2008 AIME I Problems/Problem 14
Contents
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 .
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 (Calculus)
Proceed as follows for Solution 1.
Once you approach the function , find the maximum value by setting .
Simplifying to take the derivative, we have , so . Setting , we have .
Solving, we obtain as the critical value. Hence, has the maximum value of . Since , the maximum value of occurs at , so has a maximum value of .
Note: Please edit this solution if it feels inadequate.
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 (Use Pythagorean on and then using ). Similarly, . Using the Pythagorean Theorem again and , . Using the Pythagorean Theorem , . 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 .
Solution 4
(Diagram credit goes to Solution 2)
Let . The only constraint on is that it must be greater than . Using similar triangles, we can deduce that . Now, apply law of cosines on . We can see that . We can find . Plugging this into our equation, we get: Eventually, We want to maximize . There are many ways to maximize this expression, discussed here: https://artofproblemsolving.com/community/c4h2292700_maximization. The maximum result of that expression is . Finally, evaluating for this value .
~superagh
Solution 5 (Clean)
Let be the distance from to . Observe that takes any value from to , where is the radius of the circle.
Let be the foot of the altitude from to . It is clear that is the midpoint of , and so the length is the average of and . It follows thus that .
We compute and so . This is . Equality is attained, so thus we extract the answer of
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 |
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