Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 
[[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>.
 
[[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==
+
== 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
+
==Solution 1==
  
 
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 length of 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>.   
 
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 length of 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>.   
Line 12: Line 12:
  
  
−
Solution 2
+
==Solution 2==
  
 
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 = 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}</math>.
 
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 = 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}</math>.

Latest revision as of 19:29, 31 August 2020

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 $\boxed{012}$.


Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_4_2006-2007_Problems/Problem_15&oldid=132897"