Puzzle time - making a triangle
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wrote on 30 Dec 2020, 23:37 last edited by Klaus
I think I found a nice way to illustrate the solution of the problem in one word that doesn't give away anything:
Butterfly.
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wrote on 30 Dec 2020, 23:53 last edited by
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This is my butterfly.
Obviously it occupies 25% of the square.
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wrote on 31 Dec 2020, 00:18 last edited by
Give your quick reasoning? Funny, I’ve thought of two other ways to solve it.
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wrote on 31 Dec 2020, 08:28 last edited by Klaus
Well, the x and y axis of my graph denote the randomly drawn numbers. The butterfly denotes the pairs of random numbers that result in pieces that satisfy the triangle inequality.
It’s the graph of the inequalities a+b >=c, a+c>= b, b+c >= a where a = min(x,y), b = max(x,y)-min(x,y) and c = 1-max(x,y). That is, a,b,c are the lengths of the pieces given cuts at x and y.
I guessed that this would be the obvious way to solve it hence my guess that you’d “get” the butterfly reference. Now I’m curious about your solution.
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wrote on 31 Dec 2020, 13:09 last edited by jon-nyc
I did get it, was just thinking about others that might see it.
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The official solution is like yours but in three dimensions. Consider x, y, and z the length of the three segments.You need to satisfy the equations:
x+y+z=1
x+y>z
x+z>y
y+z>x
(with x,y, and x only being between 0 and 1)Put them on a unit cube in 3d cartesian coordinates. You see all the triplets satisfying the first equation is a diagonal plane that slices off a corner of this cube, and the values between 0 and 1 lie in an equilateral triangle on this plane with vertices (0,0,1) (0,1,0) (1,0,0). (That's the universe of probabilities that is equivalent to the square from which your butterfly is cut).
To apply the inequalities, just see where the three planes (x+y-z=0), (x+z-y=0), (y+z-x=0) intersect with our plane. You find they go from midpoint to midpoint of its sides, cutting our equilateral triangle into four smaller identical ones. The region satisfying the inequalities is the innermost triangle, thus 1/4.
Now here's an easier method:
Picture the rod as a circle, with the first point being the two ends of the rod.
Now the exercise is choosing two points at random, and the test is that there can't be an arc of pi radians between any two points.
Anchor the first point at 0, and the second point at theta <=pi (WOLOG). Then the third point must be between pi and pi+theta.
So the probability is the expected vale of theta/2pi as theta ranges from 0 to pi: 1/4
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wrote on 31 Dec 2020, 14:02 last edited by
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I did get it, was just thinking about others that might see it.
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The official solution is like yours but in three dimensions. Consider x, y, and z the length of the three segments.You need to satisfy the equations:
x+y+z=1
x+y>z
x+z>y
y+z>x
(with x,y, and x only being between 0 and 1)Put them on a unit cube in 3d cartesian coordinates. You see all the triplets satisfying the first equation is a diagonal plane that slices off a corner of this cube, and the values between 0 and 1 lie in an equilateral triangle on this plane with vertices (0,0,1) (0,1,0) (1,0,0). (That's the universe of probabilities that is equivalent to the square from which your butterfly is cut).
To apply the inequalities, just see where the three planes (x+y-z=0), (x+z-y=0), (y+z-x=0) intersect with our plane. You find they go from midpoint to midpoint of its sides, cutting our equilateral triangle into four smaller identical ones. The region satisfying the inequalities is the innermost triangle, thus 1/4.
Now here's an easier method:
Picture the rod as a circle, with the first point being the two ends of the rod.
Now the exercise is choosing two points at random, and the test is that there can't be an arc of pi radians between any two points.
Anchor the first point at 0, and the second point at theta <=pi (WOLOG). Then the third point must be between pi and pi+theta.
So the probability is the expected vale of theta/2pi as theta ranges from 0 to pi: 1/4
:::wrote on 31 Dec 2020, 15:46 last edited by Klaus@jon-nyc said in Puzzle time - making a triangle:
Picture the rod as a circle
Looks like a conceptual move, not an algebraic one.
Link to videoEh, time stamps don't seem to work in embedded videos. Forward to 3:25.
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wrote on 31 Dec 2020, 17:04 last edited by
I thought this might be a good excuse to toy around with Maple a little, which I have wanted to do for a while.
Works nicely.
