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

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Default 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
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Old 2014-08-22, 22:41   #2
EdH
 
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"Ed Hall"
Dec 2009
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Sounds like a 3-4-5 right triangle...

d=1
lengths 3,4,5
angles 90, ~36.86, ~53.13
t=3*4/2
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Old 2014-08-23, 01:15   #3
Batalov
 
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"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

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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...
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Old 2014-08-25, 05:42   #4
MattcAnderson
 
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"Matthew Anderson"
Dec 2010
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Lightbulb

So I think the answer involves an expression in t.

Regards,
Matt
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Old 2014-08-25, 06:19   #5
Batalov
 
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Phi(4,2^7658614+1)/2

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Quote:
Originally Posted by MattcAnderson View Post
So I think the answer involves an expression in t.
So, you think it doesn't depend on d?
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Old 2014-08-25, 06:57   #6
rajula
 
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"Tapio Rajala"
Feb 2010
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Quote:
Originally Posted by Batalov View Post
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).
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Old 2014-08-25, 21:53   #7
MattcAnderson
 
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"Matthew Anderson"
Dec 2010
Oregon, USA

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Talking

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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