Finding two points B and C such that the perimeter of triangle ABC is minimal

Question

Given an acute angle XOY and an interior point A. Find a point B on the side OX and a point C on the side OY such that the perimeter of the triangle ABC is minimal. Hint: Introduce points symmetric to A with respect to the sides of the angle.

I found this problem in Kiselev's Geometry: Planimetry. I do not know what a minimum perimeter looks like or how it appears in algebraic form.

Answer 1

That is a variation of some very common questions about minimal distance.

Take $A_x$ symmetric of $A$ w.r.t $OX$ and $A_y$ symmetric of $A$ w.r.t $OY$.

Now take the line $A_xA_y$ and see that $A_xB=AB$ (because $A_x$ symmetric of $A$) and $A_yC=AC$.

Once $A_xA_y$ is the minimal distante from $A_x$ to $A_y$ then $$AB+BC+AC=A_xB+BC+A_yC=A_xA_y$$ is also minimal.

Answer 2

Hint:

$B$ and $C$ have to be such that $AB$ and $AC$ are the minimal distances from $A$ to the sides of the angle. This means that they are orthogonal to the sides.