onstruction of a Triangle from the Feet of Its ngle isectors Paul Yiu bstract. We study the problem of construction of a triangle from the feet of its internal angle bisectors. conic solution is possible. The analysis leads to consideration of the feet of external angle bisectors as well. 1. Introduction In this note we address the problem of construction of a triangle from the endpoints of its angle bisectors. This is Problem 138 in Wernick s list [3]. The corresponding problem of determining a triangle from the lengths of its angle bisectors have been settled by Mironescu and Panaitopol [2]. Given a triangle we seek a triangle such that the lines,, bisect the angles, and respectively. With reference to triangle, is the anticevian triangle of a point P, which is the incenter of triangle. We shall work in homogeneous barycentric coordinates with respect to. P Figure 1. Proposition 1. The locus of a point Q for which Q is a bisector of the angles between Q and Q, i.e., (Q,Q) + (Q,Q) = 0, is the cubic F a := x(c 2 y 2 b 2 z 2 ) + yz((c 2 + a 2 b 2 )y (a 2 + b 2 c 2 )z) = 0. (1) Proof. The cosine of the directed angle (Q, Q) is given by S x(y + z) + S y(z + x) S z 2 ; Q Q 1
2 P. Yiu similarly for (Q, Q). Equating the two and simplifying, we obtain the cubic equation in 1. learly, Q does not lie on the line x = 0. If P = (x : y : z), then = ( x : y : z) lies on the cubic (1). This means that P lies on F a := x(c 2 y 2 b 2 z 2 ) + yz((c 2 + a 2 b 2 )y (a 2 + b 2 c 2 )z) = 0. (2) This is the isogonal conjugate of the conic a : f a (x,y,z) := a 2 (c 2 y 2 b 2 z 2 )+b 2 (c 2 +a 2 b 2 )zx c 2 (a 2 +b 2 c 2 )xy = 0. See Figure 2. F a = 0 P a P Figure 2. The cubic F a = 0 and its isogonal conjugate conic a Proposition 2. The conic a is the hyperbola through the following points: the vertex, the endpoints of the two bisectors of angle, the point X which divides the -altitude in the ratio 2 : 1, and its traces on sidelines b and c. Proof. This is clear for the vertex and the endpoints of the -bisectors, namely, (0 : b : c) and (0 : b : c). The conic intersects the lines b and c at Y = (a 2 : 0 : c 2 + a 2 b 2 ) and Z = (a 2 : a 2 + b 2 c 2 : 0) respectively. These are the traces of X = (a 2 : a 2 + b 2 c 2 : c 2 + a 2 b 2 ), which divides the -altitude H a in the ratio X : XH a = 2 : 1. Remark. The tangents at (0 : b : ±c) pass through the midpoint of the -altitude. The tangent at passes through the circumcenter O. The tangents at and X intersect on the sideline a.
Triangle from the feet of angle bisectors 3 X Figure 3. The conic a 2. Suppose now P is a point which is the incenter of its own anticevian triangle (with respect to ). From the analysis of the preceding section, the isogonal conjugate of P lies on the a as well as the two analogous conics b : f b (x,y,z) := b 2 (a 2 z 2 c 2 x 2 )+c 2 (a 2 +b 2 c 2 )xy a 2 (b 2 +c 2 a 2 )yz = 0, and c : f c (x,y,z) := c 2 (b 2 x 2 a 2 y 2 )+a 2 (b 2 +c 2 a 2 )yz b 2 (c 2 +a 2 b 2 )zx = 0. Since f a + f b + f c = 0, the three conics generate a pencil. Note that Proposition 1 does not distinguish between internal and external angle bisectors. common point of the three conics (or of the pencil they generate) is the isogonal conjugate (with respect to ) of a point which is the incenter or an excenter of its anticevian triangle (with respect to ). + + + + + + + + + + + + Figure 4. To distinguish between the incenter and the excenter cases, we note that a nondegenerate triangle divides the planes into seven regions (see Figure 4), which
4 P. Yiu 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. In the remainder of this section, we shall denote by ε a, ε b, ε c a triple of plus and minus signs, not all minuses. Lemma 3. point is in the ε a ε b ε c region of its own anticevian triangle (with respect to ) if and only if it is in the ε a ε b ε c region of the medial triangle of. The isogonal conjugates (with respect to ) of the sidelines of the medial triangle divide the plane into seven regions, which we also label ε a ε b ε c, so that the isogonal conjugates of points in the ε a ε b ε c region are in the corresponding region partitioned by the lines of the medial triangle. See Figure 5. Figure 5. Proposition 4. Let Q be a common point of the conics a, b, c in the ε a ε b ε c region of the partitioned by the hyperbolas. The isogonal conjugate of Q is a point whose anticevian triangle has P as incenter or excenter according as all or not of ε a, ε b, ε c are plus signs. Figure 6 shows an example in which the conics have four common points, one in each of the regions + + +, + +, + +, + +. The point P 0 is in the region + + +. It is the incenter of its own anticevian triangle (with respect to ). See Figure 7.
