Prove it

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mf92

You have two tirangles ABC and A'B'C' prove that 2*BC*B'C'>=AB*A'B'+CA*C'A' if the corner BAC=B'A'C'=60°

This was one of the problems in our school math competition :)

Elroch

Nice problem. Probably to be found somewhere in Euclid? Never found trigonometry easy myself at school.

mf92

Me neither... we had 5 assignements, this and the following one was a problem to solve (if I did it even correctly:)):  

cos3x*cos(pi/2-3x)+sin3x*sin(pi/2-3x) > 3*sqr(3)/8

I'm guessing this one was to be solved with rules of addition, however it got me nowhere, probably did it wrong...

Elroch

The latter one, I would first try calculus.

mf92

with addition I got the result of 3sinxcosx*(cos4x+sin4x)> 3*sqr(3)/8 and got stuck

PS the half-"solution" :) above is probably wrong

s7silver

wolfram alpha gives this result

http://www.wolframalpha.com/input/?i=%28%28cos+x%29^3%29*%28cos%28pi%2F2-3x%29%29%2B%28%28sin+x%29^3%29*%28sin%28pi%2F2-3x%29%29%3D3*%283%2F8%29^%281%2F2%29

pawn_slayer666

I couldn't quite finish it, but I feel like I'm left at an obvious last step that could be solved by someone experienced in number theory, so here's what I did:

 

Let BC=a, B`C`=a`, AB=y, A`B`=z, AC=w, A`C`=x.

 

Law of cosines on the value of a and a` gives:

a^2=w^2+y^2-wy and a`^2=x^2+z^2-xz

Multiply the two equations and get

(aa`)^2=w^2x^2+y^2z^2=wxyz+w^2z^2+x^2y^2-x^2wy-wyz^2-w^2yz-xy^2z

We want to show

2aa`>=yz+wx or

(aa`)^2>=(y^2z^2+w^2x^2+2wxyz)/4

Substitute the equation 4 lines up into this one, and with some manipulation get:

3w^2x^2+3y^2z^2+2wxyz+4w^2z^2+4x^2y^2 >= 4x^2wy+4wyz^2+4w^2xz+4xy^2z

 

I've no idea how to prove the above, but I feel like it has something to do with some of these equations:

x^2+y^2>=2xy

x^3+y^3+z^3>=x^2y+y^2z+z^2x>=3xyz

That big long equation has fewer components in the products on the left than on the right side, so it seems intuitive, but I can't prove it.

mf92
offtherook wrote:

 

It's fairly obvious once you recognize that length BC is at least as large as the average of the two other sides.  (this applies to the prime version as well). You get this fact from the angle measure. Once you have that it's just a basic result from arithmetic.


?? that doesn't prove it, you just said " that it's true" :D

I'll post the solution when I get 'em

fireballz

mind my curiosity, but i never had any questions that was beyond the day to day eat work and sleep.

the answer is with the person that ask the question.  They say that most people ask questions of wich they already know the answer.

my question is: Should we prepare ourselves for answers, so that when someone do, ask us, then we can confirm that the answer are indeed the one that he had in mind:D

my answer would be wrong, simply because i am missing something.  But what am i missing here?

when i hear a song, i can get the value of the music.  When i look at a sunset, i can enjoy the beauty, etc...every sense can enjoy something...but to give an answer to something like a question that is totally mindblowing....it must give someone theory thrill...where does such maths fit in with day to day life?

is it the study of weirdness? quantum, cause then i feel at home, haha:D