# 1998 USAMO Problems/Problem 2

## 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); F=intersectionpoints(A--Fo,C2); 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

## Solution 2

We will use signed lengths. WLOG let $AC = 4$, $AM = x$, and $AE = y > 0$. Then $AB = 2$ and $AD = 1$. Therefore, $DM = x - 1$ and $MC = 4 - x$. By Power of a Point, $AE \cdot AF = AB^2 = 4$, so $AF = \frac{4}{AE} = \frac{4}{y}$, and $EF = \frac{4}{y} - y = \frac{4 - y^2}{y}$. Since $M$ is on the perpendicular bisectors of $DE$ and $CF$, we have $|DM| = |EM| = |x - 1|$ and $|CM| = |FM| = |4 - x|$.

By Stewart's Theorem on $\triangle MFA$ and cevian $DM$, $$\frac{4}{y}(y(\frac{4-y^2}{y}) + (x-1)^2) = y(4-x)^2 + \frac{4-y^2}{y} \cdot x^2;$$ $$4(4 - y^2 + (x-1)^2) = y^2(4-x)^2 + (4-y^2)x^2;$$ $$16 - 4y^2 + 4x^2 - 8x + 4 = x^2y^2 - 8xy^2 + 16y^2 + 4x^2 - x^2y^2;$$ $$8xy^2 - 8x- 20y^2 + 20 = (8x - 20)(y^2 - 1) = 0.$$

So either $8x - 20 = 0$ or $y^2 - 1 = 0$. If $y^2 - 1 = 0$, then $AE = y = 1 = AD$ and $AF = AC = 4$, so the perpendicular bisectors of $CF$ and $DE$ are the same line, and they do not intersect at a point. Therefore, $AM = x = \frac{5}{2}$ and $MC = \frac{3}{2}$, so $\frac{AM}{MC} = \boxed{\frac{5}{3}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 