Advanced Euclidean Geometry
Paul Yiu
Summer 2013
Department of Mathematics
Florida Atlantic University
A
b
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B
a
August 2, 2013
Summer 2013
C
Contents
1 Some Basic Theorems
1.1
The Pythagorean Theorem . . . . . . . . .
1.2
Constructions of geometric mean . . . . .
1.3
The golden ratio . . . . . . . . . . . . . .
1.3.1 The regular pentagon . . . . . . . .
1.4
Basic construction principles . . . . . . .
1.4.1 Perpendicular bisector locus . . . .
1.4.2 Angle bisector locus . . . . . . . .
1.4.3 Tangency of circles . . . . . . . . .
1.4.4 Construction of tangents of a circle
1.5
The intersecting chords theorem . . . . . .
1.6
Ptolemy’s theorem . . . . . . . . . . . . .
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101
101
104
106
106
108
108
109
110
110
112
114
2 The laws of sines and cosines
2.1
The law of sines . . . . . . . . . .
2.2
The orthocenter . . . . . . . . . .
2.3
The law of cosines . . . . . . . . .
2.4
The centroid . . . . . . . . . . . .
2.5
The angle bisector theorem . . . .
2.5.1 The lengths of the bisectors
2.6
The circle of Apollonius . . . . . .
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115
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116
117
120
121
121
123
3 The tritangent circles
3.1
The incircle . . . . . . . . . . . . . . .
3.2
Euler’s formula . . . . . . . . . . . . .
3.3
Steiner’s porism . . . . . . . . . . . . .
3.4
The excircles . . . . . . . . . . . . . . .
3.5
Heron’s formula for the area of a triangle
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125
125
128
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130
131
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133
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134
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135
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138
140
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4 The arbelos
4.1
Archimedes’ twin circles . . . . . . . . . . . . . .
4.1.1 Harmonic mean and the equation a1 + 1b = 1t
4.1.2 Construction of the Archimedean twin circles
4.2
The incircle . . . . . . . . . . . . . . . . . . . . .
4.2.1 Construction of the incircle of the arbelos . .
4.2.2 Alternative constructions of the incircle . . .
4.3
Archimedean circles . . . . . . . . . . . . . . . . .
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iv
5 Menelaus’ theorem
5.1
Menelaus’ theorem . . . . . . . .
5.2
Centers of similitude of two circles
5.2.1 Desargue’s theorem . . . . .
5.3
Ceva’s theorem . . . . . . . . . . .
5.4
Some triangle centers . . . . . . .
5.4.1 The centroid . . . . . . . .
5.4.2 The incenter . . . . . . . .
5.4.3 The Gergonne point . . . .
5.4.4 The Nagel point . . . . . . .
5.5
Isotomic conjugates . . . . . . . .
CONTENTS
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201
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205
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206
207
208
6 The Euler line and the nine-point circle
6.1
The Euler line . . . . . . . . . . . . . . . . . . . . . . . .
6.1.1 Inferior and superior triangles . . . . . . . . . . . .
6.1.2 The orthocenter and the Euler line . . . . . . . . . .
6.2
The nine-point circle . . . . . . . . . . . . . . . . . . . . .
6.3
Distances between triangle centers . . . . . . . . . . . . .
6.3.1 Distance between the circumcenter and orthocenter .
6.3.2 Distance between circumcenter and tritangent centers
6.3.3 Distance between orthocenter and tritangent centers .
6.4
Feuerbach’s theorem . . . . . . . . . . . . . . . . . . . . .
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211
211
211
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215
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216
217
219
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221
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233
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235
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238
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301
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306
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7 Isogonal conjugates
7.1
Directed angles . . . . . . . . . . . . . . . . . . . .
7.2
Isogonal conjugates . . . . . . . . . . . . . . . . . .
7.3
The symmedian point and the centroid . . . . . . . .
7.4
Isogonal conjugates of the Gergonne and Nagel points
7.4.1 The Gergonne point and the insimilicenter T+ .
7.4.2 The Nagel point and the exsimilicenter T− . .
7.5
The Brocard points . . . . . . . . . . . . . . . . . .
7.6
Kariya’s theorem . . . . . . . . . . . . . . . . . . . .
7.7
Isogonal conjugate of an infinite point . . . . . . . .
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8 The excentral and intouch triangles
8.1
The excentral triangle . . . . . . . . . . . . . . . . . . . . . . . .
8.1.1 The circumcenter of the excentral triangle . . . . . . . . . .
8.1.2 Relation between circumradius and radii of tritangent circles
8.2
The homothetic center of the intouch and excentral triangles . . . .
9 Homogeneous Barycentric Coordinates
9.1
Absolute and homogeneous barycentric coordinates
9.1.1 The centroid . . . . . . . . . . . . . . . . .
9.1.2 The incenter . . . . . . . . . . . . . . . . .
9.1.3 The barycenter of the perimeter . . . . . . .
9.1.4 The Gergonne point . . . . . . . . . . . . .
9.2
Cevian triangle . . . . . . . . . . . . . . . . . . . .
9.2.1 The circumcenter . . . . . . . . . . . . . . .
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CONTENTS
9.3
9.4
v
9.2.2 The Nagel point and the extouch triangle . . . . .
9.2.3 The orthocenter and the orthic triangle . . . . . .
Homotheties . . . . . . . . . . . . . . . . . . . . . . . .
9.3.1 Superiors and inferiors . . . . . . . . . . . . . . .
9.3.2 Triangles bounded by lines parallel to the sidelines
Infinite points . . . . . . . . . . . . . . . . . . . . . . .
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10 Some applications of barycentric coordinates
10.1 Construction of mixtilinear incircles . . . . . . . . . . . . . . . . . . . . . . .
10.1.1 The insimilicenter and the exsimilicenter of the circumcircle and incircle
10.1.2 Mixtilinear incircles . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 Isotomic and isogonal conjugates . . . . . . . . . . . . . . . . . . . . . . . . .
10.3 Isotomic conjugates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.4 Equal-parallelians point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 Computation of barycentric coordinates
11.1 The Feuerbach point . . . . . . . . . . . . . . . . . . . . . . . . .
11.2 The OI line . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.2.1 The circumcenter of the excentral triangle . . . . . . . . . .
11.2.2 The centers of similitude of the circumcircle and the incircle
11.2.3 The homothetic center T of excentral and intouch triangles .
11.3 The excentral triangle . . . . . . . . . . . . . . . . . . . . . . . .
11.3.1 The centroid . . . . . . . . . . . . . . . . . . . . . . . . .
11.3.2 The incenter . . . . . . . . . . . . . . . . . . . . . . . . .
12 Some interesting circles
12.1 A fundamental principle on 6 concyclic points
12.1.1 The radical axis of two circles . . . . .
12.1.2 Test for 6 concyclic points . . . . . . .
12.2 The Taylor circle . . . . . . . . . . . . . . . .
12.3 Two Lemoine circles . . . . . . . . . . . . . .
12.3.1 The first Lemoine circle . . . . . . . .
12.3.2 The second Lemoine circle . . . . . . .
12.3.3 Construction of K . . . . . . . . . . .
12.3.4 The center of the first Lemoine circle .
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13 Straight line equations
13.1 Area and barycentric coordinates . . . . . . . . . . . . .
13.2 Equations of straight lines . . . . . . . . . . . . . . . . .
13.2.1 Two-point form . . . . . . . . . . . . . . . . . . .
13.2.2 Intersection of two lines . . . . . . . . . . . . . .
13.3 Perspective triangles . . . . . . . . . . . . . . . . . . . .
13.3.1 The Conway configuration . . . . . . . . . . . . .
13.4 Perspectivity . . . . . . . . . . . . . . . . . . . . . . . .
13.4.1 The Schiffler point: intersection of four Euler lines
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307
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312
314
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315
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319
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321
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327
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332
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401
401
402
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403
404
405
407
408
vi
14 Cevian nest theorem
14.1 Trilinear pole and polar . . . . . . . . . .
14.1.1 Trilinear polar of a point . . . . . .
14.1.2 Tripole of a line . . . . . . . . . . .
14.2 Anticevian triangles . . . . . . . . . . . .
14.2.1 Construction of anticevian triangle .
14.3 Cevian quotients . . . . . . . . . . . . . .
14.3.1 The cevian nest theorem . . . . . .
14.3.2 Basic properties of cevian quotients
CONTENTS
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409
409
409
411
412
412
415
415
419
15 Circle equations
15.1 The power of a point with respect to a circle .
15.2 Circle equation . . . . . . . . . . . . . . . . .
15.3 Points on the circumcircle . . . . . . . . . . .
15.3.1 X(101) . . . . . . . . . . . . . . . . .
15.3.2 X(100) . . . . . . . . . . . . . . . . .
15.3.3 The Steiner point X(99) . . . . . . . .
15.3.4 The Euler reflection point E = X(110)
15.4 Circumcevian triangle . . . . . . . . . . . . .
15.5 The third Lemoine circle . . . . . . . . . . .
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421
421
422
423
423
423
424
424
425
426
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Chapter 1
Some Basic Theorems
1.1
The Pythagorean Theorem
Theorem 1.1 (Pythagoras). The lengths a ≤ b < c of the sides of a right triangle satisfy
the relation
a2 + b 2 = c 2 .
Proof.
b
a
3
2
2
a
b
b
4
b
a
b
3
a
4
a
1
1
a
b
Theorem 1.2 (Converse of Pythagoras’ theorem). If the lengths of the sides of △ABC
satisfy a2 + b2 = c2 , then the triangle has a right angle at C.
B
Y
c
a
C
b
a
A
Z
b
X
Proof. Consider a right triangle XY Z with ∠Z = 90◦ , Y Z = a, and XZ = b. By the
Pythagorean theorem, XY 2 = Y Z 2 + XZ 2 = a2 + b2 = c2 = AB 2 . It follows that
XY = AB, and △ABC ≡ △XY Z by the SSS test, and ∠C = ∠Z = 90◦ .
102
Some Basic Theorems
Example 1.1. Trigonometric ratios of 30◦ , 45◦ , and 60◦ :
√
2
√
3
60◦
1
2
45◦
1
1
θ
sin θ cos θ tan θ
◦
30
45◦
60◦
√
√
3
2
√
2
2
1
2
1
√2
2
√2
3
2
3
3
1
√
3
Example 1.2. For an arbitrary point P on the minor arc BC of the circumcircle of an
equilateral triangle ABC, AP = BP + CP .
A
Q
B
C
P
Proof. If Q is the point on AP such that P Q = P C, then the isosceles triangle CP Q is
equilateral since ∠CP Q = ∠CBA = 60◦ . Note that ∠ACQ = ∠BCP . Thus, △ACQ ≡
△BCP by the SAS congruence test. From this, AQ = BP , and AP = AQ + QP =
BP + CP .
1.1 The Pythagorean Theorem
103
Example 1.3. Given a rectangle ABCD, to construct points P on BC and Q on CD such
that triangle AP Q is equilateral.
Let BCY and CDX are equilateral triangles inside the rectangle ABCD. Extend the
lines AX and AY to intersect BC and CD respectively at P and Q. AP Q is equilateral.
Q
D
C
Y
P
X
A
B
√
Proof. Suppose AB = 2a and BC = 2b. The
distance of X above
AB = 2b − 3a. By
√
√
2
2
2
the Pythagorean theorem, AX 2 = a2 +(2b−
√ 3a) = 4(a +b − 2 3ab). X is2 the midpoint
2
2
2
of AP . Therefore,
AP = 16(a +√b − 3ab). Similarly, AQ = (2AY
) =√
4AY 2 =
√
√
2
2
2
4·(b2 +(2a−
CP
2( 3a−b),
√ 3b) ) = 16(a 2+b − 23ab). Finally,
√ = 2b−2(2b−
√ 3a) =
2
2
2
2
CQ =
√ 2( 3b − a), and P Q = CP + CQ = 4(( 3a − b) + ( 3b − a) ) = 16(a +
2
b − 3ab). It follows that AP = AQ = P Q, and triangle AP Q is equilateral.
104
Some Basic Theorems
1.2
Constructions of geometric mean
We present two ruler-and-compass constructions of the geometric means of two quantities given as lengths of segments. These are based on Euclid’s proof of the Pythagorean
theorem.
Construct the altitude at the right angle to meet AB at P and the opposite side ZZ ′ of
the square ABZZ ′ at Q. Note that the area of the rectangle AZQP is twice of the area
of triangle AZC. By rotating this triangle about A through a right angle, we obtain the
congruent triangle ABY , whose area is half of the area of the square on AC. It follows that
the area of rectangle AZQP is equal to the area of the square on AC. For the same reason,
the area of rectangle BZ ′ QP is equal to that of the square on BC. From these, the area of
the square on AB is equal to the sum of the areas of the squares on BC and CA.
X′
Y′
X
C
Y
A
B
P
Z
Q
Z′
Construction. Given two segments of length a < b, mark three points P , A, B on a line
such that P A = a, P B = b, and A, B are on the same side of P . Describe a semicircle
with P B as diameter, and let the perpendicular through A intersect the semicircle at Q.
Then P Q2 = P A · P B, so that the length of P Q is the geometric mean of a and b.
Q
Q
√
ab
P
A
B
a
b
A
P
B
1.2 Constructions of geometric mean
105
Construction. Given two segments of length a, b, mark three points A, P , B on a line (P
between A and B) such that AP = a, P B = b. Describe a semicircle with AB as diameter,
and let the perpendicular through P intersect the semicircle at Q. Then P Q2 = P A · P B,
so that the length of P Q is the geometric mean of a and b.
Q
Q
√
ab
A
a
B
P
b
P
A
B
106
1.3
Some Basic Theorems
The golden ratio
Given a segment AB, a point P in the √segment is said to divide it in the golden ratio if
AP 2 = P B · AB. Equivalently, PAPB = 5+1
. We shall denote this golden ratio by ϕ. It is
2
2
the positive root of the quadratic equation x = x + 1.
M
Q
A
B
P
A
P
B
Construction (Division of a segment in the golden ratio). Given a segment AB,
(1) draw a right triangle ABM with BM perpendicular to AB and half in length,
(2) mark a point Q on the hypotenuse AM such that M Q = M B,
(3) mark a point P on the segment AB such that AP = AQ.
Then P divides AB into the golden ratio.
Suppose P B has unit length. The length ϕ of AP satisfies
ϕ2 = ϕ + 1.
This equation can be rearranged as
1
ϕ−
2
Since ϕ > 1, we have
ϕ=
Note that
2
5
= .
4
1 √
5+1 .
2
AP
ϕ
1
2
=
=
= =√
AB
ϕ+1
ϕ
5+1
√
5−1
.
2
This explains the construction above.
1.3.1 The regular pentagon
Consider a regular pentagon ACBDE. It is clear that the five diagonals all have equal
lengths. Note that
(1) ∠ACB = 108◦ ,
1.3 The golden ratio
107
C
A
B
P
E
D
(2) triangle CAB is isosceles, and
(3) ∠CAB = ∠CBA = (180◦ − 108◦ ) ÷ 2 = 36◦ .
In fact, each diagonal makes a 36◦ angle with one side, and a 72◦ angle with another.
It follows that
(4) triangle P BC is isosceles with ∠P BC = ∠P CB = 36◦ ,
(5) ∠BP C = 180◦ − 2 × 36◦ = 108◦ , and
(6) triangles CAB and P BC are similar.
Note that triangle ACP is also isosceles since
(7) ∠ACP = ∠AP C = 72◦ . This means that AP = AC.
Now, from the similarity of CAB and P BC, we have AB : AC = BC : P B. In other
words AB · AP = AP · P B, or AP 2 = AB · P B. This means that P divides AB in the
golden ratio.
Construction. Given a segment AB, we construct a regular pentagon ACBDE with AB
as a diagonal.
(1) Divide AB in the golden ratio at P .
(2) Construct the circles A(P ) and P (B), and let C be an intersection of these two circles.
(3) Construct the circles A(AB) and B(C) to intersect at a point D on the same side of
BC as A.
(4) Construct the circles A(P ) and D(P ) to intersect at E.
Then ACBDE is a regular pentagon with AB as a diagonal.
108
1.4
Some Basic Theorems
Basic construction principles
1.4.1 Perpendicular bisector locus
A variable point P is equidistant from two fixed points A and B if and only if P lies on the
perpendicular bisector of the segment AB.
The perpendicular bisectors of the three sides of a triangle are concurrent at the circumcenter of the triangle. This is the center of the circumcircle, the circle passing through the
three vertices of the triangle.
A
A
B
O
B
C
O
C
The circumcenter of a right triangle is the midpoint of its hypotenuse.
1.4 Basic construction principles
109
1.4.2 Angle bisector locus
A variable point P is equidistant from two fixed lines ℓ and ℓ′ if and only if P lies on the
bisector of one of the angles between ℓ and ℓ′ .
The bisectors of the three angles of a triangle are concurrent at a point which is at equal
distances from the three sides. With this point as center, a circle can be constructed tangent
to the sides of the triangle. This is the incircle of the triangle. The center is the incenter.
A
Y
Z
B
I
X
C
Proof. Let the bisectors of angles B and C intersect at I. Consider the pedals of I on the
three sides. Since I is on the bisector of angle B, IX = IZ. Since I is also on the bisector
of angle C, IX = IY . It follows IX = IY = IZ, and the circle, center I, constructed
through X, also passes through Y and Z, and is tangent to the three sides of the triangle.
110
Some Basic Theorems
1.4.3 Tangency of circles
Two circles (O) and (O′ ) are tangent to each other if they are tangent to a line ℓ at the
same line P , which is a common point of the circles. The tangency is internal or external
according as the circles are on the same or different sides of the common tangent ℓ.
O
T
I
O
I
T′
The line joining their centers passes through the point of tangency.
The distance between their centers is the sum or difference of their radii, according as
the tangency is external or internal.
1.4.4 Construction of tangents of a circle
A tangent to a circle is a line which intersects the circle at only one point. Given a circle
O(A), the tangent to a circle at A is the perpendicular to the radius OA at A.
A
A
O
O
M
P
B
If P is a point outside a circle (O), there are two lines through P tangent to the circle.
Construct the circle with OP as diameter to intersect (O) at two points. These are the
points of tangency.
The two tangents have equal lengths since the triangles OAP and OBP are congruent
by the RHS test.
1.4 Basic construction principles
111
Example 1.4. Given two congruent circles each with center on the other circle, to construct
a circle tangent to the center line, and also to the given circles, one internally and the other
externally.
X
K
Y
T
A
x
B
Let AB = a. Suppose the required circle has radius r, and AT = x, where T is the
point of tangency T with the center line.
(a + x)2 + r2 = (a + r)2 ,
x2 + r2 = (a − r)2 .
From these, we have
√
3
a
· a,
x+ =
2
2
a
+ x = 2r.
2
√
This means that if M is the midpoint of AB, then M T = 23 · a, which is the height of
the equilateral triangle on AB. In other words, if C is an intersection of the two given circles, then CM and M T are two adjacent sides of a square. Furthermore, the side opposite
to CM is a diameter of the required circle!
C
X
√
3a
2
Y
K
T
x
A
M
a
2
D
B
112
1.5
Some Basic Theorems
The intersecting chords theorem
Theorem 1.3. Given a point P and a circle O(r), if a line through P intersects the circle
at two points A and B, then P A · P B = OP 2 − r2 , independent of the line.
O
O
P
B
B
P
M
M
A
A
Proof. Let M be the midpoint of AB. Note that OM is perpendicular to AB. If P is
outside the circle, then
P A · P B = (P M + M A)(P M − BM )
= (P M + M A)(P M − M A)
= P M 2 − M A2
= (OM 2 + P M 2 ) − (OM 2 + M A2 )
= OP 2 − r2 .
The same calculation applies to the case when P is inside or on the circle, provided that
the lengths of the directed segments are signed.
The quantity OP 2 − r2 is called the power of P with respect to the circle. It is positive,
zero, or negative according as P is outside, on, or inside the circle.
Corollary 1.4 (Intersecting chords theorem). If two chords AB and CD of a circle intersect, extended if necessary, at a point P , then P A · P B = P C · P D.
In particular, if the tangent at T intersects AB at P , then P A · P B = P T 2 .
C
C
D
O
O
P
B
P
A
T
A
The converse of the intersecting chords theorem is also true.
B
D
1.5 The intersecting chords theorem
113
Theorem 1.5. Given four points A, B, C, D, if the lines AB and CD intersect at a point
P such that P A · P B = P C · P D (as signed products), then A, B, C, D are concyclic.
In particular, if P is a point on a line AB, and T is a point outside the line AB such
that P A · P B = P T 2 , then P T is tangent to the circle through A, B, T .
Example 1.5. Let ABC be a triangle with B = 2C. Then b2 = c(c + a).
A
E
B
C
F
Construct a parallel through C to the bisector BE, to intersect the extension of AB at
F . Then
1
∠AF C = ∠ABE = · ∠ABC = ∠ACB.
2
This means that AC is tangent to the circle through B, C, F . By the intersecting chord
theorem, AC 2 = AB · AF , i.e., b2 = c(c + a).
114
1.6
Some Basic Theorems
Ptolemy’s theorem
Theorem 1.6 (Ptolemy). A convex quadrilateral ABCD is cyclic if and only if
AB · CD + AD · BC = AC · BD.
