Complex number bash, finally possible
by math_explorer, Mar 7, 2011, 11:21 AM
Tada. Here's the proof by Stifler stuffed to be a badly written tutorial to geometry with complex numbers.
Background knowledge. When we're using complex numbers in geometry you can think of them as superpowered plane vectors. The point/vector
becomes complex number 
, and you can add, subtract, and take the norm the same way you do with vectors. Instead of a lousy scalar dot product, you get multiplication/division and conjugation.
Conjugation (denoted by an overline, as in
, meaning to negate the imaginary part, hopefully you know that) is a very nice operation; it distributes (that's probably not the right word but w/e) over most operations you can think of, due to the symmetry between 
and 
. Geometrically it's equivalent to reflection about the x-axis. A number is equal to its conjugate iff it's real, and equal to its conjugate's negation iff it's pure imaginary. 
. 
is 
normalized, and therefore the unit circle number with twice the argument angle of 
. 
is 's real part. As a corollary, 
is the dot product of and 
as vectors.
Standard notation is to denote a point with a capital letter, and its complex number with the same lowercase letter. Primes and other hovering aliasing things float around with both.
If you're going to use the complex numbers, it's important to pick a good origin. The same goes for fitting normal rectangular coordinates on a problem. With complex numbers though you also get bonuses for points on the unit circle, because they're nicer to work with, the same way points on the x-axis might be nicer to work with in a rectangular coordinate system.
For example, the unit circle points are closed (in fact, form a group) under multiplication. Essentially equivalent to addition
. Also, conjugates and reciprocals on them become the same thing.
Anyway, circumcenters are often a good place to put the center.
This observation was frighteningly nontrivial for me.
, hence 
, and isosceles triangles are nice.
Whoa that was fast. There are a lot of unit circle theorems before this. Let's try to find
, the foot of the altitude from 
to 
.
First, for points in general: remember how
is a unit circle number with twice the argument angle of 
? Well, helpfully, we can use it as a sort of slope to compare lines; that is, 
if and only if the line formed by 
is parallel or coincident to the line formed by 
. The doubled angle means we'll also match lines pointing the opposite way, which is good, and that all we have to do to match perpendicular lines is add a negative sign, which is also good. Now, if happen to be on the unit circle, that ugly slopefractionmonster turns out to be 
. Yay.
Now plug
in. The result is 
. Thrash around; you'll find 
.
Next, since
is perpendicular to 
we also have 
. Plug 
in and solve straightforwardly to find 
. In fact, 
doesn't have to be on the unit circle for this equation to hold.
Now, the midpoint of
and 
is . We can revert to vectorish ways: 
so 
. Since is a unit circle point, 
and 
are the same, of course.
I have absolutely no idea where this lemma comes from or how to prove it (other than a blind bash). No matter, because it was patched.
MEMORIZE.
The derivation of such a messy formula is of course messy, but pretty straightforward.
and 
are perpendicular, hence 
. Cross-multiply and move terms around nicely, and you'll find 
. Quite similarly 
. Now you can just solve like a normal two-variable set of equations.
To get this you have to realize that
, because 
(the centroid), and 
(circumcenter), and their distances are related by the Euler line.
Also, that
Expaaaaanding. Straightforward but nasty, which means I don't have to say a thing.
if and only if: 
. But this is obvious, because 
.
To prove this we need to see that
and 
"point the same way". Geometrically, that's true since 
and 
are both length 1, and thus 
form a rhombus. Algebraically, you can conjugate aka reciprocal the whole thing (since and are on the unit circle so are , ,
![\[\frac{\frac{1}{b^2} + \frac{1}{c^2}}{\frac{1}{b^2} + \frac{1}{bc} + \frac{1}{c^2}}\]](//latex.artofproblemsolving.com/0/2/6/0269d4d7b640f94dfa424f0c85b169a59524c983.png)
which is the original thing, so we're done.
Background knowledge. When we're using complex numbers in geometry you can think of them as superpowered plane vectors. The point/vector
becomes complex number 
, and you can add, subtract, and take the norm the same way you do with vectors. Instead of a lousy scalar dot product, you get multiplication/division and conjugation.Conjugation (denoted by an overline, as in
, meaning to negate the imaginary part, hopefully you know that) is a very nice operation; it distributes (that's probably not the right word but w/e) over most operations you can think of, due to the symmetry between 
and 
. Geometrically it's equivalent to reflection about the x-axis. A number is equal to its conjugate iff it's real, and equal to its conjugate's negation iff it's pure imaginary. 
. 
is 
normalized, and therefore the unit circle number with twice the argument angle of 
. 
is 's real part. As a corollary, 
is the dot product of and 
as vectors.Standard notation is to denote a point with a capital letter, and its complex number with the same lowercase letter. Primes and other hovering aliasing things float around with both.
Stifler wrote:
It can be done nicely using complex numbers. Let 
be the origin of the complex plane, and 
.
be the origin of the complex plane, and 
.If you're going to use the complex numbers, it's important to pick a good origin. The same goes for fitting normal rectangular coordinates on a problem. With complex numbers though you also get bonuses for points on the unit circle, because they're nicer to work with, the same way points on the x-axis might be nicer to work with in a rectangular coordinate system.
For example, the unit circle points are closed (in fact, form a group) under multiplication. Essentially equivalent to addition
. Also, conjugates and reciprocals on them become the same thing.Anyway, circumcenters are often a good place to put the center.
Stifler wrote:
Observe that is the refflection of wrt 
.
is the refflection of 
wrt 
.This observation was frighteningly nontrivial for me.
, hence 
, and isosceles triangles are nice.Stifler wrote:
So: 
.
