Find the perimeter of the smallest triangle with integer sides that are in non-constant geometric progression.
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Let a be the smallest side. Then it will follow that the other sides are a r and a r 2 for some positive number r .
We'll have to make sure that the sum a ( 1 + r + r 2 ) is an integer. Also, we'll have to make sure that a , a r and a r 2 are all integers. Finally, we'll have to make sure that a + a r > a r 2 , or 1 + r > r 2 .
obviously, 1 < r < 2 , as the inequality above fails to hold otherwise. So we select the rational number between with the lowest possible denominator, so that 1 + r + r 2 can be multiplied with the least integer possible. With that we get r = 2 3 .
Now, substituting r = 2 3 , we have to set a such that a , a r , and a r 2 are integers. By inspection we see that the least possible value for a would be 4 .
And with that the total perimeter is 4 ( 1 + 2 3 + ( 2 3 ) 2 ) = 1 9 .
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