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 $\frac{1}{4}.$

We'll prove that it's a maximal area.

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!

  • $\begingroup$ Can you please provide an explanation for the second step. $\endgroup$ – dragonking Mar 21 at 1:37
  • $\begingroup$ @dragonking Do you mean why $\frac{S_{\Delta MUN}}{a}=\frac{MU}{UC}=\frac{y}{x}$ is true? $\endgroup$ – Michael Rozenberg Mar 21 at 5:22

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