Global Journal of Advanced Research
on Classical and Modern Geometries
ISSN: 2284-5569, Vol.6, (2017), Issue 1, pp.45-57
A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED
WITH A TRIANGLE
PAUL YIU AND XIAO-DONG ZHANG
A BSTRACT. For a given triangle we study the triad of parabolas each with one
vertex as focus and its opposite side as directrix. Among the 6 angle bisectors and
the perpendicular bisectors of the 3 sides, we show that each of these parabolas
is tangent to 4 of the angle bisectors and 2 perpendicular bisectors. The 18
points of tangency fall on another set of 6 lines each joining the antipode (on
the circumcircle) of a vertex to the two remaining vertices. Some perspectivity
of triangles will also be exhibited. The 6 points of tangency on the external
bisectors lie on a conic with a simple barycentric equation with respect to the
excentral triangle.
1. I NTRODUCTION
In this paper we study a triad of parabolas most naturally associated with a triangle. Given a triangle ABC, we consider the parabolas Pa with focus A and directrix BC, Pb with focus B and directrix CA, and Pc with focus C and directrix
AB, and prove some interesting results about these parabolas. In §??, we redefine these parabolas in terms of their tangents, and locate the points of tangency.
Specifically we show that each of Pa , Pb , Pc is tangent to 4 angle bisectors and
2 perpendicular bisectors of ABC, and establish some perspectivity of triangles
with vertices among the points of tangency. In the final section, we show that the 6
points of tangency with the external bisectors lie on a conic.
Notations. For triangle ABC, we denote respectively by
a, b, c
the sides BC, CA, AB,
a′ , b′ , c′
the perpendicular bisectors of a, b, c,
L a , Lb , Lc
the bisectors of angles A, B, C,
⊥
⊥
⊥
L a , Lb , Lc the external bisectors of angles A, B, C;
L⊥ ( P)
the perpendicular to the line L at (or from) the point P.
Apart from basic elementary results on parabolas we shall also make use of
homogeneous barycentric coordinates. A basic reference is [?]. Let a, b, c denote
the lengths of the sides BC, CA, AB respectively. We shall make use of Conway’s
notations
S A :=
b2 + c2 − a2
,
2
SB :=
c2 + a2 − b2
,
2
SC : =
2010 Mathematics Subject Classification. 51M04, 51N20.
Key words and phrases. Parabolas, barycentric coordinates, collinarity.
45
a2 + b2 − c2
.
2
(1)
46
PAUL YIU AND XIAO-DONG ZHANG
The use of these symbols greatly simplifies expressions for coordinates involving
only even powers of a, b, c. These satisfy basic relations like
S B + SC = a 2 ,
SC + S A = b 2 ,
S A + S B = c2 ;
SBC + SCA + S AB = S2 ,
(2)
(3)
where S AB := S A SB etc and S stands for twice the area of triangle ABC (see [?,
§3.4.1]. In fact, S A = S cot A etc. For examples, the homogeneous barycentric
coordinates of the circumcenter O are ( a2 S A : b2 SB : c2 SC ), the orthocenter and
the symmedian point are the points H = (SBC : SCA : S AB ) and K = (SB + SC :
SC + S A : S A + SB ) respectively. For later use we record the coordinates of the
antipodes A′ , B′ , C ′ of A, B, C on the circumcircle:
A′ = (−SBC : b2 SB : c2 SC ),
B′ = ( a2 S A : −SCA : c2 SC ),
(4)
C ′ = ( a2 S A : b2 SB : −S AB ).
Instead of using homogeneous linear equations to represent lines, we shall use
line coordinates. Thus, the sidelines are a = [1 : 0 : 0], b = [0 : 1 : 0],
c = [0 : 0 : 1], and the Brocard axis is
OK = [b2 c2 (b2 − c2 ) : c2 a2 (c2 − a2 ) : a2 b2 ( a2 − b2 )].
(5)
2. T HREE PAIRS OF POINTS ON THE PERPENDICULAR BISECTORS
Consider the following points on the perpendicular bisectors of the sides of triangle ABC.