D
A
d
A
D
P′
P
B
C
B
C
Proof. (Necessity) Assume, without loss of generality, that ∠BAD > ∠ABD. Choose a
point P on the diagonal BD such that ∠BAP = ∠CAD. Triangles BAP and CAD are
similar, since ∠ABP = ∠ACD. It follows that AB : AC = BP : CD, and
AB · CD = AC · BP.
Now, triangles ABC and AP D are also similar, since ∠BAC = ∠BAP + ∠P AC =
∠DAC + ∠P AC = ∠P AD, and ∠ACB = ∠ADP . It follows that AC : BC = AD :
P D, and
BC · AD = AC · P D.
Combining the two equations, we have
AB · CD + BC · AD = AC(BP + P D) = AC · BD.
(Sufficiency). Let ABCD be a quadrilateral satisfying (**). Locate a point P ′ such
that ∠BAP ′ = ∠CAD and ∠ABP ′ = ∠ACD. Then the triangles ABP and ACD are
similar. It follows that
AB : AP ′ : BP ′ = AC : AD : CD.
From this we conclude that
(i) AB · CD = AC · BP ′ , and
(ii) triangles ABC and AP ′ D are similar since ∠BAC = ∠P ′ AD and AB : AC = AP ′ :
AD.
Consequently, AC : BC = AD : P ′ D, and
Combining the two equations,
AD · BC = AC · P ′ D.
AC(BP ′ + P ′ D) = AB · CD + AD · BC = AC · BD.
It follows that BP ′ + P ′ D = BD, and the point P ′ lies on diagonal BD. From this,
∠ABD = ∠ABP ′ = ∠ACD, and the points A, B, C, D are concyclic.
Chapter 2
The laws of sines and cosines
2.1
The law of sines
Theorem 2.1 (The law of sines). Let R denote the circumradius of a triangle ABC.
2R =
a
b
c
=
=
.
sin α
sin β
sin γ
A
A
α
F
O
O
E
α
B
C
B
D
C
D
Since the area of a triangle is given by ∆ = 21 bc sin α, the circumradius can be written
as
R=
abc
.
4∆
116
2.2
The laws of sines and cosines
The orthocenter
Why are the three altitudes of a triangle concurrent?
Let ABC be a given triangle. Through each vertex of the triangle we construct a line
parallel to its opposite side. These three parallel lines bound a larger triangle A′ B ′ C ′ . Note
that ABCB ′ and ACBC ′ are both parallelograms since each has two pairs of parallel sides.
It follows that B ′ A = BC = AC ′ and A is the midpoint of B ′ C ′ .
A
C′
B′
Y
Z
B
H
C
X
A′
Consider the altitude AX of triangle ABC. Seen in triangle A′ B ′ C ′ , this line is the
perpendicular bisector of B ′ C ′ since it is perpendicular to B ′ C ′ through its midpoint A.
Similarly, the altitudes BY and CZ of triangle ABC are perpendicular bisectors of C ′ A′
and A′ B ′ . As such, the three lines AX, BY , CZ concur at a point H. This is called the
orthocenter of triangle ABC.
Proposition 2.2. The reflections of the orthocenter in the sidelines lie on the circumcircle.
A
O
H
B
C
Ha
Oa
Proof. It is enough to show that the reflection Ha of H in BC lies on the circumcircle.
Consider also the reflection Oa of O in BC. Since AH and OOa are parallel and have the
same length (2R cos α), AOOa H is a parallelogram. On the other hand, HOOa Ha is a
isosceles trapezoid. It follows that OHa = HOa = AO, and Ha lies on the circumcircle.
2.3 The law of cosines
2.3
117
The law of cosines
Given a triangle ABC, we denote by a, b, c the lengths of the sides BC, CA, AB respectively.
Theorem 2.3 (The law of cosines).
c2 = a2 + b2 − 2ab cos γ.
A
A
B
c
b
c
b
X
C
a
a
B
C
X
Proof. Let AX be the altitude on BC.
c2 = BX 2 + AX 2
= (a − b cos γ)2 + (b sin γ)2
= a2 − 2ab cos γ + b2 (cos2 γ + sin2 γ)
= a2 + b2 − 2ab cos γ.
Theorem 2.4 (Stewart). Let X be a point on the sideline BC of triangle ABC.
a · AX 2 = BX · b2 + XC · c2 − a · BX · XC.
Here, the lengths of the directed segments on the line BC are signed. Equivalently, if
BX : XC = λ : µ, then
AX 2 =
λµa2
λb2 + µc2
−
.
λ+µ
(λ + µ)2
A
b
c
B
X
a
C
Proof. Use the cosine formula to compute the cosines of the angles AXB and AXC, and
note that cos AXC = − cos AXB.
118
The laws of sines and cosines
Example 2.1. (Napoleon’s theorem). If similar isosceles triangles XBC, Y CA and ZAB
(of base angle θ) are constructed externally on the sides of triangle ABC, the lengths
of the segments Y Z, ZX, XZ can be computed easily. For example, in triangle AY Z,
AY = 2b sec θ, AZ = 2c sec θ and ∠Y AZ = α + 2θ.
A
θ
θ
Y
Z
θ
B
θ
θ
θ
C
X
By the law of cosines,
Y Z 2 = AY 2 + AZ 2 − 2AY · AZ · cos Y AZ
sec2 θ 2
=
(b + c2 − 2bc cos(α + 2θ))
4
sec2 θ 2
(b + c2 − 2bc cos α cos 2θ + 2bc sin α sin 2θ)
=
4
sec2 θ 2
=
(b + c2 − (b2 + c2 − a2 ) cos 2θ + 4∆ sin 2θ)
4
sec2 θ 2
=
(a cos 2θ + (b2 + c2 )(1 − cos 2θ) + 4∆ sin 2θ).
4
Likewise, we have
sec2 θ 2
(b cos 2θ + (c2 + a2 )(1 − cos 2θ) + 4∆ sin 2θ),
4
2
sec
θ 2
(c cos 2θ + (a2 + b2 )(1 − cos 2θ) + 4∆ sin 2θ).
XY 2 =
4
ZX 2 =
It is easy to note that Y Z = ZX = XY if and only if cos 2θ = 21 , i.e., θ = 30◦ .
In this case, the points X, Y , Z are the centers of equilateral triangles erected externally
on BC, CA, AB respectively. The same conclusion holds if the equilateral triangles are
constructed internally on the sides. This is the famous Napoleon theorem.
Theorem 2.5 (Napoleon). If equilateral triangles are constructed on the sides of a triangle,
either all externally or all internally, then their centers are the vertices of an equilateral
triangle.
2.3 The law of cosines
119
A
A
Y
Z
Z′
X′
B
C
Y′
X
C
B
Example 2.2. (Orthogonal circles) Given three points A, B, C that form an acute-angled
triangle, construct three circles with these points as centers that are mutually orthogonal to
each other.
Y
A
E
Z
H
F
B
D
C
X
Let BC = a, CA = b, and AB = c. If these circles have radii Ra , Rb , Rc respectively,
then
Rb2 + Rc2 = a2 ,
Rc2 + Ra2 = b2 ,
Ra2 + Rb2 = c2 .
From these,
1
Ra2 = (b2 + c2 − a2 ),
2
1
Rb2 = (c2 + a2 − b2 ),
2
1
Rc2 = (a2 + b2 − c2 ).
2
These are all positive since ABC is an acute triangle. Consider the perpendicular foot E of
B on AC. Note that AE = c cos A, so that Ra2 = 12 (b2 + c2 − a2 ) = bc cos A = AC · AE. It
follows if we extend BE to intersect at Y the semicircle constructed externally on the side
AC as diameter, then, AY 2 = AC · AE = Ra2 . Therefore we have the following simple
construction of these circles.
(1) With each side as diameter, construct a semicircle externally of the triangle.
(2) Extend the altitudes of the triangle to intersect the semicircles on the same side. Label
these X, Y , Z on the semicircles on BC, CA, AB respectively. These satisfy AY = AZ,
BZ = BX, and CX = CY .
(3) The circles A(Y ), B(Z) and C(X) are mutually orthogonal to each other.
120
2.4
The laws of sines and cosines
The centroid
Let E and F be the midpoints of AC and AB respectively, and G the intersection of the
medians BE and CF .
Construct the parallel through C to BE, and extend AG to intersect BC at D, and this
parallel at H. .
A
E
F
G
B
D
C
H
By the converse of the midpoint theorem, G is the midpoint of AH, and HC = 2 · GE
Join BH. By the midpoint theorem, BH//CF . It follows that BHCG is a parallelogram. Therefore, D is the midpoint of (the diagonal) BC, and AD is also a median of
triangle ABC. We have shown that the three medians of triangle ABC intersect at G,
which we call the centroid of the triangle.
Furthermore,
AG =GH = 2GD,
BG =HC = 2GE,
CG =HB = 2GF.
The centroid G divides each median in the ratio 2 : 1.
Theorem 2.6 (Apollonius). If ma denotes the length of the median on the side BC,
1
m2a = (2b2 + 2c2 − a2 ).
4
Example 2.3. Suppose the medians BE and CF of triangle ABC are perpendicular. This
means that BG2 + CG2 = BC 2 , where G is the centroid of the triangle. In terms of the
lengths, we have 49 m2b + 49 m2c = a2 ; 4(m2b +m2c ) = 9a2 ; (2c2 +2a2 −b2 )+(2a2 +2b2 −c2 ) =
9a2 ; b2 + c2 = 5a2 .
This relation is enough to describe, given points B and C, the locus of A for which the
medians BE and CF of triangle ABC are perpendicular. Here, however, is a very easy
construction: From b2 + c2 = 5a2 , we have m2a = 41 (2b2 + 2c2 − a2 ) = 49 a2 ; ma = 32 a. The
locus of A is the circle with center at the midpoint of BC, and radius 32 · BC.
2.5 The angle bisector theorem
2.5
121
The angle bisector theorem
Theorem 2.7 (Angle bisector theorem). The bisectors of an angle of a triangle divide its
opposite side in the ratio of the remaining sides. If AX and AX ′ respectively the internal
and external bisectors of angle BAC, then BX : XC = c : b and BX ′ : X ′ C = c : −b.
Z
A
Z′
b
c
B
X
C
X′
Proof. Construct lines through C parallel to the bisectors AX and AX ′ to intersect the line
AB at Z and Z ′ .
(1) Note that ∠AZC = ∠BAX = ∠XAC = ∠ACZ. This means AZ = AC. Clearly,
BX : XC = BA : AZ = BA : AC = c : b.
(2) Similarly, AZ ′ = AC, and BX ′ : X ′ C = BA : AZ ′ = BA : −AC = c : −b.
2.5.1 The lengths of the bisectors
Proposition 2.8. (a) The lengths of the internal and external bisectors of angle A are
respectively
α
2bc
α
2bc
cos
and t′a =
sin .
ta =
b+c
2
|b − c|
2
A
c
B
ta
X
C
b
t′a
X′
Proof. Let AX and AX ′ be the bisectors of angle A.
(1) Consider the area of triangle ABC as the sum of those of triangles AXC and ABX.
We have
1
α
1
ta (b + c) sin = bc sin α.
2
2
2
122
The laws of sines and cosines
From this,
ta =
bc
sin α
α
2bc
·
· cos .
α =
b + c sin 2
b+c
2
(2) Consider the area of triangle as the difference between those of ABX ′ and ACX ′ .
2bc
Remarks. (1) b+c
is the harmonic mean of b and c. It can be constructed as follows. If the
2bc
perpendicular to AX at X intersects AC and AB at Y and Z, then AY = AZ = b+c
.
A
ta
b
c
Z
C
B
X
Y
(2) Applying Stewart’s Theorem with λ = c and µ = ±b, we also obtain the following
expressions for the lengths of the angle bisectors:
2 !
a
,
t2a = bc 1 −
b+c
!
2
a
′2
ta = bc
−1 .
b−c
Example 2.4. (Steiner-Lehmus theorem). A triangle with two equal angle bisectors is
isosceles. More precisely, if the bisectors of two angles of a triangle have equal lengths,
then the two angles are equal.
Proof. We show that if a < b, then ta > tb . Note that from a < b we conclude
(i) α < β and cos α2 > cos β2 ;
b
a
2bc
2ca
(ii) bc > ac, b(c + a) > a(b + c); b+c
> c+a
; b+c
> c+a
.
From (i) and (ii), we have
ta =
2bc
α
2ca
β
· cos >
· cos = tb .
b+c
2
c+a
2
The same reasoning shows that a > b ⇒ ta < tb . It follows that if ta = tb , then a = b.
2.6 The circle of Apollonius
2.6
123
The circle of Apollonius
Theorem 2.9. A and B are two fixed points. For a given positive number k 6= 1, 1 the
locus of points P satisfying AP : P B = k : 1 is the circle with diameter XY , where X
and Y are points on the line AB such that AX : XB = k : 1 and AY : Y B = k : −1.
P
Y
O A
X
B
B′
Proof. Since k 6= 1, points X and Y can be found on the line AB satisfying the above
conditions.
Consider a point P not on the line AB with AP : P B = k : 1. Note that P X and P Y
are respectively the internal and external bisectors of angle AP B. This means that angle
XP Y is a right angle, and P lies on the circle with XY as diameter.
Conversely, let P be a point on this circle. We show that AP : BP = k : 1. Let
′
B be a point on the line AB such that P X bisects angle AP B ′ . Since P A and P B are
perpendicular to each other, the line P B is the external bisector of angle AP B ′ , and
AY
AX
XA
AY − XA
.
=−
=
=
′
′
′
YB
XB
XB
YX
On the other hand,
AY
AX
XA
AY − XA
=−
=
=
.
YB
XB
XB
YX
Comparison of the two expressions shows that B ′ coincides with B, and P X is the bisector
AX
A
= XB
= k.
of angle AP B. It follows that PP B
1
If k = 1, the locus is clearly the perpendicular bisector of the segment AB.
124
The laws of sines and cosines
Chapter 3
The tritangent circles
3.1
The incircle
Let the incircle of triangle ABC touch its sides BC, CA, AB at X, Y , Z respectively.
A
s−a
s−a
Y
Z
I
s−c
s−b
B
s−b
X
s−c
C
If s denotes the semiperimeter of triangle ABC, then
AY =AZ = s − a,
BZ =BX = s − b,
CX =CY = s − c.
The inradius of triangle ABC is the radius of its incircle. It is given by
r=
∆
2∆
= .
a+b+c
s
126
The tritangent circles
Example 3.1. If triangle ABC has a right angle at C, then the inradius r = s − c.
B
s−b
s−b
I
r
s−a
r
r
s−c
C
s−c
A
s−a
It follows that if d is the diameter of the incircle, then a + b = c + d.
Example 3.2. An equilateral triangle of side 2a is partitioned symmetrically into a quadrilateral, an isosceles triangle, and two other congruent triangles. If the inradii of the quadrilateral and the isosceles triangle are equal, find this inradius. 1
A
r
X
θ
Y
r
B
C
Suppose each side of the equilateral triangle has length 2, each of the congruent circles
has radius r, and ∠ACX = θ.
(i) From triangle AXC, r = cot 30◦2+cot θ .
(ii) Note that ∠BCY = 12 (60◦ − 2θ) = 30◦ − θ. It follows that r = tan(30◦ − θ) =
cot θ−cot 30◦
1
= cot
.
cot(30◦ −θ)
30◦ cot θ+1
√
√
√
3
2
2
2
3x
+
2;
x
−
2
3x − 5 = 0,
By putting cot θ = x, we have √3+x
= √x−
;
x
−
3
=
2
3x+1
√
√
2
and x = 3 + 2 2. (The negative root is rejected). From this, r = √3+x = √3+1 √2 =
√
√
3 − 2.
To √
construct the
√ circles, it is enough to mark Y on the altitude through A such that
AY = 3 − r = 2. The construction is now evident.
1
√
√
( 3 − 2)a.
3.1 The incircle
127
Example 3.3. (Conway’s circle). Given triangle ABC, extend
(i) CA and BA to Ya and Za such that AYa = AZa = a,
(ii) AB and CB to Zb and Xb such that BZb = BXb = b,
(iii) BC and AC to Xc and Yc such that CXc = CYc = c.
The six points X
√b , Xc , Yc , Ya , Za , Zb are concyclic. The circle containing them has
center I and radius r2 + s2 .
Za
Ya
a
a
A
Y
c
Z
b
I
r
Xb
B
b
X
a
C
Xc
c
c
b
Yc
Zb
Proof. Let the incircle be tangent to BC at X. BX = s − b =⇒ Xb X = b + (s − b) = s.
From this, IXb2 = r2 + s2 . Similarly, each of the remaining
five points is at the same
√
2
2
distance from I. They lie on a circle, center I, radius r + s .
128
The tritangent circles
3.2
Euler’s formula
Theorem 3.1 (Euler’s formula). OI 2 = R2 − 2Rr
A
O
I
C
B
M
Proof. Extend AI to intersect the circumcircle at M . Since ∠BAM = ∠M AC =
is the midpoint of the arc BC of the circumcircle. Note that
A
,
2
M
∠M BI = ∠M BC + ∠CBI = ∠M AC + ∠CBI = ∠BAI + ∠IBA = ∠BIM.
It follows that IM = BM . By the law of sines, BM = 2R sin A2 . Since AI = sinr A ,
2
AI · IM = 2Rr. This is equal to the power of I with respect to the circumcircle, namely,
= R2 − OI 2 . Therefore, OI 2 = R2 − 2Rr.
Corollary 3.2. R ≥ 2r; equality holds if and only if the triangle is equilateral.
3.3 Steiner’s porism
3.3
129
Steiner’s porism
Given a triangle ABC with incircle (I) and circumcircle O), let A′ be an arbitrary point
on circumcircle. Join A′ to I, to intersect the circumcircle again at M ′ , and let A′ Y ′ , A′ Z ′
be the tangents to the incircle. Construct a circle, center M ′ , through I to intersect the
circumcircle at B ′ and C ′ .
A′
A
Z′
O
B′
I
Y′
B
C
M′
C′
Denote by θ between A′ I and each tangent.
It is known that A′ I · IM ′ = 2Rr (power of incenter in (O)). Since A′ I = sinr θ , we
have IM ′ = 2R sin θ. Therefore, M ′ B ′ = M ′ C ′ = 2R sin θ, and by the law of sines,
∠B ′ A′ M = ∠C ′ A′ M = θ. It follows that A′ B ′ and A′ C ′ are tangent to the incircle (I).
Since M ′ is the midpoint of the arc B ′ C ′ , the circle M ′ (B ′ ) passes through the incenter of
triangle A′ B ′ C ′ . This means that I is the incenter, and (I) the incircle of triangle A′ B ′ C ′ .
130
3.4
The tritangent circles
The excircles
The internal bisector of each angle and the external bisectors of the remaining two angles
are concurrent at an excenter of the triangle. An excircle can be constructed with this as
center, tangent to the lines containing the three sides of the triangle.
Ib
A
Ic
X
C
B
Y
ra
ra
ra
Z
Ia
The exradii of a triangle with sides a, b, c are given by
ra =
∆
,
s−a
rb =
∆
,
s−b
rc =
∆
.
s−c
The areas of the triangles Ia BC, Ia CA, and Ia AB are 21 ara , 12 bra , and 12 cra respectively.
Since
∆ = −∆Ia BC + ∆Ia CA + ∆Ia AB,
we have
from which ra =
1
∆ = ra (−a + b + c) = ra (s − a),
2
∆
.
s−a
3.5 Heron’s formula for the area of a triangle
3.5
131
Heron’s formula for the area of a triangle
Consider a triangle ABC with area ∆. Denote by r the inradius, and ra the radius of the
excircle on the side BC of triangle ABC. It is convenient to introduce the semiperimeter
s = 21 (a + b + c).
B
Ia
I
ra
r
C
Y′
A
Y
(1) From the similarity of triangles AIY and AI ′ Y ′ ,
r
s−a
.
=
ra
s
(2) From the similarity of triangles CIY and I ′ CY ′ ,
r · ra = (s − b)(s − c).
(3) From these,
r
(s − a)(s − b)(s − c)
,
s
r
s(s − b)(s − c)
.
ra =
s−a
r=
Theorem 3.3 (Heron’s formula).
∆=
Proof. ∆ = rs.
Proposition 3.4.
s
α
(s − b)(s − c)
tan =
,
2
s(s − a)
p
s(s − a)(s − b)(s − c).
α
cos =
2
r
s(s − a)
,
bc
α
sin =
2
r
(s − b)(s − c)
.
bc
132
The tritangent circles
Chapter 4
The arbelos
4.1
Archimedes’ twin circles
Theorem 4.1 (Archimedes). The two circles each tangent to CP , the largest semicircle
AB and one of the smaller semicircles have equal radii t, given by
t=
ab
.
a+b
Q
A
O1
O
P
O2
B
A
O1
O
P
O2
B
Proof. Consider the circle tangent to the semicircles O(a + b), O1 (a), and the line P Q.
Denote by t the radius of this circle. Calculating in two ways the height of the center of
this circle above the line AB, we have
(a + b − t)2 − (a − b − t)2 = (a + t)2 − (a − t)2 .
