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Difference between revisions of "2001 USAMO Problems/Problem 2"

(solution, will do asy soon)
(→‎Solution: asymptote)
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Now consider the [[homothety]] that carries the incircle of <math>\triangle ABC</math> to its excircle. The homothety also carries <math>Q</math> to <math>D_2</math> (since <math>A,Q,D_2</math> are collinear), and carries the tangency points <math>E_1</math> to <math>G</math>. It follows that <math>\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}</math>.  
 
Now consider the [[homothety]] that carries the incircle of <math>\triangle ABC</math> to its excircle. The homothety also carries <math>Q</math> to <math>D_2</math> (since <math>A,Q,D_2</math> are collinear), and carries the tangency points <math>E_1</math> to <math>G</math>. It follows that <math>\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}</math>.  
  
−
<center><asy></asy></center>
+
<center><asy>
 +
pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6);
 +
pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP("D_1",foot(O,B,C))),E1 = D(MP("E_1",foot(O,A,C),NE)),E2 = D(MP("E_2",C+A-E1,NE)); /* arbitrary points */
 +
 
 +
/* ugly construction for OA */
 +
pair Ca = 2C-A, Cb = bisectorpoint(Ca,C,B), OA = IP(A--A+10*(O-A),C--C+50*(Cb-C)), D2 = D(MP("D_2",foot(OA,B,C))), Fa=2B-A, Ga=2C-A, F=MP("F",D(foot(OA,B,Fa)),NW), G=MP("G",D(foot(OA,C,Ga)),NE); D(OA);
 +
D(MP("A",A,N)--MP("B",B,NW)--MP("C",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D(B--Fa,linewidth(0.6)); D(C--Ga,linewidth(0.6)); D(MP("P",IP(D(A--D2),D(B--E2)),NNE)); D(MP("Q",IP(incircle(A,B,C),A--D2),SW));
 +
 
 +
clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle);
 +
</asy></center>
  
 
By [[Menelaus' Theorem]] it follows that <math>\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(c - (s-a)) \cdot a}{(a-(s-c)) \cdot AE_1} = \frac{a}{s-a}</math>. It easily follows that <math>AQ = D_2P</math>. <math>\blacksquare</math>
 
By [[Menelaus' Theorem]] it follows that <math>\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(c - (s-a)) \cdot a}{(a-(s-c)) \cdot AE_1} = \frac{a}{s-a}</math>. It easily follows that <math>AQ = D_2P</math>. <math>\blacksquare</math>

Revision as of 21:07, 24 September 2008

Problem

Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ = D_2P$.

Solution

It is well known that the excircle opposite $A$ is tangent to $\overline{BC}$ at the point $D_2$. (Proof: let the points of tangency of the excircle with the lines $BC, AB, AC$ be $D_3, F,G$ respectively. Then $AB+BD_3=AB + BF=AF = AG = AC + AG=AC + CD_3$. It follows that $2CD_3 = AB + BC - AC$, and $CD_3 = s-b = BD_1 = CD_2$, so $D_3 \equiv D_2$.)

Now consider the homothety that carries the incircle of $\triangle ABC$ to its excircle. The homothety also carries $Q$ to $D_2$ (since $A,Q,D_2$ are collinear), and carries the tangency points $E_1$ to $G$. It follows that $\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}$.

[asy] pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP("D_1",foot(O,B,C))),E1 = D(MP("E_1",foot(O,A,C),NE)),E2 = D(MP("E_2",C+A-E1,NE)); /* arbitrary points */ /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca,C,B), OA = IP(A--A+10*(O-A),C--C+50*(Cb-C)), D2 = D(MP("D_2",foot(OA,B,C))), Fa=2B-A, Ga=2C-A, F=MP("F",D(foot(OA,B,Fa)),NW), G=MP("G",D(foot(OA,C,Ga)),NE); D(OA); D(MP("A",A,N)--MP("B",B,NW)--MP("C",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D(B--Fa,linewidth(0.6)); D(C--Ga,linewidth(0.6)); D(MP("P",IP(D(A--D2),D(B--E2)),NNE)); D(MP("Q",IP(incircle(A,B,C),A--D2),SW)); clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle); [/asy]

By Menelaus' Theorem it follows that $\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(c - (s-a)) \cdot a}{(a-(s-c)) \cdot AE_1} = \frac{a}{s-a}$. It easily follows that $AQ = D_2P$. $\blacksquare$

See also

2001 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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