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Topic: Simultaneous Triangle Bisection (Read 3238 times) |
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william wu
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Simultaneous Triangle Bisection
« on: May 27th, 2008, 3:12am » |
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For any pair of triangles, prove that there exists a line which bisects them simultaneously.
Notes: I don't know how to argue it. I'd also like to know whether its relevant that the polygons mentioned in the question are *triangles*.
Source: Problem-Solving Through Problems
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| « Last Edit: May 27th, 2008, 3:14am by william wu » |
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pex
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Re: Simultaneous Triangle Bisection
« Reply #1 on: May 27th, 2008, 4:12am » |
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It isn't all that hard.
| hidden: | Name the triangles A and B. For any given direction t (measured counterclockwise from the positive x axis), we can find a line with that direction that bisects triangle A. (Consider parallel lines with direction t, one on one side of A and one on the other side; somewhere in between, the line we need exists, by continuity.)
For any direction t with 0 < t < pi, let l(t) be the line defined above, and let f(t) be the fraction of the surface of triangle B to the west of l(t). (For the sake of continuity, f(0) is the fraction of B to the north of l(0), and f(pi) is the fraction of B to the south of l(pi).)
Clearly l(0) and l(pi) coincide, so f(0) + f(pi) = 1. If both are equal to 1/2, we are done; otherwise, one of f(0) and f(pi) is below 1/2 and the other over 1/2; again by continuity, there exists t* such that f(t*) = 1/2.
Line l(t*) bisects both triangles. |
Regarding your note, no, that's irrelevant. Any two figures with finite perimeters would do; and more generally, in n dimensions, any n objects with finite outer measure can be simultaneously bisected by an (n-1) dimensional hyperplane. Background: here.
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rmsgrey
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Re: Simultaneous Triangle Bisection
« Reply #2 on: May 27th, 2008, 9:14am » |
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I took a subtly different approach:
| hidden: | Any given (bounded) shape has a center of gravity - a point through which all bisectors pass (if made of a material of uniform density, and suspended in a uniform gravitational field, there will be an equal mass on either side of that center for any orientation of cutting hyperplane).
The two centers of two 2D shapes define a common bisector, similarly for the n centers of n nD shapes defining an (n-1)hyperplane |
Obviously, the key point could stand to be proved a bit more (or at all) rigorously - particularly since the rest of the proof is entirely trivial.
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Eigenray
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Re: Simultaneous Triangle Bisection
« Reply #3 on: May 27th, 2008, 9:40am » |
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on May 27th, 2008, 9:14am, rmsgrey wrote:| Any given (bounded) shape has a center of gravity - a point through which all bisectors pass (if made of a material of uniform density, and suspended in a uniform gravitational field, there will be an equal mass on either side of that center for any orientation of cutting hyperplane). |
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But this is not true, even for an equilateral triangle. A line through the centroid parallel to a side cuts it in a 4:5 ratio.
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| « Last Edit: May 27th, 2008, 9:44am by Eigenray » |
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towr
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Re: Simultaneous Triangle Bisection
« Reply #4 on: May 27th, 2008, 9:48am » |
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We had a thread a while back in which Grimbal drew a nice picture of the closed curve inside a triangle which all the bisectors are tangent to (also to show they don't all go through the center of gravity).
Given this, it's trivial to show that for most pairs of triangles we can find two bisectors that bisect both triangles.
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| « Last Edit: May 27th, 2008, 10:00am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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rmsgrey
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Re: Simultaneous Triangle Bisection
« Reply #5 on: May 27th, 2008, 9:49am » |
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on May 27th, 2008, 9:40am, Eigenray wrote:
But this is not true, even for an equilateral triangle. A line through the centroid parallel to a side cuts it in a 4:5 ratio. |
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Thought it needed a better proof 
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