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

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

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

Let C' be the foot of the perpendicular from C to MN. We compute XY as follows.

(a) By the Extended Law of Sines in triangle ABC, we have

CA

= sin<ABC

= sin((arc AN + arc NC)/2)

= sin((pi/2 + (pi-t))/2)

= sin(3pi/4 - t/2)

= sin(pi/4 + t/2)

(b) Note that CC' = COsin(t) = (1/2)sin(t) and AO = 1/2. Since CC'Y and AOY are similar right triangles, we have CY/AY = CC'/AO = sin(t), and hence,

CY/CA

= CY/(CY + AY)

= sin(t) / (1 + sin(t))

= sin(t) / (sin(pi/2) + sin(t))

= sin(t) / (2sin(pi/4 + t/2)cos(pi/4 - t/2))

(c) We have <XCY = (arc AB)/2 = a and <CXY = (arc MB + arc CN)/2 = ((pi/2 - 2a) + (pi - t))/2 = 3pi/4 - a - t/2, and hence by the Law of Sines,

XY/CY

= sin<XCY / sin<CXY

= sin(a) / sin(3pi/4 - a - t/2)

= sin(a) / sin(pi/4 + a + t/2).

(d) Multiplying (a), (b), and (c), we have

XY

= CA * (CY/CA) * (XY/CY)

= sin(t)sin(a) / (2cos(pi/4 - t/2)sin(pi/4 + a + t/2))

= sin(t)sin(a) / (sin(pi/2 + a) + sin(a + t))

= sin(a) * sin(t) / (sin(t + a) + cos(a)),

which is a function of t (and the constant a). Differentiating this with respect to t yields

sin(a) * (cos(t)(sin(t + a) + cos(a)) - sin(t)cos(t + a)) / (sin(t + a) + cos(a))^2,

and the numerator of this is

sin(a) * (sin(t + a)cos(t) - cos(t + a)sin(t) + cos(a)cos(t)) = sin(a) * (sin(a) + cos(a)cos(t)),

which vanishes when sin(a) + cos(a)cos(t) = 0. Therefore, the length of XY is maximized when t=t', where t' is the value in [0, pi] that satisfies cos(t') = -tan(a).

Note that

(1 - tan(a)) / (1 + tan(a)) = tan(pi/4 - a) = tan((arc MB)/2) = tan<MNB = 3/4,

so tan(a) = 1/7. We compute

sin(a) = sqrt(2)/10

cos(a) = 7sqrt(2)/10

cos(t') = -tan(a) = -1/7

sin(t') = 4sqrt(3)/7

sin(t' + a) = sin(t')cos(a) + cos(t')sin(a) = (28sqrt(6) - sqrt(2))/70,

so the maximum length of XY is sin(a) * sin(t') / (sin(t' + a) + cos(a)) = 7 - 4sqrt(3), and the answer is 7 + 4 + 3 = 014.