Kvant Math Problem 1477
The phrase “cut off a similar one” is interpreted in the standard geometric sense used in olympiad problems: from a convex polygon one performs cuts along straight lines intersecting adjacent sides, p…
Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m51s
Source on kvant.digital
Problem
M1477. Does there exist a convex a) pentagon, b) $n$-gon, from which one can cut off a similar one?
Exploration
The phrase “cut off a similar one” is interpreted in the standard geometric sense used in olympiad problems: from a convex polygon one performs cuts along straight lines intersecting adjacent sides, producing a smaller convex polygon similar to the original. Each vertex is truncated by a line segment joining points on its two incident sides, so that the remaining inner polygon has the same number of sides and the same cyclic order.
A natural candidate construction is a polygon whose inward parallel shift produces a similar figure. For a general convex polygon this fails because shifting each side inward by a fixed distance preserves angles but does not preserve ratios of adjacent side lengths unless all angles and all side-length relations are rigidly uniform. A regular polygon is the unique symmetric candidate where such uniformity holds.
For a regular $n$-gon, truncating each vertex by a line parallel to the opposite side produces a smaller regular $n$-gon, hence a similar one. This works for every $n \ge 3$, including $n=5$.
The key potential pitfall is whether the truncation procedure can be interpreted consistently so that all sides scale by a common factor. For a regular polygon this follows from rotational symmetry about its center, which forces any construction respecting cyclic adjacency to be invariant under the dihedral group, hence homothetic.
Thus the existence question reduces to the existence of regular polygons.
Problem Understanding
The problem asks whether there exists a convex pentagon, and more generally a convex $n$-gon, from which one can cut off a smaller polygon similar to the original.
This is a Type A classification problem.
A natural construction suggests that regular polygons should work, since inward truncation preserves similarity. This immediately gives examples for every $n \ge 3$, including $n=5$.
The conclusion expected is that such polygons do exist for both cases.
Proof Architecture
The proof will proceed by explicit construction.
First, a lemma establishes that truncating a regular $n$-gon by lines parallel to its sides at a fixed distance produces another regular $n$-gon.
Second, a lemma shows that any two regular $n$-gons are similar.
Third, these facts combine to show that the inner polygon obtained by cutting off corners is similar to the original.
The only delicate point is verifying that all side lengths of the inner polygon are equal, which requires explicit symmetry or trigonometric justification.
Solution
Let a regular $n$-gon $A_1A_2\ldots A_n$ be given, centered at a point $O$. Each side is obtained from the previous one by rotation around $O$ by angle $2\pi/n$.
Fix a number $t$ with $0<t<1$. On each side $A_iA_{i+1}$ choose a point $B_i$ such that
$$\frac{A_iB_i}{A_iA_{i+1}} = t.$$
Connect consecutive points $B_i$ and $B_{i+1}$. This produces a convex $n$-gon $B_1B_2\ldots B_n$ inside the original.
Consider the rotation $R$ around $O$ by angle $2\pi/n$. It maps $A_i$ to $A_{i+1}$ and preserves ratios along corresponding sides, hence it maps $B_i$ to $B_{i+1}$. Therefore the polygon $B_1B_2\ldots B_n$ is invariant under the same rotational symmetry group as the original polygon.
A convex polygon invariant under rotation by $2\pi/n$ around its center must have all vertices on a circle centered at $O$, and successive vertices are obtained by the same rotation. Consequently all sides of $B_1B_2\ldots B_n$ are equal, and all its angles are equal, so it is a regular $n$-gon.
Since any two regular $n$-gons are similar, the polygon $B_1B_2\ldots B_n$ is similar to $A_1A_2\ldots A_n$.
The construction uses only truncations of vertices by straight segments $B_iB_{i+1}$, so the inner polygon is obtained from the original by cutting off $n$ corner regions. Hence from the original polygon one can cut off a similar one.
This construction works for every $n \ge 3$, in particular for $n=5$ and for arbitrary $n$.
Thus such a convex pentagon exists, and such a convex $n$-gon exists for all $n \ge 5$.
Verification of Key Steps
The critical point is that rotational symmetry forces regularity of the inner polygon. If a convex polygon is invariant under rotation by $2\pi/n$ about a point $O$, then applying this rotation repeatedly sends any vertex to all other vertices, so all vertices lie on a common circle centered at $O$. Equal angular spacing implies equal chord lengths between consecutive vertices, hence all sides are equal.
A potential failure would occur if the constructed points $B_i$ did not respect the rotational symmetry. However, since each $B_i$ is defined by the same ratio $t$ on corresponding sides $A_iA_{i+1}$, the rotation carries each defining configuration to the next, ensuring invariance.
Another delicate point is whether connecting consecutive $B_i$ produces a convex polygon. Convexity follows because each $B_i$ lies on a boundary segment of a convex set and the construction preserves cyclic order without introducing intersections.
Alternative Approaches
An alternative proof uses affine coordinates with the regular $n$-gon centered at the origin in the complex plane. Writing vertices as $z_k = e^{2\pi i k/n}$, the points $B_k$ are affine combinations of $z_k$ and $z_{k+1}$, which yields a factorization showing that $B_k$ differ from $z_k$ by a common complex scalar depending only on $t$ and $n$. This directly establishes similarity via a homothety composed with a rotation.