Difference between revisions of "2016 AIME II Problems/Problem 10"
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==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{ | + | [[File:2016 AIME II 10c.png|400px|right]] |
+ | 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{043}</math>. | ||
==Solution 3== | ==Solution 3== | ||
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==Solution 4 == | ==Solution 4 == | ||
− | 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)( | + | 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)(QT) = (AQ)(QB) = 42</math>. Thus, <math>(PQ)(QX) = 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>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>. | 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>. | ||
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<cmath>ST = \frac {7\cdot 5 \cdot 3}{7\cdot 9 – 13 \cdot 3 } = \frac {35}{8} \implies 35 + 8 = \boxed {43}.</cmath> | <cmath>ST = \frac {7\cdot 5 \cdot 3}{7\cdot 9 – 13 \cdot 3 } = \frac {35}{8} \implies 35 + 8 = \boxed {43}.</cmath> | ||
− | ''' | + | '''vladimir.shelomovskii@gmail.com, vvsss''' |
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Connect <math>AT</math> and <math>\angle{SCT}=\angle{SAT}, \angle{ACS}=\angle{ATS}, \frac{ST}{\sin \angle{SAT}}=\frac{AS}{\sin \angle{ATS}}</math> | ||
+ | |||
+ | So we need to get the ratio of <math>\frac{\sin \angle{ACS}}{\sin \angle{SCT}}</math> | ||
+ | |||
+ | By clear observation <math>\triangle{CAQ}\sim \triangle{BTQ}</math>, we have <math>\frac{CQ}{AC}=\frac{6}{5}</math>, LOS tells <math>\frac{AC}{\sin \angle{CPA}}=\frac{4}{\sin \angle{ACS}}; \frac{CQ}{\sin \angle{CPQ}}=\frac{3}{\sin \angle{PCQ}}</math> so we get <math>\frac{\sin \angle{PCQ}}{\sin \angle{ACS}}=\frac{5}{8}</math>, the desired answer is <math>7\cdot \frac{\sin \angle{SAT}}{\sin \angle{ATS}}=\frac{35}{8}</math> leads to <math>\boxed{043}</math> | ||
+ | |||
+ | ~blusoul | ||
== 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}} |
Latest revision as of 22:09, 31 January 2024
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
vladimir.shelomovskii@gmail.com, vvsss
Solution 6
Connect and
So we need to get the ratio of
By clear observation , we have
, LOS tells
so we get
, the desired answer is
leads to
~blusoul
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 |
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