1986 AIME Problems/Problem 3
Contents
Problem
If and , what is ?
Solution 1
Since is the reciprocal function of :
Thus,
Using the tangent addition formula:
.
Solution 2
Using the formula for tangent of a sum, . We only need to find .
We know that . Cross multiplying, we have .
Similarly, we have .
Dividing:
. Plugging in to the earlier formula, we have .
Solution 3 (less trig required, use of quadratic formula)
Let and . This simplifies the equations to:
Taking the tangent of a sum formula from Solution 2, we get .
We can use substitution to solve the system of equations from above: , so .
Multiplying by , we get , which is . Dividing everything by 5 and shifting everything to one side gives .
Using the quadratic formula gives . Since this looks too hard to simplify, we can solve for using , which turns out to also be , provided that the sign of the radical in is opposite the one in .
WLOG, assume and . Multiplying them gives which simplifies to .
THe denominator of ends up being , so multiplying both numerator and denominator by 6 gives .
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Remark: The quadratic need not be solved. The value of can be found through Vieta's.
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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