2008 AIME I Problems/Problem 14
Problem
Let be a diameter of circle . Extend through to . Point lies on so that line is tangent to . Point is the foot of the perpendicular from to line . Suppose , and let denote the maximum possible length of segment . Find .
Contents
Solution
Solution 1
Let . Since , it follows easily that . Thus . By the Law of Cosines on , where , so: Let ; this is a quadratic, and its discriminant must be nonnegative: . Thus, Equality holds when .
Solution 1.1
Proceed as follows for solution 1.
Once you approach the function m=(2x-27)/x^2 find the maximum value by setting dm/dx=0.
Simplifying m to take the derivative, we have 2/x-27/x^2, so dm/dx=-2/x^2+54/x^3.
Setting dm/dx=0, we have 2/x^2=54/x^3.
Solving, we obtain x=27 as the critical value.
Hence, m has the maximum value of (2*27-27)/27^2=1/27.
Since BP^2=405+729m, the maximum value of BP occurs at m=1/27, so BP^2 has a maximum value of 405+729/27=432
Note: Please edit this solution if it feels inadequate. I am not familiar with LaTeX and would greatly appreciate it if someone converted the "solution" into LaTeX.
Solution 2
From the diagram, we see that , and that .
This is a quadratic equation, maximized when . Thus, .
Solution 3 (Calculus Bash)
(Diagram credit goes to Solution 2)
We let . From similar triangles, we have that . Similarly, . Using the Pythagorean Theorem, . Using the Pythagorean Theorem once again, . After a large bashful simplification, . The fraction is equivalent to . Taking the derivative of the fraction and solving for x, we get that . Plugging back into the expression for yields , so the answer is .
See also
2008 AIME I (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.