In triangle $ABC$, $a^2+c^2=3b^2$

Question

In triangle $ABC$, we have $a=BC$, $b=CA$ and $c=AB$ as usual. What is a necessary and sufficient condition for $a^2+c^2=3b^2$ to hold?

I created this problem as a generalization of $a^2+c^2=2b^2$ which I solved yesterday as a result of my solution to a problem by David Monk. Of course, $a^2+c^2=b^2$ is just right triangles.

I think the general problem $a^2+c^2=nb^2$ is hard hence the $n=3$ case I posted, but, I have the sneaking suspicion that work has already been done on this problem (perhaps it is even solved) and I would appreciate any references.

EDIT: Thank you for the solutions, this addition is in no way meant to detract from them.

Is it possible to find a nice synthetic condition for $a^2+c^2=3b^2$ to hold like the angle condition here: Thanks! Is it possible to translate this into a synthetic condition like the angle condition here: http://www.artofproblemsolving.com/community/c6h1071579p4662340 I thought that was nice.

So far I have:

$a^2+c^2=b^2$ is equivalent to $\angle ABC=90^o$.

$a^2+c^2=2b^2$ is equivalent to $\angle LAC=\angle ABM$ and $\angle ANC = \angle ALB$, where $L,M,N$ are midpoints of the sides $BC,CA,AB$. (Is there a nicer angle condition than this? The "dual" nature of it is slightly unsettling in comparison to the simple "$\angle ABC=90^o$" of the $a^2+c^2=b^2$ case.)

$a^2+c^2=3b^2$. The answer below show that this is equivalent to a length condition in the triangle. How does this length condition convert into a nice angle condition within the triangle, like the ones above?

Answer 1

We can assume, without loss of generality, that $C=(-1,0)$ and $A=(1,0)$.

Then assuming $B=(x,y)$ the condition $a^2+c^2=n b^2$ translates into:

$$ (x+1)^2 + (x-1)^2 + 2y^2 = 4n $$ that is the equation of a circle with center in the origin and radius $\sqrt{2n-1}$.

So we have that $a^2+c^2 = nb^2$ is equivalent to:

The distance between $B$ and the midpoint of $AC$ equals $\frac{1}{2}\sqrt{2n-1}$ times the length of $AC$.

Answer 2

Let's simplify by scaling everything up/down to make $b = 1$. Then the law of cosines says that \begin{align} b^2 &= a^2 + c^2 - 2ac \cos B \\\ b^2 &= 3b^2 - 2ac \cos B \\ -2b^2 &= - 2ac \cos B \\ b^2 &= ac \cos B \\ 1 &= ac \cos B \end{align} That means that $B < \pi/2$ (so that the cosine is positive), and that the length of the perpendicular projection of edge $AB$ on edge $BC$ is the reciprocal of the length of edge $BC$.