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>. | ||
+ | |||
+ | 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 00:04, 10 April 2018
Contents
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.