# Areal Co-ordinate Geometry Question

Let $P$ be an internal point of triangle $ABC$. The line through $P$ parallel to $AB$ meets $BC$ at $L$, the line through $P$ parallel to $BC$ meets $CA$ at $M$, and the line through $P$ parallel to $CA$ meets $AB$ at $N$. Prove that $\frac{BL}{LC}×\frac{CM}{MA}×\frac{AN}{NB}≤\frac{1}{8}$ and locate the position of $P$ in triangle $ABC$ when equality holds.

Can anyone solve this problem using areal co-ordinates.

• "Areal coordinates" are best known now as "barycentric coordinates" Commented May 13 at 9:41

Sure, it's possible with areal coordinates. Here's an outline:

Let $P$ have normalised areal coordinates $(p, q, r)$. We'll calculate the coordinates of $L$.

The line through $P$ parallel to $AB$ has areal equation $z = r$, and the side $BC$ has equation $x = 0$. The intersection of these two lines is at the point with areal coordinates $(0, 1 - r, r)$, so these are the coordinates of $L$.

It follows that the ratio $\displaystyle\frac{BL}{LC} = \frac{r}{1 - r}$, and similarly for the other two.

So you are asked to prove that, given positive $p, q, r$ with $p + q + r = 1$, we have:

$8pqr \leq (1-p)(1-q)(1-r)$

This is equivalent to $8pqr \leq (p+q)(q+r)(q+p)$, which is true, since by AM-GM, we have:

$2\sqrt{pq} \leq p + q$

and two similar inequalities. Multiplying them all together gives the result.

Equality holds when $p = q = r$, which is the centroid of the triangle.