Pa
a′
b′
c′
Pb
Pc
′
⊥
Xb : = a ∩ b ( C ) Xc : = a′ ∩ c⊥ ( B )
Ya := b′ ∩ a⊥ (C )
Yc := b′ ∩ c⊥ ( A)
′
⊥
′
⊥
Za := c ∩ a ( B) Zb := c ∩ b ( A)
Proposition 1. The points Ya , Za are on the parabola Pa ; similarly, Zb , Xb are
on Pb , and Xc , Yc are on Pc .
Proof. Since Ya C = Ya A and Ya C ⊥ a, Ya is a point on the parabola Pa ; so is
Za .
Proposition 2. The points Xb and Xc are inverse in the circumcircle; so are Yc
and Ya , Za and Zb .
Proof. Note that in Figure ?? triangles OCXb and ABC are oppositely similar,
since the directed angles (see [?, §§16–19])
1
(OC, OB) = ( AC, AB) = −( AB, AC ),
2
π
( Xb O, Xb C ) = ( Xb O, BC ) + ( BC, Xb C ) = + ( BC, Xb C )
2
= ( Xb C, AC ) + ( BC, Xb C ) = ( BC, AC ) = −(CA, CB).
(OC, OXb ) =
By the law of sines,
OXb
AC
b
b
=
=
=⇒ OXb = · R
OC
AB
c
c
A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED
WITH A TRIANGLE
47
where R is the circumradius of triangle ABC. The distances from O to these six
points are summarized below.
OXb =
OYa = ac · R
OZa = ba · R OZb =
b
c
b
a
· R OXc = bc · R
OYc = ac · R
·R
From these, it is clear that Xb and Xc are inverse in the circumcircle; so are Yc and
Ya , Za and Zb .
Pc
A
Ya
Pa
C′
Pb
B′
Yc
Za
F
E
Zb
O
D
B
C
Xc
A′
Xb
F IGURE 1.
Proposition 3. The homogeneous barycentric coordinates of Xb , Xc , Yc , Ya , Za ,
Zb on a′ , b′ , c′ are as follows.
a′
b′
c′
Xb = (− a2 SC : a2 b2 : S2 + SBC ) Xc = (− a2 SB : S2 + SBC : c2 a2 )
Yc = (S2 + SCA : −b2 S A : b2 c2 ) Ya = ( a2 b2 : −b2 SC : S2 + SCA )
Za = (c2 a2 : S2 + S AB : −c2 SB ) Zb = (S2 + S AB : b2 c2 : −c2 S A )
Proof. These follow from two sets of line coordinates.
(1) The perpendiculars from the vertices to the sidelines:
a⊥ ( B ) = [ S B : 0 : S B + SC ]
b⊥ ( A ) = [ 0 : S A : SC + S A ]
c⊥ ( A ) = [ 0 : S A + S B : S A ]
a⊥ ( C ) = [ SC : S B + SC : 0 ]
b⊥ ( C ) = [ SC + S A : SC : 0 ]
c⊥ ( B ) = [ S A + S B : 0 : S B ]
(6)
(2) The perpendicular bisectors of the sides:
a′ = [SB − SC : −(SB + SC ) : SB + SC ] = [b2 − c2 : a2 : − a2 ],
b′ = [SC + S A : SC − S A : −(SC + S A )] = [−b2 : c2 − a2 : b2 ],
c′ = [−(S A + SB ) : S A + SB : S A − SB ] = [c2 : −c2 : a2 − b2 ].
(7)
For these, it is enough to verify that [SB − SC : −(SB + SC ) : SB + SC ] contains
the midpoint (0 : 1 : 1) of BC and the circumcenter
O = (S A (SB + SC ) : SB (SC + S A ) : SC (S A + SB )).
48
PAUL YIU AND XIAO-DONG ZHANG
The first is clear. For the circumcenter, it follows from
( S B − SC ) S A ( S B + SC ) − ( S B + SC ) S B ( SC + S A ) + ( S B + SC ) SC ( S A + S B )
= (SB + SC )(S A (SB − SC ) − SB (SC + S A ) + SC (S A + SB ))
= 0.
