Bashing Geometry with Complex Numbers
Evan Chen《陳誼廷》
29 August 2015
This is a (quick) English translation of the complex numbers note I wrote
for Taiwan IMO 2014 training. Incidentally I was also working on an airplane.
1 The Complex Plane
Let C and R denote the set of complex and real numbers, respectively.
Each z ∈ C can be expressed as
z = a + bi = r (cos θ + i sin θ) = re
iθ
where a, b, r, θ ∈ R and 0 ≤ θ < 2π. We write |z| = r =
√
a
2
+ b
2
and arg z = θ.
More importantly, each
z
is associated with a conjugate
z = a − bi
. It satises the
properties
w ±z = w ± z
w · z = w · z
w/z = w/z
|z|
2
= z · z
Note that z ∈ R ⇐⇒ z = z and z ∈ iR ⇐⇒ z + z = 0.
Im
Re
0
z = 3 + 4i
z = 3 − 4i
−1 − 2i
|z| = 5
θ
Figure 1: Points z = 3 + 4i and −1 − 2i; z = 3 − 4i is the conjugate.
We represent every point in the plane by a complex number. In particular, we’ll use a
capital letter (like Z) to denote the point associated to a complex number (like z).
1
Evan Chen《陳誼廷》 — 29 August 2015 Bashing Geometry with Complex Numbers
Complex numbers add in the same way as vectors. The multiplication is more interest-
ing: for each z
1
, z
2
∈ C we have
|z
1
z
2
| = |z
1
||z
2
| and arg z
1
z
2
= arg z
1
+ arg z
2
.
This multiplication lets us capture a geometric structure. For example, for any points
Z
and W we can express rotation of Z at W by 90
◦
as
z 7→ i(z − w) + w.
Im
Re
0
z
w
i(z − w) + w
z − w
i(z − w)
Im
Re
0
z = 3 + 4i
iz = −4 + 3i
Figure 2: z 7→ i(z − w) + w.
2 Elementary Propositions
First, some fundamental formulas:
Proposition 1. Let
A
,
B
,
C
,
D
be pairwise distinct points. Then
AB ⊥ CD
if and
only if
d−c
b−a
∈ iR; i.e.
d − c
b − a
+
d − c
b − a
= 0.
Proof. It’s equivalent to
d−c
b−a
∈ iR ⇐⇒ arg
d−c
b−a
≡ ±90
◦
⇐⇒ AB ⊥ CD.
Proposition 2. Let
A
,
B
,
C
be pairwise distinct points. Then
A
,
B
,
C
are collinear if
and only if
c−a
c−b
∈ R; i.e.
c − a
c − b
=
c − a
c − b
.
Proof. Similar to the previous one.
Proposition 3. Let
A
,
B
,
C
,
D
be pairwise distinct points. Then
A
,
B
,
C
,
D
are
concyclic if and only if
c − a
c − b
:
d − a
d − b
∈ R.
Proof.
It’s not hard to see that
arg
c−a
c−b
= ∠ACB
and
arg
d−a
d−b
= ∠ADB
. (Here
angles are directed).
2
Evan Chen《陳誼廷》 — 29 August 2015 Bashing Geometry with Complex Numbers
Im
Re
Im
Re
0
a
b
c
d
0
b − a
d − c
Figure 3: AB ⊥ CD ⇐⇒
d−c
b−a
∈ iR.
Now, let’s state a more commonly used formula.
Lemma 4 (Reection About a Segment). Let
W
be the reection of
Z
across
AB
. Then
w =
(a − b)z + ab − ab
a − b
.
Of course, it then follows that the foot from Z to AB is exactly
1
2
(w + z).
Im
Re
0
1
a
b
z
w
Im
Re
0
1
b − a
z − a
w − a
Im
Re
0
1
z−a
b−a
w−a
b−a
Figure 4: The reection of Z across AB.
Proof. According to Figure 4 we obtain
w − a
b − a
=
z − a
b − a
=
z − a
b − a
.
From this we derive w =
(a−b)z+ab−ab
a−b
.
Here are two more formulas.
