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Chapter 1
Pythagoras Theorem
and Its Applications
1
...
1
...
a
b
b
a
c
b
c
a
c
c
b
b
c
b
c
a
b
1
...
2
a
a
a
b
a
Converse Theorem
If the lengths of the sides of a triangles satisfy the relation a2 + b2 = c2 , then
the triangle contains a right angle
...
Let ABC be a triangle with BC = a, CA = b, and AB = c satisfying a2 + b2 = c2
...
By the Pythagorean theorem, XY 2 = a2 + b2 = c2 , so that XY = c
...
This means that
6 ACB = 6 XZY is a right angle
...
Dissect two given squares into triangles and quadrilaterals and rearrange the pieces into a square
...
BCX and CDY are equilateral triangles inside a rectangle ABCD
...
Show that
(a) AP Q is an equilateral triangle;
(b) 4AP B + 4ADQ = 4CP Q
...
ABC is a triangle with a right angle at C
...
4
...
Show that 0 < k < 1, and
a : b : c = 1 − k2 : 2k : 1 + k 2
...
5
...
If the median on the side c
is the geometric mean of the sides a and b, show that one of the acute
angles is 15◦
...
Let ABC be a right triangle with a right angle at vertex C
...
Show that the length t of a side of this square is given by
1 1
1
= +
...
1/a^2 + 1/b^2 = 1/d^2
...
More generally, for h ≤ k, there is, up to
similarity, a unique right triangle satisfying c = ha + kb provided
(i) h√ 1 ≤ k, or
<
(ii) 22 ≤ h = k < 1, or
(iii) h, k > 0, h2 + k 2 = 1
...
YIU: Euclidean Geometry
4
7
...
If d is
the height of on the hypotenuse, show that
1
1
1
+ 2 = 2
...
(Construction of integer right triangles) It is known that every right
triangle of integer sides (without common divisor) can be obtained by
choosing two relatively prime positive integers m and n, one odd, one
even, and setting
a = m2 − n2 ,
c = m2 + n 2
...
(b) Complete the following table to find all such right triangles with
sides < 100:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
(xiii)
(xiv)
(xv)
(xvi)
m
2
3
4
4
5
5
6
6
7
7
7
8
8
8
9
9
n
1
2
1
3
2
4
1
5
2
4
6
1
3
5
2
4
a = m2 − n2
3
b = 2mn
4
c = m2 + n2
5
65
72
97
YIU: Euclidean Geometry
1
...
2
...
2
...
Describe
a semicircle with P B as diameter, and let the perpendicular through A
intersect the semicircle at Q
...
Q
Q
x
P
A
PA = a, PB = b; PQ^2 = ab
...
Describe a semicircle
with AB as diameter, and let the perpendicular through P intersect the
semicircle at Q
...
Q
x
A
a
P
y
b
x
a
B
b
a
y ^2 = a(a+b) = a^2 + ab,
y ^2 = a^2 + x^2
...
Example
To cut a given rectangle of sides a < b into three pieces that can be rearranged into a square
...
4
Phillips and Fisher, p
...
YIU: Euclidean Geometry
7
Exercise
1
...
Show that the
2
2
diameter of the circle is a +d
...
Two parallel chords of a circle has lengths 168 and 72, and are at a
distance 64 apart
...
3
3
...
The distance between the two endpoints A and B is a
...
Find the radius of the circles
...
ABP Q is a rectangle constructed on the hypotenuse of a right triangle ABC
...
3
Answer: The distance from the center to the longer chord is 13
...
More generally, if these chords has lengths 2a and 2b, and the distance
between them is d, the radius r of the circle is given by
r2 =
[d2 + (a − b)2 ][d2 + (a + b)2 ]
...
√
(b) If AB = 2 · AQ, show that AX 2 + BY 2 = AB 2
...
3
Construction of regular polygons
1
...
1
Equilateral triangle, regular hexagon, and square
||
==
==
||
==
==
||
Given a circle of radius a, we denote by
an inscribed
zn
the length of a side of
regular n−gon
...
3
YIU: Euclidean Geometry
9
Exercise
1
...
C is the midpoint of the
minor arc AB and M the midpoint of the chord AB
...
1 √ √
sin 15◦ = ( 6− 2),
4
cos 15◦ =
1 √ √
( 6+ 2)
...
4
1
...
1
10
The regular pentagon and its construction
The regular pentagon
C
Y
X
A
B
P
Q
Z
Q
P
D
E
A
X
Since XB = XC by symmetry, the isosceles triangles CAB and XCB are
similar
...
It follows that
AX 2 = AB · XB
...
4
...
(1) Draw a right triangle ABP with BP perpendicular to AB and half
in length
...
(3) Mark a point X on the segment AB such that AX = AQ
...
B
YIU: Euclidean Geometry
11
Exercise
1
...
618 · · ·
...
AB
2
2
...
3
...
Show that
(i) 6 BAC = 36◦ ;
(ii) AX : XB = φ : 1
...
Let E be the midpoint of the side AC
...
4
Deduce that
√
q
q
√
√
5+1
1
◦
◦
◦
cos 36 =
10 − 2 5,
tan 36 = 5 − 2 5
...
ABC is an isosceles triangle with AB = AC = 4
...
Let D be the midpoint of BC
...
,
cos 18 =
sin 18 =
4
4
5
C
YIU: Euclidean Geometry
1
...
3
12
Construction of a regular pentagon
1
...
2
...
3
...
Then, ACBED is a regular pentagon
...
Justify the following construction of an inscribed regular pentagon
...
5
1
...
1
The cosine formula and its applications
The cosine formula
c2 = a2 + b2 − 2ab cos γ
...
Show that the (4,5,6) triangle has one angle equal to twice of another
...
If γ = 2β, show that c2 = (a + b)b
...
Find a simple relation between the sum of the areas of the three squares
S1 , S2 , S3 , and that of the squares T1 , T2 , T3
...
ABC is a triangle with a = 12, b + c = 18, and cos α =
4
a3 = b3 + c3
...
A
...
Here, b = 9 −
a
√
√
5, and c = 9 + 5
...
Show that
YIU: Euclidean Geometry
1
...
2
14
Stewart’s Theorem
If X is a point on the side BC (or its extension) such that BX : XC = λ : µ,
then
λb2 + µc2
λµa2
...
Use the cosine formula to compute the cosines of the angles AXB
and AXC, and note that cos ABC = − cos AXB
...
5
...
1 2
(2b + 2c2 − a2 )
...
Exercise
1
...
2
...
a
c
b
4
3
...
Calculate
the lengths of its medians
...
Suppose c2 = a +b
...
Give a euclidean construc2
tion of triangles satisfying this condition
...
C
YIU: Euclidean Geometry
15
5
...
6
...
Show that either
(i) b = c, or
(ii) the quadrilateral AEGF is cyclic
...
6
7
...
7
8
...
1
...
4
Length of angle bisector
The length wa of the (internal) bisector of angle A is given by
2
wa = bc[1 − (
Proof
...
b+c
Apply Stewart’s Theorem with λ = c and µ = b
...
wa =
4bcs(s−a)
(b+c)2
...
The lengths of the sides of a triangle are 84, 125, 169
...
8
3
...
9
4
...
Show that the triangle is either isosceles, or
γ = 60◦
...
In fact, b2 m2 − c2 m2 = 1 (c − b)(c + b)(b2 + c2 − 2a2 )
...
8
Answers: 975 , 26208 , 12600
...
2
2
(b + c)
(c + a)
(b + c)2 (c + a)2
10 2
2
2
a wa − b2 wb =
abc(b−a)(a+b+c)2
[a2
(a+c)2 (b+c)2
− ab + b2 − c2 ]
...
Show that the length of the external angle bisector is given by
0
wa2 = bc[(
a 2
4bc(s − b)(s − c)
...
In triangle ABC, α = 12◦ , and β = 36◦
...
11
1
...
6
...
12
Suppose β < γ in triangle ABC
...
A
N
LO
B
L
M
LO
C
Choose a point L on BM such that 6 N CL = 1 β
...
Note that
6
1
N BC = β < (β + γ) = 6 LCB,
2
and both are acute angles
...
It follows that CN < BM
...
The counterpart of the Steiner - Lehmus theorem does not hold
...
2 (1976) pp
...
D
...
MacKay (AMM E312): if the external angle
bisectors of B and C of a scalene triangle ABC are equal, then s−a is the geometric mean
a
of s−b and s−c
...
12
Gilbert - McDonnell, American Mathematical Monthly, vol
...
YIU: Euclidean Geometry
1
...
2
Second proof
...
We shall
show that β = γ
...
Compare the triangles CBM and
BCN
...
Their opposite sides therefore
satisfy the relation CM < BN
...
This is isosceles since CN = BM = N G
...
2
Since β < γ, we conclude that 6 CGM > 6 GCM
...
This contradicts the relation CM < BN obtained above
...
The bisectors of angles B and C of triangle ABC intersect the median
AD at E and F respectively
...
Show that triangle
ABC is isosceles
...
Descube, 1880
...
Chapter 2
The circumcircle and
the incircle
2
...
1
...
This is the center of the circumcircle,
the circle passing through the three vertices of the triangle
...
1
...
18
YIU: Euclidean Geometry
19
a
b
c
=
=
= 2R
...
The internal bisectors of angles B and C intersect the circumcircle of
4ABC at B 0 and C 0
...
(ii) If BB 0 = CC 0 , does it follow that β = γ?
1
A
C'
B'
B
C
2
...
3
...
Suppose the sides
BZX = 60
of XY Z have unit length
...
sin φ
(b) In triangle ABZ, show that 6 ZAB = θ and 6 ZBA = φ
...
BB 0 = CC 0 if and only if β = γ or α =
2π
...
Show that BX, BZ are
that
CX, CY
the trisectors of the angles of triangle ABC
...
(e) Suppose the extensions of BX and AY intersect at P
...
2
...
3
Johnson’s Theorem
Suppose three circles A(r), B(r), and C(r) have a common point P
...
C
Y
X
Y P
X
P
B
A
Z
A
B
Z
C
C
P
X
Y
A
B
Z
YIU: Euclidean Geometry
21
Proof
...
Thus, AY and
BX are parallel, each being parallel to P C
...
(2) Similarly, Y Z = BC and ZX = CA
...
(3) Since triangle ABC has circumradius r, the circumcenter being P ,
the circumradius of XY Z is also r
...
Show that AX, BY and CZ have a common midpoint
...
2
The incircle
2
...
1
The incenter
The internal angle bisectors of a triangle are concurrent at the incenter of
the triangle
...
If the incircle touches the sides BC, CA and AB respectively at X, Y ,
and Z,
AY = AZ = s − a,
_
BX = BZ = s − b,
CX = CY = s − c
...
2
...
4
24
=
...
Show that the three small circles are equal
...
The incenter of a right triangle is equidistant from the midpoint of the
hypotenuse and the vertex of the right angle
...
I
3
...
4
...
Mark the point Q on the side AB such that BQ = BP
...
YIU: Euclidean Geometry
23
A
R
Q
R'
I
Q'
B
P'
P
C
Continue to mark R on AC such that AR = AQ, P 0 on BC such that
CP 0 = CR, Q0 on AB such that BQ0 = BP 0 , R0 on AC such that
AR0 = AQ0
...
5
...
6
...
Suppose BY = CZ
...
7
...
The segments included between I and the
sides AC and BC have lengths 3 and 4
...
Z is a point on a segment AB such that AZ = u and ZB = v
...
Show that the area of the triangle is equal to uv
...
2
2
Solution
...
Since r = s − c for a right triangle, a = r + u and
v
B
YIU: Euclidean Geometry
24
9
...
Find the radius of the
circle tangent to the arc and the radii through A and B
...
A semicircle with diameter BC is constructed outside an equilateral
triangle ABC
...
Show that the lines AX and AY divide the side BC into
three equal parts
...
Suppose each side of equilateral triangle has length 2a
...
4
√
12
...
P XY Q is a square inscribed in
the semicircle
...
b = r + v
...
If h is the height on the hypotenuse, then 1 (u + v)h = uv
...
