Outline Mathematics
Geometry
Triangle Areas in a Parallelogram II
This is a generalization of a one hundred years old problem.
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If from any point in a diagonal of a parallelogram lines are be drawn to the opposite angles the parallelogram will be divided into two pairs of equivalent triangles.
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The problem appeared in a periodical publication Mathematical Visitor and then included into a recent collection under the same name. It took the broadcast power of internet for the problem to reach overseas and a generalization emerge. Floor van Lamoen, The Netherlands, has noticed that point M does not have to lie on a diagonal of the parallelogram for there to be two pairs of triangles with equal sum of areas.
So let M be any point inside parallelogram ABCD. Triangles AMB and CMD have equal bases: AB =
. The sum of their altitudes ME + MF =
is fixed and does not depend on point M. We have
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| Area(AMB) + Area(CMD) | = AB·ME/2 + CD·MF/2 |
| | = AB·ME/2 + AB·MF/2 |
| | = AB·(ME + MF)/2 |
| | = AB·EF/2 |
| | =
. |
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It follows that the remaining area, Area(BMC) + Area(AMD) is also Area(ABCD)/2. So that the two are equal.
References
- S. Rabinowitz (ed), Problems and Solutions from the Mathematical Visitor (1877-1896), MathPro Press, 1996
Copyright © 1996-2009 Alexander Bogomolny
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