Can You Squeeze the Squares?

A square board has side length $A$. Your friend cleverly places one unit square anywhere on the board. You must then place a second unit square, entirely on the board and not overlapping the first (touching is allowed), no matter how adversarially the first was placed. What is the smallest $A$ for which you can always succeed?

The Fiddler, Zach Wissner-Gross, August 1, 2025(original post)

Solution

The friend’s strongest move is to centre the unit square, leaving a symmetric frame of width $\tfrac{A-1}{2}$ on every side. A unit square needs a clear run of at least $1$ in some direction, and a rotated square only needs more room (its minimum width is $1$, achieved axis-aligned), so the largest square that fits in a corner has side exactly $\tfrac{A-1}{2}$. You can always place the second square precisely when $$\frac{A-1}{2}\ge 1\quad\Longleftrightarrow\quad A\ge\boxed{3}.$$ At $A=3$ the corner square $[0,1]^2$ is tangent to the centre square at $(1,1)$, which counts; below $3$ the centred square blocks every position.

The computation

Encode the placement test with the separating-axis theorem: two squares miss exactly when some edge-normal separates them. Then sweep a candidate second square over positions and angles to see whether any placement avoids the centred blocker and stays on the board. It fails just below $A=3$ and succeeds just above.

import numpy as np
def corners(cx, cy, th):
    h = .5; P = np.array([[-h,-h],[h,-h],[h,h],[-h,h]])
    R = np.array([[np.cos(th),-np.sin(th)],[np.sin(th),np.cos(th)]]); return P@R.T + [cx,cy]
def overlap(A, B, e=1e-9):
    for P in (A, B):
        for i in range(4):
            E = P[(i+1)%4] - P[i]; ax = np.array([-E[1], E[0]]); a = A@ax; b = B@ax
            if a.min() > b.max() - e or b.min() > a.max() - e: return False
    return True
def fits(bl, L, ng=80, na=16):
    cs = np.linspace(0, L, ng)
    for th in np.linspace(0, np.pi/2, na, endpoint=False):
        for cx in cs:
            for cy in cs:
                C = corners(cx, cy, th)
                if C.min() >= -1e-9 and C.max() <= L + 1e-9 and all(not overlap(C, b) for b in bl):
                    return True
    return False
for A in (2.9, 3.1):                  # 3.0 is the exact tangent threshold
    print(A, fits([corners(A/2, A/2, 0)], A))    # 2.9 -> False, 3.1 -> True

Extra Credit

Now the friend places three unit squares (which may touch but not overlap) on a board of side $B$, and you must add a fourth. What is the smallest $B$ for which you can always succeed?

Solution

Three blockers cannot shrink all four corners at once, so a purely central defence no longer pins the answer; the optimal arrangement is an off-centre, pinwheel-like trio. This is a delicate continuous min–max. A search gives a verified blocking arrangement at $B=3.3$ (no fourth square fits even under a fine position-and-angle sweep), while the defender succeeds throughout an extensive search by about $B\approx3.4$, so $$B\approx\boxed{3.4}\quad(\text{numerical estimate}).$$ (The exact value is paywalled; the bound below is my own.)

The computation

Reuse the same fits test on the pinwheel blocker at $B=3.3$: sweeping a $200\times200$ grid of positions over $60$ angles finds no legal spot for a fourth square.

bl = [corners(2.181, 2.233, 0.802), corners(1.629, 0.957, 0.835), corners(0.711, 2.559, 0.516)]
print(fits(bl, 3.3, ng=200, na=60))   # False -> no fourth square fits anywhere