Thanks, but this is only putting geometrically what I put algebraically, and doesn't really show what's going on, because your eventual equation is one between four-dimensional volumes. There ought to be a solution which stays in at most two dimensions. R. On Fri, 14 Dec 2007, Michael Reid wrote:

...

but how does one SEE what is going on ???

hopefully this "explains" where it comes from:

draws lines through P and through Q parallel to AD and AB respectively; suppose they intersect at point X and meet the opposite sides of the parallelogram at E and F respectively. my attempt to mark up the original ascii art:

B----F---------C /|\_ / / / / /_ / / | / \_ / / / / \_ / / | / \/ E----X------_--P / |/ __-- / / /_-- / A----Q---------D

then we have [ABCD] - 2 [BPQ] = [EBFX] , where i use [ABC...Z] to denote the area of the polygon ABC...Z . we also have [EBFX] / [ABFQ] = EB / AB = [EBCP] / [ABCD] , so that [EBFX] [ABCD] = [ABFQ] [EBCP] = 4 [ABQ] [BCP] . multiply the first equation by [ABCD] and substitute this last one to get [ABCD]^2 - 2 [ABCD] [BPQ] = 4 [ABQ] [BCP] , which is ([ABCD] - [BPQ])^2 - [BPQ]^2 = 4 [ABQ] [BCP] , so that [BPQ]^2 = ([ABCD] - [BPQ])^2 - 4 [ABQ] [BCP] = ([ABQ] + [BCP] + [PDQ])^2 - 4 [ABQ] [BCP] .

for the numbers in the original problem, we have [BPQ]^2 = (6 + 29 + 17)^2 - 4 * 6 * 29 = 2008 .

mike

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