Consider a convex quadrilateral with vertices at a,b,c,and d and on each side draw a square lying outside the given quadrilateral, as in the picture below. Let p,q,r,s be the centers of those squares:

what i did so far:

We want to find $p$ in terms of $a$ and $b$. We can translate point $a$ to the origin. We can notice that the magnitude of $p_1$ is $\frac{\sqrt{2}}{2}$ times the magnitude of $b_1$. Also, the argument of $p_1$ is $\frac{7\pi}{4}$ more than the argument of $b_1$. We can get that $p_1$ is $b_1$ scaled $\frac{\sqrt{2}}{2}$, and rotated $\frac{7\pi}{4}$ clockwise. So, $p_1=\frac{\sqrt{2}}{2}e^{\frac{7\pi}{4} i} b_1$, or $p_1=(0.5-0.5i)b_1$ when point $a$ is translated to the origin. Translating back, we can get $p=(0.5-0.5i)(b-a)+a$. Simplifying, we can get $\boxed{p=\frac{b-bi+a-ai}{2}}$. Similarly, we can get $\boxed{q=\frac{c-ci+b-bi}{2}}$, $\boxed{r=\frac{d-di+c-ci}{2}}$ and $\boxed{s=\frac{a-ai+d-di}{2}}$

b):

We want to prove that the line segment between $p$ and $r$ is perpendicular and equal in length to the line segment between $q$ and $s$. We can start by considering what it means for the line segment between $\overline{pr}$ is perpendicular and equal in length to the line segment between $\overline{qs}$ algebraically. This means that we can translate where the two lines intersect to the origin, then rotate one of the lines to get the other. That means that $e^{\pi/2}(r-p)=s-q$. Notice how if $e^{\pi/2}(r-p)=s-q$, $|r-p|=|s-q|$, so the lines have equal length. The lines are perpendicular and have equal length if $e^{\pi/2}(r-p)=s-q$ is true. Plugging in the values we got in a), we can see that $e^{\pi/2}(r-p)=s-q$, or $i(r-p)=s-q$ becomes $i\left(\frac{d-di+c-ci}{2}-\frac{b-bi+a-ai}{2}\right)=\frac{a-ai+d-di}{2}-\frac{c-ci+b-bi}{2}$. We can simplify this to get $i(d-di+c-ci-b+bi-a+ai)=a-ai+d-di-c+ci-b+bi$, or $di+d+ci+c-bi-b-a-ai=a-ai+d-di-c+ci-b+bi$. We can further simplify to get $2di+2c=2a+2bi$, or $di+c=a+bi$. So, if $di+c=a+bi$, then the line segment between $\overline{pr}$ is perpendicular and equal in length to the line segment between $\overline{qs}$.

Davis Sep 15, 2019