2009 AIME II Problems/Problem 15
Problem
Let be a diameter of a circle with diameter 1. Let and be points on one of the semicircular arcs determined by such that is the midpoint of the semicircle and . Point lies on the other semicircular arc. Let be the length of the line segment whose endpoints are the intersections of diameter with chords and . The largest possible value of can be written in the form , where and are positive integers and is not divisible by the square of any prime. Find .
Solution
Let be the center of the circle. Define , , and let and intersect at points and , respectively. We will express the length of as a function of and maximize that function in the interval .
Let be the foot of the perpendicular from to . We compute as follows.
(a) By the Extended Law of Sines in triangle , we have
(b) Note that and . Since and are similar right triangles, we have , and hence,
(c) We have and , and hence by the Law of Sines,
(d) Multiplying (a), (b), and (c), we have
,
which is a function of (and the constant ). Differentiating this with respect to yields
,
and the numerator of this is
,
which vanishes when . Therefore, the length of is maximized when , where is the value in that satisfies .
Note that
,
so . We compute
,
so the maximum length of is , and the answer is .
See Also
2009 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.