[math-fun] A New Year's Puzzle
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
Where is The Book solution? Writing AB = a, BC = b, DP = x, DQ = y, and noting that an affine transformation needn't change the areas, so that we can assume that the parallelogram is a rectangle, we have a(b-y) = 12, b(a-x) = 58, xy = 34, ab = 52+A, where A is the required area. The first two (with the last) give ay = A + 40, bx = A - 6 Multiplying the last two pairs of equations, 34(A+52) = (A+40)(A-6), A^2 = 2008 but how does one SEE what is going on ??? [the last equation is 4-dimensional!] R. On Thu, 13 Dec 2007, 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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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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Apologies for the hasty rubbish, below. At risk of sending more, how about: B = rt, C = su, D = (r-u)(s-t) - 1 which yield(?) A = tu - rs - 1, I hope! R. On Fri, 14 Dec 2007, Richard Guy wrote:
A^2 = (6+17+29)^2 - 4*6*17 = 52^2 - 408 = 50^2
On Fri, 14 Dec 2007, Robert Baillie wrote:
can the numbers 6, 17, 29 be changed so that the area of "?" becomes an integer?
Forget those minus ones and try B = 256, C = 320, D = 1512. R. [r = 32, s = 64, t = 8, u = 5] On Fri, 14 Dec 2007, Richard Guy wrote:
Apologies for the hasty rubbish, below. At risk of sending more, how about:
B = rt, C = su, D = (r-u)(s-t) - 1 which
yield(?) A = tu - rs - 1, I hope! R.
On Fri, 14 Dec 2007, Richard Guy wrote:
A^2 = (6+17+29)^2 - 4*6*17 = 52^2 - 408 = 50^2
On Fri, 14 Dec 2007, Robert Baillie wrote:
can the numbers 6, 17, 29 be changed so that the area of "?" becomes an integer?
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.
What I thought of as the uninsightful version of the answer is quadratic in edge lengths -- will that do? You can un-shear the parallelogram and re-scale the axes, both with area-preserving linear maps -- and this point of view is certainly linear in area, so I don't think I'm cheating here. Anyway, this lets you canonicalize the diagram to leave rectangle ABCD with lengths AB=12, AQ=1. B--------C |\\__ | | \ \__ | | \ \P | \ /| | \ / | A-----Q--D Now there is only one free parameter, the length of QD. Say QD=x, so DP=34/x, and BC=x+1. So the length of PC is both 12-34/x and 58/(x+1), and we get x=(20+sqrt(502))/6. The area of the whole rectangle is 12(x+1), and then subtract off 6+17+29. I wouldn't say it's particularly insightful, but it does get the answer by solving for a quadratic in edge lengths, not in areas. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
participants (5)
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Edwin Clark -
Michael Kleber -
Richard Guy -
Robert Baillie -
Schroeppel, Richard