PQR is an acute-angled triangle with PQ > PR and angle QPR = 60^o.

The respective circumcenter and orthocenter of this triangle are denoted by O and H. It is known that OH intersects PQ at X and PR at Y.

Find the ratio XO/HY.

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 2015-07-20 09:35:48