1991 AIME Problems/Problem 9
Problem
Suppose that and that
where
is in lowest terms. Find
Contents
Solution
Solution 1
Use the two trigonometric Pythagorean identities and
.
If we square the given , we find that
This yields .
Let . Then squaring,
Substituting yields a quadratic equation:
. It turns out that only the positive root will work, so the value of
and
.
Solution 2
Recall that , from which we find that
. Adding the equations
together and dividing by 2 gives , and subtracting the equations and dividing by 2 gives
. Hence,
and
. Thus,
and
. Finally,
so .
Solution 3
(least computation) By the given,
and
.
Multiplying the two, we have
Subtracting both of the two given equations from this, and simpliyfing with the identity , we get
Solving yields , and
Solution 4
Make the substitution (a substitution commonly used in calculus).
, so
.
Now note the following:
Plugging these into our equality gives:
This simplifies to , and solving for
gives
, and
. Finally,
.
Solution 5
We are given that
, or equivalently,
. Note that what we want is just
.
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 |
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