Let $5a+12b$ and $12a+5b$ be the side lengths of a right-angled triangle and $13a+kb$ be the hypotenuse, where $a$ , $b$ and $k$ are positive integers. How do you find the smallest possible value of $k$ and the smallest values of $a$ and $b$ for that $k$ ?

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Answer 1 · 2018-02-25T01:45:57

$k = 10$ , $a=69$ , $b=20$

By Pythagoras' theorem, we have:

$(13a+kb)^2 = (5a+12b)^2+(12a+5b)^2$

That is:

$169a^2+26kab+k^2b^2 = 25a^2+120ab+144b^2+144a^2+120ab+25b^2$

$color(white)(169a^2+26kab+k^2b^2) = 169a^2+240ab+169b^2$

Subtract the left hand side from both ends to find:

$0 = (240-26k)ab + (169-k^2)b^2$

$color(white)(0) = b((240-26k)a+(169-k^2)b)$

Since $b > 0$ we require:

$(240-26k)a+(169-k^2)b = 0$

Then since $a, b > 0$ we require $(240-26k)$ and $(169-k^2)$ to have opposite signs.

When $k in [1, 9]$ both $240-26k$ and $169-k^2$ are positive.

When $k in [10, 12]$ we find $240-26k < 0$ and $169-k^2 > 0$ as required.

So the minimum possible value of $k$ is $10$ .

Then:

$-20a+69b = 0$

Then since $20$ and $69$ have no common factor larger than $1$ , the minimum values of $a$ and $b$ are $69$ and $20$ respectively.

Let 5a+12b5a+12b and 12a+5b12a+5b be the side lengths of a right-angled triangle and 13a+kb13a+kb be the hypotenuse, where aa, bb and kk are positive integers. How do you find the smallest possible value of kk and the smallest values of aa and bb for that kk?