Difference between revisions of "2009 AIME II Problems/Problem 15"
| Line 12: | Line 12: | ||
CA | CA | ||
| + | |||
= sin<ABC | = sin<ABC | ||
| + | |||
= sin((arc AN + arc NC)/2) | = sin((arc AN + arc NC)/2) | ||
| + | |||
= sin((pi/2 + (pi-t))/2) | = sin((pi/2 + (pi-t))/2) | ||
| + | |||
= sin(3pi/4 - t/2) | = sin(3pi/4 - t/2) | ||
| + | |||
= sin(pi/4 + t/2) | = sin(pi/4 + t/2) | ||
| Line 21: | Line 26: | ||
CY/CA | CY/CA | ||
| + | |||
= CY/(CY + AY) | = CY/(CY + AY) | ||
| + | |||
= sin(t) / (1 + sin(t)) | = sin(t) / (1 + sin(t)) | ||
| + | |||
= sin(t) / (sin(pi/2) + sin(t)) | = sin(t) / (sin(pi/2) + sin(t)) | ||
| + | |||
= sin(t) / (2sin(pi/4 + t/2)cos(pi/4 - t/2)) | = sin(t) / (2sin(pi/4 + t/2)cos(pi/4 - t/2)) | ||
| Line 29: | Line 38: | ||
XY/CY | XY/CY | ||
| + | |||
= sin<XCY / sin<CXY | = sin<XCY / sin<CXY | ||
| + | |||
= sin(a) / sin(3pi/4 - a - t/2) | = sin(a) / sin(3pi/4 - a - t/2) | ||
| + | |||
= sin(a) / sin(pi/4 + a + t/2). | = sin(a) / sin(pi/4 + a + t/2). | ||
| − | Multiplying (a), (b), and (c), we have | + | (d) Multiplying (a), (b), and (c), we have |
XY | XY | ||
| + | |||
= CA * (CY/CA) * (XY/CY) | = CA * (CY/CA) * (XY/CY) | ||
| + | |||
= sin(t)sin(a) / (2cos(pi/4 - t/2)sin(pi/4 + a + t/2)) | = 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(t)sin(a) / (sin(pi/2 + a) + sin(a + t)) | ||
| + | |||
= sin(a) * sin(t) / (sin(t + a) + cos(a)), | = sin(a) * sin(t) / (sin(t + a) + cos(a)), | ||
| Line 58: | Line 74: | ||
sin(a) = sqrt(2)/10 | sin(a) = sqrt(2)/10 | ||
| + | |||
cos(a) = 7sqrt(2)/10 | cos(a) = 7sqrt(2)/10 | ||
| + | |||
cos(t') = -tan(a) = -1/7 | cos(t') = -tan(a) = -1/7 | ||
| + | |||
sin(t') = 4sqrt(3)/7 | sin(t') = 4sqrt(3)/7 | ||
| + | |||
sin(t' + a) = sin(t')cos(a) + cos(t')sin(a) = (28sqrt(6) - sqrt(2))/70, | sin(t' + a) = sin(t')cos(a) + cos(t')sin(a) = (28sqrt(6) - sqrt(2))/70, | ||
| − | so the maximum length of XY is | + | 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. |
| − | |||
| − | sin(a) * sin(t') / (sin(t' + a) + cos(a)) | ||
| − | = 7 - 4sqrt(3), | ||
| − | |||
| − | and the answer is 7 + 4 + 3 = 014. | ||
Revision as of 18:34, 20 September 2010
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) = 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.
