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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 13"
 
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Solving for <math>r_1</math> we get <math>r_1=1</math>
 
Solving for <math>r_1</math> we get <math>r_1=1</math>
  
 +
Since <math>r_1</math> is the radius of the incircle of <math>\Delta PBA</math> then,
  
 +
<math>r_1=\frac{A_1}{s}</math> where <math>A_1</math> is the area of <math>\Delta PBA</math> and <math>s</math> is half the perimeter of <math>\Delta PBA</math>
  
 +
Then, <math>A_1=r_1s=s=\frac{|AB|+(\sqrt{(h+d)^2-r_2^2}-k)+(\sqrt{h^2-r_1^2}+k)}{2}=\frac{20\sqrt{2}+\left( \frac{r_2}{r_1}+1 \right)\sqrt{h^2-r_1^2}}{2}</math>
 +
 +
<math>A_1=\frac{20\sqrt{2}+12\sqrt{3^2-1^2}}{2}=10\sqrt{2}+12\sqrt{2}=22\sqrt{2}</math>
 +
 +
<math>A_1^2=(22\sqrt{2})^2=\boxed{968}</math>
  
 
~Tomas Diaz. orders@tomasdiaz.com
 
~Tomas Diaz. orders@tomasdiaz.com
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Latest revision as of 02:36, 26 November 2023

Problem

Consider two circles of different sizes that do not intersect. The smaller circle has center $O$. Label the intersection of their common external tangents $P$. A common internal tangent intersects the common external tangents at points $A$ and $B$. Given that the radius of the larger circle is $11$, $PO=3$, and $AB=20\sqrt{2}$, what is the square of the area of triangle $PBA$?

Solution

Mock AIME 6 P13a.png

$|AB|=|CD|=20\sqrt{2}$

Let $d$ be the distance between centers, and $h=|PO|=3$

$|CD|^2+(r_2-r_1)^2=d^2$

$(20\sqrt{2})^2+(r_2-r_1)^2=d^2$

$800+(r_2-r_1)^2=d^2$ [Equation 1]

By similar triangles,

$\frac{r_1}{h}=\frac{r_2}{h+d}$

solving for $d$ we have:

$d=\frac{(r_2-r_1)h}{r_1}$ [Equation 2]

Substituting [Equation 2] into [Equation 1] we have:

$800+(r_2-r_1)^2=\frac{(r_2-r_1)^2h^2}{r_1^2}$

$800+(11-r_1)^2=\frac{9(11-r_1)^2}{r_1^2}$

Solving for $r_1$ we get $r_1=1$

Since $r_1$ is the radius of the incircle of $\Delta PBA$ then,

$r_1=\frac{A_1}{s}$ where $A_1$ is the area of $\Delta PBA$ and $s$ is half the perimeter of $\Delta PBA$

Then, $A_1=r_1s=s=\frac{|AB|+(\sqrt{(h+d)^2-r_2^2}-k)+(\sqrt{h^2-r_1^2}+k)}{2}=\frac{20\sqrt{2}+\left( \frac{r_2}{r_1}+1 \right)\sqrt{h^2-r_1^2}}{2}$

$A_1=\frac{20\sqrt{2}+12\sqrt{3^2-1^2}}{2}=10\sqrt{2}+12\sqrt{2}=22\sqrt{2}$

$A_1^2=(22\sqrt{2})^2=\boxed{968}$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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