Difference between revisions of "2010 AIME II Problems/Problem 15"
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Source: [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1831745#p1831745] by Zhero | Source: [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1831745#p1831745] by Zhero | ||
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+ | == Extension == | ||
+ | The work done in this problem leads to a nice extension of this problem: | ||
+ | |||
+ | Given a <math>\triangle ABC</math> and points <math>A_1</math>, <math>A_2</math>, <math>B_1</math>, <math>B_2</math>, <math>C_1</math>, <math>C_2</math>, such that <math>A_1</math>, <math>A_2</math> <math>\in BC</math>, <math>B_1</math>, <math>B_2</math> <math>\in AC</math>, and <math>C_1</math>, <math>C_2</math> <math>\in AB</math>, then let <math>\omega_1</math> be the circumcircle of <math>\triangle AB_1C_1</math> and <math>\omega_2</math> be the circumcircle of <math>\triangle AB_2C_2</math>. Let <math>A'</math> be the intersection point of <math>\omega_1</math> and <math>\omega_2</math> distinct from <math>A</math>. Define <math>B'</math> and <math>C'</math> similarly. Then <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> concur. | ||
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+ | This can be proven using Ceva's theorem and the work done in this problem, which effectively allows us to compute the ratio that line <math>AA'</math> divides the opposite side <math>BC</math> into and similarly for the other two sides. | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2010|n=II|num-b=14|after=Last Problem}} | {{AIME box|year=2010|n=II|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:01, 10 December 2016
Contents
Problem 15
In triangle , , , and . Points and lie on with and . Points and lie on B with and . Let be the point, other than , of intersection of the circumcircles of and . Ray meets at . The ratio can be written in the form , where and are relatively prime positive integers. Find .
Solution
Let . since . Since quadrilateral is cyclic, and , yielding and . Multiplying these together yields .
. Also, is the center of spiral similarity of segments and , so . Therefore, , which can easily be computed by the angle bisector theorem to be . It follows that , giving us an answer of .
Note: Spiral similarities may sound complex, but they're really not. The fact that is really just a result of simple angle chasing.
Source: [1] by Zhero
Extension
The work done in this problem leads to a nice extension of this problem:
Given a and points , , , , , , such that , , , , and , , then let be the circumcircle of and be the circumcircle of . Let be the intersection point of and distinct from . Define and similarly. Then , , and concur.
This can be proven using Ceva's theorem and the work done in this problem, which effectively allows us to compute the ratio that line divides the opposite side into and similarly for the other two sides.
See Also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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.