Date: 02/27/2001 at 06:08:17 From: Vmadhulika 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/
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