3. Consider the L-shaped tromino below with 3 attached unit squares. It is cut into exactly two pieces of equal area by a line segment whose endpoints lie on the perimeter of the tromino. What is the longest possible length of the line segment?
Answer: $\square$
Solution: Let the line segment have endpoints $A$ and $B$. Without loss of generality, let $A$ lie below the lines $x+y=\sqrt{3}$ (as this will cause $B$ to be above the line $x+y=\sqrt{3}$ ) and $y=x$ (we can reflect about $y=x$ to get the rest of the cases):

Now, note that as $A$ ranges from $(0,0)$ to $(1.5,0), B$ will range from $(1,1)$ to $(1,2)$ to $(0,2)$, as indicated by the red line segments. Note that these line segments are contained in a rectangle bounded by $x=0$, $y=0, x=1.5$, and $y=2$, and so the longest line segment in this case has length $\sqrt{2^{2}+1.5^{2}}=\frac{5}{2}$.
As for the rest of the cases, as $A=(x, 0)$ ranges from $(1.5,0)$ to $(\sqrt{3}, 0), B$ will be the point $\left(0, \frac{3}{x}\right)$, so it suffices to maximize $\sqrt{x^{2}+\frac{9}{x^{2}}}$ given $1.5 \leq x \leq \sqrt{3}$. Note that the further away $x^{2}$ is from 3 , the larger $x^{2}+\frac{9}{x^{2}}$ gets, and so the maximum is achieved when $x=1.5$, which gives us the same length as before.
Thus, the maximum length is $\frac{5}{2}$.