Download Geometry in the Complex Plane

Introduction
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Geometry in the Complex Plane
Hongyi Chen
o
n
UNC Awards Banquet 2016
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“All Geometry is Algebra”
Many geometry problems can be solved using a purely algebraic
approach - by placing the geometric diagram on a coordinate plane,
assigning each point an x/y coordinate, writing out the equations
of lines and circles, and solving these equations. This method of
solving geo problems (often called coordinate bashing) can be quite
powerful given the right conditions, but it has some problems.
Issues with coordinate bash
Equations for circles are ugly
Two variables are necessary for each random point
Rotations are extremely painful
Attempting to solve the equations may result in massive 5th
degree polynomials in 8 variables...
Fortunately, these problems can be fixed by replacing the Cartesian
plane with the complex plane...
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Quick Introduction to Complex Numbers
A complex number (in rectangular form) is a number of the
form a + bi, where a and b are real and i 2 = −1.
We define the real and imaginary parts of a complex
z = a + bi as Re(z) = a and Im(z) = bi.
Complex numbers can be plotted on the complex plane. The
number a + bi is placed where the coordinate (a, b) is placed
on the Cartesian plane. The horizontal axis is called the real
axis and the vertical axis is called the imaginary axis.
The conjugate of a complex number z, denoted by z̄, is its
reflection about the real axis. For any z = a + bi we have
z̄ = a − bi.
ab = ā · b̄ and a + b = ā + b̄.
z − z̄
z + z̄
Re(z) =
and Im(z) =
.
2
2
z is real if and only if Im(z) = 0, which occurs when z = z̄.
Similarly a number z is pure imaginary iff z = −z̄.
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Quick Introduction to Complex Numbers
The magnitude of z = a + bi, denoted by |z|, is its distance
from the
√ origin in the complex plane. If z = a + bi then
|z| = a2 + b 2 .
Notice that for any complex z, z z̄ = |z|2 .
|a − b| is the distance between a and b.
A complex number z can also be expressed in polar form as
r (cos θ + i sin θ) for a real r and angle θ, where r = |z| and θ
is the angle formed by the positive real axis and the ray
starting at the origin pointing towards z, measured
counterclockwise.
For simplicity we shall let cis θ = cos θ + i sin θ.
The set of possible values of cis θ forms the unit circle on the
complex plane - a circle centered at the origin with radius 1.
1
For any angle θ we have cis θ =
= cis (−θ)
cis θ
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Complex Bash
We can put entire geometry diagrams onto the complex plane.
Each point is represented by a complex number, and each line or
circle is represented by an equation in terms of some complex z
and possibly its conjugate z̄. By standard, the complex number
corresponding to a point is denoted by the lowercase character of
of the point’s label (for example, if A = (2, 3), then a = 2 + 3i).
This is called complex bashing, and can be extremely powerful.
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Translations
The greatest advantage of using complex bash as opposed to any
other type of bash is the simplicity of performing the basic
transformations of translation, rotation, and dilation.
To perform a translation of a units right and b units up to some
complex number z, notice that complex number addition works the
same way as vector addition, so one can simply add a + bi to z.
To translate an object such that some point p becomes the origin,
simply subtract p from all the points on the object.
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Homothety (Dilation)
To find the image of some point z after a homothety centered at
the origin with scale factor k, simply multiply z by k.
If the center is some point p instead, one can find the image by
translating everything such that p is at the origin, then performing
the homothety centered at the origin, then translating everything
again such that p is back at its original place.
Hence the image of z under a homothety centered at p with scale
factor k is k(z − p) + p.
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Rotations
A well-known theorem states that (cis θ)(cis φ) = cis (θ + φ).
Thus, to rotate a complex number z = r cis φ by θ about the
origin, multiply it by cis θ. To rotate z about an arbitrary point p,
we can use the same manipulation used in the previous slide:
translate everything by −p, then perform the rotation around the
origin, then translate everything +p.
Hence the result of rotating z by θ about p is (z − p)cis (θ) + p.
Introduction
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A Quick Example - Problem
The points (0, 0), (a, 11), and (b, 37) are the vertices of an
equilateral triangle. Find the value of ab. (1994 AIME #8)
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A Quick Example - Solution
Let O = 0, P = a + 11i, and Q = b + 37i on the complex plane.
Since 4OPQ is an equilateral triangle, we know ∠POQ = 60◦ , so
Q is the result of rotating P around O by 60◦ . But then we know
that
!