Theorem 5 (Complex Shoelace). Let
A
,
B
,
C
be points. Then
4ABC
has signed area
i
4
a a 1
b b 1
c c 1
.
In particular,
A
,
B
,
C
are collinear if and only if this determinant vanishes.
3
Evan Chen《陳誼廷》 — 29 August 2015 Bashing Geometry with Complex Numbers
Proof. Cartesian coordinates.
Often, Theorem 5 is easier to use than Proposition 2.
Actually, we can even write down the formula for an arbitrary intersection of lines.
Proposition 6. Let A, B, C, D be points. Then lines AB and CD intersect at
(¯ab − a
¯
b)(c − d) − (a − b)(¯cd − c
¯
d)
(¯a −
¯
b)(c − d) − (a − b)(¯c −
¯
d)
.
But unless
d = 0
or
a
,
b
,
c
,
d
are on the unit circle, this formula is often too messy to
use.
3 The Unit Circle, and Triangle Centers
On the complex plane, the unit circle is of critical importance. Indeed if
|z| = 1
we have
z =
1
z
.
Using the above, we can derive the following lemmas.
Lemma 7. If
|a| = |b| = 1
and
z ∈ C
, then the reection of
Z
across
AB
is
a + b − abz
,
and the foot from Z to AB is
1
2
(z + a + b − abz) .
Lemma 8. If
A
,
B
,
C
,
D
lie on the unit circle then the intersection of
AB
and
CD
is
given by
ab(c + d) − cd(a + b)
ab − cd
.
These are much easier to work with than the corresponding formulas in general. We
can also obtain the triangle centers immediately:
Theorem 9. Let
ABC
be a triangle center, and assume that the circumcircle of
ABC
coincides with the unit circle of the complex plane. Then the circumcenter, centroid, and
orthocenter of ABC are given by 0,
1
3
(a + b + c), a + b + c, respectively.
Observe that the Euler line follows from this.
Proof.
The results for the circumcenter and centroid are immediate. Let
h = a + b + c
.
By symmetry it suces to prove AH ⊥ BC. We may set
z =
h − a
b − c
=
b + c
b − c
.
Then
z =
b + c
b − c
=
b + c
b − c
=
1
b
+
1
c
1
b
−
1
c
=
c + b
c − b
= −z
so z ∈ iR as desired.
We can actually even get the formula for the incenter.
Theorem 10. Let triangle
ABC
have incenter
I
and circumcircle
Γ
. Lines
AI
,
BI
,
CI
meet
Γ
again at
D
,
E
,
F
. If
Γ
is the unit circle of the complex plane then there exists
x, y, z ∈ C satisfying
a = x
2
, b = y
2
, c = z
2
and d = −yz, e = −zx, f = −xy.
Note that |x| = |y| = |z| = 1. Moreover, the incenter I is given by −(xy + yz + zx).
Proof. Show that I is the orthocenter of 4DEF .
4
Evan Chen《陳誼廷》 — 29 August 2015 Bashing Geometry with Complex Numbers
4 Some Other Lemmas
Lemma 11. Let A, B be on the unit circle and select P so that P A, P B are tangents.
Then
p =
2
a + b
=
2ab
a + b
.
Proof.
Let
M
be the midpoint of
AB
and set
O = 0
. One can show
OM · OP = 1
and
that O, M, P are collinear; the result follows from this.
a
b
2ab
a+b
Figure 5: Two tangents. p =
2
a+b
.
Lemma 12. For any x, y, z, the circumcenter of 4XY Z is given by
x x¯x 1
y y¯y 1
z z¯z 1
÷
x ¯x 1
y ¯y 1
z ¯z 1
.
This formula is often easier to apply if we shift
z
to the point
0
rst, then shift back
afterwards.
5 Examples
Example 13 (MOP 2006). Let
H
be the orthocenter of triangle
ABC
. Let
D
,
E
,
F
lie on the circumcircle of
ABC
such that
AD k BE k CF
. Let
S
,
T
,
U
respectively
denote the reections of
D
,
E
,
F
across
BC
,
CA
,
AB
. Prove that points
S
,
T
,
U
,
H
are concyclic.
Proof.