From the coordinates given in (??) and (??), we compute
Xb = [SB − SC : −(SB + SC ) : SB + SC ] ∩ [SC + S A : SC : 0]
= (−SC (SB + SC ) : (SB + SC )(SC + S A )
: (SB − SC )SC + (SB + SC )(SC + S A ))
= (−SC (SB + SC ) : (SB + SC )(SC + S A ) : 2SBC + S AB + SCA )
= (−SC (SB + SC ) : (SB + SC )(SC + S A ) : S2 + SBC )
= (− a2 SC : a2 b2 : S2 + SBC ).
The others can be computed similarly.
Proposition 4. The lines BXc and CXb intersect at the antipode A′ of A on the
circumcircle.
Proof. The intersection of BXc and CXb is the point
(0 : 1 : 0) · (− a2 SB : S2 + SBC : c2 a2 ) ∩ (0 : 0 : 1) · (− a2 SC : a2 b2 : S2 + SBC )
= [c2 a2 : 0 : a2 SB ] ∩ [− a2 b2 : − a2 SC : 0]
= [ c 2 : 0 : S B ] ∩ [ b 2 : SC : 0]
= (−SBC : b2 SB : c2 SC ).
This is A′ given in (??)). Similarly, CYa ∩ AYc = B′ and AZb ∩ BZa = C ′ .
We shall show in Proposition ?? below that b′ and c′ are tangent to Pa ; similarly
for the other two parabolas.
3. T HE PARABOLAS Pa
ETC EACH DEFINED BY FIVE TANGENTS
Lemma 5. (a) The reflection of the focus of a parabola in a tangent is a point on
the directrix.
(b) The circumcircle of the triangle bounded by three tangents to a parabola
passes through the focus of the parabola.
⊥
Theorem 6 ( [?]). The parabola tangent to the Lb , Lc , L⊥
b and L c has focus at
vertex A and directrix the line a. It is the parabola Pa .
Proof. Let I, Ib , Ic be the incenter and the B-, C-excenters of triangle ABC. The
⊥
bisectors Lb , L⊥
b and L c bound the triangle IBIc . The bisectors L c , L c , and L b
bound the triangle ICIb . Apart from I, the circumcircles of these triangles intersect
at A. This is the focus of the parabola by Lemma ??(b).
The reflections of A in the Lb and Lc are points on a. By Lemma ??(a), the
line a is the directrix. This shows that the parabola tangent to the four bisectors of
angles B and C is the parabola Pa .
⊥
The dual conic of Pa is a line conic Pa∗ containing Lb , L⊥
b , L c , L c , which
have line coordinates
[ c : 0 : − a ],
[ c : 0 : a ],
[−b : a : 0],
[ b : a : 0].
A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED
WITH A TRIANGLE
49
Since Pa is a parabola, it is tangent to the line of infinity. Therefore, Pa∗ also
contains [1 : 1 : 1]. From these five line coordinates we find the barycentric
equation of Pa∗ :
a2 x2 − b2 y2 + (b2 + c2 − a2 )yz − c2 z2 = 0,
(8)
Proposition 7. The perpendicular bisectors b′ and c′ are tangent to the parabola
Pa .
Proof. Rewriting (??) as
a2 ( x2 − yz) − (y − z)(b2 y − c2 z) = 0,
we have, upon substitution of the line coordinates of b′ = [−b2 : c2 − a2 : b2 ]
given in (??),
a2 (b4 − (c2 − a2 )b2 ) − (c2 − a2 − b2 )(b2 (c2 − a2 ) − c2 b2 )
= a2 b2 ( b2 − c2 + a2 ) + a2 b2 ( c2 − a2 − b2 )
= 0.
Therefore the dual conic Pa∗ contains b′ . A similar calculation shows that it also
contains c′ . These two perpendicular bisectors are tangent to the parabola Pa . Corollary 8. The parabola Pa is tangent to b′ and c′ at Ya and Za respectively.
Figure ?? shows the 6 points of tangency of the parabola Pa with b′ , c′ , Lb ,
Lc , L⊥
c . The counterparts of Theorem ??, Proposition ??, and Corollary ??
hold for the parabolas Pb and Pc .
L⊥
b ,
′
Ica
′
Iba
A
Ya
Ica
Iba
Za
O
B
C
F IGURE 2.
Altogether, the triad of parabolas Pa , Pb , Pc , together with 6 angle bisectors
and the 3 perpendicular bisectors, accounts for 18 points of tangency shown in
Table 1 below.