From this,
ab
.
a+b
The symmetry of this expression in a and b means that the circle tangent to O(a+b), O2 (b),
and P Q has the same radius t.
t=
134
The arbelos
1
a
4.1.1 Harmonic mean and the equation
+
1
b
=
1
t
2ab
The harmonic mean of two quantities a and b is a+b
. In a trapezoid of parallel sides a and b,
the parallel through the intersection of the diagonals intercepts a segment whose length is
the harmonic mean of a and b. We shall write this harmonic mean as 2t, so that a1 + 1b = 1t .
a
D
C
b
a
t
A
B
b
Here is another construction of t, making use of the formula for the length of an angle
bisector in a triangle. If BC = a, AC = b, then the angle bisector CZ has length
tc =
C
C
2ab
cos = 2t cos .
a+b
2
2
The length t can therefore be constructed by completing the rhombus CXZY (by constructing the perpendicular bisector of CZ to intersect BC at X and AC at Y ). In particular, if the triangle contains a right angle, this trapezoid is a square.
C
X
b
Y
t
M
t
t
t
A
Z
B
a
4.1.2 Construction of the Archimedean twin circles
Construct the circle P (C3 ) to intersect the diameter AB at P1 and P2 so that P1 is on AP
and P2 is on P B. The center C1 respectively C2 is the intersection of the circle O1 (P2 )
respectively O2 (P1 ) and the perpendicular to AB at P1 respectively P2 .
Q
C1
Q1
C2
Q2
C3
A
O1
P1
OP
PO
22
B
4.2 The incircle
4.2
135
The incircle
Theorem 4.2 (Archimedes). The incircle of the arbelos has radius
ρ=
r1 r2 (r1 + r2 )
.
r12 + r1 r2 + r22
O3
X
Y
θ
A
O1
O
C
O2
B
Proof. Let ∠COO3 = θ. By the law of cosines, we have
(r1 + ρ)2 =(r1 + r2 − ρ)2 + r22 + 2r2 (r1 + r2 − ρ) cos θ,
(r2 + ρ)2 =(r1 + r2 − ρ)2 + r12 − 2r1 (r1 + r2 − ρ) cos θ.
Eliminating θ, we have
r1 (r1 + ρ)2 + r2 (r2 + ρ)2 = (r1 + r2 )(r1 + r2 − ρ)2 + r1 r22 + r2 r12 .
The coefficients of ρ2 on both sides are clearly the same. This is a linear equation in ρ:
r13 + r23 + 2(r12 + r22 )ρ = (r1 + r2 )3 + r1 r2 (r1 + r2 ) − 2(r1 + r2 )2 ρ,
from which
4(r12 + r1 r2 + r22 )ρ = (r1 + r2 )3 + r1 r2 (r1 + r2 ) − (r13 + r23 ) = 4r1 r2 (r1 + r2 ),
and ρ is as above.
136
The arbelos
4.2.1 Construction of the incircle of the arbelos
In [?], Bankoff published the following remarkable theorem which gives a construction of
the incircle of the arbelos of the incircle, much simpler than the one we designed before
from Archimedes’ proof. The simplicity of the construction is due to the existence of a
circle congruent to Archimedes’ twin circles.
Theorem 4.3 (Bankoff). The points of tangency of the incircle of the arbelos with the
semicircles (AC) and (CB), together with C, are the points of tangency of the incircle
(W3 ) of triangle O1 O2 O3 with the sides of the triangle. This circle (W3 ) is congruent to
Archimedes’ twin circles (W1 ) and (W2 ).
P
O3
Q
R
W3
A
O1
O
C
O2
B
Proof. Since O1 Q = O1 C, O2 C = O2 R, and O3 R = O3 Q, the points C, Q, R are the
points of tangency of the incircle of triangle O1 O2 O3 with its sides. The semi-perimeter of
the triangle is
(r1 + r2 )3
(r1 + r2 )2 ρ
r1 r2 (r1 + r2 )
.
= 2
=
s = r1 + r2 + ρ = r1 + r2 + 2
r1 + r2 r2 + r22
r1 + r2 r2 + r22
r1 r2
The inradius of the triangle is the square root of
r1 r2 ρ
r12 r22
=
= t2 .
s
(r1 + r2 )2
It follows that this inradius is t. The incircle of triangle O1 O2 O3 is congruent to Archimedes’
twin circles.
4.2 The incircle
137
Construction. Let M and N be the midpoints of the semicircles (AC) and (CB) respectively. Construct
(1) the lines O1 N and O2 M to intersect at W3 ,
(2) the circle with center W3 , passing through C to intersect the semicircle (AC) at Q and
(CB) at R,
(3) the lines O1 Q and O2 R to intersect at O3 .
The circle with center O3 passing through Q touches the semicircle (CB) at R and also
the semicircle (AB).
P
O3
M
Q
R
N
W3
A
O1
O
C
O2
B
138
The arbelos
4.2.2 Alternative constructions of the incircle
Theorem 4.4 (Bankoff). Let P be the intersection (apart from C) of the circumcircles of
the squares on AC and CB. Let Q be the intersection (apart from C) of the circumcircle of
the square on CB and the semicircle (AC), and R that of the circumcircle of the square on
AB and the semicircle (CB). The points P , Q, R are the points of tangency of the incircle
of the arbelos with the semicircles.
L
P
O3
M
N
Q
A
O1
O
R
C
O2
B
Proposition 4.5. The intersection S of the lines AN and BM also lies on the incircle of
the arbelos, and the line CS intersects (AB) at P .
L
P
O3
M
N
Q
S
A
O1
O
C
R
O2
B
4.2 The incircle
139
P
O3
Q
R
A
B
O1
O
O2
C
L′
Construction. Let L′ be the “lowest point” of the circle (AB). Construct
(1) the line L′ C to intersect the semicircle (AB) at P ,
(2) the circle, center L′ , through A and B, to intersect the semicircles (AC) and (CB) at
Q and R.
Proposition 4.6. Let X be the midpoint of the side of the square on AC opposite to AC,
and Y that of the side of the square on CB opposite to CB. The center O3 of the incircle
of the arbelos is the intersection of the lines AY and BX.
X
L
P
Y
O3
M
N
Q
A
O1
O
R
C
O2
B
140
The arbelos
4.3
Archimedean circles
We shall call a circle Archimedean if it is congruent to Archimedes’ twin circle, i.e., with
r2
, and has further remarkable geometric properties.
radius t = rr11+r
2
1. (van Lamoen) The circle (W3 ) is tangent internally to the midway semicircle (O1 O2 )
at a point on the segment M N . 1
M
D
M1
M′
M2
W3
A
O1
O2
O O′ C
B
2. (van Lamoem) The circle tangent to AB at O and to the midway semicircle is
Archimedean. 2
M
M1
M2
W?
A
O1
O2
O O′ C
B
3. (Schoch) Let M N intersect CD and OL at Q and K respectively. The smallest
circles through Q and K tangent to the semicircle (AB) are Archimedean.
L
M
D
W20
W21
K
A
1
2
van Lamoen, June 10, 1999.
van Lamoen, June 10, 1999.
O1
O
Q
C
N
O2
B
4.3 Archimedean circles
141
4. (a) The circle tangent to (AB) and to the common tangent of (AC) and (CB) is
Archimedean.
(b) The smallest circle through C tangent to AB is Archimedean.
D
W4
E
F
W11
A
O1
O
B
O2
C
5. Let EF be the common tangent of the semicircles (AC) and (CB). The smallest
circles through E and F tangent to CD are Archimedean.
D
E
W9
W10
A
O1
O
F
B
O2
C
6. (Schoch) Let X and Y be the intersections of the semicircle (AB) with the circles
through C, with centers A and B respectively. The smallest circle through X and Y
tangent to CD are Archimedean.
D
X
W13
W14
Y
A
O1
O
C
O2Y ′
B
142
The arbelos
7. (van Lamoen) Let Y and Z be the intersections of the midway semicircle with the
semicircles (AC) and (CB). The circles with centers Y and Z, each tangent to the
line CD, are Archimedean.
D
Y
Z
A
O1
O
C
X 2 O2
B
8. (Schoch) (a) The circle tangent to the semicircle (AB) and the circular arcs, with
centers A and B respectively, each passing through C, is Archimedean.
(b) The circle with center on the Schoch line and tangent to both semicircles (AC)
and (CB) is Archimedean.
W15
W16
A
O1
O
O2
C
B
9. (Woo) Let α be a positive real number. Consider the two circular arcs, each passing
through C and with centers (−αr1 , 0) and (αr2 , 0) respectively. The circle with
center Uα on the Schoch line tangent to both of these arcs is Archimedean.
The Woo circle (Uα ) which is tangent externally to the semicircle (AB) touches it at
D (the intersection with the common tangent of (AC) and (CB)).
W28
D
W15
A
O1
O
C
O2
B
4.3 Archimedean circles
143
10. (Power) Consider an arbelos with inner semicircles C1 and C2 of radii a and b, and
outer semicircle C of radius a + b. It is known the Archimedean circles have radius
ab
. Let Q1 and Q2 be the “highest” points of C1 and C2 respectively.
t = a+b
A circle tangent to (O) internally and to OQ1 at Q1 (or OQ2 at Q2 ) is Archimedean.
C1
r
C2
Q1
C1′
a+b−r
a
O1
A
b
O
Q2
C2′
O2
P
B
11. (van Lamoen)
D
M1 U 1
M
M2
U2
A
O
O1
C
B
O2
12. (Bui)
D
D
T1
T1
M
M
T2
T2
A
O1
O O′ C
O2
B
A
O1
O
C
O2
B
Chapter 5
Menelaus’ theorem
5.1
Menelaus’ theorem
Theorem 5.1 (Menelaus). Given a triangle ABC with points X, Y , Z on the side lines
BC, CA, AB respectively, the points X, Y , Z are collinear if and only if
BX CY AZ
·
·
= −1.
XC Y A ZB
A
Y
Z
W
B
X
C
Proof. (=⇒) Let W be the point on AC such that BW//XY . Then,
BX
WY
=
,
XC
YC
and
AZ
AY
=
.
ZB
YW
It follows that
W Y CY AY
CY AY W Y
BX CY AZ
·
·
=
·
·
=
·
·
= −1.
XC Y A ZB
YC YA YW
YC YA YW
(⇐=) Suppose the line joining X and Z intersects AC at Y ′ . From above,
BX CY ′ AZ
BX CY AZ
· ′ ·
= −1 =
·
·
.
XC Y A ZB
XC Y A ZB
It follows that
CY ′
CY
=
.
′
Y A
YA
The points Y ′ and Y divide the segment CA in the same ratio. These must be the same
point, and X, Y , Z are collinear.
202
Menelaus’ theorem
Example 5.1. The external angle bisectors of a triangle intersect their opposite sides at
three collinear points.
Y′
A
c
b
C
X′
B
a
Z′
Proof. If the external bisectors are AX ′ , BY ′ , CZ ′ with X ′ , Y ′ , Z ′ on BC, CA, AB
respectively, then
BX ′
c
CY ′
a
AZ ′
b
=
−
,
=
−
,
=− .
′
′
′
XC
b
Y A
c
ZB
a
It follows that
BX ′
X′C
·
CY ′
Y ′A
·
AZ ′
Z′B
= −1 and the points X ′ , Y ′ , Z ′ are collinear.
5.2 Centers of similitude of two circles
5.2
203
Centers of similitude of two circles
Given two circles O(R) and I(r), whose centers O and I are at a distance d apart, we
animate a point X on O(R) and construct a ray through I oppositely parallel to the ray
OX to intersect the circle I(r) at a point Y . The line joining X and Y intersects the line
OI of centers at a point T which satisfies
OT : IT = OX : IY = R : r.
This point T is independent of the choice of X. It is called the internal center of similitude,
or simply the insimilicenter, of the two circles.
Y
T′
T
O
I
Y′
X
If, on the other hand, we construct a ray through I directly parallel to the ray OX to
intersect the circle I(r) at Y ′ , the line XY ′ always intersects OI at another point T ′ . This is
the external center of similitude, or simply the exsimilicenter, of the two circles. It divides
the segment OI in the ratio OT ′ : T ′ I = R : −r.
5.2.1 Desargue’s theorem
Given three circles with centers A, B, C and distinct radii, show that the exsimilicenters of
the three pairs of circles are collinear.
A
B
Z
C
Y
X
204
5.3
Menelaus’ theorem
Ceva’s theorem
Theorem 5.2 (Ceva). Given a triangle ABC with points X, Y , Z on the side lines BC,
CA, AB respectively, the lines AX, BY , CZ are concurrent if and only if
BX CY AZ
·
·
= +1.
XC Y A ZB
A
Z
Y
P
B
X
C
Proof. (=⇒) Suppose the lines AX, BY , CZ intersect at a point P . Consider the line
BP Y cutting the sides of triangle CAX. By Menelaus’ theorem,
CY AP XB
·
·
= −1,
Y A P X BC
or
CY P A BX
·
·
= +1.
Y A XP BC
Also, consider the line CP Z cutting the sides of triangle ABX. By Menelaus’ theorem
again,
AZ BC XP
AZ BC XP
·
·
= −1, or
·
·
= +1.
ZB CX P A
ZB XC P A
Multiplying the two equations together, we have
CY AZ BX
·
·
= +1.
Y A ZB XC
(⇐=) Exercise.
5.4 Some triangle centers
5.4
205
Some triangle centers
5.4.1 The centroid
If D, E, F are the midpoints of the sides BC, CA, AB of triangle ABC, then clearly
AF BD CE
·
·
= 1.
F B DC EA
The medians AD, BE, CF are therefore concurrent. Their intersection is the centroid G
of the triangle.
Consider triangle ADC with transversal BGE. By Menelaus’ theorem,
−1 =
AG −1 1
AG DB CE
·
·
=
·
· .
GD BC EA
GD 2 1
It follows that AG : GD = 2 : 1. The centroid of a triangle divides each median in the
ratio 2:1.
A
A
F
Y
E
Z
G
B
I
C
D
B
X
C
5.4.2 The incenter
Let X, Y , Z be points on BC, CA, AB such that AX, BY , CZ bisect angles BAC, CBA
and ACB respectively. Then
b
AZ
= ,
ZB
a
BX
c
= ,
XC
b
CY
a
= .
YA
c
It follows that
b c a
AZ BX CY
·
·
= · · = +1,
ZB XC Y A
a b c
and AX, BY , CZ are concurrent. Their intersection is the incenter of the triangle.
Applying Menelaus’ theorem to triangle ABX with transversal CIZ, we have
−1 =
AI XC BZ
AI
−b a
AI
b+c
·
·
=
·
· =⇒
=
.
IX CB ZA
IX b + c b
IX
a
206
Menelaus’ theorem
5.4.3 The Gergonne point
Let the incircle of triangle ABC be tangent to the sides BC at X, CA at Y , and AB at Z
respectively. Since AY = AZ = s − a, BZ = BX = s − b, and CX = CY = s − c, we
have
s−b s−c s−a
BX CY AZ
·
·
=
·
·
= 1.
XC Y A ZB
s−c s−a s−b
By Ceva’s theorem, the lines AX, BY , CZ are concurrent. The intersection is called the
Gergonne point Ge of the triangle.
A
s−a
s−a
Y
Z
I
Ge
s−c
s−b
B
s−b
X
s−c
C
Lemma 5.3. The Gergonne point Ge divides the cevian AX in the ratio
a(s − a)
AGe
=
.
Ge X
(s − b)(s − c)
Proof. Applying Menelaus’ theorem to triangle ABX with transversal CGe Z, we have
−1 =
AGe −(s − c) s − b
AGe
a(s − a)
AGe XC BZ
·
·
=
·
·
=⇒
=
.
Ge X CB ZA
Ge X
a
s−a
Ge X
(s − b)(s − c)
5.4 Some triangle centers
207
5.4.4 The Nagel point
If X ′ , Y ′ , Z ′ are the points of tangency of the excircles with the respective sidelines, the
lines AX ′ , BY ′ , CZ ′ are concurrent by Ceva’s theorem:
s−c s−c s−a
BX ′ CY ′ AZ ′
· ′ · ′ =
·
·
= 1.
′
XC Y A ZB
s−b s−a s−b
The point of concurrency is the Nagel point Na .
Ib
A
Ic
s−b
s−c
Z′
Y′
Na
s−a
B
s−c
s−a
X′
s−b
C
Ia
Lemma 5.4. If the A-excircle of triangle ABC touches BC at X ′ , then the Nagel point
divides the cevian AX ′ in the ratio
a
ANa
=
.
′
Na X
s−a
Proof. Applying Menelaus’ theorem to triangle ACX ′ with transversal BNa Y ′ , we have
−1 =
ANa X ′ B CY ′
ANa −(s − c) s − a
ANa
a
·
· ′ =
·
·
=⇒
=
.
′
′
′
Na X BC Y A
Na X
a
s−c
Na X
s−a
208
5.5
Menelaus’ theorem
Isotomic conjugates
Given points X on BC, Y on CA, and Z on AB, we consider their reflections in the
midpoints of the respective sides. These are the points X ′ on BC, Y ′ on CA and Z ′ on AB
satisfying
BX ′ = XC, BX = X ′ C;
CY ′ = Y A, CY = Y ′ A;
AZ ′ = ZB, AZ = Z ′ B.
Clearly, AX, BY , CZ are concurrent if and only if AX ′ , BY ′ , CZ ′ are concurrent.
A
Z′
Y
Y′
Z
P•
P
B
X
C
X′
Proof.
BX CY AZ
·
·
XC Y A ZB
BX ′ CY ′ AZ ′
·
·
X ′C Y ′A Z ′B
=
BX BX ′
·
XC X ′ C
CY CY ′
·
Y A Y ′A
AZ AZ ′
·
ZB Z ′ B
The points of concurrency of the two triads of lines are called isotomic conjugates.
= 1.
5.5 Isotomic conjugates
209
Example 5.2. (The Gergonne and Nagel points)
Ib
A
Ic
Z′
Y
I
Z
Ge
B
Y′
Na
X
C
X′
Ia
Example 5.3. (The isotomic conjugate of the orthocenter) Let H • denote the isotomic
conjugate of the orthocenter H. Its traces are the pedals of the reflection of H in O. This
latter point is the deLongchamps point Lo .
A
Z′
Lo
Y
O
Y′
H•
Z
B
H
X
X′
C
210
Menelaus’ theorem
Example 5.4. (Yff-Brocard points) Consider a point P = (u : v : w) with traces X, Y , Z
satisfying BX = CY = AZ = µ. This means that
w
u
v
a=
b=
c = µ.
v+w
w+u
u+v
Elimination of u, v, w leads to
0=
0
−µ a − µ
b−µ
0
−µ
−µ c − µ
0
= (a − µ)(b − µ)(c − µ) − µ3 .
Indeed, µ is the unique positive root of the cubic polynomial
(a − t)(b − t)(c − t) − t3 .
This gives the point
P =
c−µ
b−µ
13 1 1 !
a−µ 3
b−µ 3
.
:
:
c−µ
a−µ
A
A
Z′
Y′
Y
Z
P•
P
B
C
X
B
X′
C
The isotomic conjugate
•
P =
b−µ
c−µ
13 1 1 !
c−µ 3
a−µ 3
:
:
a−µ
b−µ
has traces X ′ , Y ′ , Z ′ that satisfy
CX ′ = AY ′ = BZ ′ = µ.
These points are called the Yff-Brocard points.
L. Crelle. 2
1
2
1
They were briefly considered by A.
P. Yff, An analogue of the Brocard points, Amer. Math. Monthly, 70 (1963) 495 – 501.
A. L. Crelle, 1815.
Chapter 6
The Euler line and the nine-point circle
6.1
The Euler line
6.1.1 Inferior and superior triangles
A
C′
B′
A
G
B
F
C
E
G
A′
B
D
C
The inferior triangle of ABC is the triangle DEF whose vertices are the midpoints of
the sides BC, CA, AB.
The two triangles share the same centroid G, and are homothetic at G with ratio −1 : 2.
The superior triangle of ABC is the triangle A′ B ′ C ′ bounded by the parallels of the
sides through the opposite vertices.
The two triangles also share the same centroid G, and are homothetic at G with ratio
2 : −1.
212
The Euler line and the nine-point circle
6.1.2 The orthocenter and the Euler line
The three altitudes of a triangle are concurrent. This is because the line containing an
altitude of triangle ABC is the perpendicular bisector of a side of its superior triangle.
The three lines therefore intersect at the circumcenter of the superior triangle. This is the
orthocenter of the given triangle.
A
C′
B′
G
O
H
B
C
A′
The circumcenter, centroid, and orthocenter of a triangle are collinear. This is because
the orthocenter, being the circumcenter of the superior triangle, is the image of the circumcenter under the homothety h(G, −2). The line containing them is called the Euler line of
the reference triangle (provided it is non-equilateral).
The orthocenter of an acute (obtuse) triangle lies in the interior (exterior) of the triangle.
The orthocenter of a right triangle is the right angle vertex.
6.2 The nine-point circle
6.2
213
The nine-point circle
Theorem 6.1. The following nine points associated with a triangle are on a circle whose
center is the midpoint between the circumcenter and the orthocenter:
(i) the midpoints of the three sides,
(ii) the pedals (orthogonal projections) of the three vertices on their opposite sides,
(iii) the midpoints between the orthocenter and the three vertices.
