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Fox 324 - 345

Here is an old one for re-consideration:

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

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




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

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

but pure,
simple,
childish forgiveness.

-- Dervish Fox




Ahmet Arduç submitted this one. Geometric solutions will be appreciated.
 
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See three solutions below...





By Six:




By Bleaug:



By Henk Reuling:

A general case of an old one...

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This is a little related to the recent ones, but much simpler.
Here we'd appreciate a pure geometric solution if one found.
(ADDITIONAL INFORMATION IS ADDED BELOW!)

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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!

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




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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.





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.

DISCUSS HERE...

Good luck!

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



This has been extensively discussed in the blog.  See here...

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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: Radical.



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There is a nice animation in Joe Wilson's page.

 

 


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This ancient fox was the general case of Fox 187, which was successfully-solved.
And now, Bleaug claims that this has been solved, too.
Click here for his solution.

http://bleaug.free.fr/8foxes/8foxes187.svg
To see the animation, you will need any of the following browsers:
Chrome, Firefox, Opera, Safari, or IE 9.0.

 





CLOSE ENOUGH
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

See the checkerboard solution below.

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)
  3. Draw horizontal and vertical lines separated by 1/2=0.5 units starting from the origin.
  4. This will create 0.5 x 0.5 squares (and possibly rectangles around 2 edges -top and RHS-)
  5. 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.
  6. Each TILE will have equal areas of black and white (Why?)
  7. Therefore the overall rectangle will have equal areas of black and white.
  8. So overall rectangle has at least one integer side. (Why not? :)

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

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.



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 tough)
  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.

   

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

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


Hard-core 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

Answer (D)

 
 

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.

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!

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