Kvant Math Problem 283
Consider small examples of convex polygons, starting with triangles and quadrilaterals, and examine what happens when each side is shifted outward by a fixed distance.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m07s
Source on kvant.digital
Problem
A convex polygon has the following property: if all its sides are shifted outward by one unit, then the resulting lines form a polygon similar to the original one. Prove that a circle can be inscribed in this polygon.
Moscow Mathematical Olympiad (1974)
Exploration
Consider small examples of convex polygons, starting with triangles and quadrilaterals, and examine what happens when each side is shifted outward by a fixed distance. For a triangle, moving each side outward by one unit produces a larger triangle with parallel sides, which is always similar to the original. For a square, shifting each side outward by one unit produces another square with the same orientation, hence similar. For a rectangle that is not a square, shifting each side outward yields a rectangle of larger side lengths, which is similar but with a different aspect ratio, so similarity may fail unless the sides are equal. This suggests that for polygons with more than three sides, equal angles and proportional side growth are necessary conditions.
The key difficulty lies in connecting the property of side shifts preserving similarity to the existence of an incircle. One can suspect that uniform proportional growth in all directions implies that the polygon is tangential, that is, admits an inscribed circle. Testing this on a square or regular polygon confirms that a circle can be inscribed. Attempting a non-tangential quadrilateral, such as an arbitrary convex quadrilateral with unequal sums of opposite sides, reveals that shifting the sides outward does not yield a polygon similar to the original, suggesting that the tangency condition is forced by the similarity property.
The core insight is that outward shifts correspond to moving the sides along their normals and that similarity requires proportional distances from a common center, which is exactly the condition for tangency to a circle.
Problem Understanding
The problem asks to prove that a convex polygon whose sides, when shifted outward by one unit, form a polygon similar to the original, must admit an inscribed circle. This is a Type B problem, "Prove that [statement]", because the polygon is given with a property, and the task is to establish the existence of an incircle. The core difficulty is to rigorously show that the similarity of the shifted polygon forces equal distances from a center to each side, which is the defining property of a tangential polygon. The intuitive reason the statement should hold is that a polygon whose sides expand uniformly in all directions while preserving similarity must have sides arranged symmetrically around a center, precisely as in a tangential polygon.
Proof Architecture
Lemma 1: For any convex polygon, shifting all sides outward by a fixed distance produces a polygon whose side directions are unchanged, and each side moves along its outward normal. This is true by the geometric definition of translating lines.
Lemma 2: If the shifted polygon is similar to the original, then all vertices scale from a common center with a positive similarity ratio. This follows from the definition of similarity: corresponding angles are equal, and corresponding distances are proportional.
Lemma 3: For a convex polygon, if shifting sides outward along their normals preserves similarity, then the distances from the center of similarity to each side are equal. This is because the scaling factor must be uniform, and the shift is the same for all sides, so the initial distances from the center to the sides must be equal to maintain proportionality.
Lemma 4: A convex polygon with equal distances from a common point to all sides admits an inscribed circle centered at that point. This is the standard characterization of tangential polygons.
The hardest step is Lemma 3, where it must be proved that uniform outward shifts and similarity force equality of distances to a center. A careless argument could assume this without showing the algebraic constraints on the side normals and vertices.
Solution
Let $P$ be a convex polygon with vertices $A_1, A_2, \dots, A_n$ in order and sides $S_i = A_i A_{i+1}$ with $A_{n+1} = A_1$. Consider the operation of shifting each side $S_i$ outward by one unit along its outward normal, producing a new polygon $P'$. By construction, each side $S_i$ moves along a line perpendicular to $S_i$, and the direction of $S_i$ remains unchanged. Therefore the sides of $P'$ are parallel to the corresponding sides of $P$, as asserted in Lemma 1.
Since $P'$ is similar to $P$ by hypothesis, there exists a point $O$ and a positive scaling factor $k$ such that each vertex $A_i'$ of $P'$ is obtained from $A_i$ by the similarity transformation: $A_i' = O + k(A_i - O)$, as stated in Lemma 2. The position of each $A_i'$ also satisfies $A_i'$ lying on the line obtained by shifting $S_i$ outward by one unit along its normal. Denote the perpendicular distance from $O$ to side $S_i$ as $d_i$. The shift moves $S_i$ by one unit, while similarity scales all distances from $O$ by $k$. Therefore, the distance from $O$ to $S_i'$ equals $k d_i$. On the other hand, the shift along the normal increases the distance from $O$ by exactly one unit: $d_i + 1 = k d_i$. Solving this equation for $d_i$ gives
$$d_i + 1 = k d_i \implies d_i (k - 1) = 1 \implies d_i = \frac{1}{k - 1}.$$
Since this equation holds for every side $S_i$, all distances $d_i$ are equal. This establishes Lemma 3, as it shows that a single point $O$ exists from which all sides are equidistant.
Finally, Lemma 4 implies that a convex polygon whose sides are all at equal distance from a point $O$ admits an inscribed circle centered at $O$ with radius $r = d_i$. Therefore, $P$ admits an incircle.
This completes the proof.
∎
Verification of Key Steps
The most delicate step is showing $d_i + 1 = k d_i$ for each side. This is verified by considering the geometric constraints: the outward shift adds exactly one unit along the normal, and similarity scales all distances from the center by $k$. Testing with a square of side length $a$, if $O$ is the center and $d = a/2$, shifting by one unit produces a square similar to the original. Then $d + 1 = k d$ gives $k = 1 + 2/a$, consistent with the geometric construction. This confirms the equation works in at least one concrete example and justifies the general argument.
Another step is ensuring that the solution $d_i = 1/(k-1)$ is positive. Since the polygon is convex and the shift outward increases distances, $k > 1$, which guarantees $d_i > 0$ and a proper circle can be inscribed. A failure to check $k > 1$ could result in an invalid negative radius.
Alternative Approaches
An alternative approach is to consider the polygon as the intersection of half-planes corresponding to its sides and use linear algebra to describe the outward shift and similarity as affine transformations. By examining the system of equations defining distances from a common center to each side, one can directly prove that equality of distances is necessary. This approach emphasizes coordinates and algebraic manipulation, but the main geometric method is preferable for clarity and minimal computation, as it directly connects similarity and outward shifts to the standard characterization of tangential polygons.