A semi-circle of radius $8 \mathrm{~cm}$, rocks back and forth along a line. The distance between the line on which the semicircle sits and the line above is $12 \mathrm{~cm}$. As it rocks without slipping, the semi-circle touches the line above at two points. (When the semi-circle hits the line above, it immediately rocks back in the other direction.) The distance between these two points, in millimetres, is closest to
(A) 55
(B) 53
(D) 49
(E) 47
(C) 51
In its initial position, suppose the semi-circle touches the bottom line at $X$, with point $P$ directly above $X$ on the top line.

Consider when the semi-circle rocks to the right.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-199.jpg?height=524&width=857&top_left_y=665&top_left_x=688)

Suppose now the semi-circle touches the bottom line at $Y$ (with $O$ the point on the top of the semi-circle directly above $Y$, and $Z$ the point on the top line directly above $Y$ ) and touches the top line at $Q$. Note that $X Y=P Z$.

$Q$ is one of the desired points where the semi-circle touches the line above. Because the diagram is symmetrical, the other point will be the mirror image of $Q$ in line $X P$. Thus, the required distance is 2 times the length of $P Q$.

Now $P Q=Q Z-P Z=Q Z-X Y$.

Since the semi-circle is tangent to the bottom line, $Y O$ is perpendicular to the bottom line, and $O$ lies on a diameter, then $O$ is the centre of the circle, so $O Y=O Q=8 \mathrm{~cm}$, since both are radii (or since the centre always lies on a line parallel to the bottom line and a distance of the radius away).

Also, $O Z=4 \mathrm{~cm}$, since the distance between the two lines is $12 \mathrm{~cm}$

By the Pythagorean Theorem (since $\angle Q Z O=90^{\circ}$ ), then

$$
Q Z^{2}=Q O^{2}-Z O^{2}=8^{2}-4^{2}=64-16=48
$$

so $Q Z=4 \sqrt{3} \mathrm{~cm}$.

Also, since $Q Z: Z O=\sqrt{3}: 1$, then $\angle Q O Z=60^{\circ}$.

Thus, the angle from $Q O$ to the horizontal is $30^{\circ}$, so the semi-circle has rocked through an angle of $30^{\circ}$, ie. has rocked through $\frac{1}{12}$ of a full revolution (if it was a full circle).

Thus, the distance of $Y$ from $X$ is $\frac{1}{12}$ of the circumference of what would be the full circle of radius 8 , or $X Y=\frac{1}{12}(2 \pi(8))=\frac{4}{3} \pi \mathrm{cm}$. (We can think of a wheel turning through $30^{\circ}$ and the related horizontal distance through which it travels.)

Thus, $P Q=Q Z-X Y=4 \sqrt{3}-\frac{4}{3} \pi \mathrm{cm}$.

Therefore, the required distance double this, or $8 \sqrt{3}-\frac{8}{3} \pi \mathrm{cm}$ or about $5.4788 \mathrm{~cm}$, which is closest to $55 \mathrm{~mm}$.