Seven coins of three different sizes are placed flat on a table, arranged as shown in the diagram. Each coin, except the centre one, touches three other coins. The centre coin touches all of the other coins. If the coins labelled $C_{3}$ have a radius of $3 \mathrm{~cm}$, and those labelled $C_{2}$ have radius $2 \mathrm{~cm}$, then the radius of the coin labelled $X$ is closest to
(A) $0.615 \mathrm{~cm}$
(B) $0.620 \mathrm{~cm}$
(C) $0.610 \mathrm{~cm}$
(D) $0.605 \mathrm{~cm}$
(E) $0.625 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-047.jpg?height=379&width=333&top_left_y=1033&top_left_x=1292)
We label the centres of the coins $A, B, D, E, F, G, H$, as shown, and we join $A B, A D, A E, A F$, $A G, A H, B D, D E, E F, F G, G H$, and $H B$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-093.jpg?height=569&width=499&top_left_y=301&top_left_x=867)

The radius of the circles with centres $D, E$ and $F$ is $3 \mathrm{~cm}$ and the radius of the circles with centres $A, B$ and $G$ is $2 \mathrm{~cm}$. (From this point until the very last step of the problem, we will not include the units, which are always centimetres.)

Let the radius of the circle with centre $H$ be $r$. We want to determine $r$.

When we join the centres of two circles that are just touching, the resulting line segment passes through the point at which the circles touch and the length of this line segment is the sum of the radii of the two circles.

Thus, $A B=2+2=4$. Similarly, $A G=4, B D=A D=A E=A F=G F=5, D E=E F=6$, and $H A=H B=H G=r+2$.

Now $\triangle A D E$ and $\triangle A F E$ each have side lengths 5,5 and 6 , which means that these triangles are congruent.

Similarly, $\triangle A D B$ and $\triangle A F G$ are congruent, and $\triangle A B H$ and $\triangle A G H$ are congruent.

Since corresponding angles in congruent triangles are equal, then $\angle H A B+\angle B A D+\angle D A E$ is equal to $\angle H A G+\angle G A F+\angle F A E$.

But these six angles surround point $A$, so the sum of these six angles is $360^{\circ}$.

Thus, $\angle H A B+\angle B A D+\angle D A E=\angle H A G+\angle G A F+\angle F A E=180^{\circ}$.

Now, we remove the circles from the diagram and focus on the top half of the diagram.

We join $A$ to $K$, the midpoint of $D E$, and $D$ and $H$ to $L$, the midpoint of $A B$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-093.jpg?height=276&width=371&top_left_y=1868&top_left_x=931)

Consider $\triangle A D E$.

Since $D E=6$ and $K$ is the midpoint of $D E$, then $D K=K E=3$.

Also, since $A D=A E=5$, then $\triangle A D E$ is isosceles and so $A K$ is perpendicular to $D E$ and $A K$ bisects $\angle D A E$.

Similarly, $\triangle A H B$ is isosceles with $H A=H B$ and $\triangle A B D$ is isosceles with $D A=D B$.

Since $L$ is the midpoint of $A B$ and $A B=4$, then $A L=L B=2$ and $D L$ and $H L$ are perpendicular to $A B$ at $L$.

We know that $\angle H A B+\angle B A D+\angle D A E=\angle H A L+\angle L A D+2 \angle E A K=180^{\circ}$.

But $\sin (\angle E A K)=\frac{E K}{A E}=\frac{3}{5}$, so $\angle E A K \approx 36.87^{\circ}$.

Also, $\cos (\angle L A D)=\frac{A L}{A D}=\frac{2}{5}$, so $\angle L A D \approx 66.42^{\circ}$.

Thus, $\angle H A L=180^{\circ}-\angle L A D-2 \angle E A K \approx 180^{\circ}-66.42^{\circ}-2\left(36.87^{\circ}\right) \approx 39.84^{\circ}$.

Finally, we know that $\cos (\angle H A L)=\frac{A L}{H A}=\frac{2}{r+2}$.

Since $\cos \left(39.84^{\circ}\right) \approx 0.7679$, then $\frac{2}{r+2} \approx 0.7679$.

From this we obtain $r \approx \frac{2}{0.7679}-2$.

Since $\frac{2}{0.7679}-2 \approx 0.60451$, then of the given choices, this radius is closest to $0.605 \mathrm{~cm}$.

(If more decimal places are carried through from earlier calculations, we obtain $r \approx 0.60466$, which is still closest to $0.605 \mathrm{~cm}$.)

ANswer: (D)