Difference between revisions of "2010 AIME II Problems/Problem 15"
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In triangle <math>ABC</math>, <math>AC = 13</math>, <math>BC = 14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math>AC</math> with <math>AM=MC</math> and <math>\angle ABD = \angle DBC</math>. Points <math>N</math> and <math>E</math> lie on <math>AB</math> with <math>AN=NB</math> and <math>\angle ACE = \angle ECB</math>. Let <math>P</math> be the point, other than <math>A</math>, of intersection of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ratio <math>\frac{BQ}{CQ}</math> can be written in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m-n</math>. | In triangle <math>ABC</math>, <math>AC = 13</math>, <math>BC = 14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math>AC</math> with <math>AM=MC</math> and <math>\angle ABD = \angle DBC</math>. Points <math>N</math> and <math>E</math> lie on <math>AB</math> with <math>AN=NB</math> and <math>\angle ACE = \angle ECB</math>. Let <math>P</math> be the point, other than <math>A</math>, of intersection of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ratio <math>\frac{BQ}{CQ}</math> can be written in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m-n</math>. | ||
− | == Solution == | + | == Solution 1 == |
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>. | ||
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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. | 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. | ||
+ | |||
+ | == Solution 2 == | ||
+ | This problem can be solved with barycentric coordinates. Let triangle <math>ABC</math> be the reference triangle with <math>A=(1,0,0)</math>, <math>B=(0,1,0)</math>, and <math>C=(0,0,1)</math>. Thus, <math>N=(1:1:0)</math> and <math>M=(1:0:1)</math>. Using the Angle Bisector Theorem, we can deduce that <math>D=(14:0:15)</math> and <math>(14:13:0)</math>. Plugging the coordinates for triangles <math>ANM</math> and <math>AED</math> into the circle formula, we deduce that the equation for triangle <math>ANM</math> is <math>-a^2yz-b^2zx-c^2xy+(\frac{c^2}{2}y+\frac{b^2}{2}z)(x+y+z)=0</math> and the equation for triangle <math>AED</math> is <math>-a^2yz-b^2zx-c^2xy+(\frac{14c^2}{27}y+\frac{14b^2}{29}z)(x+y+z)=0</math>. Solving the system of equations, we get that <math>\frac{c^2y}{54}=\frac{b^2z}{58}</math>. This equation determines the radical axis of circles <math>ANM</math> and <math>AED</math>, on which points <math>P</math> and <math>Q</math> lie. Thus, solving for <math>\frac{z}{y}</math> gets the desired ratio of lengths, and <math>\frac{z}{y}=\frac{58c^2}{54b^2}</math> and plugging in the lengths <math>b=13</math> and <math>c=15</math> gets <math>\frac{725}{507}</math>. From this we get the desired answer of <math>725-507=\boxed{218}</math>. | ||
+ | -wertguk | ||
==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 20:15, 21 May 2020
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 1
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
.
- These two ratios are the same thing and can also be derived from the Ratio Lemma.
Ratio Lemma :, for any cevian AD of a triangle ABC.
For the sine ratios use Law of Sines on triangles APM and APN,
. The information needed to use the Ratio Lemma can be found from the similar triangle section above.
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.
Solution 2
This problem can be solved with barycentric coordinates. Let triangle be the reference triangle with
,
, and
. Thus,
and
. Using the Angle Bisector Theorem, we can deduce that
and
. Plugging the coordinates for triangles
and
into the circle formula, we deduce that the equation for triangle
is
and the equation for triangle
is
. Solving the system of equations, we get that
. This equation determines the radical axis of circles
and
, on which points
and
lie. Thus, solving for
gets the desired ratio of lengths, and
and plugging in the lengths
and
gets
. From this we get the desired answer of
.
-wertguk
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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