Some Constructions Related to the Kiepert Hyperbola

Transcription

1 Forum eometricorum Volume FORUM EOM ISSN Some onstructions Related to the Kiepert yperbola Paul Yiu bstract. iven a reference triangle and its Kiepert hyperbola K, we study several construction problems related to the triangles which have K as their own Kiepert hyperbolas. Such triangles necessarily have their vertices on K, and are called special Kiepert inscribed triangles. mong other results, we show that the family of special Kiepert inscribed triangles all with the same centroid form part of a poristic family between K and an inscribed conic with center which is the inferior of the Kiepert center.. Special Kiepert inscribed triangles iven a triangle and its Kiepert hyperbola K, consisting of the Kiepert perspectors Kt = S + t S + t, t R { }, S + t we study triangles with vertices on K having K as their own Kiepert hyperbolas. We shall work with homogeneous barycentric coordinates and make use of standard notations of triangle geometry as in [2]. asic results on triangle geometry can be found in [3]. The Kiepert hyperbola has equation Kx, y, z =S S yz +S S zx +S S xy =0 in homogeneous barycentric coordinates. Its center, the Kiepert center K i =S S 2 S S 2 S S 2, lies on the Steiner inellipse. In this paper we shall mean by a Kiepert inscribed triangle one whose vertices are on the Kiepert hyperbola K. If a Kiepert inscribed triangle is perspective with, it is called the Kiepert cevian triangle of its perspector. Since the Kiepert hyperbola of a triangle can be characterized as the rectangular circum-hyperbola containing the centroid, our objects of interest are Kiepert inscribed triangles whose centroids are Kiepert perspectors. We shall assume the vertices to be finite points on K, and call such triangles special Kiepert inscribed triangles. We shall make frequent use of the following notations. Publication Date December 28, ommunicating Editor Jean-Pierre Ehrmann.

2 344 P. Yiu P t = S S S + t S S S + t S S S + t Qt = S S 2 S + t S S 2 S + t S S 2 S + t f 2 = S + S + S S S S f 3 = S S S 2 + S S S 2 + S S S 2 f 4 =S S S +S S S +S S S g 3 =S S S S S S ere, P t is a typical infinite point, and Qt is a typical point on the tangent of the Steiner inellipse through K i.fork =2, 3, 4, the function f k, is a symmetric function in S, S, S of degree k. Proposition. The area of a triangle with vertices Kt i, i =, 2, 3, is g 3 t t 2 t 2 t 3 t 3 t. S 2 +2S +S +S t i +3t 2 i Proposition 2. Kiepert inscribed triangle with vertices Kt i, i =, 2, 3, is special, i.e., with centroid on the Kiepert hyperbola, if and only if S 2 f 2 +S + S + S f 3 3f 4 =0, where f 2, f 3, f 4 are the functions f 2, f 3, f 4 with S, S, S replaced by t, t 2, t 3. We shall make use of the following simple construction. polar of M P M Q Figure. onstruction of chord of conic with given midpoint onstruction 3. iven a conic and a point M, to construct the chord of with M as midpoint, draw i the polar of M with respect to, ii the parallel through M to the line in i. If the line in ii intersects at the two real points P and Q, then the midpoint of PQis M.

3 Some constructions related to the Kiepert hyperbola 345 K K i Q polar of M K 3 M K 2 Figure 2. onstruction of Kiepert inscribed triangle with prescribed centroid and one vertex simple application of onstruction 3 gives a Kiepert inscribed triangle with prescribed centroid Q and one vertex K simply take M to be the point dividing K Q in the ratio K M MQ =3. See Figure 2. ere is an interesting family of Kiepert inscribed triangles with prescribed centroids on K. onstruction 4. iven a Kiepert perspector Kt, construct i K on K and M such that the segment K M is trisected at K i and Kt, ii the parallel through M to the tangent of K at Kt, iii the intersections K 2 and K 3 of K with the line in ii. Then K K 2 K 3 is a special Kiepert inscribed triangle with centroid Kt. See Figure 3. K K i Kt K 3 M K 2 Figure 3. Kiepert inscribed triangle with centroid Kt

