Just a slice, please .....

Probability Level 5

Consider a rectangle with vertices $(0,0), (7,0), (0,5)$ and $(7,5)$ .

Next, of the set of lines that slice this rectangle into two sections, choose a line at random, (uniformly distributed over the angle $\theta$ the line makes with the positive $x$ -axis).

The expected value for the area of the smaller of the two sections that the rectangle is divided into by the chosen line can be expressed as $\frac{a}{b}$ , where $a$ and $b$ are positive coprime integers. Find $a + b$ .

The answer is 39.

1 solution

Paul Josephson
Aug 9, 2014

The expected value for the smaller half of this figure can be intuited as follows

The values of the smaller area of this rectangle will range from 0 to 1/2 the rectangle's area.
Half the rectangle's area = (1/2) * 5 * 7 = (35/2)
The chance of any one given value of this area is equal (i.e. the value of the smaller area is equally likely to be 0, (35/2), 4 etc.
So the expected value would be 1/2 way between the maximum value (35/2) and the minimum (0).
To calculate (1/2) * (35/2) = (35/4)
35 and 4 have no common whole value integer denominators. 35 and 4 are co-prime

A = 35 B = 4

A + B = 39

It's a little harder to prove that the expected area is 1/4 that of the rectangle. You can' t assume that there's an uniform distribution of the "values of the smaller area" ranging "from 0 to 1/2 of the rectangle's area", and then say the average works out to 1/4 of it.

Michael Mendrin - 6 years, 10 months ago

I definitely didn't do the rigorous proof. I'm looking forward to seeing it.

Paul Josephson - 6 years, 10 months ago

All right, give me some time to get it together, and I'll post it at top. It's a pretty interesting problem.

Michael Mendrin - 6 years, 10 months ago

@Michael Mendrin – @Michael Mendrin To save you some time, I'll provide the link to the answer you gave to the more general question I posted on Yahoo seven months ago. An alternative solution is provided on this link as well. link

Hope that you don't mind me doing this. Your solution is so elegant and intuitive that I thought it was worthwhile making reference to it here on Brilliant. :)

Brian Charlesworth - 6 years, 10 months ago

@Brian Charlesworth – Oh, you're THAT Brian! Hey, how do you like it here now in Brilliant? It's too bad that Y!A kind of wrecked the community we had going for a while (stupid market-driven corporate machinations!). I wonder what's happened to the rest of the regulars?

Michael Mendrin - 6 years, 10 months ago

@Michael Mendrin – Haha. Yes, 'tis me. I think we both emigrated to Brilliant about the same time, once 'the Troubles' at Y!A became too much to bear. (I recognized that you were "Scythian" early on since your actual name was on the Photobucket diagrams you posted when answering my Y!A questions.)

Anyway, I really enjoy Brilliant; @Calvin Lin et al are doing a great job of maintaining and constantly improving the site and there seems to be a genuine camaraderie amongst the rank and file membership. The global representation is amazing, (Mongolia, Albania, Eritrea, .....), and we're all here because we share a passion for the art of mathematical and scientific problem solving, plain and simple.

As for our former Y!A regulars, I think a few, (Fred, Rita the Dog), still answer questions, but for the most part our community of 'recreational mathematicians' has disbanded. Sad indeed, but it was fun while it lasted. I always looked forward to seeing what kind of solutions you, gianlino, String, the 'other' Michael, etc., could come up with to my questions. (More than a few solutions were worthy of publishing, in my opinion.) I'm going through my old questions and posting adapted versions here where suitable, (some of them would be 'Level 6' on Brilliant if left unaltered).

You have an impressive following now, enough to warrant your own Messageboard. It must be satisfying to get such a response from all the budding young mathematicians, scientists and engineers who frequent this site. :)

Brian Charlesworth - 6 years, 10 months ago

@Brian Charlesworth – I'm doing the same thing, going through my old files to see what can used here. But most of them involve proofs, and that's kind of hard to implement here in Brilliant as rated problems. I put up my messageboard mainly to handle questions or complaints about my problems, and that's another problem with the Brilliant format---it's hard to just jump onto a posted problem, and then start taking (or complaining) to the Asker about what's wrong with it before the complete solution is finally worked out. Y!A was more collaborative, Brilliant is more competitive. It's just too bad that Y!A just killed the collaborative aspects of it, "Either ask or answer questions, no fraternizing here now!" I think Brilliant continues to evolve in the right directions, and I hope to see more of that collaborative atmosphere here.

Anyhow, I do have plenty more problems to choose from my files to put up in Brilliant. One little artwork at a time. Happy solving!

Michael Mendrin - 6 years, 10 months ago

@Michael Mendrin – You make some good points. There was less of a barrier between the asker and potential solvers on Y!A than here, (at least initially). I've found that the reporting system on Brilliant is quite effective now, but only if there is a posted solution can we have a more interactive dialogue with the asker. I sometimes wonder if it would be better if it were required of a poster of a question that they also provide some kind of initial solution, but that would probably result in far fewer questions being posted, (which is clearly not ideal, even if it did result in the opportunity to immediately comment or express concerns about the question and its putative solution).

The collaborative aspect of our small Y!A community was its hallmark, but then again we were generally more mature of age and hence didn't have anything to 'prove', (pun intended). The average user here is much younger and hence more eager to demonstrate that they have the 'right stuff'. That being said, the likes of Sharky Kesa, Daniel Liu, Sean Ty, etc., show a remarkable mix of enthusiasm, maturity and ability that I find most encouraging. The Brilliant staff would do well to further foster the skill of collaboration, as the problems of the future, scientific and otherwise, will require many fine minds working in dynamic harmony towards a common goal. (I'll step down from my soapbox now. :) )

Brian Charlesworth - 6 years, 10 months ago

As mentioned below, I know I haven't provided a rigorous solution. However, to me, it seems obvious that the expected value of the area of the smaller half of the rectangle would be 1/2 of 1/2 of the area of the rectangle. I see no reason why the distribution of the area would be non-uniform. It's a flat rectangular space. If I were to find the expected value of the area by integration, it would be a one-dimensional equation (the area is flat bounded by straight edges) it has no thickness or curves that would lead to higher order equations.

Again, I look forward to seeing a rigorous proof.

Paul Josephson - 6 years, 10 months ago

Just a slice, please .....

The answer is 39.

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