Consider two triangles PQG and PEF, where:
E is the mid-point of PQ, and:
Points P, G and F lie on the same line, and:
QG and EF intersect at the point R
RE = 1 and, PR = PE = FG

Given that PG = x, determine PQ in terms of x.

Since E divides PQ internally and FG = EQ, it follows that R and G
must divide EF and PF internally. So I think there is only one
configuration to consider.

Let PR = PE = FG = d  and let FR  = s

Using the isosceles triangle PER:      cos/PER = 1/(2d)        (1)       

Using Menelaus’s theorem in triangle FPE with transversal GRQ:

                        (FR/RE)(EQ/QP)(PG/GF) = - 1

and taking the absolute value of each side:

     (s/1)(d/2d)(x/d) = 1,    giving    s = 2d/x                     (2)

The cosine rule in triangle EFP gives

            (d + x)² = (s + 1)² + d² – 2d(s + 1)cos/PER

and using (1), this simplifies to:     x² + 2dx = s² + s

Substituting for s, from (2), and simplifying now gives:

            4d² + 2dx – x⁴ – 2dx³ = 0

            (2d – x³)(2d + x) = 0

Thus     2d = x³  is the  only solution for d, x > 0.

i.e.         PQ = x³

Posted by Harry on 2016-07-05 08:34:56