Mirror, Mirror

Let $A$ be the vertex $(3,9)$ , $B$ be the vertex on the $y$ -axis and $C$ be the vertex on the line $y = x$ . Also let $D(-3,9)$ be the reflection of $A$ in the $y$ -axis and $E(9,3)$ be the reflection of $A$ in the line $y = x$ . Then $AB = BD$ and $AC = CE$ , and thus the perimeter of $\Delta ABC$ is equal to $DB + BC + CE.$

But the shortest distance between two points is a straight line, so

$DB + BC + CE \ge DE = \sqrt{(9 - (-3))^{2} + (3 - 9)^{2}} = \sqrt{180} = 6\sqrt{5}.$

This minimum can be obtained by then choosing $B$ and $C$ as the points of intersection of the line $DE$ with the $y$ -axis and the line $y = x$ , respectively. This gives us the points $B(0,\frac{15}{2})$ and $C(5,5)$ . The reader can confirm that this will yield a perimeter for $\Delta ABC$ of $6\sqrt{5}$ .

The desired sum is then $6 + 5 = \boxed{11}$ .

Comment: This problem is an adaptation of a Putnam contest question.

Reflection of V(3,9) in y = x is (9,3). Reflection of (9,3) in y = 0 is (-9,3). Now its distance from V gives minimum perimeter. The tell-tale title is a pointer to the solution. It is a Brilliant extension of my problem "The minimum you can do."