2008 AIME II Problems/Problem 14
Contents
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 .
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
Solution 5
The given system is equivalent to the points and forming an equilateral triangle with the origin. WLOG let this triangle have side length , so . The condition implies lies to the left of , so is the top vertex. Now we can compute (by complex numbers, or the sine angle addition identity) that , so . Minimizing this is equivalent to minimizing the denominator, which happens when and thus , resulting in , so and the answer is .
As a remark, expressing the condition that the triangle is equilateral purely algebraically instead of using trig eliminates the need for calculus or analyzing the behavior of sine.
Solution 6 (Geometry and Trigonometry)
Notice that by Pythagorean theorem, if we take a triangle with vertices and forming an equilateral triangle. Now, take a rectangle with vertices Notice that and are on the sides. Let be the angle formed by the points Then, we have that where is the side of the equilateral triangle. Also, we have that is the angle formed by the points and so Thus, we have that We see that this expression is maximized when is maximized (at least when is in the interval which it is). Then, so ew have that the maximum of is and so our answer is
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 | ||
All AIME Problems and Solutions |
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