Each exterior angle of a regular hexagon is $60^{\circ} (360^{\circ}/6)$, so when sides HB and IC are extended to meet at A , an equilateral triangle, ABC is created.

Let the sides of this triangle be of length x. As BC, DE and FG are all parallel, triangles ABC, ADE and AFG are all equilateral. So, DE = DA = $p$ + $x$; and FG = FA = $q$ + $p$ + $x$.

The perimeter of trapezium BCED = $x + p + x + 2p = 2x + 3p$;

The perimeter of trapezium DEGF = $(p + x) + (q + p + x ) + 2q = 2x + 2p + 3q$;

The perimeter of hexagon FGIKJH = $2((q + p + x) + 2r ) = 2x + 2p + 2q + 4r$.

So, $2x$ + $3p$ = $2x$ + $2p$ + $3q$ ; hence $p$ = $3q$.

Also $2x$ + $2p$ + $3q$ = $2x$ + $2p$ + $2q$ + $4r$ ; hence $q$ = $4r$ .

So, $p$ : $q$ : $r$ = $12r$ : $4r$ : $r$ = $12$ : $4$ :$1$

*This problem is taken from the UKMT Mathematical Challenges.*