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

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.

@@ Line 38: / Line 38: @@
 </center>
-== Solution ==
+== Solution 1 ==
 Let the incircle of <math>ACD</math> and the incircle of <math>BCD</math> touch line <math>AB</math> at points <math>D_a,D_b</math>, respectively; let these circles touch <math>CD</math> at <math>C_a</math>, <math>C_b</math>, respectively; and let them touch their common external tangent containing <math>E</math> at <math>T_a,T_b</math>, respectively, as shown in the diagram below.
@@ Line 89: / Line 89: @@
 We note that
-<cmath> CE = CC_a - EC_a = CC_b - EB_b = \frac{CC_a + CC_b - (EC_a + EC_b)}{2} . </cmath>
+<cmath> CE = CC_a - EC_a = CC_b - EC_b = \frac{CC_a + CC_b - (EC_a + EC_b)}{2} . </cmath>
 On the other hand, since <math>EC_a</math> and <math>ET_a</math> are tangents from the same point to a common circle, <math>EC_a = T_aE</math>, and similarly <math>EC_b = ET_b</math>, so
 <cmath> EC_a + EC_b = T_aE + ET_b = T_a T_b . </cmath>
@@ Line 99: / Line 99: @@
 <cmath> CC_b - DD_b = BC - DB, </cmath>
 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>
+== Solution 2 ==
+Define the same points as in the first solution.  First extend <math>T_aT_b</math> to intersect <math>AB</math> at a point <math>P</math>; without loss of generality let <math>A</math> lie in between <math>B</math> and <math>P</math>.  Then the incircle of <math>\triangle ACD</math> is also the incircle of <math>\triangle PED</math>, while the incircle of <math>\triangle BCD</math> is the <math>P</math>-excircle of <math>\triangle PED</math>.  It follows that <math>EC_a = C_bD</math>; denote this equality by <math>(*)</math>.
+Now remark that
+<cmath>
+CC_a + CC_b = \frac{AC+AD-CD}2 + \frac{BC+DC-BD}2 = \frac{AC+BC-AB}2 + CD.
+</cmath>
+Hence
+<cmath>
+CC_a + CC_b - CD = CC_a - C_bD \stackrel{(*)}= CC_a - C_aE = CE
+</cmath>
+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>.
+== Video Solution ==
+https://www.youtube.com/watch?v=G4UVUZ7UemY
 {{alternate solutions}}
-== Resources ==
+== See Also ==
 {{USAMO box|year=1991|num-b=4|after=Last Question}}
 * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356427#356427 Discussion on AoPS/MathLinks]
 * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=175552#175552 Further Discussion]
+{{MAA Notice}}
 [[Category:Olympiad Geometry Problems]]

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

Page

Toolbox

Search