Mock AIME 4 2006-2007 Problems/Problem 15

Problem

Triangle $ABC$ has sides $\overline{AB}$ , $\overline{BC}$ , and $\overline{CA}$ of length 43, 13, and 48, respectively. Let $\omega$ be the circle circumscribed around $\triangle ABC$ and let $D$ be the intersection of $\omega$ and the perpendicular bisector of $\overline{AC}$ that is not on the same side of $\overline{AC}$ as $B$ . The length of $\overline{AD}$ can be expressed as $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find the greatest integer less than or equal to $m + \sqrt{n}$ .

Solution

The perpendicular bisector of any chord of any circle passes through the center of that circle. Let $M$ be the midpoint of $\overline{AC}$ , and $R$ be the radius of $\omega$ . By the Power of a Point Theorem, $MD \cdot (2R - MD) = AM \cdot MC = 24^2$ or $0 = MD^2 -2R\cdot MD 24^2$ . By the Pythagorean Theorem, $AD^2 = MD^2 + AM^2 = MD^2 + 24^2$ .

Let's compute the circumradius $R$ : By the Law of Cosines, $\cos B = \frac{AB^2 + BC^2 - CA^2}{2\cdot AB\cdot BC} = \frac{43^2 + 13^2 - 48^2}{2\cdot43\cdot13} = -\frac{11}{43}$ . By the Law of Sines, $2R = \frac{AC}{\sin B} = \frac{48}{\sqrt{1 - \left(-\frac{11}{43}\right)^2}} = \frac{86}{\sqrt 3}$ so $R = \frac{43}{\sqrt 3}$ .

Now we can use this to compute $MD$ and thus $AD$ . By the quadratic formula, $MD = \frac{2R + \sqrt{4R^2 - 4*24^2}}{2} = \frac{43}{\sqrt 3} + \frac{11}{\sqrt3} = 18\sqrt{3}$ . (We only take the positive sign because angle $B$ is obtuse so $\overline{MD}$ is the longer of the two segments into which the chord $\overline{AC}$ divides the diameter.) Then $AD^2 = MD^2 + 24^2 = 1548$ so $AD = 6\sqrt{43}$ , and $12 < 6 + \sqrt{43} < 13$ so the answer is $012$ .

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Mock AIME 4 2006-2007 Problems/Problem 15

Problem

Solution