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 2014-08-22, 21:59 #1 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 967 Posts A triangle problem HI Math People, This problem comes from Tarence Tao's book, Solving Mathematical Problems. It is problem 1.1. A triangle has its lengths in an arithmetic progression, with difference d. The area of the triangle is t. Find the lengths and angles of the triangle. Best of luck. Regards, Matt
 2014-08-22, 22:41 #2 EdH     "Ed Hall" Dec 2009 Adirondack Mtns 24×257 Posts Sounds like a 3-4-5 right triangle... d=1 lengths 3,4,5 angles 90, ~36.86, ~53.13 t=3*4/2
 2014-08-23, 01:15 #3 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·3·7·229 Posts All you need is Heron's formula. [ TEX ] b=\sqrt{2(d^2+\sqrt{d^4+4/3 t^2})}[/ TEX ] a=b-d c=b+d A=acos((b^2+c^2-a^2)/(2bc))=acos((b+4*d)/2/(b+d)) B=acos((a^2+c^2-b^2)/(2ac))=acos((b^2+2*d^2)/2/(b^2-d^2)) C=acos((a^2+b^2-c^2)/(2ab))=acos((b-4*d)/2/(b-d)) Last fiddled with by Batalov on 2014-08-23 at 01:24 Reason: forgot the denominator...
 2014-08-25, 05:42 #4 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 96710 Posts So I think the answer involves an expression in t. Regards, Matt
2014-08-25, 06:19   #5
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

2·3·7·229 Posts

Quote:
 Originally Posted by MattcAnderson So I think the answer involves an expression in t.
So, you think it doesn't depend on d?

2014-08-25, 06:57   #6
rajula

"Tapio Rajala"
Feb 2010
Finland

13B16 Posts

Quote:
 Originally Posted by Batalov So, you think it doesn't depend on d?
I don't think he thinks that. (at least his comment does not imply it.)
EdH assumed the side-lengths to be integer and did not take t as an input. I understood the comment to be on this. For even with integer lengths the solution is not unique.

Batalov's solution seems correct (or at least looks like a correct one, I did not do the math).

 2014-08-25, 21:53 #7 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 967 Posts ok, to post the solution from the book defining the side lengths as b-d, b, and b+d then, applying Heron's formula gives - t^2 = 3b/2 * (3b/2 - b + d) * ( 3b/2 - b) * (3b/2 - b -d) and simplifying leads to b = sqrt(2d^2 + sqrt(4d^4 + 16*t^2/3)) simplifying for an equilateral triangle, and letting d = 0, b = 2*t^(1/2)/(3^(1/4)) Thanks for the good answers, all

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