3
Hint: The circle is tangent to the arc at its midpoint
...
3 (1 +
YIU: Euclidean Geometry
25
(a) Show that the right triangle ABC has the same area as the square
P XY Q
...
5
(c) Show that the incenter of 4ABC is the intersection of P X and
BY
...
A square of side a is partitioned into 4 congruent right triangles and
a small square, all with equal inradii r
...
14
...
If the inradii of the quadrilateral and the isosceles triangle are equal,
5
r = (3 −
√
5)a
...
What is the inradius of each of the remaining two
triangles? 6
15
...
The bisectors
AI and BI intersect the circle Z(I) at the points M and N
...
Show that
(i) XN = Y M = r;
(ii) M and N are the incenters of the right triangles ABR and BCR
respectively
...
CR is the altitude on the hypotenuse AB of a right triangle ABC
...
7
c
17
...
Suppose the large circle has radius R
...
8
√
√
( 3 − 2)a
...
8
Let θ be the semi-vertical angle of the isosceles triangle
...
If this is equal to R (1 − sin θ), then sin θ = 1
...
8
6
YIU: Euclidean Geometry
27
18
...
The four small circles have equal radii
...
9
2
...
3
...
An excircle can
be constructed with this as center, tangent to the lines containing the three
sides of the triangle
...
The inradius of the right
2R cos θ sin
triangle is 1+sin θ+cosθ
...
From this,
θ
2
5
4
sin θ = 13 , and the inradius is 13 R
...
3
...
s−c
rc =
Proof
...
Since
2
4 = −4IA BC + 4IA CA + 4IA AB,
we have
1
4 = ra (−a + b + c) = ra (s − a),
2
from which ra =
4
s−a
...
If the incenter is equidistant from the three excenters, show that the
triangle is equilateral
...
Show that the circumradius of 4IA IB IC is 2R, and the area is
abc
2r
...
Show that for triangle ABC, if any two of the points O, I, H are
concyclic with the vertices B and C, then the five points are concyclic
...
4
...
Show that IO = IH
...
Suppose α = 60◦
...
6
...
2
2
C
C
P
A
B
A
B
YIU: Euclidean Geometry
29
7
...
Denote by
and exradius of triangle
ABP
...
8
...
Show that M is also the midpoint of the segment IIA
...
Let M 0 be the midpoint of the arc BAC of the circumcircle of triangle
ABC
...
Deduce that M 0 is indeed the midpoint of the segment IB IC
...
The circle BIC intersects the sides AC, AB at E and F respectively
...
10
10
Hint: Show that IF bisects angle AF E
...
The incircle of triangle ABC touches the side BC at X
...
If D is the midpoint
of BC, show that DK = rC
...
4
Heron’s formula for the area of a triangle
Consider a triangle ABC with area 4
...
It is convenient to
introduce the semiperimeter s = 1 (a + b + c)
...
s -a
X
ra
s -c
I
r
Y
s -c
C s -b
Y'
YIU: Euclidean Geometry
31
• From the similarity of triangles AIZ and AI 0 Z 0 ,
r
s−a
=
...
• From these,
r =
4 =
s
q
(s − a)(s − b)(s − c)
,
s
s(s − a)(s − b)(s − c)
...
Exercise
1
...
What is the area of the
triangle ? 11
2
...
3
...
Calculate the radii of the other two
circles
...
The lengths of the sides are 25, 20 and 15
...
If one of the ex-radii of a triangle is equal to its semiperimeter, then
the triangle contains a right angle
...
+
1
rb
+
1
rc
= 1
...
ra rb rc = r2 s
...
Show that
(i) ra + rb + rc =
−s3 +(ab+bc+ca)s
;
4
(ii) (s − a)(s − b)(s − c) = −s3 + (ab + bc + ca)s
...
2
...
1
Appendix: A synthetic proof of ra + rb + rc = 4R + r
IB
M'
A
IC
I
O
D
Z'
B
X
X'
M
Q
IA
C
Y'
YIU: Euclidean Geometry
33
Proof
...
(2) The midpoint M 0 of IB IC is also on the circumcircle
...
(4) If D is the midpoint of BC, then DM 0 = 1 (rb + rc )
...
(6) Since M is the midpoint of IIA , M D is parallel to IA Q and is half
in length
...
2
(7) It now follows from M M 0 = 2R that ra + rb + rc − r = 4R
...
1
The orthocenter
3
...
1
The three altitudes of a triangle are concurrent
...
A
H
B'
C'
H
A
B
C
B
A'
The orthocenter is a triangle is the circumcenter of the triangle bounded
by the lines through the vertices parallel to their opposite sides
...
1
...
34
C
YIU: Euclidean Geometry
35
If the triangle is obtuse, say, α > 90◦ , then the orthocenter H is outside
the triangle
...
3
...
3
Orthocentric quadrangle
More generally, if A, B, C, D are four points one of which is the orthocenter
of the triangle formed by the other three, then each of these points is the
orthocenter of the triangle whose vertices are the remaining three points
...
3
...
4
Orthic triangle
The orthic triangle of ABC has as vertices the traces of the orthocenter
H on the sides
...
A
A
Y
Z
Z
H
C
X
B
B
C
X
Y
H
If ABC is an obtuse triangle, with γ > 90◦ , then ABH is acute, with
angles 90◦ − β, 90◦ − α, and 180◦ − γ
...
Exercise
1
...
How should this be
modified if α > 90◦ ?
2
...
YIU: Euclidean Geometry
36
3
...
AH = 2R · cos α, and
HX = 2R · cos β cos γ, where R is the circumradius
...
If an obtuse triangle is similar to its orthic triangle, find the angles of
the triangle
...
2
3
...
1
The Euler line
Theorem
The circumcenter O, the orthocenter H and the median point M of a nonequilateral triangle are always collinear
...
Proof
...
A
Y
H
G'
O
C
B
The Euler line
1
...
2
...
3
...
4
...
The line OGH is called the Euler line of the triangle
...
YIU: Euclidean Geometry
37
Exercise
1
...
circumcenter, incenter, centroid, orthocenter
...
Show that the incenter I of a non-equilateral triangle lies on the Euler
line if and only if the triangle is isosceles
...
Let O be the circumcenter of 4ABC
...
DEF is called the
medial triangle of ABC
...
(b) Show that the centroid of DEF is the centroid of 4ABC
...
Furthermore,
OG : GN : N H = 2 : 1 : 3
...
Let H be the orthocenter of triangle ABC
...
2
5
...
3
3
6
...
Show
that the Euler lines of the triangles AEF , BF D and CDE form an
equilateral triangle symmetrically congruent to ABC, the center of
symmetry lying on the diameter d
...
Th´bault, AMM E547
...
The Euler lines of triangles IBC, ICA, IAB are concurrent
...
3
4
The nine-point circle
Let ABC be a given triangle, with
(i) D, E, F the midpoints of the sides BC, CA, AB,
(ii) P , Q, R the projections of the vertices A, B, C on their opposite
sides, the altitudes AP , BQ, CR concurring at the orthocenter H,
(iii) X, Y , Z the midpoints of the segments AH, BH, CH
...
This is called the nine-point circle of 4ABC
...
It is indeed the circumcircle of the medial triangle
DEF
...
4
Crux 1018
...
YIU: Euclidean Geometry
39
A
Q
R
H
E
F
N
O
B
C
D
P
The nine-point circle of a triangle
Exercise
1
...
AP and
BQ intersect at C, and the tangents at P and Q intersect at X
...
C
X
P
Q
A
B
2
...
The midpoint of P H lies on the nine-point circle of the triangle
...
(a) Let ABC be an isosceles triangle with a = 2 and b = c = 9
...
(b) Suppose there is a circle with center I tangent externally to each
of the excircles
...
(c) Suppose there is a circle with center I tangent internally to each
of the excircles
...
4
...
3
...
This number is positive, zero, or negative according as P is outside, on,
or inside the circle
...
4
...
O
A
M
O
P
B
A
M
O
B
P
P
T
If P is outside the circle, (O)P is the square of the tangent from P to
(O)
...
4
...
A
A
D
B
P
P
B
D
C
Proof
...
Exercise
1
...
2
...
P is a point on CE, and F Q is parallel to AE
...
YIU: Euclidean Geometry
42
Q
D
C
P
E
F
A
B
3
...
P Y and QX are two chords through M
...
(i) Use the sine formula to show that
KY · KQ
HX · HP
=
...
Y
X
O
a-x
A
y
x
H
M
P
a-y
K
B
Q
4
...
K(T ) is
the circle tangent to the semicircle and the perpendiculars to AB at P
and Q
...
YIU: Euclidean Geometry
43
T
X
Y
K
A
3
...
5
...
A
r
I
O
C
B
X
Proof
...
Note that X is the midpoint
of the arc BC
...
IX = XB = XC = 2R sin α ,
2
2
...
R2 −d2 = power of I with respect to the circumcircle = IA·IX = 2Rr
...
5
...
2
2
2
YIU: Euclidean Geometry
44
Proof
...
This means
IC = 2XC · sin β = 4R sin α sin β
...
5
...
2
2
2
2
Distance between O and excenters
2
OIA = R2 + 2Rra
...
Given the circumcenter, the incenter, and a vertex of a triangle, to
construct the triangle
...
Given a circle O(R) and r < 1 R, construct a point I inside O(R) so
2
that O(R) and I(r) are the circumcircle and incircle of a triangle?
3
...
Given a circle I(r) and a point O, construct a circle O(R) so that
O(R) and I(r) are the circumcircle and incircle of a triangle?
5
...
(a) One of the angles has cosine
(b) s2 =
r
R;
(2R−r)2 (R+r)
...
The power of I with respect to the circumcircle is
abc
6
a+b+c
...
AIO ≤ 90◦ if and only if 2a ≤ b + c
...
Make use of the relation
a = r(cot
β
γ
+ cot )
2
2
to give an alternative proof of the formula r = 4R sin α sin β sin γ
...
Show that XIA = XI
...
This power is OI 2 − R2 = 2Rr =
abc
24
·
4
s
=
abc
...
1
Tests for concyclic points
4
...
1
Let A, B, C, D be four points such that the lines AB and CD intersect
(extended if necessary) at P
...
C
D
C
D
P
A
B
B
A
P
A
B
C
4
...
2
Let P be a point on the line containing the side AB of triangle ABC such
that AP · BP = CP 2
...
Exercise
1
...
If Z is the point on
2
the side AB such that BZ = BC = a, then the circumcircle of triangle
45
P
YIU: Euclidean Geometry
46
BCZ touches the side AC at C
...
Let ABC be a triangle satisfying γ = 90◦ + 1 β
...
Prove that M D = AB
...
Suppose that ABC is a triangle satisfying γ = 90◦ + 1 β, that the
2
exterior bisector of angle A intersects BC at D, and that the side AB
touches the incircle of triangle ABC at F
...
4
...
Suppose d > a + b so that the two circles do not intersect
...
D
YIU: Euclidean Geometry
47
X
P
Y
A
A
B
B
Q
Suppose d > |a − b| so that none of the circle contains the other
...
Exercise
1
...
A
B
D
1
A
C
D
2
...
The
common tangent at P intersects the two external common tangents
XY , X 0 Y 0 at K, K 0 respectively
...
(b) What is the length P K?
(c) Find the lengths of the common tangents XY and KK 0
...
B
C
YIU: Euclidean Geometry
48
X
K
A
Y
P
B
Y'
K'
X'
3
...
AP and AQ are the tangents from A to circle B(b)
...
Calculate the length of HK in
terms of d, a, and b
...
Tangents are drawn from the center of two given circles to the other
circles
...
5
...
From the extremity A0 of the diameter of A(a) on the line AB, tangents
are constructed to the circle B(b)
...
3
2
3
Answer:
Answer:
2ab
d
...
d+a+b
YIU: Euclidean Geometry
49
b
è
A'
a
d
A
B
A
B
6
...
7
...
P is a point on BC so that the incircle
of triangle ABP and the circle tangent to the lines AP , P C and CD
have equal radii
...
