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1998 USAMO Problems/Problem 2: Difference between revisions

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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,
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 AED</math>, so <math>\angle ACF = \angle AED</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,


<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>
<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>

Revision as of 14:37, 13 August 2016

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, $\triangle ACF \sim \triangle AED$, so $\angle ACF = \angle AED$. 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,

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

Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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