Conic Construction of a Triangle from the Feet of Its Angle Bisectors

 (or an excenter) of its own an-
ticevian triangle with respect to ABC . From the analysis of the preceding section,
its isogonal conjugate lies on the hyperbola Ca as well as the two analogous hyper-
bolas
Cb : fb(x, y, z) := b
2(a2z2−c2x2)+c2(a2+b2−c2)xy−a2(b2+c2−a2)yz = 0,
6 P. Yiu
and
Cc : fc(x, y, z) := c
2(b2x2−a2y2)+a2(b2+c2−a2)yz−b2(c2+a2−b2)zx = 0.
Since fa + fb + fc = 0, the three hyperbolas generate a pencil. The isogonal
conjugates of the common points of the pencil are the points that solve the angle
bisectors problem. Theorem 2 guarantees the existence of common points. To dis-
tinguish between the incenter and the excenter cases, we note that a nondegenerate
triangle ABC divides the planes into seven regions (see Figure 6), which we label
in accordance with the signs of the homogeneous barycentric coordinates of points
in the regions:
+ + +, −+ +, −+−, + +−, +−−, +−+, −−+
In each case, the sum of the homogeneous barycentric coordinates of a point is
adjusted to be positive.
A
B C
+ + +
−+ +−+−
+ +−
+−−
+−+
−−+
Figure 6. Partition of the plane by the sidelines of a triangle
In the remainder of this section, we shall denote by εa, εb, εc a triple of plus and
minus signs, not all minuses.
Lemma 4. A point is in the εaεbεc region of its own anticevian triangle (with
respect to ABC) if and only if it is in the εaεbεc region of the medial triangle of
ABC .
The isogonal conjugates (with respect to ABC) of the sidelines of the medial
triangle divide the plane into seven regions, which we also label εaεbεc, so that the
isogonal conjugates of points in the εaεbεc region are in the corresponding region
partitioned by the lines of the medial triangle. See Figure 7.
Proposition 5. Let Q be a common point of the conics Ca, Cb, Cc in the εaεbεc
region of the partitioned by the hyperbolas. The isogonal conjugate of Q is a point
whose anticevian triangle A′B′C ′ has P as incenter or excenter according as all
or not of εa, εb, εc are plus signs.
Conic construction of a triangle from the feet of its angle bisectors 7
A
B C
Figure 7. Partition of the plane by three branches of hyperbolas
6. Examples
Figure 8 shows an example in which the hyperbolas Ca, Cb, Cc have four com-
mon points Q0, Qa, Qb, Qc, one in each of the regions + + +, − + +, + − +,
+ + −. The isogonal conjugate P0 of Q0 is the incenter of its own anticevian
triangle with respect to ABC . See Figure 9.
A
B C
Qa
Q0
Qc
Qb
Figure 8. Pencil of hyperbolas with four real intersections
8 P. Yiu
A
B C
A′
Q0
B′
C′
P0
Figure 9. P0 as incenter of its own anticevian triangle
Figure 10 shows the hyperbolas Ca, Cb, Cc corresponding to the cubics in Figure
3. They have only two real intersections Q1 and Q2, none of which is in the region
+ + +. This means that there is no triangle A′B′C ′ for which A, B, C are the feet
of the internal angle bisectors. The isogonal conjugate P1 of Q1 has anticevian
triangle A1B1C1 and is its A1-excenter. Likewise, P2 is the isogonal conjugate of
Q2, with anticevian triangle A2B2C2, and is its B2-excenter.
A
B C
Q1
Q2
P1
C1
A1
B1
P2
C2
A2 B2
Figure 10. Pencil of hyperbolas with two real intersections
Conic construction of a triangle from the feet of its angle bisectors 9
7. The angle bisectors problem for a right triangle
If the given triangle ABC contains a right angle, say, at vertex C , then the point
P can be constructed by ruler and compass. Here is an easy construction. In fact,
if c2 = a2 + b2, the cubics Fa = 0, Fb = 0, Fc = 0 are the curves
x((a2 + b2)y2 − b2z2)− 2a2y2z = 0,
y((a2 + b2)x2 − a2z2)− 2b2x2z = 0,
z(b2x2 − a2y2)− 2xy(b2x− a2y) = 0.
A simple calculation shows that there are two real intersections
P1 =(a(
√
3a− b) : b(
√
3b− a) : (
√
3a− b)(
√
3b− a)),
P2 =(a(
√
3a+ b) : b(
√
3b+ a) : −(
√
3a+ b)(
√
3b+ a)).
These two points can be easily constructed as follows. Let ABC1 and ABC2
be equilateral triangles on the hypotenuse AB of the given triangle (with C1 and
C on opposite sides of AB). Then P1 and P2 are the reflections of C1 and C2 in
C . See Figure 11. Each of these points is an excenter of its own anticevian triangle
with respect toABC , except that in the case of P1, it is the incenter when the acute
angles A and B are in the range arctan
√
3
2 < A,B < arctan
2√
3
.
A
B
C
C2 C1
P1
P2
B′
A′
C′
A′′
C′′
B′′
Figure 11. The angle bisectors problems for a right triangle
Remark. The cevian triangle of the incenter contains a right angle if and only if the
triangle contains an interior angle of 120◦ angle (see [1]).