D
C
D
C
y
P
Q
x
B
A
B
A
8
...
Q is a point on BC so that the incircle
of triangle ABQ and the circle tangent to AQ, QC, CD touch each
other at a point on AQ
...
y = 1
...
3
Tangent circles
4
...
1
A basic formula
Let AB be a chord of a circle O(R) at a distance h from the center O, and
K(r)
P a point on AB
...
x
A
M
h
H
K
r
B
P
R - rh + r
O
r
A
B
P
h
O
r'
K
A
M
K'
C
Proof
...
Let K(r) be the circle
tangent to AB at P and to the minor arc AB
...
2(R + h)
2(R + h)
2(R + h)
The case for the major arc is similar
...
3
...
Mark a point Q on the circle so that
P Q = CM
...
Then r = 1 P H,
2
from this the center K can be located easily
...
2R
(2) Note that the ratio r : r0 = R − h : R + h is independent of the
position of P on the chord AB
...
3
...
Clearly,
1−
r0 − r
=
sin θ = 0
r +r
1+
r
r0
r
r0
=
1−
1+
R−h
R+h
R−h
R+h
=
h
...
Since the
center line KK 0 is perpendicular to the chord AB, the common tangent is
perpendicular to the radius OA
...
é
D
K
KP
r
A
é
B
P
h
T
A
P
KQ
Q
B
O
r'
O
K'
C
4
...
4
Let P and Q be points on a chord AB such that the circles (KP ) and (KQ ),
minor
each being tangent to the chord and the
arc AB, are also tangent to
major
YIU: Euclidean Geometry
52
each other externally
...
minor
Proof
...
Regarding these
two circles are tangent to the chord CD, and AB as an external common
tangent, we conclude that C is the midpoint of the arc AB
...
3
...
Given a circle (KP ) tangent to (O) and a chord AB, let C be the midpoint of the arc not containing KP
...
CT
and AB to intersect
(2) Construct the bisector of the angle between
CT 0
K
K T
the ray P 0 at Q
...
AB,
and to the arc AB containing KP
...
Let C be the midpoint of the major arc AB
...
T
A
Q M
B
P
O
C
2
...
CD is parallel to AB, and is at a
distance b
...
Calculate the radius of the
inscribed circle
...
If each side of the square has length a, calculate the radii of the two
small circles
...
Given a chord AB of a circle (O) which is not a diameter, locate the
0
points P on AB such that the radius of (KP ) is equal to 1 (R − h)
...
YIU: Euclidean Geometry
54
5
...
Find the radius of the circle tangent to one of the circles
internally, the other externally, and the line AB
...
A(a) and B(b) are two semicircles tangent internally to each other at
O
...
Show that
BX =
b(3a − b)
,
a+b
and r =
K
r
r
a
O
A
4ab(b − a)
...
3
...
Let M be the midpoint
of the chord AB, at a distance h from the center O
...
rP =
2(R + h)
5
√
3
r,
4
r = radius of A(B)
...
(1) Let C be the midpoint of the major arc AB
...
(2) Let the perpendicular to AB through P intersect the circle (O) at
P1 and P2
...
Q
P1
A'
A
Q'
M
P
B
Q
B'
O
C
P2
D
Then the circles tangent to the minor arc and to the chord AB at Q and
Q0 are also tangent to the circle (KP )
...
Let (KP ) and (KQ ) be two circles each tangent to the minor arc
and the chord AB, and are tangent to each other externally
...
(x − y)2 = 4rP rQ =
(R + h)2
Solving this equation for y in terms of x, we have
y−x= √
R2 − h2 − x2
...
This justifies the above
construction
...
4
Mixtilinear incircles
L
...
Let K(ρ) be the circle
tangent to the sides AB, AC, and the circumcircle at X3 , X2 , and A0 respectively
...
Also, AX2 = ρ cot α , and AE = 1 b = R sin β
...
2
Solving this equation, we obtain
·
¸
α
α
cot sin β − 1 + cos β
...
9 (1983) pp
...
¸
YIU: Euclidean Geometry
57
·
sin α sin β
α
β
α
β
2
2
= 4R
cos cos − sin sin
2 α
cos 2
2
2
2
2
= 4R
sin α sin β
α+β
2
2
cos
2 α
cos 2
2
¸
sin α sin β sin γ
2
2
2
cos2 α
2
r
...
4
...
1
The radius of the mixtilinear incircle in the angle A is given by
α
ρ1 = r · sec2
...
Note that the segment X2 X3 contains the incenter I as its midpoint, and
the mixtilinear incenter K1 is the intersection of the perpendiculars to AB
and AC at X3 and X2 respectively
...
In each of the following cases, the largest circle is the circumcircle of
the triangle (respectively equilateral and right)
...
Compute the ratio of the radii of
the two smaller circles
...
ABC is a right triangle for which the mixtilinear incircle (K) of the
right angle touches the circumcircle at a point P such that KP is
parallel to a leg of the triangle
...
7
P
B
K
C
A
3
...
Show
that the ρ1 = 2ρ2
...
ABC is an isosceles triangle with AB = AC
...
8
5
...
The three small circles have equal radii
...
9
4
...
2
Consider also the mixtilinear incircles in the angles B and C
...
8
If ρ2 = kρ1 , then tan
9
Answer:
√
3− 5
2 R
...
2
YIU: Euclidean Geometry
59
and Y3 respectively, and that in angle C touch the sides BC and AC at Z1
and Z2 respectively
...
It follows that the triangles IY1 Z1 and IY3 Z2 are congruent, and
the segment Y3 Z2 is parallel to the side BC containing the segment Y1 Z1 ,
and is tangent to the incircle
...
Sima
a
Z
plifying this, we obtain Y3a 2 = s−a
...
Similarly, the inradii of the triangles BZ1 X3
s
and CX2 BY1 are rb = s−b · r and rc = s−c · r respectively
...
We summarize this in the following proposition
...
C
YIU: Euclidean Geometry
4
...
The mixtilinear exradius
in the angle A is given by
ρA = ra sec2
where ra =
4
s−a
is the corresponding exradius
...
1
The shoemaker’s knife
Let P be a point on a segment AB
...
Suppose the smaller semicircles have radii a and
b respectively
...
This perpendicular is an internal common
tangent of the smaller semicircles
...
Show that the area of the shoemaker’s knife is πab
...
Let U V be the external common tangent of the smaller semicircles
...
3
...
61
B
YIU: Euclidean Geometry
5
...
1
62
Archimedes’ Theorem
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
...
Consider the circle tangent to the semicircles O(a + b), O1 (a), and
the line P Q
...
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
...
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=
5
...
2
Construction of the Archimedean circles
Let Q1 and Q2 be points on the semicircles O1 (a) and O2 (b) respectively
such that O1 Q1 and O2 Q2 are perpendicular to AB
...
a+b
Note that C3 P = t, the radius of the Archimedean circles
...
The center C1 of the
P
B
YIU: Euclidean Geometry
63
Archimedean circle C1 (t) is the intersection of the circle O1 (M2 ) and the
perpendicular through M1 to AB
...
Q
C1
Q1
C2
Q2
C3
O1
A
5
...
3
O
M1
P
M 2 O2
B
Incircle of the shoemaker’s knife
The circle tangent to each of the three semicircles has radius given by
ρ=
Proof
...
+ ab + b2
a2
Let 6 COO2 = θ
...
Eliminating ρ, we have
a(a + ρ)2 + b(b + ρ)2 = (a + b)(a + b − ρ)2 + ab2 + ba2
...
This is a linear
equation in ρ:
a3 + b3 + 2(a2 + b2 )ρ = (a + b)3 + ab(a + b) − 2(a + b)2 ρ,
YIU: Euclidean Geometry
64
from which
4(a2 + ab + b2 )ρ = (a + b)3 + ab(a + b) − (a3 + b3 ) = 4ab(a + b),
and ρ is as above
...
1
...
Proof
...
The semiperimeter of the triangle CO1 O2 is
a + b + ρ = (a + b) +
(a + b)3
ab(a + b)
= 2
...
3
(a + b)
a+b
This is the same as t, the radius of the Archimedean circles
...
1
...
Construct circle C3 (P ) to intersect O1 (a)
and O2 (b) at X and Y respectively
...
Then C(X) is the incircle of the shoemaker’s knife
...
Exercise
1
...
a2 + ab + b2
2
...
3
...
B
YIU: Euclidean Geometry
66
è
C
Z
X
Q1
Y
a
b
O1
A
O
P
O2
Q2
X
B
Y
A
P
4
...
5
...
5
...
5
...
1
Let U V be the external common tangent of the semicircles O1 (a) and O2 (b),
which extends to a chord HK of the semicircle O(a + b)
...
Since
O1 U = a,
O2 V = b,
and O1 P : P O2 = a : b,
ab
C4 P = a+b = t
...
B
YIU: Euclidean Geometry
67
H
Q
t
C5
U
M
N
a
V
K
C4
b
t
A
O1
b
O
a -b
P
b
K
Let M be the midpoint of the chord HK
...
it follows that (a + b) − QM = P N = 2t
...
This circle touches the
minor arc at the point Q
...
2
...
Let the perpendiculars to AB through O and P intersect Q1 Q2 at
I and J respectively
...
It follows that II 0 = 2t
...
Since I and J are isotomic
conjugates with respect to Q1 Q2 , we have JJ 0 = II 0 = 2t
...
1
5
...
3
M2 (t)
C1 (t)
and
have two internal common tangents, one of
C2 (t)
M1 (t)
which is the line P Q
...
2
A
The circles
C1
C2
A
1
2
O1
O M
1
M 2O 2
B
A
These circles are discovered by Thomas Schoch of Essen, Germany
...
O1
C M1
P
M2 2
O
B
YIU: Euclidean Geometry
69
5
...
4
The external common tangent of P (t) and
5
...
O1
O2 (b)
The Schoch line
5
...
1
The incircle of the curvilinear triangle bounded by the semicircle O(a + b)
ab
and the circles A(2a) and B(2b) has radius t = a+b
...
Denote this circle by S(x)
...
By Apollonius theorem,
(2a + x)2 + (2b + x)2 = 2[(a + b)2 + (a + b − x)2 ]
...
a+b
x S
x
S
2a
2b
a + b -x
a
A
5
...
2
b
O1
2b
a -b
O
P
O2
B
A
O1
O
Theorem (Schoch)
ab
If a circle of radius t = a+b is tangent externally to each of the semicircles
O1 (a) and O2 (b), its center lies on the perpendicular to AB through S
...
3
...
If a circle of radius
ab
t = a+b is tangent externally to both of them, then its center lies on the
Schoch line, the perpendicular to AB through S
...
Let Ak (ka) and Bk (ka) be two circles tangent externally at P , and
Sk (t) the circle tangent externally to each of these
...
a+b
Remark
For k = 2, this is the circle in the preceding proposition
...
5
...
4
Proposition
The circle Sk (t) tangent externally to the semicircle O(a + b) touches the
latter at Q
...
then
O1
O
P K O2
B
If the ray OQ is extended to meet the Schoch line at a point W ,
QW
PK
=
,
OQ
OP
YIU: Euclidean Geometry
71
and
QW =
OQ
a+b a−b
· PK =
·
t = t
...
The height of the center Sk above AB is
2ab q
k(a + b)2 + ab
...
Find the value of k for the circle in Proposition ??
...
3
...
Let
2
intersection of Schoch line with the semicircle
S0
be the
S 00
O(a + b),
, and T the inter(M )
section of (M ) with the radius OS 0
...
Show that PT is perpendicular to AB
...
Show that S 00 is
3
Answer: k =
p
a2 +4ab+b2
...
YIU: Euclidean Geometry
72
3
...
4
4
...
2(a + b)
From this, deduce that T S 0 = 2t
...
S 0 is the point Sk for k =
2a2 +11ab+2b2
...
P T S 0 S 00 is a parallelogram
...
1
Review on complex numbers
A complex number z = x + yi has a real part x and an imaginary part
y
...
