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Difference between revisions of "1991 USAMO Problems/Problem 5"

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dot(Tb,linewidth(4));
 
dot(Tb,linewidth(4));
  
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label("<math>A</math>",A,SW);
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label("\(A\)",A,SW);
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label("<math>B</math>",B,SE);
+
label("\(B\)",B,SE);
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label("<math>C</math>",C,W);
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label("\(C\)",C,W);
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label("<math>D</math>",D,S);
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label("\(D\)",D,S);
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label("<math>E</math>",E,NNE);
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label("\(E\)",E,NNE);
 
</asy>
 
</asy>
 
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</center>
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dot(Db,linewidth(4));
 
dot(Db,linewidth(4));
  
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label("<math>A</math>",A,SW);
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label("\(A\)",A,SW);
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label("<math>B</math>",B,SE);
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label("\(B\)",B,SE);
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label("<math>C</math>",C,W);
+
label("\(C\)",C,W);
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label("<math>D</math>",D,S);
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label("\(D\)",D,S);
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label("<math>E</math>",E,NNE);
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label("\(E\)",E,NNE);
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label("<math>T_a</math>",Ta,N);
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label("\(T_a\)",Ta,N);
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label("<math>T_b</math>",Tb,WNW);
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label("\(T_b\)",Tb,WNW);
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label("<math>D_a</math>",Da,S);
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label("\(D_a\)",Da,S);
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label("<math>D_b</math>",Db,S);
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label("\(D_b\)",Db,S);
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label("<math>C_a</math>",Ca,WSW);
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label("\(C_a\)",Ca,WSW);
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label("<math>C_b</math>",Cb,ENE);
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label("\(C_b\)",Cb,ENE);
 
</asy>
 
</asy>
 
</center>
 
</center>

Revision as of 13:00, 13 January 2008

Problem

Let $\, D \,$ be an arbitrary point on side $\, AB \,$ of a given triangle $\, ABC, \,$ and let $\, E \,$ be the interior point where $\, CD \,$ intersects the external common tangent to the incircles of triangles $\, ACD \,$ and $\, BCD$. As $\, D \,$ assumes all positions between $\, A \,$ and $\, B \,$, prove that the point $\, E \,$ traces the arc of a circle.

[asy] size(220); defaultpen(1); pair A=(0,0), B=(220,0), C=(18.7723,118.523); pair D=(72.6,0); pair Ia=incenter(A,D,C), Ib=incenter(B,D,C); pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129); pair E=IntersectionPoint((Ta--Tb),(C--D)); path Oa=circle(Ia,inradius(A,D,C)); path Ob=circle(Ib,inradius(B,D,C)); pair Da=IP(Oa,A--B), Db=IP(Ob,A--B); draw(D--C--A--B--C); draw(Ta--Tb); draw(Oa); draw(Ob); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(Ta,linewidth(4)); dot(Tb,linewidth(4)); label("\(A\)",A,SW); label("\(B\)",B,SE); label("\(C\)",C,W); label("\(D\)",D,S); label("\(E\)",E,NNE); [/asy]

Solution

Let the incircle of $ACD$ and the incircle of $BCD$ touch line $AB$ at points $D_a,D_b$, respectively; let these circles touch $CD$ at $C_a$, $C_b$, respectively; and let them touch their common external tangent containing $E$ at $T_a,T_b$, respectively, as shown in the diagram below.

[asy] size(220); defaultpen(1); pair A=(0,0), B=(220,0), C=(18.7723,118.523); pair D=(72.6,0); pair Ia=incenter(A,D,C), Ib=incenter(B,D,C); pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129); pair E=IntersectionPoint((Ta--Tb),(C--D)); path Oa=circle(Ia,inradius(A,D,C)); path Ob=circle(Ib,inradius(B,D,C)); pair Da=IP(Oa,A--B), Db=IP(Ob,A--B); pair Ca=IP(Oa,C--D), Cb=IP(Ob,C--D); draw(D--C--A--B--C); draw(Ta--Tb); draw(Oa); draw(Ob); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(Ta,linewidth(4)); dot(Tb,linewidth(4)); dot(Ca,linewidth(4)); dot(Cb,linewidth(4)); dot(Da,linewidth(4)); dot(Db,linewidth(4)); label("\(A\)",A,SW); label("\(B\)",B,SE); label("\(C\)",C,W); label("\(D\)",D,S); label("\(E\)",E,NNE); label("\(T_a\)",Ta,N); label("\(T_b\)",Tb,WNW); label("\(D_a\)",Da,S); label("\(D_b\)",Db,S); label("\(C_a\)",Ca,WSW); label("\(C_b\)",Cb,ENE); [/asy]

We note that \[CE = CC_a - EC_a = CC_b - EB_b = \frac{CC_a + CC_b - (EC_a + EC_b)}{2} .\] On the other hand, since $EC_a$ and $ET_a$ are tangents from the same point to a common circle, $EC_a = T_aE$, and similarly $EC_b = ET_b$, so \[EC_a + EC_b = T_aE + ET_b = T_a T_b .\] On the other hand, the segments $T_a T_b$ and $D_a D_b$ evidently have the same length, and $D_a D_b = D_aD + DD_b$, so $EC_a + EC_b = D_aD + DD_b$. Thus \[CE = \frac{CC_a + CC_b - (EC_a + EC_b)}{2} = \frac{CC_a + CC_b - D_aD - DD_b}{2} .\] If we let $s_a$ be the semiperimeter of triangle $ACD$, then $CC_a = s_a - AD$, and $D_aD = s_a - AC$, so \[CC_a - D_aD = (s_a - AD) - (s_a - AC) = AC - AD .\] Similarly, \[CC_b - DD_b = BC - DB,\] so that \[CE = \frac{CC_a + CC_b - D_aD - DD_b}{2} = \frac{AC + BC - (AD+DB)}{2} = \frac{AC + BC - AB}{2} .\] Thus $E$ lies on the arc of the circle with center $C$ and radius $(AB+BC-AB)/2$ intercepted by segments $CA$ and $CB$. If we choose an arbitrary point $X$ on this arc and let $D$ be the intersection of lines $CX$ and $AB$, then $X$ becomes point $E$ in the diagram, so every point on this arc is in the locus of $E$. $\blacksquare$


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

1991 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Last Question
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions
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