# Difference between revisions of "2009 AIME II Problems/Problem 15"

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Let <math>O</math> be the center of the circle. Define <math>\angle{MOC}=t</math>, <math>\angle{BOA}=2a</math>, and let <math>BC</math> and <math>AC</math> intersect <math>MN</math> at points <math>X</math> and <math>Y</math>, respectively. We will express the length of <math>XY</math> as a function of <math>t</math> and maximize that function in the interval <math>[0, \pi]</math>. | Let <math>O</math> be the center of the circle. Define <math>\angle{MOC}=t</math>, <math>\angle{BOA}=2a</math>, and let <math>BC</math> and <math>AC</math> intersect <math>MN</math> at points <math>X</math> and <math>Y</math>, respectively. We will express the length of <math>XY</math> as a function of <math>t</math> and maximize that function in the interval <math>[0, \pi]</math>. |

## Revision as of 03:00, 3 January 2014

## 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 . The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.