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Thread ID: 57185 2005-04-25 22:29:00 Weekly maths question Renmoo (66) PC World Chat
Post ID Timestamp Content User
348824 2005-04-27 12:16:00 Because of extreme brilliant brain by all people, I decided to reveal the answer earlier, A = 3, C = 7. Next:

Simplify the following equation:

(x-a)*(x-b)*(x-c)*(x-d).........(x-z) = ?

Note: * means multiply

Cheers :)
Oh man.. took me so long to figure it out.... somepeople could do with

(x-a)*(x-b)*(x-c)*(x-d).........(x-x)*(x-y)*(x-z) =

My answer... zero.
techie (7177)
348825 2005-04-27 12:17:00 Let us not get too complicated. By way of contrast, try this :-
Using spin$^c$ structure we prove that K\"ahler-Einstein metrics with nonpositive scalar curvature are stable (in the direction of changes in conformal structures) as the critical points of the total scalar curvature functional. Moreover if all infinitesimal complex deformation of the complex structure are integrable, then the K\"ahler-Einstein metric is a local maximal of the Yamabe invariant, and its volume is a local minimum among all metrics with scalar curvature bigger or equal to the scalar curvature of the K\"ahler-Einstein metric.
All quite plain to those who know ....( and not me anymore - have forgotten all that stuff..)
TonyF (246)
348826 2005-04-27 12:17:00 it kind of takes the wind from your sails doesnt it.....

the fact that Plod got it on the first post.....hooray for google and its infinite uses.

Good Show Plod.
Jams (1051)
348827 2005-04-27 12:41:00 Do you want the real or imaginery coordinates of this complex conjugate? (members.ispwest.com)

Joking aside, brute force won't get you very far, the best I could get on brute force after about 5 minutes was 591,709,009 / 9,306 - although Python should be quicker than Excel. Google turned up that page - although I'm not sure how to apply it.....

Yeah, brute force will not turn up the precise answer, but algebra will. Answer tomorrow.
vinref (6194)
348828 2005-04-27 22:07:00 A certain maths teacher had a piece of straight cane with which to beat those of his class who misbehaved. The class got their revenge by cutting the cane into three pieces. What is the probability that the pieces can be placed together to form a triangle if the cuts were random Dally (6292)
348829 2005-04-27 22:35:00 A certain maths teacher had a piece of straight cane with which to beat those of his class who misbehaved. The class got their revenge by cutting the cane into three pieces. What is the probability that the pieces can be placed together to form a triangle if the cuts were random
Using logic, if the lengths of the 2 shorter pieces added together is longer than the length of the longest piece, then a triangle can always be made. The first cut would result in 2 pieces of, most probably, unequal length. (What is the probability of a perfect match on the first cut?) The probablity of cutting the longer piece again (to give three pieces where the sum of the 2 shorter lengths is more than the length of the longer piece, such that a triangle can be formed) is 50%. So, assuming the cuts are purely random, my guess is 50%.
andrew93 (249)
348830 2005-04-27 22:45:00 Really good try Andrew but not quite right Dally (6292)
348831 2005-04-28 09:03:00 When you say 'not quite right' are you referring to the 50% or the logic? andrew93 (249)
348832 2005-04-28 20:41:00 After the first cut, the second cannot be in the same half which has a probability of 50% but anywhere in the other half which is more than half the length away from the first cut will not produce a triangle. So the probability of producing a triangle is less than 50% - by how much I'll let you have a go at working out. Dally (6292)
348833 2005-04-29 04:26:00 Ah, of course! The sum of the 2 shortest lengths (not just after the 1st cut but after the 2nd cut too) must be more than the length of the longest side. So for example, a first cut of 40% and 60% with a subsquent cut of the longer piece to 5% and 55% will not yield a triangle even though the longest piece was cut a 2nd time (per my original post). D'Oh!

Answer coming soon.....
andrew93 (249)
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