Let
(ABC)
be the unit circle and
h = a + b + c
. WLOG,
AD
,
BE
,
CF
are
perpendicular to the real axis (rotate appropriately); thus
d = a
and so on. Thus
s = b + c − bcd = b + c − abc and so on; we now have
s − t
s − u
=
b − a
c − a
and
h − t
h − u
=
b + abc
c + abc
.
Compute
s − t
s − u
:
h − t
h − u
=
(b − a)(c + abc)
(c − a)(b + abc)
=
1
b
−
1
a
1
c
+
1
abc
1
c
−
1
a
1
b
+
1
abc
=⇒
s − t
s − u
:
h − t
h − u
∈ R
as desired.
5
Evan Chen《陳誼廷》 — 29 August 2015 Bashing Geometry with Complex Numbers
A
B C
I
E
F
D
G
H
Q M
Example 14 (Taiwan TST 2014). In
4ABC
with incenter
I
, the incircle is tangent to
CA
,
AB
at
E
,
F
. The reections of
E
,
F
across
I
are
G
,
H
. Let
Q
be the intersection of
GH
and
BC
, and let
M
be the midpoint of
BC
. Prove that
IQ
and
IM
are perpendicular.
Solution.
Let
D
be the foot from
I
to
BC
, and set
(DEF )
as the unit circle. (This lets
us exploit the results of Section 3.) Thus
|d| = |e| = |f | = 1
, and moreover
g = −e
,
h = −f. Let x = d =
1
d
and dene y, z similarly. Then
b =
2
d + f
=
2
x + z
.
Similarly, c =
2
x+y
, so
m =
1
2
(b + c) =
1
x + y
+
1
x + z
=
2x + y + z
(x + y)(x + z)
.
Next, we have Q = DD ∩ GH, which implies
q =
dd(g + h) − gh(d + d)
d
2
− gh
=
1
x
2
−
1
y
−
1
z
−
1
yz
2
x
1
x
2
−
1
yz
=
2x + y + z
x
2
− yz
.
so
m/q =
x
2
− yz
(x + y)(x + z)
.
Now,
m/q =
1
x
2
−
1
yz
1
x
+
1
y
1
x
+
1
z
=
yz − x
2
(x + y)(x + z)
= −m/q
thus m/q ∈ iR, as desired.
Example 15 (USAMO 2012). Let
P
be a point in the plane of
4ABC
, and
γ
a line
through
P
. Let
A
0
, B
0
, C
0
be the points where the reections of lines
P A, P B, P C
with
respect to γ intersect lines BC, AC, AB respectively. Prove that A
0
, B
0
, C
0
are collinear.
Solution.
Let
p = 0
and set
γ
as the real line. Then
A
0
is the intersection of
bc
and
p¯a
.
So, using Proposition 6 we get
a
0
=
¯a(
¯
bc − b ¯c)
(
¯
b − ¯c)¯a − (b − c)a
.
6
Evan Chen《陳誼廷》 — 29 August 2015 Bashing Geometry with Complex Numbers
A
B C
P
A
′
Note that
¯a
0
=
a(b¯c −
¯
bc)
(b − c)a − (
¯
b − ¯c)¯a
.
Thus by Theorem 5, it suces to prove
0 =
¯a(
¯
bc−b¯c)
(
¯
b−¯c)¯a−(b−c)a
a(b¯c−
¯
bc)
(b−c)a−(
¯
b−¯c)¯a
1
¯
b(¯ca−c¯a)
(¯c−¯a)
¯
b−(c−a)b
b(c¯a−¯ca)
(c−a)b−(¯c−¯a)
¯
b
1
¯c(¯ab−a
¯
b)
(¯a−
¯
b)¯c−(a−b)c
c(a
¯
b−¯ab)
(a−b)c−(¯a−
¯
b)¯c
1
.
This is equivalent to
0 =
¯a(
¯
bc − b ¯c) a(
¯
bc − b ¯c) (
¯
b − ¯c)¯a − (b − c)a
¯
b(¯ca − c¯a) b( ¯ca − c¯a) ( ¯c − ¯a)
¯
b − (c − a)b
¯c(¯ab − a
¯
b) c(¯ab − a
¯
b) (¯a −
¯
b)¯c − (a − b)c
.
