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| Four Equal Triangles (Posted on 2013-06-16) |
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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.
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Submitted by Bractals
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Rating: 5.0000 (1 votes)
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Solution:
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(Hide)
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Since ABCD is convex, the diagonals AC and BD will
intersect. Let I be this intersection point.
"The diagonals bisect each other" is equivalent to
"I is the midpoint of AC and BD".
If this is true,
then ABCD is a parallelogram. Let
P = I and then the length of the bases of each of the
four triangles is ½|AC|, the length of the altitudes is
½|BD|sin(∠AIB), and the areas are
(1/8)|AC||BD|sin(∠AIB).
Therefore, there exists a point P inside ABCD
such that [PAB]=[PBC]=[PCD]=[PDA],
if
the diagonals bisect each other.
Is true.
Staying with this configuration with points
points A, B, and C fixed and P the midpoint
of AC, move point D (see note below) and the
areas of the four triangles will stay the
same, but I will only be the midpoint of BD
not AC.
Therefore, there exists a point P inside ABCD
such that [PAB]=[PBC]=[PCD]=[PDA],
only if
the diagonals bisect each other.
Is false.
QED
Note:
Let m be the line through D parallel to AC.
Let DC and DA be the points of intersection of line m with
lines BC and BA respectively. Move point D on line m
between points DC and DA.
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