Difference between revisions of "2003 USAMO Problems/Problem 4"
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| − | Let's prove the first direction: if MD * MB = | + | Let's prove the first direction: if <math>MD * MB = 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. | 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. | ||
Revision as of 16:22, 5 April 2014
Contents
Problem
Let 
be a triangle. A circle passing through 
and 
intersects segments 
and 
at 
and 
, respectively. Lines 
and 
intersect at 
, while lines 
and 
intersect at 
. Prove that 
if and only if 
.
Solution
by April
Take 
. We have:
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]](http://latex.artofproblemsolving.com/2/3/1/231b5e7a9bbf63b98064894a36059570da4b8180.png)
Solution 2
by suli
Let's prove the first direction: if 
, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
