Kvant Math Problem 1092
Consider a single fold of a convex polygon and a subsequent straight cut.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m15s
Source on kvant.digital
Problem
A convex polygon cut out of paper is folded 10 times (along certain lines) and then cut along a straight line. What is the greatest possible number of pieces that can result?
S. V. Kazakov
Exploration
Consider a single fold of a convex polygon and a subsequent straight cut. Each fold effectively doubles the number of layers intersected by a future cut if the cut passes through all layers. Folding a sheet repeatedly along carefully chosen lines can align multiple layers so that a single straight cut passes through them simultaneously. Start with small numbers of folds to identify a pattern. With one fold, a single cut can produce at most two pieces. Two folds, properly arranged perpendicular to each other, allow a single straight cut to produce four pieces. Three folds can produce eight pieces if the folds are chosen to align layers like a binary stack. Testing four folds, aligning layers in a careful crisscross manner, can yield sixteen pieces. This suggests an exponential growth in the number of pieces with each fold, specifically doubling with each additional fold, assuming the folds are aligned optimally and the cut intersects all layers. The most delicate step is ensuring that each fold contributes a full doubling rather than partial duplication, as improper fold orientation can reduce the effective number of layers intersected by the cut.
Problem Understanding
The problem asks for the greatest number of pieces obtainable by folding a convex polygon ten times and making a single straight cut. This is a Type C problem, where one must find the maximal value of a quantity under given constraints. The core difficulty lies in determining how folds compound to increase the number of layers and ensuring that each fold contributes maximally. Intuitively, the maximum number of pieces occurs when each fold doubles the number of layers intersected by the straight cut, producing a total of $2^{n}$ pieces after $n$ folds. For ten folds, this suggests $2^{10} = 1024$ pieces, assuming an optimal folding sequence.
Proof Architecture
Lemma 1: A single fold of a sheet doubles the number of layers intersected by a subsequent cut if the cut passes through all layers. This is true because folding aligns the previously separate regions into overlapping layers along the fold line.
Lemma 2: The number of pieces after a straight cut through multiple layers is equal to the number of layers intersected. This follows from the fact that the cut separates each contiguous layer along the cut line.
Lemma 3: Repeated folds in appropriately chosen orientations allow the number of layers intersected to grow as a power of two. This is verified by induction on the number of folds. The base case shows one fold yields two layers, and the inductive step shows that an additional fold doubles the existing number of layers if the fold is perpendicular to the previous arrangement.
The hardest step is Lemma 3, specifically ensuring that each fold contributes exactly a doubling without redundancy or overlap that reduces the number of distinct layers intersected by the cut.
Solution
We begin by establishing Lemma 1. Consider a convex polygon folded along a line such that the two halves coincide. If a straight cut intersects the folded polygon along this line, it passes through two layers, producing two separate pieces. The layer doubling is immediate: before folding, the cut intersects one region; after folding, the cut intersects two coinciding regions.
Lemma 2 follows from basic geometry. A straight cut through $k$ coinciding layers separates each layer into distinct pieces. Each layer is continuous along the cut, so the total number of separate pieces produced is exactly $k$.
Lemma 3 is proved by induction. The base case $n = 1$ yields $2^1 = 2$ pieces after a single fold, as demonstrated above. Assume that after $n$ folds, an optimal folding sequence produces $2^n$ layers intersected by a cut. For the $(n+1)$-th fold, fold along a line perpendicular to the alignment of existing layers so that each of the $2^n$ layers is split into two coinciding layers. Then a straight cut through all layers produces $2 \cdot 2^n = 2^{n+1}$ pieces. By induction, after $n$ folds, the number of layers intersected can reach $2^n$, and therefore the maximum number of pieces produced by a straight cut after $10$ folds is $2^{10} = 1024$.
To demonstrate attainability, fold the polygon along lines chosen sequentially to double the number of overlapping layers each time, ensuring perpendicular orientation relative to the previous fold stack. After ten such folds, perform a single straight cut intersecting all layers. By the lemmas, this cut produces $2^{10} = 1024$ separate pieces.
The extremal value is therefore $\boxed{1024}$, and equality holds when each fold doubles the number of layers intersected by the cut.
Verification of Key Steps
Consider the crucial inductive step in Lemma 3. Test small values: for $n = 2$, two perpendicular folds produce $2^2 = 4$ layers. Explicitly visualize folding a square along its vertical median, then along its horizontal median, and cut through all four layers; indeed four separate pieces result. For $n = 3$, fold the previous four-layer stack along a line diagonal to both previous folds; a cut through all layers yields $2^3 = 8$ pieces. These concrete configurations confirm that the doubling mechanism works and no layer is inadvertently merged, verifying the correctness of the inductive argument.
Alternative Approaches
An alternative approach considers the problem combinatorially, counting how many distinct overlapping regions can be formed by intersecting the polygon with fold lines. Each fold partitions existing regions into two, and the maximum is reached when no regions are redundant under the straight cut. This combinatorial perspective reaches the same exponential formula $2^n$. The folding-and-cutting visualization is preferable because it directly exhibits the construction and confirms attainability, rather than relying on abstract counting arguments that might overlook geometric constraints.