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MIT PRIMES/Art of Problem Solving

CROWDMATH 2017: The Broken Stick Problem

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i arXiv'ed report on broken stick project
v_Enhance   0
May 18, 2018
Hi everyone,

Apologies this is ages late, but I've at last finished the final report for this project, which is now available on the arXiv. You can download it at:

https://arxiv.org/abs/1805.06512

Thanks to everyone for their contributions to this CrowdMath, no matter how small!

Best wishes,
Evan
0 replies
v_Enhance
May 18, 2018
0 replies
i The 2017 CrowdMath PRIMES prizes
copeland   14
N Jan 5, 2018 by v_Enhance
The MIT Program for Research in Math, Engineering, and Science (PRIMES) will award PRIMES internships to the two highest ranked participants in CrowdMath 2017 who are still in high school in Spring 2018. These students will have the opportunity to work on a research project on Skype with a mentor at MIT through PRIMES in 2018. One student will be chosen from each of the Graph Algorithms and Applications and Broken Stick projects.

CrowdMath is an online research program open to high school and college students all over the world. High school students can improve their CrowdMath rankings by posting questions, answers, ideas, or proofs on the CrowdMath message board. The rankings for the 2017 CrowdMath PRIMES prizes will be based on CrowdMath posts during the year 2017 and will be finalized on December 31, 2017. Rankings will be posted below and updated monthly.
14 replies
copeland
Feb 6, 2017
v_Enhance
Jan 5, 2018
Is this dead
NoDealsHere   16
N Mar 13, 2024 by Technodoggo
Is it really?
16 replies
NoDealsHere
Jun 7, 2019
Technodoggo
Mar 13, 2024
Nice work!
IdanCarre   2
N Jun 18, 2019 by MathClassStudent
Just wanted to say this is utterly the coolest doc I've ever seen and the people who put it together did such a nice job! Great work everyone.
-Idan
2 replies
IdanCarre
Apr 5, 2017
MathClassStudent
Jun 18, 2019
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
Question about publishing the solution
myfavouritenumberis73   0
Nov 27, 2018
I have a little question about putting up a solution, does I have to put it up here or can I put it up somewhere else?

0 replies
myfavouritenumberis73
Nov 27, 2018
0 replies
Strategy on computing the probability to (c)
myfavouritenumberis73   1
N Nov 24, 2018 by Nirmay
(C)

A simpler case should be used, where n=1.

Since the equation of a line is y=mx+c, all the possible values of m and c using an inequality should be plotted on a graph, where the x-axis is c and y-axis is m. Then another region should be drawn which points inside, (c,m), that would give the equation of a line that would cut the square into to unequal regions. Then the ratio of these two regions should be calculated. This would probably require one to look at individual cases of what some of the lines would look like.

1 reply
myfavouritenumberis73
Oct 16, 2018
Nirmay
Nov 24, 2018
Problem(ii) Possible Solution Path
mathcrazymj   25
N Nov 24, 2018 by Nirmay
Look at my solution to problem 5 in this pdf: http://submissions.usamts.org/Year28/Round1/34279-1476743929-USAMTS%202016%20Matthew%20Roth.pdf

I found some useful results about cyclic quadrilaterals that may generalize to Problem(ii)

EDIT: I'll think about it some more but I think this problem can be solved by breaking n-gons into a bunch of quadrilaterals and maybe 1 triangle (depending on n) and maximizing the areas by making all the quadrilaterals cyclic and the triangle equilateral
25 replies
mathcrazymj
Mar 1, 2017
Nirmay
Nov 24, 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
Question on Progress
mmaotmh   0
Sep 17, 2018
It appears to me that the thread you are looking at is using n as the number of breaks, but the question defines n as the number of pieces.
0 replies
mmaotmh
Sep 17, 2018
0 replies
A generalization of Problem 7
GeronimoStilton   0
Dec 29, 2017
$a+b$ players play the following game consisting of $n - 1$ turns. Alice's $a$ clones and Bob's $b$ clones go in some order. After all turns have passed, there are $n$ pieces. What is the maximum number of triangles that Alice's clones can guarantee forming given that Bob's clones are working against Alice's clones?

Clarification: One possible ordering might be $a_1,b_1,a_2,b_2,b_3$, which would repeat, and so for the first $6$ turns in this example, Alice's clones and Bob's clones would follow the sequence $a_1,b_1,a_2,b_2,b_3,a_1$.
0 replies
GeronimoStilton
Dec 29, 2017
0 replies
Finding the Minimum
RadioActive   6
N Dec 27, 2017 by RadioActive
Instead of asking for the maximum number of triangles Alice can make, I think it might also be interesting if we ask for the minimum. What do you guys think?
6 replies
RadioActive
Dec 14, 2017
RadioActive
Dec 27, 2017
Problem(ii) Possible Solution Path
mathcrazymj   25
N Nov 24, 2018 by Nirmay
Look at my solution to problem 5 in this pdf: http://submissions.usamts.org/Year28/Round1/34279-1476743929-USAMTS%202016%20Matthew%20Roth.pdf

I found some useful results about cyclic quadrilaterals that may generalize to Problem(ii)

EDIT: I'll think about it some more but I think this problem can be solved by breaking n-gons into a bunch of quadrilaterals and maybe 1 triangle (depending on n) and maximizing the areas by making all the quadrilaterals cyclic and the triangle equilateral
25 replies
mathcrazymj
Mar 1, 2017
Nirmay
Nov 24, 2018
a