2008 AIME II Problems/Problem 14
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 .
Notice that the given equation implies
We have , so .
Then, notice , so .
The solution satisfies the equation, so , and the answer is .
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.)
, 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 .
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
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
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