The norm is the
nonnegative number |z| given by |z|2 = x2 + y 2
...
z is called a unit complex number if |z| = 1
...
z
Identifying the complex number z = x + yi with the point (x, y) in the
plane, we note that |z1 − z2 | measures the distance between z1 and z2
...
Note also that z
is the mirror image of z in the horizontal axis
...
1
...
6
...
2
Polar form
Each complex number z can be expressed in the form z = |z|(cos θ + i sin θ),
where θ is unique up to a multiple of 2π, and is called the argument of z
...
1
...
In particular,
(cos θ + i sin θ)n = cos nθ + i sin nθ
...
2
Coordinatization
Given 4ABC, we set up a coordinate system such that the circumcenter
O corresponds to the complex number 0, and the vertices A, B, C correspond to unit complex numbers z1 , z2 , z3 respectively
...
A
z1
Y
Z
0 GF
z2
H
z3
C
B
X
Exercise
1
...
3
2
...
3
...
2
4
...
Prove that AX is perpendicular
to Y Z
...
Show that these are
parallel to AC and AB respectively
...
YIU: Euclidean Geometry
6
...
1
75
The incenter
Now, we try to identify the coordinate of the incenter I
...
It is possible to choose unit complex numbers t1 , t2 , t3 such that
z1 = t2 ,
1
z 2 = t2 ,
2
z3 = t2 ,
3
and X, Y, Z are respectively the points −t2 t3 , −t3 t1 and −t1 t2
...
Exercise
1
...
6
...
Proof
...
2
Since the circumradius R = 1, the radius of the nine-point circle is
We apply Theorem 3
...
1 to calculate the inradius r:
1
2
...
r =
=
=
=
1
(1 − OI 2 )
2
1
(1 − | − t1 t2 t3 (t1 + t2 + t3 )|2 )
2
1
(1 − |t1 + t2 + t3 |2 )
2
1
− IF
...
These two circles are therefore tangent
internally
...
1
...
2
2
...
3
...
2
6
...
1
The Feuerbach point
Indeed, the three lines each joining the point of contact of the nine-point
with an excircle to the opposite vertex of the triangle are concurrent
...
Let D be the midpoint of the side BC of triangle ABC
...
A
I
B
N
D
C
2
...
3
...
In that case, they intersect at an angle arccos(1 +
2
2 cos α cos β cos γ)
...
4
The shape and orientation of a triangle with vertices z1 , z2 ,z3 is determined
by the ratio
z3 − z1
...
4
...
z2 − z1
w2 − w1
Equivalently,
z1
det z2
z3
w1
w2
w3
1
1 = 0
...
Three distinct points z1 , z2 , z3 are collinear if and only if z3 −z1 is a real
z2 −z1
number
...
Three distinct points z1 , z2 , z3 are collinear if and only if
z1
det z2
z3
z1
z2
z3
1
1 = 0
...
The equation of the line joining two distinct points z1 and z2 is
z = Az + B,
where
A=
z1 − z2
,
z1 − z2
B=
z1 z2 − z2 z1
...
Show that if z = Az + B represents a line of slope λ, then A is a unit
complex number, and λ = − 1−A i
...
The mirror image of a point z in the line z = Az + B is the point
Az + B
...
4
...
2
This is a root of the quadratic equation x2 + x = 1 = 0, the other root being
√
1
ω 2 = ω = (−1 − 3i)
...
6
...
3
z1 , z2 , z3 are the vertices of an equilateral triangle (with counter clockwise
orientation) if and only if
z1 + ωz2 + ω 2 z3 = 0
...
If u and v are two vertices of an equilateral triangle, find the third
vertex
...
If z1 , z2 , z3 and w1 , w2 , w3 are the vertices of equilateral triangles
(with counter clockwise orientation), then so are the midpoints of the
segments z1 w1 , z2 w2 , and z3 w3
...
If z1 , z2 are two adjancent vertices of a square, find the coordinates of
the remaining two vertices, and of the center of the square
...
On the three sides of triangle ABC, construct outward squares
...
5
...
Show that the latter two have the same center
...
If it has clockwise orientation, w = (1 + ω)u − ωv
...
The triangles CAX (counterclockwise) and DY B (clockwise), both similar to the first
triangle, have the same circumcenter
...
Dou, AMME 2866, 2974)
...
4
...
Proof
...
The center of this
YIU: Euclidean Geometry
81
equilateral triangle is
1
1−ω
1−ω
w3 = ((1 − ω)z1 + (1 − ω2 )z2 ) =
[z1 + (1 + ω)z2 ] =
[z1 − ω2 z2 ]
...
w1 =
3
3
These form an equilateral triangle since
w1 + ωw2 + ω 2 w3
=
1−ω
[z3 + ωz2 + ω 2 z1 − ω 2 (z1 + ωz3 + ω 2 z 2 )]
3
= 0
...
(Fukuta’s generalization of Napoleon’s Theorem) 3 Given triangle ABC,
BC
X1
let Y1 be points dividing the sides CA in the same ratio 1−k : k
...
Complete
Z2
the following equilateral triangles, all with the same orientation,
0
0
X1 X2 X3 , Y1 Y2 Y3 , Z1 Z2 Z3 , Y2 Z1 X3 , Z2 X1 Y30 , X2 Y1 Z3
...
0
0
(b) Consider the hexagon X3 Z3 Y3 X3 Z3 Y30
...
3
Mathematics Magazine, Problem 1493
...
5
Concyclic points
Four non-collinear points z1 , z2 , z3 , z4 are concyclic if and only if the cross
ratio
z4 − z1 z4 − z2
(z3 − z2 )(z4 − z1 )
/
=
(z1 , z2 ; z3 , z4 ) :=
z3 − z1 z3 − z2
(z3 − z1 )(z4 − z2 )
is a real number
...
Suppose z1 and z2 are on the same side of z3 z4
...
In this case, the ratio
z4 − z1 z4 − z2
/
z3 − z1 z3 − z2
of the complex numbers is real, (and indeed positive)
...
6
...
6
...
Then the remaining 16 vertices are the roots of the
equation
x17 − 1
= x16 + x15 + · · · + x + 1 = 0
...
, ω 15 , ω 16
...
) Geometrically, if A0 , A1 are two distinct vertices of a
regular 17−gon, then successively marking vertices A2 , A3 ,
...
= A14 A15 = A15 A16 ,
we obtain all 17 vertices
...
It follows that the regular 17−gon can be constructed if one can
construct the number ω +ω 16
...
, 16, can be separated into two “groups” of eight, each with
a sum constructible using only ruler and compass
...
But once this is done, two more applications of the same idea
eventually isolate ω + ω16 as a constructible number, thereby completing the
task of construction
...
Note that x1 + x2 = a and x1 x2 = b
...
, 16:
k
3k
0 1 2
1 3 9
3
10
4 5
13 5
6
15
7 8 9 10
11 16 14 8
11
7
12 13 14 15
4 12 2 6
YIU: Euclidean Geometry
84
Let
y1 = ω + ω 9 + ω 13 + ω 15 + ω16 + ω 8 + ω 4 + ω 2 ,
y2 = ω 3 + ω 10 + ω 5 + ω 11 + ω 14 + ω 7 + ω 12 + ω 6
...
Most crucial, however, is the fact that the product y1 y2 does not depend
on the choice of ω
...
Below each power ω k , we enter a number j (from 1 to
8 meaning that ω k can be obtained by multiplying the jth term of y1 by an
appropriate term of y2 (unspecified in the table but easy to determine):
ω
3
4
6
7
ω2
2
3
5
6
ω3
2
3
4
6
ω4
1
2
4
5
ω5
4
5
6
8
ω6
1
2
3
5
ω7
1
3
7
8
ω8
1
3
4
8
ω9
4
5
7
8
ω 10
3
4
5
7
ω 11
1
5
6
7
ω 12
1
2
4
8
ω 13
1
5
6
8
ω 14
2
6
7
8
ω 15
1
2
6
7
From this it is clear that
y1 y2 = 4(ω + ω 2 + · · · + ω 16 ) = −4
...
We may take
√
−1 + 17
y1 =
,
2
√
−1 − 17
y2 =
...
Clearly, z1 + z2 = y1
...
It follows that z1 and z2 are the roots of the quadratic equation
z 2 − y1 z − 1 = 0,
ω16
2
3
7
8
YIU: Euclidean Geometry
85
and are constructible, since y1 is constructible
...
Finally, further separating the terms of z1 into two pairs, by putting
t1 = ω + ω 16 ,
t2 = ω 13 + ω4 ,
we obtain
t1 + t2 = z1 ,
t1 t2 = (ω + ω 16 )(ω 13 + ω 4 ) = ω 14 + ω5 + ω 12 + ω3 = z3
...
6
...
2
Explicit construction of a regular 17-gon
4
To construct two vertices of the regular 17-gon inscribed in a given circle
O(A)
...
On the radius OB perpendicular to OA, mark a point J such that
OJ = 1 OA
...
Mark a point E on the segment OA such that 6 OJE =
16
4
OJA
...
Mark a point F on the diameter through A such that O is between E
and F and 6 EJF = 45◦
...
With AF as diameter, construct a circle intersecting the radius OB
at K
...
S
...
Coxeter, Introduction to Geometry, 2nd ed
...
27
...
Mark the intersections of the circle E(K) with the diameter of O(A)
through A
...
6
...
5
B
A6
A4
K
J
P6
F
O E
P4
Then A4 , A6 are two vertices of a regular 17-gon inscribed in O(A)
...
A16 , A1 = A, A3 , A5 ,
...
5
Note that P4 is not the midpoint of AF
...
1
7
...
1
Sign convention
Let A and B be two distinct points
...
A
P
B
A
-1 < AP/PB < 0
...
1
...
B
A
B
AP/PB < -1
...
PB
QB
We shall also say that P and Q are harmonic conjugates with respect to the
segment AB
...
1
...
If AB = d,
AP = p, and AQ = q, then d is the harmonic mean of p and q, namely,
2
1 1
+ =
...
This follows from
p
q
=−
...
1
...
Proposition
If (A, B; P, Q), then (A, B; Q, P ), (B, A; P, Q), and (P, Q; A, B)
...
Exercise
1
...
C
P'
M
A
P
B
Q
Given AB, construct a right triangle ABC with a right angle at B
and BC = AB
...
YIU: Euclidean Geometry
89
For every point P (except the midpoint of AB), let P 0 be the point on
AC such that P P 0 ⊥ AB
...
7
...
2
...
BX : XC = c : b;
A
b
c
B
X
c
C
B
C
BX : XC = c : -b
...
7
...
2
b
Example
The points X and X 0 are harmonic conjugates with respect to BC, since
BX : XC = c : b,
and
BX 0 : X 0 C = c : −b
...
2
...
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
...
X'
YIU: Euclidean Geometry
A
X
90
P
Y
B
Proof
...
Consider a point P not on the line AB with AP : P B = k : 1
...
This means that angle XP Y is a right angle
...
The bisectors of the angles intersect the sides BC, CA, AB respectively at P , Q, and R
...
Show
that
µ
¶
1
1 1 1
1
1
+
+
=2
...
Suppose ABC is a triangle with AB 6= AC, and let D, E, F, G be
points on the line BC defined as follows: D is the midpoint of BC,
AE is the bisector of 6 BAC, F is the foot of the perpeandicular from
YIU: Euclidean Geometry
91
A to BC, and AG is perpendicular to AE (i
...
AG bisects one of the
exterior angles at A)
...
A
B
G
E
F
C
D
3
...
4
...
2
7
...
The points
X, Y , Z are collinear if and only if
BX CY AZ
·
·
= −1
...
Suppose each of these angles is 2θ
...
From this, it is clear that the locus of P is the
circle with the segment joining the centers of similitude of (A) and (B) as diameter
...
92
(=⇒) Let W be the point on AB such that CW//XY
...
YA
ZA
It follows that
BZ W Z AZ
BZ W Z AZ
BX CY AZ
·
·
=
·
·
=
·
·
= (−1)(−1)(−1) = −1
...
From
above,
BX CY 0 AZ
BX CY AZ
· 0 ·
= −1 =
·
·
...
