2009 AIME II Problems/Problem 15

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

Let $O$ be the center of the circle. Define $\angle{MOC}=t$ , $\angle{BOA}=2a$ , 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

$CA$

$= \sin\angle{ABC}$

$= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)$

$= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)$

$= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)$

$= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)$

(b) Note that $CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)$ and $AO = \frac{1}{2}$ . Since $CC'Y$ and $AOY$ are similar right triangles, we have $CY/AY = CC'/AO = \sin(t)$ , and hence,

$CY/CA$

$= \frac{CY}{CY + AY}$

$= \frac{\sin(t)}{1 + \sin(t)}$

$= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}$

$= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}$

(c) We have $\angle{XCY} = \frac{\widehat{AB}}{2}=a$ and $\angle{CXY} = \frac{\widehat{MB}+\widehat{CN}}{2} = \frac{\left(\frac{\pi}{2} - 2a\right) + (\pi - t)}{2} = \frac{3\pi}{4} - a - \frac{t}{2}$ , and hence by the Law of Sines,

$XY/CY$

$= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}$

$= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}$

$= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}$

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

$XY$

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

$= \frac{\sin(t)\sin(a)}{2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}$

$= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}$

$= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}$ ,

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

$\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}$ ,

and the numerator of this is

$\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))$ $= \sin(a) \times (\sin(a) + \cos(a)\cos(t))$ ,

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

$\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}$ ,

so $\tan(a) = \frac{1}{7}$ . We compute

$\sin(a) = \frac{\sqrt{2}}{10}$

$\cos(a) = \frac{7\sqrt{2}}{10}$

$\cos(t') = -\tan(a) = -\frac{1}{7}$

$\sin(t') = \frac{4\sqrt{3}}{7}$

$\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}$ ,

so the maximum length of $XY$ is $\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}$ , and the answer is $7 + 4 + 3 = \boxed{014}$ . The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2009 AIME II Problems/Problem 15

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