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Difference between revisions of "1991 USAMO Problems/Problem 5"
 
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is a constant equal to <math>r := \tfrac{AC+BC-AB}2</math>, and so <math>C</math> lies on the circle with center <math>C</math> and radius <math>r</math>.
 
is a constant equal to <math>r := \tfrac{AC+BC-AB}2</math>, and so <math>C</math> lies on the circle with center <math>C</math> and radius <math>r</math>.
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== Video Solution ==
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https://www.youtube.com/watch?v=G4UVUZ7UemY
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{{alternate solutions}}
 
{{alternate solutions}}

Latest revision as of 19:53, 27 October 2024

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 1

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 \[CE = CC_a - EC_a = CC_b - EC_b = \frac{CC_a + CC_b - (EC_a + EC_b)}{2} .\] 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 $E$. $\blacksquare$

Solution 2

Define the same points as in the first solution. First extend $T_aT_b$ to intersect $AB$ at a point $P$; without loss of generality let $A$ lie in between $B$ and $P$. Then the incircle of $\triangle ACD$ is also the incircle of $\triangle PED$, while the incircle of $\triangle BCD$ is the $P$-excircle of $\triangle PED$. It follows that $EC_a = C_bD$; denote this equality by $(*)$.

Now remark that \[CC_a + CC_b = \frac{AC+AD-CD}2 + \frac{BC+DC-BD}2 = \frac{AC+BC-AB}2 + CD.\] Hence \[CC_a + CC_b - CD = CC_a - C_bD \stackrel{(*)}= CC_a - C_aE = CE\] is a constant equal to $r := \tfrac{AC+BC-AB}2$, and so $C$ lies on the circle with center $C$ and radius $r$.

Video Solution

https://www.youtube.com/watch?v=G4UVUZ7UemY


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

See Also

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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