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Home » Solution: In a right triangle, the circumradius $ R $ is half the hypotenuse: - DNSFLEX

Solution: In a right triangle, the circumradius $ R $ is half the hypotenuse: - DNSFLEX

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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First, compute \( I'(t) \) using the quotient rule, where \( u = t^2 \...
I'(t) = rac{u'v - uv'}{v^2} = rac{(2t)(1 + t^3) - t^2(3t^2)}{(1 + t^...
I'(t) = rac{2t + 2t^4 - 3t^4}{(1 + t^3)^2} = rac{2t - t^4}{(1 + t^3)...
Now, find \( I''(t) \) to locate where \( I'(t) \) has a maximum:
Let \( N(t) = 2t - t^4 \) and \( D(t) = (1 + t^3)^2 \), then:
I''(t) = rac{N'(t)D(t) - N(t)D'(t)}{[D(t)]^2}.
Compute derivatives:
N'(t) = 2 - 4t^3, \quad D'(t) = 2(1 + t^3)(3t^2) = 6t^2(1 + t^3).
I''(t) = rac{(2 - 4t^3)(1 + t^3)^2 - (2t - t^4)(6t^2(1 + t^3))}{(1 +...
Factor \( (1 + t^3) \) from numerator:

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