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%I #34 Sep 09 2017 20:14:58
%S 6,24,48,84,150,192,294,336,432,540,726,756,1014,1134,1170,1344,1734,
%T 1710,2166,2100,2310,2640,3174,3000,3750,4056,3888,4116,5046,4680,
%U 5766,5376,5808,6936,6510,6804,8214,8664,8112,8400,10086,9240,11094,10164
%N Smallest area of any triangle with integer sides a <= b <= c and inradius n.
%C a(n) == 0 (mod 6).
%C 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.
%C From _Michel Lagneau_, Mar 02 2012: (Start)
%C Subset of A188158.
%C 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.
%C 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).
%C 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:
%C 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.
%C 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.
%C 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.
%C 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).
%C 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.
%C 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.
%C (End)
%H David W. Wilson, <a href="/A120572/b120572.txt">Table of n, a(n) for n = 1..10000</a>
%H Mohammad K. Azarian, <a href="http://www.jstor.org/stable/25678790">Solution of problem 125: Circumradius and Inradius</a>, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.
%e 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.
%p T:=array(1..500):nn:=70: for n from 1 to 16 do:k:=0:ii:=0:for a from 1
%p 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
%Y See A120062 for sequences related to integer-sided triangles with integer inradius n.
%K nonn
%O 1,1
%A _David W. Wilson_, Jun 17 2006
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