I came up with a math problem a few days ago: is it possible to draw an equilateral triangle in a coordinate system so that each corner of the triangle is placed on an integer coordinate?
Please try to help me solve this, it's driving me mad.
By Pick's theorem, the area of any polygon whose vertices have integer coordinates must be half an integer. The area of an equilateral triangle is s^2 * sqrt(3)/4, which is irrational (since s^2 is an integer). Hence, it is impossible to have an equilateral triangle in R^2 with integer coordinates.
Since translations don't change anything, we can define point A to be at (0, 0). Now, for any given point B at (bx, by), there is a formula that determines the position of point C - just rotate B clockwise by 60°.
By Pick's theorem, the area of any polygon whose vertices have integer coordinates must be half an integer. The area of an equilateral triangle is s^2 * sqrt(3)/4, which is irrational (since s^2 is an integer). Hence, it is impossible to have an equilateral triangle in R^2 with integer coordinates.
Thank you! I'm glad to be able to think about something else now.
no, it would be three vertices (all three at the same location) all three connected by straight lines. If you ask the next guy to define a triangle, he'd say "three points connected by lines", that simple definition is fulfilled. Since all edges have the same length, it's an equilateral triangle.
It violates some other properties usually assumed of triangles (angles aren't defined, the points are collinear, ...), but as it is in mathematics, you can define your objects as you need them. If three vertices and three same-length edges are all you need, it's possible. Otherwise, it's not.
no, the fact that equilateral triangles have 60° angles is a property that's only valid/provable for triangles with nonzero area. The angle is defined as sin(a) = a/h or rad = s/r or something, which isn't defined when all significant lengths are zero. Division by zero is bad, hmkay?
The area of an equilateral triangle is s^2 * sqrt(3)/4
Should be sqrt(3/4), but yeah... you are right. It's impossible.
no, it actually is sqrt(3)/4
the height of an equilateral triangle of base s is s*sqrt(3)/2, so the area is (1/2)s*s*sqrt(3)/2=s^2*sqrt(3)/4
One way to think about this issue is to remember the Cartesian and coordinate-free formulas for the dot product of vectors: u.v=u1v1+u2v2=|u||v|cos(q), where u=(u1,u2), v=(v1,v2), and q is the angle between them. If q=60 degrees, then cos(q)=1/2, and |u|=|v|=sqrt(u1^2+u2^2), so the equation on the right simplifies to u1v1+u2v2=(u1^2+u2^2)/2.
Now let u1=n, u2=m, v1=p, and v2=q be integers, with 0<m<n and with n and m, and p and q, relatively prime; then it can first be noted that both n and m must be odd, because they cannot both be even (else they would not be relatively prime), and if only one were even then (n^2+m^2)/2 would not be an integer even though (n^2+m^2)/2=np+mq, which is an integer. Also, because n^2+m^2=p^2+q^2 (because |u|=|v|), the same holds for p and q.
To break the symmetry I will attempt to eliminate p between these two equations and then show that q cannot be an integer.
From the first equation, p=(n^2+m^2-2mq)/(2n), so p^2=(n^4+m^4+4m^2*q^2+2n^2*m^2-4mqn^2-4qm^3)/(4n^2), and substitution into that second equation yields
n^2+m^2=(n^4+m^4+4m^2*q^2+2n^2*m^2-4mqn^2-4qm^3+4n^2*q^2)/(4n^2)
which simplifies to
4n^4+4n^2*m^2=n^4+m^4+4m^2*q^2+2n^2*m^2-4mqn^2-4qm^3+4n^2*q^2
from which
4(n^2+m^2)q^2-4m(n^2+m^2)q-(3n^4+2n^2*m^2-m^4)=0
which has a discriminant (with respect to q) of
16m^2(n^2+m^2)^2+16(n^2+m^2)(3n^4+2n^2*m^2-m^4)
which simplifies to 3*16n^2(n^2+m^2)^2, with square root 4n(n^2+m^2)sqrt(3), so there is no way for q, which in the end turns out to be
(n±m*sqrt(3))/2, to be an integer after all, a contradiction.
Pick's Theorem is more elegant though.
I think that this is a beautiful practical demonstration of the difference between rational and irrational numbers, something which might otherwise feel like a rather abstract distinction.
(Now, if someone could come up with an equally beautiful example of the difference between algebraic and transcendental numbers...)
no, the fact that equilateral triangles have 60° angles is a property that's only valid/provable for triangles with nonzero area. The angle is defined as sin(a) = a/h or rad = s/r or something, which isn't defined when all significant lengths are zero. Division by zero is bad, hmkay?
Then it's not an equilateral triangle as you erroneously pointed out.
An equilateral triangle is defined by all sides being equal in length. So it's an equilateral triangle.
I think that the mathematical definition of a triangle contains the requirement that the three points must not be collinear. The three points being at the same place makes them collinear.
Please try to help me solve this, it's driving me mad.
