Difference between revisions of "2016 AIME II Problems/Problem 10"
(→See also) |
(→Solution 5 (5 = 2 + 3)) |
||
Line 79: | Line 79: | ||
==Solution 5 (5 = 2 + 3)== | ==Solution 5 (5 = 2 + 3)== | ||
[[File:2016 AIME II 10.png|500px|right]] | [[File:2016 AIME II 10.png|500px|right]] | ||
− | + | By Ptolemy's Theorem applied to quadrilateral <math>ASTB</math>, we find | |
− | + | <cmath>AS\cdot BT+AB\cdot ST=AT\cdot BS.</cmath> | |
− | + | Projecting through <math>C</math> we have | |
+ | <cmath>\frac{AQ \cdot PB}{PQ \cdot AB} = (A,Q; P,B)\stackrel{C}{=}(A,T; S,B)=\frac{AT \cdot BS}{ST \cdot AB}. </cmath> | ||
+ | Therefore <math>AT \cdot BS = \frac {AQ \cdot PB}{PQ} \times ST \implies</math> | ||
+ | <math>\left(\frac {AQ \cdot PB}{PQ} - AB\right)\times ST = AS \cdot BT \implies</math> | ||
+ | <math>ST = \frac {AS \cdot BT \cdot PQ}{AQ \cdot PB – AB \cdot PQ}</math> | ||
+ | <math>ST = \frac {7\cdot 5 \cdot 3}{7\cdot 9 – 13 \cdot 3 } = \frac {35}{8} \implies 35 + 8 = \boxed {43}.</math> | ||
+ | |||
+ | '''Shelomovskii, vvsss, www.deoma-cmd.ru''' | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=9|num-a=11}} | {{AIME box|year=2016|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:32, 23 June 2022
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
Solution 5 (5 = 2 + 3)
By Ptolemy's Theorem applied to quadrilateral , we find
Projecting through
we have
Therefore
Shelomovskii, vvsss, www.deoma-cmd.ru
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.