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According to this site: Sum of four cubes n = 100 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 000000 (201...

8. Straight from the question angle DBC = 180-60-24 = 96 angle EBC = 138-96 = 42 Extend CB to produce another point F (closer to B than C). angle EBF = 180-138 = 42 So EB bisects angle DBF, ED bisects angle BDA Hence E is an excentre (look it up :-) of triangle BDC and EC bisects angle BCD...

I approve!

*rofl Like you've been to any of them to know that. But anyway, back to the point, olympiad camps are for training and selecting students for competitions. NMSS is more of a social gathering of good maths students with the purpose of exposing them to level of maths/style of education...

Still browse here sometimes when I'm bored, very rarely though. Also I wasn't going to pass on a nice, cool, decent, interesting (ok I'll stop now) Euclidean geometry problem(ie no trig etc) :-p, don't see too many of those around here. Oh and another note about that question, its actually...

Okey here's a way with no trig: construct an equilateral triangle with AB as one of the basis, call it ABE. E should lie just outside of triangle ABC, near C. (in the following bit, whenever i say something like ABC, i mean angle ABC unless specified otherwise) by construction...

bored, plus templar is clueless, so 1. (50+1-7)C7 2.1/(1-x^5)*1/(1-x^10)*1/(1-x^50)*1/(1-x^100)*1/(1-x^200)

if w=cos(pi/5)+i.sin(pi/5), then w^5+1=0 and w^4-w^3+w^2-w+1=0 (z-(w+1/w))(z+(w+1/w))(z-(w^2+1/w^2)(z+(w^2+1/w^2)) =(z^2-(w+1/w)^2)(z^2-(w^2+1/w^2)^2) =(z^2-(w^2+1/w^2+2))(z^2-(w^4+1/w^4+2)) =z^2-(w^2-w^3+w^4-w+4)z^2+(w^2+1/w^2+2)(w^4+1/w^4+2)...

p(exactly one happening)=p(at least one happening) - p(both happening) so it is 0.02 but withoutaface had the wrong reasoning the chance of at least one happening is not a+b it's 1-p(none happening) = 1-(1-a)(1-b) = a+b-ab while the chance of exactly one happening is not a+b-ab it is...

nup, exam mark was 99.

i think most hsc "high level" questions are broken down into more accessible bits anyway so you shouldnt be worrying too much. as of techniques in general, a lot of it really comes under experience i guess. the more you've seen, the more techniques you'll have.

no, logic is not my problem. the problem being the fact that the actual issue was addressed perfectly well in the first few posts. yet others like to sustain an ongoing trend of stating the obvious and the irrelevant. (like people should read all previous posts before asking the same question)...

why is this thread still going...??

...er you still havent answered why it's wrong to use the terms "Euclidean geometry" in this instance here. Using Euclidean geometry just means that the implied bit can be worked out using the methods such as angle chasing etc, not algebraic bashing. Btw its a lot less boring than what you...

What's wrong with it?

i agree....

its not likely that anyone posting on the forum will ever need to worry about it, so why bother.

bash!! lhs = (2a-b+2b-c)((2a-b)^2-(2a-b)(2b-c)+(2b-c)^2)+(2c-a)^3 (sum of cubes) = (2a+b-c)(...same stuff)+... = (a-2c)(...)+... (using a+b=-c) = (a-2c)((2a-b)^2-(2a-b)(2b-c)+(2b-c)^2 - (2c-a)^2 ) = (a-2c)((2a-b+2c-a)(2a-b-2c+a)-(2a-b)(2b-c)+(2b-c)^2) (sum of squares) =...

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