# 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.

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.

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

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