Dido's Problem/Variant 2

Problem

Consider the corner of a room which is separated from the rest of the room by a screen of two identical halves, $a$ and $b$ where the length of $a$ equals the length of $b$.

The screen meets the walls of the room at $A$ and $B$.



How can the screen be arranged so that the area enclosed by the screen is a maximum?


Solution

When the screen forms two sides of a regular octagon:


Thus the sides of the screen must make an angle of $67 \frac 1 2 \degrees$ with the walls.


Proof

When the area enclosed by the screen is a maximum, so will be the area of the octagon so formed by the construction above.


This needs considerable tedious hard slog to complete it.
In particular: Invoke (and first create) the result about maximising the area of a polygon
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The angles the screen makes with the walls is half the internal angle of the regular octagon.

From Internal Angles of Regular Polygon, that is:

$A = \dfrac 1 2 \paren {\dfrac {\paren {n - 2} 180 \degrees} n}$

where:

$A$ is the angle the screen makes with the wall
$n$ is the number of sides of the regular polygon, in this case $8$.

Hence the result.

$\blacksquare$


Sources

  • 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): The Area Enclosed Against The Seashore: $32$
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