Triangle from the feet of angle bisectors 5 Q c Q 0 Q b Q a Figure 6. Q 0 P 0 Figure 7. Figure 8 shows an example in which the conics a, b, c have only two real intersections Q 1 and Q 2, none of which in the region +++. This means that there is no triangle for which,, are the feet of the internal angle bisectors. The isogonal conjugate P 1 of Q 1 has anticevian triangle such that and are the feet of the external bisectors of angles and, while is the foot of the internal bisector of angle.
6 P. Yiu P 1 Q 2 Q 1 3. Figure 8. If the given triangle contains a right angle, say, at vertex, then the point P can be constructed by ruler and compass. Here is an easy construction. Let 1 and 2 be equilateral triangles on the hypotenuse of the given triangle. If P is the reflection of one of 1, 2 in, then the anticevian triangle of P with respect to is one with P as incenter or an excenter. Proposition 5. The cevian triangle of the incenter is a right triangle if and only if contains a 120 angle. Proof. (2) If = 120, the sidelengths can be parametrized as a = t 2 1, b = 2t + 1, c = t 2 + t + 1 for t > 1. The cevian triangle has lengths a : b : c = 3(1 + t) : 1 + 2t : 4 + 10t + 7t 2. The half-tangents of are tan 2 = t 1 3(t + 1), tan 2 = 3 1 + 2t, tan 2 = 3.
Triangle from the feet of angle bisectors 7 P 2 P 1 2 1 Figure 9. If we put k := tan = 3(1+t) 1+2t, then t = 3 k 2k 3, and tan 2 = 2 k 3, tan 2 = 2k 3. ( tan 60 ) 3k 1 =, 2 3 k ( tan 60 ) = 2 3 k 3k 1. Since = 120, a necessary and sufficient condition for the existence of is 3 2 < k < 2 3. onstruction 1. Given a right triangle with = 90 and arctan, < arctan 2 3, 4. Triangles from the feet of external angle bisectors 3 2 < In this section we make a change of notations. Since the feet of the external angle bisectors are collinear, we start with three points X, Y, Z on a line, and seek a triangle whose external angle bisectors intersect their respective opposite sides at X, Y, Z. Without loss of generality we assume Y between X and Z. Figure 10 shows the collinearity of the feet X, Y, Z of the external bisectors of triangle. The line l containing them is the trilinear polar of the incenter, namely, x a + y b + z c = 0.
8 P. Yiu Z X Z Y X I Y Figure 10. 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. See Figure 10. Since the angles XX, Y Y and ZZ are right angles, the vertices,, lie on the circles with diameters XX, Y Y, ZZ respectively. Given three distinct points X, Y, Z on a line l, (assuming Y in between, nearer X than Z), let X, Y, Z be the harmonic conjugates of X, Y, Z in Y Z, ZX, XY respectively. onstruct the circles with diameters XX and ZZ. Note that the circle (XX ) is the locus of points for which the bisectors of angle Y Z pass through X and X. Since X is between Y and Z, the internal bisector of angle Y Z passes through X and the external bisector through X. Let the half-line Y intersect the circle (ZZ ) at. Then Z is the external bisector of angle XY. Let be the intersection of the lines Z and X. pplying Menelaus theorem to triangle and the transversal XY Z (with X on
Triangle from the feet of angle bisectors 9, Y on, Z on ), we have Y Y X X Z Z = 1 = Y Z X Y Z X = 1 = Y X X Z XZ Z Y Z X = 1 ( = Y X )( XZ ) Z XZ ZY X = 1 = X Z = XY Y Z. Therefore, Y bisects angle XZ, and is on the circle with diameter Y Y. The facts that X, Y, Z are on the lines,,, and that X, Y, Z are bisectors show that X, Y, Z are the external bisectors of triangle. Y X Z Y X Z Figure 11. Let be a point on the circle (XX ). onstruct the line Y to intersect the circle (ZZ ) at and (so that, are on the same side of Y ). The line Z intersects X and X at points and on the circle (Y Y ). The triangle has X, Y, Z as external angle bisectors. t the same time, has internal bisectors X, Y, and external bisector Z. See Figure 11. Proposition 6. The triangles with external bisectors X, Y, Z are characterizaed by a b : b c : a c = XY : Y Z : XZ. Proof. Without loss of generality, we assume a > b > c. The point Y is between X and Z. pplying Menelaus theorem to triangle XZ with transversal Y Z,
10 P. Yiu we have From this, XY References Y Z = a b b c XY Y Z Z X = 1.. The remaining ratios follow similarly. [1]. Kimberling, Encyclopedia of Triangle enters, available at http://faculty.evansville.edu/ck6/encyclopedia/et.html. [2] P. Mironescu and L. Panaitopol, The existence of a triangle with prescribed angle bisector lengths, mer. Math. Monthly, 101 (1994) 58 60. [3] W. Wernick, Triangle constructions with three located points, Math. Mag., 55 (1982) 227 230. Paul Yiu: Department of Mathematical Sciences, Florida tlantic University, oca Raton, Florida 33431-0991, US E-mail address: yiu@fau.edu