Inscribed Circle!

Geometry Level pending

The circle above is centered at the origin ( 0 , 0 ) (0,0) and inscribed in the two curves f ( x ) = 1 x 2 + 1 \\ f(x) = \dfrac{1}{x^2 + 1} and g ( x ) = 1 x 2 + 1 g(x) = -\dfrac{1}{x^2 + 1} . Find the area of the red region above.


The answer is 0.06041693771.

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1 solution

Rocco Dalto
Apr 14, 2020

Let P : ( x , 1 1 + x 2 ) P: (x,\dfrac{1}{1 + x^2}) and O : ( 0 , 0 ) O: (0,0)

To minimize O P : \overline{OP} :

D = ( O P ) 2 = x 2 + ( 1 + x 2 ) 2 D = (\overline{OP})^2 = x^2 + (1 + x^2)^{-2} \implies d D d x = 2 x ( ( 1 + x 2 ) 3 2 ( 1 + x 2 ) 3 ) = 0 \dfrac{dD}{dx} = 2x(\dfrac{(1 + x^2)^3 - 2}{(1 + x^2)^3}) = 0

x 0 x = ± 2 1 3 1 x \neq 0 \implies x = \pm\sqrt{2^{\frac{1}{3}} - 1} \implies y = 1 2 1 3 y = \dfrac{1}{2^{\frac{1}{3}}} \implies O P = 3 2 2 3 2 1 3 \overline{OP} = \dfrac{\sqrt{3 - 2^{\frac{2}{3}}}}{2^{\frac{1}{3}}}

Let d = O P d = \overline{OP}

I = 2 1 3 1 2 1 3 1 ( 1 x 2 + 1 d 2 x 2 ) d x I = \displaystyle\int_{-\sqrt{2^{\frac{1}{3}} - 1}}^{\sqrt{2^{\frac{1}{3}} - 1}} (\dfrac{1}{x^2 + 1} - \sqrt{d^2 - x^2}) \:\ dx

Let x = d sin ( θ ) d x = d cos ( θ ) x = d\sin(\theta) \implies dx = d\cos(\theta) \implies

I = ( 2 arctan ( x ) d 2 arcsin ( x d ) + x d 2 x 2 ) 2 1 3 1 2 1 3 1 = I = (2\arctan(x) -d^2\arcsin(\dfrac{x}{d}) + x\sqrt{d^2 - x^2})|_{-\sqrt{2^{\frac{1}{3}} - 1}}^{\sqrt{2^{\frac{1}{3}} - 1}} =

2 arctan ( 2 1 3 1 ) ( 3 2 2 3 2 2 3 ) 2\arctan(\sqrt{2^{\frac{1}{3}} - 1}) - (\dfrac{3 - 2^{\frac{2}{3}}}{2^{\frac{2}{3}}}) arcsin ( 2 1 3 2 1 3 1 3 2 2 3 ) \arcsin(\dfrac{2^{\frac{1}{3}}\sqrt{2^{\frac{1}{3}} - 1}}{\sqrt{3 - 2^{\frac{2}{3}}}}) 2 1 3 1 2 1 3 - \dfrac{\sqrt{2^{\frac{1}{3}} - 1}}{2^{\frac{1}{3}}}

\implies The desired area A = 2 I 0.06041693771 A = 2I \approx \boxed{0.06041693771} .

Note:

0 < x < 2 1 3 1 d D d x < 0 0 < x < \sqrt{2^{\frac{1}{3}} - 1} \implies \dfrac{dD}{dx} < 0 and x > 2 1 3 1 d D d x > 0 x > \sqrt{2^{\frac{1}{3}} - 1} \implies \dfrac{dD}{dx} > 0

\implies relative min at x = 2 1 3 1 x = \sqrt{2^{\frac{1}{3}} - 1}

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