Henry Ernest Dudeney/Modern Puzzles/129 - The Circle and Discs/Solution

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Modern Puzzles by Henry Ernest Dudeney: $129$

The Circle and Discs
During a recent visit to a fair we saw a man with a table,
on the oilcloth covering of which was painted a large red circle,
and he invited the public to cover this circle entirely with five tin discs which he provided,
and offered a substantial prize to anyone who was successful.
The circular discs were all of the same size, and each, of course, smaller than the red circle.
he showed that it was "quite easy when you know how," by covering up the circle himself without any apparent difficulty,
but many tried over and over again and failed every time.
It was a condition that when once you had placed any disc you were not allowed to shift it,
otherwise, by sliding them about after they had been placed, it might be tolerably easy to do.
Let us assume that the red circle is six units in diameter.
Now, what is the smallest possible diameter for the five discs in order to make a solution possible?


Solution

Dudeney-Modern-Puzzles-129-solution.png

The dotted lines represent the red circle and a regular pentagon inscribed within it.

The center of this circle is the point $C$.

Find the point $D$ which is equidistant from $A$, $B$ and $C$.

With radius $AD$, draw the circle $ABC$.

Similarly construct the points $E$, $F$, $G$ and $H$.

These points are the centers of the circles representing the discs with which the red circle is covered.


We are given that the red circle is $6$ units in diameter.

The diameter of the discs is then just under $4$ units in diameter.


Covering is possible if the ratio of the two diameters is greater than approximately $0.6094185$.

In the case above, where all five discs reach the center of the red circle, the ratio is the golden mean $\approx 0.6180340$.


Proof




Sources