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A185210 Areas A of the triangles such that A, the sides, the inradius and the radius of the three excircles are integers.

%I A185210 #58 Feb 16 2025 08:33:13
%S A185210 6,24,30,42,48,54,60,84,96,108,120,144,150,156,168,180,192,210,216,
%T A185210 240,270,294,330,336,378,384,390,420,432,462,480,486,504,510,528,540,
%U A185210 546,576,594,600,624,630,672,714,720,726,750,756,768,810,840,864,924,930
%N A185210 Areas A of the triangles such that A, the sides, the inradius and the radius of the three excircles are integers.
%C A185210 Theorem 1: Consider a triangle whose area A, sides (a,b,c), inradius r and the radius of whose three excircles r1, r2, r3 are integers. Then the sum a^2 + b^2 + c^2 + r^2 + r1^2 + r2^2 + r3^2 is a perfect square equal to 16R^2, where R is the circumradius.
%C A185210 Proof: (r1 + r2 + r3 - r)^2 = r1^2 + r2^2 + r3^2 + r^2 + a^2 + b^2 + c^2 because: r1*r2 + r2*r3 + r3*r1 - r*r1 - r*r2 - r*r3 = (a^2 + b^2 + c^2)/2 (formula from Feuerbach - see the link). But r1 + r2 + r3 - r = 4*R (see the reference: Johnson 1929, pp. 190-191), hence the result. Remark: R is not necessarily an integer; for example, at a(1) = 6 with (a,b,c) = (3, 4, 5) we obtain r = 1, r1 = 2, r2 = 3, r3 = 6 and R = 5/2. Then 3^2 + 4^2 + 5^2 + 1^2 + 2^2 + 3^2 + 6^2 = 16*(5/2)^2 = 10^2. Nevertheless, if R is an integer, then r, r1, r2 and r3 are necessarily integers (see the following theorem). The subset of a(n) with R integer is A208984 = {24, 96, 120, 168, 216, 240, 336, 384, 432, 480, 600, ...}
%C A185210 Theorem 2: Consider a triangle whose area A, sides (a,b,c) and circumradius R are integers. Then the inradius r and the radius of the three excircles r1, r2, r3 are also integers.
%C A185210 Proof: Let s be the semiperimeter, let s*A = r1*r2*r3 be an integer, and let r*r1*r2*r3 = A^2 also be an integer => r is an integer. r1 = A/(s-a), r2 = A/(s-b), r3 = A/(s-c) => r1*r2 = s*(s-c), r1*r3=s*(s-b), r2*r3 = s*(s-a) are integers. Because r1*r2*r3 is an integer => r1, r2, r3 are integers.
%D A185210 Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.
%D A185210 Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.
%H A185210 Michel Lagneau, <a href="/A185210/b185210.txt">Table of n, a(n) for n = 1..54</a>
%H A185210 Eric W. Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Excircles.html">Excircles</a>
%H A185210 Eric W. Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Exradius.html">Exradius</a>
%H A185210 Eric W. Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Inradius.html">Inradius</a>
%F A185210 A = sqrt(s*(p-a)*(s-b)*(s-c)) with s = (a+b+c)/2 (Heron's formula);
%F A185210 the inradius is r=A/s;
%F A185210 the exradii of the excircles are r1 = 2*A/(-a+b+c), x2 = 2*A*b/(a-b+c), and x3 = 2*A*c/(a+b-c).
%e A185210 24 is in the sequence because for (a, b, c) = (6, 8, 10) => s =(6+8+10)/2 = 12; A = sqrt(12(12-6)(12-8)(12-10)) = sqrt(576) = 24; r = A/s = 2; r1 = 2*24(-6+8+10) = 4; r2 = 2*24(6-8+10) = 6; r3 = 2*24(6+8-10) = 12.
%t A185210 nn = 1000; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[area2]/s] && IntegerQ[2*Sqrt[area2]/(-a+b+c)] &&  IntegerQ[2*Sqrt[area2]/(a-b+c)] && IntegerQ[2*Sqrt[area2]/(a+b-c)], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]
%Y A185210 Cf. A120572, A188158, A208984, A210207.
%K A185210 nonn
%O A185210 1,1
%A A185210 _Michel Lagneau_, Mar 21 2012