Geometry with Complex Numbers Mihai Caragiu Ohio Northern University Abstract: In the last three years, Ohio Northern University hosted a Summer Honors Institute for gifted high school students. The week- long "Geometry with Complex Numbers" course was offered in 2006 and 2007. The students were not assumed to have prior knowledge of complex numbers. In this talk I would like to share the experience we had with introducing geometrical transformations (such as rotations, reflections and projections) with complex numbers to talented high- school students, and we will explore ways in which they can be used to quickly derive elegant geometrical results including (but not limited to) the Simson's Line and the Nine Point Circle
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Geometry with Complex Numbers Mihai Caragiu Ohio Northern University Abstract: In the last three years, Ohio Northern University hosted a Summer Honors.
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Geometry with Complex Numbers
Mihai Caragiu
Ohio Northern University
Abstract: In the last three years, Ohio Northern University hosted a Summer Honors Institute for gifted high school students. The week-long "Geometry
with Complex Numbers" course was offered in 2006 and 2007. The students were not assumed to have prior knowledge of complex numbers. In this talk I
would like to share the experience we had with introducing geometrical transformations (such as rotations, reflections and projections) with complex numbers to talented high-school students, and we will explore ways in which they can be used to quickly derive elegant geometrical results including (but
not limited to) the Simson's Line and the Nine Point Circle
“The course blends algebra and geometry together in order to help students
understand the interconnections between the two subjects. Students will use
experimental activities, projects and mathematical software systems to demonstrate
how geometric shapes and concepts can be realized in the complex plane.”
ACKNOWLEDGEMENTS
Dr. Donald Hunt
Dr. Harold Putt
Dr. Rich Daquila
ONU undergraduates which helped with the Summer
Camp activities, both mathematical and recreational.
CAMP ACTIVITIES (June 10-15, 2007) – OUTLINE
Trigonometry and Geometry - basics
Introduction to Complex Variables
Geometrical Transformations with Complex Numbers
Group Projects: The Nine Point Circle and The Simson Line
2
First we argue for the necessity of extending the set of real numbers to create
a domain that contains solutions of quadratic equations as simple as +1=0
x
IMAGINARY UNIT
2 1i
Getting used with "imaginary numbers"
is hard for non-mathematicians!...
"Not only the practical man, but also men of letters
and philosophers have expressed bewilderment at
the devotion of mathematicians to mysterious entities
which by their very name are confessed to be imaginary."
(A. N. Whitehead, ,
Oxford University Press, 1958, Ch.7)
An Introduction to Mathematics
SUMMARY COMPLEX NUMBERS AND TRANSFORMATIONS ALTITUDES AND ORTHOCENTERS
THE NINE-POINT CIRCLE INSCRIPTIBLE QUADILATERALS
THE SIMSON'S LINE
A GENERALIZATION OF THE
THEOREM ABOUT THE SIMSON'S LINE
SIMSON LINES AND EULER CIRCLES
ON A PUTNAM PROBLEM ON PLANE ROTATIONS
1 1 2 2 1 2 1 2
1 1 2 2 1 2 1 2 1 2 2 1
= | ,x iy x y
x iy x iy x x i y y
x iy x iy x x y y i x y x y
C R
2 2
For define
Re( ) , Im( ) , ,
z x iy
z x z y z x iy z x y z z
so that...
1 1 22
2 2
Re( ) , Im( ) , 2 2
z z zz z z zz z
i z z
FIRST CONTACT: RECTANGULAR FORM
REACTANGULAR REPRESENTATION:
GEOMETRICAL CONNECTIONS
may be seen asz x iy The point , in the plane x y
The plane vector , x y
1z
2z 2 1 z z
Thus the transformation given by
is a TRANSLATION with the vector corresponding
to the complex number .
z w w z k
k
reflection about the axis.w z x reflection about the origin.w z
reflection about the axis.w z y
RECTANGULAR REPRESENTATION ADDITION FRIENDLY in the sense
that the addition of vectors has an obvious geometrical meaning (addition of vectors).
However, there is a nice geometrical meaning involving
multiplication and the complex conjugate function for
complex numbers in rectangular form:
1 2
1 2
1 1
2 1 2 2 1 1 1 2 1 2 1 2 2 1
2 22
OP OPArea OP P
P zz z x iy x iy x x y y i x y x y
P z
1 1 2 2CONSEQUENCES: Let A ,B ,..., , ,...a b P z P z
* The general eq. of a line perpendicular to is Re constant. AB z b a
* The general eq. of a line parallel to is Im constant. AB z b a
1 2 3 4 4 3 2 1* The segments and are perpedicular Re z =0. PP P P z z z4 3
1 2 3 42 1
z* and are perpedicular purely imaginary.
z
PP P Pz z
4 31 2 3 4
2 1
z* and are parallel purely real.
z
PP P Pz z
POLAR AND EXPONENTIAL FORMS
cos sin 0, 2iz r i re r R Z
"MULTIPLICATIVE FRIENDLY":
1 2
1 2 1 2 1 2 1 2 1 2cos sin . iz z r r i r r e
SPECIAL CASE
CONSIDER COMPLEX NUMBERS WITH u 1, cos sin . i
u u i e
iii i
i
u ew uz e re re
z re
THE TRANSFORMATION
IS A ROTATION WITH ANGLE ABOUT THE ORIGIN w uz
"POMPEIU's PROBLEM"With the distances PA , PB , PC from an arbitrary point P to the vertices
of an equilateral triangle ABC one can build a triangle.
