A Trigonometric Solution to a Difficult Sangaku Problem
Michel Cabart
Feb. 25, 2010
PROBLEM. In ΔABC, AB = BC. If one chooses D on AB and J on CD such that
SOLUTION
Let's say ∠ADC = x, AD = 1 and tan(x/2) = t. We calculate ρ, the radius of the incircle of ΔADJ by a known formula for the radius of the incircle:
ρ = AD·tan(∠DAJ)·tan(∠ADJ) / (tan(∠DAJ) + tan(∠ADJ)),
with: tan(∠ADJ) = t, tan(∠DAJ) = tan(45°  x/2) = (1  t)/(1 + t). So that
ρ = t(1  t)/(t² + 1) .
Similarly r2 = CD·tan(∠CDO_{2})·tan(∠DCO_{2}) / (tan(∠CDO_{2}) + tan(∠DCO_{2})),
with CD = 2 cos(x) = 2 (1  t²)/(1 + t²). Further,
tan(∠CDO_{2} = tan(90°  x/2) = 1/t,
tan DCO_{2} = tan (90°  3x/2) = (1  3t^{2})/(3t  t^{3}),
so that r_{2} = ½(1  3t²)/t(1 + t²).
Equating ρ and r_{2} yields the equation:
2t^{3}  5t^{2} + 1 = (2t  1)·(t^{2}  2t  1) = 0.
The second (quadratic) factor cannot be zero as it would lead to
The solution is then t = 1/2, leading to ρ = r_{2} = 1/5,
Sangaku

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