# Draw a triangle in the region bounded by a polynomial curve and a straight line. What is the supremum of the ratio of their areas?

Draw a triangle in the region bounded by a polynomial curve and a straight line. What is the supremum of the ratio of the triangle's area to the region's area?

Context

I was thinking about the quadrature of the parabola. If we draw a triangle in the region bounded by a parabola and straight line, the maximum ratio of the area of the triangle to the area of the region, is $$3/4$$. Then I wondered, if we can replace the parabola with any polynomial curve, what is the maximum or supremum of the ratio?

My attempt

I assume that one side of the triangle of largest area, is coincident with the straight line that bounds the region.

We can assume that the supremum of the ratio can be attained when the straight line is $$y=0$$, because if the supremum is attained with polynomial $$y=f(x)$$ and line $$y=mx+c$$, then we can draw a new polynomial $$y=f(x)-(mx+c)$$ and line $$y=0$$, and the ratio will be the same as before.

I conjecture that when the supremum is attained or approached, in the region of interest bounded by the curve and line, the leftmost (or rightmost) point is a stationary point of the curve. If this were not true, then it seems like the ratio of areas could be increased by vertically translating the horizontal line (such that the area of the region increases), which would be a contradiction.

I experimented with various polynomials, and so far, the maximum ratio that I've found is approximately $$0.8405$$, using the region bounded by $$y=-x^2\left(x+\frac23\right)^2(x-1)$$ and $$y=0$$ from $$x=0$$ to $$x=1$$.

Increasing the degree of the polynomial does not seem to allow for a larger ratio, but I could be wrong. (If the polynomial is a cubic, I believe the maximum ratio is exactly $$\frac{200}{243}\approx 0.8230$$.)

If there are no constraints on the polynomial's degree, the supremum is $$1$$. Take the region bounded by $$y=x^n$$ and $$y=x$$ (in case of an odd $$n$$, where there are two such bounded regions, take the one in $$x\geq 0$$), and the triangle with vertices $$(0,0)$$, $$(1,1)$$ and $$(0.5^{\frac{1}{\sqrt{n}}}, 0.5^{\sqrt{n}})$$. The triangle lies within the region, because the region is convex and the triangle's vertices lie on its boundary. As $$n\to\infty$$, $$(0.5^{\frac{1}{\sqrt{n}}}, 0.5^{\sqrt{n}})$$ gets arbitrarily close to $$(1, 0)$$, so the trianle's area goes to $$1/2$$. The region's area goes to $$1/2$$ as well, so the ratio goes to $$1$$.