A
Y
D′
F
E
NG
O
Z
H
E′
B
F′
X
D
C
Proof. (1) Let N be the circumcenter of the inferior triangle DEF . Since DEF and ABC
are homothetic at G in the ratio 1 : 2, N , G, O are collinear, and N G : GO = 1 : 2. Since
HG : GO = 2 : 1, the four are collinear, and
HN : N G : GO = 3 : 1 : 2,
and N is the midpoint of OH.
(2) Let X be the pedal of H on BC. Since N is the midpoint of OH, the pedal of
N is the midpoint of DX. Therefore, N lies on the perpendicular bisector of DX, and
N X = N D. Similarly, N E = N Y , and N F = N Z for the pedals of H on CA and AB
respectively. This means that the circumcircle of DEF also contains X, Y , Z.
(3) Let D′ , E ′ , F ′ be the midpoints of AH, BH, CH respectively. The triangle D′ E ′ F ′
is homothetic to ABC at H in the ratio 1 : 2. Denote by N ′ its circumcenter. The points N ′ ,
G, O are collinear, and N ′ G : GO = 1 : 2. It follows that N ′ = N , and the circumcircle of
DEF also contains D′ , E ′ , F ′ .
This circle is called the nine-point circle of triangle ABC. Its center N is called the
nine-point center. Its radius is half of the circumradius of ABC.
214
The Euler line and the nine-point circle
Theorem 6.2. Let Oa , Ob , Oc be the reflections of the circumcenter O in the sidelines
BC, CA, AB respectively.
(1) The circle through Oa , Ob , Oc is congruent to the circumcircle and has center at the
orthocenter H.
(2) The reflections of H in the sidelines lie on the circumcircle.
A
O
H
B
D
C
Ha
Oa
Proof. (1) If D is the midpoint of BC, OOa = 2OD = AH. This means that AHOa O is
a parallelogram, and HOa = AO. Similarly, HOb = BO and HOc = CO for the other
two reflections. Therefore, and H is the center of the circle through Oa , Ob , Oc , and the
circle is congruent to the circumcircle.
(2) If Ha is the reflection of H in BC, then HHa OA O is a trapezoid symmetric in the
line BC. Therefore, OHa = HOa = OA. This means that Ha lies on the circumcircle; so
do Hb and Hc .
6.3 Distances between triangle centers
6.3
215
Distances between triangle centers
6.3.1 Distance between the circumcenter and orthocenter
A
Y
F
E
O
Z
N
H
B
X
D
C
Proposition 6.3. OH 2 = R2 (1 − 8 cos α cos β cos γ).
Proof. In triangle AOH, AO = R, AH = 2R cos α, and ∠OAH = |β − γ|. By the law of
cosines,
OH 2 = R2 (1 + 4 cos2 α − 4 cos α cos(β − γ))
= R2 (1 − 4 cos α(cos(β + γ) + cos(β − γ))
= R2 (1 − 8 cos α cos β cos γ).
216
The Euler line and the nine-point circle
6.3.2 Distance between circumcenter and tritangent centers
Lemma 6.4. If the bisector of angle A intersects the circumcircle at M , then M is the
center of the circle through B, I, C, and Ia .
Proof. (1) Since M is the midpoint of the arc BC, ∠M BC = ∠M CB = ∠M AB. Therefore,
∠M BI = ∠M BC + ∠CBI = ∠M AB + ∠IBA = ∠M IB,
and M B = M I. Similarly, M C = M I.
(2) On the other hand, since ∠IBIa and ICIa are both right angles, the four points B,
I, C, Ia M are concyclic, with center at the midpoint of IIA . This is the point M .
Theorem 6.5 (Euler). (a) OI 2 = R2 − 2Rr.
(b) OIa2 = R2 + 2Rra .
A
Y
O
I
C
B
Y′
M
ra
Ia
Proof. (a) Considering the power of I in the circumcircle, we have
R2 − OI 2 = AI · IM = AI · M B =
α
r
= 2Rr.
α · 2R · sin
sin 2
2
(b) Consider the power of Ia in the circumcircle.
Note that Ia A = sinraα . Also, Ia M = M B = 2R sin α2 .
2
OIa2 = R2 + Ia A · Ia M
α
ra
= R2 +
α · 2R sin
sin 2
2
= R2 + 2Rra .
6.3 Distances between triangle centers
217
6.3.3 Distance between orthocenter and tritangent centers
Proposition 6.6.
HI 2 = 2r2 − 4R2 cos α cos β cos γ,
HIa2 = 2ra2 − 4R2 cos α cos β cos γ.
A
I
H
B
X
C
Proof. In triangle AIH, we have AH = 2R cos α, AI = 4R sin β2 sin γ2 and ∠HAI =
|β−γ|
. By the law of cosines,
2
β−γ
HI 2 = AH 2 + AI 2 − 2AI · AH · cos
2
β
γ
β
γ
β−γ
2
2
2
2
= 4R cos α + 4 sin
sin
− 4 cos α sin sin cos
2
2
2
2
2
β
γ
β
β
γ
β
γ
= 4R2 cos2 α + 4 sin2 sin2 − 4 cos α sin sin cos cos − 4 cos α sin2
2
2
2
2
2
2
2
β
β
γ
α
= 4R2 cos2 α + 4 sin2 sin2 − cos α sin β sin γ − 4 1 − 2 sin2
sin2 sin2
2
2
2
2
α
β
γ
= 4R2 cos α(cos α − sin β sin γ) + 8 sin2 sin2 sin2
2
2
2
α
β
γ
= 4R2 − cos α cos β cos γ + 8 sin2 sin2 sin2
2
2
2
= 2r2 − 4R2 cos α cos β cos γ.
γ
sin
2
γ
2
2
218
The Euler line and the nine-point circle
(2) In triangle AHIa , AIa = 4R cos β2 cos γ2 .
A
H
B
I
C
Aa
Ba
Ca
Ia
By the law of cosines, we have
β−γ
HIa2 = AH 2 + AIa2 − 2AIa · AH · cos
2
β
β
γ
β−γ
γ
2
2
2
2
= 4R cos α + 4 cos
cos
− 4 cos α cos cos cos
2
2
2
2
2
β
γ
β
γ
2
2 β
2
2 γ
2 β
2 γ
= 4R cos α + 4 cos
cos
− 4 cos α cos
cos
− 4 cos α cos cos sin sin
2
2
2
2
2
2
2
2
β
β
α
γ
γ
= 4R2 cos2 α + 4 cos2 cos2 − 4 1 − 2 sin2
cos2 cos2 − cos α sin β sin γ
2
2
2
2
2
α
β
γ
= 4R2 cos α(cos α − sin β sin γ) + 8 sin2 cos2 cos2
2
2
2
α
β
γ
= 4R2 − cos α cos β cos γ + 8 sin2 cos2 cos2
2
2
2
= 2ra2 − 4R2 cos α cos β cos γ.
6.4 Feuerbach’s theorem
6.4
219
Feuerbach’s theorem
Theorem 6.7 (Feuerbach). The nine-point circle is tangent internally to the incircle and
externally to each of the excircles.
I
H
O
N
Ia
Proof. (1) Since N is the midpoint of OH, IN is a median of triangle IOH. By Apollonius’ theorem,
1
1
N I 2 = (IH 2 + OI 2 ) − OH 2
2
4
1 2
2
= R − Rr + r
4
2
R
=
−r .
2
Therefore, N I is the difference between the radii of the nine-point circle and the incircle. This shows that the two circles are tangent to each other internally.
(2) Similarly, in triangle Ia OH,
1
1
N Ia2 = (HIa2 + OIa2 ) − OH 2
2
4
1 2
2
= R + Rra + ra
4
2
R
=
+ ra .
2
This shows that the distance between the centers of the nine-point and an excircle is the
sum of their radii. The two circles are tangent externally.
The point of tangency Fe of the incircle and the nine-point circle is called the Feuerbach
point.
220
The Euler line and the nine-point circle
Cb
Ib
Bc
A
Ic
Cc
Fe
Fb
Bb
Fc
I
N
B
Ac
Fa
Aa
C
Ab
Ba
Ca
Ia
Chapter 7
Isogonal conjugates
7.1
Directed angles
A reference triangle ABC in a plane induces an orientation of the plane, with respect to
which all angles are signed. For two given lines L and L′ , the directed angle ∠(L, L′ )
between them is the angle of rotation from L to L′ in the induced orientation of the plane.
It takes values of modulo π. The following basic properties of directed angles make many
geometric reasoning simple without the reference of a diagram.
Theorem 7.1. (1) ∠(L′ , L) = −∠(L, L′ ).
(2) ∠(L1 , L2 ) + ∠(L2 , L3 ) = ∠(L1 , L3 ) for any three lines L1 , L2 and L3 .
(3) Four points P , Q, X, Y are concyclic if and only if ∠(P X, XQ) = ∠(P Y, Y Q).
Remark. In calculations with directed angles, we shall slightly abuse notations by using the
equality sign instead of the sign for congruence modulo π. It is understood that directed
angles are defined up to multiples of π. For example, we shall write β + γ = −α even
though it should be more properly β + γ = π − α or β + γ ≡ −α mod π.
Exercise
1. If a, b, c are the sidelines of triangle ABC, then ∠(a, b) = −γ etc.
222
7.2
Isogonal conjugates
Isogonal conjugates
Let P be a given point. Consider the reflections of the cevians AP , BP , CP in the respective bisectors of angles A, B, C, i.e., By Ceva’s theorem, these reflections are concurrent.
Their intersection is the isogonal conjugate of P .
Let P and Q be isogonal conjugates, AP and AQ intersecting BC at X and X ′ respectively. Then
BX BX ′
c2
· ′ = 2.
XC X C
b
Example 7.1. The incenter is the isogonal conjugate of itself. The same is true for the
excenters.
Example 7.2. (The circumcenter and orthocenter) For a given triangle with circumcenter
O, the line OA and the altitude through A are isogonal lines, similarly for the circumradii
and altitudes through B and C. Since the circumradii are concurrent at O, the altitudes also
are concurrent. Their intersection is the orthocenter H, which is the isogonal conjugate of
O.
A
Y
F
E
O
Z
B
H
X
D
C
7.3 The symmedian point and the centroid
7.3
223
The symmedian point and the centroid
The isogonal lines of the medians are called the symmedians. The isogonal conjugate of
the centroid G is called the symmedian point K of the triangle.
A
G
K
B
C
Consider triangle ABC together with its tangential triangle A′ B ′ C ′ , the triangle bounded
by the tangents of the circumcircle at the vertices.
C′
A
B′
O
K
B
D
C
Y
A′
Z
Since A′ is equidistant from B and C, we construct the circle A′ (B) = A′ (C) and
extend the sides AB and AC to meet this circle again at Z and Y respectively. Note that
∠(A′ Y, A′ B ′ ) = π − 2(π − α − γ) = π − 2β,
224
Isogonal conjugates
and similarly, ∠(A′ C ′ , A′ Z ′ ) = π − 2γ. Since ∠(A′ B ′ , A′ C ′ ) = π − 2α, we have
∠(A′ Y, A′ Z) =∠(A′ Y, A′ B ′ ) + ∠(A′ B ′ , A′ C ′ ) + ∠(A′ C ′ , A′ Z)
=(π − 2β) + (π − 2α) + (π − 2γ)
=π
≡0 mod π.
This shows that Y , A′ and Z ′ are collinear, so that
(i) AA′ is a median of triangle AY Z,
(ii) AY Z and ABC are similar.
It follows that AA′ is the isogonal line of the A-median, i.e., a symmedian. Similarly, the
BB ′ and CC ′ are the symmedians isogonal to B- and C-medians. The lines AA′ , BB ′ ,
CC ′ therefore intersect at the isogonal conjugate of the centroid G.
7.4 Isogonal conjugates of the Gergonne and Nagel points
7.4
225
Isogonal conjugates of the Gergonne and Nagel points
7.4.1 The Gergonne point and the insimilicenter T+
Consider the intouch triangle DEF of triangle ABC.
(1) If D′ is the reflection of D in the bisector AI, then
(i) D′ is a point on the incircle, and
(ii) the lines AD and AD′ are isogonal with respect to A.
A
A
E
E
F′
F
F
I
E′
Ge
I P
D′
B
D′
C
D
B
C
D
(2) Likewise, E ′ and F ′ are the reflections of E and F in the bisectors BI and CI
respectively, then
(i) these are points on the incircle,
(ii) the lines BE ′ and CF ′ are isogonals of BE and CF with respect to angles B and C.
Therefore, the lines AD′ , BE ′ , and CF ′ concur at the isogonal conjugate of the Gergonne point.
(3) In fact, E ′ F ′ is parallel to BC.
A
E
F′
F
E′
I
D′
B
This follows from
D
C
226
Isogonal conjugates
(ID, IE ′ ) =(ID, IE) + (IE, IE ′ )
=(ID, IE) + 2(IE, IB)
=(ID, IE) + 2((IE, AC) + (AC, IB))
=(ID, IE) + 2(AC, IB)
β
=(π − γ) + 2 γ +
2
=β + γ = −α (mod π);
since (IE, AC) =
π
2
(ID, IF ′ ) =(ID, IF ) + (IF, IF ′ )
=(ID, IF ) + 2(IF, IC)
=(ID, IF ) + 2((IF, AB) + (AB, IC))
=(ID, IF ) + 2(AB, IC)
γ
= − (π − β) − 2 β +
2
= − (β + γ) = α (mod π)
since (IF, AB) =
π
2
Since E ′ and F ′ are on the incircle, and ID ⊥ BC, it follows that E ′ F ′ is parallel to
BC.
(4) Similarly, F ′ D′ and D′ E ′ are parallel to CA and AB respectively. It follows that
D′ E ′ F ′ is homothetic to ABC.
The ratio of homothety is r : R. Therefore, the center of homothety is the point T+
which divides OI in the ratio R : r. This is the internal center of similitude, or simply the
insimilicenter of (O) and (I).
A
E
F′
E′
F
T+
I
O
Ge
D′
B
D
C
7.4 Isogonal conjugates of the Gergonne and Nagel points
227
7.4.2 The Nagel point and the exsimilicenter T−
The isogonal conjugate of the Nagel point is the point T− which divides OI in the ratio
OT− : T− I = R : −r. This is the external center of similitude (or exsimilicenter) of the
circumcircle and the incircle.
Ib
A
Ic
Cc
D1′
O
T−
E1′
B
I
Bb
Na
F1′
Aa
Ia
C
228
7.5
Isogonal conjugates
The Brocard points
Analogous to the Crelles points, we may ask if there are concurrent lines through the vertices making equal angles with the sidelines. More precisely, given triangle ABC, does
there exist a point P satisfying
∠BAP = ∠CBP = ∠ACP = ω.
It turns out that is one such unique configuration.
A
ω
P
ω
ω
B
C
Note that if P is a point satisfying ∠BAP = ∠CBP , then the circle through P , A, B
is tangent to BC at B. This circle is unique and can be constructed as follows. Its center is
the intersection of the perpendicular bisector of AB and the perpendicular to BC at B.
A
ω
P
ω
B
C
Likewise, if ∠CBP = ∠ACP , then the circle through P , B, C is tangent to CA at
C. It follows that P is the intersection of these two circles. With this P , the circle P CA is
tangent to AB at A.
By Ceva’s theorem, the angle ω satisfies the equation
sin3 ω = sin(β − ω) sin(α − ω) sin(γ − ω).
It also follows that with the same ω, there is another triad of circles intersecting at
another point Q such that
∠CAQ = ∠ABQ = ∠BCQ = ω.
7.5 The Brocard points
229
A
A
ω
ω
Q
ω
ω
B
P
ω
ω
C
B
C
The points P and Q are isogonal conjugates. They are called the Brocard points of
triangle ABC.
A
ω
ω
Q
ω
P
ω
B
ω
ω
C
230
7.6
Isogonal conjugates
Kariya’s theorem
Given a triangle ABC with incenter I, consider a point X on the perpendicular from I to
BC, such that IX = t. We regard t > 0 if X and the point of tangency of the incircle with
the side BC are on the same side of I.
Theorem 7.2 (Kariya). Let I be the incenter of triangle ABC. If points X, Y , Z are chosen
on the perpendiculars from I to BC, CA, AB respectively such that IX = IY = IZ, then
the lines AX, BY , CZ are concurrent.
A
Y
Z
I
Q
B
C
X
Proof. (1) We compute the length of AX. Let the perpendicular from A to BC and the
parallel from X to the same line intersect at X ′ . In the right triangle AXX ′ ,
AX ′ =
XX ′ =
=
=
=
2∆
2rs
r(b + c)
−r+t=
−r+t=
+ t,
a
a
a
(s − b) − c cos B
1
1
(c + a − b) − (c2 + a2 − b2 )
2
2a
a(c + a − b) − (c2 + a2 − b2 )
2a
2
2
b − c − a(b − c)
(b − c)(b + c − a)
(b − c)(s − a)
=
=
.
2a
2a
a
Applying the Pythagorean theorem to the right triangle AXX ′ , we have
7.6 Kariya’s theorem
231
M′
A
A
I
B
C
X′
2
AX =
=
=
=
=
=
=
=
O
I
X
P
B
C
X
2
2 r(b + c)
(b − c)(s − a)
+t +
a
a
2
2
2
2
r (b + c) + (b − c) (s − a)
2r(b + c)t
+
+ t2
2
a
a
(s−a)(s−b)(s−c)(b+c)2
+ (s − a)2 (b − c)2 2r(b + c)t + at2
s
+
a2
a
2
2
(s − a)((s − b)(s − c)(b + c) + s(s − a)(b − c) ) 2r(b + c)t + at2
+
a2 s
a
(s − a) · a2 bc 2r(b + c)t + at2
+
a2 s
a
abc(s − a) 2r(b + c)t + at2
+
as
a
4Rr(s − a) 2r(b + c)t + at2
+
a
a
4Rr(s − a) + 2r(b + c)t + at2
.
a
(2) Let M ′ be the midpoint of the arc BAC of the circumcircle, and let M X intersect
OI at P . We shall prove that angle IAP = angle IAX.
First of all,
AI 2 =
(s − a)2
bc(s − a)
4Rr(s − a)
bc
2
=
=
.
=
(s
−
a)
·
A
s(s − a)
s
a
cos2 2
For later use, we also establish
AI 2 + 2Rr =
4Rr(s − a)
2Rr(2s − 2a) + 2Rr · a
2Rr(b + c)
+ 2Rr =
=
.
a
a
a
Since IP : P O = t : R, IP =
t
R+t
· OI. Recall that OI 2 = R2 − 2Rr. Applying the law
232
Isogonal conjugates
of cosines to triangle AIP , we have
AP 2 = AI 2 + IP 2 − 2 · AI · IP cos AIP
t
t2
2
(R2 − 2Rr) −
(AI 2 + OI 2 − R2 )
= AI +
2
(R + t)
R+t
t2
t
2
2
= AI +
(R
−
2Rr)
−
(AI 2 − 2Rr)
2
(R + t)
R+t
2
2 2
R
2R rt
R t
=
· AI 2 +
+
2
R+t
(R + t)
(R + t)2
R(R + t)AI 2 + 2R2 rt + R2 t2
=
(R + t)2
R2 · AI 2 + R(AI 2 + 2Rr)t + R2 t2
=
(R + t)2
2
+ 2R r(b+c)
t + R 2 t2
a
=
(R + t)2
R2 (4Rr(s − a) + 2r(b + c)t + at2 )
=
.
a(R + t)2
R2 ·
4Rr(s−a)
a
2
R
R
2
Note that AP 2 = (R+t)
2 · AX . This means that AP = R+t · AX.
(3) Let AI intersect P X at X ′′ and the circumcircle again at M , the antipode of M ′ .
M′
= −2 and OI
= − R+t
.
Note that M
M ′O
IP
t
M′
A
O
I
P
X ′′
B
C
X
M
Applying Menelaus’ theorem to triangle M ′ OP and transversal IX ′′ M , we have
−1 =
Now
XP
P M′
=
P X ′′
−t
P X ′′
−t
P X ′′ M ′ M OI
·
·
=⇒
=
=⇒
=
.
′′
′
′′
′
′
X M
M O IP
X M
2(R + t)
PM
2R + t
t
.
R
Therefore,
XP
P X ′′
=
2R+t
,
−R
and
R+t
AX
XX ′′
=
=
.
′′
X P
R
AP
7.7 Isogonal conjugate of an infinite point
233
This shows that AX ′′ bisects angle XAP . Since AX ′′ is the bisector of angle A, the lines
AP and AX are isogonal with respect to angle A.
(4) Likewise, if points Y and Z are chosen on the perpendiculars from I to CA and
AB such that IY = IZ = t = IX, then with the same point P on OI, the lines BP and
BY are isogonal with respect to angle B, and CP , CZ isogonal with respect to angle C.
Therefore the three lines AX, BY , CZ intersect at the isogonal conjugate of P (which
divides OI in the ratio OP : P I = R : t).