4 346 P. Yiu It is interesting to note that the area of the Kiepert inscribed triangle is independent of t. Itis g 3 f times that of triangle. This result and many others in the present paper are obtained with the help of a computer algebra system. 2. Special Kiepert cevian triangles iven a point P =u v w, the vertices of its Kiepert cevian triangle are S S vw P = S S v +S S w v w, S S wu P = u S S w +S S u w, S S uv P = u v. S S u +S S v These are Kiepert perspectors with parameters t, t, t given by t = S v S w v w, t = S w S u, t = S u S v. w u u v learly, if P is on the Kiepert hyperbola, the Kiepert cevian triangle P P P degenerates into the point P. Theorem 5. The centroid of the Kiepert cevian triangle of P lies on the Kiepert hyperbola if and only if P is i an infinite point, or ii on the tangent at K i to the Steiner inellipse. Proof. Let P =u v w in homogeneous barycentric coordinates. pplying Proposition 2, we find that the centroid of P P P lies on the Kiepert hyperbola if and only if u + v + wku, v, w 2 Lu, v, wp u, v, w =0, where u v w Lu, v, w = + +, S S S S S S P u, v, w = S S v 2 2S S vw +S S w 2. The factors u + v + w and Ku, v, w clearly define the line at infinity and the Kiepert hyperbola K respectively. On the other hand, the factor Lu, v, w defines the line x y z + + =0, 2 S S S S S S which is the tangent of the Steiner inellipse at K i. Each factor of P u, v, w defines two points on a sideline of triangle. If we set x, y, z = v + w,v,w in, the equation reduces to S S v 2 2S S vw +S S w 2. This shows that the two points on the line are the intercepts of lines through parallel to the asymptotes of K, and the corresponding Kiepert cevian triangles have vertices at infinite points. This is similarly the case for the other two factors of P u, v, w.

5 Some constructions related to the Kiepert hyperbola 347 Remark. ltogether, the six points defined by P u, v, w above determine a conic with equation x, y, z = x 2 2S S yz S S S S S S =0. Since g 3 x, y, z = f 2 x + y + z 2 + S S 2 x 2 2S S S S yz, this conic is a translation of the inscribed conic S S 2 x 2 2S S S S yz =0, which is the Kiepert parabola. See Figure 4. Ki Figure 4. Translation of Kiepert parabola 3. Kiepert cevian triangles of infinite points onsider a typical infinite point P t =S S S + t S S S + t S S S + t in homogeneous barycentric coordinates. It can be easily verified that P t is the infinite point of perpendiculars to the line joining the Kiepert perspector Kt to the orthocenter. The Kiepert cevian triangle of P t has vertices This is the line SS S S + tx =0.

6 348 P. Yiu S S S + ts + t t = S + S +2t t = t = S S S + ts + t S + S +2t S S S + t S S S + t S S S + t S S S + t S S S + t, S S S + t,. S S S + ts + t S + S +2t t t t Kt Figure 5. The Kiepert cevian triangle of P t is the same as the Kiepert parallelian triangle of Kt It is also true that the line joining t to Kt is parallel to ; 2 similarly for t and t. Thus, we say that the Kiepert cevian triangle of the infinite point P t is the same as the Kiepert parallelian triangle of the Kiepert perspector Kt. See Figure 5. It is interesting to note that the area of triangle ttt is equal to that of triangle, but the triangles have opposite orientations. Now, the centroid of triangle ttt is the point S S S + S 2S S + S 2S t which, by Theorem 5, is a Kiepert perspector. It is Ks where s is given by 2f 2 st + f 3 s + t 2f 4 =0. 3 Proposition 6. Two distinct Kiepert perspectors have parameters satisfying 3 if and only if the line joining them is parallel to the orthic axis. 2 This is the line S + ts + S +2tx +S + ts + ty + z =0.,

