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Geometry Level 5

Suppose we have three concentric circles with respective radii $1, 2$ and $3.$ Choose one point, uniformly and at random, on each of these circles and form a triangle with these three points as vertices.

Let $S$ be the expected area of this triangle. Find $\lfloor 10000S \rfloor$ .

Image credit: Wikimedia Dartboard

The answer is 20829.

1 solution

Brian Charlesworth
Mar 3, 2015

I have asked this question on another site, and this was the response. This may have been an unfair question to ask on this site, but I have seen interest expressed in having "level 6" questions, and I felt that this would fill that role. I also have a great deal of respect for the talent on this site, so I'm interested to see what alternative solutions others may have.

Outstanding problem, Sir! +1

User 123 - 6 years, 3 months ago

Thanks, Simran. Congrats for solving it. :)

Brian Charlesworth - 6 years, 3 months ago

Sorry to say but I wolfram alpha'd it.(being honest)

Ronak Agarwal - 6 years, 3 months ago

@Ronak Agarwal – That's fine; it was supposed to involve some analysis and then lots of number-crunching, which is why I tagged it with computer science as well.

Brian Charlesworth - 6 years, 3 months ago

Thanks Sir. The pleasure was mine!

User 123 - 6 years, 3 months ago

Sir, I have a doubt, please. Sir for this question , in the second part of Pi Han Goh's solution, how were the limits of the definite integral decided? And in general, for such questions, how do we decide the limits of a definite integral? Thanks very much, Sir.

User 123 - 6 years, 3 months ago

Pi Han Goh was working with a unit sphere, and was able to write the distance between a fixed point and a point chosen at random as a function of one variable, namely $X$ . For a unit sphere, $X$ ranges from $-1$ to $1$ , hence the limits of the integral. (Note that, to find the average, he had to divide by $1 - (-1) = 2$ .) Also, when he used the substitution $w = 2 - 2X, dw = -2 dX$ he had to change the limits of integration since as $X$ goes from $-1$ to $1$ , $w$ goes from $4$ to $0.$

(In the discussion after the solution, Michael Mendrin makes some good points as to why this method was valid from a probability standpoint, so it's a good idea to read through his comments.)

As for limits in general, we generally need to choose limits that correspond to the full range of the variable of integration. Sometimes we can exploit symmetries inherent to a problem, however, and just integrate over half the range of a variable because the other half will yield the exact same result.

Brian Charlesworth - 6 years, 3 months ago

Alright Sir. Thanks so much for your consideration and for taking the trouble to explain it to me. It helped me a lot! Just as a confirmation, Sir, assuming the sphere in question was not a unit sphere, but a sphere with $x^2+y^2 + z^2 =3$ , then wouldn't the range of X be $[-\sqrt 3,\sqrt3]$ , and the limits of the definite integral with respect to $X$ be from $-\sqrt3$ to $\sqrt3$ ? Thank you very much Sir.

User 123 - 6 years, 3 months ago

@User 123 – Yes, that's correct. In general, say we have the sphere $x^{2} + y^{2} + z^{2} = r^{2}.$ Then the distance between $(r,0,0)$ and $(X,Y,Z)$ would be

$\sqrt{(r - X)^{2} + Y^{2} + Z^{2}} = \sqrt{r^{2} - 2rX + X^{2} + Y^{2} + Z^{2}} = \sqrt{2r^{2} - 2rX}.$

The limits of integration would then be $-r$ to $r$ , and to find the average we will need to divide our integral by $r - (-r) = 2r.$ So the average distance will be

$\dfrac{1}{2r}\displaystyle\int_{-r}^{r} \sqrt{2r^{2} - 2rX} dX.$

Now we let $w = 2r^{2} - 2rX$ , and so $dw = -2r dX.$ As for the limits, as $X$ goes from $-r$ to $r$ we have $w$ going from $4r^{2}$ to $0.$ Thus the integral becomes

$-\dfrac{1}{4r^{2}}\displaystyle\int_{4r^{2}}^{0} \sqrt{w} dw = \dfrac{1}{4r^{2}} \int_{0}^{4r^{2}} \sqrt{w} dw =$

$\dfrac{1}{4r^{2}} * \dfrac{2}{3} * w\sqrt{w}$ evaluated from $w = 0$ to $w = 4r^{2}.$

This comes out to $\dfrac{1}{6r^{2}} * 4r^{2} * \sqrt{4r^{2}} = \dfrac{4r}{3}$ as expected.

Brian Charlesworth - 6 years, 3 months ago

@Brian Charlesworth – Sir, I understood whatever you explained. Thanks a ton, Sir! I'm really, really grateful to you for your help. Thanks you so much!

User 123 - 6 years, 3 months ago

@User 123 – You're welcome. Glad I could help. :)

Brian Charlesworth - 6 years, 3 months ago

@Brian Charlesworth – Sir, I'm really sorry to go off topic, but could you please help me with a comment I had posted to this ? You've really helped me in the past, Sir, and thus I digressed from this question. Thanks very much, Sir.

User 123 - 6 years, 3 months ago

@User 123 – I just wrote a reply to your comment on the other question. Hope it helps. :)

Brian Charlesworth - 6 years, 3 months ago

@Brian Charlesworth – Sir, it really helped me! I remember I was in a state of utter confusion as to why Julian Sir had taken the lower limit as $0$ instead of $\frac{\pi}{180}$ . Thank you so much for your help, Sir! I'm really grateful to you.

User 123 - 6 years, 3 months ago

@User 123 – You're welcome. :)

Brian Charlesworth - 6 years, 3 months ago

you have solved it using numerical integration, I have solved it by taking uniform random distribution over theta1, theta2 and theta3, n = 1000000, and my answer is 20835.. actually, 2.08 is the limit of significant figures, and last two digits vary with the numerical method chosen. You should increase the tolerance and put [100S].

N = 10000000;
t1 = rand(1,N)*2*pi;
t2 = rand(1,N)*2*pi;
t3 = rand(1,N)*2*pi;
a = abs(sin(t2-t1)+1.5*sin(t1-t3)+3*sin(t3-t2));
mean(a);

mean = 2.0835

Mahek Mehta - 6 years, 3 months ago

I did consider having a higher tolerance when I first posted the question, but as seen in the link the four main contributors, upon refining their analytical approaches, eventually reached agreement on the value $2.082911......$ , which is why I went with the lower tolerance. I may have been unfair here in requiring so many significant figures, but I believe that the posted answer is nevertheless correct.

Brian Charlesworth - 6 years, 3 months ago

yeah..., random angles method has slower convergence, at N = 100M, it converges to 2.0829, thanks!

Mahek Mehta - 6 years, 3 months ago

Ready, Aim, Fire!

Image credit: Wikimedia Dartboard

The answer is 20829.

1 solution

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