Kvant Math Problem 1018
Consider a regular $n$-gon $A_1 A_2 \dots A_n$ with center $O$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 9m33s
Source on kvant.digital
Problem
Let $A$ and $B$ be adjacent vertices of a regular $n$-gon with center $O$. The triangle $XYZ$, congruent to triangle $OAB$, initially coincides with it and then moves in the plane of the $n$-gon so that the points $Y$ and $Z$ remain on the boundary, while $X$ remains inside the $n$-gon. What figure will be traced out by the point $X$ when $Y$ and $Z$ make one complete circuit along the boundary of the $n$-gon?
International Mathematical Olympiad for School Students (XXVII)
Exploration
Consider a regular $n$-gon $A_1 A_2 \dots A_n$ with center $O$. Take two adjacent vertices $A = A_1$ and $B = A_2$, and the triangle $OAB$. Construct a congruent triangle $XYZ$ initially coinciding with $OAB$. If $Y$ and $Z$ are constrained to move along the polygon's boundary while $X$ remains inside, it is natural to think of $X$ as a point determined by the positions of $Y$ and $Z$ via a rigid motion, since congruence preserves distances.
For small $n$, say $n=3$ or $n=4$, tracing the motion of $X$ as $Y$ and $Z$ move along the edges gives curves that are clearly segments connecting centers of rotations induced by each edge. The symmetry suggests that $X$ traces out a scaled and rotated copy of the $n$-gon, smaller and centrally aligned. The most delicate point is ensuring that $X$ stays inside the $n$-gon for all placements of $Y$ and $Z$, which requires careful treatment of reflections when $Y$ and $Z$ cross a vertex.
Testing a triangle similar to $OAB$ with $n=4$, we see that as $Y$ moves along one edge and $Z$ along the adjacent edge, $X$ traces a straight line from the center to the midpoint of the edge opposite, hinting that in general $X$ traces another regular $n$-gon similar to the original, scaled by the ratio of distances from the center to the vertices in the initial triangle. The crux is proving this scaling and rotation rigorously for arbitrary $n$.
Problem Understanding
The problem asks for the locus of a point $X$ inside a regular $n$-gon when two points $Y$ and $Z$ of a triangle congruent to $OAB$ move along the polygon boundary. The problem is Type D because it requires constructing and identifying the figure traced by $X$ as $Y$ and $Z$ complete a circuit. The core difficulty lies in managing the congruence constraint while $Y$ and $Z$ traverse the polygon, especially at vertices where the edges meet. Intuitively, the traced figure should be a regular $n$-gon similar and centrally aligned to the original, because the initial triangle is rigid and the motion preserves symmetry.
Proof Architecture
Lemma 1: Any rigid motion of a triangle that keeps two vertices on a regular $n$-gon boundary determines the position of the third vertex uniquely. This is true because congruence preserves distances, and a rigid motion is determined by two points.
Lemma 2: As $Y$ and $Z$ traverse one side of the polygon, $X$ traces a straight line segment. This follows from Euclidean geometry: moving two points along a line while maintaining triangle congruence forces the third point to move along a parallel line segment.
Lemma 3: When $Y$ and $Z$ move around the polygon continuously, the locus of $X$ is composed of $n$ congruent linear segments joined end to end, forming a regular $n$-gon. This uses the rotational symmetry of the original polygon and the rigidity of the triangle.
Lemma 4: The traced $n$-gon is centrally aligned with the original polygon and scaled by the ratio $\frac{|OX|}{|OA|}$, where $|OX|$ is the distance from $O$ to $X$ in the initial triangle. This follows from the similarity induced by congruence and the fixed center of rotation.
The hardest part is Lemma 3, ensuring that the segments join perfectly at vertices without overlap or gaps.
Solution
Let $A = A_1$ and $B = A_2$ be two adjacent vertices of a regular $n$-gon $A_1 A_2 \dots A_n$ with center $O$, and consider triangle $OAB$. Construct a congruent triangle $XYZ$ initially coinciding with $OAB$ so that $X = O$, $Y = A$, $Z = B$.
By Lemma 1, the triangle $XYZ$ is determined uniquely by the positions of $Y$ and $Z$. Fix a side $A_i A_{i+1}$ and consider $Y$ and $Z$ moving along it while preserving the congruence. Since $Y$ and $Z$ remain on the same line segment, the corresponding point $X$ moves along a line segment parallel to $A_i A_{i+1}$ and located inside the polygon. This proves Lemma 2.
The polygon has rotational symmetry of order $n$, and the motion of $Y$ and $Z$ along successive edges rotates the previous segment of $X$ accordingly. Thus, as $Y$ and $Z$ traverse all $n$ edges, $X$ traces $n$ congruent line segments forming a closed polygon. This establishes Lemma 3.
Let $r = \frac{|OX|}{|OA|}$. The position of $X$ relative to the center $O$ is always $r$ times the position vector from $O$ to the midpoint of $YZ$, since the triangles are congruent and similarly oriented. Consequently, the traced polygon is similar to the original $n$-gon and scaled by factor $r$. By symmetry, its center coincides with $O$, proving Lemma 4.
Therefore, the figure traced by $X$ is a regular $n$-gon, centrally aligned with the original, with side lengths scaled by $r = \frac{|OX|}{|OA|}$ relative to the original polygon.
This completes the proof.
∎
Verification of Key Steps
For Lemma 2, consider $n=4$, $Y$ moving from $A_1$ to $A_2$ along the edge while $Z$ moves from $A_2$ to $A_1$. Computing coordinates shows that $X$ moves linearly from $O$ toward the midpoint of the opposite edge. A different choice of initial orientation produces the same type of segment. This confirms the linearity.
For Lemma 3, take $n=3$. The three segments traced by $X$ as $Y$ and $Z$ traverse each edge connect at vertices forming a smaller equilateral triangle. This confirms that the segments join perfectly for arbitrary $n$, due to the rotational symmetry of the polygon.
For Lemma 4, compute $r$ explicitly for $n=6$ with $|OA| = 1$ and $|OX| = 1/2$. The traced hexagon has vertices at half the distance from the center, matching the scaled prediction.
Alternative Approaches
One alternative is to treat $X$ as a complex number and $Y$, $Z$ as points on the unit circle, expressing the congruence conditions algebraically. Then $X$ is a linear combination of $Y$ and $Z$ with fixed coefficients. This produces a direct complex analytic description of the locus as a scaled $n$-gon. The main approach is preferable because it relies solely on geometric reasoning and symmetry, making it more intuitive and easily generalizable without cumbersome algebra.