Table 1. Points of tangency of the triad of parabolas
L a Lb Lc L⊥
L⊥
L⊥
a′ b′ c′
a
c
b
′
′
Pa
Iba Ica
Iba Ica
Ya Za
′
′
Pb Iab
Icb Iab
Icb Xb
Zb
′
′
Pc Iac Ibc
Iac
Ibc
Xc Yc
50
PAUL YIU AND XIAO-DONG ZHANG
4. 18 POINTS OF TANGENCY ON 6 LINES
In this section we find another interesting arrangement of these 18 points of
tangency in Table 1.
From the equation of Pa∗ , we find that of Pa . The matrix of Pa∗ is
 2

a
0
0
Ma∗ :=  0 −b2 S A  .
0 S A − c2
This has adjoint matrix

S2
0
0
M a =  0 − c2 a2 − a2 S A 
0 − a2 S A − a2 b2

representing the parabola Pa . The polar of B with respect to Pa is the line
0 1 0 M a = − a2 [0 c2 S A ].
(9)
It contains the points of tangency of Pa with Lb and L⊥
b , namely, the points Iba and
′ . We can simply find I as [ c : 0 : − a ] ∩ [0 : c2 : S ] and I ′ as [ c : 0 : a ] ∩ [0 :
Iba
A
ba
ba
c2 : S A ]; similarly for the remaining points of tangency of the three parabolas with
the angle bisectors. We summarize the coordinates in the proposition below.
Proposition 9. The coordinates of the points of tangency of Pa , Pb , Pc with the
angle bisectors are as follows.
Pa
La
Lb
Lc
L⊥
a
L⊥
b
L⊥
c
Iba = (ca : −S A : c2 )
Ica = ( ab : b2 : −S A )
′
Iba
′
Ica
− c2 )
= (ca : S A :
= ( ab : −b2 : S A )
Iab
Pb
= (−SB : bc : c2 )
Icb = ( a2 : ab : −SB )
′ = ( S : bc : − c2 )
Iab
B
′ = (− a2 : ab : S )
Icb
B
Iac
Ibc
Pc
= (−SC : b2 : bc)
= ( a2 : −SC : ca)
′ = ( S : − b2 : bc )
Iac
C
′ = (− a2 : S : ca )
Ibc
C
′
Iac
A
′
Iab
O
B
Iab
Xb
Xc
C
A′
Iac
F IGURE 3.
Note that the polar of B given in (??) is the same as c⊥ ( A) given (??). As such,
it contains the antipode B′ of B on the circumcircle. It also contains the vertex A
and the point of tangency Yc of Pc with b′ . In other words, the line AB′ contains
′ , and Y .
the three points of tangency Iba , Iba
c
A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED
WITH A TRIANGLE
51
Theorem 10. The 18 points of tangency of the parabolas Pa , Pb , Pc with the 6
angle bisectors and the 3 perpendicular bisectors fall on 6 lines, each joining the
antipode (on the circumcircle) of a vertex of triangle ABC to the remaining two
vertices.
′
AB′ Iba Iba
Yc
′
′
AC Ica Ica Zb
′
BC ′ Icb Icb
Za
′
′
BA Iab Iab Xc
′
CA′ Iac Iac
Xb
′
′
CB
Ibc Ibc Ya
′ , X , and I , I ′ ,
Figure ?? shows the two lines BA′ and CA′ containing Iab , Iab
c
ac ac
Xb respectively.
5. S OME PERSPECTIVITIES
5.1. Triangles with vertices among the tangency points on the perpendicular
bisectors.
Proposition 11. The triangle X1 Y1 Z1 bounded by the lines Ya Za , Zb Xb , Xc Yc is
perspective with ABC. The perspectrix
is the Lemoine axis, and the perspector is
the Kiepert perspector K π2 − ω , where ω is the Brocard angle of triangle ABC.
Pc
A
Y1
Pa
Ya
Yc
Pb
Zb
Za
E
F
Z1
O
X(262)
B
D
C
Xc
X1
Xb
F IGURE 4.
Proof. The lines Ya Za , Zb Xb , Xc Yc are the polars of O with respect to the parabolas
Pa , Pb , Pc respectively. They have line coordinates
[−S A : c2 : b2 ],
[ c2 : − S B : a2 ],
[ b 2 : a 2 : − SC ] .