√ !
√
a
11
3
a
3
11
q = p · cis 60◦ =⇒ b + 37i =
−
+
+
i
2
2
2
2
Notice that the imaginary part of the left side of this equation
must equal the imaginary part of the right side, so
!
√
√
a 3 11
37i =
+
i =⇒ a = 21 3.
2
2
Similarly we can equate the real parts, which gives us
√
√
a 11 3
= 5 3.
b= −
2
2
√ √ Thus the answer is 21 3 5 3 = 315 .
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Lines and Collinearity
The following theorems are commonly used in complex bash:
Three distinct complex numbers a, b, and c are collinear if
c −a
and only if
is real.
b−a
Since a complex number is real if and only if it is equal to its
conjugate, the above means the equation for a line passing
through a and b, in terms of z is
z −a
=
b−a
z −a
.
b−a
If a point Z divides a segment AB into segments AZ and BZ
in the ratio k/(1 − k) then z = (b − a)k + a.
In particular, the midpoint of AB is
a+b
.
2
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The Centroid - Problem
Let ABC be a triangle in the complex plane. Find a formula for the
centroid of 4ABC in terms of the complex numbers a, b, and c.
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The Centroid - Solution
b+c
Let M be the midpoint of BC . Then we know m =
. A
2
well-known theorem states that the centroid G divides MA in the
ratio 13 : 23 . Hence,
1
b+c 1 b+c
a+b+c
g = (a − m) + m = a −
+
=
.
3
2
3
2
3
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Parallelism and Perpendicularity
Given lines AB and CD, one can determine whether or not they
are parallel or perpendicular with the following theorems.
b−a
b−a
b−a
AB||CD ⇐⇒
is real ⇐⇒
=
.
d −c
d −c
d −c
b−a
b−a
b−a
AB ⊥ CD ⇐⇒
is imaginary ⇐⇒
=−
.
d −c
d −c
d −c
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Reflections - Problem
Given points A, B, and C , let Z be the reflection of C about AB.
Find z in terms of a, b, and c.
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Reflections - Solution
Let M be the midpoint of ZC . We know m =
z +c
.
2
If Z is the reflection of C about AB, then M should lie on AB.
Hence we have
m−b
m−b
z + c − 2b
z̄ + c̄ − 2b̄
=
=
=⇒
.
a−b
a−b
2(a − b)
2(ā − b̄)
Also ZC should be perpendicular to AB, so
z −c
z −c
c̄ − z̄
=−
=
.
a−b
a−b
ā − b̄
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Reflections - Solution
We now have
z + c − 2b
z̄ + c̄ − 2b̄
c̄ − z̄
z −c
=
=
and
.
2(a − b)
a−b
2(ā − b̄)
ā − b̄
This is a system of equations in z and z̄. Solving for z and z̄, we
see that
ac̄ + b ā − ab̄ − b c̄
z=
.
ā − b̄
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Using the Unit Circle
If z is on the unit circle, |z| = 1, so by the theorem z z̄ = |z|2
1
we get z̄ = .
z
This substitution is useful.
Scaling a diagram does not change length ratios or angles
Scale diagram so that an important circle or some important
points are on the unit circle.
If problem involves a “central” triangle, let the three vertices
of the triangle be on the unit circle.
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The Orthocenter - Problem
Let ABC be a triangle inscribed in the complex unit circle, and let
h = a + b + c. Prove that H is the orthocenter of 4ABC .
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The Orthocenter - Solution
Let us define
k=
h−a
(a + b + c) − a
b+c
=
=
.
b−c
b−c
b−c
Notice that
k̄ =
b+c
b−c
=
b̄ + c̄
1/b + 1/c
c +b
b+c
=
=
=−
= −k.
1/b − 1/c
c −b
b−c
b̄ − c̄
Hence k is imaginary, so HA ⊥ BC . Similarly, HB ⊥ CA and
HC ⊥ AB, so H is the orthocenter of 4ABC , as desired.
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Euler Line
A famous theorem by Euler states that in any triangle, the
circumcenter, the centroid, and the orthocenter are collinear. The
line passing through these three points is called the Euler Line of
the triangle.
To prove this, take any 4ABC and
inscribe it in the complex unit circle.
Let H be the orthocenter, O be the
circumcenter, and G be the centroid.
o = 0.
a+b+c
g=
.
3
h = a + b + c.
