🔷 Interactive Polygon Bisector 🔷

Step-by-Step Equal Area Partition
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Step 1: Generate a Polygon
Click "Generate Polygon" to create a random convex polygon

Area Information

Total Polygon Area
0.00
Left Area (Red Region)
0.00
Right Area (Green Region)
0.00

The Algorithm Explained

This interactive demonstration proves that any convex polygon can be divided into two regions of equal area by a straight line.

Interactive Steps:

  1. Generate a Polygon: Click the button to create a random convex polygon.
  2. Choose Mode: Select Manual Mode to explore yourself, or Auto Mode to watch the computer solve it.
  3. Select Pivot Vertex $v$: Click on any vertex to use as your pivot point (shown in red).
  4. Explore Each Line: Draw lines from $v$ to each other vertex $u_1, u_2, \ldots, u_{n-1}$ in sequence. Watch how the left area changes!
  5. Watch the Areas: The left area starts at 0 and grows. When it crosses half the total area, we've found our transition point!
  6. Find the Bisector: Once you've found where the area transitions, compute the precise point $w$ on the edge that creates perfect equal areas.

Mathematical Foundation:

The key insight is that as we rotate a line through vertex $v$, sweeping across all other vertices, the area to one side of the line changes continuously from 0 to the full polygon area. By the Intermediate Value Theorem, there must be some orientation where the area equals exactly half!

Area Function: Let $A(\theta)$ be the area to the left of a line through $v$ at angle $\theta$. Then:

$$A(0) = 0 \quad \text{and} \quad A(2\pi) = A_{\text{total}}$$

By continuity, there exists some angle $\theta^*$ where:

$$A(\theta^*) = \frac{A_{\text{total}}}{2}$$

When we find vertices $u_k$ and $u_{k+1}$ where the area transitions across the halfway point, we can find the exact point $w$ on edge $\overline{u_k u_{k+1}}$ using linear interpolation:

$$w = u_k + t \cdot (u_{k+1} - u_k)$$

where $t \in [0,1]$ is chosen so that the area of the left region equals exactly $\frac{A_{\text{total}}}{2}$.

The Ham Sandwich Theorem:

This result generalizes to higher dimensions through the celebrated Ham Sandwich Theorem, which states that for any $n$ measurable sets in $\mathbb{R}^n$, there exists a hyperplane that simultaneously bisects each set.

The theorem gets its whimsical name from the 3D case: given a sandwich consisting of two slices of bread and a piece of ham (possibly irregularly shaped), it's always possible to cut the sandwich with a single planar slice so that each component is exactly halved!

Historical Note: The Ham Sandwich Theorem was first proposed by Hugo Steinhaus and proved in full generality by Stefan Banach in 1938. Banach's contribution to this problem showcases his profound influence on functional analysis and geometric measure theory. The proof uses the Borsuk-Ulam theorem, which states that any continuous map from an $n$-sphere to $\mathbb{R}^n$ must identify antipodal points.

This works for any polygon, even non-convex ones! The principle extends beautifully across mathematics, demonstrating the power of continuity arguments and topological methods in solving geometric problems.