I need help to solve the following problem.
Consider a square in the Euclidean plane with vertices $A,B,C,D\in\mathbb{R}^2$ (named in counterclockwise direction). Moreover, let $P$ be a point inside the square, such that it has the following distances to the vertices: $\vert AP\vert=1, \vert BP\vert=2, \vert CP\vert=3$. Question: How large is the angle $\angle (APB)$?
I tried to compute the length of any side of the square, as well as the distances of $P$ from the sides of the square. To do so, one can draw the perpendicular from $P$ to the sides of the square und use Pythagorean theorem to derive some equations. I obtained:
$$ h^2+y^2=4\\
(x-h)^2+y^2 = 1\\
(x-y)^2+ h^2= 9,$$
where $x$ denotes the side length of the square, $y:=d(P, AB)$ is the distance of $P$ to the side $AB$, and $h:=d(P,BC)$ is the distance from $P$ to the side $BC$.
We now have three equations with three unknowns. But I couldn't solve these equations by hand. Is that possible? Or am I on the wrong track, is there a better approach?
I would appreciate your help a lot!
Best wishes
Best Answer
Rotate $P$ around $B$ for $-90^{\circ}$ to $Q$. Then triangle $PBQ$ is isosceles riht triangle, so $\angle PQB = 45^{\circ}$ and $PQ =\sqrt{8}$. Then triangle $PQC$ is right at $Q$, since $PC^2 = CQ^2+PQ^2$. Thus $\angle BQC =135^{\circ}$. Finaly $$\angle APB = \angle BQC =135^{\circ}$$