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

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==Problem==
 
==Problem==
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Triangle <math>ABC</math> has sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math> of length 43, 13, and 48, respectively. Let <math>\omega</math> be the circle circumscribed around <math>\triangle ABC</math> and let <math>D</math> be the intersection of <math>\omega</math> and the perpendicular bisector of <math>\overline{AC}</math> that is not on the same side of <math>\overline{AC}</math> as <math>B</math>. The length of <math>\overline{AD}</math> can be expressed as <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find the greatest integer less than or equal to <math>m + \sqrt{n}</math>.
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[[Triangle]] <math>ABC</math> has sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math> of [[length]] 43, 13, and 48, respectively. Let <math>\omega</math> be the [[circle]] [[circumscribe]]d around <math>\triangle ABC</math> and let <math>D</math> be the [[intersection]] of <math>\omega</math> and the [[perpendicular bisector]] of <math>\overline{AC}</math> that is not on the same side of <math>\overline{AC}</math> as <math>B</math>. The length of <math>\overline{AD}</math> can be expressed as <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are [[positive integer]]s and <math>n</math> is not [[divisibility | divisible]] by the [[square]] of any [[prime]]. Find the greatest [[integer]] less than or equal to <math>m + \sqrt{n}</math>.
 
==Solution==
 
==Solution==
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The perpendicular bisector of any [[chord]] of any circle passes through the [[center]] of that circle.  Let <math>M</math> be the midpoint of <math>\overline{AC}</math>, and <math>R</math> be the radius of <math>\omega</math>.  By the [[Power of a Point Theorem]], <math>MD \cdot (2R - MD) = AM \cdot MC = 24^2</math> or <math>0 = MD^2 -2R\cdot MD 24^2</math>.  By the [[Pythagorean Theorem]], <math>AD^2 = MD^2 + AM^2 = MD^2 + 24^2</math>. 
  
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{{solution}}
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Let's compute the [[circumradius]] <math>R</math>: By the [[Law of Cosines]], <math>\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}</math>.  By the [[Law of Sines]], <math>2R = \frac{AC}{\sin B} = \frac{48}{\sqrt{1 - \left(-\frac{11}{43}\right)^2}} = \frac{86}{\sqrt 3}</math> so <math>R = \frac{43}{\sqrt 3}</math>.
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Now we can use this to compute <math>MD</math> and thus <math>AD</math>.  By the [[quadratic formula]], <math>MD = \frac{2R + \sqrt{4R^2 - 4*24^2}}{2} = \frac{43}{\sqrt 3} + \frac{11}{\sqrt3} = 18\sqrt{3}</math>.  (We only take the positive sign because [[angle]] <math>B</math> is [[obtuse]] so <math>\overline{MD}</math> is the longer of the two [[segment]]s into which the chord <math>\overline{AC}</math> divides the diameter.)  Then <math>AD^2 = MD^2 + 24^2 = 1548</math> so <math>AD = 6\sqrt{43}</math>, and <math>12 < 6 + \sqrt{43} < 13</math> so the answer is <math>012</math>.
  
  
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*[[Mock AIME 4 2006-2007 Problems/Problem 14| Previous Problem]]
 
*[[Mock AIME 4 2006-2007 Problems/Problem 14| Previous Problem]]
 
*[[Mock AIME 4 2006-2007 Problems]]
 
*[[Mock AIME 4 2006-2007 Problems]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 15:31, 12 February 2007

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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