1992 USAMO Problems/Problem 2
Contents
Problem
Prove
Solution
Solution 1
Consider the points in the coordinate plane with origin , for integers .
Evidently, the angle between segments and is , and the length of segment is . It then follows that the area of triangle is . Therefore so as desired.
Solution 2
First multiply both sides of the equation by , so the right hand side is . Now by rewriting , we can derive the identity . Then the left hand side of the equation simplifies to as desired.
Solution 3
Multiply by . We get:
we can write this as:
This is an identity
Therefore;
, because of telescoping.
but since we multiplied in the beginning, we need to divide by . So we get that:
as desired. QED
Solution 4
Let .
Multiplying by gives
Notice that after expanding the sine, and so so
See Also
1992 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.