# Maximizing Area of a quadrilateral inside of a square

The square ABCD has point M located on side AB and point N on side CD. Lines CM and BN intersect at point U. Lines DM and AN intersect at point V. Determine where points M and N should be placed to maximize the area of MUNV.

I attempted to solve to problem by placing the square at (0,0) and making the square 1x1. I then tried approaching it as a polygon of constraints and found the rules of the lines but this gave me too many variables. Should I use a trig approach instead? I do not see where to begin in this case.

Let $$AB=1$$, $$CN=x$$, $$BM=y$$ and $$S_{\Delta CNU}=a$$.
Thus, $$S_{\Delta MCN}=\frac{x\cdot1}{2}=\frac{x}{2}=a+S_{\Delta MUN}.$$ But $$\frac{S_{\Delta MUN}}{a}=\frac{MU}{UC}=\frac{y}{x},$$ which says $$S_{\Delta MUN}=\frac{ay}{x}.$$ Thus, $$\frac{x}{2}=a+\frac{ay}{x},$$ which gives $$a=\frac{x^2}{2(x+y)}$$ and $$S_{\Delta MUN}=\frac{xy}{2(x+y)}.$$ By the same way we obtain: $$S_{\Delta MNV}=\frac{(1-x)(1-y)}{2(2-x-y)}$$ and $$S_{MUNV}=\frac{xy}{2(x+y)}+\frac{(1-x)(1-y)}{2(2-x-y)},$$ which for $$x=y=\frac{1}{2}$$ gives a value $$\frac{1}{4}.$$
Indeed, we need to prove that $$\frac{xy}{2(x+y)}+\frac{(1-x)(1-y)}{2(2-x-y)}\leq\frac{1}{4},$$ which is just $$(x-y)^2\geq0.$$ Done!
• @dragonking Do you mean why $\frac{S_{\Delta MUN}}{a}=\frac{MU}{UC}=\frac{y}{x}$ is true? – Michael Rozenberg Mar 21 at 5:22