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 $FG\parallel CD$ .

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 tangent cord theorem.

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 $MB\cdot MD = MC^2$ .

Solution 3

We will construct a series of equivalent statements. First: $\begin{align} MB\cdot MD = MC^2 &\equiv \frac{MB}{MC} = \frac{MC}{MD} \\ &\equiv \triangle MCD\sim \triangle MBC \\ &\equiv \angle CBA = \angle FCA \end{align}$ where the second equivalence follows from SAS similarity (because $\angle CMD = \angle BMC$ ).

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2003 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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