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bonanova

Member Since --
Offline Last Active Yesterday, 09:12 PM
*****

#338563 Paint Color Picker

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.

 

Spoiler for Except that


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#338527 DENSA entrance exam question

Posted by bonanova on 02 June 2014 - 04:31 AM

Until this posting, it was a fairly well-kept secret that I am a card-carrying member of the DENSA society. That is not a typo. DENSA is an acronym for Diversely Educated, Not Seriously Affected.
 
DENSA membership requires supplying the correct answer to the following multiple-choice question.
 
1. How many stars are there?
 
      a. All of the above.
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#338421 The "aha!" problems 8: Reflect on this

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 22.5o, 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.

mirrors.jpg


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#338401 shortest distance

Posted by bonanova on 21 May 2014 - 06:59 AM

Spoiler for For starters


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#338228 Adding really simple fractions

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?


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#338150 The "aha!" problems 5. Four-point square

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.

. . . . . . . . . . . . . . . 4point_square_a.jpg

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.


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#338131 The "aha!" problems - 3 minimum areas

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 (x1) and f (x2).

A vertical line is drawn through a point on the curve (black dot) between x1 and x2.

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.

jp21.gif


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#337991 More triangles in circles

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.


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#337815 More triangles in circles

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?


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#337814 More random numbers

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?


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#337738 How big was the dinner table?

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.


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#337737 Dueling cards

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.


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#337727 Orderly numbers

Posted by bonanova on 07 March 2014 - 03:00 PM

Spoiler for Hint


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#337617 Triangles inside circles

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?


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#337573 Orderly numbers

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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