k-man

This belongs in the "Optical Illusions" section of the site.

Sure, gavinksong. You're right. I had a brain fart with the second part.

Wouldn't a triangle with one side = 0 collapse into a line segment or 2 co-located line segments ?? I don't have an answer, but I suspect that is NOT the answer that BMAD is looking for - but it made me chuckle [thanks]. Either a degenerate triangle, as gavinksong suggests, where a2 + 0 = c2, Hmm... technically any 3 numbers a, b, and c can be part of this equation a2+b2+x=c2. They don't even have to be sides of a triangle and x=c2-a2-b2. But x=0 only when a, b and c are sides of a right triangle and c is the hypotenuse.

hmm... Being a mathematician, logician and a perfectionist (all-in-one), I disagree

3. Any local extremums for f(x) will occur when the value of g(x)=sqrt(b/a) and plugging sqrt(b/a) into the formula for f(x) yields f(x) = 2*sqrt(ab) 4. Inserting 9 and 4 in place of a and b finds the minumum of 12.

Not sure if this qualifies for c), but should be sufficient for a) and b)

but how do you 'know' these can't be further optimized? Well.... I guess I don't. Your assumption (highlighted in the quote) is not that difficult to prove.

Right on!

k-man always has such elegant solutions to these problems. I'm sorry but I couldn't contain myself. Thanks, gavinksong, but the real credit should be given to BMAD for posting these great puzzles.

P.S. I see that you already recognized it.

Hah! Who would know that better than k-man?

Is this claim for any nxm table or only when n = m ? For any nxm. I don't have a rigorous proof, but I think it would go along these lines...

Interesting analogy, but I'm not convinced it applies here.

Using the interpretation of "crossing" as "having a common point with"...

Both #1 and #2 can be true at the same time. I guess you're looking for a proof that #1 and #2 cannot be both false.

a 2-digit number

Oops, you're right. I skipped a pour.