Triangle Inequalities in Right Angled triangle.

Question

In $\triangle{ABC}$, $\angle{ABC}=90^{\circ}$, $AB=BC$ and $AC=\sqrt{3}-1$. Suppose there exist a point $P_0$ in the plane of $\triangle{ABC}$ such that $AP_0+BP_0+CP_0 \leq AP+BP+CP$ for all points $P$ in the plane of $\triangle{ABC}$.How do I find value of $AP_0+BP_0+CP_0$. Please provide figure for this question too.

Answer 1

That point is called the Fermat Point.

From it's properties we have that: $\angle BP_0A = \angle BP_0C = \angle AP_0C = 120^{\circ}$. Also from the way you construct it, it's easy to notice that $P_0$ lies on the angle bisector of $\angle ABC$, since the triangle is isoscelec. Also this implies that $AP_0 = CP_0$. Then using Sine Theorem one can easily find the values of all $AP_0, CP_0$ and $BP_0$.