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

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<cmath>\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}</cmath>, | <cmath>\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}</cmath>, | ||

− | so the maximum length of <math>XY</math> is <math>\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt | + | so the maximum length of <math>XY</math> is <math>\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}</math>, and the answer is <math>7 + 4 + 3 = \boxed{014}</math>. |

## Revision as of 04:19, 26 February 2012

## 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

(For some reason, I can't submit LaTeX for this page.)

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 .