Friday Extra

Re: Friday Extra

About 14" square?

Re: Friday Extra

About. Did you get that by calculation or experimentation?

Re: Friday Extra

Calculation. Attempting to hide the answer that involved one radical.

13.92" would be closer, if I am right.

Re: Friday Extra

My answer is to the north side of 14". But that's no guarantee that it is right.

Hmmm

I had the squares awith their centerlines along the diagonals of the big square, with sides = sqrt(2)/4. Is there another arrangement that betters that? Here are my conclusions:

1) All interior squares will be oriented in the same direction (I can't prove that to myself, and this might be an area that I need to explore if you tell me my solution is not optimal.)

2) Given #1, looking at the internal square sizes as a function of rotation, zero and 45 degrees of rotation are local maxima, and the sqares are larger at 45 than at 0.

3) That's as far as I took it. Is there a larger maximum between those two, or is premise #1 wrong?

Interesting problem.

Re: Hmmm

I get 13.92" also.

I set up an equation for the length of one side of the 1 meter outer square. Its side = 39.37". There are three line segments making up each side of the 1 meter square, and these line segments are all simple trig functions of X, where X is the side length of the 5 interior squares. Two of the line segments are 0.707X, and the middle segment is X/0.707 = 1.414X.

Eqn: 2(0.707X) + 1.414X = 2.828X = 39.37". Hence, X = 13.92".

Wiley

Not quite optimal

Our solutions diverge at #1

I'll be darned

That took a while to see!

2/(4+sqrt(2)) meters on a side?

That'll do it!