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Hi all,

Technically, that simplification is correct only if y is not negative. Otherwise, you'd need to have |y| instead of y in the numerator there.

"No problem".

Hej bobbym

Inget problem!

Hej,

You are usign the highest sum you can find in the grid. In the 2x2 case, that is 6. If you arrange the numbers any other way, you'll get a higher maximum sum. For example:

Here, the highest sum is 7, which is higher than the 6 for the previous grid, so we do not want that.

For the second, 4x4 grid, the highest sum inside it is 19. To prove that this is the minimal highest sum, we would need to prove that for all other arrangements of numbers from 1 to 16 into the grid, there will be two adjacent numbers which sum to 19 or higher.

Hej bobbym

You look at all possible fillings of the grid and for each filling you see what the maximum sum of adjacent ellements is. Then you find the minimum of those numbers.

For example, for n=1, you are filling a 2x2 grid with numbers 1,2,3,4. The lowest adjacent sum, which is 6, can be obtained with the filling:

or any equivalent filling, (obtained by rotation or symmetrical mapping, t.ex.)

For n=2, a filling with S=19 *can* be found, but I am not sure if there is a lower one (probably not):

The minimum of S across all possible ways to fill the grid up, I am guessing?

Well, t.ex. and osv. are abbreviations from Swedish, not English, so that is not too surprising.

I shall take a look at that page.

Well, "ex-" sure sounds much better there. "e-" does not seem like a productive suffix.

And besides, you cannot expect a suffix to recognize mathematical terminology, can you? I doubt it would recpgnize "homomorphism" and the like. xD

Hej Bob!

While I'd love to do that, I don't think I'll have time in the near future.

Oh and yes, inscribed circles are a must!

Hej Bob!

These threads are a fantastic idea! Nice work.

An intersting fact about the Nine Point Circle (I also know it under the name of Euler's Circle) is that when you apply inversion around the Nine Point Circle, the circumcircle gets mapped into the incircle and vice versa!

No problem! That's the only workaround I'd ever found.

Does entering an empty math tag somewhere before help?

**UIOAE**

unimposable

Hi mrpace

You can prove that there is at least one number c in (a,b) such that f(c)=C by the Intermediate Value Theorem. To prove that there is only one such number, we will take any two numbers, c1 and c2, such that f(c1)=f(c2)=C. If c1 were less than c2, than f(c1) would be less than f(c2) which is not possible. Also, if c1 were greater than c2, then f(c1) would be greater than f(c2), which is impossible as well. Thus, c1 must be equal to c2, which proves that there is exactly one such number.

You can do it by finding the tangents from (0,0) to the circle.

Yes, I am going by that idea as well.

Thank you! Nice to see you again!

I cannot get it completely simplified yet. (without f^-1, that is).

That is not correct. That would not get the wanted integral. It would give:

Then, the coordinates of C should be (f^-1(pi/2), pi/2).

Is that the graph of f or f^-1?

Hi bobbym

The y-coordinate of C needs to be pi/2, though, not the x coordinate.