A Distance Property of the Feuerbach Point and Its Extension

Transcription

1 Forum Geometricorum Volume 16 (016) FOUM GEOM ISSN A Distance Property of the Feuerbach Point and Its Extension Sándor Nagydobai Kiss Abstract. We prove that among the distances from the inner Feuerbach point of a triangle to the midpoints of the three sides, one is equal to the sum of the remaining two. The same is true if the inner Feuerbach point is replaced by any one of the outer Feuerbach points. 1. Introduction The famous Feuerbach theorem asserts that the nine-point circle of a triangle is tangent to the incircle and each of the excircles. The point of tangency with the incircle is the Feuerbach point ; it is in the interior of the triangle. We call it the inner Feuerbach point. The points of tangency with the excircles are exterior to the triangle, and are called the exterior Feuerbach points. In this note we give an interesting distance property of F e (Theorem 1 and Figure 1 below), and its analogues for the exterior Feuerbach points. A F e Y Z Y Z I O N B X X C Figure 1 Theorem 1. If F e is the inner Feuerbach point of triangle ABC, and X, Y, Z are the midpoints of the sides BC, CA, AB, respectively, then one of the distances F e X, F e Y, F e Z is equal of the sum of the two others. Publication Date: July 15, 016. Communicating Editor: Paul Yiu.

2 84 S. N. Kiss We make use of standard notations in triangle geometry (see [3]). Given triangle ABC, denote by a, b, c the lengths of the sides BC, CA, AB respectively, s the semiperimeter, Δ the area, and, r the circumradius and inradius respectively. In homogeneous barycentric coordinates, the inner Feuerbach point is F e ((s a)(b c) : (s b)(c a) : (s c)(a b) ). (1) We shall simplify calculations in this paper by employing the notations u : s a b + c a, v : s b c + a b, w : s c a + b c. In terms of u, v, w, F e (u(v w) : v(w u) : w(u v) ), () with coordinate sum σ e : u(v w) + v(w u) + w(u v) (v + w)(w + u)(u + v) 8uvw (3) abc 8r s 4rs 8r s 4rs( r) 4Δ ( r) 4Δ OI (4) by Euler s formula ([1, Theorem 97]), where O and I are the circumcenter and incenter of the triangle. Working with the distance formula, we also make use of S A : b + c a, S B : c + a b, S C : a + b c. Lemma. (1) v + w a, w + u b, u + v c, and u + v + w s. () The inradius and the exradii are r Δ s, r a Δ u, r b Δ v, r c Δ w. (3) uvw r s rδ. (4) vw + wu + uv r(4 + r). (5) S A us vw, S B vs wu, S C ws uv. Proof. (4) vw + wu + uv uvw u + uvw + uvw v w rδ u + rδ v + rδ w r(r a + r b + r c ) r(4 + r), since r a + r b + r c 4 + r (see [1, 98 (c)]). (5) S A b +c a (w+u) +(u+v) (v+w) u(u+v+w) vw us vw.

3 A distance property of the Feuerbach point and its extension 85 Proposition 3. F e X OI b c. Proof. We make use of the distance formula (see [3, 7.1]) for two points P (u, v, w) and Q (u,v,w ) in absolute barycentric coordinates: PQ S A (u u ) + S B (v v ) + S C (w w ). The absolute barycentric coordinates of F e are (x e,y e,z e ) 1 σ e (u(v w),v(w u),w(u v) ). Since the midpoint X of BC has absolute barycentric coordinates ( 0, 1, 1 ), and y e 1 1 ( v(w u) ) σ e σ e 1 ( u(v w) v(w u) + w(u v) ) σ e 1 (v w)(w + u)(u v); σ e z e 1 1 (v w)(w u)(u + v), σ e the distance of F e X is given by ( F e X S A x e + S B y e 1 ) ( + S C z e 1 ) (v w) 4σ e ( 4(us vw)u (v w) +(vs wu)(w + u) (u v) +(ws uv)(w u) (u + v) ) (v w) ( s(4u 3 4σe (v w) + v(w + u) (u v) + w(w u) (u + v) ) u(4uvw(v w) + w(w + u) (u v) + v(w u) (u + v) ) ). It turns out that both 4u 3 (v w) + v(w + u) (u v) + w(w u) (u + v) and 4uvw(v w) + w(w + u) (u v) + v(w u) (u + v) are equal to (w + u)(u + v)((v + w)(w + u)(u + v) 8uvw) (w + u)(u + v)σ e. Therefore, F e X (v w) (s u)(w + u)(u + v)σ e 4σ e (v + w)(w + u)(u + v) 4σ e (v w) abc 4σ e (b c) 4 OI (b c), with σ e given by (4) and abc 4Δ. This gives F e X OI b c.

