Below, you find a nice geometry puzzle.
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Solution 1 by Bruce
- a circle with radius r1 and inside
- two circles with radius r2 and r3 which contact circle 1 at points A and B
- the centers of circles 1,2 and 3.
- the intersection point P of circles 2 and 3
Prove the following theorem:
For APB to be a straight line, we need to show that the sum of angles
APD + DPE + EPB = 180
Notice that CD + AD = r1 also CE + EB = r1
But, our condition is r1 = r2 + r3
So, this means that DPEC is a parallelogram.
So, CD = r3 and CE = r2
Finally, we use the parallelogram to relate angles
DAP = EPB and ADP = DPE
Since angles of triangle ADP sum to 180deg
Then so do the angles on the line APB.