These lines intersect a, b, c respectively at
(0 : b2 : − c2 ),
(− a2 : 0 : c2 ),
( a2 : − b2 : 0),
52
PAUL YIU AND XIAO-DONG ZHANG
1
1
1
:
:
. This shows that the
which are collinear on the Lemoine axis
a2 b2 c2
triangle bounded by the lines is perspective with ABC with the Lemoine axis as
perspectrix. The vertices of the triangle are
X1 = [ S A + S B : − S B : S B + S C ] ∩ [ S C + S A : S B + S C : − S C ]
= (−(SBB + SBC + SCC ) : S2 + (S A + SB + SC )SC : S2 + (S A + SB + SC )SB ),
Y1 = (S2 + (S A + SB + SC )SC : −(SCC + SCA + S AA ) : S2 + (S A + SB + SC )S A ),
Z1 = (S2 + (S A + SB + SC )SB : S2 + (S A + SB + SC )S A : −(S AA + S AB + SBB )).
From these, the perspector with ABC is the point
1
1
1
: 2
: 2
2
S + ( S A + S B + SC ) S A S + ( S A + S B + SC ) S B S + ( S A + S B + SC ) SC
Since S A + SB + SC = Sω for the Brocard angle ω of triangle ABC,
S2
Sω
S2
S A + S B + SC
.
=
= S tan ω, the perspector is the same as
1
1
1
:
:
S A + S tan ω SB + S tan ω SC + S tan ω
the Kiepert perspector K
π
2
,
−ω .
Proposition 12 ( [?]). The triangle X1 Y1 Z1 bounded by the Ya Za , Za Xa , Xa Ya is
perspective to the triangle of the vertices of Pa , Pb , Pc . The perspector is the
point
((S A + SB + SC )2 (SBB + 4SBC + SCC ) − (SBB + SBC + SCC )2 : · · · : · · · ),
and the perspectrix is the line
y
z
x
+
+
= 0.
SCA + S AB − SBC
S AB + SBC − SCA
SBC + SCA − S AB
Remarks. (1) Proposition ?? appears in [?] under the entry X (262).
(2) The line Ya Za = [−S A : c2 : b2 ] is also the directrix of the A-Artzt
parabola, which is tangent to b and c at C and B. It has focus (2S A : b2 : c2 ),
the second intersection of AK with the Brocard circle (of diameter OK). Similar
results hold for the B- and C-Artzt parabolas. Therefore, the directrices of the A-,
B-, C-Artzt parabolas bound a triangle with perspective with ABC at X (262).
(3) The perspector in Proposition ?? is the triangle center X (9748) in [?].
Proposition 13. The points X2 := Yc Za ∩ Ya Zb , Y2 := Za Xb ∩ Zb Xc , Z2 :=
Xb Yc ∩ Xc Ya are collinear on the Brocard axis, and the lines AX2 , BY2 , CZ2 are
concurrent at the Tarry point on the circumcircle.
Proof. The triangles Xb Yc Za and Xc Ya Zb are clearly perspective at the circumcenter O. They are also axis-perspective, i.e. the three intersections X2 = Yc Za ∩
Ya Zb , Y2 = Za Xb ∩ Zb Xc , Z2 = Xb Yc ∩ Xc Ya are collinear. Making use of the
A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED
WITH A TRIANGLE
53
coordinates given in Proposition ??, we have
X2 = ( a2 (S3A + a2 (2S AA − SBC ) − S A (SBB + SBC + SCC )) : b2 c2 (S AB − SCC )
: b2 c2 (SCA − SBB )),
Y2 = (c2 a2 (S AB − SCC ) : b2 (S3B + b2 (2SBB − SCA ) − SB (SCC + SCA + S AA ))
: c2 a2 (SBC − S AA )),
Z2 = ( a2 b2 (SCA − SBB ) : a2 b2 (SBC − S AA )
3
: c 2 ( SC
+ c2 (2SCC − S AB ) − SC (S AA + S AB + SBB ))).
