# 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 polygonTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

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: $32$