Edited by Richard Walker, Tuesday, 26 May 2020, 22:54
Here's my solution to the A4 Paper Puzzle I posted a couple of days back. As Jan Pinfield correctly said (well done Jan!) the outlined perimeter has length 4, which is rather neat.
Here's how we can convince ourselves of this. The explanation is not mathematically rigorous because I wanted to keep it reasonably short and intuitive. Filling in the details is not so very hard but it might obscure the main argument, which is a rather elegant one.
In the graphic below I have reproduced the original puzzle, in which the starting rectangle had dimensions 1 by square root 2, and the question was to find the length of the outlined perimeter. I've added three more diagrams to help explain the argument.
The explanation goes like this.
Diagram (1) shows the perimeter is a kite, with AC a mirror line about which it is symmetrical. So distance AEDC is half the length we are trying to find.
In diagram (2) the rectangle outlined in blue is a scaled-down copy of the 1 by square root 2 rectangle we began with, and because it has the same proportions DC must be square root 2 times as long as ED.
In diagram (3) ED is calculated as the difference between square root 2 and 1, and multiplying ED by square root 2 gives the length of DC as 2 - square root 2. Adding AD and DC as shown in the box gives 2, and the whole perimeter is twice this, so it is 4.
A4 Paper Puzzle - My Solution
Here's my solution to the A4 Paper Puzzle I posted a couple of days back. As Jan Pinfield correctly said (well done Jan!) the outlined perimeter has length 4, which is rather neat.
Here's how we can convince ourselves of this. The explanation is not mathematically rigorous because I wanted to keep it reasonably short and intuitive. Filling in the details is not so very hard but it might obscure the main argument, which is a rather elegant one.
In the graphic below I have reproduced the original puzzle, in which the starting rectangle had dimensions 1 by square root 2, and the question was to find the length of the outlined perimeter. I've added three more diagrams to help explain the argument.
The explanation goes like this.
Diagram (1) shows the perimeter is a kite, with AC a mirror line about which it is symmetrical. So distance AEDC is half the length we are trying to find.
In diagram (2) the rectangle outlined in blue is a scaled-down copy of the 1 by square root 2 rectangle we began with, and because it has the same proportions DC must be square root 2 times as long as ED.
In diagram (3) ED is calculated as the difference between square root 2 and 1, and multiplying ED by square root 2 gives the length of DC as 2 - square root 2. Adding AD and DC as shown in the box gives 2, and the whole perimeter is twice this, so it is 4.