Consider a triangle $ABC$ such that $A((\frac{2}{√3})e^{iπ/2})$, $B((\frac{2}{√3})e^{-iπ/6})$, $C((\frac{2}{√3})e^{i5π/6})$. Let P be any point on the incircle of $ABC$. Prove that $PA^2+PB^2+PC^2=5$
My Progress:
$ABC$ is right angled at $A$ with $\angle B=30^\circ$ and $\angle C=60^\circ$. I proceeded to calculate all the side lengths and the inradius. After this I didn't know exactly what to do so I tried taking a new coordinate system with axis parallel to $AB$ and $AC$ and center coinciding with center of in-circle. Then I took an arbitrary point $(x,y)$ on the in-circle whose equation was $x^2+y^2=r^2=\frac{{(√3-1)}^2}{3}$ and used the Pythagoras theorem to calculate $PA^2,PB^2,PC^2$ and was hoping for the $(x,y)$ terms to maybe magically cancel out but of course, they didn't. Can someone please give me a clue as to how I should proceed? Thanks
Let's proceed from where you're stuck. Note that $x^2 + y^2 = \frac{(\sqrt3-1)^2}{3}$. So substitute $x^2 = \frac{(\sqrt3-1)^2}{3} - y^2$. This should make $y^2$ magically cancel out, unless you made an error in calculation.
Indeed, you should not only hope that they cancel out, you should be absolutely confident that they will cancel out. This is because you definitely used every single piece of information exhaustively (i.e. you drained all value out): By making the above substitution, you eliminated $x$ completely from the scene. And you know that $y$ can vary within a given interval (since $P$ lies on a circle), so if there are still remaining $y$'s it would mean that $PA^2+PB^2+PC^2$ is not constant, after all!
I hope this kind of thinking helps you solve further problems :)