Subfield: Geometry


Question: A triangulation of a convex polygon $\Pi$ is a partitioning of $\Pi$ into triangles by diagonals having no common points other than the vertices of the polygon. We say that a triangulation is a Thaiangulation if all triangles in it have the same area.

Prove that any two different Thaiangulations of a convex polygon $\Pi$ differ by exactly two triangles. (In other words, prove that it is possible to replace one pair of triangles in the first Thaiangulation with a different pair of triangles so as to obtain the second Thaiangulation.)

Solution:
We denote by $[S]$ the area of a polygon $S$.

Recall that each triangulation of a convex $n$-gon has exactly $n-2$ triangles. This means that all triangles in any two Thaiangulations of a convex polygon $\Pi$ have the same area.

Let $\mathcal{T}$ be a triangulation of a convex polygon $\Pi$. If four vertices $A, B, C$, and $D$ of $\Pi$ form a parallelogram, and $\mathcal{T}$ contains two triangles whose union is this parallelogram, then we say that $\mathcal{T}$ contains parallelogram $A B C D$. Notice here that if two Thaiangulations $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ of $\Pi$ differ by two triangles, then the union of these triangles is a quadrilateral each of whose diagonals bisects its area, i.e., a parallelogram.

We start with proving two properties of triangulations.

Lemma 1. A triangulation of a convex polygon $\Pi$ cannot contain two parallelograms.

Proof. Arguing indirectly, assume that $P_{1}$ and $P_{2}$ are two parallelograms contained in some triangulation $\mathcal{T}$. If they have a common triangle in $\mathcal{T}$, then we may assume that $P_{1}$ consists of triangles $A B C$ and $A D C$ of $\mathcal{T}$, while $P_{2}$ consists of triangles $A D C$ and $C D E$ (see Figure 1). But then $B C\|A D\| C E$, so the three vertices $B, C$, and $E$ of $\Pi$ are collinear, which is absurd.

Assume now that $P_{1}$ and $P_{2}$ contain no common triangle. Let $P_{1}=A B C D$. The sides $A B$, $B C, C D$, and $D A$ partition $\Pi$ into several parts, and $P_{2}$ is contained in one of them; we may assume that this part is cut off from $P_{1}$ by $A D$. Then one may label the vertices of $P_{2}$ by $X$, $Y, Z$, and $T$ so that the polygon $A B C D X Y Z T$ is convex (see Figure 2; it may happen that $D=X$ and/or $T=A$, but still this polygon has at least six vertices). But the sum of the external angles of this polygon at $B, C, Y$, and $Z$ is already $360^{\circ}$, which is impossible. A final contradiction.

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Figure 1

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Figure 2

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Figure 3

Lemma 2. Every triangle in a Thaiangulation $\mathcal{T}$ of $\Pi$ contains a side of $\Pi$.

Proof. Let $A B C$ be a triangle in $\mathcal{T}$. Apply an affine transform such that $A B C$ maps to an equilateral triangle; let $A^{\prime} B^{\prime} C^{\prime}$ be the image of this triangle, and $\Pi^{\prime}$ be the image of $\Pi$. Clearly, $\mathcal{T}$ maps into a Thaiangulation $\mathcal{T}^{\prime}$ of $\Pi^{\prime}$.

Assume that none of the sides of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is a side of $\Pi^{\prime}$. Then $\mathcal{T}^{\prime}$ contains some other triangles with these sides, say, $A^{\prime} B^{\prime} Z, C^{\prime} A^{\prime} Y$, and $B^{\prime} C^{\prime} X$; notice that $A^{\prime} Z B^{\prime} X C^{\prime} Y$ is a convex hexagon (see Figure 3). The sum of its external angles at $X, Y$, and $Z$ is less than $360^{\circ}$. So one of these angles (say, at $Z$ ) is less than $120^{\circ}$, hence $\angle A^{\prime} Z B^{\prime}>60^{\circ}$. Then $Z$ lies on a circular arc subtended by $A^{\prime} B^{\prime}$ and having angular measure less than $240^{\circ}$; consequently, the altitude $Z H$ of $\triangle A^{\prime} B^{\prime} Z$ is less than $\sqrt{3} A^{\prime} B^{\prime} / 2$. Thus $\left[A^{\prime} B^{\prime} Z\right]<\left[A^{\prime} B^{\prime} C^{\prime}\right]$, and $\mathcal{T}^{\prime}$ is not a Thaiangulation. A contradiction.



Now we pass to the solution. We say that a triangle in a triangulation of $\Pi$ is an ear if it contains two sides of $\Pi$. Note that each triangulation of a polygon contains some ear.

Arguing indirectly, we choose a convex polygon $\Pi$ with the least possible number of sides such that some two Thaiangulations $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ of $\Pi$ violate the problem statement (thus $\Pi$ has at least five sides). Consider now any ear $A B C$ in $\mathcal{T}_{1}$, with $A C$ being a diagonal of $\Pi$. If $\mathcal{T}_{2}$ also contains $\triangle A B C$, then one may cut $\triangle A B C$ off from $\Pi$, getting a polygon with a smaller number of sides which also violates the problem statement. This is impossible; thus $\mathcal{T}_{2}$ does not contain $\triangle A B C$.

Next, $\mathcal{T}_{1}$ contains also another triangle with side $A C$, say $\triangle A C D$. By Lemma 2, this triangle contains a side of $\Pi$, so $D$ is adjacent to either $A$ or $C$ on the boundary of $\Pi$. We may assume that $D$ is adjacent to $C$.

Assume that $\mathcal{T}_{2}$ does not contain the triangle $B C D$. Then it contains two different triangles $B C X$ and $C D Y$ (possibly, with $X=Y$ ); since these triangles have no common interior points, the polygon $A B C D Y X$ is convex (see Figure 4). But, since $[A B C]=[B C X]=$ $[A C D]=[C D Y]$, we get $A X \| B C$ and $A Y \| C D$ which is impossible. Thus $\mathcal{T}_{2}$ contains $\triangle B C D$.

Therefore, $[A B D]=[A B C]+[A C D]-[B C D]=[A B C]$, and $A B C D$ is a parallelogram contained in $\mathcal{T}_{1}$. Let $\mathcal{T}^{\prime}$ be the Thaiangulation of $\Pi$ obtained from $\mathcal{T}_{1}$ by replacing the diagonal $A C$ with $B D$; then $\mathcal{T}^{\prime}$ is distinct from $\mathcal{T}_{2}$ (otherwise $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ would differ by two triangles). Moreover, $\mathcal{T}^{\prime}$ shares a common ear $B C D$ with $\mathcal{T}_{2}$. As above, cutting this ear away we obtain that $\mathcal{T}_{2}$ and $\mathcal{T}^{\prime}$ differ by two triangles forming a parallelogram different from $A B C D$. Thus $\mathcal{T}^{\prime}$ contains two parallelograms, which contradicts Lemma 1.

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Figure 4

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Figure 5