The complexity and beauty of the Affine Plank Conjecture

Zaz Brown

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Convex bodies

The Affine Plank Conjecture is about planks covering convex bodies.

A convex body is a shape with no inward curves, like the one shown in red on the leftbelow.

to make it convex.

Planks and relative width

A plank is a region between two parallel lines (or two parallel hyperplanes in higher dimensions).

A plank's relative width is its width divided by the width of the convex body, measured perpendicular to the plank.

The Affine Plank Conjecture

In 1951 Thøger Bang conjectured that if planks cover a convex body, their relative widths must sum to at least $100%$ .

But nobody has been able to prove this for 75 years.

40 % + 39 % + 37 % = 116 %

Grab planks to , , and .

1 parameter

x position · w relative width · θ angle

Gap cells

We call the uncovered regions gap cells.

There are 8 possible gap cells.

The Affine Plank Conjecture is equivalent to saying that when the relative widths of the planks sum to less than one, there is always a gap cell with positive area.

Gap cells

Grab planks to , , and .

g_i(x₁, w₁, θ₁, x₂, w₂, θ₂, x₃, w₃, θ₃) is the area of gap i.

Each triple is a plank’s position, relative width, and angle.

% of triangle covered

Gap-cell formulas

The slide shows one variable at a time, with the other eight fixed. The full formula accounts for all nine.

Parameters

Each plank has a position x, a relative width w, and an angle θ. Subscripts identify the purple, green, and red planks. Formula angles are in radians; the live inputs display degrees.

q = (x_{1}, w_{1}, θ_{1}, x_{2}, w_{2}, θ_{2}, x_{3}, w_{3}, θ_{3})

For the rational angle formulas, $t_{j} = \tan (\frac{θ_{j}}{2})$ , for plank $j$ .

Full explicit formula

These eight equations cover all configurations, including empty gaps. Each value is the fraction of the triangle occupied by that gap.

F and every symbol it uses are defined below.

1. Plank boundaries

A triple (a, b, c) means the half-plane au + bv ≤ c. P_j⁰ and P_j¹ describe the two sides of plank j. Including the boundary does not change its area.

n_{j} = (- \sin θ_{j}, \cos θ_{j})

W_{j} = \max_{k \in {1, 2, 3}} (n_{j} \cdot v_{k}) - \min_{k \in {1, 2, 3}} (n_{j} \cdot v_{k})

s_{0} = 1, s_{1} = - 1

P_{j}^{σ} = (s_{σ} n_{jx}, s_{σ} n_{jy}, s_{σ} x_{j} - \frac{w_{j} W_{j}}{2})

2. Triangle boundaries

The vertices v₁, v₂, v₃ run counterclockwise. Set v₄ = v₁. Its three outward boundary triples are:

α_{r} = v_{r+1,y} - v_{r,y}, β_{r} = v_{r,x} - v_{r+1,x}

T_{r} = (α_{r}, β_{r}, α_{r} v_{r,x} + β_{r} v_{r,y})

3. Shared area formula and all cases

For F(P₁, P₂, P₃), take the six triples in the order (T₁, T₂, T₃, P₁, P₂, P₃). Write the rth triple as (a_r, b_r, c_r).

F (P_{1}, P_{2}, P_{3}) = \frac{\sum_{r = 1}^{6} c_{r} e_{r}}{2 | K |}

The shared expressions below give e_r. Both r and s range from 1 to 6.

Q_{r} = {a_{r}}^{2} + {b_{r}}^{2}

D_{rs} = a_{r} b_{s} - a_{s} b_{r}

B_{rs} = c_{s} - \frac{c_{r} (a_{r} a_{s} + b_{r} b_{s})}{Q_{r}}

Each of the other inequalities puts a lower or upper bound on the part of line r that can remain. Take the strongest bounds:

t_{r}^{-} = \max_{s : D_{rs} < 0} \frac{B_{rs}}{D_{rs}}

t_{r}^{+} = \min_{s : D_{rs} > 0} \frac{B_{rs}}{D_{rs}}

Both sets have a member because the triangle is bounded. Let X_r mean a parallel inequality excludes line r, and R_r mean an earlier line repeats the same outward boundary:

X_{r} \Leftrightarrow \exists s : D_{rs} = 0 \land B_{rs} < 0

R_{r} \Leftrightarrow \exists s < r : (\begin{array}{l} D_{rs} = 0 \land B_{rs} = 0 \\ \land a_{r} a_{s} + b_{r} b_{s} > 0 \end{array})

The complete piecewise rule is:

e_{r} = {\begin{cases} 0 & if X_{r} \lor R_{r} \\ \max (0, t_{r}^{+} - t_{r}^{-}) & otherwise \end{cases}

R_r counts coincident outward boundaries only once. Opposite boundaries are retained: if the gap collapses to a line, their area contributions cancel.

This uses only arithmetic, trigonometric functions, finite minima and maxima, and the written conditions. With angles fixed, it is piecewise quadratic in positions and widths.

Why this gives the area

Parameterize boundary r by:

z_{r} (t) = \frac{c_{r}}{Q_{r}} (a_{r}, b_{r}) + t (- b_{r}, a_{r})

Substitution in inequality s gives D_rst ≤ B_rs, which explains the bounds. A surviving edge runs from t_r⁻ to t_r⁺. Its oriented shoelace contribution is c_re_r. Summing the edges, dividing by 2, and dividing by |K| gives the area fraction. An empty intersection has no surviving edges; a point or line has zero area.

Current pieces and their boundaries

These are the pieces active at the current configuration. Their boundary conditions are listed below each formula.

Symbols

For each gap, lines 1–3 are the triangle boundaries and lines 4–6 are the chosen sides of the three planks. Line r is a_ru + b_rv = c_r. Its corners are intersections of the listed lines, inside all six half-planes.

D_{rs} = a_{r} b_{s} - a_{s} b_{r}

U_{rs} = \frac{c_{r} b_{s} - c_{s} b_{r}}{D_{rs}}, V_{rs} = \frac{a_{r} c_{s} - a_{s} c_{r}}{D_{rs}}

E_{rs,tu} = U_{rs} V_{tu} - V_{rs} U_{tu}

|K| is the triangle's area. The coefficients depend on the angles; with angles fixed, each piece is quadratic in positions and widths.

Solving the equations

Grab planks to , , and .

So all that remained was to set the area of every gap cell equal to zero and solve.

As a computer scientist, I thought: finding an intersection of sets is easy.

For finite sets, yes, but not so for sets of real numbers.

For polynomial inequalities, a standard general method is cylindrical algebraic decomposition (CAD)…

But its worst-case running time is doubly exponential in the number of variables, so the naive approach is not feasible, even on the triangle.

3 planks

Thank you for exploring the
Affine Plank Conjecture

Read the proof of the Affine Plank Conjecture for three planks in the plane

Zaz Brown

The complexity and beauty of the Affine Plank Conjecture

Convex bodies

Planks and relative width

The Affine Plank Conjecture

Gap cells

Gap-cell formulas

1. Plank boundaries

2. Triangle boundaries

3. Shared area formula and all cases

Why this gives the area

Symbols

Solving the equations

Thank you for exploring theAffine Plank Conjecture

Thank you for exploring the
Affine Plank Conjecture