Triangle Identities Via Elimination Theory

Triangle Identities via Elimination Theory

Dam Van Nhi and Luu Ba Thang

Abstract The aim of this paper is to present some new identities via resultants and elimination theory, by presenting the sidelengths, the lengths of the altitudes and the exradii of a triangle as roots of cubic polynomials.

1 Introduction

Finding identities between elements of a triangle represents a classical topic in elementary geometry. In Recent Advances in Geometric Inequalities [3], Mitrinovic et al presents the sidelengths, the lengths of the altitudes and the various radii as roots of cubic polynomials, and collects many interesting equalities and inequalities between these quantities. As far as we know, almost of them are built using only elementary algebra and geometry. In this paper, we use transformations of rational functions and resultants to build similar nice equalities and inequalities might be diﬃcult to prove with diﬀerent methods.

2 Main results

To ﬁx notations, suppose we are given a triangle ∆ABC with sidelengh a, b, c. Denote the radius of the circumcircle by R, the radius of incircle by r, the area of ∆ABC by S, the semiperimeter as p, the radii of the excircles as r1, r2, r3, and the altitudes from sides a, b and c, respectively, as ha, hb and hc. We have the results from [3, Chapter 1]. Theorem 1. [3] Using the above notations, (i) a, b, c are the roots of the cubic polynomial

x3 − 2px2 + (p2 + r2 + 4Rr)x − 4Rrp. (1)

(ii) ha, hb, hc are the roots of the cubic polynomial S2 + 4Rr3 + r4 2S2 2S2 x3 − x2 + x − . (2) 2Rr2 Rr R

(iii) r1, r2, r3 are the roots of the cubic polynomial x3 − (4R + r)x2 + p2x − p2r. (3)

Next, we present how to use transformations of rational functions to build new geometric equalities ax + b and inequalities. The transformation that we use here is given by y = , a, b, c, d ∈ . cx + d R Proposition 2. Let

3r1 − 2r 3r2 − 2r 3r3 − 2r T1 = + + 2r1 + r 2r2 + r 2r3 + r 3r1 − 2r 3r2 − 2r 3r2 − 2r 3r3 − 2r 3r3 − 2r 3r1 − 2r T2 = + + 2r1 + r 2r2 + r 2r2 + r 2r3 + r 2r3 + r 2r1 + r 3r1 − 2r 3r2 − 2r 3r3 − 2r T3 = . 2r1 + r 2r2 + r 2r3 + r

1 We have 2 2 3r1 − 2r 3r2 − 2r 3r3 − 2r 100r + 9p T3 = > 2 2 . 2r1 + r 2r2 + r 2r3 + r 19r + 12p

3ri − 2r 3 r(yi + 2) Proof. Let yi = , i = 1, 2, 3 we have yi 6= and ri = . Since r1, r2, r3 are the roots of 2x + r 2 3 − 2yi the polynomial (3), if we substitute ri into the polynomial (3), we obtain

2 2 3 2 2 2 2 2 2 2 (3r + 8Rr + 12p )yi + (11r + 20Rr − 40p )yi + (8r − 16Rr + 39p )yi − (4r + 48Rr + 9p ) = 0; therefore y1, y2, y3 are the roots of (3r2 + 8Rr + 12p2)y3 + (11r2 + 20Rr − 40p2)y2 + (8r2 − 16Rr + 39p2)y − (4r2 + 48Rr + 9p2).

By Viete’s formula, we have 40p2 − 20Rr − 11r2 T = y + y + y = 1 1 2 3 3r2 + 8Rr + 12p2 8r2 − 16Rr + 39p2 T = y y + y y + y y = 2 1 2 2 3 3 1 3r2 + 8Rr + 12p2 4r2 + 48Rr + 9p2 T = y y y = . 3 1 2 3 3r2 + 8Rr + 12p2 4r2 + 48rt + 9p2 Let f(t) = then f is an increasing function over [2r, +∞]. Therefore, from R ≥ 2r 3r2 + 8rt + 12p2 we have T3 ≥ f(2r) i.e 2 2 3r1 − 2r 3r2 − 2r 3r3 − 2r 100r + 9p > 2 2 . 2r1 + r 2r2 + r 2r3 + r 19r + 12p

Proposition 3. Using the above notations, let

2r1 − r 2r2 − r 2r3 − r T1 = + + r1 + r r2 + r r3 + r 2r1 − r 2r2 − r 2r2 − r 2r3 − r 2r3 − r 2r1 − r T2 = + + r1 + r r2 + r r2 + r r3 + r r3 + r r1 + r 2r1 − r 2r2 − r 2r3 − r T3 = . r1 + r r2 + r r3 + r then we have an equality 4T3 + T2 − 2T1 = 5.

