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Consider a convex quadrilateral with vertices at $๐‘Ž,~๐‘,~๐‘$ and $๐‘‘$ and on each side draw a square lying outside the given quadrilateral, as in the picture below. Let $๐‘,~๐‘ž,~๐‘Ÿ$ and $๐‘ $ be the centers of those squares:

enter image description here

a) Find expressions for $๐‘,~๐‘ž,~๐‘Ÿ$ and $๐‘ $ in terms of $๐‘Ž,~๐‘,~๐‘$ and $๐‘‘$.

b) Prove that the line segment between $๐‘$ and $๐‘Ÿ$ is perpendicular and equal in length to the line segment between $๐‘ž$ and $๐‘ $.

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 $bโˆ’a$ and $p$ would be $pโˆ’a$, right? $pโˆ’a$ is half of the diagonal. So then $(pโˆ’a)=(bโˆ’a)\cdot\frac{\sqrt2}{2}$.

Rotating by $-\frac{\pi}{4}$ would give us $$\frac{(bโˆ’a)2}{โˆš2}\cdot e^{โˆ’i\frac{\pi}{4}}=\frac{(bโˆ’a)\sqrt2}{2}\cdot\left(\frac{\sqrt2}{2}โˆ’i\frac{\sqrt2}{2}\right)=\left(\frac{(bโˆ’a)}{2}โˆ’i\frac{(bโˆ’a)}{2}\right)=\frac{bโˆ’aโˆ’bi+ai}{2}$$ Therefore, $pโˆ’a=bโˆ’aโˆ’bi+ai2$ and when we translate everything back we get $$p=bโˆ’aโˆ’bi+ai2+aโŸนp=bโˆ’aโˆ’bi+ai+(2a)2โŸนp=b+aโˆ’bi+ai2.$$ I can do a similar process for the rest of the points, right?

Does it matter which point I translate to the origin?

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  • $\begingroup$ We can prove $b)$ without $a)$. If you want to see my solution, show please your attempts. $\endgroup$ Sep 27 '20 at 2:21
  • $\begingroup$ 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. $\endgroup$ Sep 27 '20 at 16:18
  • $\begingroup$ Show, how exactly you made it. $\endgroup$ Sep 27 '20 at 17:22
  • $\begingroup$ 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 $\frac{(b-a)\sqrt{2}}{2} \cdot e^{-i\frac{\pi}{4}} = \frac{(b-a)\sqrt{2}}{2} \cdot (\frac{\sqrt{2}}{2} -i\frac{\sqrt{2}}{2}) = (\frac{(b-a)}{2} - i\frac{(b-a)}{2}) = \frac{b-a-bi+ai}{2}$. $\endgroup$ Sep 27 '20 at 22:03
  • $\begingroup$ 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? $\endgroup$ Sep 27 '20 at 22:03
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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,

$$p=a+\frac{b-a}{2}+(\hbox{right turn by $\frac{b-a}{2}$)}.$$

So how do you get a right turn by a complex number? How do complex numbers relate to rotations?

For reference, this is van Aubel's theorem; a proof can be found in Chapter 1 of Tristan Needham's wonderful text Visual Complex Analysis.

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  • $\begingroup$ Multiply by $e^{i\theta}$? Would $\theta = \frac{\pi}{2}$? $\endgroup$ Sep 27 '20 at 16:28
  • $\begingroup$ Almost; rotation in the complex plane is clockwise by convention so $e^{i\pi/2}$ would represent a $90^\circ$ left turn $\endgroup$ Sep 28 '20 at 0:02

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