## Tuesday, July 19, 2011

### 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...

## Thursday, July 14, 2011

### Fox 345

A 2-circle covering problem would have been easier. But Let's start with this one.

## Sunday, June 26, 2011

### Fox 344

Two flow together
one rises one falls
one after another
one shines one glooms

## Wednesday, June 22, 2011

### Fox 343

x, y, and z are perpendicular to the chord.

What makes a man a man
is not the fire
nor the bullets,

but pure,
simple,
childish forgiveness.

-- Dervish Fox

## Tuesday, June 14, 2011

### Fox 342

Ahmet Arduç submitted this one. Geometric solutions will be appreciated.

http://www.8foxes.com/

## Saturday, June 4, 2011

### Fox 339 - Solutions

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

## Friday, April 29, 2011

### Fox 330 - Solutions

Below are three solutions to Fox 330.

By Six:

By Bleaug:

By Henk Reuling:

http://www.8foxes.com/

## Saturday, April 16, 2011

### Fox 339

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

## Monday, April 11, 2011

### 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.

## Thursday, April 7, 2011

### Fox 337

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

## Saturday, March 26, 2011

### Fox 336

Let's mess around little bit of what we got, with a little twist of Physics. Don't blame us college kids! It is the truth that's calling. What an overwhelming call that is!

Do we need to know the object's mass? Probably not.

http://www.8foxes.com/

## Thursday, March 24, 2011

### Fox 335

Constructing the Centroid of a Pentagon

## Monday, March 21, 2011

### Fox 334

This is more like a google search than a geometry problem, but it is nice to have a reference in our blog as well. Good luck to those who want to try!

## Monday, March 14, 2011

### Fox 333

Red Fox: Happy Pi+1 Day!
Himalayan Fox: What is that exactly?
Red: Well, March 14 was the Pi Day, you know, 3.14, and today is March 15th.
Himalayan: But Pi+1 would be 4.14 which is April 14th, isn't it?
Red: Oh, I didn't think that way. Whatever it is, happy March 15th to you brother.
Himalayan: OK, I'll mechanically say "to you too", but can we really happy while thousands of souls swept away with the water?
Red: Well, we can not die with the dead.
Himalayan: But we can help the living.
Red: I hear you brother.
Himalayan: Then we can celebrate 2Pi Day a few months later.

## Wednesday, March 9, 2011

### Balanced Mancala Problem

We rarely ask pure math problems but here is one with a "very little" touch of geometry. This does not need to be defined as a mancala game, but here it goes:

We have stones coming in batches. Each stone has a color and a weight. If the color of a stone is:

Yellow: It must be placed every other pit (every 2 pits) Batch size is always: 6 Total batches: y Total number = 6y

Red: Must be placed every 3 pits Batch size: 4 Total batches: r Total number = 4r

Green: Must be placed every 4 pits Batch size: 3 Total batches: g Total numbers = 3g

Blue: Must be placed every 6 pits Batch size: 2 Total batches: b Total number = 2b

Purple: Must be placed only once in 12 pits Batch size: 1 Total batches: p Total numbers = p

So there are N=6y+4r+3g+2b+p many stones. The stones in the same batch have the same weight. Different batches may have different weights. WLOG, assume that all weights are integers. We have a proof that ending up with the best well-balanced mancala is very difficult (NP-Hard). Here "well-balanced" means that the pit with the maximum weight is minimized when all stones are distributed. Let's call this maximum pit weight as W.

Consider the following heuristic process:

Step 1. Sort the batches with respect to their weights (batches with the high-weight stones go first) Step 2. Insert the first batch starting from pit number 1. Step 3. Insert the next batch in a way that the total maximum weight throughout 12 pits remains minimum. Step 4. Repeat Step 3 until all batches are placed in the mancala. Let H be the maximum weight throughout 12 pits.

A simple Example: Suppose we have only 4 batches: Yellow (6 stones, each 45 grams) Blue (2 stones, each 40 grams) Yellow (6 stones, each 30 grams) Green (3 stones, each 20 grams) First batch (Yellow) goes to pits: 1, 3, 5, 7, 9, and 11. H=45. Second batch (Blue) goes to pits: 2 and 8. H=45. Third batch (Yellow) goes to pits: 2, 4, 6, 8, 10, and 12. H=70. Fourth batch (Green) goes to pits: 3, 7, and 11. H=70.

