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        riddles >> hard >> Simultaneous Triangle Bisection
        
(Message started by: william wu on May 27^th, 2008, 3:12am)

Title: Simultaneous Triangle Bisection
Post by william wu on May 27^th, 2008, 3:12am

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

Title: Re: Simultaneous Triangle Bisection
Post by pex on May 27^th, 2008, 4:12am

It isn't all that hard.

[hideb]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.[/hideb]

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 (http://mathworld.wolfram.com/HamSandwichTheorem.html).

Title: Re: Simultaneous Triangle Bisection
Post by rmsgrey on May 27^th, 2008, 9:14am

I took a subtly different approach:

[hideb]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[/hideb]

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.

Title: Re: Simultaneous Triangle Bisection
Post by Eigenray on May 27^th, 2008, 9:40am

on 05/27/08 at 09:14:22, 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).

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.

Title: Re: Simultaneous Triangle Bisection
Post by towr on May 27^th, 2008, 9:48am

We had a thread (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1195111221;start=16#16)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.

Title: Re: Simultaneous Triangle Bisection
Post by rmsgrey on May 27^th, 2008, 9:49am

on 05/27/08 at 09:40:12, 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.

Thought it needed a better proof :)

Title: Re: Simultaneous Triangle Bisection
Post by william wu on May 27^th, 2008, 1:02pm

Thanks guys! I was pleased to learn about the "Ham Sandwich Theorem" :)