Three circles in a triangle: can the circles move? Three unit circles are in an isosceles right triangle. In the beginning, each circle is tangent to the other circles and one edge of the triangle; there is a vertical line of symmetry.


Can the circles move without overlapping?

This is a variation of another question of mine.
My attempt

I superimposed a cartesian coordinate system. I assumed that the red circle moves left a small distance left, and the only possible loss of tangency is between the red and green circles.
I let the coordinates of the red circle's centre be $(t,0)$, then found that the  distance between the red circle's centre and the green circle's centre is
$\sqrt{\left(t-a\right)^2+\left(a-\sqrt2-\sqrt6\right)^2}$ where $a=\frac12 \left(\sqrt2+\sqrt6-\sqrt{4-t^2}+\sqrt{2(\sqrt2+\sqrt6)\sqrt{4-t^2}+t^2-4-4\sqrt3}\right)$.
I tried to show algebraically that this distance is greater than $2$, which would show that the circles can move, but this is rather difficult.
This desmos animation suggests that the circles can move.
I am hoping for an intuitive answer.
 A: Draw perpendiculars on the sides of the triangle at the points where the circles touch. Those three lines intersect at a single point P due to symmetry.
Consider the three circles as if they form a single object, and rotate that object around that point P. Clearly every circle can move sideways relative to its touching side because the tangents are all perpendicular to the radial lines from P. As the rotation continues the circles even move away from the sides and no longer touch them.
A: In line with the earlier question, about the mobility of five circles in a rectangle, whose centers form a regular pentagon, I wondered whether four circles would be mobile in an isosceles right triangle when the centers of the circles form a square and two circles are tangent to the base, as in the figure below.
For the unit circle, I find $AC=4\sqrt 2+2=7.657$, and $AB=8+2\sqrt 2=10.828$.
However, if the four circles are arranged as in the next figure, I find that $AC=6+\sqrt 2=7.414$ and $AB=6\sqrt 2+2=10.485$

Thus in the second position (three circles in a corner as in the pentagon case) the circles fit in a smaller isosceles right triangle, suggesting to me that, as in the pentagon/rectangle case, they might be moveable into that second position if they begin in the somewhat larger triangle required for the first position.
