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

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==Solution 3==
 
==Solution 3==
  
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Let the midpoint of <math>AC</math> be <math>M</math>. Extend the perpendicular bisector of <math>AC</math> to meet <math>\omega</math> at <math>E</math>. Note <math>\angle ACE = \angle ABC</math> as <math>ACBE</math> is cyclic. Further, note <math>\angle ABC = \angle AEC = \angle AED + \angle DEC = \angle ACD + \angle DAC</math> as <math>ADCE</math> is cyclic.  
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Let the midpoint of <math>AC</math> be <math>M</math>. Extend the perpendicular bisector of <math>AC</math> to meet <math>\omega</math> at <math>E</math>. Note <math>\angle ACE = \angle ABC</math> as <math>ACBE</math> is cyclic. Further, note <math>\angle ABC = \angle AEC = \angle AED + \angle DEC = \angle ACD + \angle DAC</math> as <math>ADCE</math> is cyclic.
  
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Note <math>AD = AC</math> as <math>D</math> is on the perpendicular bisector of <math>AC</math>. Hence, <math>\angle ACD = \angle DAC = \frac{\angle ABC}{2}</math>. The problem then boils down to finding <math>\cos\left(\frac{B}{2}\right)</math>, which we know we can do. \\
+
Note <math>AD = AC</math> as <math>D</math> is on the perpendicular bisector of <math>AC</math>. Hence, <math>\angle ACD = \angle DAC = \frac{\angle ABC}{2}</math>. The problem then boils down to finding <math>\cos\left(\frac{B}{2}\right)</math>, which we know we can do.
  
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By Heron's formula and <math>rs = A</math>, we yield that the inradius is <math>3\sqrt{3}</math>. Let <math>I</math> and <math>P</math> be the incenter and the foot of the perpendicular from the incenter to <math>BC</math> respectively. Then, <math>BP = s - b = 4</math>, and <math>BI = \sqrt{43}</math> by the Pythagorean theorem. Then, <math>\cos\left(\frac{B}{2}\right) = \frac{4}{\sqrt{43}}</math>. Therefore, <math>AD = 6\sqrt{43}</math>, by examining right triangle <math>AMD</math>. \\
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By Heron's formula and <math>rs = A</math>, we yield that the inradius is <math>3\sqrt{3}</math>. Let <math>I</math> and <math>P</math> be the incenter and the foot of the perpendicular from the incenter to <math>BC</math> respectively. Then, <math>BP = s - b = 4</math>, and <math>BI = \sqrt{43}</math> by the Pythagorean theorem. Then, <math>\cos\left(\frac{B}{2}\right) = \frac{4}{\sqrt{43}}</math>. Therefore, <math>AD = 6\sqrt{43}</math>, by examining right triangle <math>AMD</math>.
  
 
As <math>12 < 6 + \sqrt{43} < 13</math>, our answer is <math>\boxed{012}</math>.
 
As <math>12 < 6 + \sqrt{43} < 13</math>, our answer is <math>\boxed{012}</math>.

Latest revision as of 13:46, 27 December 2024

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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