The time is now - Spring classes are filling up!

MIT PRIMES/Art of Problem Solving

CROWDMATH 2017: The Broken Stick Problem

G
Topic
First Poster
Last Poster
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
No more topics!
a