Below, you find a nice geometry puzzle. Send your solution , with your name and e-mail address, to so I can publish it here. Solution 1 by Bruce Description Given are: - a circle with radius r _{1} and inside - two circles with radius r _{2} and r_{3} 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: -
If P is on line AB then
r must be true
_{1} = r_{2} + r_{3}success...! Solution 1 (Bruce) 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. |
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