Good question! If you write 6=B, 29=C, 17=D in my solution, it boils down to A^2 = (B+C+D)^2 - 4BC the general integer solution of which presumably involves two or three parameters. The most likely intention was to swap C and D: B = 6, C = 17, D = 29 and A^2 = (6+17+29)^2 - 4*6*17 = 52^2 - 408 = 50^2 I'd still like to see a solution that's linear in the areas, rather than quadratic, but perhaps there's no such animal?? R. On Fri, 14 Dec 2007, Robert Baillie wrote:
can the numbers 6, 17, 29 be changed so that the area of "?" becomes an integer?
bob ---
Schroeppel, Richard wrote:
A New Year's Puzzle that's been circulating here ...
B--------------C /|\_ / / / \_ 29 / / | \_ / / / \_ / / | ? \/ / / __--P / 6 | __-- / / /_-- 17 / A----Q---------D
ABCD is a parallelogram. P is on CD, Q is on AD. The areas of the triangles ABQ, BCP, and PDQ, are 6, 29, and 17. What's the area of triangle BPQ?
Rich
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