Difference between revisions of "2010 AIME II Problems/Problem 15"
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Let <math>Y = MN \cap AQ</math>. <math>\frac {BQ}{QC} = \frac {NY}{MY}</math> since <math>\triangle AMN \sim \triangle ACB</math>. Since quadrilateral <math>AMPN</math> is cyclic, <math>\triangle MYA \sim \triangle PYN</math> and <math>\triangle MYP \sim \triangle AYN</math>, yielding <math>\frac {YM}{YA} = \frac {MP}{AN}</math> and <math>\frac {YA}{YN} = \frac {AM}{PN}</math>. Multiplying these together yields <math>\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)</math>. | Let <math>Y = MN \cap AQ</math>. <math>\frac {BQ}{QC} = \frac {NY}{MY}</math> since <math>\triangle AMN \sim \triangle ACB</math>. Since quadrilateral <math>AMPN</math> is cyclic, <math>\triangle MYA \sim \triangle PYN</math> and <math>\triangle MYP \sim \triangle AYN</math>, yielding <math>\frac {YM}{YA} = \frac {MP}{AN}</math> and <math>\frac {YA}{YN} = \frac {AM}{PN}</math>. Multiplying these together yields <math>\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)</math>. | ||
− | <math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13 | + | <math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}</math>. |
− | + | Now we claim that <math>\triangle PMD \sim \triangle PNE</math>. To prove this, we can use cyclic quadrilaterals. | |
+ | |||
+ | From <math>AMPN</math>, <math>\angle PNE \cong \angle PAM</math> and <math>\angle ANM \cong \angle APM</math>. So, <math>m\angle PNA = m\angle PNE + m\angle ANM = m\angle PAM + m\angle APM = 180-m\angle PMA</math> and <math>\angle PNA \cong \angle PMD</math>. | ||
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+ | From <math>ADPE</math>, <math>\angle PDE \cong \angle PAE</math> and <math>\angle EDA \cong \angle EPA</math>. Thus, <math>m\angle MDP = m\angle PDE + m\angle EDA = m\angle PAE + m\angle EPA = 180-m\angle PEA</math> and <math>\angle PDM \cong \angle PEN</math>. | ||
+ | |||
+ | Thus, from AA similarity, <math>\triangle PMD \sim \triangle PNE</math>. | ||
+ | |||
+ | Therefore, <math>\frac {PN}{PM} = \frac {NE}{MD}</math>, which can easily be computed by the angle bisector theorem to be <math>\frac {145}{117}</math>. It follows that <math>\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>. | ||
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 |
Revision as of 23:04, 9 April 2018
Contents
[hide]Problem 15
In triangle , , , and . Points and lie on with and . Points and lie on 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 .
.
Now we claim that . To prove this, we can use cyclic quadrilaterals.
From , and . So, and .
From , and . Thus, and .
Thus, from AA similarity, .
Therefore, , which can easily be computed by the angle bisector theorem to be . It follows that , giving us an answer of .
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 |
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