19th Turkish Mathematical Olympiad

We will show that $K=K(M,N)$ is a regular polygon if and only if either $N$ is homothetic to $M$, or $N$ is obtained from $M$ by a $180/m$ degree rotation about the center of $M$ followed by a translation where $m$ is the number of edges of $M$.

In the following when we say an edge $XY$ of a convex polygon it will always be implied that $Y$ is the vertex coming after $X$ counterclockwise along the boundary of the polygon.

Let $AB$ be an edge of $M$. Then there is a unique line $\ell$ such that $\ell \cap N$ is either (1) an edge $A^{\prime }{B}^{\prime }$ or (2) a vertex $C$ of $N$, and the closed half-planes determined by the lines $AB$ and $\ell$ and containing $M$ and $N$, respectively, have nonempty intersection. Let $e(AB)$ denote the central median of the trapezoid $AB{B}^{\prime }{A}^{\prime }$ in Case 1 and the midline parallel to the side $AB$ of the triangle $ABC$ in Case 2. Let $h(AB)$ denote the closed half-plane defined by the line containing $e(AB)$ which has nonempty intersection with the half-planes mentioned above. The same notation will also be used when the roles of $M$ and $N$ are interchanged.

Let $K_1$ be the intersection of the half-planes $h(AB)$ for all edges $AB$ of $M$ and $N$. $K_1$ is a convex polygon and its boundary is the union of the line segments $e(AB)$ for all edges $AB$ of $M$ and $N$. Since $h(AB)$ contains $K$ for all edges $AB$ of $M$ or $N$, $K_1$ contains $K$. In fact, $K_1=K$: If $X$ is in $K_1$, then (by the Intermediate Value Theorem) $X$ is the midpoint of a line segment $YZ$ where $Y$ and $Z$ are on the edges of $K_1$. There exists $D$ and $E$ on the edges of $M$, and $D^{\prime }$ and $E^{\prime }$ on the edges of $N$ such that $Y$ and $Z$ are the midpoints of the line segments $D{D}^{\prime }$ and $E{E}^{\prime }$, respectively. Let $F$ be the midpoint of the line segment $DE$, and let $F^{\prime }$ be the midpoints of the line segment $D^{\prime }{E}^{\prime }$. Then $F$ is in $M$, $F^{\prime }$ is in $N$, and $X$ is the midpoint of $F{F}^{\prime }$. This shows that $K_1=K$.

An edge of $K$ will be called t-type in Case 1 and M-type in Case 2. N-type is defined similarly. Let $d_M$ and $d_N$ be the edge lengths of $M$ and $N$, respectively. Then M-type, N-type and t-type edges have lengths $d_M/2$, $d_N/2$ and $(d_M+d_N)/2$, respectively.

Assume that $K$ is a regular polygon. Since $K$ is equilateral, it cannot have $M$-type and $t$-type edges or $N$-type and $t$-type edges at the same time. If all edges are $t$-type, then edges of $M$ and $N$ are parallel in pairs, and since $M$ and $N$ are regular polygons, this means that they are homothetic. In this case the reverse implication is obvious.

Now we will consider the remaining case when $K$ is regular: $K$ has both $M$-type edges and $N$-type edges, but no $t$-type edge. Then $d_M=d_N$. Choose adjacent edges of different types: Let $AB$ be an $M$-type edge of $K$ and let $BC$ be an $N$-type edge of $K$. Then $AB=e(A_1{B}_1)$ is the midline of the triangle $A_1{B}_1{B}_2$, and $BC=e(B_2{C}_2)$ is the midline of the triangle $B_2{C}_2{B}_1$ where $A_1{B}_1$ is an edge of $M$ and $B_2{C}_2$ is an edge of $N$. Let $B_1{C}_1$ be the other edge of $M$ at vertex $B_1$, and let $A_2{B}_2$ be the other edge of $N$ at vertex $B_2$.

Since $AB//A_1{B}_1$ and $BC//B_2{C}_2$, the angle $ABC$ is greater than both the angle $A_1{B}_1{C}_1$ and the angle $A_2{B}_2{C}_2$. On the other hand, the angle between two adjacent $M$-type edges of $K$ is equal to an interior angle of $M$, and the angle between two adjacent $N$-type edges is equal to an interior angle of $N$. Since $K$ is regular, we conclude that $K$ cannot have adjacent edges of the same type. Hence $M$-type and $N$-type edges must alternate and are equal in number. As the $M$-type edges are in bijection with the edges of $M$, and the $N$-type edges with the edges of $N$, $N$ is also an $m$-gon and $K$ is an equilateral $2m$-gon.

Let $\theta$ be the acute angle between the lines $B_1{C}_1$ and $B_2{C}_2$. Then we have the equality $\angle ABC=\angle A_1{B}_1{C}_1+\theta$. This means $180^{\circ }\cdot (2m-2)/(2m)=180^{\circ }\cdot (m-2)/m+\theta$. That is, $\theta =180^{\circ }/m$. This completes the proof of the remaining case. In this case the reverse implication follows immediately from the same equality and the relation among the side lengths.