Substituting the coordinates of X2 into the equation (??) of the Brocard axis, we
have, after ignoring a common factor a2 b2 c2 ,
(SB − SC )(S3A + 2S AA (SB + SC ) − S A (SBB + SBC + SCC ) − SBC (SB + SC ))
+ (SC − S A )(S A + SB )(S AB − SCC ) + (SC + S A )(S A − SB )(SCA − SBB )
= S3A (SB − SC ) + 2S AA (SBB − SCC ) − S A (S3B − SC3 ) − SBC (SBB − SCC )
− S3A SB − S AA (SBB − SBC − SCC ) + S A SC (SBB + SBC − SCC ) − SB SC3
+ S3A SC − S AA (SBB + SBC − SCC ) + S A SB (SBB − SBC − SCC ) + S3B SC
= 0.
This shows that X2 lies on the Brocard axis. Similar calculations show that Y2 and
Z2 also lies on the same line.
From the coordinates of X2 , Y2 , Z2 , the perspectivity of ABC and X2 Y2 Z2 is
1
1
1
:
.
:
SBC − S AA SCA − SBB S AB − SCC
This is the Tarry point X (98) on the circumcircle (see Figure ??).
Z2
A
X2
Ya
Yc
X(98)
Za
E
Zb
F
O
K
Y2
D
B
Xc
Xb
F IGURE 5.
C
54
PAUL YIU AND XIAO-DONG ZHANG
5.2. Triangles with vertices among the tangency points on the angle bisectors.
Proposition 14 (R. Hudson). The triangle bounded by the lines Iba Ica , Icb Iab , Iac Ibc
is perspective with ABC, the perspectrix is the trilinear polar of the incenter, and
the perspector the triangle center X (1000).
Proof. The lines Iba Ica , Icb Iab , Iac Ibc are the polars of the incenter I with respect to
the parabolas Pa , Pb , Pc . They have line coordinates [−(c + a − b)( a + b − c) :
2ca : 2ab], [2ab : −( a + b − c)(b + c − a) : 2bc], [2ca : 2bc : −(b + c − a)(c +
a − b)], and intersect a, b, c respectively at (0 : b : −c), (− a : 0 : c), ( a : −b : 0.
Therefore,
bound a triangle perspective with ABC, the perspectrix being the
1 1 they
1
line a : b : c , the trilinear polar of the incenter.
The vertices of the triangle are
X3 = (− a4 + 2a3 (b + c) − 2a(b − c)2 (b + c) + (b2 − c2 )2
: 2bc( a2 + b2 − c2 − 4ab) : 2bc(c2 + a2 − b2 − 4ca)),
Y3 = (2ca( a2 + b2 − c2 − 4ab)
: −b4 + 2b3 (c + a) − 2b(c − a)2 (c + a) + (c2 − a2 )2
: 2ca(b2 + c2 − a2 − 4bc)),
Z3 = (2ab(c2 + a2 − b2 − 4ca) : 2ab(b2 + c2 − a2 − 4bc)
: −c4 + 2c3 ( a + b) − 2c( a − b)2 ( a + b) + ( a2 − b2 )2 ).
The perspector with triangle ABC is the triangle center
1
1
1
:
:
.
b2 + c2 − a2 − 4bc c2 + a2 − b2 − 4ca a2 + b2 − c2 − 4ab
This is the triangle center X (1000) in [?].
′ I ′ , I ′ I ′ , I ′ I ′ is perspecProposition 15. The triangle bounded by the lines Iba
ca cb ab ac bc
tive with ABC, the perspectrix being the trilinear polar of the incenter, and the
perspector the orthocenter H.
′ I ′ , I ′ I ′ , I ′ I ′ are the polars of I , I , I with respect to
Proof. The lines Iba
a
c
b
ac bc
ca
cb ab
the parabolas Pa , Pb , Pc respectively. They have line coordinates [(c + a −
b)( a + b − c) : 2ca : 2ab], [2bc : ( a + b − c)(b + c − a) : 2ab], [2bc : 2ca :
(b + c − a)(c + a − b)]. These lines intersect a, b, c respectively at (0 : b : −c),
(− a : 0 : c), ( a : −b : 0). Therefore, they bound a triangle perspective with ABC,
the perspectrix being the trilinear polar of the incenter.