7.7
Isogonal conjugate of an infinite point
Proposition 7.3. Given a triangle ABC and a line ℓ, let ℓa , ℓb , ℓc be the parallels to ℓ
through A, B, C respectively, and ℓ′a , ℓ′b , ℓ′c their reflections in the angle bisectors AI, BI,
CI respectively. The lines ℓ′a , ℓ′b , ℓ′c intersect at a point on the circumcircle of triangle
ABC.
A
O
P
ℓ′b
I
ℓ′a
B
ℓ
ℓa
ℓb
C
ℓc
ℓ′c
Proof. Let P be the intersection of ℓ′b and ℓ′c .
(BP, P C) =(ℓ′b , ℓ′c )
=(ℓ′b , IB) + (IB, IC) + (IC, ℓ′c )
=(IB, ℓb ) + (IB, IC) + (ℓc , IC)
=(IB, ℓ) + (IB, IC) + (ℓ, IC)
=2(IB, IC)
π A
+
=2
2
2
=(BA, AC) (mod π).
Therefore, ℓ′b and ℓ′c intersect at a point on the circumcircle of triangle ABC.
Similarly, ℓ′a and ℓ′b intersect at a point P ′ on the circumcircle. Clearly, P and P ′ are
the same point since they are both on the reflection of ℓb in the bisector IB. Therefore, the
three reflections ℓ′a , ℓ′b , and ℓ′c intersect at the same point on the circumcircle.
234
Isogonal conjugates
Proposition 7.4. The isogonal conjugates of the infinite points of two perpendicular lines
are antipodal points on the circumcircle.
A
ℓ′
Q
ℓ
I
O
P
B
C
Proof. If P and Q are the isogonal conjugates of the infinite points of two perpendicular
lines ℓ and ℓ′ through A, then AP and AQ are the reflections of ℓ and ℓ′ in the bisector AI.
(AP, AQ) = (AP, IA) + (IA, AQ) = −(ℓ, IA) − (IA, ℓ′ ) = −(ℓ, ℓ′ ) =
Therefore, P and Q are antipodal points.
π
.
2
Chapter 8
The excentral and intouch triangles
8.1
The excentral triangle
8.1.1 The circumcenter of the excentral triangle
The excentral triangle of ABC has vertices the excenters IA , Ib , IC .
Since AIa ⊥ Ib Ic , BIb ⊥ Ic Ia , and CIc ⊥ Ia Ib ,
AIa , BIb , CIc are the altitudes, and
I is the orthocenter of the excentral triangle.
It follows that the circumcircle of triangle ABC is the nine-point circle of the excentral
triangle.
Ib
A
Ic
O
I′
I
C
B
Ia
Corollaries:
(1) Each side of the excentral triangle intersects the circumcircle (O) at its own midpoint
(apart from the vertices of triangle ABC).
(2) The circumcenter of the excentral triangle is the reflection of I in O.
236
The excentral and intouch triangles
Let the bisector AIa intersect the circumcircle at M . M is the midpoint of the arc BC
of the circumcircle (on the opposite side of A). Therefore, OM is perpendicular to BC.
Ib
A
Ic
O
I′
I
B
X′
C
M
Ia
The circumcenter I ′ is the reflection of I in O. Join I ′ to Ia . Since IO = OI ′ and
IM = M Ia , I ′ Ia is parallel to OM (and I ′ Ia = 2 · OM = 2R). Therefore, I ′ Ia is
perpendicular to BC, and it passes through the point of tangency of X ′ of BC with the
A-excircle.
Corollary. The perpendiculars from the excenters to the corresponding sides of a triangle
are concurrent. The point of concurrency is the circumcenter of the excentral triangle.
8.1 The excentral triangle
237
8.1.2 Relation between circumradius and radii of tritangent circles
Proposition 8.1. ra + rb + rc = 4R + r.
Ib
M′
A
Ic
I′
O
I
Ac
B
X
D
Aa
C
Ab
M
Ia
Proof. (1) Consider the four points Ib , B, C, Ic . They are concyclic since both Ib BIc and
Ib CIc are right angles. The center of the circle must be the midpoint M ′ of Ib Ic , which
must also lie on the perpendicular bisector of BC, which is the line DM ′ . Therefore, M ′
is the antipode of M on the circumcircle.
(2) Therefore, rb + rc = 2 · DM ′ = 2(R + OD).
(3) On the other hand, 2 · OD = r + I ′ Aa = r + 2R − ra .
(4) It follows that ra + rb + rc = 4R + r.
238
The excentral and intouch triangles
8.2
The homothetic center of the intouch and excentral
triangles
The intouch triangle has vertices the points of the tangency of the incircle with the sidelines.
Proposition 8.2. The excentral and the intouch triangles are homothetic at a point T dividing OI in the ratio OT : T I = 2R + r : −2r.
Ib
A
Ic
Y
I′
Z
T
I
B
O
C
X
Ia
Proof. (1) The segments Ib Ic and Y Z are parallel, since they are both perpendicular to the
bisector of angle A. Similarly, Ic Ia //ZX and Ia Ib //XY . Therefore the two triangles are
homothetic.
(2) Since the excentral triangle has circumradius 2R and the intouch triangle has circumradius r, the homothetic ratio is 2R : r.
(3) The homothetic center is the point dividing the segment I ′ I externally in the ratio
I ′ T : T I = 2R : −r.
(4) Since I ′ is the reflection of I in O, the homothetic center T divides OI in the ratio
OT : T I = 2R + r : −2r.
Chapter 9
Homogeneous Barycentric Coordinates
9.1
Absolute and homogeneous barycentric coordinates
The notion of barycentric coordinates dates back to A. F. Möbius (–). Given a reference
triangle ABC, we put at the vertices A, B, C masses u, v, w respectively, and determine
the balance point. The masses at B and C can be replaced by a single mass v + w at the
point X = v·B+w·C
. Together with the mass at A, this can be replaced by a mass u + v + w
v+w
at the point P which divides AX in the ratio AP : P X = v + w : u. This is the point with
, provided u + v + w 6= 0. 1 We also say that
absolute barycentric coordinate u·A+v·B+w·C
u+v+w
the balance point P has homogeneous barycentric coordinates (u : v : w) with reference
to ABC.
9.1.1 The centroid
The midpoints of the sides are
D=
B+C
,
2
E=
C +A
,
2
F =
A+B
.
2
The centroid G divides each median in the ratio 2 : 1. Thus,
G=
A + 2D
A+B+C
=
.
3
3
This is the absolute barycentric coordinate of G (with reference to ABC). Its homogeneous
barycentric coordinates are simply
G = (1 : 1 : 1).
1
A triple (u : v : w) with u + v + w = 0 does not represent any finite point on the plane. We shall say
that it represents an infinite point. See §9.4.
302
Homogeneous Barycentric Coordinates
A
A
F
Y
E
Z
I
G
B
C
D
B
X
C
9.1.2 The incenter
The bisector AX divides the side BC in the ratio BX : XC = c : b. This gives X =
bB+cC
ca
. Note that BX has length b+c
. Now, in triangle ABX, the bisector BI divides AX
b+c
ca
in the ratio AI : IX = c : b+c = b + c : a. It follows that
I=
aA + (b + c)X
aA + bB + cC
=
.
a+b+c
a+b+c
The homogeneous barycentric coordinates of the incenter are
I = (a : b : c).
9.1 Absolute and homogeneous barycentric coordinates
303
9.1.3 The barycenter of the perimeter
Consider the barycenter (center of mass) of the perimeter of triangle ABC. The edges
BC, CA, AB can be replaced respectively by masses a, b, c at their midpoint D = B+C
,
2
A+B
,
and
F
=
.
With
reference
to
the
medial
triangle
DEF
,
this
has
coordinates
E = C+A
2
2
a : b : c. Since the sidelengths of triangle DEF are in the same proportions, this barycenter
is the incenter of the medial triangle, also called the Spieker center Sp of ABC.
A
F
E
Sp
B
D
C
The center of mass of the perimeter is therefore the point
a·D+b·E+c·F
a+b+c
a · B+C
+ b · C+A
+ c · A+B
2
2
2
=
a+b+c
(b + c)A + (c + a)B + (a + b)C
.
=
2(a + b + c)
Sp =
In homogeneous barycentric coordinates,
Sp = (b + c : c + a : a + b) .
304
Homogeneous Barycentric Coordinates
9.1.4 The Gergonne point
We follow the same method to compute the coordinates of the Gergonne point Ge . Here,
BX = s − b and XC = s − c, so that
X=
(s − b)b + (s − c)C
.
a
A
Y
Z
I
Ge
B
X
C
The ratio AGe : Ge X, however, is not immediate obvious. It can nevertheless be found
by applying the Menelaus theorem to triangle ABX with transversal CZ. Thus,
AGe XC BZ
·
·
= −1.
Ge X CB ZA
From this,
Therefore,
CB ZA
−a s − a
a(s − a)
AGe
=−
·
=−
·
=
.
Ge X
XC BZ
s−c s−b
(s − b)(s − c)
(s − b)(s − c)A + a(s − a)X
(s − b)(s − c) + a(s − a)
(s − b)(s − c)A + (s − a)(s − c)B + (s − a)(s − b)C
=
.
(s − b)(s − c) + a(s − a)
Ge =
The homogeneous barycentric coordinates of the Gergonne point are
Ge = (s − b)(s − c) : (s − c)(s − a) : (s − a)(s − b)
1
1
1
: s−b
: s−c
.
= s−a
9.2 Cevian triangle
9.2
305
Cevian triangle
It is clear that the calculations in the preceding section applies in the general case. We
summarize the results in the following useful alternative of the Ceva theorem.
Theorem 9.1 (Ceva). Let X, Y , Z be points on the lines BC, CA, AB respectively. The
lines AX, BY , CZ are collinear if and only if the given points have coordinates of the
form
X = (0 : y : z),
Y = (x : 0 : z),
Z = (x : y : 0),
for some x, y, z. If this condition is satisfied, the common point of the lines AX, BY , CZ
is P = (x : y : z).
A
A
BP
CP
P
P
B
C
AP
B
X
C
Remarks. (1) The points X, Y , Z are called the traces of P . We also say that XY Z is
the cevian triangle of P (with reference to triangle ABC). Sometimes, we shall adopt the
more functional notation for the cevian triangle and its vertices:
cev(P ) :
AP = (0 : y : z),
BP = (x : 0 : z),
CP = (x : y : 0).
(2) The point P divides the segment AX in the ratio P X : AX = x : x + y + z.
(3) It follows that the areas of the oriented triangles P BC and ABC are in the ratio
∆(P BC) : ∆(ABC) = x : x + y + z. This leads to the following interpretation of
homogeneous barycentric coordinates: the homogeneous barycentric coordinates of a point
P can be taken as the proportions of (signed) areas of oriented triangles:
P = ∆(P BC) : ∆(P CA) : ∆(P AB).
306
Homogeneous Barycentric Coordinates
9.2.1 The circumcenter
Consider the circumcenter O of triangle ABC.
A
O
B
C
Since ∠BOC = 2A, the area of triangle OBC is
1
1
· OB · OC · sin BOC = R2 sin 2A.
2
2
Similarly, the areas of triangles OCA and OAB are respectively 21 R2 sin 2B and 12 R2 sin 2C.
It follows that the circumcenter O has homogeneous barycentric coordinates
∆OBC : ∆OCA : ∆OAB
1
1
1
= R2 sin 2A : R2 sin 2B : R2 sin 2C
2
2
2
= sin 2A : sin 2B : sin 2C
= a cos A : b cos B : c cos C
c 2 + a2 − b 2
a2 + b 2 − c 2
b 2 + c 2 − a2
:b·
:c·
=a·
2bc
2ca
2ab
= a2 (b2 + c2 − a2 ) : b2 (c2 + a2 − b2 ) : c2 (a2 + b2 − c2 ).
9.2 Cevian triangle
307
9.2.2 The Nagel point and the extouch triangle
The A-excircle touches the side BC at a point X ′ such that BX ′ = s − c and X ′ C = s − b.
From this, the homogeneous barycentric coordinates of X ′ are 0 : s − b : s − c; similarly
for the points of tangency Y ′ and Z of the B- and C-excircles:
Ib
A
Ic
Z′
Na
Y′
C
B
s−c
X′ s − b
s−b
s−c
Ia
X ′ =(0 : s − b : s − c),
Y ′ =(s − a : 0 : s − c),
Z ′ =(s − a : s − b : 0),
From these we conclude that AX ′ , BY ′ , and CZ ′ concur. Their common point is called
the Nagel point and has coordinates
Na = (s − a : s − b : s − c).
The triangle X ′ Y ′ Z ′ is called the extouch triangle.
308
Homogeneous Barycentric Coordinates
9.2.3 The orthocenter and the orthic triangle
For the orthocenter H with traces X, Y , Z on BC, CA, AB respectively, we have BX =
c cos B, XC = b cos C. This gives
BX : XC = c cos B : b cos C =
cos B cos C
:
;
b
c
similarly for the other two traces.
X
Y
Z
H
=
=
=
=
0
a
cos α
a
cos α
a
cos α
:
:
:
:
b
cos β
0
b
cos β
b
cos β
A
c
cos γ
c
cos γ
:
:
:
:
0
c
cos γ
A
Y
O
Z
B
H
X
C
B
The triangle XY Z is called the orthic triangle.
C
9.3 Homotheties
9.3
309
Homotheties
Let P be a given point, and k a real number. The homothety with center P and ratio k is
−→
−−→
the transformation h(P, k) which maps a point X to the point Y such that P Y = k · P X.
Equivalently, Y divides P X in the ratio P Y : Y X = k : 1 − k, and
h(P, k)(X) = (1 − k)P + kX.
k
P
1−k
Y
X
9.3.1 Superiors and inferiors
The homotheties h(G, −2) and h G, − 21 are called the superior and inferior operations
respectively. Thus, sup(P ) and inf(P ) are the points dividing P and the centroid G according to the ratios
P G : Gsup(P ) = 1 : 2,
P G : Ginf(P ) = 2 : 1.
P
sup(P )
G
inf(P )
Proposition 9.2. If P = (u : v : w) in homogeneous barycentric coordinates, then
sup(P ) = (v + w − u : w + u − v : u + v − w),
inf(P ) = (v + w : w + u : u + v).
Proof. In absolute barycentric coordinates,
sup(P ) = 3G − 2P
2(uA + vB + wC)
u+v+w
(u + v + w)(A + B + C) 2(uA + vB + wC)
−
=
u+v+w
u+v+w
(v + w − u)A + (w + u − v)B + (u + v − w)C
.
=
u+v+w
= (A + B + C) −
Therefore, sup(P ) = (v + w − u : w + u − v : u + v − w) in homogeneous barycentric
coordinates.
The case for inferior is similar.
310
Homogeneous Barycentric Coordinates
Example 9.1. (1) The superior of the incenter is the Nagel point. The inferior of the incenter is the Spieker center, the barycenter of the perimeter of the triangle.
(2) The inferior and superior of the Gergonne point
Ge = ((s − b)(s − c) : (s − c)(s − a) : (s − a)(s − b))
are
inf(Ge ) = ((s − c)(s − a) + (s − a)(s − b) : (s − a)(s − b) + (s − b)(s − c)
: (s − b)(s − c) + (s − c)(s − a))
= ((s − a)(s − c + s − b) : (s − b)(s − a + s − c) : (s − c)(s − b + s − a))
= (a(s − a) : b(s − b) : c(s − c)),
and
sup(Ge ) = ((s − c)(s − a) + (s − a)(s − b) − (s − b)(s − c)
: (s − a)(s − b) + (s − b)(s − c) − (s − c)(s − a)
: (s − b)(s − c) + (s − c)(s − a) − (s − a)(s − b))
= (a(s − a) − (s − b)(s − c) : b(s − b) − (s − c)(s − a)
: c(s − c) − (s − a)(s − b))
= (−s2 + (a + b + c)s − a2 − bc : −s2 + (a + b + c)s − b2 − ca
: −s2 + (a + b + c)s − ab)
= (s2 − a2 − bc : s2 − b2 − ca : s2 − c2 − ab)
= ((a + b + c)2 − 4a2 − 4bc : (a + b + c)2 − 4b2 − 4ca
: (a + b + c)2 − 4c2 − 4ab)
= (−3a2 + 2a(b + c) + (b − c)2 : −3b2 + 2b(c + a) + (c − a)2
: −3c2 + 2c(a + b) + (a − b)2 ).
(2) The nine-point center, being the midpoint of O and H, is the inferior of O.
Lo
N
H
G
O
From the homogeneous barycentric of O, we obtain
N = b2 (c2 + a2 − b2 ) + c2 (a2 + b2 − c2 ) : · · · : · · ·
= a2 (b2 + c2 ) − (b2 − c2 )2 : · · · : · · · .
Similarly, the deLongchamps point, being the reflection of H in O, is the superior of
H. It has coordinates
Lo = (−3a4 + 2a2 (b2 + c2 ) + (b2 − c2 )2 : −3b4 + 2b2 (c2 + a2 ) + (c2 − a2 )2
: −3c4 + 2c2 (a2 + b2 ) + (a2 − b2 )2 ).
9.3 Homotheties
311
A
Y
Na
X
H
Z
B
C
Example 9.2. The distance between O and Na is equal to R − 2r.
Proof. Since O = sup(N ) and Na = sup(I), ONa = 2 · N I = 2
R
2
− r = R − 2r.
Example 9.3. Pedals on altitudes equidistant from vertices We locate the point P whose
pedals on the altitudes are equidistant from the vertices.
Let P = (u : v : w). We require
v+w
2∆
w+u
2∆
u+v
2∆
v+w
w+u
u+v
·
=
·
=
·
=⇒
=
=
.
u+v+w a
u+v+w b
u+v+w c
a
b
c
This means that inf(P ) = I and P = sup(I) = Na , the Nagel point.
With P = Na , the common distance from the vertices to the pedals is
(c + a − b) + (a + b − c)
2∆
4∆
·
=
= 2r,
(b + c − a) + (c + a − b) + (a + b − c) a
a+b+c
the diameter of the incircle. The three pedals lie on the circle with diameter Na H, called
the Fuhrmann circle.
312
Homogeneous Barycentric Coordinates
9.3.2 Triangles bounded by lines parallel to the sidelines
Theorem 9.3 (Homothetic center theorem). If parallel lines Xb Xc , Yc Ya , Za Zb to the sides
BC, CA, AB of triangle ABC are constructed such that
AB : BXc = AC : CXb =1 : t1 ,
BC : CYa = BA : AYc =1 : t2 ,
CA : AZb = CB : BZa =1 : t3 ,
these lines bound a triangle A∗ B ∗ C ∗ homothetic to ABC with homothety ratio 1 + t1 +
t2 + t3 . The homothetic center is a point P with homogeneous barycentric coordinates
t1 : t2 : t3 .
A∗
A∗
Zb
Zb
Yc
Yc
A
A
P
Za
B∗
B
Xc
P
Ya
C
Xb
Za
C ∗B ∗
B
Ya
C
Xc
Xb
C∗
Proof. Let P be the intersection of B ∗ B and C ∗ C. Since
B ∗ C ∗ = B ∗ Xc + Xc Xb + Xb C ∗
= t3 a + (1 + t1 )a + t2 a = (1 + t1 + t2 + t3 )a,
we we have
P B : P B ∗ = P C : P C ∗ = 1 : 1 + t1 + t2 + t3 .
A similar calculation shows that AA∗ and BB ∗ intersect at the same point P . This shows
that A∗ B ∗ C ∗ is the image of ABC under the homothety h(P, 1 + t1 + t2 + t3 ).
Now we compare areas. Note that
c
a
(1) ∆(BZa Xc ) = BX
· BZ
· ∆(ABC) = t1 t3 ∆(ABC),
AB
CB
∆(P BC)
PB
CB
(2) ∆(BZa B ∗ ) = BB ∗ · BZa = t1 +t12 +t3 · t13 = t3 (t1 +t1 2 +t3 ) .
Since ∆(BZa B ∗ ) = ∆(BZa Xc ), we have ∆(P BC) = t1 +tt12 +t3 · ∆(ABC).
Similarly, ∆(P CA) = t1 +tt22 +t3 · ∆(ABC) and ∆(P AB) = t1 +tt32 +t3 · ∆(ABC). It
follows that
∆(P BC) : ∆(P CA) : ∆(P AB) = t1 : t2 : t3 .
9.3 Homotheties
313
The Grebe symmedian point
Consider the square erected externally on the side BC of triangle ABC, The line containing
the outer edge of the square is the image of BC under the homothety h(A, 1 + t1 ), where
2∆
+a
2
2
a
a
1 + t1 = a2∆ = 1 + 2∆
, i.e., t1 = 2∆
. Similarly, if we erect squares externally on the
a
other two sides, the outer edges of these squares are on the lines which are the images of
b2
c2
CA, AB under the homotheties h(B, 1 + t2 ) and h(C, 1 + t3 ) with t2 = 2∆
and t3 = 2∆
.
A∗
Ba
Ca
A
Bc
K
Cb
B
B∗
Ab
C
C∗
Ac
The triangle bounded by the lines containing
edges is called the Grebe
2these2 outer
b
c2
a
triangle of ABC. It is homothetic to ABC at 2∆ : 2∆ : 2∆ = (a2 : b2 : c2 ), and the
ratio of homothety is
1 + (t1 + t2 + t3 ) =
2∆ + a2 + b2 + c2
.
