Problem with four equal-sized Circles I think the following is true but haven't managed to prove it yet.
Consider a circle with center $O$ and radius $r$. Choose three points $A,B,C$ on or inside the circle such that all sides of $\triangle ABC$ have length greater than $r$. Show that the three circles of radius $r$ centered at $A,B,C$ cannot intersect outside $C(O,r)$.
 A: Wlog I assume the main disc $O, r$ to be centered at the origin, i.e. $O=(0,0)$.
Wlog (by symmetry arguments) one can define the intersection of the circles around $A,B,C$ as a point on the positive $x$-axis
$$
x := \text{intersection}(A,B,C) = (l, 0)\,, \quad l > r \,.
$$
As this point is an intersection of circles of radius $r$ with centers inside $\text{disc}(O,r)$ we know an upper limit on $l$
$$
r < l \leq 2r \,.
$$
Now, this part is important (and would be best supplied with a scetch..).
The points $A,B,C$ must lie on the circle: center $x$, radius $r$.
To achieve the maximal distances between all 3 points we need to set one point on the $x$-axis.
Wlog i choose $A$ to be that point. Thus, center of $A$: $c_A = (l-r, 0)$.
By the same argument we need to set the centers of $B,C$ on the intersections of the circle($x$, $r$), and circle($O$, $r$)
$$
y_\pm = \text{intersections}(\text{circle}(x, r), \text{circle}(O, r))
= (\frac{l}{2}, \pm d)\,,
\quad
d = \sqrt{r^2-\frac{l^2}{4}}\,.
$$
Wlog: the center of $B$ is $c_B = y_+$ and center of $C$ is $c_C = y_-$.
Their distances can be computed to
$$
|c_A - c_{B/C}|^2 = (\frac{l}{2}-r)^2 + (r^2-\frac{l^2}{4})
= 2r^2-lr = r^2(2-\frac{l}{r})
\\
\implies
|c_A - c_{B/C}| = r \sqrt{2-\frac{l}{r}} < r \,.
$$
Note that from $r < l \leq 2r$ we can follow $0 \leq \sqrt{2-\frac{l}{r}} < 1$.
Thus, we have shown that even for optimal placement of $A,B,C$ the sides of $\Delta ABC$ are smaller than $r$.

Terminoligy

*

*disc: all points no farther away from the center than its radius (the whole 2d area)

*circle: a disc's boundary (just the 1D line)

A: With $r = 1$, start by fixing the intersection point, $X = (0,0)$, and $C = (1, 0) $ in the plane. $A$ and $B$ lie on a (red) circle with radius 1, centred at $X$. Let $\alpha = \angle AXC$ and $\beta = \angle BXC$. By a relabelling argument, we assume $\alpha, \beta \leq \pi$. Note, also that $\alpha, \beta > \frac{\pi}{3}$, as we assume $|AC|,|BC| > 1 $. Finally, consider a (blue) circle of radius 1 and centre $O$ that contains the three points $A,B$ and $C$.

Our aim now is to show that $O$ must lie inside the red circle.
Let $A'$ and $B'$ represent the points of minimum angle for $A$ and $B$ respectively and mark the position diametrically opposite $C$ as $D$. By construction, we have that the blue circle must contain a portion of the arc $DA'$ (purple) and a portion of the arc $DB'$ (green) and the point $C$. Label the arc $A'B'$ pink.
Clearly, the red and blue circles cannot intersect 0 or 1 times. If they were to intersect in two places of the same colour (green, purple,pink) we can see that they must coincide. The only possibility to intersect at two points of different colours, is when the colours are purple and green and in which case the points $A'$ and $B'$ are contained in the blue circle. But in this case, the point $O$ must lie within distance 1 of both $A'$ and $B'$, which is shown by the orange shape. In all cases the point $O$ lies in the red circle.

A: You're right. First, if three circles of radius $r$ all intersect at point $X$, then their centers must all lie on a circle $\mathcal{C}(X,r)$:

So let's just envision where that circle $\mathcal{C}(X,r)$ might lie, relative  to the fourth circle. We'll pick the intersection point $X$ and work backwards to pick three centers whose intersection is $X$. You can pick any three points within the fourth circle that lie along the circumference of $\mathcal{C}(X,r)$ to get the centers of three circles that intersect at $X$ (You can pick any three points along the dashed blue line; the three circles with those centers will intersect at $X$):

The final requirement is that the three centers must be separated by at least distance $r$. When the intersection point $X$ is too far from the fourth circle, the three points will be forced too close together. As $X$ gets closer to the fourth circle, the three points can spread out. The very first moment they can be separated by distance $r$ or more is when $\triangle ABX$  and $\triangle BCX$ are equilateral triangles with side length $r$:

At that point, the center $O$ and the intersection point $X$ are separated with distance $r$; hence the intersection point $X$ is exactly on the edge of the circle. And if you bring $X$ any closer, it will lie inside the circle.
A: Take the point $O=(0,0); A=(r,0); B=(0,r); C=(-r,0)$. Now we have that (length) $L_{ab}=L_{bc}=\sqrt{2}r$ and $L_{ac}=2r$ which are obviously $>r$. Write the equation for circle A as $\left(x-r\right)^{2}+y^{2}=r^{2}$ and circle B $x^{2}+\left(y-r\right)^{2}=r^{2}$ and the intersection are the solution to the equation $\left(x-r\right)^{2}+y^{2}=x^{2}+\left(y-r\right)^{2}$. By simple calculations you get a line $y=x$. Substitute in one of the prevous equation and you get (for $y$) $\left(y-r\right)^{2}+y^{2}=r^{2}\rightarrow y\left(y-r\right)=0$ whose solutions are $y=0; y=r$. But wait, we said before that the intersection must be in the line $y=x$ so the points of intersection must be $(0,0)$ and $(r,r)$, where the last one is obviously outside the circle centered in $O=(0,0)$.
The case limit for $A=(r,0); B=\left(r\cos\theta,r\sin\theta\right)$ and $C$ somewhere else is when $L_{ab}=r$. Since $L_{oa}=L_{ob}=L_{ab}=r$, we have an equilateral triangle △OAB, where $\theta$ must be $\pi/3$. A bit of calculations and we have that the intersections are $(0,0)$ and $\left(\frac{3}{2}r,\frac{\sqrt{3}}{2}r\right)$ and the line is $y=\frac{x}{\sqrt{3}}$.