Evaluating the determinant gives
X
cyc
((
¯
b − ¯c)¯a − (b − c)a) · −
b
¯
b
c ¯c
· (¯ca − c¯a)
¯ab − a
¯
b
or, noting the determinant is b¯c −
¯
bc and factoring it out,
(
¯
bc − c
¯
b)(¯ca − c¯a)(¯ab − a
¯
b)
X
cyc
ab − ac + ¯c¯a −
¯
b¯a
= 0.
Example 16 (Taiwan TST Quiz 2014). Let
I
and
O
be the incenter and circumcenter
of
ABC
. A line
`
is drawn parallel to
BC
and tangent to the incircle of
ABC
. Let
X
,
Y
be on
`
so that
I
,
O
,
X
are collinear and
∠XIY = 90
◦
. Show that
A
,
X
,
O
,
Y
are
concyclic.
Solution.
Let
X
0
and
Y
0
respectively denote the reections of
X
and
Y
across
I
. Note
that X, Y lie on BC. Also, let P , Q be the intersections of IY with the circumcircle.
Of course,
(ABC)
is the unit circle. Let
j
be the complex number corresponding to
I
(to avoid confusion with i =
√
−1). Thus,
x
0
=
bc − bc
(j − 0) −
j0 − j0
(b − c)
(b − c)(j − 0) − (b − c)(j −0)
=
j ·
c
2
−b
2
bc
j ·
c−b
bc
− (b − c)j
=
j(b + c)
j + bcj
.
We seek y
0
now. Consider the quadratic equation in z given by
z − j
j
+
1
z
− j
j
= 0 ⇐⇒ z
2
− 2jz + j/j = 0.
7
Evan Chen《陳誼廷》 — 29 August 2015 Bashing Geometry with Complex Numbers
Its zeros in
z
are
p
and
q
, which implies that
p + q = 2j
and
pq = j/j
(by Vieta!). From
this we can compute
y
0
=
pq(b + c) − bc(p + q)
pq − bc
=
j(b + c) − 2bcjj
j − bcj
=
j(b + c) − 2bcjj
j − bcj
.
which gives
x = 2j − x
0
=
j(2j − b − c + 2bcj)
j + bcj
and y = 2j − y
0
=
j(2j − b − c)
j − bcj
.
From this we can obtain
y − x = j ·
(2j − b − c)(j + bcj) − (2j −b − c + 2bcj)(j − bcj)
(j − bcj)(j + bcj)
= j ·
2bcj(2j − b − c) − 2bcj(j − bcj)
(j − bcj)(j + bcj)
= j ·
2bcj
j − b − c + bcj
(j − bcj)(j + bcj)
X =
y − x
x
=
2bcj
j − b − c + bcj
(j − bcj)(2j − b − c + 2bcj)
A =
y − a
a
=
j(2j − b − c − a) + abcj
a(j − bcj)
We need to prove
X/A = X/A
. Now set
a = x
2
,
b = y
2
,
c = z
2
,
j = −(xy + yz + zx)
,
j = −
x+y+z
xyz
(this is a dierent
x
,
y
than the points
X
and
Y
.) So, the above rewrites as
X =
2
yz
x
(x + y + z)(
yz
x
(x + y + z) + y
2
+ z
2
+ xy + yz + zx)
−
yz
x
(x + y + z) + xy + yz + zx
y
2
+ z
2
+ 2(xy + yz + zx) + 2
yz
x
(x + y + z)
=
2yz(x + y + z)
2xyz +
P
sym
x
2
y
(y + z)(x
2
− yz) ( x(y + z)(2x + y + z) + 2yz(x + y + z))
8
Evan Chen《陳誼廷》 — 29 August 2015 Bashing Geometry with Complex Numbers
=
2yz(x + y + z)(x + y)(x + z)
(x
2
− yz) (( x
2
+ yz)(y + z) + (xy + yz + zx)(x + y + z))
and
A =
(xy + yz + zx)(x + y + z)
2
− xyz(x + y + z)
x
2
(−(xy + yz + zx) +
yz
x
(x + y + z))
=
(x + y + z)(x + y)(y + z)(z + x)
x(yz − x
2
)(y + z)
thus
X/A =
−2xyz
(x
2
+ yz)(y + z) + (x + y + z)(xy + yz + zx)
=
−
2
xyz
(
1
x
2
+
1
yz
)(
1
y
+
1
z
) + (
1
x
+
1
y
+
1
z
)(
1
xy
+
1
yz
+
1
zx
)
= X/A.
