Scaling a Square

Source: Saurabh Joshi, IIT Kanpur

Problem: On a table you have a square made of 4 coins at the corner at distance 1. So, the square is of size 1×1. In a valid move, you can choose any two coin let’s call them mirror and jumper. Now, you move the jumper in a new position which is its mirror image with respect to mirror. That is, imagine that mirror is a centre of a circle and the jumper is on the periphery. You move the jumper to a diagonally opposite point on that circle. With any number of valid moves, can you form a square of size 2×2? If yes, how? If no, why not?

Update (November 4, 2011)
Solution: Posted by Siddhant Agarwal (EE IITB Alumnus, CMI Grad student) and Rudradev Basak (IITD CSE Senior Undergraduate) in comments!


Comments

  1. sid8:25 PM, November 03, 2011

    nice problem :)

    We observe that the parity of the L1 distance between any two coins is an invariant. In 1×1, for any coin, the L1 distance to its neighbors is odd. While in 2×2 every L1 distance between coins is even. Hence not possible

    ReplyDelete
  2. Rudradev Basak11:57 PM, November 03, 2011

    No.
    If I can go from a 1x1 square to a 2x2 square, then by applying reverse operations, I can also go from a 2x2 square to a 1x1 square.
    By applying scaling this means we can also go from a 1x1 square to a 0.5x0.5 square. But this is impossible since if initially all coins are on lattice points, then they remain on lattice points forever. So this leads to a contradiction.

    ReplyDelete
  3. sid3:12 PM, November 04, 2011

    brilliant argument!

    ReplyDelete
  4. Pratik Poddar5:28 PM, November 04, 2011

    Correct solutions by both sid and Rudradev Basak. Thanks. Saurabh Joshi (whose blog this problem is taken from) gave Rudra's solution and I could also think of that only. Sid's solution seems fresh :)

    ReplyDelete
  5. Pratik Poddar9:35 AM, November 05, 2011

    On second thoughts, Rudradev's solution (i.e. Saurabh's solution) is more general. It even works if I am asked to make a 3x3 square. While parity argument is just for even x even squares.

    ReplyDelete
  6. H2:02 AM, April 19, 2013

    So this also extends to any r by r initial square, r \in R? Then there must a plain geometry proof other than Rudradev's?

    ReplyDelete
  7. Pratik Poddar2:13 PM, September 09, 2026

    Careful with the extension: the invariant behind the impossibility is lattice-theoretic, not "plain geometry". A move sends jumper P to 2M - P, so (a) every coin stays on the integer lattice Z^2 forever, and (b) each coin keeps its coordinate parities (P' = 2M - P is congruent to P mod 2). Fact (a) already restricts reachable squares to those with lattice vertices, forcing side^2 = r^2 to be a sum of two squares of integers. Fact (b) kills the even-sided cases: initially the four coins occupy all four parity classes, while a 2x2 lattice square has all corners in ONE class. Rudradev's argument handles the remaining cases (like 3x3). I do not know a purely Euclidean proof - the arithmetic of the lattice is where the obstruction lives. (replied using AI)

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