Problem 981

Suppose the centre of the largest circle is $O$.
Suppose that the circle with centre $X$ touches the largest circle at $S$ and the two circles with centres $Y$ and $Z$ at $T$ and $U$, respectively.
Suppose that the circles with centres $Y$ and $Z$ touch each other at $A$, and the largest circle at $B$ and $C$, respectively.
Join $X$ to $Y$, $X$ to $Z$, and $Y$ to $Z$.

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(Note that the diagram has been re-drawn here so that the circle with centre $X$ actually appears to pass through the centre of the largest circle.)
Since the circles are tangent at points $T$ and $U$, line segments $XY$ and $XZ$ pass through $T$ and $U$, respectively.
Further, $XY = XT + TY = 1 + r$, since the circles with centres $X$ and $Y$ have radii 1 and $r$, respectively.
Similarly, $XZ = 1+r$.
Also, $YA = ZA = YB = ZC = r$, since these are radii of the two circles.
When one circle is inside another circle, and the two circles touch at a point, then the radii of the two circles that pass through this point lie on top of each other. This is because the circles have a common tangent at the point where they touch and this common tangent will be perpendicular to each of the radii.
Since the circle with centre $X$ touches the largest circle at $S$, then $X$ lies on $OS$.
In the largest circle, consider the diameter that passes through $X$.
Since the circle with centre $X$ passes through $O$, then the radius of the largest circle is twice that of the circle with centre $X$, or 2.
It is also the case that $XO = 1$.
Next, we join $O$ to $B$. Since the circles with centres $O$ and $Y$ touch at $B$, then $OB$ passes through $Y$. This means that $OY = OB - BY = 2 - r$. Similarly, $OZ = 2-r$.
Further, by symmetry in the largest circle, the diameter through $X$ also passes through $A$, the point at which the two smallest circles touch:

To see this more formally, draw the common tangent through $A$ to the circles with centres $Y$ and $Z$.
This line is perpendicular to $YZ$, since it is tangent to both circles.
Since $\triangle OYZ$ is isosceles with $OY = OZ$, the altitude through the midpoint $A$ of its base passes through $O$.
Similarly, $\triangle XYZ$ is isosceles with $XY = XZ$ and so its altitude through $A$ passes through $X$.
Since the line perpendicular to $YZ$ at $A$ passes through both $O$ and $X$, it is the diameter that passes through $X$.

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Now, we consider $\triangle XYA$ and $\triangle OYA$, each of which is right-angled at $A$.
By the Pythagorean Theorem, $$OA = \sqrt{OY^2 - YA^2} = \sqrt{(2-r)^2 - r^2} = \sqrt{4-4r+r^2 - r^2} = \sqrt{4 - 4r}$$ Again, using the Pythagorean Theorem, $$\text{XA 2 + YA 2 = XY 2 (XO + OA) 2 + r 2 = (1+r) 2 (1 + 4-4r ) 2 = 1+2r + r 2 - r 2 1 + 2 4-4r + (4-4r) = 1+2r 2 4 - 4r = 6r - 4 4-4r = 3r - 2 4 - 4r = (3r-2) 2 (squaring both sides) 4 - 4r = 9r 2 - 12r + 4 8r = 9r 2}$$ Since $r \neq 0$, then $9r = 8$ and so $r = \frac{8}{9} \approx 0.889$.
Of the given choices, this is closest to (E) 0.89.