1 3Let = cos sin .
2 2 3 3
i i
3 21, 1. Consider the equilateral
triangle with vertices A 0 ,B 1 , and
an arbitrary point P .
C
z
If we rotate P with about the origin, we get P . Then the sides of the triangle 3
with vertices , , are identical with the distances z , 1 and from
to the vertices 0,1, of the equila
z z P PC
z z zz zteral triangle we considered initially.
221 and 1 1 z z z z z z z z z
1 2 3 1 2 3 1 2 3 1 2 3
1 3 1 2
Let , , , , , be complex numbers. Then the triangles z and
are similar if the way of obtaining the complex segment z z from z z is the same
as the way of obtaining the complex s
z z z w w w z z w w w
1 3 1 2egment from .w w w w
1 3 1 2
1 1 1 3
1 2
For example, if can be obtained from by a rotation with an angle of around5
followed by a dilation by a factor of 2 with respect to , then can be obtained
from by a rotation
z z z z
z z w w
w w1
1
with an angle of around followed by a dilation by a 5
factor of 2 with respect to .
w
w
SIMILARITY - a problem of rotations and homotheties...
3 1 3 11 2 3 1 2 3
2 1 2 1
z zIf z z z ~ and if , then z z
w ww w w a aw w
3 1 3 11 2 3 1 2 3
2 1 2 1
z zz z z ~ if and only if z z
w ww w w w w
GENERAL ROTATIONS IN THE PLANE
0 0 0
The transformation formula for a counterclockwise rotation by an
angle about a point is . iz w z e z z
01 i iw e z e z
GENERAL FORM OF DIRECT MOTIONS
(orientation-preserving isometries)
a translation, if 2 , or otherwise
is a rotation of angle around 1
i
i
w e z c c
e
Z
1 2
1 2
1 2 2 1 1 2
3 1 21 2
1 2
, if 2 R
(translation) if 2
COMPOSITIONS: R R ; k
k k
k k k k k k
R zz R z
T
z T z T R z R z
T T T T T
Z
Z
Let , , . Let . We need to find the 1 2 1 2
reflection of about the line through and . 1 2
z z z z z
w z z z
C C
REFLECTIONS: orientation-reversing isometries
1 2 1Thus 1 2 1 2 1 12 1 2 1
z z z zw z z z z z z zz z z z
1Let = . Then and .1 2 1 1 2 12 1
z zz z z z w z z zz z
2 11 12 1
z zw z z zz z
GENERAL FORM OF INVERSE MOTIONS
(orientation-reversing isometries)
iw e z cDirect motions - composition of two reflections.
Inverse motions - composition of three reflections.
2 1Back to the reflection formula 1 12 1
z zw z z zz z
1 2
1 2 1 21 2
It is particularly useful to consider the case in which , are points on the unit
1 1circle, that is, 1, or and . The reflection of becomes
z z
z z z z w zz z
1 2 1 2 w z z z z z
1 2 1 2As a corollary, if 1, the projection of onto the line through , is z z z z z
1 2 1 2
1
2 2
z wp z z z z z z
A PRACTICAL ADVICE
1 11 2
2 1 2 1
Line through , : z z z z
z zz z z z
2 1 2 1 1 2 2 1z z z z z z z z z z
1 2We get a simpler form if we can assume z 1, since the conjugate
function for numbers of modulus one is the same with the inverse:
z
1 2 1 2 z z z z z z
1 2Let 1. Then if we set , it turns out that the equation
of the tangent line to the unit circle at the point reduces to:
a z z a
a
2 2 z a z a
Equations become simpler if we can make the assumption that some
of the complex numbers involved have modulus 1.
ORTHOCENTERSThe effectiveness of complex numbers in solving problems of triangle geometry
may increase if we assume that the vertices of the triangle ABC under discussion
are points on the unit circle: 1. a b c
THEOREM. Let be the orthocenter of the triangle ABC,
with 1. Then .
H h
a b c h a b c
Check that : h a c b
_______ _______
_______
1 1
1 1
h a b c b c h ab cc b c b c bc b
c b
Similarly and . h b c a h c b a
THE FEET OF THE ALTITUDES...
