Newton's semicircle problem: The diameter is a positive real root of d^3 - (a^2 + b^2 + c^2)d - 2abc = 0. By Descartes' Sign Rule, this cubic equation has exactly one positive real root. Let A(a,b,c) = (a + b + c)/3 the arithmetic mean, R(a,b,c) = sqrt((a^2 + b^2 + c^2)/3) the root mean square, and G(a,b,c) = (abc)^(1/3) the geometric mean of the numbers a,b,c. By the arithmetic geometric mean inequality we have G <= A <= R. The cubic equation is already in reduced form x^3 + 3px + 2q = 0 with p = - (a^2 + b^2 + c^2)/3 and q = - abc. The discriminant D = p^3 + q^2 = G^6 - R^6 < 0 unless a=b=c. Now we try to find good upper and lower bounds for d. Let f(x) = x^3 - (a^2 + b^2 + c^2)x - 2abc. As f(2R) = 8R^3 - 6R^3 - 2G^3 = 2(R^3 - G^3) >= 0 and f(2R(sqrt(3/4)) = -2G^3 <= 0 we get 2 sqrt((a^2 + b^2 + c^2)/3) >= d >= 2 sqrt((a^2 + b^2 + c^2)/4). We can find other lower bounds, namely 2G which is easy and 2A which is hard. f(2G) = 8G^3 - 6R^2G - 2G^3 = 6G(G^2 - R^2) <= 0. f(2A) = 8A^3 - 6R^2A - 2G^3 = (9A^3 - 6R^2A - 3G^3) + (G^3 - A^3) <= 0. The first term is not positive according to the Lehmus inequality. Note that we could have used the theorem of Lagrange to find this upper bound. As the discriminant is negative we knew that all roots are real. The roots of the polynomial x^n + s1 x^(n-1) + s2 x^(n-2) + ... + sn are contained in the interval with endpoints - s1/n +- (n-1)/n sqrt( s1^2 - 2n/(n-1) s2 ). Plugging in n=3, s1=0, and s2 = -(a^2 + b^2 + c^2) we get x <= 2R. Sharp power mean bounds: Let M_p(a,b,c) = ((a^p + b^p + c^p)/3)^(1/p) the power mean. Upto now we found the 2M_2 >= d >= 2M_1. As M_p is an increasing function in the parameter p we can ask for the smallest p in the upper and the largest p in the lower bound. Numerical evidence indicates. Best possible upper power mean seans to be p = log(3)/log(2) = 1.5849625007211561815 with equality iff (a,b,c)=(a,0,0). Best possible upper power mean seans to be p = 2(log(3)/log(2)-1) = 1.169925001442312363 with equality iff (a,b,c)=(a,a,0). Solving the cubic equation: The positive solution of the "Casus irreducibilis" is d = 2R cos( phi/3 ) with cos(phi) = (G/R)^3. Again we see that d <= 2R. We can make a series expansion in z = (R^6 - G^6)/R^6 = -D/R^6. d = 2R (1 - 1/18 z - 35/1944 z^2 - 1001/104976 z^3 + O(z^4)). Lehmus Inequality: Let a, b, c, be the lengths of a triangle with s = (a+b+c)/2. Then we have several equivalent inequalities abc >= 8(s-a)(s-b)(s-c) abc >= (b + c - a)(c + a - b)(a + b - c) a^3 + b^3 + c^3 + 3abc >= a^2(b+c) + b^2(c+a) + c^2(a+b) 2(a^3 + b^3 + c^3) + 3abc >= (a^2 + b^2 + c^2)(a + b + c) 3abc >= a(b^2 + c^2 - a^2) + b(c^2 + a^2 - b^2) + c(a^2 + b^2 - c^2) (a + b + c)^3 <= 2(a^2 + b^2 + c^2)(a + b + c) + 9abc (a + b + c)^3 + 9abc >= 4(ab + bc + ca)(a + b + c) a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) >= 0 cos A + cos B + cos C <= 3/2 and the equality holds if and only if a=b=c. The polynomial inequalities are valid for positive a, b, c. It is a special case of Schur's inequality. Integral Configurations with d<=25: a b c d ------------------- 1 1 1 2 2 2 7 8 1 6 6 9 3 3 7 9 2 7 11 14 * 2 9 12 16 * 3 3 17 18 6 11 14 21 * 5 5 23 25 7 15 15 25 10 10 17 25 Lines marked with * have two equal sides. If a=c the cubic reduces to a quadratic