2 thoughts on “Circle-Step Puzzle”

1. sanjay godse March 15, 2026 at 1:49 am

   Another approach.

   Let AB=4, BC=3 and CD=2.
   Extend BC and DC to meet the circumference at E and F respectively.
   If we complete the rectangle ABCH, the point H will be on FD.
   It is easy to prove that FH=2 since HC=4 and CD=2. Thus FC=2+4=6
   BCE and DCF are intersecting chords.
   BC×CE = FC×CD
   3×CE = 6×2
   CE= 4
   BE=BC+CE=3+4=7
   From ABE right angled triangle AE^2 = AB^2 + BE^2
   Diameter AE^2 = 4^2 + 7^2 = 16+49 = 65

   Circle area = PI*(diameter square)/4
   = PI*65/4

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2. Jim Stevenson Post authorMarch 15, 2026 at 9:14 am

   Very nice. The more geometry you know (and can remember) the more possibilities are available. And this solution is plane geometry and not analytic geometry.

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