2∆
This homothetic center is called the Grebe symmedian point
K = (a2 : b2 : c2 ).
Remark. Note that the homothetic center remains unchanged if we replaced t1 , t2 , t3 by
kt1 , kt2 , kt3 for the same nonzero k. This means that if similar rectangles are constructed
on the sides of triangle ABC, the lines containing their outer edges always bound a triangle
with homothetic center K.
314
9.4
Homogeneous Barycentric Coordinates
Infinite points
A point with zero sum of homogeneous coordinates is called an infinite point. It can be
expressed as the difference of the absolute barycentric coordinates of two points (with
equal nonzero sums of homogeneous coordinates). As such, an infinite point defines the
direction of a family of parallel lines. All infinite points form the line at infinity: consisting
of points (x : y : z) satisfying x + y + z = 0. The infinite point of a line is defined uniquely
up to a nonzero scalar multiple.
Example 9.4.
1. The infinite points of the side lines BC, CA, AB are (0 : −1 : 1),
(1 : 0 : −1), (−1 : 1 : 0) respectively.
2. The infinite point of the A−altitude has homogeneous coordinates
(0 : Sγ : Sβ ) − a2 (1 : 0 : 0) = (−a2 : Sγ : Sβ ).
3. The infinite point of the Euler line is the point
(Sα (Sβ + Sγ ), Sβ (Sγ + Sα ), Sγ (Sα + Sβ ) (Sβγ , Sγα , Sαβ )
−
2S 2
S2
(Sα (Sβ + Sγ ) − 2Sβγ , Sβ (Sγ + Sα ) − 2Sγα , Sγ (Sα + Sβ ) − 2Sαβ
.
=
2S 2
O−H =
In homogeneous coordinates, this is
(Sα (Sβ + Sγ ) − 2Sβγ : Sβ (Sγ + Sα ) − 2Sγα : Sγ (Sα + Sβ ) − 2Sαβ ).
4. The infinite point of the OI-line is
(ca(c − a)(s − b) − ab(a − b)(s − c) : · · · : · · · )
∼ (a(a2 (b + c) − 2abc − (b + c)(b − c)2 ) : · · · : · · · ).
Chapter 10
Some applications of barycentric
coordinates
10.1
Construction of mixtilinear incircles
10.1.1 The insimilicenter and the exsimilicenter of the circumcircle
and incircle
The centers of similatitude of two circles are the points dividing the centers in the ratio of
their radii, either internally or externally. For the circumcircle and the incircle, these are
1
(r · O + R · I),
R+r
1
(−r · O + R · I).
T− =
R−r
T+ =
M′
A
Y
Z
B
T−
I
O
T+
C
X
M
We give an interesting application of these centers of similitude.
316
Some applications of barycentric coordinates
10.1.2 Mixtilinear incircles
A mixtilinear incircle of triangle ABC is one that is tangent to two sides of the triangle
and to the circumcircle internally. Denote by A′ the point of tangency of the mixtilinear
incircle K(ρ) in angle A with the circumcircle. The center K clearly lies on the bisector of
angle A, and AK : KI = ρ : −(ρ − r). In terms of barycentric coordinates,
K=
1
(−(ρ − r)A + ρI) .
r
Also, since the circumcircle O(A′ ) and the mixtilinear incircle K(A′ ) touch each other at
A′ , we have OK : KA′ = R − ρ : ρ, where R is the circumradius. From this,
K=
1
(ρO + (R − ρ)A′ ) .
R
A
A
O
I
I
T−
K
B
KA
C
A′
O
B
C
A′
M
Comparing these two equations, we obtain, by rearranging terms,
R(ρ − r)A + r(R − ρ)A′
RI − rO
=
.
R−r
ρ(R − r)
We note some interesting consequences of this formula. First of all, it gives the intersection
of the lines joining AA′ and OI. Note that the point on the line OI represented by the left
hand side is T− , the exsimilicenter of the circumcircle and the incircle.
This leads to a simple construction of the mixtilinear incircle. Given a triangle ABC,
extend AT− to intersect the circumcircle at A′ . The intersection of AI and A′ O is the center
KA of the mixtilinear incircle in angle A.
The other two mixtilinear incircles can be constructed similarly.
10.2 Isotomic and isogonal conjugates
317
10.2
Isotomic and isogonal conjugates
10.3
Isotomic conjugates
The Gergonne and Nagel points are examples of isotomic conjugates. Two points P and
Q (not on any of the side lines of the reference triangle) are said to be isotomic conjugates
if their respective traces are symmetric with respect to the midpoints of the corresponding
sides. Thus,
BX = X ′ C,
CY = Y ′ A,
AZ = Z ′ B.
B
X′
Z
Z′
X
P•
P
C
Y
Y′
A
We shall denote the isotomic conjugate of P by P • . If P = (x : y : z), then
1 1 1
•
= (yz : zx : xy).
: :
P =
x y z
318
Some applications of barycentric coordinates
Example 10.1. (The Gergonne and Nagel points)
Ge =
1
s−a
1
s−b
:
:
1
s−c
,
Na = (s − a : s − b : s − c).
Ib
A
Ic
Z′
Y
Z
Ge
B
I
Na
X
Y′
C
X′
Ia
Example 10.2. (The isotomic conjugate of the orthocenter) The isotomic conjugate of the
orthocenter is the point
H • = (b2 + c2 − a2 : c2 + a2 − b2 : a2 + b2 − c2 ).
Its traces are the pedals of the deLongchamps point Lo , the reflection of H in O.
A
Z′
Lo
Y
O
Y′
Z
B
H•
H
X
X′
C
10.4 Equal-parallelians point
10.4
319
Equal-parallelians point
Given triangle ABC, we want to construct a point P the three lines through which parallel
to the sides cut out equal intercepts. Let P = xA + yB + zC in absolute barycentric
coordinates. The parallel to BC cuts out an intercept of length (1 − x)a. It follows that the
three intercepts parallel to the sides are equal if and only if
1−x:1−y :1−z =
1 1 1
: : .
a b c
The right hand side clearly gives the homogeneous barycentric coordinates of I • , the isotomic conjugate of the incenter I. 1 This is a point we can easily construct. Now, translating
into absolute barycentric coordinates:
1
1
I • = [(1 − x)A + (1 − y)B + (1 − z)C] = (3G − P ).
2
2
we obtain P = 3G − 2I • , and can be easily constructed as the point dividing the segment
I • G externally in the ratio I • P : P G = 3 : −2. The point P is called the congruentparallelians point of triangle ABC.
B
P
G
I•
C
1
The isotomic conjugate of the incenter appears in ETC as the point X75 .
A
320
Some applications of barycentric coordinates
Chapter 11
Computation of barycentric coordinates
11.1
The Feuerbach point
Proposition 11.1. The homogeneous barycentric coordinates of the Feuerbach point are
((b + c − a)(b − c)2 : (c + a − b)(c − a)2 : (a + b − c)(a − b)2 ).
Proof. The Feuerbach point Fe is the point of (internal) tangency of the incircle and the
nine-point circle. It divides N I in the ratio N Fe : Fe I = R2 : −r = R : −2r. Therefore,
Fe =
R · I − 2r · N
R − 2r
in absolute barycentric coordinates.
A
Fe
I
H
O
N
B
C
From the homogeneous barycentric coordinates of N ,
(a2 (b2 + c2 ) − (b2 − c2 )2 , b2 (c2 + a2 ) − (c2 − a2 )2 , c2 (a2 + b2 ) − (a2 − b2 )2 ) = 32∆2 · N.
we have
r
a2 (b2 + c2 ) − (b2 − c2 )2 , · · · , · · ·
2
16∆
R
=
a2 (b2 + c2 ) − (b2 − c2 )2 , · · · , · · · ,
4sabc
R
R
(a, b, c) =
· 2abc(a, b, c).
R·I =
2s
4sabc
2r · N =
322
Computation of barycentric coordinates
Therefore,
Fe ∼
∼
=
=
=
=
∼
R · I − 2r · N
2abc · a − (a2 (b2 + c2 ) − (b2 − c2 )2 )
(a2 (2bc − b2 − c2 ) + (b − c)2 (b + c)2 , · · · , · · · )
(−a2 (b − c)2 + (b − c)2 (b + c)2 , · · · , · · · )
(((b + c)2 − a2 )(b − c)2 , · · · , · · · )
((a + b + c)(b + c − a)(b − c)2 , · · · , · · · )
((b + c − a)(b − c)2 , · · · , · · · ).
Proposition 11.2. (a) ONa is parallel to N Fe .
(b) ONa = R − 2r.
(c) Na H is parallel to OI.
(d) The reflection of H in I and the reflection of Na in O are the same point.
(e) IN and OSp intersect at the midpoint of Na H.
A
Fe
I
N
Sp
O
Na
H
B
C
11.2 The OI line
11.2
323
The OI line
11.2.1 The circumcenter of the excentral triangle
Let I ′ be the reflection of I in O. Show that I ′ is also the midpoint of Na Lo .
Let N ′ be the midpoint of OLo . The triangles OI ′ N ′ and OIN are congruent. I ′ N ′ is
parallel to IN and hence Na O. Furthermore, I ′ N ′ = IN = 12 Na O. It follows that I ′ is the
midpoint of Na Lo .
I ′ = (a(a3 + a2 (b + c) − a(b + c)2 − (b + c)(b − c)2 ) : · · · : · · · ).
11.2.2 The centers of similitude of the circumcircle and the incircle
T+ = (a2 (s − a) : b2 (s − b) : c2 (s − c)).
Proof.
T+ ∼ r · O + R · I
r
R
=
(a2 (b2 + c2 − a2 ), b2 (c2 + a2 − b2 ), c2 (a2 + b2 − c2 )) + (a, b, c)
2
16∆
2s
2 2
2
2
2 2
2
2
2 2
2
2
∼ (a (b + c − a ), b (c + a − b ), c (a + b − c )) + 8Rrs(a, b, c)
= (a2 (b2 + c2 − a2 ), b2 (c2 + a2 − b2 ), c2 (a2 + b2 − c2 )) + 2abc(a, b, c)
= (a2 (b2 + 2bc + c2 − a2 ), · · · , · · · )
= (a2 (a + b + c)(b + c − a), · · · , · · · )
∼ (a2 (b + c − a), · · · , · · · ).
T− =
Proof.
a2
s−a
:
b2
s−b
:
c2
s−c
.
T− ∼ r · O − R · I
R
r
(a2 (b2 + c2 − a2 ), b2 (c2 + a2 − b2 ), c2 (a2 + b2 − c2 )) − (a, b, c)
=
2
16∆
2s
∼ (a2 (b2 + c2 − a2 ), b2 (c2 + a2 − b2 ), c2 (a2 + b2 − c2 )) − 8Rrs(a, b, c)
= (a2 (b2 + c2 − a2 ), b2 (c2 + a2 − b2 ), c2 (a2 + b2 − c2 )) − 2abc(a, b, c)
= (a2 (b2 − 2bc + c2 − a2 ), · · · , · · · )
= (a2 (b − c + a)(b − c − a), · · · , · · · )
∼ (a2 (c + a − b)(a + b − c), · · · , · · · ).
324
Computation of barycentric coordinates
Example 11.1. (a) G, T+ , Fe are collinear.
(b) H, T− , Fe are collinear.
Proof. (a) This follows from
(a2 (b + c − a), b2 (c + a − b), c2 (a + b − c))
− ((b + c − a)(b − c)2 , (c + a − b)(c − a)2 , (a + b − c)(a − b)2 )
= ((b + c − a)(a2 − (b − c)2 ), (c + a − b)(b2 − (c − a)2 ), (a + b − c)(c2 − (a − b)2 )
= ((b + c − a)(a − b + c)(a + b − c), (c + a − b)(b − c + a)(b + c − a),
(a + b − c)(c − a + b)(c + a − b))
= (b + c − a)(c + a − b)(a + b − c)(1, 1, 1).
11.2.3 The homothetic center T of excentral and intouch triangles
The two triangles are homothetic since their corresponding sides are perpendicular to the
angle bisectors of triangle ABC. Denote by T the homothetic center. This is clearly the
exsimilicenter of their circumcircles. It is therefore the point dividing I ′ and I in the ratio
I ′ T : IT = 2R : r. It follows that OT : T I = 2R + r : −2r.
I′
I
T
O
T =
a
s−a
:
b
s−b
:
c
s−c
.
Proof.
T ∼ −2r · O + (2R + r)I
−2r
2R + r
=
(a, b, c)
(a2 (b2 + c2 − a2 ), · · · , · · · ) +
2
2
16r s
2s
∼ −(a2 (b2 + c2 − a2 ), · · · , · · · ) + (2R + r)(4rs)(a, b, c)
= −(a2 (b2 + c2 − a2 ), · · · , · · · ) + (2abc + 4(s − a)(s − b)(s − c))(a, b, c)
= (a(−a(b2 + c2 − a2 ) + (2abc + 4(s − a)(s − b)(s − c))), · · · , · · · )
1
=
a(a + b + c)(c + a − b)(a + b − c), · · · , · · ·
2
a
,··· ,··· .
∼
b+c−a
11.3 The excentral triangle
325
Exercise
1. Show that T I : IO = 2r : 2R − r.
2. Find the ratio of division T+ T : T T− .
3. Show that G, Ge , and T are collinear by finding p, q satisfying
p(1, 1, 1) + q((c + a − b)(a + b − c), (a + b − c)(b + c − a), (b + c − a)(c + a − b))
= 2(a(c + a − b)(a + b − c), b(a + b − c)(b + c − a), c(b + c − a)(c + a − b)).
Answer: p = −(b + c − a)(c + a − b)(a + b − c) and q = a + b + c.
4. Given that GGe : Ge T = 2(2R − r) : 3r, show that Ge , I, Lo are collinear.
Apply the converse of Menelaus’ theorem to triangle OGT with Ge on GT , I on
T O, and Lo on OG.
GGe T I OLo
2(2R − r)
2r
−3
·
·
=
·
·
= −1.
Ge T IO Lo G
3r
2R − r 4
11.3
The excentral triangle
11.3.1 The centroid
The centroid of the excentral triangle is the point
Ia + Ib + Ic
1 (−a, b, c) (a, −b, c) (a, b, −c)
=
+
+
3
3 b+c−a c+a−b a+b−c
∼ (c + a − b)(a + b − c)(−a, b, c) + (a + b − c)(b + c − a)(a, −b, c)
+ (b + c − a)(c + a − b)(a, b, −c)
∼ (a(−(s − b)(s − c) + (s − c)(s − a) + (s − a)(s − b)), · · · , · · · )
∼ (a(s2 − 2as − bc + ca + ab), · · · , · · · )
∼ (a(s2 − a2 − bc), · · · , · · · )
∼ (a(−3a2 + 2a(b + c) + (b − c)2 ), · · · , · · · ).
11.3.2 The incenter
B (a, −b, c)
C (a, b, −c)
A (−a, b, c)
·
+ cos ·
+ cos ·
2 b+c−a
2 c+a−b
2 a+b−c
B
A
cos 2
cos C2
cos 2
(−a,
b,
c)
+
(a,
−b,
c)
+
(a, b, −c)
∼
2r cot A2
2r cot B2
2r cot C2
A
B
C
∼ sin (−a, b, c) + sin (a, −b, c) + sin (a, b, −c)
2
2 2
B
C
A
, ··· ,··· .
∼ a − sin + sin + sin
2
2
2
∼ cos
326
Computation of barycentric coordinates
Chapter 12
Some interesting circles
12.1
A fundamental principle on 6 concyclic points
12.1.1 The radical axis of two circles
Given two nonconcentric circles C1 and C2 . The locus of points of equal powers with
respect to the circle is a straight line perpendicular to the line joining their centers. In fact,
if the circles are concentric, there is no finite point with equal powers with respect to the
circles. On the other hand, if the centers are distinct points A and B at a distance d apart,
there is a unique point P with distances AP = x and P B = d − x such that
r12 − x2 = r22 − (d − x)2 .
If this common value is m, then every point Q on the perpendicular to AB at P has power
m − P Q2 with respect to each of the circles. This line is called the radical axis of the two
circles.
If the two circles intersect at two distinct points, then the radical axis is the line joining
these common points. If the circles are tangent to each other, then the radical axis is the
common tangent.
Theorem 12.1. Given three circles with distinct centers, the radical axes of the three pairs
of circles are either concurrent or are parallel.
Proof. (1) If any two of the circles are concentric, there is no finite point with equal powers
with respect to the three circles.
(2) If the centers of the circles are distinct and noncollinear, then two of the radical
axes, being perpendiculars to two distinct lines with a common point, intersect at a point.
This intersection has equal powers with respect to all three circles, and also lies on the third
radical axis.
(3) If the three centers are distinct but collinear, then the three radical axes three parallel
lines, which coincide if any two of them do. This is the case if and only if the three circles
two points in common, or at mutually tangent at a point. In this case we say that the circles
are coaxial.
328
Some interesting circles
12.1.2 Test for 6 concyclic points
Proposition 12.2. Let X, X ′ be points on the sideline a, Y , Y ′ on b, and Z, Z ′ on c. The six
points are on a circle if and only if the four points on each pair of sidelines are concyclic.
Z
Y′
A
Z′
B
C
X
Y
X
Proof. It is enough to prove the sufficiency part. Let Ca be the circle through the four points
Y , Y ′ , Z, Z ′ on b and c, and Cb the one through Z, Z ′ , X, X (on c and a), and Cc through
X, X, Y , Y ′ (on a and (b). We claim that these three circles are identical. If not, then they
are pairwise distinct. The three pairs among them have radical axes a (for Cb and Cc ), b (for
Cc and Ca ), and c (for Ca and Cb ) respectively. Now, the three radical axes of three distinct
circles either intersect at a common point (the radical center), or are parallel (when their
centers are on a line), or coincide (when, in additon, the three circles are coaxial). In no
case can the three radical axes form a triangle (with sidelines a, b, c. This shows that the
three circles coincide.
12.2 The Taylor circle
12.2
329
The Taylor circle
Consider the orthic triangle XY Z, and the pedals of each of the points X, Y , Z on the two
sides not containing it. Thus,
Sideline Pedals of X Pedals of Y
a
Xb
b
Ya
c
Za
Zb
Pedals of Z
Xc
Yc
A
Zb
Yc
Y
Ya
Z
H
Za
B
Xc
Xb
X
C
It is easy to write down the lengths of various segments. From these we easily determine
the coordinates of these pedals. For example, from Ya C = b cos2 γ, we have AYa =
b − b cos2 γ = b sin2 γ. Note that
AYa · AYc =(b − b cos2 γ)(b cos2 α) = b2 cos2 α sin2 γ = 4R2 cos2 α sin2 β sin2 γ,
AZa · AZb =(c − c cos2 β)(c cos2 α) = c2 cos2 α sin2 β = 4R2 cos2 α sin2 β sin2 γ,
S 2 ·S
2
αα
giving AYa ·AYc = AZa ·AZb = S4Rαα2 = Sa2·S
. Similarly, BXb ·BXc = BZb ·BZa = a2 b2ββ
b 2 c2
c2
2
S ·Sγγ
and CYc · CYa = CXc · CXb = a2 b2 c2 . By Proposition 12.2, the six points Xb , Xc , Yc , Ya ,
Za , Zb are concyclic. The circle containing them is called the Taylor circle.
Exercise
1. Calculate the length of Xb Xc .
2. Find the equation of the line Xb Xc .
1
3. The three lines Xb Xc , Yc Ya , Za Zb bound a triangle with perspector K.
1
−S 2 x + SBB y + SCC z = 0.
330
12.3
Some interesting circles
Two Lemoine circles
12.3.1 The first Lemoine circle
Given triangle ABC, how can one choose a point K so that when parallel lines are constructed through it to intersect each sideline at two points, the resulting six points are on a
circle?
A
Yc
Zb
K
Za
B
Xc
L
Ya
Xb
C
Analysis. By the intersecting chords theorem, AYa · AYc = AZa · AZb . We have
AZa AZb
AYa AYc
·
= c2 ·
·
AC AC
AB AB
AYc
AZb
=⇒ b2 ·
= c2 ·
AC
AB
CXb
Xb C
BX
c
= c2 ·
= c2 ·
=⇒ b2 ·
BC
CB
BC
BXc
c2
=⇒
= 2.
Xb C
b
b2 ·
Similarly,
BZa
Zb A
=
a2
,
b2
and
Xc Xb
a2
Xc K
BZa
Xc Xb
=
= 2.
=
=
Xb C
KYa
KYc
Zb A
b
Therefore, BXc : Xc Xb : Xb C = c2 : a2 : b2 .
These proportions determine the points Xb and Xc on BC, and subsequently the other
points: the parallels Xc Yc and Xb Zb (to c and b respectively) intersect at K, and the parallel through K to a determines the points Ya and Za . The point K is called the Lemoine
symmedian point of triangle ABC, and the circle containing these six points is called the
first Lemoine circle.
