bonanova

We place a square inside a triangle so that a side of the square remains flush with one of the sides of the triangle, but is free to slide along it. As the square is enlarged, eventually two corners of the square will touch the other two sides of the triangle, making further enlargement of the square impossible. Every triangle is thus associated with three "largest" squares. In general the areas of these squares are distinct. Calabi I have a particular triangle whose three largest squares are congruent. What can we say with certainty about this triangle?

Agree with k-man. Moving.

Re-reading the previous dialog makes me miss SuperPrismatic. And Cap'n Ed.

That's it. Someone pointed out in an article I read recently that AM>GM solves a whole slew of optimization problems. An application easily but not normally made.

Yeah, I had to think that point through for a moment. I saw it as having a sum that I didn't know how to evaluate. So I added and subtracted the term that would make the sum zero. That step simultaneously evaluated the sum and added 2 to the mix. Your approach is neater, reasoning that the answer is not changed if you add the squared length of the nth ray. That's a nice insight.

I had to think about this point: does it matter there are only n-1 rays?

Show that the sum of the squares of the lengths of all sides and diagonals emanating from a vertex of a regular n-gon inscribed in the unit circle is 2n.

Your solution fits the OP so I'll mark it solved. Persistence paid off. Anyone see another approach that fits the title? What means are available here, with a little manipulation? What do we know about various means? Know of any inequalities that have to do with means? Can I think of even more clues?

Yeah. But they get solved. Smart people up there at MIT. I dated one of them back in the day.

Great job.

There is a function f (n) on the positive integers n = 1, 2, 3, ... It is defined solely by these properties: f (n) is itself a positive integer f is strictly increasing: f (n+1) > f (n) applying f twice returns the argument multiplied by 3: f ( f (n) ) = 3n Work out the values of f for the first 30 integers or so: f (1), f (2), f (3), … , f (30). Start, of course, with 1 and 2. When you see some patterns, find f (2015). Credit: Recent Math Olympiad.

Are you saying plasmid's post is the correct answer?

No correct answers yet. @gs I see where you're coming from, but another answer better fits the clue. @plas, a dimension I did not think of, nice. Not the answer I had in mind.

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) That is the key finding. 4. Inserting 9 and 4 in place of a and b finds the minumum of 12. You're flirting with derivatives. Suppose this were an algebra problem. What if the title were not chosen arbitrarily?