Contribution for Puzzlet #064
From: Denis Borris [email@example.com]
smaller triangle: 15 by a by b
larger triangle : 15 by 18 by c
S = sqrt[k(k-15)(k-a)(k-b)] where k = (15+a+b)/2
L = sqrt[j(j-15)(j-18)(j-c)] where j = (15+18+c)/2 = (33+c)/2
since 9S = 7L, squaring gives:
81[k(k-15)(k-a)(k-b)] = 49[j(j-15)(j-18)(j-c)]
I said what the hell, tizz Sunday and as penance for my sins this year,
lets multiply this out:
[(15+a+b)/2] * [(15+a+b)/2 - 15] * [(15+a+b)/2 - a] * [(15+a+b)/2 - b]
= [-a^4 + 2a^2b^2 + 450a^2 - b^4 + 450b^2 - 50625] /16
[(33+c)/2] * [(33+c)/2 - 15] * [(33+c)/2 - 18)] * [(33+c)/2 - c]
= [-c^4 + 1098c^2 - 9801] / 16
So, not giving up, cancelling out the "/16" and changing the signs:
c^4 - 1098c^2 + 9801 = [81(a^4 - 2a^2b^2 - 450a^2 + b^4 - 450b^2 + 50625)]/49
well, let y = "the right side"; so c^4 - 1098c^2 + 9801 = y
and let x = c^2; so x^2 - 1098x + 9801 - y = 0
quadraticize that to get: x = 549 +- sqrt(291600 + y)
No can go no farther!
At least, looping reduced to a and b;
at a=13 and b= 14, success: x = 225 or 873;
so c = sqrt(225) = 15 or sqrt(873)= non-integer.
Recap: Sir Ellis' triangles are 13-14-15 and 15-15-18 !
Next time you put up something similar, please warn me ahead of time!
Nice one, Denis! And, as ever, correct. Looks like this one made you think a little more than usual ...