Difference between revisions of "2003 USAMO Problems/Problem 4"

(Solution)
Line 13: Line 13:
 
\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>
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 +
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));
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draw((-2.48,0.52)--(3.78,0.52));
 +
draw((3.78,0.52)--(-0.16,3.1));
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draw(circle((-0.42,1), 2.12));
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draw((1.04,4.43)--(3.78,0.52));
 +
draw((1.04,4.43)--(-2.48,0.52));
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draw((1.04,4.43)--(1.64,0.52));
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draw((2.41,2.48)--(-2.48,0.52));
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dot((-0.16,3.1),dotstyle);
 +
label("$A$", (-0.4,3.3), NE * labelscalefactor);
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dot((-2.48,0.52),dotstyle);
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label("$B$", (-2.84,0.54), SW * labelscalefactor);
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dot((3.78,0.52),dotstyle);
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label("$C$", (3.86,0.64), NE * labelscalefactor);
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dot((1.4,2.08),dotstyle);
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label("$D$", (1.48,2.2), NE * labelscalefactor);
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dot((1.64,0.52),dotstyle);
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label("$E$", (1.72,0.64), NE * labelscalefactor);
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dot((1.04,4.43),dotstyle);
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label("$F$", (1.12,4.56), NE * labelscalefactor);
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dot((2.41,2.48),dotstyle);
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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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