Re: Friday Puzzler -- Circle's radius

Larry Barrett
I don't have an exact solution; in fact so far I don't have any solution.
And I am not sure this sheds any light on the path to an exact solution. But in geometry there is a theorem called the Intersection Chords theorem. This theorem says that if two chords intersect and the two segments of one chord are a and b, and the two segments of the other chord are c and d, then a x b = c x d.
In this problem, the two segments of the horizontal chord are 2 and 6; one segment of the vertical chord is 3 but the other segment is not labeled; call it y. The theorem says that 2 x 6 = 3 x y, so y must be 4.

And the two chords are perpendicular, so that must also be important in finding R.

That's my two cents.

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