s09 - circle. Then angle EDA is a right angle, angle BAC is...

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Solution to Problem 9 Congratulations to this week’s winner Mike Fitzpatrick There were 4 solutions received. Correct solutions were also received from Kevin Bourrillion, Jeffrey Decker, and Daniel Reeves. Call the centers of the big circles A, B, and C and denote by D the midpoint of AB. Clearly, by symmetry, D is the point of contact between two of the big circles. Let E be the center of the little
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Unformatted text preview: circle. Then angle EDA is a right angle, angle BAC is 60 degrees and, since EA bisects BAC, BAE is 30 degrees. We are trying to find the length of DA, call it R. The hypotenuse of the triangle EDA is EA and has length r + R. The two sides DA and EA of the 30-60-90 triangle EDA will be in the ratio 3 (1/2) /2, that is, R/(r + R) = 3 (1/2) /2. Therefore R = r / (2/3 (1/2)- 1)....
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s09 - circle. Then angle EDA is a right angle, angle BAC is...

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