Expected Distance on a Circle

Calculus Level 5

An angle θ \theta is chosen randomly in the interval [ 0 , 2 π ) [0,2\pi ) , and its corresponding point on the unit circle is plotted. Let this point be P P . Now let A = ( 1 , 0 ) A=(1,0) and B = ( − 1 , 0 ) B=(-1,0) .

Let the expected value of P A + P B − 2 PA+PB-2 be E E .

What is ⌊ 1000 E ⌋ \lfloor 1000E\rfloor ?


The answer is 546.

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Daniel Liu by Daniel Liu , 21, USA

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2 solutions

Discussions for this problem are now closed

Jatin Yadav
May 4, 2014

The point can be taken as cos ⁡ θ , sin ⁡ θ \cos \theta, \sin \theta

PA + PB = ( 1 − cos ⁡ θ ) 2 + sin ⁡ 2 θ + ( 1 + cos ⁡ θ ) 2 + sin ⁡ 2 θ = 2 − 2 cos ⁡ θ + 2 + 2 cos ⁡ θ \sqrt{(1-\cos \theta)^2 + \sin^2 \theta} + \sqrt{(1+\cos \theta)^2 + \sin^2 \theta} = \sqrt{2 - 2\cos \theta } + \sqrt{2 + 2 \cos \theta}

= 2 ( ∣ sin ⁡ θ 2 ∣ + ∣ cos ⁡ θ 2 ∣ ) 2(|\sin \frac{\theta}{2}|+|\cos \frac{\theta}{2}|)

The probability that the point is chosen between θ \theta , and θ + d θ \theta + d \theta is d θ 2 π \dfrac{d \theta}{2 \pi}

Hence, expected value of P A + P B − 2 PA+PB-2 is ∫ 0 2 π 2 ( ∣ sin ⁡ θ 2 ∣ + ∣ cos ⁡ θ 2 ∣ − 1 ) d θ 2 π = 8 π − 2 \dfrac{\int_{0}^{2\pi} 2(|\sin \frac{\theta}{2}|+|\cos \frac{\theta}{2}|-1) d \theta}{2 \pi} = \dfrac{8}{\pi} - 2

Hence, our answer is ⌊ 1000 ( 8 π − 2 ) ⌋ = 546 \bigg \lfloor 1000(\frac{8}{\pi} - 2) \bigg \rfloor = 546

24 Helpful 0 Interesting 0 Brilliant 0 Confused

How many points are there between x and x+dx?(here theta)

A Former Brilliant Member - 7 years, 1 month ago

@Anup Navin

The notation d x dx means taking the limit as d x dx approaches 0. So as d x dx approaches 0, x + d x x + dx approaches x x and so, taking the limit, we find something very interesting.

Even as d x dx becomes smaller and smaller, there are still an infinite number of points between x x and x + d x x + dx . However, when d x = 0 dx = 0 , we find no points between x x and x + d x x + dx . So what is the number of points between x x and x + d x x + dx ? Is this question very well-defined?

Colin Tang - 7 years, 1 month ago

so, assuming that the probability that the point is chosen between x and x+dx is still an approximation?

A Former Brilliant Member - 7 years, 1 month ago
Trevor B.
May 4, 2014

The point P P is going to be located at ( cos ⁡ θ , sin ⁡ θ ) . (\cos\theta,\sin\theta). By appling the distance formula to both points, we find this. P A + P B = ( 1 − cos ⁡ θ ) 2 + sin ⁡ 2 θ + ( − 1 − cos ⁡ θ ) 2 + sin ⁡ 2 θ = 1 − 2 cos ⁡ θ + cos ⁡ 2 θ + sin ⁡ 2 θ + 1 + 2 cos ⁡ θ + cos ⁡ 2 θ + sin ⁡ 2 θ = 2 − 2 cos ⁡ θ + 2 + 2 cos ⁡ θ \begin{aligned} PA+PB&=\sqrt{(1-\cos\theta)^2+\sin^2\theta}+\sqrt{(-1-\cos\theta)^2+\sin^2\theta}\\ &=\sqrt{1-2\cos\theta+\cos^2\theta+\sin^2\theta}+\sqrt{1+2\cos\theta+\cos^2\theta+\sin^2\theta}\\ &=\sqrt{2-2\cos\theta}+\sqrt{2+2\cos\theta} \end{aligned} Now we have to find the average value of this as theta ranges from 0 0 to 2 π . 2\pi. Start by simplifying the integral a little bit. 1 2 π ∫ 0 2 π 2 − 2 cos ⁡ θ + 2 + 2 cos ⁡ θ d θ = 2 2 π ∫ 0 2 π 1 − cos ⁡ θ + 1 + cos ⁡ θ d θ = 2 2 π ( ∫ 0 2 π 1 − cos ⁡ θ d θ + ∫ 0 2 π 1 + cos ⁡ θ d θ ) \begin{aligned} \dfrac{1}{2\pi}\int_0^{2\pi}\sqrt{2-2\cos\theta}+\sqrt{2+2\cos\theta}\text{ }d\theta&=\dfrac{\sqrt{2}}{2\pi}\int_0^{2\pi}\sqrt{1-\cos\theta}+\sqrt{1+\cos\theta}\text{ }d\theta\\ &=\dfrac{\sqrt{2}}{2\pi}\left(\int_0^{2\pi}\sqrt{1-\cos\theta}\text{ }d\theta+\int_0^{2\pi}\sqrt{1+\cos\theta}\text{ }d\theta\right) \end{aligned} Those integrals aren't going to be easy to solve directly, so simplify the radicals to something we can make a u -substitution u\text{-substitution} out of.