equation d^2 - bd - 2a^2 = 0. The two factorisations (d - r^2)(d + 2s^2) = d^2 - (r^2 - 2s^2)d - 2(rs)^2 and (d + r^2)(d - 2s^2) = d^2 - (2s^2 - r^2)d - 2(rs)^2 give the solutions a = c = rs, b = r^2 - 2s^2, d = r^2 and a = c = rs, b = 2s^2 - r^2, d = 2s^2 If d=2b the cubic reduces to a quadratic equation a^2 + ac + c^2 = 3b^2. a b c d b<=200 ------------------- 2 7 11 14 1 13 22 26 11 19 26 38 13 31 46 62 26 37 47 74 22 43 61 86 23 49 71 98 47 61 74 122 13 67 109 134 46 73 97 146 11 79 131 158 61 91 118 182 74 91 107 182 2 97 167 194 37 103 157 206 71 109 143 218 107 127 146 254 23 133 218 266 97 133 166 266 22 139 229 278 59 151 227 302 118 157 193 314 37 163 262 326 146 169 191 338 1 181 313 362 143 193 239 386 109 199 277 398 The b sequence are cevians (products of primes == 1 (mod 6)). A parametrisation is a = 3r^2 - 6rs - s^2 b = 3r^2 + s^2 c = 3r^2 + 6rs - s^2 d = 2b Sastry (2002, example 4) gave a series of Brahmagupta quadrilaterals. These have rational diagonals and area too. a = (m^2 - n^2)(p^2 + q^2) b = ((m+n)p-(m-n)q)((m+n)q-(m-n)p) c = (m^2 + n^2)(p^2 - q^2) d = (m^2 + n^2)(p^2 + q^2) diameter e = 2pq(m^2 + n^2) diagonal f = 2mn(p^2 + q^2) diagonal d a b c f e area --------------------------------- 25 7 15 15 24 20 192 65 25 33 39 60 52 1344 65 16 25 52 63 39 1134 References: Newton's semicircle problem Heinrich Dörrie; Kubische und biquadatische Gleichungen, Leibniz Verl., München 1948 - section 28: Planimetrische Aufgaben 5. Newtons Halbkreisaufgabe: Den Durchmesser zu ermitteln, in den sich drei gegebene Strecken sukzessiv als Sehnen eintragen lassen. Nick Hobson; 63. Cyclic hexagon, http://www.qbyte.org/puzzles/puzzle07.html (Problem) http://www.qbyte.org/puzzles/p063s.html (Solution) References: inequalities Coxeter, H.S.M.; The Lehmus Inequality. Aequationes Math. 28(1985) 4-12. Zbl 0554.51010 This is a survey on the Lehmus inequality. The main facts reported are: The cubic polynomial $\Gamma =abc-(b+c-a)(c+a-b)(a+b-c)$ has an interesting history. D. C. L. Lehmus proved in 1820 that the inequality $\Gamma >0$ holds whenever a, b, c are positive and not all equal. Other proofs were devised by G. Peano in 1902, A. Padoa in 1925, G. N. Watson in 1953, and O. Bottema in 1982. In 1746, W. Chapple showed that the distance OI between the circumcentre and the incentre of a triangle is given by $OI\sp2=R(R-2r)$, where R and r are the circumradius and inradius. In 1870, R. Baltzer remarked that the consequent inequality $R>2r$ (for a triangle whose sides a, b, c are not all equal) is equivalent to $\Gamma >0.$ \par In 1891, E. W. Hobson expressed the difference R-2r as 2NI, where N is the centre of Feuerbach's nine-point circle, thus providing a simple proof that this circle is tangent to the incircle. In 1967, L. Bankoff used the inequality $R>2r$ to prove that the angle GIH (where G and H are the barycentre and orthocentre) is always obtuse. Finally, J. C. Linders and J. H. H. Chalk have contributed new proofs that, in terms of barycentric coordinates a, b, c the cubic curve given by the parametric equations $a=\alpha (\beta -\gamma)\sp 2$, $b=\beta (\gamma -\alpha)\sp 2$, $c=\gamma (\alpha -\beta)\sp 2$, $\alpha +\beta +\gamma =0$ has the explicit equation $\Gamma =0$. Hojoo Lee; Topics in Inqualities, 2005-10-30 (Version 0.5) - chapter 2.1: Euler's Theorem and the Ravi Substitution - theorem 2, theorem 3 - chapter 3.2: Schur's Theorem - theorem 10, remark 2 K. Wu; The Rearrangement Inquality, - example 3 (IMO 64) A. Padoa; Period. Mat. (4)5 (1925), 80-85 Arthur Engel; Problem-solving strategies, Problem Books in Mathematics. New York, NY: Springer. x, 403 p. (1998) ISBN 0-387-98219-1/hbk Zbl 887.00002 - chapter 7: Inequalities - example E10, problem 43 Eckard Specht; geometria - scientiae atlantis. 300+ Aufgaben zur Geometrie und zu Ungleichungen insbesondere zur Vorbereitung auf Mathematik-Olympiaden, Otto-von-Guericke-Universität Magdeburg, 2001, ISBN 3-929757-39-7 - Kapitel U.2: Fundamentale Ungleichungen - problem U.25: Schursche Ungleichung - Kapitel G.1: Ungleichungen im Dreieck - problem G.11: Lehmus Ungleichung - problem G.13: IMO 1964 Eckard Specht; 470+ Mathematik-Aufgaben zum Training in Vorbereitung auf Olympiaden und Wettbewerbe http://www.math4u.de/ (online version) - Kapitel U.2: Fundamentale Ungleichungen - problem U.25: Schursche Ungleichung - Kapitel G.1: Ungleichungen im Dreieck - problem G.11: Lehmus Ungleichung - problem G.13: IMO 1964 References: cubic equation Heinrich Dörrie; Kubische und biquadatische Gleichungen, Leibniz Verl., München 1948 L. Gegenbauer, Über den sogenannten Casus irreducibilis der Cardanischen Formel, Monatsh. Math. Phys. 4 (1893) 155-158. W. D. Lambert, A generalized trigonometric solution of the cubic equation, American Mathematical Monthly 13 (1906) 73-76. A. Loewy, Über algebraische Gleichungen mit reellen Wurzeln und den sog. Casus irreducibilis bei kubischen Gleichungen, Math. Z. 11 (1921) 108-114. References: Heron triangles, Brahmagupta quadrilaterals K.R.S. Sastry; Brahmagupta Quadrilaterals, Forum Geometricorum 2 (2002) 167-173 http://forumgeom.fau.edu/FG2002volume2/FG200221.pdf List of a, b, c <= 100. a b c d ----------------- 1 1 1 2 1 6 6 9 1 13 22 26 1 22 50 55 1 33 93 99 1 35 35 50 2 2 7 8 2 7 11 14 2 9 12 16 3 3 7 9 3 3 17 18 3 14 25 30 3 17 47 51 3 26 66 72 4 4 31 32 4 62 91 112 5 5 23 25 5 5 49 50 5 33 65 75 6 6 71 72 6 11 14 21 6 25 63 70 6 39 94 104 7 7 47 49 7 7 97 98 7 15 15 25 7 18 58 63 7 19 25 35 7 19 51 57 7 20 20 32 7 58 92 112 7 88 88 128 8 17 22 32 9 9 79 81 9 35 79 90 10 10 17 25 10 21 45 54 10 55 62 88 11 17 21 33 11 19 26 38 11 24 81 88 11 25 43 55 11 38 74 88 11 39 46 66 12 12 23 32 12 19 33 44 13 31 46 62 13 33 63 77 13 37 77 91 13 40 68 85 13 43 57 78 14 14 41 49 14 23 91 98 14 27 66 77 15 15 41 50 15 34 99 110 16 25 52 65 17 27 69 81 17 28 28 49 17 28 53 68 17 38 87 102 17 63 63 98 18 18 73 81 19 44 49 76 19 83 91 133 21 21 31 49 21 21 89 98 21 39 71 91 22 43 61 86 23 35 55 77 23 42 42 72 23 43 68 92 23 49 71 98 23 77 77 121 24 87 100 145 25 33 39 65 25 38 74 95 25 51 77 105 25 59 77 110 26 37 47 74 28 41 50 80 28 91 96 147 29 36 69 92 30 30 47 72 31 45 45 81 33 38 58 87 33 65 75 117 34 43 50 85 35 35 73 98 35 42 57 90 35 44 100 125 36 36 49 81 36 40 51 85 41 69 73 123 41 99 99 162 42 57 68 112 44 44 89 121 46 73 97 146 47 61 74 122 47 72 72 128 49 66 66 121 52 63 98 144 52 73 88 143 55 55 71 121 56 56 79 128 56 67 76 133 69 81 91 161 71 91 91 169 78 78 97 169 -- mailto:Sillke@mathematik.uni-bielefeld.de http://www.mathematik.uni-bielefeld.de/~sillke/