# 2003 USAMO Problems/Problem 4

## Problem

Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$.

## Solutions

### Solution 1

Extend segment $DM$ through $M$ to $G$ such that $FG\parallel CD$. $[asy] defaultpen(fontsize(10)+0.6); size(250); var theta=22, r=0.58; pair B=origin, A=dir(theta), C=A+(rotate(78)*0.8*A), O=IP(CR(B,r),CR(A,r)); path c=CR(O,r); pair D=IP(c,A--C), E=IP(c,B--C), F=extension(A,B,D,E), M=extension(B,D,C,F), G=extension(D,M,F,F+C-D); draw(A--B--C--A^^E--F--C^^A--F^^B--M^^E--M); draw(c); draw(F--G--C^^M--G,gray+0.4); dot("A",A,dir(F-E)); dot("B",B,2*dir(B-A)); dot("C",C,1.5*dir(C-A)); dot("D",D,2.5*dir(250)); dot("E",E,2.5*dir(C-A)); dot("F",F,dir(F-E)); dot("M",M,2.5*dir(255)); dot("G",G,dir(G-M)); [/asy]$ Then $MF = MC$ if and only if quadrilateral $CDFG$ is a parallelogram, or, $FD\parallel CG$. Hence $MC = MF$ if and only if $\angle GCD = \angle FDA$, that is, $\angle FDA + \angle CGF = 180^\circ$.

Because quadrilateral $ABED$ is cyclic, $\angle FDA = \angle ABE$. It follows that $MC = MF$ if and only if $$180^\circ = \angle FDA + \angle CGF = \angle ABE + \angle CGF,$$ that is, quadrilateral $CBFG$ is cyclic, which is equivalent to $$\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM.$$ Because $\angle DMC = \angle CMB$, $\angle CBM = \angle DCM$ if and only if triangles $BCM$ and $CDM$ are similar, that is $$\frac{CM}{BM} = \frac{DM}{CM},$$ or $MB\cdot MD = MC^2$.

### Solution 2

We first assume that $MB\cdot MD = MC^2$. Because $\frac{MC}{MD} = \frac{MB}{MC}$ and $\angle CMD = \angle BMC$, triangles $CMD$ and $BMC$ are similar. Consequently, $\angle MCD = \angle MBC$. This exact condition can also be visualized through power of M with respect to the circumcircle of triangle $BDC$ and getting the angle condition from the alternate segment theorem. $[asy] defaultpen(fontsize(10)+0.6); size(250); var theta=22, r=0.58; pair B=origin, A=dir(theta), C=A+(rotate(78)*0.8*A), O=IP(CR(B,r),CR(A,r)); path c=CR(O,r); pair D=IP(c,A--C), E=IP(c,B--C), F=extension(A,B,D,E), M=extension(B,D,C,F), G=extension(D,M,F,F+C-D); draw(A--B--C--A^^E--F--C^^A--F^^B--M^^E--M); draw(c); draw(A--E,gray+0.4); dot("A",A,dir(F-E)); dot("B",B,2*dir(B-A)); dot("C",C,1.5*dir(C-A)); dot("D",D,2.5*dir(250)); dot("E",E,2.5*dir(C-A)); dot("F",F,dir(F-E)); dot("M",M,2.5*dir(255)); [/asy]$ Because quadrilateral $ABED$ is cyclic, $\angle DAE = \angle DBE$. Hence $$\angle FCA = \angle MCD = \angle MBC = \angle DBE = \angle DAE = \angle CAE,$$ implying that $AE\parallel CF$, so $\angle AEF = \angle CFE$. Because quadrilateral $ABED$ is cyclic, $\angle ABD = \angle AED$. Hence $$\angle FBM = \angle ABD = \angle AED = \angle AEF = \angle CFE = \angle MFD.$$ Because $\angle FBM = \angle DFM$ and $\angle FMB = \angle DMF$, triangles $BFM$ and $FDM$ are similar. Consequently, $\frac{FM}{DM} = \frac{BM}{FM}$, or $FM^2 = BM\cdot DM = CM^2$. Therefore $MC^2 = MB\cdot MD$ implies $MC = MF$.

Now we assume that $MC = MF$. Applying Ceva's Theorem to triangle $BCF$ and cevians $BM, CA, FE$ gives $$\frac{BA}{AF}\cdot\frac{FM}{MC}\cdot\frac{CE}{EB} = 1,$$ implying that $\frac{BA}{AF} = \frac{BE}{EC}$, so $AE\parallel CF$.

Consequently, $\angle DCM = \angle DAE$. Because quadrilateral $ABED$ is cyclic, $\angle DAE = \angle DBE$. Hence $$\angle DCM = \angle DAE = \angle DBE = \angle CBM.$$ Because $\angle CBM = \angle DCM$ and $\angle CMB = \angle DMC$, triangles $BCM$ and $CDM$ are similar. Consequently, $\frac{CM}{DM} = \frac{BM}{CM}$, or $CM^2 = BM\cdot DM$.

Combining the above, we conclude that $MF = MC$ if and only if $MB\cdot MD = MC^2$.

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