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

(rescuing diagrams)
m (TeX)
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<cmath> CC_b - DD_b = BC - DB, </cmath>
 
<cmath> CC_b - DD_b = BC - DB, </cmath>
 
so that
 
so that
−
<cmath> CE = \frac{CC_a + CC_b - D_aD - DD_b}{2} = \frac{AC + BC - (AD+DB)}{2} = \frac{AC + BC - AB}{2} . </cmath>
+
<cmath> \begin{align*}
 +
CE &= \frac{CC_a + CC_b - D_aD - DD_b}{2} = \frac{AC + BC - (AD+DB)}{2} \\
 +
&= \frac{AC + BC - AB}{2} .
 +
\end{align*} </cmath>
 
Thus <math>E</math> lies on the arc of the circle with center <math>C</math> and radius <math>(AB+BC-AB)/2</math> intercepted by segments <math>CA</math> and <math>CB</math>.  If we choose an arbitrary point <math>X</math> on this arc and let <math>D</math> be the intersection of lines <math>CX</math> and <math>AB</math>, then <math>X</math> becomes point <math>E</math> in the diagram, so every point on this arc is in the locus of <math>E</math>. <math>\blacksquare</math>
 
Thus <math>E</math> lies on the arc of the circle with center <math>C</math> and radius <math>(AB+BC-AB)/2</math> intercepted by segments <math>CA</math> and <math>CB</math>.  If we choose an arbitrary point <math>X</math> on this arc and let <math>D</math> be the intersection of lines <math>CX</math> and <math>AB</math>, then <math>X</math> becomes point <math>E</math> in the diagram, so every point on this arc is in the locus of <math>E</math>. <math>\blacksquare</math>
  

Revision as of 14:01, 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 \begin{align*} CE &= \frac{CC_a + CC_b - D_aD - DD_b}{2} = \frac{AC + BC - (AD+DB)}{2} \\ &= \frac{AC + BC - AB}{2} . \end{align*} 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)
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