0A
Y
YA
The points Y 0 and Y divide the segment CA in the same ratio
...
Exercise
1
...
The line CM intersects the side AB at N
...
3
2
...
Suppose AB 6= AC
...
Show that P and D divide BC harmonically
...
C
Z
P
B
Y
D
C
YIU: Euclidean Geometry
93
3
...
X is a point inside 4ABC such that the incircle of
4XBC touches BC at D also, and touches CX and XB at Y and Z
respectively
...
4
B
IA
X
I
X'
A
Y
C
Y'
4
...
Then the lines XY and X 0 Y 0 intersect on the bisector of
angle A, at the projection of B on this bisector
...
4
The Ceva Theorem
Let X, Y , Z be points on the lines BC, CA, AB respectively
...
XC Y A ZB
Proof
...
Consider the line BP Y cutting the sides of 4CAX
...
or
CY P A BX
·
·
= +1
...
By Menelaus’
theorem again,
AZ BC XP
·
·
= −1,
ZB CX P A
or
AZ BC XP
·
·
= +1
...
Y A ZB XC
(⇐=) Exercise
...
5
7
...
1
Examples
The centroid
If D, E, F are the midpoints of the sides BC, CA, AB of 4ABC, then
clearly
AF BD CE
·
·
= 1
...
Consider the line BGE intersecting the sides of 4ADC
...
−1 =
GD BC EA
GD 2 1
It follows that AG : GD = 2 : 1
...
X
YIU: Euclidean Geometry
7
...
2
95
The incenter
Let X, Y , Z be points on BC, CA, AB such that
AX
BY bisects
CZ
then
AZ
b
= ,
ZB
a
6
6
6
BAC
CBA ,
ACB
BX
c
= ,
XC
b
CY
a
=
...
Exercise
1
...
2
...
(i) Show that if AX bisects angle A and BX · CY = XC · BZ, then
4ABC is isosceles
...
3
...
Show that these are the medians of the triangle
...
ABC is a right triangle
...
Q
C
P
A
B
R
YIU: Euclidean Geometry
5
...
6
...
Show that 6
c
cos α =
...
Given triangle ABC, construct points A0 , B 0 , C 0 such that ABC 0 ,
BCA0 and CAB 0 are isosceles triangles satisfying
6
BCA0 = 6 CBA0 = α,
6
CAB 0 = 6 ACB 0 = β,
Show that AA0 , BB 0 , and CC 0 are concurrent
...
6
6
ABC 0 = 6 BAC 0 = γ
...
6
...
Then
BX
c sin α1
...
X
C
B
X
By the sine formula,
BX
c sin α1
BX/AX
sin α1 / sin β
sin γ sin α1
= ·
...
7 0 0 0
A B C is the tangential triangle of ABC
...
6
...
The lines
AX, BY , CZ are concurrent if and only if
sin α1 sin β1 sin γ1
·
·
= +1
...
Analogous to
BX
c sin α1
= ·
XC
b sin α2
are
CY
a sin β1
,
= ·
YA
c sin β2
AZ
b sin γ1
...
·
·
=
ZB XC Y B
sin α2 sin β2 sin γ2
Exercise
1
...
2
...
Show that AA0 , BB 0 , CC 0 are
concurrent
...
Let X be the intersection of AA0 and BC
...
sin β
YIU: Euclidean Geometry
98
3
...
9
7
...
The segments AA0 ,
BB 0 , and CC 0 are concurrent
...
We examine how the mixtilinear incircle divides the minor arc BC
of the circumcircle
...
Denote α1 := 6 A0 AB
and α2 := 6 A0 AC
...
In triangle KOC, we have
OK = R − ρ1 ,
OC = R,
6
KOC = 2α2 ,
where R is the circumradius of triangle ABC
...
Applying the cosine formula to triangle KOC, we have
1
2R(R − ρ1 ) cos 2α2 = (R − ρ1 )2 + R2 − ρ2 −
1
µ
b(s − c)
s
¶2
...
·p
s
2R(R − ρ1 )
Consider these as cevians of triangle IA IB IC
...
s
2R(R − ρ)
c(s − b)
sin α1
=
...
sin γ2
a(s − b)
From these,
sin α1 sin β1 sin γ1
a(s − c) b(s − a) c(s − b)
·
·
=
·
·
= +1
...
Exercise
1
...
Show that AA0 is a common tangent of the mixtilinear
incircles of angle A in triangle AA0 B and of angle A in triangle AA0 C
...
B
YIU: Euclidean Geometry
7
...
Y, Y
Z, Z 0
AB
The points X 0 , Y 0 , Z 0 are collinear if and only if the cevians AX, BY ,
Z
CZ are concurrent
...
Y
By assumption,
BX 0
BX
=−
,
X 0C
XC
CY 0
CY
=−
,
Y 0A
YA
AZ 0
AZ
=−
...
XC Y A ZB
The result now follows from the Menelaus and Ceva theorems
...
8
...
Choose a point C outside the line AB
...
Through P draw a line intersecting CA at Y and
CB at X
...
Finally, let Q
be the intersection of CZ with AB
...
1
1
5
2
2
2
7
4
5
4
4
2
2
4
6
3
6
3
5
A
P
B
Q
H a r m o n ic c o n j u g a te
7
...
2
O
A
H
h a rm o n i c m e a n
Harmonic mean
Let O, A, B be three collinear points such that OA = a and OB = b
...
Since this also means that (A, B; O, H), the
point H is the harmonic conjugate of O with respect to the segment AB
...
9
7
...
1
5
Triangles in perspective
Desargues Theorem
Given two triangles ABC and A0 B 0 C 0 , the lines AA0 , BB 0 , CC 0 are conAB, A0 B 0
current if and only if the intersections of the pairs of lines BC, B 0 C 0 are
CA, C 0 A0
collinear
...
Suppose AA0 , BB 0 , CC 0 intersect at a point X
...
C 0 C QA A0 X
Multiplying these three equation together, we obtain
AR BP CQ
·
·
= −1
...
A'
X
A
B'
C'
B
C
P
Q
R
7
...
2
Two triangles satisfying the conditions of the preceding theorem are said to
be perspective
...
7
...
3
Given two triangles ABC and A0 B 0 C 0 , if the lines AA0 , BB 0 , CC 0 are parAB A0 B 0
allel, then the intersections of the pairs of lines BC B 0 C 0 are collinear
...
µ
BP CQ AR
BB 0
·
·
= −
P C QA RB
CC 0
¶µ
−
CC 0
AA0
¶µ
−
AA0
BB 0
¶
= −1
...
9
...
Proof
...
Then
CX
BC
CA
= 0 0 = 0 0
...
The intersection of AA
A
C'
B'
X
A'
B
C
7
...
5
Two triangles whose sides are parallel in pairs are said to be homothetic
...
Distances of corresponding points to the homothetic center are in
the same ratio as the lengths of corresponding sides of the triangles
...
1
Coordinates of points on a line
8
...
1
Let B and C be two distinct points
...
If BX : XC = λ0 : λ, then we say that X
has homogeneous coordinates λ : λ0 with respect to the segment BC
...
In this case,
we shall normalize the homogeneous coordinates to obtain the barycentric
λ
λ0
coordinate of X : λ+λ0 B + λ+λ0 C
...
Given two distinct points B, C, and real numbers y, z, satisfying
y + z = 1, yB + zC is the point on the line BC such that BX : XC =
z : y
...
If λ 6= 1 , the harmonic conjugate of the point P = (1 − λ)A + λB is
2
the point
1−λ
λ
P0 =
A−
B
...
2
Coordinates with respect to a triangle
Given a triangle ABC (with positive orientation), every point P on the plane
has barycenteric coordinates of the form P : xA + yB + zC, x + y + z = 1
...
We shall often identify a point with its barycentric coordinates, and write
P = xA + yB + zC
...
A
A
Y
Z
P
P
C
B
B
X
C
Exercises
If P has homogeneous coordinates of the form 0 : y : z, then P lies on the
line BC
...
2
...
Show that X has
homogeneous coordinates 0 : y : z, and hence barycentric coordinates
z
y
B+
C
...
Likewise, if Y and Z are
respectively the intersections of BP with CA, and of CP with AB, then
z
x
y
x
A+
C, Z =
A+
B
...
2
...
In barycentric coordinates,
P =
8
...
3
1
(µν · A + λν · B + λµ · C)
...
The medians intersect at the centroid G, which has homogeneous coordinates
1 : 1 : 1, or
1
G = (A + B + C)
...
These bisectors intersect at the incenter I with homogeneous coordinates
1
1 1
:
:
= a : b : c
...
2
...
8
...
5
Example
Consider the tangent at A to the circumcircle of triangle ABC
...
This intersects the line BC at a point X
...
From
this,
µ
¶
µ
¶
AX 2
AB 2 c2
AX 2
BX · CX
BX
=
=
=
= 2,
=
CX
CX 2
CX 2
CX
CA
b
where we have made use of the similarity of the triangles ABX and CAX
...
A
Z
Y
A
X
B
C
X
B
Similarly, if the tangents at B and C intersect respectively the lines CA
and AB at Y and Z, we have
BX : XC =
c2 : −b2 =
1
2 :
a2 = − a2
AY :
Y C = −c
1
2 : −a2
=
AZ : ZB
=
b
a2
From this, it follows that the points X, Y , Z are collinear
...
C
YIU: Euclidean Geometry
8
...
3
...
Suppose r1 6= r2
...
The line P Q always passes a fixed point K on the line AB
...
The point K has homogeneous coordinates r2 : −r1 with respect to the
segment AB,
P
Q
A
B
H
Q'
8
...
2
Internal center of similitude
If AP and BQ0 are oppositely parallel radii of the circles, then the line P Q0
always passes a fixed point H on the line AB
...
The point H has homogeneous coordinates r2 : r1 with respect to the segment AB
...
K
YIU: Euclidean Geometry
109
Example
Consider three circles Oi (ri ), i = 1, 2, 3, whose centers are not collinear and
whose radii are all distinct
...
Since
r2 : r3 ,
O2 C1 : C1 O3 =
C2 O3 = r1 :
r3 ,
O1 C2 :
= r1 : r2
,
O1 C3 : C3 O2
the lines O1 A1 , O2 A2 , O3 A3 are concurrent, their intersection being the
point
1 1 1
:
:
r1 r2 r3
with respect to the triangle O1 O2 O3
...
Let P2 be the external center of similitude of the circles (O3 ), (O1 )
...
YIU: Euclidean Geometry
110
IB
2
...
Show that the lines AA0 , BB 0 ,
IB
C0
AB
CA
0 are concurrent
...
4
Consider a circle with center K, radius ρ, tangent to the sides AB and AC,
and the circumcircle of triangle ABC
...
Since AK : AI = ρ :
r, AK : KI = ρ : −(ρ − r),
1
K = [ρI − (ρ − r)A]
...
Since OP :
KP = R : ²ρ, we have OP : P K = R : −²ρ, and
P =
−²ρ
R
1
(R · K − ²ρ · O) =
·O+
· K
...
894
...
S905
...
Assuming A not on the line OI, it is clear that
AP intersects OI at a point with homogeneous −²r : R with respect to the
segment OI
...
external
center of similitude of the circumcircle (O) and the ininternal
−1
circle (I) according as ² =
, i
...
, the circle (K) touching the circumcircle
1
internally
of ABC
...
R − ²r
This applies to the mixtilinear incircles (excircles) at the other two vertices too
...
4
...
The three segments joining the each vertex of
incircles
the triangle to the point of contact of the corresponding mixtilinear
excircles
external
are concurrent at
center of similitude of the circumcircle and the
internal
incircle
...
5
112
Isotomic conjugates
Let X be a point on the line BC
...
Note that
BX 0
=
X 0C
µ
BX
XC
¶−1
...
5
...
Denote by X, Y , Z the intersections of the lines AP , BP ,
CP with the sides BC, CA, AB
...
If X 0 , Y 0 , and Z 0 are the isotomic conjugates of X, Y , and Z on the respective
sides, then
0
AY
AZ 0
y : z,
BX 0 : X 0 C =
:
Y 0C = x :
z,
0B
: Z
= x : y
...
