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2003 USAMO Problems/Problem 4: Difference between revisions

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\Longleftrightarrow\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM \\
\Longleftrightarrow\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM \\
\Longleftrightarrow\triangle BCM\sim\triangle CDM\Longleftrightarrow MB\cdot MD = MC^{2}</math>
\Longleftrightarrow\triangle BCM\sim\triangle CDM\Longleftrightarrow MB\cdot MD = MC^{2}</math>
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>


== Resources ==
== Resources ==

Revision as of 16:21, 12 April 2012

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]

Resources

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