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Find a value of k for which the three numbers arctan(1/2), arctan(1/2 + k) and arctan(1/2 + 2k) can be the angle measures of a triangle.

1. k = 11/4
2. k = 1/2
3. k = 1
4. k = 1/3

Correct answer: k = 11/4

Solution

For triangle angles A+B+C = pi: tan A + tan B + tan C = tan A tan B tan C. With t1 = 1/2, t2 = 1/2+k, t3 = 1/2+2k: sum = 3/2 + 3k; product = (1/2)(1/2+k)(1/2+2k) = 1/8 + (3/4)k + k². Equate: 3/2 + 3k = 1/8 + (3/4)k + k² => 8k² - 18k - 11 = 0 => k = [18 +- sqrt(324 + 352)]/16 = [18 +- 26]/16 = 11/4 or -1/2. The value k = -1/2 makes t2 = 0 (a zero angle), so it is rejected. Thus k = 11/4, giving tangents 1/2, 13/4, 6 (all positive, valid acute angles summing to pi).

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