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, 2020

Geometry Difficulty 5.4 AIME, harder Prove it United States

Problem:

Points $G$ and $N$ are chosen on the interiors of sides $ED$ and $DO$ of unit square $DOME$ , so that pentagon $GNOME$ has only two distinct side lengths. The sum of all possible areas of quadrilateral $NOME$ can be expressed as $\frac{a-b \sqrt{c}}{d}$ , where $a, b, c, d$ are positive integers such that $\operatorname{gcd}(a, b, d)=1$ and $c$ is square-free (i.e. no perfect square greater than 1 divides $c$ ). Compute $1000a+100b+10c+d$ .

Solution

Solution:

Since $MO=ME=1$ , but $ON$ and $GE$ are both less than $1$ , we must have either $ON=NG=GE=x$ (call this case 1) or $ON=GE=x$ , $NG=1$ (call this case 2).

Either way, the area of $NOME$ (a trapezoid) is $\frac{1+x}{2}$ , and triangle $NGT$ is a $45$ - $45$ - $90$ triangle. In case 1, we have $1=ON+NT=x\left(1+\frac{\sqrt{2}}{2}\right)$ , so $x=2-\sqrt{2}$ and the area of the trapezoid is $\frac{3-\sqrt{2}}{2}$ . In case 2, we have $1=ON+NT=x+\frac{\sqrt{2}}{2}$ , which yields an area of $\frac{4-\sqrt{2}}{4}$ as $x=\frac{2-\sqrt{2}}{2}$ . The sum of these two answers is $\frac{10-3\sqrt{2}}{4}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.