Inspired by AOTM: Day 1

So we went to a lecture yesterday and Sir Bill Chen, a very accomplished quantitative analyst, gave a speech on his life and how mathematics has affected him. During his speech, he talked about one problem in particular which caught my mind.

"An 8x8 chessboard has two of its opposite corners removed. For example, A1 and H8. Can this board be tiled completely by a sufficient number of 2x1 rectangles with no overlap?"

The answer turned out to be no. So I decided to give the problem some actual thought and I came up with an over complicated proof using modular arithmetic.

This inspired me to make a new problem:

How many ways can you remove two tiles from an 8x8 chess board such that the board cannot be completely tiled by 31, 2x1 rectangles without overlap?


The answer is 992.

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1 solution

This can be solved by way of Gomory's Theorem . Only if two squares of opposite colors are chosen will a 31-domino tiling exists. Since we can choose two white or two black squares in ( 32 2 ) = 496 \binom{32}{2} = 496 ways each, there will be 2 × 496 = 992 2 \times 496 = \boxed{992} "mutilated chessboards" that cannot be tiled by 31 31 dominos.

(A proof of Gomory's Theorem is given under the Theorem 1.1 on pages 12-14 here .)

Oh wow, I didn't know that this result had a name!

Calvin Lin Staff - 5 years, 10 months ago

@Trevor Arashiro Hey Trevor. How's it goin'? So is this some sort of conference you're attending?

Edit: Using magical interweb powers I have deduced that you are in Philly. Congrats on being chosen to attend AOTM this year. I bet you're meeting some amazing people. :)

Brian Charlesworth - 5 years, 10 months ago

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It's going.great! Thanks for asking.

Yes, I was lucky enough to be chosen for AOTM 2015 and I haven't stopped smiling. The math gang is growing to include people outside brilliant. Haha, we math nerds can't be stopped from working on USAMO problems during our lunch break, unfortunately Kishlaya was the only one to solve it. Agnishom Misread the problem and thought we were proving an identity.

Sorry I don't have much time for responses and stuff because this camp is keeping me super busy. I'll be posting a note if I can find the time detailing all the amazing things that have happened so far.

Trevor Arashiro - 5 years, 10 months ago

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Haha Sign #314 that you're a Math Nerd: you work on USAMO problems on your lunch break, just for fun. :P Good to hear the camp is going great; hope that you're keeping a journal on your "adventures". :)

Brian Charlesworth - 5 years, 10 months ago

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@Brian Charlesworth The first sine of being a math nerd is that you're making math puns, though

Agnishom Chattopadhyay - 5 years, 10 months ago

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@Agnishom Chattopadhyay Hahaha Yes, cos they are funnier tan other puns. :)

Brian Charlesworth - 5 years, 10 months ago

I had the same reasoning!I didn't put in the answer as i was too scared!

Adarsh Kumar - 5 years, 10 months ago

@Brian Charlesworth Sir, could you please direct me to the proof again??? The link does not seem to work.........

Aaghaz Mahajan - 2 years, 6 months ago

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I've updated the link. Hope it works for you. :)

Brian Charlesworth - 2 years, 6 months ago

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Thanks Sir!!

Aaghaz Mahajan - 2 years, 6 months ago

The solution is based on a simple idea, so it's worthwhile for you to think about how to do this.

To be fair, when I encountered the generalized problem (to chessboards of arbitrary size) ~20 years ago, I couldn't figure out how to solve this. But when I saw the idea, it made it "obvious".

Calvin Lin Staff - 2 years, 6 months ago

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Yes, Sir I know...I have solved the problem, on the basis of colouring arguments, but I thought that maybe the link given had some additional information which would be useful sometime later..........:)

Aaghaz Mahajan - 2 years, 6 months ago

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