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

Revision as of 18:32, 20 September 2010

Problem

Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$ . Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$ . The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$ , where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$ .

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).

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.

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

Revision as of 18:32, 20 September 2010

Problem

Solution

@@ Line 1: / Line 1: @@
+== Problem ==
 Let <math>\overline{MN}</math> be a diameter of a circle with diameter 1. Let <math>A</math> and <math>B</math> be points on one of the semicircular arcs determined by <math>\overline{MN}</math> such that <math>A</math> is the midpoint of the semicircle and <math>MB=\frac{3}5</math>. Point <math>C</math> lies on the other semicircular arc. Let <math>d</math> be the length of the line segment whose endpoints are the intersections of diameter <math>\overline{MN}</math> with chords <math>\overline{AC}</math> and <math>\overline{BC}</math>. The largest possible value of <math>d</math> can be written in the form <math> r-s\sqrt{t} </math>, where <math>r, s</math> and <math>t</math> are positive integers and <math>t</math> is not divisible by the square of any prime. Find <math>r+s+t</math>.
 == Solution ==
-Let <math>\angle BAC = 60^\circ</math>
+(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).
+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.