The Hardest Problem from the 1986 IMO

[@WrathofMath]

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[@chemicalbrother5743]

Me: Time to concentrate and follow the solution. My brain: bapbap

[@shadowdragon2484]

Cool video and problem! I was able to get an intuitive (if rather unrigourous) answer by considering the algorithm as a sort of "averaging" fn. Since the sum is positive we know there is more positive "mass" than negative, and so if you repeatedly distribute the smaller value amongst the positive mass, the magnitude of the negative will go down and each node will come closer to the average, which is positive.

[@Rafix_989]

But there can be 2 negative numnera next to each other and then the 2nd negative number becomes more negative

[@arekkrolak6320]

It doesnt have to be 4 -6 3. It can also be 3 -6 4. The algorithm doesnt say in which direction we assign consecutive vertices. ;)

[@luisfonseca2299]

What a nice video! That function seemed like the most random thing ever but I guess that´s why the problem is so complicated... I can only imagine the thought proccess of these students: "well, the sum is always the so that obviously doesn't work. Hum, I need to find an invariable. *somehow finds it*. Well, well, well, The difference depends on the sum and this one vertex value that I need to be negative, what a nice coincidence". I swear, this type of things blow my mind

[@GhostyOcean]

Okay so after working through an example I think the answer is yes, it always terminates. Some things I've noticed: The sum of the sides doesn't change. If we transform (x,y,z) to (x+y,-y, z+y) then we added a y to each outside coordinate and subtracted 2y from the middle coordinates. Thus the sum is unchanged. Furthermore, x+y is necessarily smaller in magnitude than both x and y (because they're different signs). I'm not sure how having multiple negatives in a problem would affect the outcome. Let's continue watching and see.

[@GhostyOcean]

Use the difference of squares pattern to help with the simplifying! a²-b² = (a+b)(a-b). It'll very quickly show that the entire expression is divisible by x_4.

[@GhostyOcean]

OH! I think what I've shown is sufficient to show that sum( |x_i| ) is a non-negative and decreasing function! On a triplet (x,y,z) ➡️ (x+y,-y,z+y), since both x➡️x+y and z➡️z+y are decreasing in magnitude, and y➡️-y is a sign change and doesn't affect magnitude, then the sum of all three absolute values must have decreased after the transformation! So the sum of absolute values across all vertices decreased, and by construction is non-negative. Therefore it must stop eventually!

[@MichaelPiz]

Something you either left out or I was oblivious to its obviousness or that it just doesn't matter is proving that consecutive values of the function cannot increase. Can we have consecutive function values of, for example, -78 and -31? Would that mean that the function is increasing?

[@WrathofMath]

We prove that the consecutive values decrease, so they certainly cannot increase. Unless I have misunderstood your question! By showing fn+1 - fn < 0, we know fn+1 < fn, that is - the function decreases.

[@coppertones7093]

the function can’t be negative, also

[@MichaelPiz]

@WrathofMath OK. So the example I gave, if it's possible, would mean that the function is decreasing but at a slower rate? IOW, the absolute values of the negative function values need not get larger. (The math to figure this out myself is within my capability but I'm a bit brain-fogged today.)

[@graf_paper]

What is the music playing in the background of this video?! Feels like a mall piano being played in the distance 😌

[@RunstarHomer]

There are a few different songs in the video, one of them is Zora's Domain from Breath of the Wild

[@WrathofMath]

Huma-Huma From Russia with Love Donkey Kong Country Aquatic Ambience Super Mario Sunshine Sirena Beach Xenoblade Chronicles Valak Mountain (Night) Breath of the Wild Zora's Domain (Night) Pikmin 2 Piklopedia Super Mario 64 End Theme

[@prodqrn]

I too hope the Texans topple the Chiefs but I am fearful.

[@acc-lab1233]

I tried to construct such function by sum(x_i x_{i+2}) but that didn't work then when you hint ( )^2 I got the correct construction, cool problem!

[@thatcatthatalwayseatsyourchees]

This guy sounds exactly like sheldon

[@dan-florinchereches4892]

I am not sure how far I can go with it but if we look at the transformation of x and z with respect with y it is similar to Euclids algorithm with repeated subtractions. I am going to consider the case where we have a,b,c,d,e numbers and b,c are negative the the transformations go (a,b,c) -> (a+b,-b,b+c) (-b,b+c,d)->(c,-b-c,b+c+d) (a+b,c,-b-c)->(a+b+c,-c,-b) pentagon at this step is (a+b+c,-c,-b,b+c+d,e) looks like all the negative numbers are migrating towards the positive poles. I think the transformation applied to consecutive nodes each turn is adding -x, 2x, -x. Now we need to determine if over time x is decreasing Another Idea is what if the transformation is like a ripple on the pentagon where the transformation sends -x to the left and right of the node until they hit a positive node large enough to swallow them. Something like the waves a waterdrop falling into the bucket creates. I think I will work out a particular example myself where I am trying to eliminate the negative from center position. (10,-1,-1,-1,-1) -> (10,-2,1,-2,-1) -> (8,2,-1,-2,-1) -> (8,1,1,-3,-1) -> (8,1,-2,3,-4)-> (4,1,-2,-1,4)->(4,-1,2,-1,4)->(3,1,1,-1,4)->(3,1,0,1,3) (5,-4,3-2,1)-> (1,4,-1,-2,1) -> (1,3,1,-3,1) -> (1,3,-2,3,-2) -> (1,1,1,1,-2)->(-1,1,1,-1,2)->(1,0,1,-1,1)->(1,0,0,1,0) well regardless what I do I end up with a symmetric result. I guess it makes sense since the transformation is symmetrical and we have an odd number of vertices? I guess I am not smart enough to get it quickly gonna watch the video

[@GhostyOcean]

Your example ends at step 1 because you don't have 2 positive vertices. The algorithm requires at least 2 positive vertices, but that's not sufficient. Notice how (1,1,-1,-1,-1) does not have a negative between two positives, so the algorithm halts immediately. Be sure to always check if your example satisfies the rules, or that you apply the rules correctly!

[@dan-florinchereches4892]

@GhostyOcean I think you have a problem there as the sum is supposed to be >0 The problem does not state how many consecutive negative number we use just that we perform the transformation around the negative number

[@Aman_iitbh]

Is it possible that sq function gets halt after certain steps but x,x2..x5 doesent ,it goes on certain cycle to maintain sq sum?

[@Aman_iitbh]

That cant happen ,i got it thnx

[@Nikolas_Davis]

Lovely problem! Near the end: you could have saved yourself quite a lot of algebra if you reproduced as many of the terms of f_{n} as possible in f_{n+1}, for example by breaking up (x1 - (x3+x4))^2 = ((x1-x3)-x4)^2 = (x1-x3)^2 + x4^2 - 2 * x4 * (x1-x3), and so on. Most of the terms would cancel out in the subtraction, except for the terms containing x4, which you could then collect with minimal fuss 🙂