Difference between revisions of "2010 AIME II Problems/Problem 15"
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== Problem 15 == | == Problem 15 == | ||
− | 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> | + | 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>. |
− | == | + | ==Diagram== |
− | + | <asy> | |
+ | size(250); | ||
+ | defaultpen(fontsize(9pt)); | ||
+ | picture pic; | ||
+ | pair A,B,C,D,E,M,N,P,Q; | ||
+ | B=MP("B",origin, SW); | ||
+ | C=MP("C", (12.5,0), SE); | ||
+ | A=MP("A", IP(CR(C,10),CR(B,15)), dir(90)); | ||
+ | N=MP("N", (A+B)/2, dir(180)); | ||
+ | M=MP("M", midpoint(C--A), dir(70)); | ||
+ | D=MP("D", extension(B,incenter(A,B,C),A,C), dir(C-B)); | ||
+ | E=MP("E", extension(C,incenter(A,B,C),A,B), dir(90)); | ||
+ | P=MP("P", OP(circumcircle(A,M,N),circumcircle(A,D,E)), dir(-70)); | ||
+ | Q = MP("Q", extension(A,P,B,C),dir(-90)); | ||
+ | draw(B--C--A--B^^M--P--N^^D--P--E^^A--Q); | ||
+ | draw(circumcircle(A,M,N), gray); | ||
+ | draw(circumcircle(A,D,E), heavygreen); | ||
+ | dot(A);dot(B);dot(C);dot(D);dot(E);dot(P);dot(Q);dot(M);dot(N); | ||
+ | </asy> | ||
− | <math>\ | + | == Solution 1 (Linearity) == |
+ | Define the function <math>f:\mathbb{R}^{2}\rightarrow\mathbb{R}</math> by <cmath>f(X)=\text{Pow}_{(AMN)}(X)-\text{Pow}_{(ADE)}(X)</cmath> for points <math>X</math> in the plane. Then <math>f</math> is linear, so <math>\frac{BQ}{CQ}=\frac{f(B)-f(Q)}{f(Q)-f(C)}</math>. But <math>f(Q)=0</math> since <math>Q</math> lies on the radical axis of <math>(AMN)</math>, <math>(ADE)</math> thus <cmath>\frac{BQ}{CQ}=-\frac{f(B)}{f(C)}=-\frac{BN\cdot BA-BE\cdot BA}{CM\cdot CA-CD\cdot CA}</cmath> Let <math>AC=b</math>, <math>BC=a</math> and <math>AB=c</math>. Note that <math>BN=\tfrac{c}{2}</math> and <math>CM=\tfrac{b}{2}</math> because they are midpoints, while <math>BE=\frac{ac}{a+b}</math> and <math>CD=\frac{ab}{a+c}</math> by Angle Bisector Theorem. Thus we can rewrite this expression as <cmath>\begin{align*}&-\frac{\tfrac{c^{2}}{2}-\tfrac{ac^{2}}{a+b}}{\tfrac{b^{2}}{2}-\tfrac{ab^{2}}{a+c}} \\ =&-\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{\tfrac{1}{2}-\tfrac{a}{a+b}}{\tfrac{1}{2}-\tfrac{a}{a+c}}\right) \\ =&\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{a+c}{a+b}\right) \\ =&\left(\frac{225}{169}\right)\left(\frac{29}{27}\right) \\ =&~\frac{725}{507}\end{align*}</cmath> so <math>m-n=\boxed{218}</math>. | ||
− | + | == Official Solution (MAA) == | |
+ | |||
+ | The Angle Bisector Theorem implies that <math>E</math> lies on <math>\overline{AN}</math> and <math>D</math> lies on <math>\overline{MC}</math> because <math>AE/EB = AC/BC < 1</math> and <math>AD/DC = AB/CB > 1</math>. The Angle Bisector Theorem furthermore implies | ||
+ | <cmath>NE=AN-AE=\frac 12\cdot AB - \frac{AC}{AC+BC}\cdot AB=\frac 5{18}</cmath> | ||
+ | and | ||
+ | <cmath>MD = CM - CD = \frac 12\cdot AC - \frac{BC}{BC+BA}\cdot AC = \frac{13}{58}.</cmath> | ||
+ | Because <math>ANPM</math> is cyclic, <math>\angle ENP = \angle ANP = \angle PMD</math>. Because <math>AEPD</math> is cyclic, <math>\angle NEP = 180^\circ-\angle AEP = \angle MDP</math>. Because <math>\angle ENP =\angle PMD</math> and <math>\angle NEP = \angle MDP</math>, triangles <math>NEP</math> and <math>MDP</math> are similar. Hence | ||
+ | <cmath>\frac{NE}{MD}=\frac{NP}{MP}.</cmath> | ||
+ | Applying the Law of Sines to <math>\triangle ANP</math> and <math>\triangle AMP</math> gives | ||
+ | <cmath>\frac{NE}{MD} = \frac{NP}{MP} = \frac{\sin \angle NAP}{\sin \angle PAM} = \frac{\sin \angle BAQ}{\sin \angle QAC}</cmath> | ||
+ | and thus | ||
+ | <cmath>\frac{\sin \angle BAQ}{\sin \angle QAC} = \frac{(\frac{5}{18})}{(\frac{13}{58})} = \frac{145}{117}.</cmath> | ||
+ | Thus | ||
+ | <cmath>\frac{BQ}{QC} = \frac{\textrm{Area}(ABQ)}{\textrm{Area}(ACQ)} = \frac{\frac{1}{2}\cdot AB \cdot AQ \cdot \sin \angle BAQ}{\frac{1}{2}\cdot AC \cdot AQ \cdot \sin \angle QAC}= \frac{15}{13} \cdot \frac{145}{117} = \frac{725}{507},</cmath> | ||
+ | and <math>m - n = 218</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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 PNY \cong \angle PAM</math> and <math>\angle ANM \cong \angle APM</math>. So, <math>m\angle PNA = m\angle PNY + 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>. | ||
+ | |||
+ | *These two ratios are the same thing and can also be derived from the Ratio Lemma. | ||
+ | Ratio Lemma :<math>\frac{BD}{DC} = \frac{AB}{AC} \cdot \frac{\sin \angle BAD}{\sin \angle CAD}</math>, for any cevian AD of a triangle ABC. | ||
+ | For the sine ratios use Law of Sines on triangles APM and APN, <cmath>\frac{PM}{\sin \angle PAM}=\frac{AP}{\sin \angle AMP}=\frac{AP}{\sin \angle ANP}=\frac{PN}{\sin \angle PAN}</cmath>. The information needed to use the Ratio Lemma can be found from the similar triangle section above. | ||
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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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 3 == | ||
+ | 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>E = (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}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:24, 21 November 2023
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
.
Diagram
Solution 1 (Linearity)
Define the function by
for points
in the plane. Then
is linear, so
. But
since
lies on the radical axis of
,
thus
Let
,
and
. Note that
and
because they are midpoints, while
and
by Angle Bisector Theorem. Thus we can rewrite this expression as
so
.
Official Solution (MAA)
The Angle Bisector Theorem implies that lies on
and
lies on
because
and
. The Angle Bisector Theorem furthermore implies
and
Because
is cyclic,
. Because
is cyclic,
. Because
and
, triangles
and
are similar. Hence
Applying the Law of Sines to
and
gives
and thus
Thus
and
.
Solution 2
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 3
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.