The intersection P 0 is
called the isotomic conjugate of P (with respect to the triangle ABC)
...
x y z
YIU: Euclidean Geometry
113
Exercise
1
...
2
...
8
...
2
Gergonne and Nagel points
Suppose the incircle I(r) of triangle ABC touches the sides BC, CA, and
AB at the points X, Y , and Z respectively
...
This means the cevians AX, BY , CZ are concurrent
...
A
A
Y
Y
Z
L
Z
I
Y'
Z'
N
B
X
C
B
X
X'
Let X 0 , Y 0 , Z 0 be the isotomic conjugates of X, Y , Z on the respective
sides
...
The cevians AX 0 , BY 0 , CZ 0 are
C
YIU: Euclidean Geometry
114
concurrent
...
This is the
isotomic conjugate of the Gergonne point L
...
Which point is the isotomic conjugate of itself with respect to a given
triangle
...
Suppose the excircle on the side BC touches this side at X 0
...
3
3
...
Let A0 , B 0 , C 0 be the points on the incircle diametrically opposite to X, Y , Z respectively
...
4
8
...
6
...
A
A
B
2
C
B
M
D
E N
C
The centroid
...
Apply the Menelaus theorem
AN
a
to 4AX 0 C and the line BNY 0 to obtain NX 0 = s−a
...
4
0
The line AX intersects the side BC at the point of contact X 0 of the excircle on this
side
...
It follows that these three lines intersect at the Nagel
point of the triangle
...
Show that
BX ∗
c2 XC
= 2·
...
Given a triangle ABC, let D and E be points on BC such that
6 BAD = 6 CAE
...
Show that
1
1
1
1
+
=
+
...
6
...
Denote by la , lb , lc their isogonal cevians
...
Their
intersection P ∗ is called the isogonal conjugate of P with respect to 4ABC
...
6
...
If the cevian BP and its isogonal cevian respectively meet the side
CP
BC
X
X∗
CA at Y and Y ∗ , then since
Z∗
AB
Z
BX : XC = z : y,
AY : Y C = z : x,
AZ : ZB = y : x,
YIU: Euclidean Geometry
116
we have
AY ∗
AZ ∗
BX ∗ : X ∗ C =
c2 y : b2 z =
:
Y ∗ C = c2 x :
a2 z =
: Z ∗B
= b2 x : a2 y
=
x
a2
x
a2
y
b2
:
:
y
b2
:
z
c2 ,
c2
z ,
...
x y z
8
...
4
Circumcenter and orthocenter as isogonal conjugates
H =O *
O
The homogeneous coordinates of the circumcenter are
a cos α : b cos β : c cos γ = a2 (b2 + c2 − a2 ) : b2 (c2 + a2 − b2 ) : c2 (a2 + b2 − c2 )
...
Show that a triangle is isosceles if its circumcenter, orthocenter, and
an excenter are collinear
...
6
...
It has
homogeneous coordinates,
K = a2 : b2 : c2
...
If α 6= 90◦ , the lines AO
and AH are isogonal with respect to the bisector AIA , if O, H, IA are collinear, then
6 OAIA = 6 HAIA = 0 or 180◦ , and the altitude AH falls along the line AIA
...
YIU: Euclidean Geometry
117
A
K
X
Y
G
B
Z
C
Exercise
1
...
8
...
6
The symmedians
If D∗ is the point on the side BC of triangle ABC such that AD∗ is the
isogonal cevian of the median AD, AD∗ is called the symmedian on the side
BC
...
ta = 2
2
b +c
b2 + c2
Exercise
1
...
2
...
6
8
...
7
The exsymmedian points
Given a triangle ABC, complete it to a parallelogram BACA0
...
Since each of the pairs BP , BA0 ,
and BA, BC is symmetric with respect to the bisector of angle B, 6 P BA =
6 A0 BC = 6 BCA
...
Similarly, the isogonal cevian of CA0 is the tangent at C to the circumcircle
of triangle ABC
...
YIU: Euclidean Geometry
118
the triangle
...
Since A0
has homogeneous coordinates −1 : 1 : 1 with respect to triangle ABC, the
exsymmedian point KA has homogeneous coordinates −a2 : b2 : c2
...
These
exsymmedian points are the vertices of the tangential triangle bounded by
the tangents to the circumcircle at the vertices
...
KC
B
A'
A
P
I
A
O
KB
C
B
C
KA
Exercise
1
...
Given λ, µ, ν, there is a (unique) point P such that
P P1 : P P2 : P P3 = λ : µ : ν
if and only if each “nontrivial” sum of aλ, bµ and cν is nonzero
...
aλ + bµ + cν
aλ + bµ + cν
aλ + bµ + cν
3
...
YIU: Euclidean Geometry
119
(a) The angles of the tangential triangle are 180◦ − 2α, 180◦ − 2β, and
180◦ − 2γ, (or 2α, 2β and 2γ − 180◦ if the angle at C is obtuse)
...
4
...
Point
Centroid
Incenter
Symbol
G
I
IA
IB
IC
L
N
K
KA
KB
KC
O
H
Excenters
Gergonne point
Nagel point
Symmedian point
Exsymmedian points
Circumcenter
Orthocenter
Homogeneous coordinates
1:1:1
a:b:c
−a : b : c
a : −b : c
a : b : −c
(s − b)(s − c) : (s − c)(s − a) : (s − a)(s − b)
s−a : s−b :s−c
a2 : b2 : c2
−a2 : b2 : c2
a2 : −b2 : c2
a2 : b2 : −c2
2 2
2
2
2 2
a (b + c − a ) : b (c + a2 − b2 ) : c2 (a2 + b2 − c2 )
(a2 + b2 − c2 )(c2 + a2 − b2 )
: (b2 + c2 − a2 )(a2 + b2 − c2 )
: (c2 + a2 − b2 )(b2 + c2 − a2 )
5
...
Furthermore, IG : GN = 1 : 2
...
6
...
7
Solution
...
2
2
7
YIU: Euclidean Geometry
120
7
...
8
...
8
9
...
10
...
Show that AGA , BGB , and
CGC are concurrent
...
If the sides of a triangle are in arithmetic progression, then the line
joining the centroid to the incenter is parallel to a side of the triangle
...
If the squares of a triangle are in arithmetic progression, then the line
joining the centroid and the symmedian point is parallel to a side of
the triangle
...
6
...
Suppose the line AA0 , BB 0 , CC 0 intersects the sides BC, CA, AB at points
X, Y , Z respectively
...
XC
b sin α2
(s − c)/c2
BX : XC =
AY :
YC =
AZ : ZB
=
8
s−c
a2
s−c
a2
s−b
b2
:
s−b
b2
:
s−c
c2 ,
s−c
c2 ,
...
It follows that the orthocenter must be a
vertex of the triangle, and the triangle must be right
...
844
...
9
At the centroid of A, B, C, P ; see MGQ781
...
YIU: Euclidean Geometry
121
These cevians therefore intersect at the point with homogeneous coordinates
b2
c2
a2
:
:
...
8
...
9
The isogonal conjugate of the Nagel point is the external center of similitude
of the circumcircle and the incircle
...
Show that the isogonal conjugate of the Gergonne point is the internal
center of similitude of the circumcircle and the incircle
...
7
Point with equal parallel intercepts
Given a triangle ABC, we locate the point P through which the parallels
to the sides of ABC make equal intercepts by the lines containing the sides
of ABC
...
For the equal - parallel - intercept point
P,
1 1 1
1−x: 1−y :1−z = : :
...
This means that 3G − P = 2I 0 , the isotomic conjugate of the incenter I
...
10
AMM E396
...
L
...
C
...
YIU: Euclidean Geometry
122
A
P G I'
C
B
Exercise
1
...
11
2
...
Supose the
parallel through P to BC intersects AC at Y and AB at Z
...
12
A
A
Z
P
Y
P
C
B
B
C
3
...
ab + bc + ca
4
...
The centroid G divides each of the segments OH, IN, and I 0 P in the ratio 1 : 2
...
The segment Y Z
has length a(y+z)
...
b
c
b
c
b
c
11
12
YIU: Euclidean Geometry
123
If the triangle ABC is completed into parallelograms ABA0 C, BCB 0 A,
and CAC 0 B, then the lines A0 X, B 0 Y , and C 0 Z are concurrent at the
point Q with homogeneous coordinates 14
1 1 1 1 1 1 1 1 1
− + + : − + : + −
...
If the triangle ABC is completed into a parallelogram ABA0 C, the fourth vertex A0 is
the point −1 : 1 : 1
...
From this it is straightforward to verify that these three lines are
concurrent at the given point
...
8
124
Area formula
If P , Q and R are respectively the points
P = x1 A + y1 B + z1 C,
Q = x2 A + y2 B + z2 C,
R = x3 A + y3 B + z3 C,
then the area of triangle PQR is given by
¯ x1
¯
4P QR = ¯ x2
¯
x3
Exercise
y1
y2
y3
z1 ¯
¯
z2 ¯4
...
Let X, Y , and Z be points on BC, CA, and AB respectively such
that
BX : XC = λ : λ0 ,
CY : Y A = µ : µ0 ,
The area of triangle XY Z is given by
4XY Z =
AZ : ZB = ν : ν 0
...
(λ + λ0 )(µ + µ0 )(ν + ν 0 )
2
...
3
...
15
Proof
...
ν + ν0
ν + ν0
By the preceding exercise,
4XY Z
=
=
¯0
¯
1
¯µ
0 )(µ + µ0 )(ν + ν 0 ) ¯
(λ + λ
ν0
λµν + λ0 µ0 ν 0
...
9
8
...
1
125
Routh’s Theorem
Intersection of two cevians
Let Y and Z be points on the lines CA and AB respectively such that
CY : Y A = µ : µ0 and AZ : ZB = ν : ν 0
...
9
...
µν + µ0 ν 0 + µν 0
Theorem
Let X, Y and Z be points on the lines BC, CA and AB respectively such
that
BX : XC = λ : λ0 ,
CY : Y A = µ : µ0 ,
AZ : ZB = ν : ν 0
...
+ µ0 ν 0 + µν 0 )(νλ + ν 0 λ0 + νλ0 )
λµ0 )(µν
Exercise
1
...
Find 4
...
The cevians AX, BY , CZ are such that BX : XC = CY : Y A =
AZ : ZB = λ : 1
...
7
YIU: Euclidean Geometry
126
3
...
Show that
16
[DEF ]
PD PE PF
=2
·
·
...
The cevians AD, BE, and CF of triangle ABC intersect at P
...
8
...
10
...
2
Proof
...
The distances from P to the sides
BC and CA are respectively P P1 = 24 · x and P P2 = 24 · y
...
a2 b2
It follows that CP =
CP 2 =
P1 P2
sin γ
=
ab·P1 P2
24
is given by
1 2 2
[x (b + c2 − a2 ) + y 2 (c2 + a2 − b2 ) + (z − 1)2 (a2 + b2 − c2 )]
...
16
Crux 2161
...
1
Given a triangle ABC, to locate a point P on the side BC so that the
incircles of triangles ABP and ACP have equal radii
...
1
...
By
Stewart’s Theorem,
x2 = kb2 + (1 − k)c2 − k(1 − k)a2
...
c + x + ka
b + x + (1 − k)a
127
YIU: Euclidean Geometry
128
This latter equation can be rewritten as
c + x + ka
b + x + (1 − k)a
=
,
k
1−k
b+x
c+x
=
,
k
1−k
or
(9
...
2)
from which
k=
x+c
...
Rearranging, we have
(x + b)(x + c)a2 =
=
=
=
=
(2x + b + c)[(x + c)b2 + (x + b)c2 − x2 [(x + b) + (x + c)]]
(2x + b + c)[(x + b)(c2 − x2 ) + (x + c)(b2 − x2 )]
(2x + b + c)(x + b)(x + c)[(c − x) + (b − x)]
(2x + b + c)(x + b)(x + c)[(b + c) − 2x]
(x + b)(x + c)[(b + c)2 − 4x2 ]
...
