The side lengths of a triangle are 15 units, 47 units and \(\frac{x}{2}\) units. How many integer values of \(x\) are possible?
+6250 \(32=(47-15)<\dfrac x 2 < (47+15) = 62 \\ 64 < x < 124 \\ x = 65, 66, \dots , 123\\ \text{which is 59 values}\)
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+130071 By the Triangle inequality, we have these possible inequalities
x/2 + 15 > 47
x/2 > 32
x > 64
47 + 15 > x/2
62 > x/2
124 > x
x < 124
x/2 + 47 > 15
x/2 > -32
x > -64
We need to take the most restictive interval that will solve all three inequalities....this is
64 < x < 124
And the number of integers in this interval is 123 - 65 + 1 = 59
