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2003 USAMO Problems/Problem 4

Revision as of 10:11, 9 January 2023 by Shihan (talk | contribs) (→‎Solution 2)

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 $FG\parallel CD$.

2003usamo4-1.png

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.


2003usamo4-2.png

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 $MB\cdot MD = MC^2$.


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

See also

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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