2ri − r r(yi + 1) Proof. Let yi = , i = 1, 2, 3 we have yi 6= 2 and x = . Since r1, r2, r3 are the roots of ri + r 2 − yi the polynomial (3), if we substitute ri into the polynomial (3), we obtain

2 2 3 2 2 2 2 2 2 2(r + 2Rr + p )yi − 3(−r + 3p )yi − 12(Rr − p )yi − (r + 8Rr + 4p ) = 0

Therefore, y1, y2, y3 are the roots of the polynomial 2(r2 + 2Rr + p2)y3 − 3(−r2 + 3p2)y2 − 12(Rr − p2)y − (r2 + 8Rr + 4p2).

Applying the Viete’s formulas for this polynomial, we obtain 4T3 + T2 − 2T1 = 5. Elimination theory [1] is one of the most eﬀective method to solve polynomial equations, so if we combine this method and Theorem 1, we can obtain some nice equalities. Consequently, we give a brief overview of the resultant of two polynomials. Given a ﬁeld K ⊆ C and two polynomials f, g ∈ K[x] of positive degree

m m−1 f = a0x + a1x + ··· + am, a0 6= 0, m > 0 n n−1 g = b0x + b1x + ··· + bn, b0 6= 0, n > 0.

2 We build the square matrix of size (m + n) × (m + n) associated to f, g (called Sylvester matrix) as follows:  a a ··· a   0 1 m     a0 a1 ··· am   . . .   ......     a a ··· a  S(f, g) := 0 1 m (4) b0 b1 ··· bn    b b ··· b   0 1 n   ......   . . .     b0 b1 ··· bn  where n first rows only depend on the coefficients of f, m last rows depend on the coefficients of g and the blank spaces are filled with 0. Then the resultant of f and g, denoted Res(f, g), is the determinant of Sylvester matrix S(f, g). If we want to emphasize the dependence on x, we write Res(f, g, x) instead of Res(f, g). Two important properties of resultants [2, Chapter 3] are

• (Common factor) Res(f, g) = 0 if and only if f and g have a non trivial common factor in K[x].

• (Elimination) There are polynomials U, V ∈ K[x] such that Uf +V g = Res(f, g). The coeﬃcients of U, V are integer polynomial in the coeﬃcients of f and g.

Proposition 4. Using the above notations,

(i) (a2 + 2Rr)(b2 + 2Rr)(c2 + 2Rr) = 2Rr(ab + bc + ca − 2Rr)2.

(ii) (a2 − 2Rr)(b2 − 2Rr)(c2 − 2Rr) = 4a2b2c2 − 2Rr(ab + bc + ca + 2Rr)2. (r + r + r + r)2 a2 + p2 + 4Rr + r2 b2 + p2 + 4Rr + r2 c2 + p2 + 4Rr + r2 (iii) 1 2 3 = . S S S S p(a2 + p2 + 4Rr + r2)(b2 + p2 + 4Rr + r2)(c2 + p2 + 4Rr + r2) (iv) S = . r1 + r2 + r3 + r r r r (v) (r2 + 2Rr)(r2 + 2Rr)(r2 + 2Rr) = r r r − 2Rr(r + r + r )2 +2Rr2Rr − 1 2 3 2. 1 2 3 1 2 3 1 2 3 r r2 r2 r2 h(2R + r)2 i (vi) 1 − 1 2 − 1 3 − 1 = 4 − 1 p2 p2 p2 p2 Proof. (i) Let f := x3 − 2px2 + (p2 + 4Rr + r2)x − 4Rrp = 0 and g := x2 + 2Rr − y. By elimination property, two polynomials f(x), g(x) have a common root if and only if resultant Res(f, g; x) = 0. So,

y3 − (4p2 + 2Rr − 2T )y2 + (T 2 − 2Rr)y − 2RrT 2 = 0 where T = p2 + 2Rr + r2. (5) 2 2 2 Let y1 = a + 2Rr, y2 = b + 2Rr and y3 = c + 2Rr. Because a, b and c are the roots of f(x), y1, y2, y3 are the roots of the equation (5). Using Viete’s formulas, we have

2 2 y1y2y3 = 2RrT = 2Rr(ab + bc + ca − 2Rr) .