4 86 S. N. Kiss Proof of Theorem 1. If Y and Z are the midpoints of CA and AB respectively, then analogous to the result of Proposition 3, F e Y OI c a and F ez a b. OI Thus, F e X, F e Y, F e Z are in the proportions of b c, c a, a b. It is clear that in the latter triad, one of the terms (the greatest) is the sum of the remaining two. The same holds for the former triad F e X, F e Y, F e Z. This completes the proof of Theorem 1. Proposition 4. Let the incircle of triangle ABC touch the sides BC, CA, AB at X, Y, Z respectively. The triangles F e XX, F e YY, F e ZZ are similar to the triangles AOI, BOI, COI respectively. Proof. It is enough to prove the similarity of triangles F e XX and AOI (see Figure 1). Since BX s b c+a b, XX a c+a b 1 b c. By Proposition 3, FeX AO XX OI b c FeX OI. It remains to show that AI b c OI also. For this, we compute the length of F e X. The absolute barycentric coordinates of X are 1 1 a (0,s c, s b) v+w (0,w,v). Now, y e w v + w see Lemma 5(a) below. Similarly, z e v(w u) w σ e v + w v(v + w)(w u) wσ e (v + w)σ e u(v w)((v + w)(w + u) 4vw), (v + w)σ e v v + w Therefore, ( F e X S A x e + S B y e where u(v w)((u + v)(v + w) 4vw) (v + w)σ e. w ) ( + S C z e v + w v ) u (v w) v + w (v + w) σe F, F S A (v + w) (v w) + S B ((v + w)(w + u) 4vw) + S C ((u + v)(v + w) 4vw). (5) We shall establish in Lemma 5(b) below that F 4vw(v + w)σ e. From this, Now, F e X u (v w) (v + w) σe 4u r s(b c) (v + w) 4rs AI r sin A uvw s 4vw(v + w)σ e 4u vw(v w) (v + w)σ e u r(b c) OI (v + w)oi (w + u)(u + v) vw u(w + u)(u + v). s

5 A distance property of the Feuerbach point and its extension 87 Therefore, F e X AI u r(b c) s (v + w)oi u(w + u)(u + v) rs(b c) (b c) 4abc OI 4 OI, and FeX AI and AOI. b c OI. The completes the proof of the similarity of triangles F exx Lemma 5. (a) v(v + w)(w u) wσ e u(v w)((v + w)(w + u) 4vw). (b) The polynomial F defined in (5) is 4vw(v + w)σ e. Proof. (a) Using (3) for σ e,wehave v(v + w)(w u) wσ e v(v + w)(w u) w((v + w)(w + u)(u + v) 8uvw) (v + w)(v(w u) w(w + u)(u + v)) + 8uvw (v + w)(u v uw u w 3uvw)+8uvw u((v + w)(uv uw w 3vw)+8vw ) u(u(v w ) (v + w)w(3v + w)+8vw ) u(u(v w ) w((v + w)(3v + w) 8vw)) u(u(v w ) w(v w)(3v w)) u(v w)(u(v + w)+w(w 3v)) u(v w)((v + w)(w + u) 4vw). (b) The polynomial F defined in (5) is F (us vw)(v + w) (v w) +(vs wu)((v + w)(w + u) 4vw) Note that the coefficient of s is +(ws uv)((u + v)(v + w) 4vw). u(v + w) (v w) + v((v + w)(w + u) 4vw) + w((u + v)(v + w) 4vw) u(v + w) (v w) + v(v + w) (w + u) 8v w(v + w)(w + u)+16v 3 w + w(u + v) (v + w) 8vw (u + v)(v + w)+16v w 3 (v + w) ( u(v + w)(v w) + v(v + w)(w + u) + w(u + v) (v + w) 8v w(w + u) 8vw (u + v)+16v w ) (v + w) ( u(v + w)(v w) + v(v + w)(w + u) + w(u + v) (v + w) 8uvw(v + w)) (v + w) ( u(v w) + v(w + u) + w(u + v) 8uvw) ) (v + w) ((v + w)(w + u)(u + v) 8uvw)) (v + w) σ e.

