Fox 347

Below is the general case submitted by Roland Sampy.
There seems to be at least 2 expressions for r.
We had a solution, but it looked a little clumsy. Moreover, we lost it :)

So, do help us to recover... Oh yes, do enjoy as well...

Fox 9

This was an old one. I think a solution was found but then lost. Yeah, it sounds stupid :)
Here, we received a claim that a solution was found again. It should not be terribly hard. Enjoy...

http://www.8foxes.com/

Fox 339 - Solutions

Two geometric solutions are below for this fox. First by bleaug, next by Bob Ryden. Enjoy...

http://www.8foxes.com/

Fox 339

A general case of an old one, as promised last week... http://www.8foxes.com/

Fox 15 - Solution

Well, we haven't received any comments. So here is the animated solution for Fox 15 by Bleaug.

"... Not an easy one though. I have tried to provide an animated SVG solution because packing all info into one single drawing or comment text would become clumsy. The proof is based on well-known (?) geometric properties of parabolas..."

To see the animation below, you will need any of the following browsers:
Chrome, Firefox, Opera, Safari, or IE Explorer 9.0.

http://bleaug.free.fr/8foxes/8foxes187.svg

Use mouse clicks or left-right arrows on your keyboard to browse thru the pictures.

http://www.8foxes.com/

Fox 320 - Solutions

This fox has been discussed extensively. See here...

1. Calculus by Bleaug

He says: "I spotted this problem in a Paul Halmos book "Problems for mathematicians, young and old, 1991", actually in a French translation. He provided the solution below (reformulated by myself) as being proposed by Hugh Montgomery in 1985 in some Math Conference."






2. Checkerboard Solution by Rochberg and Stein:

1. Call sub rectangles as TILEs.


2. Start from the lower left corner of the overall rectangle (let's call this point as the origin)Draw horizontal and vertical lines separated by 1/2=0.5 units starting from the origin.


3. This will create 0.5 x 0.5 squares (and possibly rectangles around 2 edges -top and/or RHS)


4. Color the first square (that has the origin as one of its corners) as black the next one as white, and so on to generate a checkerboard-look.


5. Each TILE will have equal areas of black and white (Why?)


6. Therefore the overall rectangle will have equal areas of black and white.


7. So overall rectangle has at least one integer side.
   

H means the horizontal side is integer, and V means that the vertical side is integer. a1 is the square at the origin. There could have been a smarter combination, but this simple one illustrates the process clearly. This looks like the closest geometric solution we can get -at this time.

When there's hardly no day nor hardly no night
There's things half in shadow and halfway in light

3. Induction by Robinson

1. Assume that each H-tile has a width of 1, and each V-tile has a height of 1. Note that any rectangle can be converted this way without distorting the original problem. (This may increase the number of tiles significantly though)


2. Chose any H-tile, say T(0). (If there is no H-tile, then the result is immediate)


3. If there are H-tiles whose lower border shares a segment with T(0)'s upper border, choose one and call it T(1).


4. Otherwise only V-tiles share this border. In this case, we can expand T(0) upward 1 unit. This does not increase the number of H-tiles. Also, the cut V-tiles still have height 1. (They are still V-tiles)


5. Continue expanding T(0) until either the top of the rectangle is reached or a choice of an adjacent H-tile T(1) is possible.


6. Then repeat the same process from T(1). (Continue upward similarly from T(1) to get T(2), and so on...)


7. This will result in a chain of T(0), T(1), T(2), ... , T(m).


8. Starting from T(0) again, work downward similarly to obtain a bigger chain:
   T(-n), T(1-n), ... , T(0), T(1), ... , T(m-1), T(m) of H-tiles stretching from bottom to top.


9. Remove these tiles and slide the rest together to get a rectangle with fewer H-tiles.


10. Induction applied to this smaller rectangle yields the result for the original rectangle.
    

http://www.8foxes.com/

Fox 15

This ancient fox was the general case of Fox 187, which was successfully-solved. And now, Bleaug claims that this has been solved, too. Let's see if we'll be able to confirm his claim.

http://www.8foxes.com/

Fox 320

Bleaug has a solution for this general case.

http://www.8foxes.com/

Fox 305

Yet again,

try to see the goodness,

see the beauty,

surrounding you.

Forget about the numbers, summations, subscripts.

Leave behind the accounts, stocks, papers, statistics.

Just leave yourselves to the arms of an ocean,

full of love and compassion.

Drift away with the blowing wind...

-- Dervish Fox

http://www.8foxes.com/

Fox 300 - Solutions

Bleaug:

Newzad:

Giannno:

http://www.8foxes.com/

Fox 300

We have been delaying this for a while, but Ajit and Vihaan have already solved it. So there is no point of keeping it a secret :)
Note that this may be a general case for Foxes: 296, 297, and 298. It is also related to Fox 301.

It would be great if there are purely geometric solutions!!

http://www.8foxes.com/

Fox 298 - Solutions

Solutions to Fox 298.

by Giannno:
by Newzad:

http://www.8foxes.com/

Fox 298

Proving this one immediately solves Foxie 296 and 297.
That's why this is the general case.

http://www.8foxes.com/

Fox 273

http://www.8foxes.com/

Fox 258

Bleaug submitted this as the general case of a series of foxes.

http://www.8foxes.com/