Coke Machine

We have Edwin Taylor, age 18 of the Leventhorpe School, Sawbridgeworth, to thank for the following solution.

This is a diagram of the 50 pence piece with lines between $AE$ and $AD$ drawn in. We will say that these lines (and the others like them) have length $R$. Also, we will call the centre of the heptagon point $O$ and say that the lines $OA$, $OB$, $OC$ etc have length $r$. Now, as the vertices of the heptagon are the furthest points away from $O$ and the distance from $O$ to any vertex is $r$, the maximum radius of a disc that could use the same chute as the 50 pence piece must also be $r$. If the radius of this disc is $r$, then its perimeter (i.e. circumference) must be $2\pi r$. This is what the segment $OED$ with the arc $ED$ looks like:



We know the angle EOD is ${{2\pi } \over 7}$ because it is one seventh of a complete revolution. The diagram also shows a point $M$ which is the midpoint of $ED$. Now if we look at the right angled triangle $EMO$, we can find out what $EM$ is:



$$ EM = r\sin {\pi \over 7} $$ If we consider the segment $AED$, the arc $ED$ subtends an angle ${{2\pi } \over 7}$ at $O$ so angle $EAD$ is ${\pi \over 7}$ and angle $EAM$ is ${\pi \over {14}}$.



Using what we know about $EM$ we can find an expression for $R$ : $$ R = {{EM} \over {\sin {\pi /{14}}}} = {{r\sin {\pi /7}} \over {\sin {\pi /{14}}}}. $$ Now, if we use the trigonometric identity $\sin 2A = 2\sin A\cos A$ we get $$ R = 2r\cos {\pi /{14}}. $$ So, using the formula $S = r\theta $ that links the arc length ($S$ ) of a circle to the angle ($\theta $ ) that subtends it and the radius ($r$ ) of the circle, the length of the arc $ED$ is given by $$ ED = \left[ {2r\cos {\pi /{14}}} \right] \times {\pi /7}. $$ As the perimeter of the 50p is made up of 7 arcs of the same length as ED, the length of the perimeter of the 50p is $2\pi r\cos {\pi /{14}}$ . The ratio of the perimeter of the disc to the perimeter of the 50p piece is: $$ 2\pi r\ :\ 2\pi r\cos {\pi /{14}} = 1\ :\ \cos {\pi /{14}}. $$ So for every revolution of the 50p, the disc has gone 97.5% of a revolution, or every revolution the 50p gains on the disc by 1/40th of the discs revolution, or the 50p would do 40 revolutions in the space it would take the disc to do 39 revolutions.