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**+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

CPhill
Aug 11, 2017