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 Four Equal Triangles (Posted on 2013-06-16)

Here's a problem I found in the Problems section of a
math journal. The way I read it, it is not true.

Let ABCD be a convex quadrilateral. Prove that

there exists a point P inside ABCD such that
[PAB]=[PBC]=[PCD]=[PDA],

if and only if

the diagonals bisect each other.

Here [XYZ] denotes the area of triangle XYZ.

 See The Solution Submitted by Bractals No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 1 of 3
This definitely works one way:
If the diagonals bisect each other then the point P exists.  In fact it is very easy to show that P is the point of intersection found.

The problem with the statement is it isn't true the first way:
The point P can exist even if the diagonals don't bisect each other.

The simplest counterexample I can think of is a kite with AB=AD and BC=CD but AB≠BC.  The diagonals don't intersect but P is the midpoint of AC we get four equal area triangles.

In general Let B and D be any points on the opposite sides of and equidistant from the line containing AC.  P will be the midpoint of AC but need not be the midpoint of BD.

I was going to try to restate the problem but I can't fathom what the author of the problem was thinking.

 Posted by Jer on 2013-06-16 22:34:58

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