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Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 15"

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

Revision as of 22:51, 26 December 2009

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

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$\boxed{012}$.


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