Well, you named this "paint color."
Paint is formed of pigment, where CMYK is relevant.
In the RGB space, I didn't find any of them remarkable.
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Posted by bonanova on 04 June 2014 - 04:01 PM
Well, you named this "paint color."
Paint is formed of pigment, where CMYK is relevant.
In the RGB space, I didn't find any of them remarkable.
All the others are multiples 0f 50.
Is that it?.
Posted by bonanova on 02 June 2014 - 04:31 AM
Posted by bonanova on 22 May 2014 - 12:22 PM
A ray of light encounters a pair of angled mirrors from the left, as shown, a distance d from one mirror and at an angle a from the other mirror. If angle a is exactly 22.5^{o}, how close will the ray get to their intersection at point O before it eventually exits again to the left? The drawing may not be to scale.
Posted by bonanova on 21 May 2014 - 06:59 AM
is meant "distance between the two closest points."
Create a half-matrix of n(n-1)/2 values of D = max (dx, dy) for each pair of points.
Find the pair for which D is minimal and equal to D_{min}.
An upper bound of the distance for that pair is D_{min }Sqrt(2) = 1.414 D_{min}.
Disregard all point pairs for which D > 1.414 D_{min}.
Now calculate using Pythagoras the actual distance d only for the remaining pairs.
Posted by bonanova on 02 May 2014 - 11:34 AM
Johnny was asked to write the sum of the two fractions p/n and q/n where p, q and n are single-digit numbers.
He remembered that the numerators were to be added, but he thought maybe the denominators should be added as well.
Also, he decided that putting digits next to each other could suggest addition.
So his first try was to say that
p/n + q/n = pq/nn. (where pq and nn are two-digit numbers.)
Then he remembered the result should not depend on the order of addition,
but he saw that p/n + q/n = qp/nn was just as bad. So he decided to use them both.
His final representation of the sum was
p/n + q/n = (pq + qp)/nn.
The puzzle asks: Let p, q and n be chosen at random from {1, 2, 3, 4, 5, 6, 7, 8, 9}.
What is the expected error in Johnny's formula?
Posted by bonanova on 28 April 2014 - 08:05 AM
Take any four points A, B, C, D in the plane, no three of which are collinear.
They describe a unique quadrilateral, if we take the points as being its vertices.
But they also describe a square, if we require only that the points lie on its sides.
Using a compass and straightedge, construct a square such that the four points lie, one each, on its sides
Edit: or the extensions of its sides.
Hint: The points are not special.
Draw four similar points and do the construction on a sheet of paper if that helps.
The answer would then be to describe the process.
Posted by bonanova on 27 April 2014 - 07:27 AM
A while back I started a series of puzzles that seem difficult but
have not-so-difficult answers. I can't find any of them, so I don't
know how many exist. There must have been at least two, so I'll
number this one 3. Have fun.
A monotonic increasing function f (x) is cut above and below by horizontal lines that intersect it at f (x_{1}) and f (x_{2}).
A vertical line is drawn through a point on the curve (black dot) between x_{1} and x_{2}.
The curve and the lines define the green and red areas shown on the figure.
You are to find the point on the curve that minimizes the sum of these areas.
Posted by bonanova on 04 April 2014 - 03:28 PM
Edit: It's actually more complicated than that hideous monstrosity... I left out that you would need to normalize the probabilities to sum to 1 instead of using raw squares of distances from the center.
You'd have the integrals you describe in the numerator of a fraction, divided by the same integrals without the HasOrigin part in the denominator.
I think that would do the normalization. Ugh, agreed.
Posted by bonanova on 16 March 2014 - 06:13 AM
While our collective brain trust ponders the nature of triangles defined
by three uniformly random points chosen inside a circle, specifically
the mode of their areas, we ask another question.
Recalling that the mean and median areal coverage of its circle's area are
about 7.4% and 5.4%, what is the probability that a random triangle covers
its circle's center?
Posted by bonanova on 16 March 2014 - 05:42 AM
Inspired by BMAD's puzzle, consider choosing four random numbers on [0,1].
If you sort them lowest to highest, what is the expected value of the third number?
Posted by bonanova on 08 March 2014 - 10:33 AM
At a family party, a grandfather, a grandmother, two fathers, two mothers, four children, three grandchildren, one brother, two sisters, two sons, two daughters, one father-in-law, one mother-in-law, and one daughter-in-law, sit at a table. What is the fewest number of chairs required to seat all of them? Each had his/her own chair.
Posted by bonanova on 08 March 2014 - 06:26 AM
Moe and Joe had some time on their hands, so they played a simple game called "Dueling Cards."
In this game, each player has a deck of cards, shuffled and without jokers.
A hand consists of a single card taken from the top of each player's deck.
Suits and face values are both ranked. No two cards in a deck have the same rank.
A player wins the hand by holding a card that outranks the card of his opponent.
The bank pays 1 chip to the winner of a hand.
It's a simple game. No poker hands. No betting. Just high card wins.
After 52 hands the decks are re-shuffled, and play continues as long as desired.
Moe and Joe played through their decks 10 times, then stopped to see who won.
As they counted their chips Moe (a statistician) said, I wonder what the most likely winning score is.
Joe (a simpler person) replied, I wonder what the most likely combined score is.
I'm sure Bushindo can answer Moe's question.
I can't. So instead this puzzle asks Joe's question.
Posted by bonanova on 07 March 2014 - 03:00 PM
Posted by bonanova on 28 February 2014 - 12:18 AM
A few puzzles posted in this forum have related to random triangles inside a circle.
By evaluating nasty integrals, or by my preferred method, simulation, it can be shown, perhaps surprisingly, that triangles constructed from sets of three uniformly chosen points within a circle cover only about 7.388% of the circle's area on average. After looking at Phil's recent problem on the subject, I simulated 1 million triangles to determine the median area. It turns out to be about 5.335% of the circle's area. Read: a random triangle has a 50% chance of being smaller.
If the distribution of random-triangle areas has a mean of about 7.4% and a median of about 5.3%, what value might you expect for the mode?
Posted by bonanova on 27 February 2014 - 08:48 AM
Consider the numbers from one to one million: 1, 2, 3, ..., 999998, 999999, 1000000.
What is remarkable about the numbers 40, 8, and 2202?
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