4
4
9
...
2
Lau 1 has proved an interesting formula which leads to a simple construction
of the point P
...
1
Solution to Crux 1097
...
Now,
the circle A(Y ) intersects the side BC at two points, one of which is the
required point P
...
1
...
Construct the circle with XY as diameter, and then the tangents
from A to this circle
...
A
X
B
Y
I
P
C
YIU: Euclidean Geometry
130
Exercise
1
...
a
2
...
2
9
...
4
Proof of Lau’s formula
Let θ be the angle between the median and the bisector of angle A
...
In triangle
AA0 C, we have
6
AA0 = 2ma ,
ACA0 = 180◦ − α,
6
AC = b,
AA0 C = α + θ,
2
6
A
A0 C
= c;
0 AC = α − θ
...
=
=
2ma
sin(180◦ − α)
sin α
cos α
2
From this it follows that
ma · cos θ =
2
Hint: AP is tangent to the circle XY P
...
2
2
YIU: Euclidean Geometry
Now, since wa =
2bc
b+c
131
cos α , we have
2
ma · wa · cos θ = bc cos2
α
= s(s − a)
...
9
...
5
Here, we make an interesting observation which leads to a simpler construction of P , bypassing the calculations, and leading to a stronger result: (3)
remains valid if instead of inradii, we equate the exradii of the same two
subtriangles on the sides BP and CP
...
A
C
B
P
Let θ = 6 AP B so that 6 AP C = 180◦ − θ
...
ρ
2
2
2
2
Since tan θ tan(90◦ − θ ) = 1, we also have
2
2
µ 0 ¶2
r
ρ
This in turn leads to
= tan
v
u
β
γ
tan
...
2
tan β
2
YIU: Euclidean Geometry
132
In terms of the sides of triangle ABC, we have
θ
tan =
2
s
s−b
=
s−c
p
(s − b)(s − c)
=
s−c
√
BX · XC
...
Let the incircle of 4ABC touch the side BC at X
...
Mark a point Q on the line BC such that AQ//Y C
...
A
Y
B
X
P
C
Q
Exercise
1
...
F is the midpoint
of AB, and the side BA is extended to a point K with AK = 1 AC
...
P is the point on BC (the one closer to B if there are two)
such that AP = AQ
...
YIU: Euclidean Geometry
133
K
A
F
Q
B
P
C
2
...
, Pn be points on BC such
that P0 = B, Pn = C and the inradii of the subtriangles APk−1 Pk ,
k = 1,
...
For k = 1, 2,
...
k
Show that tan θ2 , k = 1,
...
e
...
tan
, tan
2
2
2
2
form a geometric progression
...
Let P be a point on the side BC of triangle ABC such that the excircle
of triangle ABP on the side BP and the incircle of triangle ACP have
the same radius
...
2(b + c)
If BP : P C = k : 1 − k, and AP = x, then
(1 − k)
k
=
...
YIU: Euclidean Geometry
134
A
B
P
C
4
...
On the minor arc BC of the circle A(B), mark a point X such that
CX = CD
...
Let P be a
point on BC such that AP = AY
...
A
Y
C
B
P
P'
D
X
YIU: Euclidean Geometry
135
9
...
A
G
C
B
Let G be the common point of the circles, and X2 , X3 on the side BC,
Y1 , Y3 on CA, and Z1 , Z2 on AB, the points of contact
...
2
...
The line joining the
center to each vertex is the bisector of the angle at that vertex
...
It follows that the segments X2 X3 ,
Y3 Y1 , and Z1 Z2 are all equal in length
...
Then XX2 = XX3
...
This means that X and A are both on the radical axis of the circles (K2 )
BY
is the
and (K3 )
...
Similarly, the line
CZ
(K3 ) (K1 )
...
4
Th´bault - Eves, AMM E457
...
The other two centers K2 and K3 can be similarly located
...
3
Given a triangle, to construct three congruent circles through a common
point, each tangent to two sides of the triangle
...
3
...
Note that the lines I2 I3 and BC are parallel; so are the pairs
I3 I1 , CA, and I1 I2 , AB
...
The line joining their circumcenters
passes through I
...
This means
that T , O and I are collinear
...
R
R+r
This means I divides the segment OT in the ratio
T I : IO = −r : R + r
...
9
...
2
Construction
Let O and I be the circumcenter and the incenter of triangle ABC
...
(2) Construct the perpendicular from O to BC, intersecting the circumcircle at M (so that IX and OM are directly parallel)
...
Through their intersection P draw a line parallel to IX, intersecting OI at T , the internal center of similitude of the
circumcircle and incircle
...
(5) The circles Ij (T ), j = 1, 2, 3 are three equal circles through T each
tangent to two sides of the triangle
...
4
9
...
1
Proposition
Let I be the incenter of 4ABC, and I1 , I2 , I3 the incenters of the triangles
IBC, ICA, and IAB respectively
...
Then, the lines AA0 , BB 0 , CC 0 are concurrent at a
point 5 with homogeneous barycentric coordinates
a sec
Proof
...
2
2
2
The angles of triangle IBC are
1
π − (β + γ),
2
β
,
2
γ
...
2
2
2
This point apparently does not appear in Kimberling’s list
...
2
2
Here, we have made use of the sine formula:
a cos
b
c
2s
2s
a
...
2
2
2
2
Similarly, B 0 and C 0 have coordinates
A0
B0
C0
β
γ
0 : b sec : c sec ,
2
2
α
γ
a sec : 0 : c sec ,
2
2
α
β
a sec : b sec : 0
...
2
2
2
Exercise
1
...
Are the lines O1 I1 , O2 I2 , O3 I3 concurrent?
9
...
5
...
YIU: Euclidean Geometry
140
Construction
Let I be the incenter of triangle ABC
...
(2) Construct the external common tangents of each pair of these incircles
...
Label
the other common tangent Y1 Z1 with Y1 on CA and Z1 on AB respectively
...
) These common tangents intersect at a point
P
...
A
Y1
Z1
A'
E
Y3
F
Y'
Z'
I'
Z2
C'
B'
B
X2
X'
D
X3
Exercise
1
...
Find the lengths of the sides of the triangle bounded by their external
C
YIU: Euclidean Geometry
common tangents
...
6
9
...
1
Given a circle K(a) tangent to O(R) at A, and a point B, to construct a
circle K 0 (b) tangent externally to K(a) and internally to (O) at B
...
Construct the perpendicular bisector
of KP to intersect OB at K 0 , the center of the required circle
...
6
...
s
AB = 2R
6
Crux 618
...
R−a R−b
√
√
√
√
√
√
r
r
( r2 r3 − r3 r1 + r1 r2 ) +
( r2 r3 + r3 r1 − r1 r2 )
r − r2
r − r3
√
√
√
√
r1 + r2 + r3 + r1 + r2 + r3 √
· r1 r2 r3
...
142
Let 6 AOB = θ
...
Exercise
1
...
2
...
Show that
s
a
b
·
...
6
...
If (P ) is a circle tangent internally to (O) at C, and externally
to each of (H) and (K), then
AC : BC =
Proof
...
R−b
The lengths of AC and BC are given by
s
AC = 2R
ac
,
(R − a)(R − c)
s
BC = 2R
bc
...
By Ceva theorem,
AX : XB =
a
b
:
...
Let the perpendicular
through X to AB intersect this circle at Q and Q0
...
Note that AQ2 = AX · AB and BQ2 = XB · AB
...
It follows that
AY : Y B =
r
a
:
R−a
s
b
...
Then C and C 0 are the points of contact of the circles with (O), (H),
and (K)
...
A
A
Q
X
H
O
K B
Y
H
O
X
K
B
Q'
9
...
4
Given three points A, B, C on a circle (O), to locate a point D such that
there is a chain of 4 circles tangent to (O) internally at the points A, B, C,
D
...
Let
Y be the point diametrically opposite to X
...
YIU: Euclidean Geometry
Y
144
B
B
A
E
A
C
C
D
D
X
Beginning with any circle K(A) tangent internally to O(A), a chain of
four circles can be completed to touch (O) at each of the four points A, B,
C, D
...
Let A, B, C, D, E, F be six consecutive points on a circle
...
A4
A5
A
A3
B
A6
A2
F
A1
A7
E
C
A 12
A8
D
A9
A 10
A 11
2
...
A12 be a regular 12− gon
...
3
...
(Remark: C7 = C1 )
...
Show that A1 A4 , A2 A5 and A3 A6 are concurrent
...
7 (1987)
pp
...
The statement is still valid if each of the circles Ci , i = 1, 2, 3, 4, 5, 6, is
outside the circle C
...
1
Area formula
Consider a quadrilateral ABCD with sides
AB = a,
BC = b,
CD = c,
DA = d,
angles
6
DAB = α,
ABC = β,
6
6
BCD = γ,
6
CDA = δ,
and diagonals
AC = x,
BD = y
...
Eliminating x, we have
a2 + b2 − c2 − d2 = 2ab cos β − 2cd cos δ,
Denote by S the area of the quadrilateral
...
2
2
Combining these two equations, we have
=
=
=
=
=
16S 2 + (a2 + b2 − c2 − d2 )2
4(ab sin β + cd sin δ)2 + 4(ab cos β − cd cos δ)2
4(a2 b2 + c2 d2 ) − 8abcd(cos β cos δ − sin β sin δ)
4(a2 b2 + c2 d2 ) − 8abcd cos(β + δ)
β+δ
4(a2 b2 + c2 d2 ) − 8abcd[2 cos2
− 1]
2
β+δ
4(ab + cd)2 − 16abcd cos2
...
−16abcd cos2
2
16S 2 = 4(ab + cd)2 − (a2 + b2 − c2 − d2 )2 − 16abcd cos2
Writing
2s := a + b + c + d,
we reorganize this as
S 2 = (s − a)(s − b)(s − c)(s − d) − abcd cos2
β+δ
...
1
...
The area
2
formula becomes
S=
q
(s − a)(s − b)(s − c)(s − d),
where s = 1 (a + b + c + d)
...
If the lengths of the sides of a quadrilateral are fixed, its area is greatest
when the quadrilateral is cyclic
...
Show that the Heron formula for the area of a triangle is a special case
of this formula
...
2
Ptolemy’s Theorem
Suppose the quadrilateral ABCD is cyclic
...
It follows that
a2 + b2 − x2 c2 + d2 − x2
+
= 0,
2ab
2cd
and
(ac + bd)(ad + bc)
...
(ad + bc)
From these, we obtain
xy = ac + bd
...
We give a synthetic proof of the theorem and
its converse
...
2
...
Proof
...
Choose a point P on the diagonal BD such that 6 BAP = 6 CAD
...
It follows that
AB : AC = BP : CD, and
AB · CD = AC · BP
...
It follows that
AC : BC = AD : P D, and
BC · AD = AC · P D
...
A
D
D
O
P'
P
B
C
B
C
(Sufficiency)
...
Locate a
point P 0 such that 6 BAP 0 = 6 CAD and 6 ABP 0 = 6 ACD
...
It follows that
AB : AP 0 : BP 0 = AC : AD : CD
...
Consequently, AC : BC = AD : P 0 D, and
6
BAC =
6
P 0 AD and
AD · BC = AC · P 0 D
...
It follows that BP 0 + P 0 D = BC, and the point P 0 lies on diagonal BD
...
Exercise
1
...
Show that AP = BP + CP
...
P is a point on the incircle of an equilateral triangle ABC
...
1
1
If each side of the equilateral triangle has length 2a, then AP 2 + BP 2 + CP 2 = 5a2
...
Each diagonal of a convex quadrilateral bisects one angle and trisects
the opposite angle
...
2
4
...
5
...
Find the diameter of the circumcircle
...
The radius R of the circle containing the quadrilateral is given by
R=
(ab + cd)(ac + bd)(ad + bc)
...
2
...
(s − a)(s − d)
...
ad + bc
Answer: Either A = D = 72◦ , B = C = 108◦ , or A = D =
720 ◦
,
7
B=C=
540 ◦
...
