Quadratic Inequality

[Chris_]

Hey. I got a math competition back (I'm allowed to talk about it and I'm not cheating) and I'm stuck on a question.

Find all pairs of natural numbers (a,b) with a<b having the property that there exists a right triangle with legs of length a and b whose hypotenuse has length 1/3(ab)-a-b.

I put the numbers into a Pythagorean Theorem and reduced it to b=(-18+6a)/(a-6). I then substituted it into a<b. Solving that inequality, I get a^2-12a+18<0. I can find the roots of the equation, but where do I go from there to get my pairs of a and b?

 

[JeffM]

Hey. I got a math competition back (I'm allowed to talk about it and I'm not cheating) and I'm stuck on a question.

Find all pairs of natural numbers (a,b) with a<b having the property that there exists a right triangle with legs of length a and b whose hypotenuse has length 1/3(ab)-a-b.

I put the numbers into a Pythagorean Theorem and reduced it to b=(-18+6a)/(a-6). I then substituted it into a<b. Solving that inequality, I get a^2-12a+18<0. You made a technical error here. What if a < 6?

I can find the roots of the equation, but where do I go from there to get my pairs of a and b?

\(\displaystyle a > 0 < 0\) because a and b are lengths of the legs of a triangle. And \(\displaystyle a < b\) is given as is \(\displaystyle a,\ b \in \mathbb N.\)

Also given is: \(\displaystyle a^2+ b^2 = \left(\dfrac{ab}{3} - \{a + b\}\right)^2 = \dfrac{a^2b^2}{9} - 2 * \dfrac{ab}{3} * (a + b) + (a + b)^2 \implies\)

\(\displaystyle 9a^2 + 9b^2 = a^2b^2 - 6ab(a + b) + 9a^2 + 18ab + 9b^2 \implies ab(ab + 18 - 6a - 6b) = 0 \implies ab - 6b = 6a - 18 \implies b = \dfrac{6a - 18}{a - 6}.\)

So far so good.

Consequently, \(\displaystyle a \ne 6.\)

\(\displaystyle And\ a < \dfrac{6a - 18}{a - 6}.\) You are good to here, but I'd proceed differently.

\(\displaystyle b = \dfrac{6a - 18}{a - 6} = \dfrac{6a - 36 + 36 - 18}{a - 6} = \dfrac{6(a - 6) + 18}{a - 6} = 6 + \dfrac{18}{a - 6}.\)

Thus (a - 6) must divide 18 evenly, which entails that (a - 6) = - 18, - 9, - 6, - 3, - 2, - 1, 1, 2, 3, 6, 9, 18.
But a must be positive which entails that a = 3, 4, 5, 7, 8, 9, 12, 15, or 24.
If a = 3, b = 0 < a. No
If a = 4, b = - 3 < 0 < a. No
If a = 5, b = -12 < 0 < a. No
If a = 7, b = 24. OK
And so on.