Then
g −o
(a + b + c)/3
1
=
= ∈ R, so O, G , H are collinear.
h−o
a+b+c
3
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Outer Napoleon Triangle - Problem
Given any 4ABC , construct point EA such that BEA C is an
equilateral triangle, with EA being on the opposite side of BC as
A. Let NA be the centroid of 4BEA C . Similarly define NB and
NC . Prove that 4NA NB NC is an equilateral triangle.
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Outer Napoleon Triangle - Solution
We assume that 4ABC is oriented counter-clockwise (if it is
oriented clockwise, the solution is almost identical).
Notice that EC is the result of rotating A by 60◦ about B, so
ec = (a − b)cis (60◦ ) + b =
a + b a − b√
+
3i.
2
2
Then since NC is the centroid of AEC B,
nc =
Similarly, na =
ec + a + b
a + b a − b√
=
+
3i.
3
2
6
b −c√
c + a c − a√
b+c
+
3i and nb =
+
3i.
2
6
2
6
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Outer Napoleon Triangle - Solution
To prove NA NB NC is equilateral, it suffices to show that NB is the
result of rotating NA by 60◦ about NC . But doing so is equivalent
to showing that (na − nc )cis 60◦ + nc = nb , which can be done
with the calculations below.
◦
(na − nc )cis 60 + nc
√ !
a+b
1
b+c
b−c√
a−b√
3
a+b
a−b√
3i −
3i
3i
=
+
−
+i
+
+
2
6
2
6
2
2
2
6
!
√
c−a
2b − a − c √
1
3
a+b
a−b√
=
+
3i
+i
+
+
3i
2
6
2
2
2
6
2b − a − c √
c − a√
2b − a − c
a+b
a−b√
+
3i +
3i −
+
+
3i
4
12
4
4
2
6
(2b − a − c) + 3(c − a) + 2(a − b) √
(c − a) − (2b − a − c) + 2(a + b)
+
3i
=
4
12
c+a
c − a√
=
+
3i = nb .
2
6
=
c−a
And hence 4NA NB NC is equilateral, as desired. Introduction
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A Cyclic Kite - Problem
Quadrilateral APBQ is inscribed in circle ω with ∠P = ∠Q = 90◦
and AP = AQ < BP. Let X be a variable point on segment PQ.
Line AX meets ω again at S (other than A). Point T lies on arc
AQB of ω such that XT is perpendicular to AX . Let M denote
the midpoint of chord ST . As X varies on segment PQ, show that
M moves along a circle. (2015 USAMO #2)
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A Cyclic Kite - Solution
Let ω be the complex unit circle, and let O be the origin. Since
∠P = ∠Q = 90◦ , AB is a diameter of ω, so we can let a = 1 and
b = −1. Now notice that APBQ is a kite, so PQ ⊥ AB. But AB
coincides with the real axis, hence PQ is perpendicular to the real
axis and thus all points on PQ have the same real part. Therefore
Re(x) is constant.
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A Cyclic Kite - Solution
Since A, X , and S are collinear, we have
x −1
x −1
x̄ − 1
=
=
.
s −1
s −1
1/s − 1
Since XT is perpendicular to AS, we have
x −t
x −t
x̄ − 1/t
=−
.
=−
s −1
s −1
1/s − 1
Solving these equations for x and x̄, we get
1
s
1
1 1 t
x=
1+s +t −
and x̄ =
1+ + −
.
2
t
2
s
t
s
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A Cyclic Kite - Solution
We now see that
x + x̄
1
Re(x) =
=
2
4
1 1 s
t
2+s +t + + − −
.
s
t
t
s
1
Let z = . Notice that
2
s + t − 1 2
s +t −1
s +t −1
2
|m − z| = =
2
2
2
1 1
1
= (s + t − 1)
+ −1
4
s
t
1
1 1 s
t
5
=
3−s −t − − + +
= − Re(x).
4
s
t
t
s
4
Hence r
all possible M are located on a circle centered at Z with
5
radius
− Re(x). 4
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A Nontrivial Mess - Challenge
Let ABC be a scalene triangle. Let Ka , La and Ma be the
respective intersections with BC of the internal angle bisector,
external angle bisector, and the median from A. The circumcircle
of AKa La intersects AMa a second time at point Xa different from
A. Define Xb and Xc analogously. Prove that the circumcenter of
Xa Xb Xc lies on the Euler line of ABC. (2015 USA TSTST #2)
Good luck!