Difference between revisions of "1998 USAMO Problems/Problem 2"

(Solution)
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== Solution ==
 
== Solution ==
{{solution}}
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<center>
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<asy>
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pair O,A,B,C,D,E,F,DEb,CFb,Fo,M;
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O=(0,0);
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A=(1.732,1);
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B=(0,1);
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C=(-1.732,1);
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D=(0.866,1);
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Fo=(-1,-0.5);
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path AC,AF,DE,CF,DEbM,CFbM,C1,C2;
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C1=circle(O,2);
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C2=circle(O,1);
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E=intersectionpoints(A--Fo,C2)[0];
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F=intersectionpoints(A--Fo,C2)[1];
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DEb=((D.x+E.x)/2.0,(D.y+E.y)/2.0);
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CFb=((C.x+F.x)/2.0,(C.y+F.y)/2.0);
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M=(-0.433,1);
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path AC=A--C;
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path AF=A--F;
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path DEbM=DEb--M;
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path CFbM=CFb--M;
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path DE=D--E;
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path CF=C--F;
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draw(AC);
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draw(AF);
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draw(DE);
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draw(CF);
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draw(DEbM);
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draw(CFbM);
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draw(C1);
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draw(C2);
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label("A",A,NE);
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label("B",B,N);
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label("C",C,NW);
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label("D",D,N);
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label("E",E,SE);
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label("F",F,SW);
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label("M",M,N);
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</asy>
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</center>
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First, <math>AD=\frac{AB}{2}=\frac{AC}{4}</math>. Because <math>E</math>,<math>F</math> and <math>B</math> all lie on a circle, <math>AE \cdot AF=AB \cdot AB=\frac{AB}{2} \cdot 2AB=AD \cdot AC</math>. Therefore, <math>\triangle ACF \sim \triangle AEB</math>, so <math>\angle ACF = \angle AEB</math>. Thus, quadrilateral <math>CFED</math> is cyclic, and <math>M</math> must be the center of the circumcircle of <math>CFED</math>, which implies that <math>MC=\frac{CD}{2}</math>. Putting it all together,
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<math>\frac{AM}{MC}=\frac{AC-MC}{MC}=\frac{AC-\frac{CD}{2}}{\frac{CD}{2}}=\frac{AC-\frac{AC-AD}{2}}{\frac{AC-AD}{2}}=\frac{AC-\frac{3AC}{8}}{\frac{3AC}{8}}=\frac{\frac{5AC}{8}}{\frac{3AC}{8}}=\frac{5}{3}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 16:58, 24 June 2014

Problem

Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.

Solution

[asy]  pair O,A,B,C,D,E,F,DEb,CFb,Fo,M; O=(0,0); A=(1.732,1); B=(0,1); C=(-1.732,1); D=(0.866,1); Fo=(-1,-0.5);  path AC,AF,DE,CF,DEbM,CFbM,C1,C2; C1=circle(O,2); C2=circle(O,1);  E=intersectionpoints(A--Fo,C2)[0]; F=intersectionpoints(A--Fo,C2)[1]; DEb=((D.x+E.x)/2.0,(D.y+E.y)/2.0); CFb=((C.x+F.x)/2.0,(C.y+F.y)/2.0); M=(-0.433,1);  path AC=A--C; path AF=A--F; path DEbM=DEb--M; path CFbM=CFb--M; path DE=D--E; path CF=C--F;  draw(AC); draw(AF); draw(DE); draw(CF); draw(DEbM); draw(CFbM); draw(C1); draw(C2);  label("\(A\)",A,NE); label("\(B\)",B,N); label("\(C\)",C,NW); label("\(D\)",D,N); label("\(E\)",E,SE); label("\(F\)",F,SW); label("\(M\)",M,N);  [/asy]

First, $AD=\frac{AB}{2}=\frac{AC}{4}$. Because $E$,$F$ and $B$ all lie on a circle, $AE \cdot AF=AB \cdot AB=\frac{AB}{2} \cdot 2AB=AD \cdot AC$. Therefore, $\triangle ACF \sim \triangle AEB$, so $\angle ACF = \angle AEB$. Thus, quadrilateral $CFED$ is cyclic, and $M$ must be the center of the circumcircle of $CFED$, which implies that $MC=\frac{CD}{2}$. Putting it all together,

$\frac{AM}{MC}=\frac{AC-MC}{MC}=\frac{AC-\frac{CD}{2}}{\frac{CD}{2}}=\frac{AC-\frac{AC-AD}{2}}{\frac{AC-AD}{2}}=\frac{AC-\frac{3AC}{8}}{\frac{3AC}{8}}=\frac{\frac{5AC}{8}}{\frac{3AC}{8}}=\frac{5}{3}$

See Also

1998 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAMO Problems and Solutions

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