(ii) Let f := x3 − 2px2 + (p2 + 4Rr + r2)x − 4Rrp = 0 and g := x2 − 2Rr − y then

y3+(2Rr+2T −4p2)y2+(T 2+4RrT −32Rrp2)y+2RrT 2−64R2r2p2 = 0 where T = p2+6Rr+r2. (6)

2 2 2 Let y1 = a − 2Rr, y2 = b − 2Rr and y3 = c − 2Rr, then y1, y2, y3 are the roots of the equation (6). Using Viete’s formulas, we have

2 2 2 2 y1y2y3 = 4a b c − 2Rr(ab + bc + ca + 2Rr) .

3 (iii) Let f(x) = x3 − 2px2 + (p2 + 4Rr + r2)x − 4Rrp and g(x) = x2 + p2 + 4Rr + r2 − y, we have

y3 − (p2 + 4Rr + r2)y2 − (2py − 4Rrp − 2r2p)2 = 0. (7)

2 2 2 2 2 2 2 2 2 Let y1 = a + p + 4Rr + r , y2 = b + p + 4Rr + r and y3 = c + p + 4Rr + r , then y1, y2, y3 are 2 2 2 the roots of the equation (7). Using Viete’s formulas, we have y1y2y3 = (4R + 2r) r p . Hence,

(a2 + p2 + 4Rr + r2)(b2 + p2 + 4Rr + r2)(c2 + p2 + 4Rr + r2) = (4Rr + 2r2)2p2.

Otherwise, r1 + r2 + r3 = 4R + r, we have

(r + r + r + r)2 a2 + p2 + 4Rr + r2 b2 + p2 + 4Rr + r2 c2 + p2 + 4Rr + r2 1 2 3 = . S S S S (iv) This equality is deduced from (iii). (v) Let f := x3 − (4R + r)x2 + p2x − p2r and g := x2 + 2Rr − y. Similar to (i), we have

2 2 2 2 (r1 + 2Rr)(r2 + 2Rr)(r3 + 2Rr) = r1r2r3 − 2Rr(r1 + r2 + r3) r r r +2Rr2Rr − 1 2 3 2. r (vi) Similarly to (i), let f := x3 − (4R + r)x2 + p2x − p2r and g := x2 − p2 − y; the conclusion follows.

If we combine them with the well-known inequality R ≥ 2r [3], we can yield some interesting inequalities. In particular,

Corollary 5. Using the above notations,

2 2 2 2 2 (i) (a + 2Rr)(b + 2Rr)(c + 2Rr) 6 R (ab + bc + ca − 2Rr) . (ii) (a2 + 2Rr)(b2 + 2Rr)(c2 + 2Rr) ≤ 2R6.

2 2 2 2 2 2 2 2 2 (iii) (a − 2Rr)(b − 2Rr)(c − 2Rr) > 4a b c − R (ab + bc + ca + R ) . 2 2 2 2 2 2 2 2 2 (iv) (a − 2Rr)(b − 2Rr)(c − 2Rr) 6 4a b c − 4r (ab + bc + ca + 4r ) . Proof. (i) By Proposition 4(i) and R ≥ 2r, we are done. (ii) Let T = (a2+2Rr)(b2+2Rr)(c2+2Rr), by Proposition 4(i) we have (a2+2Rr)(b2+2Rr)(c2+2Rr) = 4p2 2Rr(ab + bc + ca − 2Rr)2. Combining with ab + bc + ca ≤ and 4p2 ≤ 27R2, we obtain 3

2 2 3 2 T ≤ 2Rr(9R − 2Rr) ⇒ T 6 R .2r(9R − 2r) .

