Polar Fox: Hey Red, these points seem to have identity issues. Are they black or white?

Red Fox: An obvious illusion! They are white for sure.

Polar Fox: So what I perceive is not the exact reality?

Red Fox: This is just a simple game of colors playing on your retina. It is perfectly explainable and part of the reality.

Polar Fox: Yes, most definitely. But, can this be an artful clue of an even bigger reality that there is truth beyond what you see?

Red Fox: Here we go again!

Polar Fox: A brush mark from a picture? A letter from a book? A note from a symphony?

Red Fox: And who is the artist, juggling with colors to deceive our perception?

Polar Fox: Deceive? I am not so sure. But looks like someone who knows us inside and out. And I think he wants to talk.

Red Fox: Why me? Why me?

Polar Fox: Yep, and you have bigger ears than mine.

I see a simpler proof. Area of equilateral triangle is s^2*sqrt(3)/4 where s is the length of any side. Since each corner of the triangle is at integer points, the area is irrational.

ReplyDeleteOn the other hand, consider the rectangle drawn around the triangle. This area minus the corner areas around the triangle must be the area of the triangle. However, the rectangle area is rational, and the corner areas are rational. This means that the triangle area is rational.

Contradiction.

I can buy this great proof. It makes sense.

ReplyDeleteRational + Irrational = Irrational.

Your WolframAlpha formulation also indicates the same fact.

This fact seems to deceive perception, doesn't it?

There are infinitely-many points and infinitely-many different equilaterals, YET, none fits on three.

Amazing!

Please also see six' WolframAlpha formulation for this problem below:

ReplyDeletea^2 + b^2 = (b + c)^2 + d^2 = (a + d)^2 + c^2

Wolframalpha seems to confirm that the only integer solutions to this equation are when each variable is 0.

http://www.wolframalpha.com/input/?i=a%5E2+%2B+b%5E2+%3D+c%5E2+%2B+d%5E2+%2B+2bc+%2B+b%5E2+%3D+c%5E2+%2B+d%5E2+%2B+2ad+%2B+a%5E2

I suppose you could generalize this to any simple polygon with an irrational area.

ReplyDeleteThat can be a conjecture I am planning to ask. That is:

ReplyDelete"No regular polygon -other than square- can be drawn on discrete points of the Cartesian plane"

which could be much harder to prove.