Question

Let the length of the sides of a triangle $△ABC$ be integers with $A$ as the origin. $(2,-1)$ and $(3,6)$ are points on the line $AB$ and $AC$ respectively (line $AB$ and $AC$ may be extended to contain these points), and lengths of exactly two sides are primes that differ by $50$. If $a$ is least possible length of the third side and $S$ is the least possible perimeter of the triangle, then $aS$ is equal to

Answer 1

Since $(AB{)}^2+(AC{)}^2=50=(BC{)}^2$ the lines AB and AC are at right angles.
It is given that all sides are integers and since the triangle is right angled, they form a pythagorean triplet,
so one leg must be even. As two sides are primes, both cannot be the legs of the right angled triangle.
So, let $AC=p$, a prime, $BC=p+50$ and $AB=a$ (even)
then $a^2=(p+50{)}^2-p^2=100(p+25)$
$\Rightarrow a=10\sqrt{p+25}$.
Since a is least possible integer and p is a prime
we get $a=60$ when $p=11$.
So smallest value of the third side is 60.
$S=60+11+61=132$
$aS=60\times 132=7920$