7 Some constructions related to the Kiepert hyperbola 349 Proof. The orthic axis S x + S y + S z =0has infinite point P =S S S S S S. The line joining Ks and Kt is parallel to the orthic axis if and only if S +s S +s S +s S +t S +t S +t S S S S S S =0. For s t, this is the same condition as 3. This leads to the following construction. t t Ks t Kt Figure 6. The Kiepert cevian triangle of P t has centroid Ks onstruction 7. iven a Kiepert perspector Ks, to construct a Kiepert cevian triangle with centroid Ks, draw i the parallel through Ks to the orthic axis to intersect the Kiepert hyperbola again at Kt, ii the parallels through Kt to the sidelines of the triangle to intersect K again at t, t, t respectively. Then, ttt has centroid Ks. See Figure Special Kiepert inscribed triangles with common centroid We construct a family of Kiepert inscribed triangles with centroid, the centroid of the reference triangle. This can be easily accomplished with the help

8 350 P. Yiu of onstruction 3. eginning with a Kiepert perspector K = Kt and Q =, we easily determine M =S +ts +S +2t S +ts +S +2t S +ts +S +2t. The line through M parallel to its own polar with respect to K 3 has equation S S S + t x + S S S + t y + S S z =0. 4 S + t s t varies, this line envelopes the conic S S 4 x 2 +S S 4 y 2 +S S 4 z 2 2S S 2 S S 2 xy 2S S 2 S S 2 yz 2S S 2 S S 2 zx =0, which is the inscribed ellipse E tangent to the sidelines of at the traces of S S 2 S S 2 S S 2, and to the Kiepert hyperbola at, and to the line 4 at the point S + t 2 S + t 2 S + t 2. It has center S S 2 +S S 2 S S 2 +S S 2 S S 2 +S S 2, the inferior of the Kiepert center K i. See Figure 7. P Ki P 3 M P 2 Figure 7. Poristic triangles with common centroid 3 The polar of M has equation S S S S 2 2S + S t 2t 2 x =0and has infinite point S + ts S + S 2t S + S S S + t.

9 Some constructions related to the Kiepert hyperbola 35 Theorem 8. poristic triangle completed from a point on the Kiepert hyperbola outside the inscribed ellipse E with center the inferior of K i has its center at and therefore has K as its Kiepert hyperbola. More generally, if we replace by a Kiepert perspector K g, the envelope is a conic with center which divides K i K g in the ratio 3. It is an ellipse inscribed in the triangle in onstruction family of special Kiepert cevian triangles 5.. Triple perspectivity. ccording to Theorem 5, there is a family of special Kiepert cevian triangles with perspectors on the line 2 which is the tangent of the Steiner inellipse at K i. Since this line also contains the Jerabek center J e =S S S 2 S S S 2 S S S 2, its points can be parametrized as Qt =S S 2 S + t S S 2 S + t S S 2 S + t. The Kiepert cevian triangle of Qt has vertices S S S S S + ts + t t = S + t t = S S 2 S SS SS + ts + t S + t S + t t = S S 2 S + t S S 2 S + t S S 2 S + t S S 2 S + t, S S 2 S + t, S SS SS + ts + t S + t. Theorem 9. The Kiepert cevian triangle of Qt is triply perspective to. The three perspectors are collinear on the tangent of the Steiner inellipse at K i. Proof. The triangles t t t and t t t are each perspective to, at the points Q S + t S + t S + t t =, S S S S S S and Q S + t S + t S + t t = S S S S S S respectively. These two points are clearly on the line Special Kiepert cevian triangles with the same area as. The area of triangle t t t is f 2 t 2 + f 3 t f 4 3 f2 S + t 2 S S 2 S S 2 mong these, four have the same area as the reference triangle.

10 352 P. Yiu t = S S +S 2S S +S 2S. The points Qt = 2S S S S S S, Q t =S S 2S S S S, Q t =S S S S 2S S, give the Kiepert cevian triangle This has centroid K f 3 = 2f 2 = S S 2S S 2S S, =2S S S S 2S S, =2S S 2S S S S. S S S + S 2S S S S + S 2S S S S + S 2S t t t is also the Kiepert cevian triangle of the infinite point P of the orthic axis. See Figure 8.. Q Q K i Q Figure 8. Oppositely oriented triangle triply perspective with at three points on tangent at K i

