The 45° Point

At 45° the triangle under the point has two 45° angles, so its legs are equal. Call each a. Pythagoras with hypotenuse 1: ${a}^{2}+{a}^{2} = 1$, so ${a}^{2} = \frac{1}{2}$ and $a = \frac{\sqrt{2}}{2}$, about 0.71.

So the point at 45° is ($\frac{\sqrt{2}}{2}$, $\frac{\sqrt{2}}{2}$): across and height are equal, which is the only place in quadrant I where that happens. The three special points of quadrant I use just three sizes: 1/2, √2/2 and √3/2.