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Circles \(S\) and \(T\) have radii \(1\) and intersect at \(A\) and \(B\). The distance between their centers is \(\sqrt{2}\). Let \(P\) be a point on circle \(S\), and let \(\overline{PA}\) and \(\overline{PB}\) intersect circle \(T\) again at \(C\) and \(D\), respectively. Show that \(\overline{CD}\) is a diameter of circle \(T\).

 

 

 

Thanks a lot!

 Apr 3, 2020
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Let L be the center of circle S, and let F be the center of circle T.  Since LA = AF = 1 and LF = sqrt(2), triangle LAF is a 45-45-90 triangle.  This means angle ALF is 45 degrees.  By symmetry, angle BLF is also 45 degrees.

 

Then angle ALB = angle ALF + angle LBF = 45 + 45 = 90 degrees, so arc AB is 90 degrees.  Since angle APB subtends arc AB, angle APB is 90/2 = 45 degrees.  Then by Poncelet's Theorem, triangle PCD is a 45-45-90 triangle, so angle PCD is 90 degrees.  Then angle ACD is 90 degrees, which implies that CD is a diameter of circle T.

 Apr 3, 2020

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