Fox 355

Roland Sampy submited this one.  Let's confirm his answer.
Happy new year to all, by the way!

Fox 338

This is a little related to the recent ones, but much simpler. Here we'd appreciate a pure geometric solution if one found.

http://www.8foxes.com/

Fox 337

This one is related to Fox 334, and should be simpler.

http://www.8foxes.com/

Fox 327

http://www.8foxes.com/

Earth is not a perfect sphere,
It's a geoid of some kind,
Not perfectly-round but close enough.

A year is a little more than 365 days.
A day is not exactly 24 hours,
little short of that.
Pi is more than 3.
Twenty-two over seven is not far off though,
often close enough.

Man hardly lives a 100 years.
Few goes beyond,
Some get close enough.

Truth ascends above digits,
distorts formulas,
bends orbits.

Man searches it with greed,
never reaches,
never conquers.

So my dear,
give up the precision,
stop running the numbers.
Instead,
smell the soil after rain,
hold a cuddling baby in your hand,
wander your vision from Vega to Orion,
let a snowflake die in your palm.
hear what morning breeze tells,
Then round up what you got.
A little more or a little less.
Even if it is still not found,
You're close enough.

--- Polar Fox

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 320

Bleaug has a solution for this general case.

http://www.8foxes.com/

Fox 313

Bleaug submitted the following fox, saying: "a problem which can be expressed in geometric terms and that is elegantly solved using pure algebraic arguments, still giving deep insight."

General case of this problem can be found in literature (with at least 14 proofs :)
But let's try this smaller version here.

http://www.8foxes.com/

Fox 295

Let the one who lives, live with an evidence,

Let the one who dies, perish with an evidence.

http://www.8foxes.com/

Fox 293 - Solutions

According to the order of submission...
by Bob Ryden: (Ajit commented a very similar one)


by Bleaug:

Two tangents to a given circle define a symmetric 'kite' figure. By construction, the two kites defined by OUV and OTU are isometric because they have identical short legs. Since angles in T, U, V are right angles, blue and green angles are supplemental. Hence, angle a is angle(OUV). From Thales theorem we get cos(a) = 1/3

by Yu:
Several steps remain implicit in the figure. More words may be needed for a formal proof. But the solution holds.


by Binary Descartes:

Common tangents from the same point have equal lengths.
See the two identical deltoids (kites) sharing the same side.
Then cos(a) = e/3e = 1/3.

http://www.8foxes.com/

Fox 294

Let's continue to mess around with this configuration...

http://www.8foxes.com/

Fox 293

The options are corrected below.
There are about 5 different solutions, that we can post in coming days.

http://www.8foxes.com/

Fox 261 - Solutions

Four good solutions for Fox 261, all drawing lines in the sand...

http://www.8foxes.com/

Fox 261

The claim below makes some information in Fox 252 redundant.

http://www.8foxes.com/

This is an intriguing construction! We can easily post half a dozen more good foxes. Keep ckeckin' !!
Also there are at least 3 purely-geometric solutions...

Fox 12

Yu sent a confirmed solution.http://www.8foxes.com/

Fox 11

An old fox resurrects since a new solution has been received. http://www.8foxes.com/

Fox 252 - Solution-2

http://www.8foxes.com/

See Fox 252

Fox 252 - Solution-1

http://www.8foxes.com/
See Fox 252

Fox 252

http://www.8foxes.com/

All along the wall, between us

I see a teacher, there for us

I look at the wall, I see right through it

I lean on the wall, there for us

Fox 243 - Solution

http://www.8foxes.com/

Bleaug claims the following:

Almost a pure geometric proof this time (only requires Pythagorean theorem).

- TIJ is the section of the pyramid by a plane orthogonal to base and parallel to BC and AD

- by construction, TI is orthogonal to AB, and TJ to CD

- hence: AT^2-BT^2 = AI^2+IT^2-BI^2-IT^2 = AI^2-BI^2, similarly DT^2-CT^2 = DJ^2-CJ^2

- since AIB is a translation of DJC: AT^2-BT^2 = DT^2-CT^2

- from which we get DT^2 = AT^2-BT^2+CT^2 = 9-2+1 = 8

Fox 244

http://www.8foxes.com/

Look into thy heart, and see,

it desireth to fly in Heavens,

whilst there are knots pulling thee

to cruelly darketh corners!
--- Dervish Fox