Determine the value of $a + b $ if the equation $\vert x^2 - 2ax +b\vert=8$ has only three real roots, which are the sides of a right triangle.

Question

Determine the value of $a + b $ if the equation $\vert x^2 - 2ax +b\vert=8$ has only three real roots, which are the sides of a right triangle.

Ive spent like an hour trying to figure out how to approach this. Please assist

From the 2018 IMC

Answer 1

In order for the equation to have three roots we need $$b=a^2-8$$ such that a local maximum occurs at the point $(x,y)=(a,8)$ and two other points intersect with the line $y=8$ at $(x,y)=(a\pm4,8)$. If these three roots are sides of a right triangle then $$(a-4)^2+a^2=(a+4)^2$$ $$a^2-8a+16+a^2=a^2+8a+16$$ $$a(a-16)=0$$ $$a=16$$ as we need $a\gt0$. Hence the value of $a+b$ is $$16+16^2-8=264$$