6 Practice Problems
1.
Let
ABCD
be cyclic. Let
H
A
,
H
B
,
H
C
,
H
D
denote the orthocenters of
BCD
,
CDA, DAB, ABC. Show that AH
A
, BH
B
, CH
C
, DH
D
are concurrent.
2.
(China TST 2011) Let
Γ
be the circumcircle of a triangle
ABC
. Assume
AA
0
,
BB
0
,
CC
0
are diameters of
Γ
. Let
P
be a point inside
ABC
and let
D
,
E
,
F
be the feet
from
P
to
BC
,
CA
,
AB
. Let
X
be the reection of
A
0
across
D
; dene
Y
and
Z
similarly. Prove that 4XY Z ∼ 4ABC.
3.
In circumscribed quadrilateral
ABCD
with incircle
ω
, Prove that the midpoint of
AC and the midpoint of BD are collinear with the center of ω.
4. (Simson Line) Let ABC be a triangle and P a point on its circumcircle.
(a)
Let
D
,
E
,
F
be the feet from
P
to
BC
,
CA
,
AB
. Show that
D
,
E
,
F
are
collinear.
(b)
Moreover, prove that the line through these points bisects
P H
, where
H
is
the orthocenter of ABC.
5.
(PUMaC Finals) Let
γ
and
I
be the incircle and incenter of triangle
ABC
. Let
D
,
E
,
F
be the tangency points of
γ
to
BC
,
CA
,
AB
and let
D
0
be the reection of
D
about
I
. Assume
EF
intersects the tangents to
γ
at
D
and
D
0
at points
P
and
Q. Show that ∠DAD
0
+ ∠P IQ = 180
◦
.
6.
(Schier Point) Let triangle
ABC
have incenter
I
. Prove that the Euler lines of
4AIB, 4BIC, 4CIA, 4ABC are concurrent.
7.
(USA TST 2014) Let
ABCD
be a cyclic quadrilateral and let
E
,
F
,
G
,
H
be the
midpoints of
AB
,
BC
,
CD
,
DA
. Call
W
,
X
,
Y
,
Z
the orthocenters of
AHE
,
BEF
,
CF G, DGH. Prove that ABCD and W XY Z have the same area.
8.
(Iran 2004) Let
O
be the circumcenter of
ABC
. A line
`
through
O
cuts
AB
and
AC
at points
X
and
Y
. Let
M
and
N
be the midpoints of
BY
,
CX
. Show that
∠MON = ∠BAC.
9.
(APMO 2010) Let
ABC
be an acute triangle, where
AB > BC
and
AC > BC
.
Denote by
O
and
H
the circumcenter and orthocenter. The circumcircle of
AHC
intersects
AB
again at
M
; the circumcircle of
AHB
intersects
AC
again at
N
.
Prove that the circumcenter of triangle MNH lies on line OH.
9
Evan Chen《陳誼廷》 — 29 August 2015 Bashing Geometry with Complex Numbers
10.
(Iran 2013) Let
ABC
be acute, and
M
the midpoint of minor arc
d
BC
. Let
N
be
on the circumcircle of ABC such that AN ⊥ BC, and let K, L lie on AB, AC so
that
OK k M B
,
OL k M C
. (Here
O
is the circumcenter of
ABC
). Prove that
NK = NL.
11.
(MOP 2006) Cyclic quadrilateral
ABCD
has circumcenter
O
. Let
P
be a point
in the plane and let
O
1
,
O
2
,
O
3
,
O
4
be the circumcenters of
P AB
,
P BC
,
P CD
,
P DA. Show that the midpoints of O
1
O
3
, O
2
O
4
, OP are concurrent.
10