## Monday Math 152

Find the point in the interior of a triangle for which the product of the distances from that point to the sides of the triangle is maximized.

Let us call the lengths of the sides a, b and c, and let the distances from our interior to these sides be x, y and z, respectively. Lastly, let A be the area of the triangle.

Now, we note that the segments connecting our point to the vertices of the triangle divide it into three smaller triangles.

We see that these triangles have bases a, b, c and heights x, y, z, respectively, and thus areas of $\frac{1}{2}ax$, $\frac{1}{2}by$ and $\frac{1}{2}cz$. But these have to add up to the area A of our (original) triangle:
$\frac{1}{2}ax+\frac{1}{2}by+\frac{1}{2}cz=A$
or, multiplying by 2:
$ax+by+cz=2A$
for any point in the interior of the triangle.

Thus, we want to find the point which maximizes the product xyz subject to the contstraint $ax+by+cz=2A$. But this is just last week’s Lagrange multiplier problem with d=2A. And thus, we see that the maximum is obtained when $ax=by=cz=\frac{2A}{3}$. However, this means that the areas of the subtriangles are equal:
$\frac{1}{2}ax=\frac{1}{2}by=\frac{1}{2}cz=\frac{A}{3}$
But we saw two weeks ago that the point for which the triangle is divided into equal area subtriangles in this fashion is the centroid of the triangle.

Thus, our answer is that the point in the interior of a triangle for which the product of the distances from that point to the sides is maximized is the centroid of the triangle.