If a line is cut in two independent random places, what are the odds that the pieces can form a triangle?
probability/two.cuts.yield.triangle Added material
Problem
If a line is cut in two independent random places, what are the odds that the pieces can form a triangle?
Show solution
1/4. Two proofs, one algebraic, one geometric: Algebraic proof: Let the length be 1 unit. To form a triangle, the longest piece must be less than .5, meaning that the two cuts must be on opposite sides of the .5 mark. The probability of this is 1/2. Now let the cut that is below the .5 mark = x in length. The other cut must be between .5 and .5 + x. For all x between 0 and .5, this probability averages 1/2 (assuming they are on opposite sides of the .5 mark). Multiplying the two probabilities gives 1/4. Geometric proof: If you pick a random point (x,y) in a square and break the stick at the points of both coordinates, this is equivalent to breaking a stick in two random places. When can you get a triangle from the remaining pieces? If x and y are both less than 1/2, then the third piece will be longer than 1/2; this has probability 1/4. If x and y are both more than 1/2, then the first piece will be longer than 1/2; this has probability 1/4. If 0<x<y-1/2 or 0<y<x-1/2, then the middle piece will be longer than 1/2; this has probability 1/4. (See the figure for the corresponding regions of the square, which are all disjoint.) The remaining probability is 1/4. +---+---+ |22/|111| |2/.|111| |/..|111| +---+---+ |333|../| |333|./2| |333|/22| +---+---+
Additional material
Another geometric solution. (Use a non-proportional font.)
/\ /\
/ \ / \
/ \ / \
/ _\ /______\
/-_ _- \ /\ /\
/ -X \ / \ / \
/ | \ / \ / \
/______|_______\ /______\/______\
The left hand illustration shows the 3 pieces placed inside an
equilateral triangle that has perpendicular height the same as the
length of the original piece of wood. Each piece touches one of the
sides, and is perpendicular to that side. The pieces will always meet
at a single point - marked X above.
The right hand illustration shows the same triangle divided into 4
equilateral triangles. Imagine placing the pieces in this triangle. If
the point X is within the central small equilateral triangle then the
pieces can be put together to form a triangle, otherwise they can't.
The small triangles have 1/4 the area of the large triangle, so the
required probability is 1/4.
-- NickGard <nickgard@aol.com>
[ Editor: the probability that the triangle is acute, from a June 1994
rec.puzzles thread "enhanced broken stick". ]
In article <199406292214.RAA22771@news.cs.utexas.edu>,
ajit pendse <CMPEND%LSUVM.BITNET@uga.cc.uga.edu> wrote:
>restating the problem: find the probability of an acute
>triangle (all angles < 90) forming from the three pieces of a stick.
>from the previous discussion the probability that a triangle
>is possible is 1/4. without a proof, i assume that
>the set of all triagles is covered by the three pieces.
Yes, any triangle with a perimeter the length of the stick is
possible...
>the problem is therefore to find the probability of an acute triangle.
>if we see the 'phase diagram' of the angles A and B,
Wrong. You are assuming an extreme amount about the probability
distribution as soon as you draw any sort of "phase diagram".
[phase diagram based on two angles A and B and the incorrect
conclusion it leads to deleted]
Using the original probability distribution phase diagram I proposed,
which was based on the assumption that the two points are chosen
independently and using the normal (Lebesgue [Editor: "Lesbegue" in the original]) norm, here is the set of
all states for which a triangle can be formed:
_______________
1m| /| |
| /::| |
| /::::| |
|/::::::| |
y |-------+-------| The area of this set is 1/4 the total area.
(m)| |::::::/|
| |::::/ |
| |::/ |
0m|_______|/______|
0m x(m) 1m
Call this set S (the set of states for which a triangle may be formed).
Call the set of states for which an acute triangle will be formed T.
A triangle is acute iff the sum of the squares of any two sides is more
than the square of the third.
Due to the symmetry of the problem, assume that x < 1/2.
Assuming (x,y) is in S, and x < 1/2, when will (x,y) be in T? The
sides have length x, y-x, and 1-y. The squares of these are xx,
xx-2xy+yy, and yy-2y+1 (forgive the unusual notation).
The set T intersected with {(x,y) | x<1/2} is thus defined by:
(a) xx < xx-2xy+yy+yy-2y+1
(b) xx-2xy+yy < xx+yy-2y+1
(c) yy-2y+1 < xx+xx-2xy+yy
(a) reduces to 2xy < 2yy-2y+1, which reduces to x < y-1+1/(2y)
(b) reduces to -2xy < -2y+1, which reduces to x > 1-1/(2y)
(c) reduces to -2y+1 < 2xx-2xy, which reduces to
-2(1-x)y < 2xx-1, which reduces to y > (.5-xx)/(1-x)
The area described leads to a set T which looks like:
_______________
1m| // |
| /| |
| /::| |
|_-::::\ |
y |--"""--+--___--| Note that the "pinching" going on here
(m)| \::::- | has absolutely no correlation to a
| |::/ | phase diagram based on two of the
| |/ | triangle's angles.
0m|_______//______|
0m x(m) 1m
The integral of 1/(2y) from .5 to 1 is .5ln2. The area of the slice
taken out of S by (a) is .375-.5ln2, around .0284.
It's not hard to see that the areas of all six slices taken out have
the same area, so the total area sliced away from S is
6*(.375-.5ln2), or around .1706.
The area of T is 1/4-6*(.375-.5ln2), which simplifies to -2+3ln2,
around .0794. This is significantly more than 1/16, which is .0625.
-- Isaac Kuo <isaackuo@math.berkeley.edu>
[ Editor: see also probability/acute.triangle. ]