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

6, 24, 30, 42, 48, 54, 60, 84, 96, 108, 120, 144, 150, 156, 168, 180, 192, 210, 216, 240, 270, 294, 330, 336, 378, 384, 390, 420, 432, 462, 480, 486, 504, 510, 528, 540, 546, 576, 594, 600, 624, 630, 672, 714, 720, 726, 750, 756, 768, 810, 840, 864, 924, 930

Offset: 1

Author

Michel Lagneau, Mar 21 2012

Keywords

• nonn

Comments

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.

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, ...}

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.

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.

Examples

			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.
		

References

• 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.
• Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.

Links

• Michel Lagneau, Table of n, a(n) for n = 1..54
• Eric W. Weisstein's World of Mathematics, Excircles
• Eric W. Weisstein's World of Mathematics, Exradius
• Eric W. Weisstein's World of Mathematics, Inradius

Crossrefs

Cf. A120572, A188158, A208984, A210207.

Programs

Mathematica

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]

Formula

A = sqrt(s*(p-a)*(s-b)*(s-c)) with s = (a+b+c)/2 (Heron's formula);

the inradius is r=A/s;

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).