10 P. Yiu
8. Triangles from the feet of external angle bisectors
In this section we make a change of notations. Figure 12 shows the collinearity
of the feetX, Y , Z of the external bisectors of triangle ABC . The line ℓ containing
them is the trilinear polar of the incenter, namely, x
a
+ y
b
+ z
c
= 0. If the internal
bisectors of the angles intersect ℓ at X ′, Y ′, Z ′ respectively, then X, X ′ divide
Y , Z harmonically, so do Y, Y ′ divide Z, X, and Z, Z ′ divide X, Y . Since
the angles XAX ′, Y BY ′ and ZCZ ′ are right angles, the vertices A, B, C lie on
the circles with diameters XX ′, Y Y ′, ZZ ′ respectively. This leads to the simple
solution of the external angle bisectors problem.
A
B C
I
X
Y
Z
X′
Y ′
Z′
Figure 12. The external angle bisectors problem
We shall make use of the angle bisector theorem in the following form. Let
ε = ±1. The ε-bisector of an angle is the internal or external bisector according as
ε = +1 or −1.
Lemma 6 (Angle bisector theorem). Given triangle ABC with a point X on the
line BC . The line AX is an ε-bisector of angle BAC if and only if
BX
XC
= ε · AB
AC
.
Here the left hand side is a signed ratio of directed segments, and the ratio AB
AC
on the right hand side is unsigned.
Given three distinct points X, Y , Z on a line ℓ (assuming, without loss of gen-
erality, Y in between, nearer to X than to Z , as shown in Figure 12), let X ′, Y ′, Z ′
be the harmonic conjugates of X, Y , Z in Y Z , ZX, XY respectively. Here is a
Conic construction of a triangle from the feet of its angle bisectors 11
very simple construction of these harmonic conjugates and the circles with diame-
ters XX ′, Y Y ′, ZZ ′. These three circles are coaxial, with two common points F
and F ′ which can be constructed as follows: if XYM and Y ZN are equilateral
triangles erected on the same side of the line XY Z , then F and F ′ are the Fermat
point of triangle YMN and its reflection in the line. See Figure 13.
X Y ZX′Y ′ Z′
M
N
F
F ′
Figure 13. Coaxial circles with diameters XX ′, Y Y ′, ZZ′
Note that the circle (XX ′) is the locus of points A for which the bisectors of
angle Y AZ pass through X and X ′. Since X ′ is between Y and Z , the internal
bisector of angle Y AZ passes through X ′ and the external bisector through X. Let
the half-line Y A intersect the circle (ZZ ′) at C . Then CZ is the external bisector
of angle XCY . Let B be the intersection of the lines AZ and CX.
Lemma 7. The point B lies on the circle with diameter Y Y ′.
Proof. Applying Menelaus’ theorem to triangle ABC and the transversal XY Z
(with X on BC , Y on CA, Z on AB), we have
AY
Y C
· CX
XB
· BZ
ZA
= −1.
Here, each component ratio is negative. See Figure 12. We rearrange the numer-
ators and denominators, keeping the signs of the ratios, but treating the lengths of
the various segments without signs:(
−AY
AZ
)(
−CX
CY
)(
−BZ
BX
)
= −1.
Applying the angle bisector theorem to the first two ratios, we have
Y X
XZ
· XZ
ZY
·
(
−BZ
BX
)
= −1.
Hence, ZY
YX
= BZ
BX
, and BY is the internal bisector of angle XBZ . This shows
that B lies on the circle with diameter Y Y ′. 
12 P. Yiu
The facts that X, Y , Z are on the lines BC , CA, AB, and that AX ′, BY , CZ ′
are bisectors show that AX, BY , CZ are the external bisectors of triangle ABC .
This leads to a solution of a generalization of the external angle bisector problem.
X Y ZX′Y ′ Z′
A
C
C′
B
B′
Figure 14. Solutions of the external angle bisectors problem
Let A be a point on the circle (XX ′). Construct the line Y A to intersect the
circle (ZZ ′) at C and C ′ (so that A, C are on the same side of Y ). The line AZ
intersects CX and C ′X at points B and B′ on the circle (Y Y ′). The triangle
ABC has AX, BY , CZ as external angle bisectors. At the same time, AB′C ′ has
internal bisectors AX, B′Y , and external bisector C ′Z . See Figure 14.
We conclude with a characterization of the solutions to the external angle bisec-
tors problem.
Proposition 8. The triangles ABC with external bisectors AX, BY , CZ are
characterized by
a− b : b− c : a− c = XY : Y Z : XZ.
Proof. Without loss of generality, we assume a > b > c. See Figure 12. The point
Y is between X and Z . Since AX and CZ are the external bisector of angles
BAC and ACB respectively, we have BX
XC
= −c
b
and AZ
ZB
= −b
a
. From these,
CX
BC
= b−(b−c) and
AB
ZA
= a−b
b
. Applying Menelaus’ theorem to triangle XZB
with transversal Y AZ , we have
XY
Y Z
· ZA
AB
· BC
CX
= −1.
Hence, XY
Y Z
= −CX
BC
· AB
ZA
= a−b
b−c . The other two ratios follow similarly. 
Conic construction of a triangle from the feet of its angle bisectors 13
References
[1] H. Demir and J. Oman, Problem 998, Math. Mag., 49 (1976) 252; solution, 51 (1978) 199–200.
[2] P. Mironescu and L. Panaitopol, The existence of a triangle with prescribed angle bisector
lengths, Amer. Math. Monthly, 101 (1994) 58–60.
[3] W. Wernick, Triangle constructions with three located points, Math. Mag., 55 (1982) 227–230.
[4] P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University Lecture Notes,
2001.
Paul Yiu: Department of Mathematical Sciences, Florida Atlantic University, 777 Glades Road,
Boca Raton, Florida 33431-0991, USA
E-mail address: yiu@fau.edu