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### Semicircle Proof

```
Date: 02/27/2001 at 06:08:17
Subject: Geometry

M is the midpoint of AB. Three semicircles with diameters AM, MB, and
AB are drawn. A circle with centre O and radius r touches all three.
Prove that r = 1/6 AB.
```

```
Date: 02/28/2001 at 14:06:21
From: Doctor Floor
Subject: Re: Geometry

Hi,

Thanks for writing to Dr. Math.

Let's sketch the situation:

X and Y are the centers of the two smaller semicircles. C is the point
where MO (perpendicular bisector of AB) meets the greater semicircle.

Let R be the radius of the greater semicircle, so R = AB/2.

We find that MX = R/2, MO = R-r, and XO = R/2+r. These three segments
form a right triangle, so the Pythagorean theorem must hold:

(R/2)^2 + (R-r)^2 = (R/2+r)^2.

We derive from this

R^2/4 + R^2 - 2rR + r^2 = R^2/4 + rR + r^2

R^2 - 3rR = 0

R(R-3r) = 0

Since R is nonzero, this means that R-3r = 0, thus r = R/3.
And since R = AB/2 this gives r = AB/6, as desired.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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