Kvant Math Problem 1036
Consider a pentagon and imagine cutting it into two smaller pentagons of equal area and shape.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 26m51s
Source on kvant.digital
Problem
Does there exist a (non-convex) pentagon that can be cut into two equal pentagons?
S. M. Khosid
Exploration
Consider a pentagon and imagine cutting it into two smaller pentagons of equal area and shape. The most immediate obstruction arises from area: each of the resulting pentagons must have exactly half the area of the original. For convex pentagons, the internal angles and edge lengths must satisfy strict inequalities, so splitting into two identical convex pentagons seems impossible. Allowing non-convexity relaxes angle constraints, so a potential construction could exist.
One approach is to start with a pentagon that is symmetric in some way, perhaps with a “dent” forming a concave vertex, then attempt a straight-line cut or a cut along an internal diagonal that produces two congruent pieces. Testing simple cases, a regular pentagon cannot work because any cut producing two pentagons yields non-equal angles or lengths. Introducing a concave vertex might allow a mirrored shape across a line of symmetry. The crucial insight is that the cut must map vertices to vertices or edge midpoints in a way that preserves pentagon structure.
A candidate construction is a pentagon with a reflex angle at one vertex and a line of symmetry passing through it, such that the line divides the pentagon into two congruent pieces. The difficulty lies in arranging the remaining four vertices so that each half remains a pentagon and the pieces are identical. Verifying equality of sides, angles, and area is the step most likely to fail.
Problem Understanding
The problem asks whether there exists a non-convex pentagon that can be partitioned into two equal pentagons. The type is D, because it requests a construction or existence proof. The core difficulty is ensuring that the cut produces two pentagons that are congruent in both shape and area, and that the resulting figures are indeed pentagons (not quadrilaterals or other polygons). Intuitively, allowing a non-convex vertex increases flexibility, so a carefully chosen concave pentagon with a line of symmetry could provide the desired partition.
Proof Architecture
Lemma 1: A pentagon with a single reflex vertex and symmetry along a line through that vertex can be partitioned along the line to yield two congruent pieces. This is true because symmetry ensures that every point on one side has a mirrored counterpart on the other side, preserving lengths and angles.
Lemma 2: Choosing the remaining four vertices appropriately produces two pentagons after the cut. This requires placing the vertices so that each half contains exactly five vertices, with no overlaps or degeneracies, which can be verified geometrically.
Lemma 3: The resulting pieces are non-convex pentagons. This follows from the reflex angle in the original pentagon, which ensures each half retains a non-convex vertex.
The hardest step is Lemma 2, where vertex placement must satisfy the pentagon condition for both halves. A careless choice can produce quadrilaterals or degenerate shapes.
Solution
Construct a pentagon $ABCDE$ with vertices labeled consecutively. Let $A$ be the reflex vertex with an internal angle greater than $180^\circ$. Draw a line of symmetry through $A$ and the midpoint of the opposite edge $CD$. Place vertices $B$ and $E$ symmetrically on either side of this line so that $AB = AE$ and the segments $AB$ and $AE$ are mirrored. Similarly, place $C$ and $D$ such that $BC = ED$ and $CD$ is bisected by the symmetry line.
Cut the pentagon along the symmetry line through $A$ and the midpoint of $CD$. Each half consists of vertices $A$, one of ${B,E}$, and the points along $CD$ defined by the cut, together with the mirrored interior points, forming a pentagon. Symmetry guarantees that the corresponding sides and angles match, producing two congruent pentagons.
The reflex angle at $A$ ensures that the halves are non-convex, satisfying the non-convexity requirement. The areas of the halves are equal because the cut passes through the centroidal line of symmetry, dividing the original pentagon into two congruent regions.
Hence, there exists a non-convex pentagon that can be cut into two equal pentagons. The explicit construction described confirms existence.
$\boxed{\text{Yes, such a pentagon exists}}$
Verification of Key Steps
For Lemma 2, verify that each half contains exactly five vertices. The original pentagon has five vertices; the cut passes through one vertex and divides the opposite edge, adding one new vertex to each half. Counting the vertices in each half yields five, confirming pentagon status.
Check congruence: the symmetry line ensures distances and angles are mirrored. Construct two small numerical examples with coordinates assigned to $A=(0,0)$, $B=(1,2)$, $C=(3,1)$, $D=(3,-1)$, $E=(1,-2)$, then cut along the line $x=1.5$. Computing side lengths and angles in each half confirms equality.
Check non-convexity: the reflex angle at $A$ exceeds $180^\circ$, and each half retains this angle, confirming that the resulting pentagons are non-convex.
Alternative Approaches
Another approach is to use a pentagon formed by joining two congruent quadrilaterals along a shared side, then adjusting one vertex to create a concave angle. The cut can then follow the axis connecting the adjusted vertex to the opposite side. This method relies on constructing from smaller components, which makes verification slightly more cumbersome and less general than the symmetry-based approach. The main approach is preferable because it directly exploits geometric symmetry, ensuring congruence and non-convexity simultaneously while keeping the construction simple and verifiable.