Now that we discussed about orthocenters, it makes senseto find out the complex numbers corresponding to the feet
of the altitudes of our triangle ABC remember, 1 . a b c
1 2 1 2
1 2 1 2
1Recall that represents the projection of
2onto the line through and where 1.
z z z z z z z
z z z z
Thus the foot of the altitude from A is
1 1
2 2
bca b c bca a b c
a
1Similarly, the foot of the altitude from B is , and
2
aca b c
b
1the foot of the altitude from C is .
2
aba b c
c
ORTHOCENTER a b c
THE NINE POINT CIRCLE
CENTROID ("center of gravity")3
a b c
So far we are aware of and 1 3
What about : other than being the midpoint of ?2
a b c a b ch g
a b ce OH
* The midpoints , , of the sides , , are at a distance 2 2 2
11 2 from . Indeed, , etc.
2 2 2 2
a b b c c aAB BC CA
a b c a b ce e
1 1 1* The feet of the altitudes , ,
2 2 2
1 1are also at a distance 1 2 from : , etc.
2 2 2
bc ac aba b c a b c a b c
a b c
bc bce e a b c
a a
* The midpoints of the three segments from the orthocenter to the vertices,
1 1 1 , and are at a distance 1 2 from .
2 2 2 a h b h c h e
We need three more points to complete a beautiful theorem!
1 1Indeed,
2 2 2
1, .
2 2
a b ce a h a a b c
aetc
THEOREM (THE NINE POINT CIRCLE, OR THE EULER'S CIRCLE)
Given any triangle , the nine points listed below all lie on the same circle
(Euler's Circle) centered at the midpoint between the circumcenter
ABC
and the
orthocenter of the triangle:
The midpoints of the three sides of the triangle,
The feet of the three altitudes of the triangle, and
The midpoints of the three segments from the orthocenter
to the vertices.
1 2 3 4
1 2 3 4
2 3 4
1 3 4 1 2 4 1 2 3
1 2 3 4
Let , , , represent the vertices of
an inscriptible quadrilateral. Let , , ,
be the Euler circles of the triangles ,
, and , respectively, with
centers , , ,
z z z z
E E E E
z z z
z z z z z z z z z
e e e e
C
1 2 3 4
1 2 3 4
.Then , , , have a
common point. For the proof we may assume,
as usual, that 1.
E E E E
z z z z
2 3 4 1 3 41 2
1 2 31 2 43 4
1 2 3 4
, ,2 2
, .2 2
Define : .2
z z z z z ze e
z z zz z ze e
z z z ze
1 2 3 4 1 2 3 4
1 belongs to each all Euler circles , , , .
2e e e e e e e e e E E E E
INSCRIPTIBLE QUADRILATERALS
1 2 3 4
1 2 3 4
, , , lie on the same (blue) circle of radius 1 2 centered at .
This will be the "Euler Circle of the inscriptible quadrilateral "
e e e e e
z z z z
1 2 1 2
1 2 1 2
RECALL: If 1, then the projection of onto the line through , is
1
2
z z z z z
z z z z z z
BACK TO PROJECTIONS!!
1 2 3 1 1 2 2 3 3 1 2 3
1 2 3
Consider a triangle with , , 1,
and let P z be an arbitrary point in the plane of the triangle .
A A A A z A z A z z z z
A A A
1 1 2 2 3 3 2 3 1 3 1 2Let , , be the projections of P onto the sides , and respectively. P p P p P p A A A A A A
1 2 3 2 3
1
2 p z z z z z z
2 1 3 1 3
1
2 p z z z z z z
3 1 2 1 2
1
2 p z z z z z z
SIMSON's THEOREM
1 2 3
1 2 3
The three projections, , , are collinear if and only if 1, that is,
if and only if is on the circumcircle of the triangle .
p p p z
P z A A A
2 1 1 3 1 3 2 3 2 3 2 1 3
1 1 11
2 2 2 p p z z z z z z z z z z z z z z z z
3 1 1 2 1 2 2 3 2 3 3 1 2
1 1 11
2 2 2 p p z z z z z z z z z z z z z z z z
___________
3 1 3 11 2 3
2 1 2 1
, , collinear
p p p pp p p
p p p p
3 1 2 3 1 2
2 1 3 2 1 3
1 1
1 1
z z z z z z z z
z z z z z z z z
3 1 2 3 1 2
2 1 3
2 1 3
1 11
1
1 1 11
zz z z z z z z
z z z z zz z z
2 2
3 3
1
1
z z z z
z z z z 2
2 3 1 0 z z z
z 1
P
R
O
O
F
SIMSON's THEOREM A GENERALIZATION
1 2 3
1 2 3
With the previous notation in place, we want to characterize the set of all
points P z such that the (oriented) area of the triangle determined