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MIT PRIMES/Art of Problem Solving

CROWDMATH 2017: The Broken Stick Problem

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Problem 3: Expected Area on "Biased Splits"
always_correct   13
N Dec 2, 2018 by Nirmay
Notice that if we do the opposite procedure, that is, place the next break in the unit length in the smaller segment, we end up with no possible triangle. This is because the larger segment must have length $> \frac{1}{2}$, and as this segment is not broken up any further, it remains larger than the semiperimeter. Thus, as given before, the area expected is $\frac{\pi}{105}$, as noted before. The same applies to the second scenario in which we place the next cut in the rightmost segment.
13 replies
always_correct
Mar 9, 2017
Nirmay
Dec 2, 2018
Solution to 3 ii
adamodoherty   5
N Sep 26, 2018 by hickory
Here I think is a solution to the question: what is the expected area when we pick the second point from inside the largest interval?

Call the small first interval $y$, call $x$ the leftmost piece from the second break. It's more useful to not define the the larger interval.

We can then say that the area of the triangle is obviously $\sqrt{\frac12\left(\frac12 - y\right)\left(\frac12-x\right)\left(\frac12-(1-y-x)\right)}$ and so the average area is $\frac1{\frac12-(\frac12-y)}\int_{\frac12-y}^{\frac12}\sqrt{\frac12\left(\frac12 - y\right)\left(\frac12-x\right)\left(\frac12-(1-y-x)\right)}\ dx$

From here we can make multiple simplifications:

$$
   \frac1y \sqrt{\tfrac12(\tfrac12-y)} \int_{\frac12-y}^{\frac12} \sqrt{\left(\frac12-x\right)\left(\frac12-(1-y-x)\right)}\ dx
$$
We now look for a substitution to solve which was initially where I got stuck but $u=x+y-\frac12$ works and makes the integral

$$\frac1y \sqrt{\tfrac12(\tfrac12-y)} \int_0^y \sqrt{u(y-u)} dx$$
It's pretty obvious when considering the substitution now that $\sqrt{u(y-u)}$ is a semicircle centered at ($\tfrac y2$, $0$). The semicircle has area $\frac12 \pi (\frac y2)^2$ and so our equation changes to $\frac\pi 8 y\sqrt{\frac12(\frac12-y)}$

The integral which is then formed, $2\int_0^{1/2}\frac\pi 8 y \sqrt{\tfrac12(\tfrac12-y)}\ dy$, would be pretty easy to solve by hand but I just plugged it into symbolab (https://www.symbolab.com/solver/definite-integral-calculator/2%5Cint_%7B0%7D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cfrac%7B%5Cpi%7D%7B8%7Dy%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7D%5Cleft(%5Cfrac%7B1%7D%7B2%7D-y%5Cright)%7Ddy) which yields

$$2\int_0^{1/2}\frac\pi 8 y \sqrt{\tfrac12(\tfrac12-y)}\ dy = {\pi\over120}$$
Which is unsurprisingly pretty close to the other value we found of ${\pi\over105}$.

I am currently working on the other distribution discussed in problem three, hope this doesn't take too long.

Anyway, this seems right to me. Let me know if you see any errors.
5 replies
adamodoherty
Nov 14, 2017
hickory
Sep 26, 2018
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