Denote by L the center of the first Lemoine circle. Note that L lies on the perpendicular
bisector of each of the parallel segments Xc Xb and Za Ya . This means that the line joining
the midpoints of these segments is the common perpendicular bisector of the segments, and
the trapezoid Xc Xb Ya Za is symmetric; so are Ya Yc Zb Xb and Zb Za Xc Yc by the same reasoning. It follows that the segments Yc Zb , Za Xc , and Xb Ya have equal lengths. (Exercise:
abc
Show that this common length is a2 +b
2 +c2 ).
12.3 Two Lemoine circles
331
12.3.2 The second Lemoine circle
Now, if we construct the parallels of these segments through K to intersect the sidelines,
we obtain another three equal segments with common midpoint K. The 6 endpoints, Xb′ ,
abc
Xc′ on a, Yc′ , Ya′ on b, and Za′ , Zb′ on c, are on a circle center K and radius a2 +b
2 +c2 . This
circle is called the second Lemoine circle of triangle ABC.
A
Za′
Ya′
Yc′
K
Zb′
B
Xb′
C
Xc′
12.3.3 Construction of K
The length of AK is twice the median of triangle AYc Zb on the side Yc Zb . If we denote by
a
ma etc the lengths of medians of triangle ABC, then AK = a22bcm
. Similarly, BK =
+b2 +c2
2camb
2abmc
and CK = a2 +b2 +c2 . This allows us to determine the angles KAB etc and the radii
a2 +b2 +c2
=
of the circles KAB etc. The radius of the circle KAB, for example, is Rc = AB·AK·BK
4∆(KAB)
abcma mb
BK
∆
. It follows that sin KAB = 2Rc = bma .
(a2 +b2 +c2 )∆(ABC)
A
A
I
K
G
KG
B
C
B
C
a mc
Consider also the circle GAC. This has radius Rb′ = AC·AG·CG
= bm3∆
. From this,
4∆(GAC)
∆
CG
sin GAC = 2R′ = bma . This shows that ∠KAB = ∠GAC, and AK and the median AG
b
are isogonal lines with respect to the sides AB and AC. Similarly, BK and CK are the
lines isogonal to the medians BG and CG respectively, and K is the symmedian point of
triangle ABC.
332
Some interesting circles
12.3.4 The center of the first Lemoine circle
We show that the center L of the first Lemoine circle is the midpoint between K and the
circumcenter O. It is clear that this center is on the perpendicular bisector of Xb Xc . Let M
be the midpoint of Xb Xc , and X the orthogonal projection of K on BC.
A
Yc
Zb
O
Za
B
K
XcX
Ya
M
DXb
C
We have
BM =
1
a3
a(a2 + 2c2 )
ac2
+
·
=
,
a2 + b 2 + c 2 2 a2 + b 2 + c 2
2(a2 + b2 + c2 )
and
BX = BXc + KXc cos B
a2 c
c 2 + a2 − b 2
ac2
+
·
= 2
a + b 2 + c 2 a2 + b 2 + c 2
2ca
ac2 + a(c2 + a2 − b2 )
=
2(a2 + b2 + c2 )
a(a2 − b2 + 3c2 )
.
=
2(a2 + b2 + c2 )
It follows that
BX + BD =
a(a2 + 2c2 )
a(a2 − b2 + 3c2 ) a(a2 + b2 + c2 )
+
=
= 2 · BM.
2(a2 + b2 + c2 )
2(a2 + b2 + c2 )
a2 + b 2 + c 2
This means that M is the midpoint of XD, and the perpendicular to a at M contains the
midpoint of OK. The same reasoning shows that the midpoint of OK also lies on the
perpendiculars to b and c respectively at the midpoints of Yc Ya and Za Zb . It is therefore the
center L of the first Lemoine circle. (Exercise. Calculate the radius of the first Lemoine
circle).
Chapter 13
Straight line equations
13.1
Area and barycentric coordinates
Theorem 13.1. If for i = 1, 2, 3, Pi = xi · A + yi · B + zi · C (in absolute barycentric
coordinates) , then the area of the oriented triangle P1 P2 P3 is
x1 y1 z 1
∆P1 P2 P3 = x2 y2 z2 · ∆ABC.
x3 y3 z 3
Example 13.1. (Area of cevian triangle) Let P = (u : v : w) be a point with cevian triangle
XY Z. The area of the cevian triangle XY Z is
0 v w
1
2uvw
u 0 w ·∆=
· ∆.
(v + w)(w + u)(u + v)
(v + w)(w + u)(u + v)
u v 0
√
point (so that u, v, w are positive), then v + w ≥ 2 vw, w + u ≥
√ If P is an interior √
2 wu, and u + v ≥ 2 uv. It follows that
2uvw
2uvw
1
√
√ = .
≤ √
(v + w)(w + u)(u + v)
4
2 vw · 2 wu · 2 uv
Equality holds if and only if u = v = w, i.e., P = (1 : 1 : 1) = G, the centroid. The
centroid is the interior point with largest cevian triangle.
402
Straight line equations
13.2
Equations of straight lines
13.2.1 Two-point form
The area formula has an easy and extremely important consequence: three points Pi =
(ui , vi , wi ) are collinear if and only if
u1 v 1 w 1
u2 v2 w2 = 0.
u3 v 3 w 3
Consequently, the equation of the line joining two points with coordinates (x1 : y1 : z1 )
and (x2 : y2 : z2 ) is
x1 y1 z 1
x2 y2 z2 = 0,
x y z
or
(y1 z2 − y2 z1 )x + (z1 x2 − z2 x1 )y + (x1 y2 − x2 y1 )z = 0.
Examples
(1) The equations of the sidelines BC, CA, AB are respectively x = 0, y = 0, z = 0.
(2) Given a point P = (u : v : w), the cevian line AP has equation wy − vz = 0;
similarly for the other two cevian lines BP and CP . These lines intersect corresponding
sidelines at the traces of P :
AP = (0 : v : w),
BP = (u : 0 : w),
CP = (u : v : 0).
(3) The equation of the line joining the centroid and the incenter is
1 1 1
a b c = 0,
x y z
or (b − c)x + (c − a)y + (a − b)z = 0.
(4) The equations of some important lines:
Euler line
OI-line
Soddy line
Brocard axis
van Aubel line
OH
OI
IGe
OK
HK
P
(b2 − c2 )(b2 + c2 − a2 )x = 0
Pcyclic
bc(b − c)(b + c − a)x = 0
Pcyclic
(b − c)(s − a)2 x = 0
Pcyclic 2 2 2
b c (b − c2 )x = 0
Pcyclic
cyclic Sαα (Sβ − Sγ )x = 0
13.2 Equations of straight lines
403
13.2.2 Intersection of two lines
The intersection of the two lines
p1 x + q1 y + r1 z = 0,
p 2 x + q 2 y + r2 z = 0
is the point
(q1 r2 − q2 r1 : r1 p2 − r2 p1 : p1 q2 − p2 q1 ).
Proposition 13.2. Three lines pi x + qi y + ri z = 0, i = 1, 2, 3, are concurrent if and only if
p 1 q 1 r1
p2 q2 r2 = 0.
p 3 q 3 r3
Examples
(1) The intersection of the Euler line and the Soddy line is the point
(c − a)(s − b)2
(a − b)(s − c)2
: ··· : ···
2
2
2
2
2
(c − a )(c + a − b ) (a − b2 )(a2 + b2 − c2 )
2
=(c − a)(a − b)
=(c − a)(a − b)
(s − b)2
(s − c)2
: ··· : ···
(c + a)(c2 + a2 − b2 ) (a + b)(a2 + b2 − c2 )
(s − b)2
a(b − c)
: ··· : ···
2
2
2
(c + a)(c + a − b ) (b − c)(a + b + c)2
=(b − c)(c − a)(a − b)
(s − b)2
a
: ··· : ···
(c + a)(c2 + a2 − b2 ) (a + b + c)2
1
(c + a − b)2
4a
= (b − c)(c − a)(a − b)
: ··· : ···
2
2
2
(c
+
a)(c
+
a
−
b
)
(a
+
b + c)2
4
1
= (b − c)(c − a)(a − b)(−3a4 + 2a2 (b2 + c2 ) + (b2 − c2 )2 ) : · · · : · · ·
4
Writing a2 = Sβ + Sγ etc., we have
− 3a4 + 2a2 (b2 + c2 ) + (b2 − c2 )2
= − 3(Sβ + Sγ )2 + 2(Sβ + Sγ )(2Sα + Sβ + Sγ ) + (Sβ − Sγ )2
=4(Sαβ + Sγα − Sβγ ).
This intersection has homogeneous barycentric coordinates
−Sβγ + Sγα + Sαβ : Sβγ − Sγα + Sαβ : Sβγ + Sγα − Sαβ .
This is the reflection of H in O, and is called the deLongchamps point Lo .
404
13.3
Straight line equations
Perspective triangles
Many interesting points and lines in triangle geometry arise from the perspectivity of triangles. We say that two triangles X1 Y1 Z1 and X2 Y2 Z2 are perspective, X1 Y1 Z1 ⊼ X2 Y2 Z2 , if
the lines X1 X2 , Y1 Y2 , Z1 Z2 are concurrent. The point of concurrency, ∧(X1 Y1 Z1 , X2 Y2 Z2 ),
is called the perspector. Along with the perspector, there is an axis of perspectivity, or the
perspectrix, which is the line joining containing
Y 1 Z2 ∩ Z1 Y 2 ,
Z 1 X 2 ∩ X 1 Z2 ,
X1 Y 2 ∩ Y 1 X2 .
We denote this line by L∧ (X1 Y1 Z1 , X2 Y2 Z2 ).
Homothetic triangles are clearly prespective. If triangles T and T′ , their perspector is
the homothetic center, which we shall denote by ∧0 (T, T′ ).
Proposition 13.3. A triangle with vertices
X = U : v
Y = u : V
Z = u : v
: w,
: w,
: W,
for some U , V , W , is perspective to ABC at ∧(XY Z) = (u : v : w). The perspectrix is
the line
x
y
z
+
+
= 0.
u−U
v−V
w−W
Proof. The line AX has equation wy − vz = 0. It intersects the sideline BC at the point
(0 : v : w). Similarly, BY intersects CA at (u : 0 : w) and CZ intersects AB at (u : v : 0).
These three are the traces of the point (u : v : w).
The line Y Z has equation −(vw − V W )x + u(w − W )y + u(v − V )z = 0. It intersects
the sideline BC at (0 : v − V : −(w − W )). Similarly, the lines ZX and XY intersect CA
and AB respectively at (−(u − U ) : 0 : w − W ) and (u − U : −(v − V ) : 0). It is easy to
see that these three points are collinear on the line
x
y
z
+
+
= 0.
u−U
v−V
w−W
13.3 Perspective triangles
405
The excentral triangle
The excentral triangle is perspective with ABC; the perspector is the incenter I:
Ia
Ib
Ic
I
= −a : b : c
= a : −b : c
= a : b : −c
= a : b : c
13.3.1 The Conway configuration
Given triangle ABC, extend
(i) CA and BA to Ya and Za such that AYa = AZa = a,
(ii) AB and CB to Zb and Xb such that BZb = BXb = b,
(iii) BC and AC to Xc and Yc such that CXc = CYc = c.
Za
Ya
a
a
A
Z
Y
?
Xb
C
B
b
X
b
Xc
c
c
Yc
Zb
These points have coordinates
Ya = (a + b : 0 : −a),
Zb = (−b : b + c : 0),
Xc = (0 : −c : c + a),
Za = (c + a : −a : 0);
Xb = (0 : a + b : −b);
Yc = (−c : 0 : b + c).
From the coordinates of Yc and Zb , we determine easily the coordinates of X = BYc ∩CZb :
Yc = −c :
0
: b + c = −bc :
0
: b(b + c)
Zb = −b : b + c :
0
= −bc : c(b + c) :
0
X =
= −bc : c(b + c) : b(b + c)
406
Straight line equations
Similarly, the coordinates of Y = CZa ∩ AXc , and Z = AXb ∩ BYa can be determined.
The following table shows that the perspector of triangles ABC and XY Z is the point with
homogeneous barycentric coordinates a1 : 1b : 1c .
X
Y
Z
?
=
−bc
: c(b + c) : b(b + c)
= c(c + a) :
−ca
: a(c + a)
= b(a + b) : a(a + b) :
−ab
=
=
=
=
=
−1
b+c
1
a
1
a
1
a
:
:
:
:
1
b
−1
c+a
1
b
1
b
:
:
:
:
1
c
1
c
−1
a+b
1
c
13.4 Perspectivity
13.4
407
Perspectivity
Many interesting points and lines in triangle geometry arise from the perspectivity of triangles. We say that two triangles X1 Y1 Z1 and X2 Y2 Z2 are perspective, X1 Y1 Z1 ⊼ X2 Y2 Z2 , if
the lines X1 X2 , Y1 Y2 , Z1 Z2 are concurrent. The point of concurrency, ∧(X1 Y1 Z1 , X2 Y2 Z2 ),
is called the perspector. Along with the perspector, there is an axis of perspectivity, or the
perspectrix, which is the line joining containing
Y1 Z2 ∩ Z1 Y2 ,
Z 1 X 2 ∩ X 1 Z2 ,
X1 Y 2 ∩ Y 1 X2 .
We denote this line by L∧ (X1 Y1 Z1 , X2 Y2 Z2 ). We justify this in §below.
If one of the triangles is the triangle of reference, it shall be omitted from the notation.
Thus, ∧(XY Z) = ∧(ABC, XY Z) and L∧ (XY Z) = L∧ (ABC, XY Z).
Homothetic triangles are clearly prespective. If triangles T and T′ , their perspector is
the homothetic center, which we shall denote by ∧0 (T, T′ ).
Proposition 13.4. A triangle with vertices
X = U : v
Y = u : V
Z = u : v
: w,
: w,
: W,
for some U , V , W , is perspective to ABC at ∧(XY Z) = (u : v : w). The perspectrix is
the line
x
y
z
+
+
= 0.
u−U
v−V
w−W
Proof. The line AX has equation wy − vz = 0. It intersects the sideline BC at the point
(0 : v : w). Similarly, BY intersects CA at (u : 0 : w) and CZ intersects AB at (u : v : 0).
These three are the traces of the point (u : v : w).
The line Y Z has equation −(vw − V W )x + u(w − W )y + u(v − V )z = 0. It intersects
the sideline BC at (0 : v − V : −(w − W )). Similarly, the lines ZX and XY intersect CA
and AB respectively at (−(u − U ) : 0 : w − W ) and (u − U : −(v − V ) : 0). These three
points are collinear on the trilinear polar of (u − U : v − V : w − W ).
The triangles XY Z and ABC are homothetic if the perspectrix is the line at infinity.
408
Straight line equations
13.4.1 The Schiffler point: intersection of four Euler lines
Theorem 13.5. Let I be the incenter of triangle ABC. The Euler lines of the triangles
IBC, ICA, IAB are concurrent at a point on the Euler line of ABC, namely, the Schiffler
point
a(b + c − a) b(c + a − b) c(a + b − c)
Sc =
.
:
:
b+c
c+a
a+b
Proof. Let I be the incenter of triangle ABC.
We first compute the equation of the Euler line of the triangle IBC.
The centroid of triangle IBC is the point (a : a + 2b + c : a + b + 2c). The circumcenter
of triangle is the midpoint of IIa . This is the point (−a2 : b(b + c) : c(b + c)). From these
we obtain the equation of the Euler line:
x
y
z
a
a
+
2b
+
c
a
+
b
+ 2c = (b − c)(b + c)x + a(c + a)y − a(a + b)z.
0=
2
−a
b(b + c)
c(b + c)
The equations of the Euler lines of the triangles ICA and IAB can be obtained by
cyclic permutations of a, b, c and x, y, z. Thus the three Euler lines are
(b − c)(b + c)x +
a(c + a)y −
a(a + b)z = 0,
−b(b + c)x + (c − a)(c + a)y +
b(a + b)z = 0,
c(b + c)x −
c(c + a)y + (a − b)(a + b)z = 0.
Computing the intersection of the latter two lines, we have the point
(b + c)x : (c + a)y : (a + b)z
=
c−a
b
−b
b
−b c − a
: −
:
−c a − b
c a−b
c
−c
= (c + a)(a − b) + bc : b(c + a − b) : c(b − (c − a))
= a(b + c − a) : b(c + a − b) : c(a + b − c).
It is easy to verify that this point also lies on the Euler line of IBC given by the first
equation:
(b − c)a(b + c − a) + ab(c + a − b) − ac(a + b − c)
= a((b − c)(b + c − a) + b(c + a − b) − c(a + b − c))
= a(b2 − c2 − ab + ca + bc + ab − b2 − ca − bc + c2 )
= 0.
It is routine to verify that this point also lies on the Euler line of ABC, with equation
(b2 − c2 )(b2 + c2 − a2 )x + (c2 − a2 )(c2 + a2 − b2 )y + (a2 − b2 )(a2 + b2 − c2 )z = 0.
This shows that the four Euler lines are concurrent.
Chapter 14
Cevian nest theorem
14.1
Trilinear pole and polar
14.1.1 Trilinear polar of a point
Given a point P with traces AP , BP , and CP on the sidelines of triangle ABC, let
X = BP CP ∩ BC,
Y = CP AP ∩ CA,
Z = AP BP ∩ AB.
These points X, Y , Z lie on a line called the trilinear polar (or simply tripolar) of P .
Y
A
BP
CP
X
B
Z
P
AP
C
410
Cevian nest theorem
If P = (u : v : w), then BP = (u : 0 : w) and CP = (u : v : 0). The line BP CP has
equation
x y
z
− + + = 0.
u v w
It intersects the sideline BC at the point X = (0 : v : −w).
Similarly, AP = (0 : v : w) and the points Y , Z are
Y = (−u : 0 : w),
Z = (u : −v : 0).
The line containing the three points X, Y , Z is
x y
z
+ + = 0.
u v w
This is the tripolar of P .
14.1 Trilinear pole and polar
411
14.1.2 Tripole of a line
Given a line L intersecting BC, CA, AB at X, Y , Z respectively, let
A′ = BY ∩ CZ,
B ′ = CZ ∩ AX,
C ′ = AX ∩ BY.
The lines AA′ , BB ′ and CC ′ are concurrent. The point of concurrency is the tripole P of
L.
A
Z
C′
Y
A′
X
B
C
P
B′
Clearly P is the tripole of L if and only if L is the tripolar of P .
412
14.2
Cevian nest theorem
Anticevian triangles
The vertices of the anticevian triangle of a point P = (u : v : w)
cev−1 (P ) :
Pa = (−u : v : w),
Pb = (u : −v : w),
Pc = (u : v : −w)
are the harmonic conjugates of P with respect to the cevian segments AAP , BBP and
CCP , i.e.,
AP : P AP = −APa : Pa AP ;
similarly for Pb and Pc . This is called the anticevian triangle of P since ABC is the cevian
triangle P in Pa Pb Pc . It is also convenient to regard P , Pa , Pb , Pc as a harmonic quadruple
in the sense that any three of the points constitute the harmonic associates of the remaining
point.
14.2.1 Construction of anticevian triangle
If the trilinear polar LP of P intersects the sidelines BC, CA, AB at X ′ , Y ′ , Z ′ respectively, then the anticevian triangle cev−1 (P ) is simply the triangle bounded by the lines
AX ′ , BY ′ , and CZ ′ .
Y
A
BP
CP
Pc
P
X
B
AP
Pa
Z
Pb
C
14.2 Anticevian triangles
413
Another construction of anticevian triangle
Here is an alternative construction of cev−1 (P ).
Let AH BH CH be the orthic triangle, and X the reflection of P in a, then the intersection
of the lines AH X and OA is the harmonic conjugate Pa of P in AAP :
APa
AP
=−
.
Pa AP
P AP
A
A
P
B
AH
P
AP
C
X
B
AH
AP
X
Pa
′
C
Pa
A′
Proof. Let A be the reflection of A in BC. Applying Menelaus’ theorem to triangle
AP AA′ with transversal AH XPa , we have
APa AP X A′ AH
·
= −1.
·
Pa AP XA′ AH A
This gives
APa
PA
AP
XA′
=−
=−
=−
,
Pa AP
AP X
AP P
P AP
showing that Pa and P divide AAP harmonically.
414
Cevian nest theorem
Examples of anticevian triangles
(1) The anticevian triangle of the centroid is the superior triangle, bounded by the lines
through the vertices parallel to the opposite sides.
(2) The anticevian triangle of the incenter is the excentral triangle whose vertices are
the excenters.
(3) The vertices of the tangential triangle being
A′ = (−a2 : b2 : c2 ),
B ′ = (a2 : −b2 : c2 ),
C ′ = (a2 : b2 : −c2 ),
these clearly form are the anticevian triangle of a point with coordinates (a2 : b2 : c2 ),
which we call the symmedian point K.
(4) The anticevian triangle of the circumcenter. Here is an interesting property of
−1
cev (O). Let the perpendiculars to AC and AB at A intersect BC at Ab and Ac respectively. We call AAb Ac an orthial triangle of ABC. The circumcenter of AAb Ac is the
vertex Oa of cev−1 (O); similarly for the other two orthial triangles. (See §??).
