# 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,

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