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Difference between revisions of "2003 USAMO Problems/Problem 4"
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</asy>
 
</asy>
  
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==Solution 2==
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by suli
 +
 +
Let's prove the first direction: if MD * MB = <math>MC^2</math>, then MF = MC.
 +
 +
We start that noticing by SAS Similarity triangles MDC and MCB are similar. Thus, <MBC = <MCD. Because they intercept the same arc, <EAD = <MBC = <MCD and so EA // CF. It can further be shown that AF / AB = EC / EB using similar triangles. Now, let us use Ceva's Theorem on FBC to deduce that MF / MC = 1, and so MF = MC.
 +
 +
The other direction follows similarly; the proof will be left as an exercise for the reader.
  
 
== See also ==
 
== See also ==

Revision as of 16:21, 5 April 2014

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 $MB\cdot MD = MC^2$.

Solution

by April

Take $G\in BD: \,FG\parallel CD$. We have:

$MF = MC\Longleftrightarrow \textrm{the quadrilateral}\; CDFG\; \textrm{is a parallelogram} \\ \Longleftrightarrow FD\parallel CG\Longleftrightarrow\angle FDA = \angle GCD\Longleftrightarrow\angle FDA + \angle CGF = 180^\circ \\ \Longleftrightarrow \angle ABE + \angle CGF = 180^\circ\Longleftrightarrow\textrm{the quadrilateral}\;CBGF\;\textrm{is cyclic} \\ \Longleftrightarrow\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM \\ \Longleftrightarrow\triangle BCM\sim\triangle CDM\Longleftrightarrow MB\cdot MD = MC^{2}$

Added diagram:

[asy] import graph; size(5cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -4.3, xmax = 6.36, ymin = -3.98, ymax = 6.3; draw((-0.16,3.1)--(-2.48,0.52)); draw((-2.48,0.52)--(3.78,0.52)); draw((3.78,0.52)--(-0.16,3.1)); draw(circle((-0.42,1), 2.12)); draw((1.04,4.43)--(3.78,0.52)); draw((1.04,4.43)--(-2.48,0.52)); draw((1.04,4.43)--(1.64,0.52)); draw((2.41,2.48)--(-2.48,0.52)); dot((-0.16,3.1),dotstyle); label("$A$", (-0.4,3.3), NE * labelscalefactor); dot((-2.48,0.52),dotstyle); label("$B$", (-2.84,0.54), SW * labelscalefactor); dot((3.78,0.52),dotstyle); label("$C$", (3.86,0.64), NE * labelscalefactor); dot((1.4,2.08),dotstyle); label("$D$", (1.48,2.2), NE * labelscalefactor); dot((1.64,0.52),dotstyle); label("$E$", (1.72,0.64), NE * labelscalefactor); dot((1.04,4.43),dotstyle); label("$F$", (1.12,4.56), NE * labelscalefactor); dot((2.41,2.48),dotstyle); label("$M$", (2.48,2.6), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

Solution 2

by suli

Let's prove the first direction: if MD * MB = $MC^2$, then MF = MC.

We start that noticing by SAS Similarity triangles MDC and MCB are similar. Thus, <MBC = <MCD. Because they intercept the same arc, <EAD = <MBC = <MCD and so EA // CF. It can further be shown that AF / AB = EC / EB using similar triangles. Now, let us use Ceva's Theorem on FBC to deduce that MF / MC = 1, and so MF = MC.

The other direction follows similarly; the proof will be left as an exercise for the reader.

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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