Chaitanya's Random Pages

February 27, 2016

Cutting a triangle in half

Filed under: mathematics — ckrao @ 9:40 pm

Here is a cute triangle result that I’m surprised I had not known previously. If we are given a point on one of the sides of a triangle, how do we find a line through the triangle that cuts its area in half?

Clearly if that point is either a midpoint or one of the vertices, the answer is a median of the triangle. A median cuts a triangle in half since the two pieces have the same length side and equal height.

median

So what if the point is not a midpoint or a vertex? Referring to the diagram below, if P is our desired point closer to A than B, the end point Q of the area-bisecting segment would need to be on side BC so that area(BPQ) = area(ABC)/2.

areabisector

In other words, we require area(BPQ) = area(BDQ), or, subtracting the areas of triangle BDQ from both sides,

\displaystyle |DPQ| = |DCQ|.

Since these two triangles share the common base DQ, this tells us that we require them to have the same height. In other words, we require CP to be parallel to DQ. This tells us how to construct the point Q given P on AB:

  1. Construct the midpoint D of AB.
  2. Draw DQ parallel to AP.

See [1] for an animation of this construction.

In turns out that the set of all area-bisecting lines are tangent to three hyperbolas and enclose a deltoid of area (3/4)\ln(2) - 1/2 \approx 0.01986 times the original triangle. [2,3,4]

References

[1] Jaime Rangel-Mondragon, “Bisecting a Triangle” http://demonstrations.wolfram.com/BisectingATriangle/ from the Wolfram Demonstrations Project Published: July 10, 2013

[2] Ismail Hammoudeh, “Triangle Area Bisectors”  http://demonstrations.wolfram.com/TriangleAreaBisectors/ from the Wolfram Demonstrations Project

[4] Henry Bottomley, Area bisectors of a triangle, January 2002
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