In the diagram, the circle with centre $X$ is tangent to the largest circle and passes through the centre of the largest circle. The circles with centres $Y$ and $Z$ are each tangent to the other three circles, as shown. The circle with centre $X$ has radius 1 . The circles with centres $Y$ and $Z$ each have radius $r$. The value of $r$ is closest to
(A) 0.93
(B) 0.91
(D) 0.87
(E) 0.89
(C) 0.95

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-023.jpg?height=360&width=341&top_left_y=899&top_left_x=1358)
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 $Z$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-045.jpg?height=349&width=377&top_left_y=1712&top_left_x=923)

(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 $X Y$ and $X Z$ pass through $T$ and $U$, respectively.

Further, $X Y=X T+T Y=1+r$, since the circles with centres $X$ and $Y$ have radii 1 and $r$, respectively.

Similarly, $X Z=1+r$.

Also, $Y A=Z A=Y B=Z C=r$, 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 $O S$.

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 $X O=1$.

Next, we join $O$ to $B$. Since the circles with centres $O$ and $Y$ touch at $B$, then $O B$ passes through $Y$. This means that $O Y=O B-B Y=2-r$. Similarly, $O Z=2-r$.

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 $Y Z$, since it is tangent to both circles.

Since $\triangle O Y Z$ is isosceles with $O Y=O Z$, the altitude through the midpoint $A$ of its base passes through $O$.

Similarly, $\triangle X Y Z$ is isosceles with $X Y=X Z$ and so its altitude through $A$ passes through $X$.

Since the line perpendicular to $Y Z$ at $A$ passes through both $O$ and $X$, it is the diameter that passes through $X$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-046.jpg?height=339&width=377&top_left_y=1229&top_left_x=923)

Now, we consider $\triangle X Y A$ and $\triangle O Y A$, each of which is right-angled at $A$. By the Pythagorean Theorem,

$$
O A=\sqrt{O Y^{2}-Y A^{2}}=\sqrt{(2-r)^{2}-r^{2}}=\sqrt{4-4 r+r^{2}-r^{2}}=\sqrt{4-4 r}
$$

Again, using the Pythagorean Theorem,

$$
\begin{aligned}
X A^{2}+Y A^{2} & =X Y^{2} \\
(X O+O A)^{2}+r^{2} & =(1+r)^{2} \\
(1+\sqrt{4-4 r})^{2} & =1+2 r+r^{2}-r^{2} \\
1+2 \sqrt{4-4 r}+(4-4 r) & =1+2 r \\
2 \sqrt{4-4 r} & =6 r-4 \\
\sqrt{4-4 r} & =3 r-2 \\
4-4 r & =(3 r-2)^{2} \quad \text { (squaring both sides) } \\
4-4 r & =9 r^{2}-12 r+4 \\
8 r & =9 r^{2}
\end{aligned}
$$

Since $r \neq 0$, then $9 r=8$ and so $r=\frac{8}{9} \approx 0.889$.

Of the given choices, this is closest to (E) 0.89 .

ANSWER: (E)