1991 AIME Problems/Problem 9
Problem
Suppose that 
and that 
where 
is in lowest terms. Find 
Solution
Solution 1
Use the two trigonometric Pythagorean identities 
and 
.
If we square 
, we find that 
, so 
. Solving shows that 
.
Call 
. Rewrite the second equation in a similar fashion: 
. Substitute in 
to get a quadratic: 
. The quadratic is factorable (though somewhat ugly); 
. It turns out that only the positive root will work, so the value of 
and 
.
Solution 2
This solution is fast for calculus students. Make the substitution 
. 
, so 
. 
Now note the following:
;
Plugging these into our equality gives:
This simplifies to 
, and solving for 
gives 
, and 
. Finally, 
.
See also
| 1991 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