.Whoa that was fast. There are a lot of unit circle theorems before this. Let's try to find
, the foot of the altitude from 
to 
.First, for points in general: remember how
is a unit circle number with twice the argument angle of 
? Well, helpfully, we can use it as a sort of slope to compare lines; that is, 
if and only if the line formed by 
is parallel or coincident to the line formed by 
. The doubled angle means we'll also match lines pointing the opposite way, which is good, and that all we have to do to match perpendicular lines is add a negative sign, which is also good. Now, if happen to be on the unit circle, that ugly slopefractionmonster turns out to be 
. Yay.Now plug
in. The result is 
. Thrash around; you'll find 
.Next, since
is perpendicular to 
we also have 
. Plug 
in and solve straightforwardly to find 
. In fact, 
doesn't have to be on the unit circle for this equation to hold.Now, the midpoint of
and is . We can revert to vectorish ways: 
so 
. Since is a unit circle point, 
and 
are the same, of course.Stifler wrote:
By the same way: 
. Now, we're supposed to find 
(the circumcenter of 
). Using the well-known lemma:
If
and 
are three distinct points in plane, then the circumcenter 
of 
satisfy:
![\[
w\left (\dfrac{\overline{x}-\overline{y}}{x-y}-\dfrac{\overline{x}-\overline{z}}{x-z}\right )=\dfrac{|x|^2-|y|^2}{x-y}-\dfrac{|x|^2-|z|^2}{x-z}
\]](//latex.artofproblemsolving.com/f/7/2/f72b4fd60ac1baf33a6824b6ed99fe2730c5e05b.png)
. Now, we're supposed to find 
(the circumcenter of 
). Using the well-known lemma:If
and 
are three distinct points in plane, then the circumcenter 
of 
satisfy:![\[
w\left (\dfrac{\overline{x}-\overline{y}}{x-y}-\dfrac{\overline{x}-\overline{z}}{x-z}\right )=\dfrac{|x|^2-|y|^2}{x-y}-\dfrac{|x|^2-|z|^2}{x-z}
\]](http://latex.artofproblemsolving.com/f/7/2/f72b4fd60ac1baf33a6824b6ed99fe2730c5e05b.png)
I have absolutely no idea where this lemma comes from or how to prove it (other than a blind bash). No matter, because it was patched.
math154 wrote:
I'd like to point out that there's a slightly simpler way to carry out this computation: first note that the circumcenter of the triangle with vertices 
is
![\[\frac{xy(\overline{x}-\overline{y})}{\overline{x}y-x\overline{y}},\]](//latex.artofproblemsolving.com/1/5/3/153516bb32e82f5e7954e7bb182edf77bfb30d17.png)
is![\[\frac{xy(\overline{x}-\overline{y})}{\overline{x}y-x\overline{y}},\]](http://latex.artofproblemsolving.com/1/5/3/153516bb32e82f5e7954e7bb182edf77bfb30d17.png)
MEMORIZE.
The derivation of such a messy formula is of course messy, but pretty straightforward.
and 
are perpendicular, hence 
. Cross-multiply and move terms around nicely, and you'll find 
. Quite similarly 
. Now you can just solve like a normal two-variable set of equations.math154 wrote:
so by translating we have
![\[o'=h+\frac{(m-h)(n-h)(\overline{m}-\overline{n})}{(\overline{m}-\overline{h})(n-h)-(m-h)(\overline{n}-\overline{h})}.\]](//latex.artofproblemsolving.com/1/5/d/15d989ad2dc61af0c602b43ad8ace7a4877ecd82.png)
Now we just use
![\[o'=h+\frac{(m-h)(n-h)(\overline{m}-\overline{n})}{(\overline{m}-\overline{h})(n-h)-(m-h)(\overline{n}-\overline{h})}.\]](http://latex.artofproblemsolving.com/1/5/d/15d989ad2dc61af0c602b43ad8ace7a4877ecd82.png)
Now we just use\[\begin{align*}
m-h=-\frac{b}{c}(a+c) &\implies \overline{m}-\overline{h}=-\frac{a+c}{ab},\\
n-h=-\frac{c}{b}(a+b) &\implies \overline{n}-\overline{h}=-\frac{a+b}{ac},\\
&\implies \overline{m}-\overline{n}=\frac{(b-c)(a+b+c)}{abc}.
\end{align*}\]To get this you have to realize that
, because 
(the centroid), and 
(circumcenter), and their distances are related by the Euler line.Also, that
Stifler wrote:
After expanding both sides we found that:
![\[
o'=\dfrac{(b^2+c^2)(a+b+c)}{b^2+bc+c^2}
\]](//latex.artofproblemsolving.com/0/9/1/0913f2558034c99f299f65f03065383cf10526df.png)
![\[
o'=\dfrac{(b^2+c^2)(a+b+c)}{b^2+bc+c^2}
\]](http://latex.artofproblemsolving.com/0/9/1/0913f2558034c99f299f65f03065383cf10526df.png)
Expaaaaanding. Straightforward but nasty, which means I don't have to say a thing.
Stifler wrote:
if and only if: 
. But this is obvious, because 
.To prove this we need to see that
and 
"point the same way". Geometrically, that's true since 
and 
are both length 1, and thus 
form a rhombus. Algebraically, you can conjugate aka reciprocal the whole thing (since and are on the unit circle so are , , ![\[\frac{\frac{1}{b^2} + \frac{1}{c^2}}{\frac{1}{b^2} + \frac{1}{bc} + \frac{1}{c^2}}\]](http://latex.artofproblemsolving.com/0/2/6/0269d4d7b640f94dfa424f0c85b169a59524c983.png)
which is the original thing, so we're done.
This post has been edited 2 times. Last edited by math_explorer, Aug 19, 2011, 9:27 AM
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