2008 AIME II Problems/Problem 14
Problem
Let and be positive real numbers with . Let be the maximum possible value of for which the system of equations has a solution in satisfying and . Then can be expressed as a fraction , where and are relatively prime positive integers. Find .
Contents
Solutions
Solution 1
Notice that the given equation implies
We have , so .
Then, notice , so .
The solution satisfies the equation, so , and the answer is .
Solution 2
Consider the points and . They form an equilateral triangle with the origin. We let the side length be , so and .
Thus and we need to maximize this for .
Taking the derivative shows that , so the maximum is at the endpoint . We then get
Then, , and the answer is .
(For a non-calculus way to maximize the function above:
Let us work with degrees. Let . We need to maximize on .
Suppose is an upper bound of on this range; in other words, assume for all in this range. Then: for all in . In particular, for , must be less than or equal to , so .
The least possible upper bound of on this interval is . This inequality must hold by the above logic, and in fact, the inequality reaches equality when . Thus, attains a maximum of on the interval.)
Solution 3
Consider a cyclic quadrilateral with , and . Then From Ptolemy's Theorem, , so Simplifying, we have .
Note the circumcircle of has radius , so and has an arc of , so . Let .
, where both and are since triangle must be acute. Since is an increasing function over , is also increasing function over .
maximizes at maximizes at . This squared is , and .
Note:
None of the above solutions point out clearly the importance of the restriction that , , and be positive. Indeed, larger values of p are obtained when the lower vertex of the equilateral triangle in Solution 2 dips below the x-axis. Take for example . This yields
Solution 4
The problem is looking for an intersection in the said range between parabola : and the hyperbola : . The vertex of is below the x-axis and it's x-coordinate is a, which is to the right of the vertex of the , which is . So for the intersection to exist with and , needs to cross x-axis between , and , meaning, Divide both side by , which can be easily solved by moving to RHS and taking square roots. Final answer
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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