11 Some constructions related to the Kiepert hyperbola t =. With the Kiepert center K i = Q, we have the points Q =S S 2 S S 2 S S 2, Q =, S S S S S S Q =, S S S S S S The points Q and Q are the intersection with the parallels through, to the line joining to the Steiner point S t = S S S S S S. These points give the Kiepert cevian triangle which is the image of under the homothety hk i, 2 =S S S S S S 2 S S 2, 2 =S S 2 S S S S S S 2, 2 =S S 2 S S 2 S S S S, which has centroid K S + S + S =. 3 S + S 2S S + S 2S S + S 2S The points Q t, Q t and 2 are on the Steiner circum-ellipse. See Figure Q Q = K i Q O 2 Figure 9. Oppositely congruent triangle triply perspective with at three points on tangent at K i S t

12 354 P. Yiu t = f 3 2f 2. Qt is the infinite point of the line 2. Qt =S S S + S 2S S S S + S 2S S S S + S 2S, Q t =S S S + S 2S S S S + S 2S S S S + S 2S, Q t =S S S + S 2S S S S + S 2S S S S + S 2S. These give the Kiepert cevian triangle 3 = S S S + S 2S 3 = S S S + S 2S 3 S S = S + S 2S S S S + S 2S S S S + S 2S S S S + S 2S S S S + S 2S S S S + S 2S S S S + S 2S,,, with centroid See Figure 0. S S. S S 2 +2S S S S 3 K i Q Q Q O Figure 0. Triangle triply perspective with with the same orientation at three points on tangent at K i

13 Some constructions related to the Kiepert hyperbola t = S. For t = S,wehave Qt =0S S S S, Q t = S S 0S S, Q t =S S S S 0. These points are the intercepts Q a, Q b, Q c of the line 2 with the sidelines,, respectively. The lines Q a, Q b, Q c are the tangents to K at the vertices. The common Kiepert cevian triangle of Q a, Q b, Q c is oppositely oriented as,,, triply perspective with at Q a, Q b, Q c respectively. 6. Special Kiepert inscribed triangles with two given vertices onstruction 0. iven two points K and K 2 on the Kiepert hyperbola K, construct i the midpoint M of K K 2, ii the polar of M with respect to K, iii the reflection of the line K K 2 in the polar in ii. If K 3 is a real intersection of K with the line in iii, then the Kiepert inscribed triangle K K 2 K 3 has centroid on K. See Figure. K 3 K3 K i K M K 2 Figure. onstruction of special Kiepert inscribed triangles given two vertices K, K 2

14 356 P. Yiu Proof. point K 3 for which triangle K K 2 K 3 has centroid on K clearly lies on the image of K under the homothety hm,3. It is therefore an intersection of K with this homothetic image. If M =u v w in homogeneous barycentric coordinates, this homothetic conic has equation u + v + w 2 Kx, y, z +2x + y + z S S vw +S S 3u + ww +S S 3u + vvx =0. The polar of M in K is the line S S v +S S wx =0. 5 The parallel through M is the line 3S S vw +S S u ww +S S u vvx =0. 6 The reflection of 6 in 5 is the radical axis of K and its homothetic image above. If there are two such real intersections K 3 and K 3, then the two triangles K K 2 K 3 and K K 2 K 3 clearly have equal area. These two intersections coincide if the line in onstruction 0 iii above is tangent to K. This is the case when K K 2 is a tangent to the hyperbola 4f 2 Kx, y, z 3g 3 x + y + z 2 =0, which is the image of K under the homothety hk i, 2. See Figure 2. K 3 K i K g K 2 M K Figure 2. Family of special Kiepert inscribed triangles with K, K 2 uniquely determining K 3

15 Some constructions related to the Kiepert hyperbola 357 The resulting family of special Kiepert inscribed triangles is the same family with centroid Kt and one vertex its antipode on K, given in onstruction 4. References []. Kimberling, Encyclopedia of Triangle enters, available at http//faculty.evansville.edu/ck6/encyclopedia/et.html. [2] F. M. van Lamoen and P. Yiu, The Kiepert pencil of Kiepert hyperbolas, Forum eom., [3] P. Yiu, Introduction to the eometry of the Triangle, Florida tlantic University lecture notes, 200. Paul Yiu Department of Mathematical Sciences, Florida tlantic University, oca Raton, Florida , US address