1

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

enter image description here

a) Find expressions for $,~,~$ and $$ in terms of $,~,~$ and $$.

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

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?

Martin Sleziak
  • 56,060
Henry Smith
  • 11
  • We can prove $b)$ without $a)$. If you want to see my solution, show please your attempts. – Michael Rozenberg Sep 27 '20 at 02:21
  • 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. – Henry Smith Sep 27 '20 at 16:18
  • Show, how exactly you made it. – Michael Rozenberg 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 $\frac{(b-a)\sqrt{2}}{2} \cdot e^{-i\frac{\pi}{4}} = \frac{(b-a)\sqrt{2}}{2} \cdot (\frac{\sqrt{2}}{2} -i\frac{\sqrt{2}}{2}) = (\frac{(b-a)}{2} - i\frac{(b-a)}{2}) = \frac{b-a-bi+ai}{2}$. – Henry Smith 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? – Henry Smith Sep 27 '20 at 22:03
  • Does it matter which point I translate to the origin? – Henry Smith Sep 27 '20 at 22:14
  • See here: https://math.stackexchange.com/questions/3705224 – Michael Rozenberg Sep 28 '20 at 01:10
  • Welcome to Math.SE! A minor quibble: "I think this problem has been asked before, but they don't give any good hints." ... If you can report "they don't give any good hints", then it would seem that you more-than-"think" the question has been asked before. :) In the future, when you know that a question has been asked before, please provide a link (like this one); this can help people avoid wasting time duplicating responses you've seen and found unsatisfactory. – Blue Sep 28 '20 at 04:21

1 Answers1

1

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.

ho boon suan
  • 2,099