A JMO question

Prove that no three points with integer coordinates can be the vertices of an equilateral triangle.

#Geometry

Note by Swapnil Das
6 years, 2 months ago

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Comments

He Andrei! @[@Andrei Golovanov] First of all thanks for following me! Do you know any solution to this problem?

Swapnil Das - 6 years, 2 months ago

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Hey swapnil...i wanna ask u something that are u able to solve such type of hard questions because seeing this I am not able to do these sums but being smaller than me u are able to...

sarvesh dubey - 6 years, 2 months ago

I had a look and came up with a solution resembling Vishnu's (above). I used the properties of right-angled triangles. Interesting problem: geometry is not my forte, and I had some fun solving it! By the way, have you tried my olympiad problem set? It is as interesting as math gets (with the exception of other problems on Brilliant, of course)! @Swapnil Das

A Former Brilliant Member - 6 years, 2 months ago

Let's say that you had such a triangle. Assuming you're talking about 2D space, let two of the points lie on the x-axis, equidistant about the origin at C(a,0) and B(-a,0), where a is a positive integer. Using symmetry, we can see that the third point must lie on the y-axis, say at A(0,b), where b is also a positive integer. Now, using distance formula, we can see that AB=a2+b2\sqrt{{a}^{2}+{b}^{2}}=BC=2a. Squaring both sides, we get b2=3a2⇒b=a3.{b}^{2}=3{a}^{2}\Rightarrow b=a\sqrt{3}. As we said that b and a are both integers, and we reached the conclusion that b and a cannot both be integers at the same time, we can say that the assumption was wrong. So, you cannot have an equilateral triangle with integer coordinates.

vishnu c - 6 years, 2 months ago

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vishnu c A will lie at (0,b)(0,b) and not (b,0)(b,0). Other than that, I believe your proof is true. I might also add that b=a3b=a \sqrt 3 can be found due to the properties of a 30−60−9030-60-90 triangle. Each angle of an equilateral triangle is 60∘60^\circ, so the height bb will equal a3a \sqrt 3.

A Former Brilliant Member - 6 years, 2 months ago

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Damn it! I've always had that problem! Dunno why I do that. I had seen the problem a while ago and I didn't bother to do it on paper now and I ended up skipping the b thing. Well, I made the change now! Thanks!

vishnu c - 6 years, 2 months ago

if two of the vertices of a equilateral stands on points with integer co-ordinates then their distance can be described by an integer. but there cant be another point in a plane with integer co-ordinates and a integer distance from both.

তানজিউজ্জামান সাকিব - 6 years, 2 months ago
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