6 88 S. N. Kiss The sum of the terms in F without s is vw(v + w) (v w) wu((v + w)(w + u) 4vw) uv((u + v)(v + w) 4vw) vw(v + w) (v w) wu(v + w) (w + u) +8uvw (v + w)(w + u) 16uv w 3 uv(u + v) (v + w) +8uv w(u + v)(v + w) 16uv 3 w (v + w) ( vw(v + w)(v w) + wu(v + w)(w + u) + uv(u + v) (v + w) 8uvw (w + u) 8uv w(u + v)+16uv w ) (v + w) ( (v + w) ( vw(v w) + wu(w + u) + uv(u + v) ) 8uvw(u(v + w)+(v w) ) ) (v + w) ( (v + w)(w + u)(u + v)(u(v + w)+(v w) ) 8uvw(u(v + w)+(v w) ) ) (v + w)(u(v + w)+(v w) )((v + w)(w + u)(u + v) 8uvw) (v + w)(u(v + w)+(v w) )σ e. Therefore, F (v + w) σ e s (v + w)(u(v + w)+(v w) )σ e ( (v + w)σ e (v + w)(u + v + w) (u(v + w)+(v w) ) ) 4vw(v + w)σ e.. Outer Feuerbach points The nine-point circle is also tangent to each of the excircles. If the excircle (I a ) touches BC at X a, and the extensions of AC and AB at Y a, Z a respectively, then the outer Feuerbach point F a is the intersection of the nine-point circle with the segment joining I a to the nine-point center N (see Figure ). In homogeneous barycentric coordinates, F a ( s(b c) : (s c)(c + a) : (s b)(a + b) ). Note that these can be obtained from the coordinates of F e by changing (a, b, c) into ( a, b, c). Under this transformation, (u, v, w, s) becomes (s, w, v, u). In terms of u, v, w, F a ( s(v w) : w(s + v) : v(s + w) ) with coordinate sum σ a which can be obtained from σ e (v + w)(w + u)(u + v)+8uvw

7 A distance property of the Feuerbach point and its extension 89 A Z N O Y B F a X X a C Y a Z a I a Figure by the transformation (u, v, w, s) becomes (s, w, v, u). Thus, σ a (v + w)(w + u)(u + v)+8vws 4Δ( +r a ) 4Δ ( +r a) 4Δ OI a, where I a is the A-excenter of the triangle (see [1, 95]). The transformations (a, b, c) ( a, b, c) and (u, v, w, s) (s, w, v, u) also wrap X and X a, Y and Y a, Z and Z a. Therefore, we can easily translate the results in the previous section about the incircle into results on the excircles. First of all, the translation of the distance formulas: F e X OI b c F ax OI a b c ; F e Y OI c a F ay OI a (c + a); F e Z OI a b F az OI a (a + b). Since a + max(b, c) (a +min(b, c)) + b c, we easily obtain the following analogue of Theorem 1. Proposition 6. If the nine-point circle touches the A-excircle at F a, then one of F a X, F a Y, F a Z is the sum of the remaining two. The analogues of Proposition 4 also hold. From the relation F e X AI b c OI,

8 90 S. N. Kiss we obtain F a X a b c XX a F ax AI a OI a OI a AO. The similarity of triangles F a XX a and AOI a follows (see Figure ). Similar results hold for the other two excircles. eferences [1]. A. Johnson, Advanced Euclidean Geometry, 199, Dover reprint 007. [] M. J. G.Scheer, A simple vector proof of Feuerbach s theorem, Forum Geom., 11 (011) [3] P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University Lecture Notes, 001; with corrections, 013, available at S. Nagydobai Kiss: Constatin Brâncuşi Technology Lyceum, Satu Mare, omania address: d.sandor.kiss@gmail.com