Difference between revisions of "2016 AIME II Problems/Problem 10"
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Edit: Note that the finish is much simpler. Once you get, <math>\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}</math>, so <math>ST=\dfrac{AS*\sin(\beta)}{\sin(\alpha)}=7*(15/24)=35/8</math>. | Edit: Note that the finish is much simpler. Once you get, <math>\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}</math>, so <math>ST=\dfrac{AS*\sin(\beta)}{\sin(\alpha)}=7*(15/24)=35/8</math>. | ||
− | ==Solution 2(Projective Geometry)== | + | ==Solution 2 (Projective Geometry)== |
Projecting through <math>C</math> we have <cmath>\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}</cmath> which easily gives <math>ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{43.}</math> | Projecting through <math>C</math> we have <cmath>\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}</cmath> which easily gives <math>ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{43.}</math> | ||
Revision as of 20:25, 5 March 2020
Problem
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
.
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 (Projective Geometry)
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.