Does the triangle inequality apply?

Find the perimeter of the smallest triangle Δ A B C \Delta{ABC} with integer sides that are in non-constant geometric progression.


The answer is 19.

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1 solution

Efren Medallo
Jun 21, 2015

Let a a be the smallest side. Then it will follow that the other sides are a r ar and a r 2 ar^{2} for some positive number r r .

We'll have to make sure that the sum a ( 1 + r + r 2 ) a( 1 + r + r^2) is an integer. Also, we'll have to make sure that a a , a r ar and a r 2 ar^2 are all integers. Finally, we'll have to make sure that a + a r > a r 2 a + ar > ar^2 , or 1 + r > r 2 1 + r > r^2 .

obviously, 1 < r < 2 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 1 + r + r^2 can be multiplied with the least integer possible. With that we get r = 3 2 r = \frac {3}{2} .

Now, substituting r = 3 2 r = \frac {3}{2} , we have to set a a such that a a , a r ar , and a r 2 ar^2 are integers. By inspection we see that the least possible value for a a would be 4 4 .

And with that the total perimeter is 4 ( 1 + 3 2 + ( 3 2 ) 2 ) = 19. 4 (1 + \frac {3}{2} + (\frac {3}{2})^2) = 19.

Don't forget to reshare! :)

The choice of r=3/2 comes from the ratio of smallest possible 2 numbers that will satisfy our condition of 1<r<2

Shib Shankar Sikder - 5 years, 11 months ago

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