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A120572 Smallest area of any triangle with integer sides a <= b <= c and inradius n. 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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)

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.

EXAMPLE

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.

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

CROSSREFS

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

Sequence in context: A253770 A002688 A083212 * A000056 A083170 A087081

Adjacent sequences:  A120569 A120570 A120571 * A120573 A120574 A120575

KEYWORD

nonn

AUTHOR

David W. Wilson, Jun 17 2006

STATUS

approved

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Last modified April 21 07:53 EDT 2019. Contains 322327 sequences. (Running on oeis4.)