In this exercise, heuristic actually finds the optimum, i.e., H=W=70 grams, observed in pits 2 and 8.

And the question: Prove that the worst-case of the heuristic solution, H, is always less than 2W. In other words, W ≤ H ≤ 2W always holds. If you disagree, then try to generate a counter-example.

Good luck!

## Monday, March 7, 2011

### Fox 332

Won't you come and see?
Some fly away,
some remain lonely.
-- Himalayan Fox

## Monday, February 28, 2011

### Fox 331

Here is a simple contruction problem.
http://www.8foxes.com/

## Sunday, February 20, 2011

### Fox 329

E[R] below can be computed depending on the random process... Try to guess what it is before computing. Randomness may be a fancy business...
See a similar concept: Bertrand Paradox

Dervish Fox: How can you cut a square randomly?

Red Fox:
Here is a way; first generate a random number to represent point A. Then another random number for point B along the three others sides. Drawn the line AB, and cut it out.

Dervish:
And how can you generate a random number?

Red:
Excel does it easily. Type "=RAND()", and boom, you get one right away.

Dervish:
But that depends on some seed numbers in the machine's memory, no?

Red:
Yes, that's why they call them pseudo-random numbers. Do you have a problem with that?

Dervish:
I have a bigger problem than that.

Red:
So... you believe that random numbers can't exist.

Dervish:
I believe that the word "random" should be removed from all world dictionaries.

Red:

http://www.8foxes.com/

## Thursday, February 17, 2011

### Fox 325, 326, 328 - Solutions

These simple ones are all related, so their solutions are posted here together.

Fox 325 by Bob Ryden:

Fox 325 by six:

Fox 326 by Bob Ryden:

Fox 326 by Binary:

Fox 328 by Bob Ryden:

http://www.8foxes.com/

## Thursday, February 10, 2011

### 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.

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

## Tuesday, February 1, 2011

### Fox 327

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.
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.
You're close enough.

--- Polar Fox

## Tuesday, January 25, 2011

### Fox 326

Hard talks much, yet simple tells more.
--- Himalayan Fox

## Sunday, January 23, 2011

### Fox 325

We've been carried away with very hard problems recently. Let's have a simple reality check with this classic. But, please try solving this without any words :)

## Sunday, January 16, 2011

### 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.

## Sunday, January 9, 2011

### 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.

## Friday, January 7, 2011

### Fox 319 - Solutions

Below are 2 distinct solutions for Fox 319...
A solution based on symmetry by Bleaug:

Trigonometry and calculus by Six:

x + y + z = 360 and sinx*siny*sinz

-> f(x,y) = sin(x)*sin(y)*sin(360-x-y)
-> f(x,y) = sin(x)*sin(y)*-sin(x+y)

Now to find the critical points of f(x,y), we just need to find the partial derivative with respect to x and y and solve for 0.

f_x(x,y) =

sin(x)sin(y)(-cos(x+y)-cos(x)sin(y)sin(x+y)

solve for 0.

sin(x)sin(y)(-cos(x+y)-cos(x)sin(y)sin(x+y)=0

-> tan(x) = -tan(x+y)

Since the function is symmetric, we should get the same partial derivative for y.

-> tan(y) = -tan(y+x)

-> tan(x)=tan(y)

-> x = y or they are opposites. However, if they are opposites, the original function just becomes 0. Thus, x = y.

Now substitute in x for y in the original equation and find its critical points.

Eventually, you will get sin(3x)=0

x = 120 degrees
y = 120 degrees
z = 120 degrees

http://www.8foxes.com/

## Monday, January 3, 2011

### Fox 324

Before starting to publish received solutions for earlier foxes, let's start with a new one. Here, "random selection" means that points P and Q are uniformly distributed along 2 parallel sides.
Ahh, yes, expected value of the "ratio of" the bigger area to the smaller one is asked - obviously. Let's give some leeway to minor mistakes in our life. "Perfect" may sometimes be boring!

## Saturday, January 1, 2011

### Happy new year everybody!

A new year,
a new beginning,
An unheard adventure - even it makes you late for dinner,
a new beginning.

Cold air that you breath in,
A warm shoulder that you can lean
A promise you've given to yourself - one more time,
a new beginning.

A morning sky painted in the East,
A colorful sunset as it fills the West,
Bread, cheese, water - and the rest,
a new beginning.