ABCD is a fixed rhombus. P lies on AB and Q on BC, so that PQ is perpendicular to BD. Similarly P' lies on AD and Q' on CD, so that P'Q' is perpendicular to BD. The distance between PQ and P'Q' is more than BD/2. Show that the perimeter of the hexagon APQCQ'P' depends only on the distance between PQ and P'Q'.
Solution
BPQ and DQ'P' are similar. Let PQ meet BD at X and P'Q' meet BD at Y. XY is fixed, so BX + DY is fixed. Hence also, BP + DQ' and BQ + DP' and PQ + P'Q' are fixed. So PQ + P'Q' - BP - BQ - DP' - DQ' is fixed, so PQ + P'Q' + (AB - BP) + (BC - BQ) + (CD - DP') + (DA - DQ') is fixed, and that is the perimeter of the hexagon.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002