Philippine Mathematical Olympiad - Area Stage #21

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High School Math Grades 9-12, Ages 13-18

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Philippine Mathematical Olympiad - Area Stage #21

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difference of squares

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#1 h Sep 24, 2014, 2:18 AM • 2 Y

Y by Adventure10, Mango247

Two circle of radius 12 have their centers on each other as shown in the figure. A smaller circle is constructed tangent to $AB$ and the two given circles, internally to the circle below and externally to the circle above, as shown. Find the radius of the smaller circle.

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#2 h Sep 24, 2014, 10:59 AM • 3 Y

Y by james digol, Adventure10, Mango247

This isn't the best solution in the world.

Let the center and radius of the small circle be $O,r$ , and let the shown intersection of the circle center $A$ and the line $AB$ be $X$ . Also let the point where the small circle touches $\overline{AB}$ be $P$ , and the point where it touches the bigger circle that is not centered at $A$ be $Q$ . Note that $|OX|=|QX|-|QO|=12-r$ . By Pythagoras,
1. $|PO|^2+|PX|^2=|OX|^2\Leftrightarrow r^2+|PX|^2=(12-r)^2$
2. $|PO|^2+|AP|^2=|OA|^2\Leftrightarrow r^2+(12+|PX|)^2=(12+r)^2$
So by difference of squares, $|PX|^2=144-24r$ and $(12+|PX|)^2=144+24r$ .
Adding up we get
$|PX|^2+(12+|PX|)^2=288\Leftrightarrow |PX|^2+12|PX|-72=0$ , hence
$2|PX|+12= \sqrt{144-4(-72)}=12\sqrt{ 3}\Leftrightarrow |PX|=6\sqrt{3}-6$ .
So $144-24r=|PX|^2=108+36-72\sqrt{3}\Leftrightarrow r=\bf 3\sqrt{3}$ .

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