1991 AIME Problems/Problem 9
Suppose that and that where is in lowest terms. Find
Use the two trigonometric Pythagorean identities and .
If we square the given , we find that
This yields .
Let . Then squaring,
Note: The problem is much easier computed if we consider what is, then find the relationship between and (using , and then computing using and then the reciprocal of .
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,
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
Make the substitution (a substitution commonly used in calculus). By the half-angle identity for tangent, , so . Also, we have Now note the following:
Plugging these into our equality gives:
This simplifies to , and solving for gives , and . Finally, .
We are given that , or equivalently, . Note that what we want is just .
Assign a right triangle with angle , hypotenuse , adjacent side , and opposite side . Then, through the given information above, we have that..
Hence, because similar right triangles can be scaled up by a factor, we can assume that this particular right triangle is indeed in simplest terms.
Furthermore, by the Pythagorean Theorem, we have that
Solving for in the first equation and plugging in into the second equation...
Now, we want
Plugging in, we find the answer is
Hence, the answer is
We know that and that where , , represent the hypotenuse, adjacent, and opposite (respectively) to angle in a right triangle. Thus we have that . We also have that . Set and csc(x)+cot(x) = . Then, notice that ( This is because of the Pythagorean Theorem, recall ). But then notice that . From the information provided in the question, we can substitute for . Thus, . Since, essentially we are asked to find the sum of the numerator and denominator of , we have .
Firstly, we write where and . This will allow us to spot factorable expressions later. Now, since , this gives us Adding this to our original expressions gives us or Now since , So we can write Upon simplification, we get We are asked to find so we can write that as Now using the fact that and yields, so
Rewriting and in terms of and , we know that
Squaring to get an expression in terms of and ,
Expanding then collecting terms yields a quadratic in
To make calculations easier, let
Upon inspection, is a root. Dividing by ,
Substituting we see that doesn't work, as , leaving undefined.
We conclude that
After checking via the given equation, we know that only the positive solution works.
Adding and , our answer is
-Benedict T (countmath1)
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