1 − cos ⁡ θ = 1 − cos ⁡ 2 θ 1 + cos ⁡ θ = ∣ sin ⁡ θ ∣ 1 + cos ⁡ θ \sqrt{1-\cos\theta}=\dfrac{\sqrt{1-\cos^2\theta}}{\sqrt{1+\cos\theta}}=\dfrac{|\sin\theta|}{\sqrt{1+\cos\theta}}

1 + cos ⁡ θ = 1 − cos ⁡ 2 θ 1 − cos ⁡ θ = ∣ sin ⁡ θ ∣ 1 − cos ⁡ θ \sqrt{1+\cos\theta}=\dfrac{\sqrt{1-\cos^2\theta}}{\sqrt{1-\cos\theta}}=\dfrac{|\sin\theta|}{\sqrt{1-\cos\theta}}

To avoid dealing with improper integrals, you can say that ∫ 0 2 π ∣ sin ⁡ θ ∣ 1 ± cos ⁡ θ d θ = ∫ 0 π 2 4 sin ⁡ θ 1 ± cos ⁡ θ d θ \displaystyle\int_0^{2\pi}\dfrac{|\sin\theta|}{\sqrt{1\pm\cos\theta}}\text{ }d\theta=\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{4\sin\theta}{\sqrt{1\pm\cos\theta}}\text{ }d\theta

2 2 π ( ∫ 0 2 π 1 − cos ⁡ θ d θ + ∫ 0 2 π 1 + cos ⁡ θ d θ ) = 2 2 π ( ∫ 0 π 2 sin ⁡ θ 1 + cos ⁡ θ d θ + ∫ 0 π 2 sin ⁡ θ 1 − cos ⁡ θ d θ ) \dfrac{\sqrt{2}}{2\pi}\left(\int_0^{2\pi}\sqrt{1-\cos\theta}\text{ }d\theta+\int_0^{2\pi}\sqrt{1+\cos\theta}\text{ }d\theta\right) =\dfrac{2\sqrt{2}}{\pi}\left(\int_0^\frac{\pi}{2}\dfrac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ }d\theta+\int_0^\frac{\pi}{2}\dfrac{\sin\theta}{\sqrt{1-\cos\theta}}\text{ }d\theta\right)

Let's solve these integrals one at a time.

∫ 0 π 2 sin ⁡ θ 1 + cos ⁡ θ d θ = ∫ 2 1 − d u u = ∫ 1 2 d u u = 2 u ∣ 1 2 = 2 2 − 2 \begin{aligned} \int_0^\frac{\pi}{2}\dfrac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ }d\theta&=\int_2^1\dfrac{-du}{\sqrt{u}}\\ &=\int_1^2\dfrac{du}{\sqrt{u}}\\ &=\left.2\sqrt{u}\right|^2_1\\ &=2\sqrt{2}-2 \end{aligned} ∫ 0 π 2 sin ⁡ θ 1 − cos ⁡ θ d θ = ∫ 0 1 d u u = 2 u ∣ 0 1 = 2 \begin{aligned} \int_0^\frac{\pi}{2}\dfrac{\sin\theta}{\sqrt{1-\cos\theta}}\text{ }d\theta&=\int_0^1\dfrac{du}{\sqrt{u}}\\ &=\left.2\sqrt{u}\right|^1_0\\ &=2 \end{aligned} Summing these together, we see this. ∫ 0 π 2 sin ⁡ θ 1 + cos ⁡ θ d θ + ∫ 0 π 2 sin ⁡ θ 1 − cos ⁡ θ d θ = 2 2 − 2 + 2 = 2 2 \int_0^\frac{\pi}{2}\dfrac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ }d\theta+\int_0^\frac{\pi}{2}\dfrac{\sin\theta}{\sqrt{1-\cos\theta}}\text{ }d\theta=2\sqrt{2}-2+2=2\sqrt{2}

Remembering the scalar from earlier, we find this. 2 2 π ( ∫ 0 π 2 sin ⁡ θ 1 + cos ⁡ θ d θ + ∫ 0 π 2 sin ⁡ θ 1 − cos ⁡ θ d θ ) = 2 2 π × 2 2 = 8 π \dfrac{2\sqrt{2}}{\pi}\left(\int_0^\frac{\pi}{2}\dfrac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ }d\theta+\int_0^\frac{\pi}{2}\dfrac{\sin\theta}{\sqrt{1-\cos\theta}}\text{ }d\theta\right)=\dfrac{2\sqrt{2}}{\pi}\times2\sqrt{2}=\boxed{\dfrac{8}{\pi}}

