In a triangle $ABC$, point $X$ is picked on $BC$ such that the sum of the areas of the circumcircles of $ABX$ and $ACX$ is minimised. Describe where $X$ would be located on $BC$, and prove that this choice of $X$ is optimal.
Edit: Original problem statement said maximised, changed to minimised.
I would assume the point chosen would be a special point: midpoint, foot of altitude/angle bisector, etc., but don't know where to begin on a problem like this.
On
This is not an answer but a query instead. It is too large to be stated in the comment.
First of all, I totally agree with @abel’s deduction.
However, when I tried to play with it using Geogebra, I have the following finding:-
A = (30, 80), B = (0, 0), C = (120, 0), X (now renamed as D) = (30, 0).
The blue circle ABD is drawn with $e: (x – 15)^2 + (y – 40)^2 = 1825$ (machine calculated, same for others)
The green circle ACD is also drawn with $f: (x – 75)^2 + (y – 40)^2 = 3625$
When the variable point D coincides with B, only one circle (the red one circumscribing ABC) needs to be drawn. Its equation is $g: (x – 60)^2 + (y – 23.13)^2 = 4134.77$
Clearly, $4134.77 < 1825 + 3625$.
This means the area of the single circle is less than the suggested minimal sum.
Would like to know what went wrong.
here is the reason why $X$ must be either $B$ or $C.$ let $X$ be any interior point on $BC$ the diameter of the circumcircle $ABX$ is ${AB \over \sin \angle AXB}$ in the same way the diameter of the circumcircle $ACX$ is ${AC \over \sin \angle AXC}$ and $\sin \angle AXB = \sin \angle AXC$ plus the sum of the areas of the circle is ${\pi \over 4}{AB^2 + AC^2 \over \sin^2 \angle AXC}$ the maximum is achieved when this angle is the smallest. this happens at one of the extreme points $B$ or $C$ and minimum is achieved when $X$ is the foot of the perpendicular from $A$ to $BC.$