Difference between revisions of "2003 USAMO Problems/Problem 4"
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+ | ==Solution 2== | ||
+ | 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. | ||
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+ | 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
Contents
[hide]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:
Solution 2
by suli
Let's prove the first direction: if MD * MB = , 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.