Oc
A
Ob
Ab
B
Ac
C
Oa
14.3 Cevian quotients
14.3
415
Cevian quotients
14.3.1 The cevian nest theorem
Theorem 14.1. For arbitrary points P and Q, the cevian triangle cev(P ) and the anticevian triangle cev−1 (Q) are always perspective. If P = (u : v : w) and Q = (u′ : v ′ : w′ ),
then
(i) the perspector is the point
∧(cev(P ), cev−1 (Q)) =
′
′
v′ w′
w ′ u′
v
w ′ u′ v ′
u
: v′ − +
+
: w′ − + +
,
u′ − + +
u
v
w
v
w
u
w
u
v
(ii) the perspectrix is the line L∧ (cev(P ), cev−1 (Q)) with equation
X 1 u′ v ′ w ′ − + +
x = 0.
u
u
v
w
cyclic
Y′
A
Z′
Y
Q
M
Z
P
B
C
X
X′
Proof. (i) Let cev(P ) = XY Z and cev−1 (Q) = X ′ Y ′ Z ′ . Since X = (0 : v : w) and
X ′ = (−u′ : v ′ : w′ ), the line XX ′ has equation
1 w′ v′
1
1
−
x − · y + · z = 0.
′
u w
v
v
w
The equations of Y Y ′ and ZZ ′ can be easily written down by cyclic permutations of
(u, v, w), (u′ , v ′ , w′ ) and (x, y, z). It is easy to check that the line XX ′ contains the point
′
′
u
v
w ′ u′ v ′
v′ w′
w ′ u′
′
′
′
u − + +
+
:v − +
:w − + +
u
v
w
v
w
u
w
u
v
416
Cevian nest theorem
whose coordinates are invariant under the above cyclic permutations. This point therefore
also lies on the lines Y Y ′ and ZZ ′ .
(ii) The lines Y Z and Y ′ Z ′ have equations
− xu +
y
v
y
v′
+
+
z
w
z
w′
= 0,
= 0.
They intersect at the point
U ′ = (u(wv ′ − vw′ ) : vwv ′ : −vww′ ).
Similarly, the lines pairs ZX, Z ′ X ′ and XY , X ′ Y ′ have intersections
V ′ = (−wuu′ : v(uw′ − wu′ ) : wuw′ )
and
W ′ = (uvu′ : −uvv ′ : w(vu′ − uv ′ )).
The three points U ′ , V ′ , W ′ lie on the line with equation given above.
Corollary 14.2. If T′ is a cevian triangle of T and T′′ is a cevian triangle of T′ , then T′′
is a cevian triangle of T.
Proof. With reference to T2 , the triangle T1 is anticevian.
Remark. Suppose T′ = cevT (P ) and T′′ = cevT′ (Q). If P = (u : v : w) with respect to
T, and Q = (u′ : v ′ : w′ ) with respect to T′ , then,
u
v
w
′′
∧(T, T ) =
(v + w) : ′ (w + u) : ′ (u + v)
u′
v
w
with respect to triangle T. The equation of the perspectrix L∧ (T, T′′ ) is
X 1 v + w w + u u + v
x = 0.
+
− ′ +
′
′
u
u
v
w
cyclic
These formulae, however, are quite difficult to use, since they involve complicated
changes of coordinates with respect to different triangles.
We shall simply write
P/Q :=
V
(cev(P ), cev−1 (Q))
and call it the cevian quotient of P by Q.
14.3 Cevian quotients
417
The cevian quotients of the centroid G/P
If P = (u : v : w),
G/P = (u(−u + v + w) : v(−v + w + u) : w(−w + u + v)).
Some common examples of G/P .
P G/P coordinates
I
Mi (a(s − a) : b(s − b) : c(s − c))
O K (a2 : b2 : c2 )
K
O (a2 Sα : b2 Sβ : c2 Sγ )
The point Mi = G/I is called the Mittenpunkt of triangle ABC. 1 It is the symmedian
point of the excentral triangle. The tangential triangle of the excentral triangle is homothetic
to ABC at T .
(i) We compute the symmedian point of the excentral triangle. Note that
Ib Ic2 =
a2 bc
,
(s − b)(s − c)
Ic Ia2 =
ab2 c
,
(s − c)(s − a)
Ia Ib2 =
abc2
.
(s − a)(s − b)
For the homogeneous barycentric coordinates of the symmedian point of the excentral triangle Ia Ib Ic , we have
Ib Ic2 · Ia + Ic Ia2 · Ib + Ia Ib2 · Ic
(−a, b, c)
ab2 c
(a, −b, c)
abc2
(a, b, −c)
a2 bc
·
+
·
+
·
=
(s − b)(s − c) 2(s − a)
(s − c)(s − a) 2(s − b)
(s − a)(s − b) 2(s − c)
abc
(a(−a, b, c) + b(a, −b, c) + c(a, b, −c))
=
2(s − a)(s − b)(s − c)
abc
=
(a(−a + b + c), b(a − b + c), c(a + b − c)) .
2(s − a)(s − b)(s − c)
From this, we obtain the Mittenpunkt Mi .
(ii) Since the excentral triangle is homothetic to the intouch triangle at T , their tangential triangles are homothetic at the same triangle center. Using the cevian nest theorem, we
have
V
a
b
c
T = 0 (cev(Ge ), cev−1 (I)) = s−a
.
: s−b
: s−c
1
This appears as X9 in ETC.
418
Cevian nest theorem
The cevian quotients of the orthocenter H/P
For P = (u : v : w),
H/P = (u(−Sα u + Sβ v + Sγ w) : v(−Sβ v + Sγ w + Sα u) : w(−Sγ w + Sα u + Sβ v)).
Examples
(1) H/G = (Sβ + Sγ − Sα : Sγ + Sα − Sβ : Sα + Sβ − Sγ ) is the superior of H • .
(2) H/I is a point on the OI-line, dividing OI in the ratio R + r : −2r. 2
H/I = (a(a3 + a2 (b + c) − a(b2 + c2 ) − (b + c)(b − c)2 ) : · · · : · · · ).
2
2
2
(3) H/K = Saα : Sbβ : Sc γ is the homothetic center of the orthic and tangential trian-
gle. 3 It is a point on the Euler line.
(4) H/O is the orthocenter of the tangential triangle. 4
(5) H/N is the orthocenter of the orthic triangle. 5
The cevian quotient Ge /K
This is the perspector of the intouch triangle and the tangential triangle 6
Ge /K = (a2 (a3 − a2 (b + c) + a(b2 + c2 ) − (b + c)(b − c)2 ) : · · · : · · · ).
2
This point appears as X46 in ETC.
This appears as X25 in ETC.
4
This appears as X155 in ETC.
5
This appears as X52 in ETC.
6
This appears as X1486 in ETC.
3
14.3 Cevian quotients
419
14.3.2 Basic properties of cevian quotients
Proposition 14.3. (1) P/P = P .
(2) If P/Q = M , then Q = P/M .
Y′
A
Z′
Y
Q
M
Z ′′
Z
P
B
C
X
X ′′
X′
Y ′′
Proof. (2) Let P = (u : v : w), Q = (u′ : v ′ : w′ ), and M = (x : y : z). We have
′
x
u
u′
v′ w′
− + +
,
=
u
u
u
v
w
v ′ u′ v ′ w ′
y
=
− +
,
v
v u
v
w
z
w ′ u′ v ′ w ′
=
+ −
.
w
w u
v
w
From these,
′
′
x y
u
z
v′ w′
v′ w′
u
− + + =
,
− +
+ −
u v w
u
v
w
u
v
w
′
′
x y
z
u
v′ w′
v′ w′
u
− + = − + +
+ −
,
u v w
u
v
w
u
v
w
′
′
z
u
v′ w′
v′ w′
u
x y
+ − = − − +
− +
,
u v w
u
v
w
u
v
w
420
Cevian nest theorem
and
z y x y
z z x y
z
x x y
− +
+ −
− + +
:
:
u
u v w
v u v w
w u v w
u′ v ′ w ′
:
:
.
=
u
v
w
It follows that
x y
x y
x y
z
z
z
u′ : v ′ : w ′ = x − + +
− +
+ −
: y
: z
.
u v w
u v w
u v w
Chapter 15
Circle equations
15.1
The power of a point with respect to a circle
The power of P with respect to a circle Q(ρ) is the quantity P(P ) := P Q2 − ρ2 . A point
P is in, on, or outside the circle according as P is negative, zero, or positive.
Proposition 15.1. Let P = xA + yB + zC in absolute barycentric coordinates.
P(P ) = xP(A) + yP(B) + zP(C) − (a2 yz + b2 zx + c2 xy).
A
P
Q
B
X
C
, so that
Proof. Let X be the trace of P on BC. In absolute coordinates, X = yB+zC
y+z
P = xA + (y + z)X. Applying Stewart’s theorem in succession to triangles QAX, QBC
and ABC, we have
P Q2 = xAQ2 + (y + z)XQ2 − x(y + z)AX 2
y
z
yz
2
2
2
2
= xAQ + (y + z)
BQ +
CQ −
BC − x(y + z)AX 2
y+z
y+z
(y + z)2
yz
BC 2 − x(y + z)AX 2
= xAQ2 + yBQ2 + zCQ2 −
y+z
yz
BC 2
= xAQ2 + yBQ2 + zCQ2 −
y+z
y
yz
z
2
2
2
AC +
AB −
· BC
− x(y + z)
y+z
y+z
(y + z)2
(1 − x)yz
= xAQ2 + yBQ2 + zCQ2 −
· BC 2 − zx · AC 2 − xy · AB 2
y+z
= xAQ2 + yBQ2 + zCQ2 − (a2 yz + b2 zx + c2 xy).
422
Circle equations
From this it follows that
P(P ) = P Q2 − ρ2
= x(AQ2 − ρ2 ) + y(BQ2 − ρ2 ) + z(CQ2 − ρ2 ) − (a2 yz + b2 zx + c2 xy)
= xP(A) + yP(B) + zP(C) − a2 yz − b2 zx − c2 xy.
15.2
Circle equation
Using homogeneous barycentric coordinates for P and writing
f := P(A),
g := P(B),
h := P(C),
for the powers of A, B, C with respect to a circle Q(ρ), we have,
f x + gy + hz a2 yz + b2 zx + c2 xy
P(P ) =
−
x+y+z
(x + y + z)2
(a2 yz + b2 zx + c2 xy) − (x + y + z)(f x + gy + hz)
.
= −
(x + y + z)2
Therefore, the equation of the circle is
(a2 yz + b2 zx + c2 xy) − (x + y + z)(f x + gy + hz) = 0.
Example 15.1. (1) The equation of the circumcircle is a2 yz + b2 zx + c2 xy = 0 since
f = g = h = 0.
(2) For the incircle, we have
f = (s − a)2 ,
g = (s − b)2 ,
h = (s − c)2 .
The equation of the incircle is
a2 yz + b2 zx + c2 xy − (x + y + z)((s − a)2 x + (s − b)2 y + (s − c)2 z) = 0.
(3) Similarly, the A-excircle has equation
a2 yz + b2 zx + c2 xy − (x + y + z)(s2 x + (s − c)2 y + (s − b)2 z) = 0.
(4) For the nine-point circle, we have f = 2b · c cos A = 2b · Sbα =
g = 21 Sβ and h = 12 Sγ . Therefore, the equation of the nine-point circle is
1
S .
2 α
2(a2 yz + b2 zx + c2 xy) − (x + y + z)(Sα x + Sβ y + Sγ z) = 0.
Similarly,
15.3 Points on the circumcircle
423
Exercise
1. Find the equation of the Conway circle.
2. Find the equations of the circles
(i) CBBC passing through B and C, and tangent to BC at B,
(ii) CBCC passing through B and C, and tangent to BC at C.
3. Compute the coordinates of the Brocard points:
(i) Ω→ as the intersection of the circles CBBC , CCCA , and CAAB ,
(ii) Ω← as the intersection of the circles CBCC , CCAA , and CABB .
4. Find the equation of the circle with diameter BC.
15.3
Points on the circumcircle
The equation of the circumcircle can be written in the form
a2 b 2 c 2
+ +
= 0.
x
y
z
This shows that the circumcircle consists of the isogonal conjugates of infinite points.
15.3.1 X(101)
The point
X(101) =
b2
c2
a2
:
:
b−c c−a a−b
a
b
c
:
:
b−c c−a a−b
is clearly on the circumcircle.
15.3.2 X(100)
The point
X(100) =
is clearly on the circumcircle. It is the isogonal conjugate of the infinite point
(a(b − c) : b(c − a) : c(a − b))
(on the trilinear polar of the incenter, namely, the line xa + yb + zc = 0).
Its inferior is a point on the nine-point circle. To find this, we rewrite
X(100) = (a(c − a)(a − b) : b(a − b)(b − c) : c(b − c)(c − a)).
424
Circle equations
From this,
inf(X(100)) = (b(a − b)(b − c) + c(b − c)(c − a) : · · · : · · · )
= ((b − c)(b(a − b) + c(c − a)) : · · · : · · · )
= ((b − c)2 (b + c − a) : · · · : · · · ),
the Feuerbach point!
1. The distance from X(100) to the Nagel point is the diameter of the incircle.
2. X(100) is the intersection of the Euler lines of the triangles Ia BC, Ib CA, Ic AB.
15.3.3 The Steiner point X(99)
The Steiner point
X(99) =
1
1
1
: 2
: 2
2
2
2
b −c
c −a
a − b2
The inferior of the Steiner point is
X(115) = ((b2 − c2 )2 : (c2 − a2 )2 : (a2 − b2 )2 ).
This is the midpoint between the Fermat points.
15.3.4 The Euler reflection point E = X(110)
Theorem 15.2. The reflections of the Euler line in the sidelines of triangle ABC are concurrent at
a2
b2
c2
E=
:
:
b 2 − c 2 c 2 − a2 a2 − b 2
on the circumcircle.
Proof. The Euler line
Sα (Sβ − Sγ )x + Sβ (Sγ − Sα )y + Sγ (Sα − Sβ )z = 0
intersects the sideline BC at
X = (0 : Sγ (Sα − Sβ ) : −Sβ (Sγ − Sα ).
We find the reflection of H in BC as follows. From the relation
(Sβγ + Sγα + Sαβ )(0, Sγ , Sβ ) + Sβγ (Sβ + Sγ , −Sγ , −Sβ )
= (Sβ + Sγ )(Sβγ , Sγα , Sαβ ),
we have
H=X+
Sβγ
(Sβ + Sγ , −Sγ , −Sβ ).
(Sβ + Sγ )(Sβγ + Sγα + Sαβ )
15.4 Circumcevian triangle
425
Therefore, its reflection in BC is the point
X′ = X −
Sβγ
(Sβ + Sγ , −Sγ , −Sβ ).
(Sβ + Sγ )(Sβγ + Sγα + Sαβ )
In homogeneous barycentric coordinates, this is
X ′ = (Sβγ + Sγα + Sαβ )(0, Sγ , Sβ ) − Sβγ (Sβ + Sγ , −Sγ , −Sβ )
= (−Sβγ (Sβ + Sγ ), Sγ (2Sβγ + Sγα + Sαβ ), Sβ (2Sβγ + Sγα + Sαβ ))
The inferior of the Euler reflection is the point
X(125) = ((b2 − c2 )2 (b2 + c2 − a2 ) : (c2 − a2 )2 (c2 + a2 − b2 ) : (a2 − b2 )2 (a2 + b2 − c2 )).
This is the intersection of the Euler lines of the triangles AY Z, BZX, CXY , where XY Z
is the orthic triangle.
15.4
Circumcevian triangle
Let P = (u : v : w). The lines AP , BP , CP intersect the circumcircle again at X, Y , Z.
The triangle XY Z is called the circumcevian triangle of P . Since X lies on the line AP ,
X = (x : v : w) for some x. This point lies on the circumcircle if and only if
a2 vw + b2 xw + c2 xv = 0.
This gives x =
−a2 vw
.
b2 w+c2 v
Therefore,
X = (−a2 vw : (b2 w + c2 v)v : (b2 w + c2 v)w).
Similarly,
Y = ((c2 u+a2 w)u : −b2 wu : (c2 u+a2 w)w),
Z = ((a2 v+b2 u)u : (a2 v+b2 u)v : −c2 uv).
Proposition 15.3. The circumcevian triangle of P = (u : v : w) is perspective with the
tangential triangle at
4
b4
c4
a
2
a − 2 + 2 + 2 : ··· : ··· .
u
v
w
Proof. The vertices of the tangential triangle are (−a2 , b2 , c2 ), (a2 , −b2 , c2 ), (a2 , b2 , −c2 ).
The line joining (−a2 , b2 , c2 ) to X is
x
y
z
−a2
b2
c2
= 0.
2
2
2
2
2
−a vw (b w + c v)v (b w + c v)w
426
Circle equations
This is
(b4 w2 − c4 v 2 )x + a2 b2 w2 y − a2 c2 v 2 z = 0.
Similarly, the lines joining (a2 , −b2 , c2 ) to Y and (a2 , b2 , −c2 ) to Z are
−a2 b2 w2 x + (c4 u2 − a4 w2 )y + b2 c2 u2 z = 0,
a2 b2 v 2 x − b2 c2 u2 y + (b4 v 2 − b4 u2 )z = 0.
These three lines concur at a point with coordinates given above.
1. G: X(22) = (a2 (−a4 + b4 + c4 ) : · · · : · · · ).
2 4 4 4 22 22
a (a +b +c −2a b −2a c )
2. H: X(24) =
: ··· : ··· .
b2 +c2 −a2
These two points are on the Euler line, and are the centers of similitude of the circumcircle and incircle of the tangential triangle.
Example 15.2.
15.5
The third Lemoine circle
Given a point P = (u : v : w), it is easy to find the equation of the circle Ca through P , B,
C. Since P(B) = P(C) = 0, the equation of the circle is
a2 yz + b2 zx + c2 xy − (x + y + z) · f x = 0
Ca :
for some f . Since the circle passes through P = (u : v : w), we must have
f=
a2 vw + b2 wu + c2 uv
.
u(u + v + w)
This circle Ca intersects the lines AC and AB each again at another point. To find the
intersection with AC, we put y = 0 in the equation of (Ca ) and obtain b2 zx−f x(x+z) = 0,
x((b2 − f )z − f x)) = 0. Therefore, apart from C = (0, 0, 1), the circle Ca intersects AC at
Ba = (b2 − f : 0 : f ) = (b2 u2 + b2 uv − a2 vw − c2 uv : 0 : a2 vw + b2 wu + c2 uv).
Similarly, the circle Ca intersects AB again at
Ca = (c2 − f : f : 0) = (c2 u2 + c2 wu − a2 vw − b2 wu : a2 vw + b2 wu + c2 uv : 0).
Similarly, with
g=
a2 vw + b2 wu + c2 uv
v(u + v + w)
and
h=
a2 vw + b2 wu + c2 uv
,
w(u + v + w)
the circles Cb through P , C, A and Cc through P , A, B intersect the sidelines again at
Cb = (g : c2 −g : 0),
Ab = (0 : a2 −g : g),
Ac = (0 : h : a2 −h),
Bc = (h : 0 : b2 −h).
15.5 The third Lemoine circle
427
Note the lengths of the segments:
ABa =
f
f
·b= ,
2
b
b
ABc =
b2 − h
b2 − h
·
b
=
,
b2
b
and
f
c2 − g
c2 − g
f
·
c
=
,
AC
=
·
c
=
.
b
c2
c
c2
c
The four points Ba , Bc , Ca , Cb are concyclic if and only if
ACa =
ABa · ABc = ACa · ACb =⇒
f (b2 − h)
f (c2 − g)
b2
h
v2
=
=⇒
=
=
.
b2
c2
c2
g
w2
2
Likewise, the four points Cb , Ca , Ab , Ac are concyclic if and only if ac 2 = wu , and the
2
four points Ac , Ab , Bc , Ba are concyclic if and only if ab2 = uv .
By the principle of 6 concyclic points, the six points Ab , Ac , Bc , Ba , Ca , Cb are concyclic if and only if
u : v : w = a2 : b 2 : c 2 ,
namely, P = (u : v : w) = (a2 : b2 : c2 ), the symmedian point. The circle C containing
these 6 points is the third Lemoine circle.
For this choice of P ,
f=
3b2 c2
,
a2 + b 2 + c 2
g=
3c2 a2
,
a2 + b 2 + c 2
h=
3a2 b2
.
a2 + b 2 + c 2
With respect to the circle C containing these 6 points, we have
P(A) =
3b2 c2 · b2 (b2 + c2 − 2a2 )
3b2 c2 (b2 + c2 − 2a2 )
f (b2 − h)
=
=
.
b2
b2 (a2 + b2 + c2 )2
(a2 + b2 + c2 )2
Similarly,
P(B) =
3c2 a2 (c2 + a2 − 2b2 )
,
(a2 + b2 + c2 )2
P(C) =
3a2 b2 (a2 + b2 − 2c2 )
.
(a2 + b2 + c2 )2
From these, we obtain the equation of the third Lemoine circle:
(a2 + b2 + c2 )2 (a2 yz + b2 zx + c2 xy)
− 3(x + y + z) b2 c2 (b2 + c2 − 2a2 )x + c2 a2 (c2 + a2 − 2b2 )y + a2 b2 (a2 + b2 − 2c2 )z = 0.