See Fig. OB = OC = CA = AB and the diagonals OA and BC are at right angles. Let OC = a. Suppose B = [x1, 4x1/3]. Equation of the side AB is y = (4x1)/3 and the coordinates of A are [a + x1, (4x1)/3]. Hence
(CA)2 = a2
x12 + 16x12 = a2
a = ± 5x1/3
Case 1: If a = 5x1/3, we have
and O = (0.0)
By hypothesis, points B, (2/3, 2/3) and C are collinear. This implies that
Now, x1 = 0 = B = (0, 0), which is actually not the origin. Hence, x1 = 3 5/ .Therefore
0 = (0,0), B = (3,5, 4/5), A = 8/5,4/5
and C = (1,0)
Case 2: a = -5x1/3