Difference between revisions of "2016 AIME II Problems/Problem 10"
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Extend <math>\overline{AB}</math> past <math>B</math> to point <math>X</math> so that <math>CPTX</math> is cyclic. Then, by Power of a Point on <math>CPTX</math>, <math>(CQ)(QT) = (PQ)(QX)</math>. By Power of a Point on <math>CATB</math>, <math>(CQ)(QB) = (AQ)(QB) = 42</math>. Thus, <math>(CQ)(QT) = 42</math>, so <math>BX = 8</math>. | Extend <math>\overline{AB}</math> past <math>B</math> to point <math>X</math> so that <math>CPTX</math> is cyclic. Then, by Power of a Point on <math>CPTX</math>, <math>(CQ)(QT) = (PQ)(QX)</math>. By Power of a Point on <math>CATB</math>, <math>(CQ)(QB) = (AQ)(QB) = 42</math>. Thus, <math>(CQ)(QT) = 42</math>, so <math>BX = 8</math>. | ||
− | By the Inscribed Angle Theorem on <math>CPTX</math>, <math>\angle SCT = \angle BXT</math>. By the Inscribed Angle Theorem on <math>ASTC</math>, <math>\angle SCT = \angle SAT</math>, so <math>\angle BXT = \angle SAT</math>. Since <math> | + | By the Inscribed Angle Theorem on <math>CPTX</math>, <math>\angle SCT = \angle BXT</math>. By the Inscribed Angle Theorem on <math>ASTC</math>, <math>\angle SCT = \angle SAT</math>, so <math>\angle BXT = \angle SAT</math>. Since <math>ASTB</math> is cyclic, <math>\angle AST = \angle TBX</math>. Thus, <math>\triangle AST \sim \triangle XBT</math>, so <math>AS/XB = ST/BT</math>. Solving for <math>ST</math> yields <math>ST = \frac{35}{8}</math>, for a final answer of <math>35+8 = \boxed{043}</math>. |
~ Leo.Euler | ~ Leo.Euler |
Revision as of 19:32, 22 December 2021
Contents
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 , you can solve quickly from there getting
.
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
.
Solution 4
Extend past
to point
so that
is cyclic. Then, by Power of a Point on
,
. By Power of a Point on
,
. Thus,
, so
.
By the Inscribed Angle Theorem on ,
. By the Inscribed Angle Theorem on
,
, so
. Since
is cyclic,
. Thus,
, so
. Solving for
yields
, for a final answer of
.
~ Leo.Euler
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.