This week the #puzzle is: Can You Squeeze the Squares? #optimal #placements #packing

There’s a square board with side length A. Your friend cleverly places a unit square on the board and challenges you to place another unit square on the board—without moving the first one—so that it too is entirely on the board and the squares don’t overlap. (The unit squares can touch each other.)

Alas, it’s impossible for you to do so! But there’s some minimum value of A for which you can always place a second unit square on the board, no matter how cleverly your friend places the first one.

What is this minimum value of A?

And for extra credit:

Now there’s another square board with side length B. This time, your friend cleverly places three unit squares (which can touch but not overlap) and issues a similar challenge, asking you to place one more unit square on the board.

Once again, it’s impossible for you to do so! But there’s some minimum value of B for which you can always place a fourth unit square on the board, no matter how cleverly your friend places the first three squares.

What is this minimum value of B?

Can You Squeeze the Squares?

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“Last Week’s Fiddler. Congratulations to the (randomly selected) winner from last week: 🎻 Lise Andreasen 🎻 from Valby, Copenhagen, Denmark.” 😁😁😁

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Highlight to reveal (possibly incorrect) solution:

Desmos 1 Desmos pictures 1 and 2.

Some assumptions:

• The placement of the 1st square has to be symmetrical in some way. It will be in the middle. If it is not in the middle, some corner or side will leave more room for the 2nd square, and of course my friend doesn’t want that.
• The 1st square will either have sides parallel to the sides of the big square or 45 degrees shifted.

Some experimentation:

• As picture 1 shows, if the 1st, green square is tilted, A is about 2.7. As picture 2 shows, if the 1st, red square isn’t tilted, A is 3. (This picture also demonstrates, that the 2nd square shouldn’t be tilted.) My friend will choose not to tilt the square. A = 3.
• If A was slightly smaller, the ring around the 1st square wouldn’t be wide enough for an extra unit square to fit.

And for extra credit:

Desmos 2

Experimentation:

• A small square in the middle of the big square, aligned with the sides of the big square, dominates all 4 corners of the big square nicely. It’s very hard to come up with something using 2 extra squares, that could dominate a bigger square better. I therefore say as a 1st guess B >= 3.
• A tilted square might dominate 2 corners. 2 other aligned squares in the 2 other corners handle the rest. (Desmos shows 2 green squares, when there’s room for them.) In that case B = 2 + √2 = 3.4142.