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Re: Chords of Concentric Circles [WARNING! SPOILER!]
Posted:
Mar 23, 1993 3:32 PM
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In article <1olds5$m16@forum.swarthmore.edu> Bill Schneider <Bill.Schneider@f32.n272.z1.fidonet.org> writes:
> There are two concentric circles such that the radius of the smaller
> plus the radius of the larger is 36. A chord of the larger circle is
> trisected by the smaller; that is, it must pass through the smaller
> circle. This chord is length 36 also. What is the radius of the large
> circle?
Let a and b denote the radii of the small and large circles, respectively.
Let L denote the length of the chord of the larger circle, and let l denote
the length of its sub-segment that forms a chord of the smaller circle.
Finally, let h denote the distance of the chord from the common center of
the circles.
By the Pythagorean theorem: (L/2)^2 + h^2 = b^2 and (l/2)^2 + h^2 = a^2.
Subtract these to obtain: L^2 - l^2 = 4*(b^2 - a^2). Therefore:
L^2 - l^2 = 4*(b-a)*(b+a).
Now, plug in L=36, l=L/3=36/3, and b+a=36, to obtain b-a=8.
Solve the system { b+a=36, b-a=8 } to arrrive at: a=14, b=22. Q.E.D.
--
Rouben Rostamian
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