Geometry in the Complex Plane - UNC Math ContestGeometry in the Complex Plane Hongyi Chen on UNC...

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Introduction Transformations Lines Unit Circle More Problems

Geometry in the Complex Plane

Hongyi Chen on UNC Awards Banquet 2016


“All Geometry is Algebra”

Many geometry problems can be solved using a purely algebraicapproach - by placing the geometric diagram on a coordinate plane,assigning each point an x/y coordinate, writing out the equationsof lines and circles, and solving these equations. This method ofsolving geo problems (often called coordinate bashing) can be quitepowerful 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 5thdegree polynomials in 8 variables...

Fortunately, these problems can be fixed by replacing the Cartesianplane with the complex plane...


Quick Introduction to Complex Numbers

A complex number (in rectangular form) is a number of theform a + bi , where a and b are real and i2 = −1.

We define the real and imaginary parts of a complexz = a + bi as Re(z) = a and Im(z) = bi .

Complex numbers can be plotted on the complex plane. Thenumber a + bi is placed where the coordinate (a, b) is placedon the Cartesian plane. The horizontal axis is called the realaxis and the vertical axis is called the imaginary axis.

The conjugate of a complex number z , denoted by z , is itsreflection about the real axis. For any z = a + bi we havez = a− bi .

ab = a · b and a + b = a + b.

Re(z) =z + z

2and Im(z) =

z − z

2.

z is real if and only if Im(z) = 0, which occurs when z = z .Similarly a number z is pure imaginary iff z = −z .


Quick Introduction to Complex Numbers

The magnitude of z = a + bi , denoted by |z |, is its distancefrom the origin in the complex plane. If z = a + bi then|z | =

√a2 + b2.

Notice that for any complex z , zz = |z |2.

|a− b| is the distance between a and b.

A complex number z can also be expressed in polar form asr(cos θ + i sin θ) for a real r and angle θ, where r = |z | and θis the angle formed by the positive real axis and the raystarting at the origin pointing towards z , measuredcounterclockwise.

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 thecomplex plane - a circle centered at the origin with radius 1.

For any angle θ we have cis θ =1

cis θ= cis (−θ)


Complex Bash

We can put entire geometry diagrams onto the complex plane.Each point is represented by a complex number, and each line orcircle is represented by an equation in terms of some complex zand possibly its conjugate z . By standard, the complex numbercorresponding to a point is denoted by the lowercase character ofof the point’s label (for example, if A = (2, 3), then a = 2 + 3i).

This is called complex bashing, and can be extremely powerful.


Translations

The greatest advantage of using complex bash as opposed to anyother type of bash is the simplicity of performing the basictransformations of translation, rotation, and dilation.

To perform a translation of a units right and b units up to somecomplex number z , notice that complex number addition works thesame 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.


Homothety (Dilation)

To find the image of some point z after a homothety centered atthe origin with scale factor k , simply multiply z by k.

If the center is some point p instead, one can find the image bytranslating everything such that p is at the origin, then performingthe homothety centered at the origin, then translating everythingagain such that p is back at its original place.

Hence the image of z under a homothety centered at p with scalefactor k is k(z − p) + p.


Rotations

A well-known theorem states that (cis θ)(cis φ) = cis (θ + φ).Thus, to rotate a complex number z = rcis φ by θ about theorigin, 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 theorigin, then translate everything +p.

Hence the result of rotating z by θ about p is (z − p)cis (θ) + p.


A Quick Example - Problem

The points (0, 0), (a, 11), and (b, 37) are the vertices of anequilateral triangle. Find the value of ab. (1994 AIME #8)


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◦, soQ is the result of rotating P around O by 60◦. But then we knowthat

q = p · cis 60◦ =⇒ b + 37i =

(a

2− 11

√3

2

)+

(a√

3

2+

11

2

)i

Notice that the imaginary part of the left side of this equationmust equal the imaginary part of the right side, so

37i =

(a√

3

2+

11

2

)i =⇒ a = 21

√3.

Similarly we can equate the real parts, which gives us

b =a

2− 11

√3

2= 5√

3.

Thus the answer is(21√

3) (

5√

3)

= 315 .


Lines and Collinearity

The following theorems are commonly used in complex bash:

Three distinct complex numbers a, b, and c are collinear if

and only ifc − a

b − ais real.

Since a complex number is real if and only if it is equal to itsconjugate, the above means the equation for a line passingthrough 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 BZin the ratio k/(1− k) then z = (b − a)k + a.

In particular, the midpoint of AB isa + b

2.


The Centroid - Problem

Let ABC be a triangle in the complex plane. Find a formula for thecentroid of 4ABC in terms of the complex numbers a, b, and c .


The Centroid - Solution

Let M be the midpoint of BC . Then we know m =b + c

2. A

well-known theorem states that the centroid G divides MA in theratio 1

3 : 23 . Hence,

g = (a−m)1

3+ m =

(a− b + c

2

)1

3+

b + c

2=

a + b + c

3.


Parallelism and Perpendicularity

Given lines AB and CD, one can determine whether or not theyare parallel or perpendicular with the following theorems.

AB||CD ⇐⇒ b − a

d − cis real⇐⇒ b − a

d − c=

(b − a

d − c

).

AB ⊥ CD ⇐⇒ b − a

d − cis imaginary⇐⇒ b − a

d − c= −

(b − a

d − c

).


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.


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

a− b=

(m − b

a− b

)=⇒ z + c − 2b

2(a− b)=

z + c − 2b

2(a− b).

