A120572 Smallest area of any triangle with integer sides a <= b <= c and inradius n.

6, 24, 48, 84, 150, 192, 294, 336, 432, 540, 726, 756, 1014, 1134, 1170, 1344, 1734, 1710, 2166, 2100, 2310, 2640, 3174, 3000, 3750, 4056, 3888, 4116, 5046, 4680, 5766, 5376, 5808, 6936, 6510, 6804, 8214, 8664, 8112, 8400, 10086, 9240, 11094, 10164

Offset: 1

Author

David W. Wilson, Jun 17 2006

Keywords

• nonn

Comments

a(n) == 0 (mod 6).

Empirically, 3*sqrt(3) < a(n)/n^2 <= 6. The lower bound is provably tight, the upper bound seems to be achieved infinitely often, e.g., for prime n >= 5.

From Michel Lagneau, Mar 02 2012: (Start)

Subset of A188158.

The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2. The radius of the incircle or inscribed circle (also known as the inradius, r) is given by r = A/s.

From n = 17, it is possible to find couples of triangles with the property: r1 > r2 and A1 < A2 where A1, A2 are the consecutive areas corresponding to the inradius r1, r2. For example, a(17) = 1734 with (a,b,c) = (51, 68, 85) and a(18) = 1710 with (a,b,c) = (57, 65, 68).

Another interesting property of this sequence is that a(n) is divisible by 6 and, except n = 3, a(n)/6 = n^2 if n prime, hence the proposition:

Among the set of triangles whose area and sides are integers and whose inradius r is a prime number other than 3, the smallest area A is given by A = 6r^2.

Example: if r = 5, the areas of the triangles are {150, 210, 270, 330, 390, 510, ...} and the smallest area of them is A = 6*5^2 = 150 because 5 is prime.

Proof: Let r be a number such that the sides of a triangle are a = 3r, b = 4r, c = 5r. Then s = (a+b+c)/2 = 6r and A = sqrt(s(s-a)(s-b)(s-c)) = sqrt(36r^4) = 6r^2 is a possible area. Is 6r^2 the smallest area? The response is no in the general case for the composite numbers.

Writing a = (m+n)/2, b = (n+l)/2, c = (l+m)/2, and using rs = A and Heron's formula for A, we find lmn = 4r^2(l+m+n). Since m, n and l have to be of the same parity for a, b and c to be integral, they must therefore be even. Setting l = 2u, m = 2v, and n = 2w, we have a = v+w, b = w+u, c = u+v, and uvw = r^2(u+v+w). Then r^2 = uvw/(u+v+w).

First case: If r = p is prime, we prove that A = 6p^2 is the smallest area of all the triangles whose inradius is p. Suppose A' < A with inradius(A') = p. The area A is the corresponding value of the triangle (u,v,w) = (1*p, 2*p, 3*p) because 6p^3/6p = p^2. However, inradius(A') = p => u'v'w'/(u'+v'+w') = p^2 => (u',v',w') = (u,v,w) and A is the smallest area.

Second case: If r = q is composite, the triangle (u,v,w) = (1*q, 2*q, 3*q) gives an area A with inradius(A) = q, but it is possible to find A' < A with inradius(A') = q; for example, if q = 10, the triangle (u,v,w) = (30, 20, 10) whose area is A = 600 gives sqrt{(30*20*10)/(30+20+10)} = sqrt(6000/60) = 10 and the triangle(u',v',w') = (24,15,15) whose area is A' = 540 gives sqrt{(24*15*15)/(24+15+15)} = sqrt(5400/54) = 10.

(End)

Examples

			a(4) = 84 because, for (a,b,c) = (13,14,15) => A = sqrt(21(21-13)(21-14)(21-15)) = 84 and r = 84/21 = 4.
		

Links

• David W. Wilson, Table of n, a(n) for n = 1..10000
• Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.

Crossrefs

See A120062 for sequences related to integer-sided triangles with integer inradius n.

Programs

Maple

T:=array(1..500):nn:=70: for n from 1 to 16 do:k:=0:ii:=0:for a from 1
to nn do: for b from a to nn  do: for c from b to nn  do: p:=(a+b+c)/2 : x:=p*(p-a)*(p-b)*(p-c): if x>0 then s:=sqrt(x) :if s=floor(s) and s/p = n then k:=k+1:T[k]:=s: else fi:fi:od:od:od: L := [seq(T[i], i=1..k)]:A:=sort(L, `<`): w:=A[1]: printf ( "%d %d \n", n, w):od: # Michel Lagneau, Mar 02 2012