C.a.R.

Here is the geometrical solution.

We remember that the parabola y=x^2 is the track of all points, with equal distance from B=(0,1/4) and the line y=1/4. Thus BY=YX for all positions of X. Thus Y is on the middle perpendicular to XB.

Move X, such that the angle at Y is 90°.

In this position the middle perpendicular of XB is tangent to the parabola at Y. It is the point of minimum. We find that SBX, BXY and YMB are congruent. Consequently, ML=BH=1/2. This means, that the y-coordinate of Y is 1/2. The x-coordinate is the square root of 1/2.

Moreover, BM=BS, so that Y is in the Thales circle on MS. This allows to construct Y.

This ends the tutorial!