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}
+ d^{2} – 2d(s + 1)cos/PER
and using (1), this simplifies to: x^{2} + 2dx = s^{2} + s
Substituting for s, from (2), and simplifying now gives:
4d^{2} + 2dx – x^{4}
– 2dx^{3} = 0
(2d – x^{3})(2d + x) =
0
Thus 2d = x^{3} is the
only solution for d, x > 0.
i.e. PQ = x^{3}

Posted by Harry
on 20160705 08:34:56 