It's possible entirely with the 'power of a point' paradigm: Let the line segment RSP intersect C again at T. By the tangent- secant theorem, RQ^2 = RS.RT, so RT = 12^2/8 = 18. Hence, ST = 10 and PT = 6. The power of P with respect to C is -(PS.PT) = -24, which is also the value of OP^2 - r^2. Since OP^2 = 16, we can deduce that r^2 = 40 and thus r = sqrt(40). Sincerely, Adam P. Goucher http://cp4space.wordpress.com

----- Original Message ----- From: davidwwilson@comcast.net Sent: 01/26/13 06:32 PM To: math-fun Subject: Re: [math-fun] Geometry puzzle

Sorry, let's try again

Let segment PR cut C at point S bete

This was a problem posed to me by a friend:

Let circle C have center O.

Let point P satisfy OP = 4.

Draw tangent line L to C at point Q. Place point R on L with QR = 12.

Let segment PR cut C at point S so that RS = 8 and RP = 12.

What is the radius of C?

Perhaps I got it right this time.

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