2 0 Consider the function f(x) = x(9R − x) where 0 < x 6 R. We have f (x) > 0 with each x ∈ (0,R], deduce that the function f is the increasing function in (0,R]. Therefore f(x) ≤ f(R) = 64R3 with each x ∈ (0,R]. Notice that R ≥ 2r, so T ≤ 64R6 and the equality holds if and only if 4ABC is equilateral. By Proposition 4(ii) and R ≥ 2r, we have the proof of (iii) and (iv).

Corollary 6. Suppose given a convex quadrilateral ABCD of AB = a, BC = b, CD = c, DA = d inscribed the circle of center O, radius R. Denoted r1, r2, r3, r4 as the radius of incircle the triangles 26R6 4ABC, 4BCD, 4CDA and 4DAB, respectively. Let T = √ then ac + bd + 4 r1r2r3r4 q 2 2 2 2 T ≥ (ab + 4r1)(bc + 4r2)(cd + 4r3)(da + 4r4).

4 Proof. Let AC = x, BD = y, by Proposition 4 we have

2 2 2 6 (a + 2Rr1)(b + 2Rr1)(x + 2Rr1) 6 64R .

2 2 2 Otherwise, we have (a + 2Rr1)(b + 2Rr1) ≥ (ab + 2Rr1) , so we deduce that

2 2 6 (ab + 2Rr1) (x + 2Rr1) ≤ 64R .

Similar, we obtain

2 2 6 (bc + 2Rr2) (y + 2Rr2) ≤ 64R 2 2 6 (cd + 2Rr3) (x + 2Rr3) ≤ 64R 2 2 6 (da + 2Rr4) (y + 2Rr4) ≤ 64R

Multiplying, we obtain

2 2 2 2 2 24 24 (ab+2Rr1)(bc+2Rr2)(cd+2Rr3)(da+2Rr4)] (x +2Rr1)(x +2Rr3)(y +2Rr2)(y +2Rr4) ≤ 2 R . (8) Consequently,

2 2 2 √ 2 (x + 2Rr1)(x + 2Rr3) > (x + 2R r1r3) 2 2 2 √ 2 (y + 2Rr2)(y + 2Rr4) > (y + 2R r2r4) and √ √ √ 2 2 4 2 (x + 2R r1r3)(y + 2R r2r4) > (xy + 2R r1r2r3r4) .

√ 4 2 2 2 2 2 ⇒ (xy + 2R r1r2r3r4) ≤ (x + 2Rr1)(x + 2Rr3)(y + 2Rr2)(y + 2Rr4) (9) From (8) and (9), it follows that √ 4 12 12 (ab + 2Rr1)(bc + 2Rr2)(cd + 2Rr3)(da + 2Rr4)(xy + 2R r1r2r3r4) ≤ 2 R . (10)

From 2ri ≤ R, i = 1, 2, 3, 4 and by Ptolemy’s theorem, we furthermore have that √ √ √ 4 4 xy + 2R r1r2r3r4 = ac + bd + 2R r1r2r3r4 ≥ ac + bd + 4 r1r2r3r4.

Therefore, we deduce that

12 12 12 12 2 2 R 2 R T = √ 2 > √ 2 (ac + bd + 4 r1r2r3r4) (ac + bd + 2R 4 r1r2r3r4) > (ab + 2Rr1)(bc + 2Rr2)(cd + 2Rr3)(da + 2Rr4).

Using again the inequality 2ri ≤ R, i = 1, 2, 3, 4, we obtain q 2 2 2 2 T > (ab + 4r1)(bc + 4r2)(cd + 4r3)(da + 4r4).

References

[1] D.A.Cox, J. B. Little, and D. O’Shea. Using Algebraic Geometry. Volume 185 of Graduate Texts in Mathematics. Springer-Verlag, NY, 499 pages, 1998.

[2] D. A. Cox, J. B. Little and D. O’Shea, Ideals, Varieties, and Algorithms 2nd edition, Springer- Verlag, New York, 1997.

[3] D.S. Mitrinovic, J.E. Pecaric and V. Volenec, Recent Advances in Geometric Inequalities, Acad. Publ., Dordrecht, Boston, London 1989.

5 Luu Ba Thang Department of Mathematics and Informatics, Hanoi National University of Education 136 Xuan Thuy, Cau Giay, Hanoi, Vietnam [email protected] Dam Van Nhi School for gifted student, Hanoi National University of Education 136 Xuan Thuy, Cau Giay, Hanoi, Vietnam