Piece of cake ?

Calculus Level 5

The red circle is the biggest circle inscribed in the circular sector of angle α \alpha and radius 100 100 .

If α \alpha is chosen uniformly at random between 0 ∘ 0^\circ and 18 0 ∘ 180^\circ , what is the expected value of the radius of the red circle? (Give your result to 3 decimals.)


The answer is 36.338.

Romain Bouchard by Romain Bouchard , 34, France

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1 solution

Let the radius of the inscribed circle be r r . We note that:

sin ⁡ α 2 = r 100 − r ⟹ r = 100 sin ⁡ α 2 1 + sin ⁡ α 2 \begin{aligned} \sin \frac \alpha 2 & = \frac r{100-r} \\ \implies r & = \frac {100\sin \frac \alpha 2}{1+\sin \frac \alpha 2} \end{aligned}

Then the expected value of the radius of the inscribed circle for 0 ∘ ≤ r ≤ 18 0 ∘ 0^\circ \le r \le 180^\circ or 0 ≤ r ≤ π 0 \le r \le \pi is given by:

E [ R ] = 1 π ∫ 0 π 100 sin ⁡ α 2 1 + sin ⁡ α 2 d α Let θ = α 2 ⟹ 2 d θ = d α = 200 π ∫ 0 π 2 sin ⁡ θ 1 + sin ⁡ θ d θ Using half-angle tangent = 200 π ∫ 0 1 2 t 1 + t 2 1 + 2 t 1 + t 2 ⋅ 2 d t 1 + t 2 substitution and let t = tan ⁡ θ 2 = 200 π ∫ 0 1 4 t ( t 2 + 1 ) ( t 2 + 2 t + 1 ) d t = 400 π ∫ 0 1 ( 1 t 2 + 1 − 1 ( t + 1 ) 2 ) d t = 400 π [ tan ⁡ − 1 t + 1 t + 1 ] 0 1 = 400 π [ π 4 + 1 2 − 1 ] = 100 − 200 π ≈ 36.338 \begin{aligned} E[R] & = \frac 1\pi \int_0^\pi \frac {100\sin \frac \alpha 2}{1+\sin \frac \alpha 2} d \alpha & \small \color{#3D99F6} \text{Let }\theta = \frac \alpha 2 \implies 2 d\theta = d\alpha \\ & = \frac {200}\pi \int_0^\frac \pi 2 \frac {\sin \theta}{1+\sin \theta} d \theta & \small \color{#3D99F6} \text{Using half-angle tangent} \\ & = \frac {200}\pi \int_0^1 \frac {\frac {2t}{1+t^2}}{1+\frac {2t}{1+t^2}} \cdot \frac {2\ dt}{1+t^2} & \small \color{#3D99F6} \text{substitution and let }t = \tan \frac \theta 2 \\ & = \frac {200}\pi \int_0^1 \frac {4t}{(t^2+1)(t^2+2t+1)} dt \\ & = \frac {400}\pi \int_0^1 \left(\frac 1{t^2+1} - \frac 1{(t+1)^2}\right) dt \\ & = \frac {400}\pi \left[\tan^{-1} t + \frac 1{t+1}\right]_0^1 \\ & = \frac {400}\pi \left[\frac \pi 4 + \frac 12 - 1 \right] \\ & = 100 - \frac {200}\pi \\ & \approx \boxed{36.338} \end{aligned}

12 Helpful 1 Interesting 0 Brilliant 0 Confused

To make things easier you could split ∫ 0 π 2 s i n θ s i n θ + 1 d θ = π 2 − ∫ 0 π 2 1 s i n θ + 1 d θ \int_0^\frac {\pi}{2} \frac {sin \theta}{sin \theta + 1} d\theta = \frac{\pi}{2}- \int_0^\frac{\pi}{2} \frac {1}{sin \theta + 1} d\theta and then use Weierstrass' substitution. This is not crucial but makes things a bit easier :)

Robert Szafarczyk - 3 years, 1 month ago

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Or you could multiply top and bottom by 1 − sin ⁡ θ 1-\sin \theta : ∫ sin ⁡ θ sin ⁡ θ + 1 d θ = ∫ sin ⁡ θ ( 1 − sin ⁡ θ ) 1 − sin ⁡ 2 θ d θ = ∫ sin ⁡ θ − sin ⁡ 2 θ cos ⁡ 2 θ d θ = 1 cos ⁡ θ − ∫ tan ⁡ 2 θ d θ = 1 cos ⁡ θ − ( tan ⁡ θ − θ ) = 1 − sin ⁡ θ cos ⁡ θ + θ . \begin{aligned} \int \frac{\sin \theta}{\sin \theta + 1} d\theta &= \int \frac{\sin \theta(1-\sin \theta)}{1-\sin^2\theta} d\theta \\ &= \int \frac{\sin\theta-\sin^2\theta}{\cos^2\theta} d\theta \\ &= \frac1{\cos\theta} - \int \tan^2 \theta \, d\theta \\ &= \frac1{\cos\theta} - (\tan\theta - \theta) \\ &= \frac{1-\sin\theta}{\cos\theta} + \theta. \end{aligned}

Then at θ = π / 2 \theta=\pi/2 this evaluates to π / 2 \pi/2 (have to use L'Hopital on the first part), and at θ = 0 \theta = 0 this evaluates to 1 , 1, so you get 200 π ( π 2 − 1 ) = 100 − 200 π . \frac{200}{\pi}\left( \frac{\pi}2 - 1\right) = 100 - \frac{200}{\pi}.

Patrick Corn - 3 years, 1 month ago

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Yeah! This simplifies it too.

Robert Szafarczyk - 3 years, 1 month ago

Since α \alpha is a random number between 180 and 0, wouldn't the expected value of α \alpha be 90°? Therefore, you could just find the radius of a circle inscribed in a quarter circle with radius 100, right?

Blan Morrison - 3 years, 1 month ago

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That's what I was thinking at first but then I realised that the radius of the red circle is not as well distributed as the angle α. That's why it can't be derived to such simplicity.

Robert Szafarczyk - 3 years, 1 month ago

The radius is a function of α , \alpha, say r = f ( α ) , r=f(\alpha), and in general it's not true that E ( f ( α ) ) = f ( E ( α ) ) . E(f(\alpha)) = f(E(\alpha)).

Patrick Corn - 3 years, 1 month ago

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