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In a right triangle, the length of hypotenuse is $c$. The centers of three circles of radius $c/5$ are found at its vertices. Find the radius of the fourth circle which touches the three given circles and doesn't enclose them.

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I found the radius. Now what? –  The Chaz 2.0 Oct 2 '11 at 1:51
    
@TheChaz: Can you please write proof –  Ramana Venkata Oct 2 '11 at 1:54
    
Why the upvote? (This is me "telling", btw) –  The Chaz 2.0 Oct 2 '11 at 5:08
    
@TheChaz: You parenthesized remark is completely cryptic. Your question is not: I upvoted it since it seems like a reasonable question. –  Michael Hardy Oct 2 '11 at 13:36
    
@Michael: I see. –  The Chaz 2.0 Oct 2 '11 at 13:54

2 Answers 2

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The midpoint of the hypotenuse is at a distance of $c/2$ from either of the endpoints of the hypotenuse. It's not hard to show that it's also at that same distance from the vertex of the right angle. Therefore a circle centered there that is just big enough to touch the two circles of radius $c/5$ centered at the endpoints of the hypotenuse will also be just big enough to touch that third circle of radius $c/5$ centered at the vertex of the right angle. So do the arithmetic.

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Hint: Show that the center of the fourth circle is equidistant to each vertex of the triangle. Hence show that the center of the fourth circle is the circumcenter of the triangle. As the circumradius of a right triangle with hypotenuse $c$ is $c/2$, the radius of the fourth circle is $c/2-c/5$.

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