5\cdot 7+13\cdot ST=AT\cdot BS.</cmath> | <cmath>5\cdot 7+13\cdot ST=AT\cdot BS.</cmath> | ||
− | Therefore, in order to find <math>ST</math>, it suffices to find <math>AT\cdot BS</math>. We do this using similar triangles. | + | Therefore, in order to find <math>ST</math>, it suffices to find <math>AT\cdot BS</math>. We do this using similar triangles, which can be found by using Power of a Point theorem. |
As <math>\triangle APS\sim \triangle CPB</math>, we find | As <math>\triangle APS\sim \triangle CPB</math>, we find |
Revision as of 19:28, 13 March 2017
Triangle is inscribed in circle
. Points
and
are on side
with
. Rays
and
meet
again at
and
(other than
), respectively. If
and
, then
, where
and
are relatively prime positive integers. Find
.
Contents
Solution 1
Let
,
, and
. Note that since
we have
, so by the Ratio Lemma
Similarly, we can deduce
and hence
.
Now Law of Sines on ,
, and
yields
Hence
so
Hence
and the requested answer is
.
Edit: Note that the finish is much simpler. Once you get, , so
.
Solution 2
Projecting through we have
which easily gives
Solution 3
By Ptolemy's Theorem applied to quadrilateral , we find
Therefore, in order to find
, it suffices to find
. We do this using similar triangles, which can be found by using Power of a Point theorem.
As , we find
Therefore,
.
As , we find
Therefore,
.
As , we find
Therefore,
.
As , we find
Therefore,
. Thus we find
But now we can substitute in our previously found values for
and
, finding
Substituting this into our original expression from Ptolemy's Theorem, we find
Thus the answer is
.
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.