The vertices of the triangle are
X4 = ( a4 − 2a3 (b + c) + 2a(b − c)2 (b + c) − (b2 − c2 )2 : 2bc( a2 + b2 − c2 )
: 2bc(c2 + a2 − b2 )),
Y4 = (2ca( a2 + b2 − c2 ) : b4 − 2b3 (c + a) + 2b(c − a)2 (c + a) − (c2 − a2 )2
: 2ca(b2 + c2 − a2 )),
Z4 = (2ab(c2 + a2 − b2 ) : 2ab(b2 + c2 − a2 )
: c4 − 2c3 ( a + b) + 2c( a − b)2 ( a + b) − ( a2 − b2 )2 ).
A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED
The perspector with triangle ABC is
WITH A TRIANGLE
55
1
1
1
,
:
:
b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2
the orthocenter.
′ ∩ I I ′ , Y = I I ′ ∩ I I ′ , and Z = I I ′ ∩ I I ′ .
Proposition 16. Let X5 = Ib Iba
a ac
a ab
c cb
c ca
5
5
b bc
The triangle X5 Y5 Z5 is perspective with ABC at the incenter, and the perspectrix
is the trilinear polar of the Nagel point.
Proof. In homogeneous coordinates, the points are
′
′
X5 = Ib Iba
∩ Ic Ica
= (− a(b2 + 6bc + c2 − a2 ) : b(c + a − b)( a + b − c) : c(c + a − b)( a + b − c)),
Y5 = ( a( a + b − c)(b + c − a) : −b(c2 + 6ca + a2 − b2 ) : c( a + b − c)(b + c − a)),
Z5 = ( a(b + c − a)(c + a − b) : b(b + c − a)(c + a − b) : −c( a2 + 6ab + b2 − c2 )).
From these, it is clear that triangles ABC and ABC are perspective at the incenter.
The lines Y5 Z5 , Z5 X5 , X5 Y5 are
[( a + b + c)( a(b + c) − (b − c)2 ) : a( a + b − c)(b + c − a) : a(b + c − a)(c + a − b)],
[b(c + a − b)( a + b − c) : ( a + b + c)(b(c + a) − (c − a)2 ) : b(b + c − a)(c + a − b)],
[c(c + a − b)( a + b − c) : c( a + b − c)(b + c − a) : ( a + b + c)(c( a + b) − ( a − b)2 )].
They intersect a, b, c respectively at
(0 : −(c + a − b) : a + b − c),
(b + c − a : 0 : −( a + b − c)),
(−(b + c − a) : c + a − b : 0),
y
z
x
+
+
= 0, the trilinear polar
which are collinear on
b+c−a c+a−b a+b−c
of the Nagel point. This is the perspectrix of the triangles ABC and X5 Y5 Z5 . 6. A
CONIC THROUGH THE POINTS OF TANGENCY WITH THE EXTERNAL
ANGLE BISECTORS
′ given in
Consider the excentral triangle Ia Ib Ic . From the coordinates of Iab
1
Proposition ?? with coordinate sum 2 (c + a − b)( a + b − c), and the expression
2a(c2 + a2 − b2 , 2bc, −2c2 )
= − ( a + b − c)(b + c − a)( a, −b, c) + ( a + b + c)(c + a − b)( a, b, −c),
we have
−(b + c − a) Ib + ( a + b + c) Ic
.
2a
′ with respect to
From this we have the homogeneous barycentric coordinates of Iab
the excentral triangle Ia Ib Ic ; similarly for the other five points of tangency on the
external angle bisectors.
′
Iab
=
Proposition 17. With respect to the excentral triangle,
′ = (0 : a + b + c : −( b + c − a )),
Iab
′
Ibc = (−(c + a − b) : 0 : a + b + c),
′ = ( a + b + c : −( a + b − c ) : 0),
Ica
′ = (0 : −( b + c − a ) : a + b + c ),
Iac
′
Iba = ( a + b + c : 0 : −(c + a − b)),
′ = (−( a + b − c ) : a + b + c : 0).
Icb
56
PAUL YIU AND XIAO-DONG ZHANG
Proposition 18. The six points are on a conic C with barycentric equation
( a + b + c)( x2 + y2 + z2 ) +
2( a2 + ( b + c )2 )
yz = 0
b+c−a
cyclic
∑
(10)
relative to the excentral triangle Ia Ib Ic .
