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For how many positive integers $p$ does there exist a triangle with sides of length 3p-1, 3p, and p^2 + 1?

AdminMod2  Aug 11, 2017
 #1
avatar+90088 
+1

 

According to the triangle inequality....we have the following system of equations

 

(3p - 1) + 3p >  (p^2 + 1 )  →   6p - 1 > p^2 + 1    →  p^2 - 6p  + 2   <  0    (1)


(p^2 + 1) + 3p > 3p - 1    →    p^2   > - 2      (2)

 

p^2 + 1 + 3p - 1 >   3p    →   p^2    >  0   (3)

 

The last two are true for any positive integer p, so we can ignore  them

 

Looking at the first equation and setting it to 0 , we have that

 

p^2 - 6p + 2  = 0        the solutions (roots) to this equation   are    p ≈ .35  and  p ≈ 5.6

 

And since this is an upward turning parabola, all  integers between these two values will make (1) true...

 

So.....the  integers that make  this true are  1,2,3,4 5

 

So.....there are five integer values for  p 

 

 

cool cool cool

CPhill  Aug 11, 2017
edited by CPhill  Aug 11, 2017

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