2ad
2(ad + bc)
In an alternative form, this can be written as
tan2
α
2
=
=
1 − cos α
(b + c)2 − (a − d)2
=
1 + cos α
(a + d)2 − (b − c)2
(−a + b + c + d)(a + b + c − d)
(s − a)(s − d)
=
...
Let Q denote an arbitrary convex quadrilateral inscribed in a fixed
circle, and let F(Q) be the set of inscribed convex quadrilaterals whose
sides are parallel to those of Q
...
3
2
...
(a) Prove that a + b > |c − d| and c + d > |a − b| are necessary
and sufficient conditions for there to exist a convex quadrilateral that
admits a circumcircle and whose side lengths, in cyclic order, are a, b,
c, d
...
4
3
...
5
10
...
3
Construction of cyclic quadrilateral of given sides
10
...
4
The anticenter of a cyclic quadrilateral
Consider a cyclic quadrilateral ABCD, with circumcenter O
...
The
midpoint of XZ is the centroid G of the quadrilateral
...
Denote by O0 the intersection of this
3
MG1472
...
(E
...
951
...
(J
...
945
...
(M
...
Klamkin)
4
YIU: Euclidean Geometry
153
perpendicular with the lien OG
...
B
B
X
A
OG
O'
A
O'
O G
C
D
Z
D
C
It follows that the perpendiculars from the midpoints of the sides to
the opposite sides of a cyclic quadrilateral are concurrent at the point O0 ,
which is the symmetric of the circumcenter in the centroid
...
10
...
5
Let P be the midpoint of the diagonal AC
...
Let X 0 and W 0 be the projections of the midpoints X
and W on their respective opposite sides
...
Clearly, O0 , W 0 , C, X 0 are concyclic
...
B
X
W '
P
O
A
O'
W
C
X'
D
YIU: Euclidean Geometry
154
It follows that the four points P , X, W , and O0 are concyclic
...
From this, we have
Proposition
The nine-point circles of the four triangles determined by the four vertices
of a cyclic quadrilateral pass through the anticenter of the quadrilateral
...
2
...
K
D
A
R
Q
P
S
C
B
H
Proof
...
Note that P and S are both on the circle
H(B) = H(C)
...
For the same
reason, it is also the perpendicular bisector of QR
...
The same reasoning also shows that the chord joining the midpoints of
the arcs AB and CD is the common perpendicular bisector of P Q and RS
...
6
Court, p
...
YIU: Euclidean Geometry
10
...
7
155
Corollary
The inradii of these triangles satisfy the relation
7
ra + rc = rb + rd
...
If AB and CD are parallel, then each is parallel to HK
...
More generally,
1
ra − rb = P Q sin (6 BDC − 6 AHD)
2
and
1
rd − rc = SR sin (6 BAC − 6 AHD)
...
Exercise
1
...
BD
Y
A
D
X
W
Z
B
Show that
7
Y
C
1
XY = ZW = |a − b + c − d|
...
127,
does not cover the case of a bicentric quadrilateral
...
3
156
Circumscriptible quadrilaterals
A quadrilateral is said to be circumscriptible if it has an incircle
...
3
...
Proof
...
D
D
X
S
A
A
Y
K
R
P
B
Q
C
B
C
(Sufficiency) Suppose AB + CD = BC + DA, and AB < AD
...
Then
BC < CD, and there are points
Y
CD
CY = CD
DX = DY
...
AK bisects angle A
since the triangles AKX and AKB are congruent
...
It follows that K is equidistant
from the sides of the quadrilateral
...
10
...
2
8
Let ABCD be a circumscriptible quadrilateral, X, Y , Z, W the points of
contact of the incircle with the sides
...
8
See Crux 199
...
Charles Trigg remarks that the Nov
...
issue of Math
...
The proof here was given by Joseph Konhauser
...
Proof
...
This is clearly
AP · P X
4AP X
=
...
It follows
that
AP · AX
4AP X
=
...
PZ
CZ
This means that the point P divides the diagonal XZ in the ratio AX : CZ
...
The same reasoning shows that Q divides XZ in the ratio BX : DZ
...
The diagonal XZ passes through the intersection of AC and BD
...
Exercise
1
...
2
YIU: Euclidean Geometry
158
In particular, if the quadrilateral is also cyclic, then
√
S = abcd
...
√ a cyclic quadrilateral with sides a, b, c, d (in order) has area S =
If
abcd, is it necessarily circumscriptible? 9
3
...
4
...
5
...
11
10
...
4
...
10
...
2
A quadrilateral is orthodiagonal if and only if the sum of squares on two
opposite sides is equal to the sum of squares on the remaining two opposite
sides
...
Consider the following
three statements for a quadrilateral
...
(b) The quadrilateral is circumscriptible
...
Apart from the exception noted above, any two of these together implies the third
...
10
Is it possible to find integers a and c so that b is also an integer?
11
PME417
...
S79S
...
W
...
Let K be the intersection of the diagonals, and 6 AKB = θ
...
Now,
BC 2 +DA2 −AB 2 −CD2 = 2 cos θ(BK·CK+DK·AK+AK·BK+CK·DK)
It is clear that this is zero if and only if θ = 90◦
...
Let ABCD be a cyclic quadrilateral with circumcenter O
...
12
2
...
Show that the projections of P on the sides of ABCD
form the vertices of a bicentric quadrilateral, and that the circumcircle
also passes through the midpoints of the sides of ABCD
...
5
Bicentric quadrilateral
A quadrilateral is bicentric if it has a circumcircle and an incircle
...
5
...
2
2
r
(R + d)
(R − d)2
The proof of this theorem is via the solution of a locus problem
...
Court called this Brahmagupta’s Theorem
...
YIU: Euclidean Geometry
10
...
2
160
Fuss problem
Given a point P inside a circle I(r), IP = c, to find the locus of the intersection of the tangents to the circle at X, Y with 6 XP Y = 90◦
...
We first find a relation between x and θ
...
Since IXQ is a right triangle and
XM ⊥ IQ, we have IM · IQ = IX 2 , and
IM =
r2
...
Since P K is perpendicular to the hypotenuse XY of the right triangle
P XY ,
P K 2 = XK · Y K = r 2 − IK 2 = r2 − IM 2 − M K 2
...
YIU: Euclidean Geometry
161
and, after rearrangement,
x2 + 2x ·
cr 2
2r 4
· cos θ = 2
...
Fuss observed that this becomes constant by choosing
d=
cr2
...
(r 2 − c2 )2
Proof of Theorem
By eliminating c, we obtain a relation connecting R, r and d
...
r − c2
On the other hand,
R 2 + d2 =
r 4 (2r2 − c2 )
c2 r4
2r 6
+ 2
= 2
...
YIU: Euclidean Geometry
10
...
3
162
Construction problem
Given a point I inside a circle O(R), to construct a circle I(r) and a bicentric
quadrilateral with circumcircle (O) and incircle (I)
...
We
shall assume I and O distinct
...
Choose a point M
such that IM is perpendicular to IK, and IK = IM
...
The circle I(P ) is the required incircle
...
5
...
The quadrilateral bounded by the tangents
to circumcircle at the vertices is cyclic if and only if Q is orthodiagonal
...
Given a cyclic quadrilateral quadrilateral XY ZW , let ABCD be
the quadrilateral bounded by the tangents to the circumcircle at X, Y , Z,
W
...
YIU: Euclidean Geometry
163
A
~
é
é
W
D
É
X
é
É
Z
l
É
é
_
10
...
5
É
Ñ
C
Y
Proposition
(a) Let ABCD be a cyclic, orthodiagonal quadrilateral
...
X
D
A
W
W
D
A
X
l
Y
B
C
Z
l
_
Y
Z
C
(b) Let ABCD be a bicentric quadrilateral
...
Furthermore, the diagonals of XY ZW intersect at a point
on the line joining the circumcenter and the incenter of ABCD
...
The diagonals of a cyclic quadrilateral are perpendicular and intersect
at P
...
15
2
...
16
3
...
Let HK be
the diameter of the circumcircle perpendicular to the diagonal AC (so
that B and H are on the same side of AC)
...
17
A
D
K
P
I
M
O
B
C
H
4
...
18
5
...
4r2
pq
15
Crux 2209
...
(CMJ 304
...
882
...
17
D
...
Smeenk, Crux 2027
...
Extend BO to N such that ON = OM
...
The one closer to the shorter side of AB
and BC is P
...
19
Crux 1376; also Crux 1203
...
5
...
10
...
Let A0 , B 0 , C 0 , D0' be the projections of K on the sides AB, BC,
D
A
CD, DA respectively
...
6
...
C
B'
D
A'
B
P
D'
C'
A
We prove this in two separate propositions
...
a
YIU: Euclidean Geometry
166
Proposition A
...
The
projections of K on the sides of ABCD form the vertices of a circumscriptible quadrilateral
...
Note that the quadrilaterals KA0 AB 0 , KB 0 BC 0 , KC 0 CD0 , and KD0 DA0
are all cyclic
...
Then
6
KA0 D0 = 6 KAD0 = 6 CAD = 6 CBD = 6 B 0 BK = 6 B 0 A0 K
...
The same reasoning
shows that K also lies on the bisectors of each of the angles B 0 , C 0 , D0
...
Proposition B
...
The perpendiculars to OA at A, OB at B, OC at C, and OD at D bound a cyclic
quadrilateral whose diagonals intersect at O
...
The quadrilaterals OAB 0 B, OBC 0 C, OCD0 D, and ODA0 A are all
cyclic
...
Similarly, 6 AOB 0 = 6 CBC 0
...
6
...
The centers of these squares form a quadrilateral whose diagonals are
equal and perpendicular to each other
...
7
Centroids
The centroid G0 is the center of
The edge-centroid G1
The face-centroid G2 :
10
...
8
...
Show that there
exists a straight line segment with ends on opposite sides dividing both the
permieter and the area into two equal parts
...
Consider the converse
...
8
...
23
10
...
21
Crux 1179
...
38?
...
(V
...
See editorial comment on 837
...
e
23
E992
...
S52?,531
...
Tan)
22
YIU: Euclidean Geometry
168
10
...
1
(a) If Q is cyclic, then Q(O) is circumscriptible
...
24
(c) If Q is cyclic, then Q(I) is a rectangle
...
25
...
Prove that the four triangles of the complete quadrangle formed by
the circumcenters of the four triangles of any complete quadrilateral
are similar to those triangles
...
Let P be a quadrilateral inscribed in a circle (O) and let Q be the
quadrilateral formed by the centers of the four circles internally touching (O) and each of the two diagonals of P
...
27
10
...
10
...
The orthocenters of
the triangle P AB, P BC, P CD, P DA form a parallelogram that is similar
to the figure formed by the centroids of these triangles
...
532
...
(V
...
444
...
(W
...
Clarke)
27
Th´bault, AMM 3887
...
S837
...
p486
...
25
YIU: Euclidean Geometry
10
...
11
...
P , Q, R, S
are the projections of K on the sides AB, BC, CD, and DA
...
29
10
...
2
The diagonals of a convex quadrilateral ABCD intersect at K
...
Prove that if
KP = KR and KQ = KS, then ABCD is a parallelogram
...
12
The quadrilateral
0
Q(center)
10
...
1
If Q0 (I) is cyclic, then Q is circumscriptible
...
12
...
The lines AB, CD intersect at E, and the lines AD, BC intersect
at F
...
Then the points L, M , N are collinear
...
Let P , Q, R be the midpoints of the segments AE, AD, DE respectively
...
Apply the
Menelaus theorem to the transversal BCF of 4EAD
...
W
...
31
Seimiya, Crux 2338
...
Suppose ABCD is a plane quadrilateral with no two sides parallel
...
If M, N, P
are the midpoints of AC, BD, EF respectively, and AE = a·AB, AF =
b · AD, where a and b are nonzero real numbers, prove that M P =
ab · M N
...
33
32
33
The Gauss-Newton line of the complete quadrilateral formed by the
four Feuerbach tangents of a triangle is the Euler line of the triangle
...
8810
...
537
...
(R
...
a