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 $m + \sqrt{n}$ .

Best Solution

We set up a trivial coordinate bash. Let A = 0,0, C = 48,0, B = 83/2, 13sqrt3/2. We find the coordinates of the circumcenter to be 24, -11sqrt3/3. The radius is 43sqrt3.Then the coordinate of point D are 24, -18sqrt3. The answer is then 6 + sqrt43, which yields 12.

Solution 1

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 length of 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, $MD = \frac{2R + \sqrt{4R^2 - 4\cdot24^2}}{2} = \frac{43}{\sqrt 3} + \frac{11}{\sqrt3} = 18\sqrt{3}$ . (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 $012$ .

Solution 2

Let angle $ABC$ = $a$ , angle $ADC = b$ , and $AD = DC = x$ . Since ABCD is a cyclic quadrilateral, $a + b = 180$ degrees. Using the Law of Cosines, $48^2 = 43^2 + 13^2 - (2)(43)(13)\cos a$ , so $\cos a = -\frac{11}{43}$ . Since $\cos a = -\frac{11}{43}$ , then $\cos b$ is $\frac{11}{43}$ . Using the Law of Cosines on triangle ADC, $48^2 = 2x^2 - 2x^2 \cos b = 2x^2(1 - \cos b) = 2x^2(\frac{32}{43})$ . Solving for $x$ , we get $x = 6 + {\sqrt 43}$ which is between $12$ and $13$ , so the answer is $\boxed{012}$ .

Solution 3

Let the midpoint of $AC$ be $M$ . Extend the perpendicular bisector of $AC$ to meet $\omega$ at $E$ . Note $\angle ACE = \angle ABC$ as $ACBE$ is cyclic. Further, note $\angle ABC = \angle AEC = \angle AED + \angle DEC = \angle ACD + \angle DAC$ as $ADCE$ is cyclic.

Note $AD = AC$ as $D$ is on the perpendicular bisector of $AC$ . Hence, $\angle ACD = \angle DAC = \frac{\angle ABC}{2}$ . The problem then boils down to finding $\cos\left(\frac{B}{2}\right)$ , which we know we can do.

By Heron's formula and $rs = A$ , we yield that the inradius is $3\sqrt{3}$ . Let $I$ and $P$ be the incenter and the foot of the perpendicular from the incenter to $BC$ respectively. Then, $BP = s - b = 4$ , and $BI = \sqrt{43}$ by the Pythagorean theorem. Then, $\cos\left(\frac{B}{2}\right) = \frac{4}{\sqrt{43}}$ . Therefore, $AD = 6\sqrt{43}$ , by examining right triangle $AMD$ .

As $12 < 6 + \sqrt{43} < 13$ , our answer is $\boxed{012}$ .

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