The problem is essentially showing that:It hatttesss you!
(s-b)(s-c)} \ $where$ \ s= \frac{1}{2} (a+b+c) \\ \\ $Show that$ \ \ A \leq \frac{a^2+b^2+c^2}{4} )
YepThe problem is essentially showing that:
Expanding the LHS basically gives the result.
^2)
Yep
Alternatively just use
and just a bit more.
Another method (using triangle inequality):Yep
Alternatively just use
and just a bit more.
These aren't that hard, but there are really intuitive solutions to them which I think are nice.
1) For integersshow that
has atleast one solution pair
for all
Where
2) Prove that there are no solutions to the above problem such that
Yep your second solution was the same as mine, I'll post up my solution for the first though (imo it's a cool method).}{2}+\frac{n(n-1)}{2}=n^2\\ \\$If $n^2=2T_k=k(k+1)$, we have:$\\ \\k<n<k+1,\\ \\$a clear contradiction.$)
Haha I know, I normally don't answer these for a while so other people have a chance. Felt like spoiling the fun today though.
Oh right cool, yeah those triangles are exactly the ones in my construction. Just numbers vs a picture.A square number can be represented visually by a square, whilst a triangular number can be represented visually by a triangle. Any square can be broken into two triangles, and thus there will always be a solution to the equation:
Yep I think it's nice to see how an equation like that can easily be done visually. I've been trying to make a harder question which uses a similar method (constructing a visual of a certain type of number - square/triangular numbers for example) which turns out to be a lot easier than any algebraic method. No luck yet though.
That could indeed be the reason why I'm having no luck haha.
You would of course have to check the b=0 case separately, but it is obvious this won't give us any issues thanks to symmetry.Alternatively:
.