Math Problem
August 7th, 2019 at 3:49:35 AM
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how many triangles are there in this image?
from the article:
"You can count and fight over who’s a better counter. But the beauty of this problem is that it provides a nifty insight into combinatorics — another area of math that figures out the number of ways things can be shuffled. And once learned, that’s a mathematical tool that can be applied to other problems.
That’s the kind of problem that makes mathematicians smile.
A hint from Dr. Loh: “You can think about counting triangles in a completely different perspective, which is very cool. A triangle is made of three lines. So if I have six lines, then the whole question becomes how can I pick three lines out of those six lines?”"
https://www.nytimes.com/2019/08/06/science/math-equation-pemdas.html?action=click&module=Well&pgtype=Homepage§ion=Science
from the article:
"You can count and fight over who’s a better counter. But the beauty of this problem is that it provides a nifty insight into combinatorics — another area of math that figures out the number of ways things can be shuffled. And once learned, that’s a mathematical tool that can be applied to other problems.
That’s the kind of problem that makes mathematicians smile.
A hint from Dr. Loh: “You can think about counting triangles in a completely different perspective, which is very cool. A triangle is made of three lines. So if I have six lines, then the whole question becomes how can I pick three lines out of those six lines?”"
https://www.nytimes.com/2019/08/06/science/math-equation-pemdas.html?action=click&module=Well&pgtype=Homepage§ion=Science
Last edited by: lilredrooster on Aug 7, 2019
the foolish sayings of a rich man often pass for words of wisdom by the fools around him
August 7th, 2019 at 5:00:18 AM
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I keep getting 16. Too early to feel confident of that, though.
For each set of 2 adjacent lines, there are 4 triangles. But some of those will be duplications, so it's less than 48. And some triangles are made from all non-adjacent lines.
So I think you maybe have to label the intersections, then eliminate all 3-matching-letter sets, regardless of order.
Feels as far from a mathematical expression as ever, though. No light bulb.
For each set of 2 adjacent lines, there are 4 triangles. But some of those will be duplications, so it's less than 48. And some triangles are made from all non-adjacent lines.
So I think you maybe have to label the intersections, then eliminate all 3-matching-letter sets, regardless of order.
Feels as far from a mathematical expression as ever, though. No light bulb.
If the House lost every hand, they wouldn't deal the game.
August 7th, 2019 at 5:55:57 AM
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There are 6 lines. Each line crosses all 5 other lines.
A triangle is formed by the intersection of three lines. Therefore, there should be c(6,3) =20 triangles.
To count them: pick any 2 lines and see how they form 4 different triangles with the other four lines. Repeat.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
August 7th, 2019 at 8:34:18 AM
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I like this approach as that's what I started to do to understand the hint...Quote: gordonm888
There are 6 lines. Each line crosses all 5 other lines.
A triangle is formed by the intersection of three lines. Therefore, there should be c(6,3) =20 triangles.
To count them: pick any 2 lines and see how they form 4 different triangles with the other four lines. Repeat.
Starting with the two lines that intersect to the most left and top, there are four other lines that are left, these all cross the pair of lines to form a triangle.
You can pick any pair of two lines and repeat this. Two lines from 6 is 6x5/2 = 15.
For each traingle you count it three times (the pairs of lines covering (AB AC) (AB BC) (AC BC).
For each pair of line (15) you count (4) triangles but count each triangle (3) times, so total = 15*4/3 = 20.
You can pick any pair of two lines and repeat this. Two lines from 6 is 6x5/2 = 15.
For each traingle you count it three times (the pairs of lines covering (AB AC) (AB BC) (AC BC).
For each pair of line (15) you count (4) triangles but count each triangle (3) times, so total = 15*4/3 = 20.
August 7th, 2019 at 10:54:45 AM
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I didn't know what the C meant in Gordon888's spoiler so I looked it up - it means "combinations"
it made me realize that I really wasn't sure how to calculate the chance of winning a lottery such as picking 6 numbers out of 49
so, anyway, I got a very good answer from Wikipedia and now I know how to do it so I learned something
here's Wiki's answer on how to do it:
"In a typical 6/49 game, each player chooses six distinct numbers from a range of 1-49. If the six numbers on a ticket match the numbers drawn by the lottery, the ticket holder is a jackpot winner—regardless of the order of the numbers. The probability of this happening is 1 in 13,983,816.
The chance of winning can be demonstrated as follows: The first number drawn has a 1 in 49 chance of matching. When the draw comes to the second number, there are now only 48 balls left in the bag, because the balls are drawn without replacement. So there is now a 1 in 48 chance of predicting this number.
Thus for each of the 49 ways of choosing the first number there are 48 different ways of choosing the second. This means that the probability of correctly predicting 2 numbers drawn from 49 in the correct order is calculated as 1 in 49 × 48. On drawing the third number there are only 47 ways of choosing the number; but of course we could have arrived at this point in any of 49 × 48 ways, so the chances of correctly predicting 3 numbers drawn from 49, again in the correct order, is 1 in 49 × 48 × 47. This continues until the sixth number has been drawn, giving the final calculation, 49 × 48 × 47 × 46 × 45 × 44, This works out to 10,068,347,520, which is much bigger than the ~14 million stated above.
However; the order of the 6 numbers is not significant. That is, if a ticket has the numbers 1, 2, 3, 4, 5, and 6, it wins as long as all the numbers 1 through 6 are drawn, no matter what order they come out in. Accordingly, given any set of 6 numbers, there are 6 × 5 × 4 × 3 × 2 × 1 = 6! or 720 orders in which they could be drawn. Dividing 10,068,347,520 by 720 gives 13,983,816"
https://en.wikipedia.org/wiki/Lottery_mathematics
it made me realize that I really wasn't sure how to calculate the chance of winning a lottery such as picking 6 numbers out of 49
so, anyway, I got a very good answer from Wikipedia and now I know how to do it so I learned something
here's Wiki's answer on how to do it:
"In a typical 6/49 game, each player chooses six distinct numbers from a range of 1-49. If the six numbers on a ticket match the numbers drawn by the lottery, the ticket holder is a jackpot winner—regardless of the order of the numbers. The probability of this happening is 1 in 13,983,816.
The chance of winning can be demonstrated as follows: The first number drawn has a 1 in 49 chance of matching. When the draw comes to the second number, there are now only 48 balls left in the bag, because the balls are drawn without replacement. So there is now a 1 in 48 chance of predicting this number.
Thus for each of the 49 ways of choosing the first number there are 48 different ways of choosing the second. This means that the probability of correctly predicting 2 numbers drawn from 49 in the correct order is calculated as 1 in 49 × 48. On drawing the third number there are only 47 ways of choosing the number; but of course we could have arrived at this point in any of 49 × 48 ways, so the chances of correctly predicting 3 numbers drawn from 49, again in the correct order, is 1 in 49 × 48 × 47. This continues until the sixth number has been drawn, giving the final calculation, 49 × 48 × 47 × 46 × 45 × 44, This works out to 10,068,347,520, which is much bigger than the ~14 million stated above.
However; the order of the 6 numbers is not significant. That is, if a ticket has the numbers 1, 2, 3, 4, 5, and 6, it wins as long as all the numbers 1 through 6 are drawn, no matter what order they come out in. Accordingly, given any set of 6 numbers, there are 6 × 5 × 4 × 3 × 2 × 1 = 6! or 720 orders in which they could be drawn. Dividing 10,068,347,520 by 720 gives 13,983,816"
https://en.wikipedia.org/wiki/Lottery_mathematics
the foolish sayings of a rich man often pass for words of wisdom by the fools around him
