Problem (Geometry):
Diagonal $W Y$ of square $W X Y Z$ has slope 2. Determine the sum of the slopes of $W X$ and $X Y$.

Solution:
Suppose that $W Y$ makes an angle of $\theta$ with the horizontal.



Since the slope of $W Y$ is 2 , then $\tan \theta=2$, since the tangent of an angle equals the slope of a line that makes this angle with the horizontal.

Since $\tan \theta=2>1=\tan 45^{\circ}$, then $\theta>45^{\circ}$.

Now $W Y$ bisects $\angle Z W X$, which is a right-angle.

Therefore, $\angle Z W Y=\angle Y W X=45^{\circ}$.

Therefore, $W X$ makes an angle of $\theta+45^{\circ}$ with the horizontal and $W Z$ makes an angle of $\theta-45^{\circ}$ with the horizontal. Since $\theta>45^{\circ}$, then $\theta-45^{\circ}>0$ and $\theta+45^{\circ}>90^{\circ}$.

We note that since $W Z$ and $X Y$ are parallel, then the slope of $X Y$ equals the slope of $W Z$.

To calculate the slopes of $W X$ and $W Z$, we can calculate $\tan \left(\theta+45^{\circ}\right)$ and $\tan \left(\theta-45^{\circ}\right)$.

Using the facts that $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$ and $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$, we obtain:

$$
\begin{aligned}
& \tan \left(\theta+45^{\circ}\right)=\frac{\tan \theta+\tan 45^{\circ}}{1-\tan \theta \tan 45^{\circ}}=\frac{2+1}{1-(2)(1)}=-3 \\
& \tan \left(\theta-45^{\circ}\right)=\frac{\tan \theta-\tan 45^{\circ}}{1-\tan \theta \tan 45^{\circ}}=\frac{2-1}{1+(2)(1)}=\frac{1}{3}
\end{aligned}
$$

Therefore, the sum of the slopes of $W X$ and $X Y$ is $-3+\frac{1}{3}=-\frac{8}{3}$.

Consider a square $W X Y Z$ whose diagonal $W Y$ has slope 2 .

Translate this square so that $W$ is at the origin $(0,0)$. Translating a shape in the plane does not affect the slopes of any line segments.

Let the coordinates of $Y$ be $(2 a, 2 b)$ for some non-zero numbers $a$ and $b$.

Since the slope of $W Y$ is 2 , then $\frac{2 b-0}{2 a-0}=2$ and so $2 b=4 a$ or $b=2 a$.

Thus, the coordinates of $Y$ can be written as $(2 a, 4 a)$.

Let $C$ be the centre of square $W X Y Z$.

Then $C$ is the midpoint of $W Y$, so $C$ has coordinates $(a, 2 a)$.

We find the slopes of $W X$ and $X Y$ by finding the coordinates of $X$.

Consider the segment $X C$.

Since the diagonals of a square are perpendicular, then $X C$ is perpendicular to $W C$.

Since the slope of $W C$ is 2 , then the slopes of $X C$ and $Z C$ are $-\frac{1}{2}$.

Since the diagonals of a square are equal in length and $C$ is the midpoint of both diagonals, then $X C=W C$.

Since $W C$ and $X C$ are perpendicular and equal in length, then the "rise/run triangle" above $X C$ will be a $90^{\circ}$ rotation of the "rise/run triangle" below $W C$.



This is because these triangles are congruent (each is right-angled, their hypotenuses are of equal length, and their remaining angles are equal) and their hypotenuses are perpendicular.

In this diagram, we have assumed that $X$ is to the left of $W$ and $Z$ is to the right of $W$. Since the slopes of parallel sides are equal, it does not matter which vertex is labelled $X$ and which is labelled $Z$. We would obtain the same two slopes, but in a different order. To get from $W(0,0)$ to $C(a, 2 a)$, we go up $2 a$ and right $a$.

Thus, to get from $C(a, 2 a)$ to $X$, we go left $2 a$ and up $a$.

Therefore, the coordinates of $X$ are $(a-2 a, 2 a+a)$ or $(-a, 3 a)$.

Thus, the slope of $W X$ is $\frac{3 a-0}{-a-0}=-3$.

Since $X Y$ is perpendicular to $W X$, then its slope is the negative reciprocal of -3 , which is $\frac{1}{3}$.

The sum of the slopes of $W X$ and $X Y$ is $-3+\frac{1}{3}=-\frac{8}{3}$.

Final answer: $-\frac{8}{3}$