1. Originally Posted by vcs
1.) There is a string consisting of 2m black beads and 2n white beads, arranged in some random order. Prove that it is possible to make at most two cuts to the string such that two people can be given m black beads and n white beads.
IIRC there is some theorem that states that for a continuous curve F(x) bounded between 2 points say X1 and X2 on x-axis, you can always find a segment of the curve with points x1 and x2 such that (x1, F(x1)] and [x2,F(x2)] can be connected with a line that has same slope as the slope line connecting [X1,F(X1)] and [X2,F(X2)] for any given length (x2-x1). Alternatively, if a car has average speed of v during a journey of duration, T you can always find a period t1 to t2 where average speed was exactly v, for any duration t2-t1<T.

Same principal can apply here. Define a function F(k) in period 0 to 2m+2n such that F(k) is difference of number of black beads and white beads in first k beads on the string for integer k. The overall slope of line connecting end points of this curve is (2m-2n)/(2m+2n) = (m-n)/(m+n). For non-integer k, F(k) will fall on a straight line connecting points at 2 nearest integers.

For this problem you have to find 2 points k1 and k2 that have exactly m black and n white beads between them i.e. k2-k1 = m+n and F(k2) - F(k1) = m-n. Line connecting end points of this segment will have a slope of exactly (m-n)/(m+n). By that theorem above it is always possible to find such points.

This is obviously dirty version of the proof because I haven't named the theorem (probably it doesn't have a name, it's a common sense conclusion) and I have not proven how this will always be possible at integer k1 and k2.

2. Originally Posted by ankitj
IIRC there is some theorem that states that for a continuous curve F(x) bounded between 2 points say X1 and X2 on x-axis, you can always find a segment of the curve with points x1 and x2 such that (x1, F(x1)] and [x2,F(x2)] can be connected with a line that has same slope as the slope line connecting [X1,F(X1)] and [X2,F(X2)] for any given length (x2-x1). Alternatively, if a car has average speed of v during a journey of duration, T you can always find a period t1 to t2 where average speed was exactly v, for any duration t2-t1<T.
isnt that just a corollary of the intermediate value theorem?

3. Correct. Good work, you guys. A more general proof in this video.

https://youtu.be/yuVqxCSsE7c

4. that theorem ankit mentioned is the mean value theorem

5. It should be possible to solve that problem using induction too. Not able to do it myself.

6. I used a similar idea but not exactly induction.

I defined a function F(k) = (number of black beads seen in the last m+n beads) - m, where k varies between m+n to 2m+2n.

If F(m+n) == 0, we are done, since we can cut the necklace at the midpoint.

If not, let us assume that F(m+n) > 0. In which case, F(2m+2n) must necessarily be < 0, since the second half of the necklace must have less than m black beads. Now we have a function whose value goes from positive to negative, and at every step the window shifts by exactly one bead. This means that there must exist a point p somewhere between m+n and 2m+2n, where F(p) = 0. We can guarantee the existence of such a point because, at every intermediate point, we drop one bead from the window at the left, and add one bead to the window at the right, as we keep moving the window from m+n to 2m+2n.

A similar argument can be applied to the case F(m+n) < 0.

7. There is a channel called MindYourDecisions on Youtube that posts Math problems. They range from fairly simple to very tricky. Ones on geometry are generally quite good. Take a look at this one for example: https://www.youtube.com/watch?v=dF67AJH9mjM

8. Originally Posted by ankitj
There is a channel called MindYourDecisions on Youtube that posts Math problems. They range from fairly simple to very tricky. Ones on geometry are generally quite good. Take a look at this one for example: https://www.youtube.com/watch?v=dF67AJH9mjM
Ha, that is excellent. I tried solving it but couldn't come up with the answer so looked at the solution.

This one is tangentially connected in terms of the idea and approach.

http://jwilson.coe.uga.edu/EMAT6680F.../AS06/WU06.htm

This is a special collection of problems that were given to select applicants during oral entrance exams to the math department of Moscow State University. These problems were designed to prevent Jews and other undesirables from getting a passing grade. Among problems that were used by the department to blackball unwanted candidate students, these problems are distinguished by having a simple solution that is difficult to find. Using problems with a simple solution protected the administration from extra complaints and appeals. This collection therefore has mathematical as well as historical value.
WTF. How can otherwise intelligent people be such morons.

10. Originally Posted by vcs

WTF. How can otherwise intelligent people be such morons.
Serious 1880s areas here.

11. Not gonna lie, when I saw this thread was bumped I thought it was gonna be Athlai trolling me about Terence Tao.

12. I googled cosine law and then wrote a complicated systems of equations. Got nowhere. Then saw the solution and was quite impressed by how elegant it was.

13. Slow day at work so tried another one out. This was fairly easy.

14. Originally Posted by vcs
Slow day at work so tried another one out. This was fairly easy.

I was quite impressed with myself for solving it under 5 minutes till I saw that some 5th graders did it under 1 minute.

15. Originally Posted by vcs
Slow day at work so tried another one out. This was fairly easy.

Thanks for sharing.

Originally Posted by Malcolm
I was quite impressed with myself for solving it under 5 minutes till I saw that some 5th graders did it under 1 minute.
It was worth the 9 minutes. I had a good refresher in geometry and arithmetic. We take these things for granted when we are young.