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 $$𝑝,~𝑞,~𝑟$$ and $$𝑠$$ be the centers of those squares: a) Find expressions for $$𝑝,~𝑞,~𝑟$$ and $$𝑠$$ in terms of $$𝑎,~𝑏,~𝑐$$ and $$𝑑$$.

b) Prove that the line segment between $$𝑝$$ and $$𝑟$$ is perpendicular and equal in length to the line segment between $$𝑞$$ and $$𝑠$$.

I think this problem has been asked before, but they don't give any good hints. I don't really know where to start. I tried finding $$p$$ first by finding $$(p-a)$$ and $$(p-b)$$. I tried another way by translating $$a$$ to the origin. I haven't been able to go farther than this.

I think I have an idea for part $$b$$ using similar triangles and things, but part a is really confusing.Thank you!

I translated the square with $$p$$ as its center so that a would be at the origin. So $$b$$ would then be $$b−a$$ and $$p$$ would be $$p−a$$, right? $$p−a$$ is half of the diagonal. So then $$(p−a)=(b−a)\cdot\frac{\sqrt2}{2}$$.

Rotating by $$-\frac{\pi}{4}$$ would give us $$\frac{(b−a)2}{√2}\cdot e^{−i\frac{\pi}{4}}=\frac{(b−a)\sqrt2}{2}\cdot\left(\frac{\sqrt2}{2}−i\frac{\sqrt2}{2}\right)=\left(\frac{(b−a)}{2}−i\frac{(b−a)}{2}\right)=\frac{b−a−bi+ai}{2}$$ Therefore, $$p−a=b−a−bi+ai2$$ and when we translate everything back we get $$p=b−a−bi+ai2+a⟹p=b−a−bi+ai+(2a)2⟹p=b+a−bi+ai2.$$ I can do a similar process for the rest of the points, right?

Does it matter which point I translate to the origin?

• We can prove $b)$ without $a)$. If you want to see my solution, show please your attempts. Sep 27 '20 at 2:21
• I tried finding p first by finding $(p-a)$ and $(p-b)$. I tried another way by translating $a$ to the origin. I haven't been able to go farther than this. Sep 27 '20 at 16:18
• Show, how exactly you made it. Sep 27 '20 at 17:22
• Ok, so I translated the square with $p$ as its center so that $a$ would be at the origin. So $b$ would then be $b-a$ and $p$ would be $p-a$, right? $p-a$ is half of the diagonal. So then $(p-a) = (b-a) \cdot \frac{\sqrt{2}}{2}$. Rotating by $-\frac{\pi}{4}$ would give us $\frac{(b-a)\sqrt{2}}{2} \cdot e^{-i\frac{\pi}{4}} = \frac{(b-a)\sqrt{2}}{2} \cdot (\frac{\sqrt{2}}{2} -i\frac{\sqrt{2}}{2}) = (\frac{(b-a)}{2} - i\frac{(b-a)}{2}) = \frac{b-a-bi+ai}{2}$. Sep 27 '20 at 22:03
• Therefore, $p-a = \frac{b-a-bi+ai}{2}$ and when we translate everything back we get $p = \frac{b-a-bi+ai}{2} + a \Longrightarrow p = \frac{b-a-bi+ai + (2a)}{2} \Longrightarrow p = \frac{b+a-bi+ai}{2}$. I can do a similar process for the rest of the points, right? Sep 27 '20 at 22:03

I'll give a hint for part (a). Points like $$a$$, $$b$$, $$c$$ and $$d$$ on the complex plane can also be thought of as vectors starting at the origin and ending at that point, so that the vector going from $$a$$ to $$c$$ is $$c-a$$; in other words, $$a+(c-a)=c$$. So to get to the point $$p$$, I have to go halfway between $$a$$ and $$b$$, then make a $$90^\circ$$ turn to the right and move by that same distance. That is,
$$p=a+\frac{b-a}{2}+(\hbox{right turn by \frac{b-a}{2})}.$$
• Multiply by $e^{i\theta}$? Would $\theta = \frac{\pi}{2}$? Sep 27 '20 at 16:28
• Almost; rotation in the complex plane is clockwise by convention so $e^{i\pi/2}$ would represent a $90^\circ$ left turn Sep 28 '20 at 0:02