A triangle problem

MattcAnderson · 2014-08-22, 21:59

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

EdH · 2014-08-22, 22:41

Sounds like a 3-4-5 right triangle...

d=1
lengths 3,4,5
angles 90, ~36.86, ~53.13
t=3*4/2

Batalov · 2014-08-23, 01:15

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))

MattcAnderson · 2014-08-25, 05:42

So I think the answer involves an expression in t.

Regards,
Matt

Batalov · 2014-08-25, 06:19

Quote:

Originally Posted by MattcAnderson

So I think the answer involves an expression in t.

So, you think it doesn't depend on d?

rajula · 2014-08-25, 06:57

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).

MattcAnderson · 2014-08-25, 21:53

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

2014-08-22, 21:59	#1
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 2⁵×5² 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

2014-08-23, 01:15	#3
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 3⁶×13 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+4d)/2/(b+d)) B=acos((a^2+c^2-b^2)/(2ac))=acos((b^2+2d^2)/2/(b^2-d^2)) C=acos((a^2+b^2-c^2)/(2ab))=acos((b-4d)/2/(b-d)) Last fiddled with by Batalov on 2014-08-23 at 01:24 Reason: forgot the denominator...*

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2014-08-22, 22:41	#2
EdH "Ed Hall" Dec 2009 Adirondack Mtns 11·347 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

2014-08-25, 05:42	#4
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 1100100000₂ Posts	So I think the answer involves an expression in t. Regards, Matt

2014-08-25, 21:53	#7
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 1440₈ 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 + 16t^2/3)) simplifying for an equilateral triangle, and letting d = 0, b = 2t^(1/2)/(3^(1/4)) Thanks for the good answers, all