Also ZC should be perpendicular to AB, so

z − c

a− b= −

(z − c

a− b

)=

c − z

a− b.


Reflections - Solution

We now have

z + c − 2b

2(a− b)=

z + c − 2b

2(a− b)and

z − c

a− b=

c − z

a− b.

This is a system of equations in z and z . Solving for z and z , wesee that

z =ac + ba− ab − bc

a− b.


Using the Unit Circle

If z is on the unit circle, |z | = 1, so by the theorem zz = |z |2

we get z =1

z.

This substitution is useful.

Scaling a diagram does not change length ratios or angles

Scale diagram so that an important circle or some importantpoints are on the unit circle.

If problem involves a “central” triangle, let the three verticesof the triangle be on the unit circle.


The Orthocenter - Problem

Let ABC be a triangle inscribed in the complex unit circle, and leth = a + b + c . Prove that H is the orthocenter of 4ABC .


The Orthocenter - Solution

Let us define

k =h − a

b − c=

(a + b + c)− a

b − c=

b + c

b − c.

Notice that

k =

(b + c

b − c

)=

b + c

b − c=

1/b + 1/c

1/b − 1/c=

c + b

c − b= −b + c

b − c= −k.

Hence k is imaginary, so HA ⊥ BC . Similarly, HB ⊥ CA andHC ⊥ AB, so H is the orthocenter of 4ABC , as desired.


Euler Line

A famous theorem by Euler states that in any triangle, thecircumcenter, the centroid, and the orthocenter are collinear. Theline passing through these three points is called the Euler Line ofthe triangle.

To prove this, take any 4ABC andinscribe it in the complex unit circle.Let H be the orthocenter, O be thecircumcenter, and G be the centroid.

o = 0.

g =a + b + c

3.

h = a + b + c .

Theng − o

h − o=

(a + b + c)/3

a + b + c=

1

3∈ R, so O, G , H are collinear.


Outer Napoleon Triangle - Problem

Given any 4ABC , construct point EA such that BEAC is anequilateral triangle, with EA being on the opposite side of BC asA. Let NA be the centroid of 4BEAC . Similarly define NB andNC . Prove that 4NANBNC is an equilateral triangle.


Outer Napoleon Triangle - Solution

We assume that 4ABC is oriented counter-clockwise (if it isoriented 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

2+

a− b

2

√3i .

Then since NC is the centroid of AECB,

nc =ec + a + b

3=

a + b

2+

a− b

6

√3i .

Similarly, na =b + c

2+

b − c

6

√3i and nb =

c + a

2+

c − a

6

√3i .


Outer Napoleon Triangle - Solution

To prove NANBNC is equilateral, it suffices to show that NB is theresult of rotating NA by 60◦ about NC . But doing so is equivalentto showing that (na − nc)cis 60◦ + nc = nb, which can be donewith the calculations below.

(na − nc )cis 60◦ + nc

=

(b + c

2+

b − c

6

√3i −

a + b

2−

a − b

6

√3i

)(1

2+ i

√3

2

)+

a + b

2+

a − b

6

√3i

=

(c − a

2+

2b − a − c

6

√3i

)(1

2+ i

√3

2

)+

a + b

2+

a − b

6

√3i

=c − a

4+

2b − a − c

12

√3i +

c − a

4

√3i −

2b − a − c

4+

a + b

2+

a − b

6

√3i

=(c − a)− (2b − a − c) + 2(a + b)

4+

(2b − a − c) + 3(c − a) + 2(a − b)

12

√3i

=c + a

2+

c − a

6

√3i = nb.

And hence 4NANBNC is equilateral, as desired. �


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 arcAQB of ω such that XT is perpendicular to AX . Let M denotethe midpoint of chord ST . As X varies on segment PQ, show thatM moves along a circle. (2015 USAMO #2)


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 andb = −1. Now notice that APBQ is a kite, so PQ ⊥ AB. But ABcoincides with the real axis, hence PQ is perpendicular to the realaxis and thus all points on PQ have the same real part. ThereforeRe(x) is constant.



Since A, X , and S are collinear, we have

x − 1

s − 1=

(x − 1

s − 1

)=

x − 1

1/s − 1.

Since XT is perpendicular to AS , we have

x − t

s − 1= −

(x − t

s − 1

)= − x − 1/t

1/s − 1.

Solving these equations for x and x , we get

x =1

2

(1 + s + t − s

t

)and x =

1

2

(1 +

1

s+

1

t− t

s

).



We now see that

Re(x) =x + x

2=

1

4

(2 + s + t +

1

s+

1

t− s

t− t

s

).

Let z =1

2. Notice that

|m − z |2 =

∣∣∣∣s + t − 1

2

∣∣∣∣2 =

(s + t − 1

2

)(s + t − 1

2

)=

1

4(s + t − 1)

(1

s+

1

t− 1

)=

1

4

(3− s − t − 1

s− 1

t+

s

t+

t

s

)=

5

4− Re(x).

Hence all possible M are located on a circle centered at Z with

radius

√5

4− Re(x). �


A Nontrivial Mess - Challenge

Let ABC be a scalene triangle. Let Ka, La and Ma be therespective intersections with BC of the internal angle bisector,external angle bisector, and the median from A. The circumcircleof AKaLa intersects AMa a second time at point Xa different fromA. Define Xb and Xc analogously. Prove that the circumcenter ofXaXbXc lies on the Euler line of ABC. (2015 USA TSTST #2)

Good luck!

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Geometry in the Complex Plane - UNC Math ContestGeometry in the Complex Plane Hongyi Chen on UNC...

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