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Select three points where two lines meet in such a way that the triangle that is formed is equilateral and has an area that is as small as possible.
Select three points where two lines meet in such a way that the triangle that is formed is equilateral and has an area that is as small as possible.
May 27th, 2009 at 2:47 am
Let points A, B, C have coordinates (x1, y1), (x2, y2), (x3, y3) espectively, and let these coordinates be integers. Suppose ABC is an equilateral triangle. Then
|AB|²=|BC|²=|CA|²=s ___ [1]
where s is positive (if s=0, there is no triangle, just one point!) and as small as possible.
Proof 1: By Cartesian coordinates and modular arithmetic.
Let AB=(x, y). Then
s=x²+y²
so
Iff x and y are even, s=0 mod 4, in which case AB, BC, CA all have both coordinates even, so there is a smaller such triangle DEF with DE=AB/2, EF=BC/2, FD=CA/2, contradicting minimality of s.
Iff x and y have different parities, s=1 mod 4 iff x+y is odd iff x2-x1+y2-y1 is odd iff x1+y1 and x2+y2 have different parities. But at least two of x1+y1, x2+y2 and x3+y3 must have the same parity, so s cannot be equal to 1 mod 4 and satisfy equation [1].
Iff x and y are odd, s=2 mod 4, in which case x1 and x2 have different parities, as do y1 and y2. But at least two of x1, x2 and x3 must have the same parity, so s cannot be equal to 2 mod 4 and satisfy equation [1].
Thus there are no such points A, B, C.
Proof 2: By trigonometry.
tan(a-b) = [tan a - tan b]/[1 - tan a tan b]
so if tan a and tan b are rational, so also is tan(a-b).
Let X=A+(1,0). tan XAB and tan XAC are rational.
Then BAC=XAC-XAB=±60° so tan BAC = ±sqrt(3) which is irrational. Thus there are no such points A, B, C.
May 27th, 2009 at 3:50 am
Another trig proof. Notation as before.
Let P be the midpoint of AB. Then the coordinates of P, and therefore of CP, are rational. But CP is perpendicular to AB and |CP| = (sqrt(3)/2) |AB|, so CP’s x and y coordinates are irrational multiples of the rational y and x coordinates (respectively) of AB, and are therefore irrational. Contradiction.
May 28th, 2009 at 1:57 am
I dunno… Sounds more like plain plane geometry to me.
Pick a square. Any square. Label the corners A, B, C, D, left to right, top to bottom. The top of the square is represented by the line AB; the left side represented by AC; the right side by BD; the bottom by CD. There are four possible solutions. First solution: The smallest triangle would be represented by the sides AC, CD, and AD and would be an equilateral right triangle. Rotate the triangle 90 degrees clock-wise (or counter clock-wise) to get the other three.
May 28th, 2009 at 1:58 pm
Nick, the triangle must be equilateral.
Typo in my first trig proof: one equation should read:
tan(a-b) = [tan a - tan b]/[1 + tan a tan b]
May 29th, 2009 at 7:14 am
Richard,
Oops…
Somehow I got fixed on the idea of an Isosceles triangle.