One base of a [trapezoid](https://artofproblemsolving.com/wiki/index.php/Trapezoid) is $100$ units longer than the other base. The segment that joins the [midpoints](https://artofproblemsolving.com/wiki/index.php/Midpoint) of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3$. Let $x$ be the length of the segment joining the legs of the trapezoid that is [parallel](https://artofproblemsolving.com/wiki/index.php/Parallel) to the bases and that divides the trapezoid into two regions of equal area. Find the [greatest integer](https://artofproblemsolving.com/wiki/index.php/Ceiling_function) that does not exceed $x^2/100$.
Let the shorter base have length $b$ (so the longer has length $b+100$), and let the height be $h$. The length of the midline of the trapezoid is the average of its bases, which is $\frac{b+b+100}{2} = b+50$. The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height $h/2$. Then,


\[\frac{\frac 12 (h/2) (b + b+50)}{\frac 12 (h/2) (b + 50 + b + 100)} = \frac{2}{3} \Longrightarrow \frac{b + 75}{b + 25} = \frac 32 \Longrightarrow b = 75\]
Construct the perpendiculars from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle and two triangles; it also splits the desired line segment into three partitions with lengths $x_1, 75, x_2$. By similar triangles, we easily find that $\frac{x - 75}{100} = \frac{x_1+x_2}{100} = \frac{h_1}{h}$.


The area of the region including the shorter base must be half of the area of the entire trapezoid, so
\[2 \cdot \frac 12 h_1 (75 + x) = \frac 12 h (75 + 175) \Longrightarrow x = 125 \cdot \frac{h}{h_1} - 75\]
Substituting our expression for $\frac h{h_1}$ from above, we find that
\[x = \frac{12500}{x-75} - 75 \Longrightarrow x^2 - 75x = 5625 + 12500 - 75x \Longrightarrow x^2 = 18125\]
The answer is $\left\lfloor\frac{x^2}{100}\right\rfloor = \boxed{181}$.