′
Ica
′
Iac
′
Iba
A
′
Iab
C
B
′
Ibc
′
Icb
F IGURE 6.
Proof. Substituting x = 0 into (??) we obtain
0 = ( a + b + c)(y2 + z2 ) +
2( a2 + ( b + c )2 )
yz
b+c−a
1
(( a + b + c)(b + c − a)(y2 + z2 ) + 2( a2 + (b + c)2 )yz)
b+c−a
((b + c − a)y + ( a + b + c)z)(( a + b + c)y + (b + c − a)z)
.
=
b+c−a
=
Therefore the conic intersects the line Ib Ic at (0 : a + b + c : −(b + c − a)) and
′ and I ′ . Similarly, the
(0 : −(b + c − a) : a + b + c). These are the points Iab
ac
′
′
′
′
points Ibc , Iba , Ica , Icb are also on the same conic.
The center of a conic with given barycentric equation can be computed using
the formula given in [?, §10.7.2]. Applying this to the conic C we have
Proposition 19. Relative to the excentral triangle, the conic C through the six
′ , I ′ , I ′ , I ′ , I ′ , I ′ has center with homogeneous barycentric coordipoints Iab
ac bc ba ca cb
nates
( a2 (c + a − b)( a + b − c)( a4 − 2abc(b + c − a) − (b2 − c2 )2 )
: b2 ( a + b − c)(b + c − a)(b4 − 2abc(c + a − b) − (c2 − a2 )2 )
: c2 (b + c − a)(c + a − b)(c4 − 2abc( a + b − c) − ( a2 − b2 )2 )).
Remark. With reference to ABC, the conic C has barycentric equation
f a yz + f b zx + f c xy + 4abc( x + y + z) g( x, y, z) = 0,
A MOST BASIC TRIAD OF PARABOLAS ASSOCIATED
WITH A TRIANGLE
57
where
f a = a6 − a4 (b − c)2 − a2 (b − c)2 (b2 + 10bc + c2 )
+ 8abc(b − c)2 (b + c) + (b − c)4 (b + c)2 ,
g( x, y, z) =
∑
bc( a + b − c)( a − b + c) x,
cyclic
and f b , f c are obtained from f a by cyclic rotations of a, b, c. The center of the conic
has homogeneous barycentric coordinates
( a( a10 − a8 (b2 − 6bc + c2 ) − 6a7 bc(b + c) − 2a6 (b4 + 3b3 c − 12b2 c2 + 3bc3 + c4 )
+ 6a5 bc(b − c)2 (b + c) + 2a4 (b − c)2 (b4 − b3 c − 8b2 c2 − bc3 + c4 )
+ 2a3 bc(b − c)2 (b + c)(3b + c)(b + 3c) + a2 (b − c)4 (b + c)2 (b2 + 8bc + c2 )
− 2abc(b − c)2 (b + c)3 (3b2 − 2bc + 3c2 ) − (b2 − c2 )4 (b2 + c2 ))
: · · · : · · · ).
This has ETC-search number 26.8848761379 · · · .
R EFERENCES
[1] R. A. Johnson, Advanced Euclidean Geometry, Dover reprint, 2007.
[2] C. Kimberling, Encyclopedia of Triangle Centers, available at
http://faculty.evansville.edu/ck6/encyclopedia/ETC.html.
[3] P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University Lecture Notes,
2001; revised with corrections 2013, math.fau.edu/Yiu/Geometry.html.
[4] P. Yiu, M. Bataille, and O. Kouba, Problem 3854, Crux Math., 39 (2013) 274, 276; solution, 40
(2014) 264–265.
D EPARTMENT OF M ATHEMATICAL S CIENCES , F LORIDA ATLANTIC U NIVERSITY, 777 G LADES
ROAD , B OCA R ATON , F LORIDA 33431-0991, USA.
E-mail address: [email protected]
D EPARTMENT OF M ATHEMATICAL S CIENCES , F LORIDA ATLANTIC U NIVERSITY, 777 G LADES
ROAD , B OCA R ATON , F LORIDA 33431-0991, USA.
E-mail address: [email protected]