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Let A(x,y,z) denote the integer area A of the triangle with integer side lengths (x,y,z). a(n) is the smallest area such that there exist exactly n distinct triangles (x_i, y_i, z_i) satisfying A(x_1, y_1, z_1) = A(x_2, y_2, z_2) = ... = A(x_n, y_n, z_n).
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%I #20 Feb 16 2025 08:33:20

%S 6,12,126,60,240,210,1080,336,1260,924,2016,2640,7980,6930,420,2520,

%T 840,3696,10080,5460,3360

%N Let A(x,y,z) denote the integer area A of the triangle with integer side lengths (x,y,z). a(n) is the smallest area such that there exist exactly n distinct triangles (x_i, y_i, z_i) satisfying A(x_1, y_1, z_1) = A(x_2, y_2, z_2) = ... = A(x_n, y_n, z_n).

%C The area A of a triangle whose sides have lengths x, y, and z is given by A188158. The area is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2. The Maple program examines all triangles having longest side <= 600.

%C Conjecture: there exists an integer corresponding to the common area of exactly n distinct triangles (x_i, y_i, z_i) for i = 1,2,...,n such that A(x_1, y_1, z_1) = A(x_2, y_2, z_2) = ... = A(x_n, y_n, z_n) for n = 1, 2, ...

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Triangle.html">Triangle</a>

%e a(6) = 210 because there exists exactly 6 triangles with the same area. We obtain:

%e (3,148,149) => A = sqrt(150*(150-3)*(150-148)*(150-149)) = 210;

%e (7,65,68) => A = sqrt(70*(70-7)*(70-65)*(70-68)) = 210;

%e (12,35,37) => A = sqrt(42*(42-12)*(42-35)*(42-37)) = 210;

%e (17,25,28) => A = sqrt(35*(35-17)*(35-25)*(35-28)) = 210;

%e (17,28,39) => A = sqrt(42*(42-17)*(42-28)*(42-39)) = 210;

%e (20,21,29) => A = sqrt(35*(35-20)*(35-21)*(35-29)) = 210.

%p with(numtheory):nn:=600:lst:={}:T:=array(1..30000):k:=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 x1:=abs(x):s:=sqrt(x1) :else fi:if s=floor(s) then lst:=lst union {s}:k:=k+1:T[k]:=s:else fi:od:od:od:k1:=nops(lst):for n from 1 to 50 do:jj:=0:for i from 1 to k1 while(jj=0) do:ii:=0:for j from 1 to k while(jj=0) do:if lst[i]=T[j] then ii:=ii+1:else fi:od: if ii = n then jj:=1:printf ( "%d %d \n",n,lst[i]):else fi:od:od:

%Y Cf. A188158, A228847.

%K nonn,hard,changed

%O 1,1

%A _Michel Lagneau_, Sep 05 2013