The expected value for P A + P B PA+PB is 8 π ≈ 2.54647 … , \frac{8}{\pi}\approx2.54647\ldots, so E ≈ 0.54647 … E\approx0.54647\ldots and ⌊ 1000 E ⌋ = 546 \lfloor1000E\rfloor=\boxed{546}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

There isn't a need to use the distance formula. The lengths are obvious from the sketch. It can be seen that ∠ P A B = π / 2 − θ / 2 \angle PAB=\pi/2-\theta/2 . Since Δ P A B \Delta PAB is a right angled triangle, P A = 2 ∣ sin ⁡ ( θ / 2 ) ∣ PA=2|\sin(\theta/2)| . Similarly, P B = 2 ∣ cos ⁡ ( θ / 2 ) ∣ PB=2|\cos(\theta/2)| .

Pranav Arora - 7 years, 1 month ago

Excellent.

Finn Hulse - 7 years, 1 month ago

Another path you can take after arriving at 1 + cos ⁡ θ \sqrt{1+\cos\theta} and 1 − cos ⁡ θ \sqrt{1-\cos\theta} is using the reverse half-angle identity to get 1 + cos ⁡ θ = 2 ∣ cos ⁡ ( θ 2 ) ∣ \sqrt{1+\cos\theta}=2\left|\cos \left(\dfrac{\theta}{2}\right)\right| and 1 − cos ⁡ θ = 2 ∣ sin ⁡ ( θ 2 ) ∣ \sqrt{1-\cos\theta}=2\left|\sin \left(\dfrac{\theta}{2}\right)\right| .

We can also get rid of the absolute value signs by noting that we just need to take the integral from 0 0 to π \pi because of symmetry, which forces cos ⁡ ( θ 2 ) \cos \left(\dfrac{\theta}{2}\right) and sin ⁡ ( θ 2 ) \sin \left(\dfrac{\theta}{2}\right) to both be positive.

You also typoed in your 4th to last equation: it should be 2 2 − 2 2\sqrt{2}-2 , not 2 2 − 1 2\sqrt{2}-1 .

Daniel Liu - 7 years, 1 month ago

You can restrict the integration still further because there is symmetry between placing the point P in the first or second quadrant; so you can just integrate from 0 to π 2 \frac{\pi}{2} and skip the absolute value brackets entirely. Also, as long as you're going to use the half-angle trigonometric terms, you can write their sum as 2 sin ⁡ θ 2 + 2 cos ⁡ θ 2 = 2 2 ( 2 2 sin ⁡ θ 2 + 2 2 cos ⁡ θ 2 ) 2 \sin \frac{\theta}{2} + 2 \cos \frac{\theta}{2} = 2\sqrt{2} \ ( \frac{\sqrt{2}}{2} \sin \frac{\theta}{2} + \frac{\sqrt{2}}{2} \cos \frac{\theta}{2} ) = 2 2 [ sin ⁡ θ 2 cos ⁡ ( π 4 ) + cos ⁡ θ 2 sin ⁡ ( π 4 ) ] = 2\sqrt{2} \ [ \ \sin \frac{\theta}{2} \cos ( \frac{\pi}{4} ) + \cos \frac{\theta}{2} \sin ( \frac{\pi}{4} ) \ ] = 2 2 sin ⁡ ( θ 2 + π 4 ) = 2\sqrt{2} \ \sin (\frac{\theta}{2} + \frac{\pi}{4} ) [which uses the "angle-addition" formulas]. This simplifies the integrand considerably.

Gregory Ruffa - 7 years, 1 month ago

Well, that certainly would have saved a bit of work! \text{ } ;)

Fixed!

Trevor B. - 7 years, 1 month ago

are you sure it REQUIRES calculus?

Nathan Ramesh - 7 years, 1 month ago

Seeing it requires you to find the average value of a continuous curve, I'd say yes. @Trevor B. knows more about this than me, I'm still a noob. What do you say?

Daniel Liu - 7 years, 1 month ago

I believe it does, but do you have a non-calculus solution?

Trevor B. - 7 years, 1 month ago

@Trevor B. – no i suck at math idk any calculus so i was just wondering

Nathan Ramesh - 7 years, 1 month ago

I must say that taking the modulus value is absolute necessary. First time I solved it, I did not take modulus and got the expected value as zero!

Avineil Jain - 7 years, 1 month ago

For any conic including circle PA+PB is always constant ...... Then what is the need of using integration ?

Kanthi Deep - 7 years, 1 month ago

No, that would be true if A A and B B were the foci. Wht you are thinking of are the foci of an ellipse. Additionally, in a circle, the only thing constant is A P 2 + B P 2 AP^2+BP^2 .

Trevor B. - 7 years, 1 month ago

Thanks I forgot it

Kanthi Deep - 7 years, 1 month ago

Search wolframalpha for the following query have the answer presented to you-- floor(1000*(integrate 2abs(sin(x/2))+2abs(cos(x/2))-2 from 0 to 2pi)/(2pi))

Connor Kenway - 7 years, 1 month ago

actually wat is integration

Suvir M - 7 years, 1 month ago

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