Complex Quadrilateral Problem

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HenryS

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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:
1601169839996.png
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 don't really know where to start. I think I have part B solved using similar triangles and things, but part a is really confusing. Thank you!
 
HenryS said:
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:
View attachment 21892
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 don't really know where to start. I think I have part B solved using similar triangles and things, but part a is really confusing. Thank you!
Click to expand...
You say:

a) Find expressions for ?,?,? and ? in terms of ?,?,? and ?.

Do you mean:

Find expressions for coordinates of ?,?,? and ? in terms of coordinates of ?,?,? and ?.

Please show us what you have tried and exactly where you are stuck.​
​
Please follow the rules of posting in this forum, as enunciated at:​
​
​
Please share your work/thoughts about this problem.​
 
HenryS said:
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:
View attachment 21892
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 don't really know where to start. I think I have part B solved using similar triangles and things, but part a is really confusing. Thank you!
Click to expand...
The picture suggests that everything is to be done in terms of complex numbers. Is that true? It should have been stated.

My first thought, until I noticed that, was to use vectors; and that may still be a good starting point. For instance, you can find p by adding half of vector ab to a, and then adding a vector perpendicular to it.
 
Dr.Peterson said:
The picture suggests that everything is to be done in terms of complex numbers. Is that true? It should have been stated.

My first thought, until I noticed that, was to use vectors; and that may still be a good starting point. For instance, you can find p by adding half of vector ab to a, and then adding a vector perpendicular to it.
Click to expand...

Yes, it is in terms of complex numbers. Sorry!
 
Subhotosh Khan said:
You say:

a) Find expressions for ?,?,? and ? in terms of ?,?,? and ?.

Do you mean:

Find expressions for coordinates of ?,?,? and ? in terms of coordinates of ?,?,? and ?.

Please show us what you have tried and exactly where you are stuck.​
​
Please follow the rules of posting in this forum, as enunciated at:​
​
​
Please share your work/thoughts about this problem.​
Click to expand...

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.
Although the problem is in complex numbers, can I still use vectors? Would I be able to multiply by e^(i theta) to rotate it?
 
Last edited:
HenryS said:
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.
Although the problem is in complex numbers, can I still use vectors? Would I be able to multiply by e^(i theta) to rotate it?
Click to expand...
If you don't see an answer directly in terms of complex numbers, I would definitely try thinking in terms of vectors. And yes, one way to rotate a vector by 90 degrees is to multiply by the appropriate complex number (which is a very simple one!).

Give it a try, and show us your work.
 
Ok, so I translated the square with [MATH]p[/MATH] as its center so that [MATH]a[/MATH] would be at the origin. So [Math]b[/Math] would then be [Math]b-a[/Math] and [Math]p[/Math] would be [Math]p-a[/Math], right? [Math]p-a[/Math] is half of the diagonal. So then [Math](p-a) = (b-a) \cdot \frac{\sqrt{2}}{2}[/Math]. Rotating by [Math]-\frac{\pi}{4}[/Math] would give us [Math]\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}[/Math].

Therefore, [Math]p-a = \frac{b-a-bi+ai}{2}[/Math] and when we translate everything back we get [Math]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}[/Math]. I can do a similar process for the rest of the points, right?

If so, does it matter which point I translate to the origin?
 
I think that looks about right.

Let's try what I have in mind (just a little different from my original suggestion), and see if I get the same result. If I call the point opposite a on its square a', then
(a' - a) = (b - a) + (a' - b) = (b - a) + (b - a)(-i) [rotating the latter 90 degrees clockwise]​
= b - a + (a - b)i;​
(p - a) = (a' - a)/2 = (b - a + (a - b)i)/2;​
so
p = a + (p - a) = a + (b - a + (a - b)i)/2 = ((a + b) + (a - b)i)/2​

And we agree.
 
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