Difference between revisions of "1992 USAMO Problems/Problem 2"
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<math>\frac {\sin{1}}{\cos{0}\cos{1}} + \frac {\sin{1}}{\cos{1}\cos{2}} + ... + \frac {\sin{1}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}</math> | <math>\frac {\sin{1}}{\cos{0}\cos{1}} + \frac {\sin{1}}{\cos{1}\cos{2}} + ... + \frac {\sin{1}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}</math> | ||
− | we can | + | we can write this as: |
<math>\frac {\sin{1 - 0}}{\cos{0}\cos{1}} + \frac {\sin{2 - 1}}{\cos{1}\cos{2}} + ... + \frac {\sin{89 - 88}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}</math> | <math>\frac {\sin{1 - 0}}{\cos{0}\cos{1}} + \frac {\sin{2 - 1}}{\cos{1}\cos{2}} + ... + \frac {\sin{89 - 88}}{\cos{88}\cos{89}} = \frac {\cos{1}}{\sin{1}}</math> |
Revision as of 12:35, 8 August 2010
Problem
Prove
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
Resources
1992 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |