I have an Isosceles triangle with slope $k$ (like the one below), which length I am cuting into $n$ equidistant parts, creating $n$ polygons (for simplicity let $n$ be odd). The first polygon, a triangle, has an area of x. The middle $(n+1)/2$ polygon has area Y. What is the maximum ratio Y/X? Is this unbounded if the slope $k$ is very large?
$X$ and $Y/2$ have base $a/n$ and $a/(2n)$ and heights proportional to $k$ (through $a$). So the ratio of their areas remains constant wrt $k$.
In fact
$X=1/2 \cdot a/n \cdot k \cdot a/n$ and $Y/2=a/(2n) \cdot 1/2 \cdot (k a/2+k(a/2-a/2n))$
(can you continue and simplify the above ?)