Our college used to run a math circle before covid hit. It was officially rejuvenated last week. A day of introductory sessions meant to encourage an interest in investigating mathematical puzzles among the potential participants was arranged. I sat through almost all of them, and boy, was it worth it.
Some of the participants, who were probably pre-high school, had more mathematical insights than I did. 😅
Anyway, before the actual sessions started, one of our teachers showed a small puzzle as sort of an appetizer.
It showcased a way of thinking about an answer to a problem like Polya mentions in How to solve it:
If you cannot solve the proposed problem, try to solve first some related problem.
The puzzle looked difficult when I looked at it, but the solution felt so obvious after the answer was revealed to us.
Solution to this puzzle may be evident for some, but it certainly wasn't the case for me at first. And it's a simple puzzle, so I thought I would write about it.
Consider a square with 3 pairs of points laid out like this:
A
+-----------------------------+
| |
| |
| |
| |
| |
| |
| |
B| C' B' |C
| |
| |
| |
| |
| |
| |
| |
+-----------------------------+
A'
A and A' are aligned along the same vertical line. And the points B, B', C, C' are aligned such that they fall on the same horizontal line.
The question is, can we connect the following pairs of points without any of the connections crossing each other?
The lines/curves connecting the point pairs cannot go outside the square.
It should be noted that the square itself doesn't connect any points. It's just there as a 'frame'.
ie, for example, A and B are not connected by the edges of the square.
When the puzzle was posed, I couldn't go beyond connecting two pairs of points.
Like this:
A
+-----------------------------+
| |
| |
| |
| |
| |
| +------+ |
| / \ |
B|------+ C B' +-------|C
| \ / |
| +-------+ |
| |
| |
| |
| |
| |
+-----------------------------+
A'
Because now the BB' and CC' seem to block the way for us to have AA'.
We had a minute or two before the solution was revealed, but I still couldn't see a way out.
All I couldn't think of way drawing a connection that went outside the square. Like:
+---------------------+
| \ A
| +--------------+--------------+
| | |
| | |
| | |
| | |
| | |
| | +------+ |
| | / \ |
| B|------+ C B' +-------|C
| | \ / |
| | +-------+ |
| | |
| | |
| | |
| | |
| | |
| +--------------+--------------+
| / A'
+---------------------+
But this isn't a solution since the connections cannot go outside the square.
Okay, we couldn't make much headway in solving the original puzzle.
But remember what was mentioned in How to solve it:
If you cannot solve the proposed problem, try to solve first some related problem.
All right, a related problem it is then.
Here is a simpler, related problem:
A
+-----------------------------+
| |
| |
| |
| |
| |
| |
| |
B| B' C' |C
| |
| |
| |
| |
| |
| |
| |
+-----------------------------+
A'
This is a cakewalk as after drawing the BB' and CC', we get some space through which we can draw AA'.
A
+-----------------------------+
| | |
| | |
| | |
| | |
| | |
| | |
| | |
B|----------B' | C'---------|C
| | |
| | |
| | |
| | |
| | |
| | |
| | |
+-----------------------------+
A'
Now we'll see if we can use the solution to this problem to find a solution to our original problem.
Okay, we got a solution to the simpler problem ready.
Let's see if we can use it to arrive at a solution for the original problem.
Try moving the points in the simpler problem a bit, adjusting the connections between the points accordingly.
First, let's move B' a bit to the top-right.
(To make the diagram clearer, I'm making the AA' connection in bold).
A
+-----------------------------+
| ┃ |
| ┖────────┒ |
| ┃ |
| +--------B' ┃ |
| | ┃ |
| | ┎────────┚ |
| | ┃ |
B|---------+ ┃ C'---------|C
| ┃ |
| ┃ |
| ┃ |
| ┃ |
| ┃ |
| ┃ |
| ┃ |
+-----------------------------+
A'
Then C' toward bottom-left.
A
+-----------------------------+
| ┃ |
| ┖────────┒ |
| ┃ |
| +--------B' ┃ |
| | ┃ |
| | ┎────────┚ |
| | ┃ |
B|---------+ ┃ +-----|C
| ┃ | |
| ┎─────────┚ | |
| ┃ | |
| ┃ C-------------+ |
| ┃ |
| ┖─────────┒ |
| ┃ |
+-----------------------------+
A'
Now, move C' to its intended position in the left part of the square:
A
+-----------------------------+
| ┃ |
| ┖────────┒ |
| ┃ |
| +---------------B' ┃ |
| | ┃ |
| | ┎────────────────┚ |
| | ┃ |
B|--+ ┃ C' +-----|C
| ┃ | | |
| ┃ | | |
| ┃ | | |
| ┃ +-------------+ |
| ┃ |
| ┖───────┒ |
| ┃ |
+-----------------------------+
A'
Likewise, B' to its intended position on the right part of square:
A
+-----------------------------+
| ┃ |
| ┖──────┒ |
| ┃ |
| +-------------+ ┃ |
| | | ┃ |
| | ┎──────┒ | ┃ |
| | ┃ ┃ | ┃ |
B|---+ ┃ C ┃ B' ┃ +---|C
| ┃ | ┃ ┃ | |
| ┃ | ┖───────┚ | |
| ┃ | | |
| ┃ +---------------+ |
| ┃ |
| ┖───────┒ |
| ┃ |
+-----------------------------+
A'
And we end up with a solution to our original problem.
That was it!
After knowing the answer, it looks too simple. But it certainly didn't look simple when we first encountered it. 😊
Acknowledgements: Thanks to Jaikrishnan sir for exposing us to this puzzle and for explaining a way (á la Polya) to think about finding an answer to it.
The original puzzle involved circle instead of square.
A A
▄▄▄▄▄▄▄▄▄ ▄▄▄▄▄▄▄▄▄
██ ██ ██ o ██
██ ██ ██ | ██
█ █ █ | █
█ +-------+ █ █ | █
█ / \ █ █ | █
B █--+ C' B' +--█ C B █------B' | C'-----█ C
█ \ / █ █ | █
█ +-------+ █ █ | █
█ █ █ | █
██ ██ ██ | ██
██ ██ ██ | ██
▀▀▀▀▀▀▀▀▀ ▀▀▀▀▀▀▀▀▀
A' A'
But I went with square because it's faster to make ASCII art of squares when compared to circles. Though the artist-mode of emacs makes it easier even for circles. 😃