Can three perpendicular bisectors determine a triangle? I would like to know if it is possible to determine the sides of a triangle given the lengths of the three perpendicular bisectors.
Assume that the three perpendicular bisectors intersect at $P$ and intersects $a, b, c$ respectively at $K, L, M$, and let the lengths of $PK, PL, PM$ be $p_a, p_b, p_c$ which is given. Is it possible to determine the lengths of $a, b, c$?
We know that
$$\frac {a^2}{4}+p_a^2=\frac {b^2}{4}+p_b^2=\frac {c^2}{4}+p_c^2=r^2$$
Where $r$ is the radius of the circumscribed circle.
Additionally, we have
$$\frac{1}{2}(ap_a+bp_b+cp_c)=\sqrt{s(s-a)(s-b)(s-c)}$$
And by substitution, we have
$$a^2+4p_a^2=\frac{a^2(a^2+4p_a^2-4p_b^2)(a^2+4p_a^2-4p_c^2)}{a^2(a^2+4p_b^2-4p_c^2)}$$
What to do next?
 A: To incorporate all possible cases it is convenient to allow the values $p_i$ be signed, negative sign meaning that the side $i$ lies against the obtuse angle. There can be only one negative $p_i$ and it has to be the least of three by the absolute value. If all three $p_i$ are positive or the least by the absolute value is negative and two other are positive the solution always exists and is unique.
The finding of $a,b,c$ can be reduced to finding the circumradius $r$. For this one should solve the equation:
$$
\arccos\frac{p_a}r+\arccos\frac{p_b}r+\arccos\frac{p_c}r=\pi.\tag1
$$
Using the formula
$$
\cos(x+y+z)=\cos x\cos y\cos z-\cos x\sin y\sin z-\sin x\cos y\sin z-\sin x\sin y\cos z
$$
one can after some straightforward algebra reduce the equation $(1)$ to:
$$
r^3-3pr-2q=0.\tag2
$$
with
$$
p=\frac{p_a^2+p_b^2+p_c^2}3,\quad q=p_a p_b p_c.\tag3
$$
This cubic equation has depressed form and can be easily solved. Since due to AM-GM inequality we have in general
$$
-p^3+q^2\le0 
$$
the equation has three real roots and we need the largest one.
The simplest way to represent the required soluton of $(2)$ is to use the trigonometric form:
$$
r=2p^{1/2}\cos\left[\frac13\arccos\left(\frac q{p^{3/2}}\right)\right].\tag4
$$
