Knowing all areas of three overlapping circles excepting the area of $A'∩B'∩C$ and the area of $C$, is it possible to find the area of $A'∩B'∩C$?

Question

Let's imagine I have three overlapping circles. I know the following information:

$1.$ the area of circle $A$ = the area of circle $B = 1$

$2.$ the area of $A\cap B =\frac 15$

$3.$ the area of $A\cap C =$ the area of $B\cap C = \frac 9{20}$

$4.$ the area of $A\cap B \cap C=x$ such that $0\lt x\lt \frac 15 $

$5.$ the area of circle $C$ is unknown.

Is it possible, using this information, to find the possible range of areas for $A'\cap B'\cap C$, and thus find the range of areas for $C$?

EDIT: I am aware of some methods whereby you can measure the overlap of three circles. I'm beginning to wonder if there's a way to do it in reverse to find $C$.

Answer 1

There is no upper limit for the area of $C$. The center of $C$ lies on the line $r$ passing through the intersection points of circumferences $A$ and $B$, and the arc of $C$ inside $A$ approaches a straight line when the radius of $C$ tends to infinity. That limiting line is perpendicular to $r$ and cuts $A$ so that the lesser part is 45% of $A$. This situation is compatible with your constraints because $\hbox{area}(A\cap B\cap C)=0.45\ \hbox{area}(A\cap B)=0.09$.

A lower limit can be found, because it corresponds to the situation when the three circumferences have a point in common. Again, the bound on the area of $A\cap B\cap C$ is unimportant, but the area of $A\cap C$ is $0.45$ and this implies the radius of $C$ cannot be too small. Experimentally the lowest area turns out to be around $0.72$.

Answer 2

From 1 and 2, the circles $A$ and $B$ are completely determined (up to movement of the whole figure). Indeed, on finds radius $$r=\sqrt{1/\pi}\approx 0.56418958354775628694807945156077258585$$ and numerically using suitable formulas a distance of circle centres $O_A$, $O_B$ of $$d\approx 0.77525158218538574803265479010075364915.$$

From 3 we know that the centres of the circle form an isosceles triangle. If we assign coordinates $O_A=(-\tfrac d2,0)$, $O_B=(\tfrac d2,0)$, we may assume that $O_C=(0,y)$ for suitable $y>0$. Note that $A$ and $B$ intersect in $(0,\pm h)$ with $$h=\sqrt{r^2-\frac14d^2}\approx 0.40994649926978848797914142299189026942.$$ From 4 we infer that the radius $r_C$ obeys $y-h<r_C<y+h$. We can solve numerically, for which $y$ the lens-shaped area of $A$ intersect the circle around $(0,y)$ with radius $y+h$ equals $0.45$ and find that this happens at $$\tag1 y=y_0\approx 0.24774217358345628137394294376637568202.$$ We should do the same (for $y>h$)with radius $y-h$ in place of $y+h$ to investigate the other end. However, the resulting lens-shape will always be less than the circle segment determined by the horizontal line through $(0,h)$, which is less than $0.45$ in area. We conclude that the only restriction comes from $(1)$, i.e., $y$ may take any value $>y_0$ and can otherwise grow arbitrarily large. At the same time, the radius $r_C$ can vary from $y_0+h$ up to infinity. In other words, all that can be said about $C$ is that $$ C>\pi(y_0+h)^2\approx 1.3589096951569625386674237564989042025.$$