Problem (Geometry):
Let $a, b, m, n$ be positive integers with $a m=b n=120$ and $a \neq b$. In the coordinate plane, let $A=(a, m), B=(b, n)$, and $O=(0,0)$. If $X$ is a point in the plane such that $A O B X$ is a parallelogram, compute the minimum area of $A O B X$.

Solution:
The area of parallelogram $A O B X$ is given by the absolute value of the cross product $|\langle a, m\rangle \times\langle b, n\rangle|=|a n-m b|$. Because $m=\frac{120}{a}$ and $n=\frac{120}{b}$, the desired area of $A O B X$ equals $120\left|\frac{a}{b}-\frac{b}{a}\right|$. Note that the function $f(x)=x-\frac{1}{x}$ is monotone increasing for $x>1$. (Proof: if $x_{1}>x_{2}>0$, then $f\left(x_{1}\right)-f\left(x_{2}\right)=\left(x_{1}-x_{2}\right)+\frac{x_{1}-x_{2}}{x_{1} x_{2}}$, where both terms are positive because $x_{1} x_{2}>0$.) So the minimum value of $[A O B X]$ is attained when $\frac{a}{b}$ is as close as possible to 1 , that is, when $a$ and $b$ are consecutive divisors of 120. By symmetry, consider only $a<b$; notice too that because $\frac{120 / a}{120 / b}=\frac{b}{a}$, only values with $b \leq \sqrt{120}$ need be considered. These observations can be used to generate the table below:

| $a, m$ | 1,120 | 2,60 | 3,40 | 4,30 | 5,24 | 6,20 | 8,15 | 10,12 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $b, n$ | 2,60 | 3,40 | 4,30 | 5,24 | 6,20 | 8,15 | 10,12 | 12,10 |
| $[A O B X]$ | 180 | 100 | 70 | 54 | 44 | 70 | 54 | 44 |

The smallest value is $\mathbf{4 4}$, achieved using $(5,24)$ and $(6,20)$, or using $(10,12)$ and $(12,10)$.

Note: The fact that $a$ and $b$ must be consecutive divisors of 120 can also be established by the following geometric argument. Notice that $[A O B X]=2[A O B]$. Suppose $C$ is a point on the hyperbola $y=120 / x$ between $A$ and $B$, as shown in the diagram below.





Because the hyperbola is concave up, $[O A C]+[O C B]<[O A B]$, so in particular, $[O A C]<$ $[O A B]$. Thus, if $[O A B]$ is minimal, there can be no point $C$ with integer coordinates between $A$ and $B$ on the hyperbola.

Final answer: 44