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%I #48 Jul 27 2023 17:30:13
%S 4,6,9,9,12,16,16,20,25,25,30,36,25,35,49,25,40,64,36,42,49,49,56,64,
%T 49,63,81,49,70,100,49,77,121,64,72,81,64,88,121,81,90,100,81,99,121,
%U 81,117,169,81,126,196,100,110,121,100,130,169,121,132,144,121,143,169
%N Primitive triples for integer-sided triangles whose sides a < b < c form a geometric progression.
%C These triangles are called "geometric triangles" in Project Euler problem 370 (see link).
%C The triples are displayed in increasing lexicographic order (a, b, c).
%C Equivalently: triples of integer-sided triangles such that b^2 = a*c with a < c and gcd(a, c) = 1.
%C When a < b < c are in geometric progression with b = a*q, c = b*q, q is the constant, then 1 < q < (1+sqrt(5))/2 = phi = A001622 = 1.6180... (this bound is used in Maple code).
%C For each triple (a, b, c), there exists (r, s), 0 < r < s such that a = r^2, b = r*s, c = s^2, q = s/r.
%C Angle C < 90 degrees if 1 < q < sqrt(phi) and angle C > 90 degrees if sqrt(phi) < q < phi with sqrt(phi) = A139339 = 1.2720...
%C For k >= 2, each triple (a, b, c) of the form (k^2, k*(k+1), (k+1)^2) is (A008133(3k+1), A008133(3k+2), A008133(3k+3)).
%C Three geometrical properties about these triangles:
%C 1) The sinus satisfy sin^2(B) = sin(A) * sin(C) with sin(A) < sin(B) < sin(C) that form a geometric progression.
%C 2) The heights satisfy h_b^2 = h_a * h_c with h_c < h_b < h_a that form a geometric progression.
%C 3) b^2 = 2 * R * h_b, with R = circumradius of the triangle ABC.
%H David A. Corneth, <a href="/A339856/b339856.txt">Table of n, a(n) for n = 1..10002</a>
%H Project Euler, <a href="https://projecteuler.net/problem=370">Problem 370: Geometric Triangles</a>.
%e The smallest such triangle is (4, 6, 9) with 4*9 = 6^2.
%e There exist four triangles with small side = 49 corresponding to triples (49, 56, 64), (49, 63, 81), (49, 70, 100) and (49, 77, 121).
%e The table begins:
%e 4, 6, 9;
%e 9, 12, 16;
%e 16, 20, 25;
%e 25, 30, 36;
%e 25, 35, 49;
%e 25, 40, 64;
%e 36, 42, 49;
%e ...
%p for a from 1 to 300 do
%p for b from a+1 to floor((1+sqrt(5))/2 * a) do
%p for c from b+1 to floor((1+sqrt(5))/2 * b) do
%p k:=a*c;
%p if k=b^2 and igcd(a,b,c)=1 then print(a,b,c); end if;
%p end do;
%p end do;
%p end do;
%o (PARI) lista(nn) = {my(phi = (1+sqrt(5))/2); for (a=1, nn, for (b=a+1, floor(a*phi), for (c=b+1, floor(b*phi), if ((a*c == b^2) && (gcd([a,b,c])==1), print([a,b,c])););););} \\ _Michel Marcus_, Dec 25 2020
%o (PARI) upto(n) = my(res=List(), phi = (sqrt(5)+1) / 2); for(i = 2, sqrtint(n), for(j = i+1, (i*phi)\1, if(gcd(i, j)==1, listput(res, [i^2, i*j, j^2])))); concat(Vec(res)) \\ _David A. Corneth_, Dec 25 2020
%Y Cf. A339857 (smallest side), A339858 (middle side), A339859 (largest side), A339860 (perimeter).
%Y Cf. A336755 (similar for sides in arithmetic progression).
%Y Cf. A335893 (similar for angles in arithmetic progression).
%Y Cf. A001622 (phi), A139339 (sqrt(phi)), A008133.
%K nonn,easy,tabf
%O 1,1
%A _Bernard Schott_, Dec 19 2020
%E Data corrected by _David A. Corneth_, Dec 25 2020
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