http://www.8foxes.com/Two flow together
one rises one falls
one after another
one shines one glooms
Do we need to know the object's mass? Probably not.
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!
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!
See a similar concept: Bertrand Paradox
Fox 325 by six:
Fox 326 by Bob Ryden:
Fox 326 by Binary:
Fox 328 by César Lozada:
Fox 328 by Bob Ryden:
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
This fox has been discussed extensively. See here...
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
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)
A new year,
