Before I begin it should be noted that if the ratio is constant it must equal 1. This is because XH/OY would also be constant and equal to the sought ratio. That the ratio is constant is implied by not giving any relative lengths.
Coordinatize: P=(0,0), Q=(1,√3), R=(r,0)
Note: for the triangle to be acute 1<r<4
I'll dispense with the algebra and give simplified results.
The orthocenter is the intersection of any two (or all three) altitudes.
The altitude to Q is the line x=r/2
The altitude to R is the line y=√3x/3 + r√3/3
The intersection H = (1 , r√3/3  √3/3)
The circumcenter is the intersection of the perpendicular bisectors of any two (or all three) sides.
The perpendicular bisector of PR is the line x=r/2
The perpendicular bisector of PQ is the line y  √3/2 = √3/3(x  1/2)
The intersection O = (r/2 , r√3/6 + 2√3/3)
The line OH is y = √3x +r√3/3 + 2√3/3
OH intersects PQ at X = (r/6 + 1/3, r√3/6 + √3/3)
OH intersects PR at Y = (r/3 + 2/3, 0)
The distances OX and OY are both the same: 2r/3  2/3.
Note: if r=2, the triangle is equilateral. The is no line OH because O=H, but the two distances are clearly still equal.

Posted by Jer
on 20150720 09:35:48 