Subfield: Geometry


Question: To each side $a$ of a convex polygon we assign the maximum area of a triangle contained in the polygon and having $a$ as one of its sides. Show that the sum of the areas assigned to all sides of the polygon is not less than twice the area of the polygon.

Solution:
Lemma. Every convex $(2 n)$-gon, of area $S$, has a side and a vertex that jointly span a triangle of area not less than $S / n$.

Proof. By main diagonals of the (2n)-gon we shall mean those which partition the (2n)-gon into two polygons with equally many sides. For any side $b$ of the $(2 n)$-gon denote by $\Delta_{b}$ the triangle $A B P$ where $A, B$ are the endpoints of $b$ and $P$ is the intersection point of the main diagonals $A A^{\prime}, B B^{\prime}$. We claim that the union of triangles $\Delta_{b}$, taken over all sides, covers the whole polygon.

To show this, choose any side $A B$ and consider the main diagonal $A A^{\prime}$ as a directed segment. Let $X$ be any point in the polygon, not on any main diagonal. For definiteness, let $X$ lie on the left side of the ray $A A^{\prime}$. Consider the sequence of main diagonals $A A^{\prime}, B B^{\prime}, C C^{\prime}, \ldots$, where $A, B, C, \ldots$ are consecutive vertices, situated right to $A A^{\prime}$.

The $n$-th item in this sequence is the diagonal $A^{\prime} A$ (i.e. $A A^{\prime}$ reversed), having $X$ on its right side. So there are two successive vertices $K, L$ in the sequence $A, B, C, \ldots$ before $A^{\prime}$ such that $X$ still lies to the left of $K K^{\prime}$ but to the right of $L L^{\prime}$. And this means that $X$ is in the triangle $\Delta_{\ell^{\prime}}, \ell^{\prime}=K^{\prime} L^{\prime}$. Analogous reasoning applies to points $X$ on the right of $A A^{\prime}$ (points lying on main diagonals can be safely ignored). Thus indeed the triangles $\Delta_{b}$ jointly cover the whole polygon.

The sum of their areas is no less than $S$. So we can find two opposite sides, say $b=A B$ and $b^{\prime}=A^{\prime} B^{\prime}$ (with $A A^{\prime}, B B^{\prime}$ main diagonals) such that $\left[\Delta_{b}\right]+\left[\Delta_{b^{\prime}}\right] \geq S / n$, where $[\cdots]$ stands for the area of a region. Let $A A^{\prime}, B B^{\prime}$ intersect at $P$; assume without loss of generality that $P B \geq P B^{\prime}$. Then

$$
\left[A B A^{\prime}\right]=[A B P]+\left[P B A^{\prime}\right] \geq[A B P]+\left[P A^{\prime} B^{\prime}\right]=\left[\Delta_{b}\right]+\left[\Delta_{b^{\prime}}\right] \geq S / n
$$

proving the lemma.

Now, let $\mathcal{P}$ be any convex polygon, of area $S$, with $m$ sides $a_{1}, \ldots, a_{m}$. Let $S_{i}$ be the area of the greatest triangle in $\mathcal{P}$ with side $a_{i}$. Suppose, contrary to the assertion, that

$$
\sum_{i=1}^{m} \frac{S_{i}}{S}<2
$$

Then there exist rational numbers $q_{1}, \ldots, q_{m}$ such that $\sum q_{i}=2$ and $q_{i}>S_{i} / S$ for each $i$.

Let $n$ be a common denominator of the $m$ fractions $q_{1}, \ldots, q_{m}$. Write $q_{i}=k_{i} / n$; so $\sum k_{i}=2 n$. Partition each side $a_{i}$ of $\mathcal{P}$ into $k_{i}$ equal segments, creating a convex (2n)-gon of area $S$ (with some angles of size $180^{\circ}$ ), to which we apply the lemma. Accordingly, this refined polygon has a side $b$ and a vertex $H$ spanning a triangle $T$ of area $[T] \geq S / n$. If $b$ is a piece of a side $a_{i}$ of $\mathcal{P}$, then the triangle $W$ with base $a_{i}$ and summit $H$ has area

$$
[W]=k_{i} \cdot[T] \geq k_{i} \cdot S / n=q_{i} \cdot S>S_{i}
$$

in contradiction with the definition of $S_{i}$. This ends the proof.
As in the first solution, we allow again angles of size $180^{\circ}$ at some vertices of the convex polygons considered.

To each convex $n$-gon $\mathcal{P}=A_{1} A_{2} \ldots A_{n}$ we assign a centrally symmetric convex $(2 n)$-gon $\mathcal{Q}$ with side vectors $\pm \overrightarrow{A_{i} A_{i+1}}, 1 \leq i \leq n$. The construction is as follows. Attach the $2 n$ vectors $\pm \overrightarrow{A_{i} A_{i+1}}$ at a common origin and label them $\overrightarrow{\mathbf{b}_{1}}, \overrightarrow{\mathbf{b}_{2}}, \ldots, \overrightarrow{\mathbf{b}_{2 n}}$ in counterclockwise direction; the choice of the first vector $\overrightarrow{\mathbf{b}_{1}}$ is irrelevant. The order of labelling is well-defined if $\mathcal{P}$ has neither parallel sides nor angles equal to $180^{\circ}$. Otherwise several collinear vectors with the same direction are labelled consecutively $\overrightarrow{\mathbf{b}_{j}}, \overrightarrow{\mathbf{b}_{j+1}}, \ldots, \overrightarrow{\mathbf{b}_{j+r}}$. One can assume that in such cases the respective opposite vectors occur in the order $-\overrightarrow{\mathbf{b}_{j}},-\overrightarrow{\mathbf{b}_{j+1}}, \ldots,-\overrightarrow{\mathbf{b}_{j+r}}$, ensuring that $\overrightarrow{\mathbf{b}_{j+n}}=-\overrightarrow{\mathbf{b}_{j}}$ for $j=1, \ldots, 2 n$. Indices are taken cyclically here and in similar situations below.

Choose points $B_{1}, B_{2}, \ldots, B_{2 n}$ satisfying $\overrightarrow{B_{j} B_{j+1}}=\overrightarrow{\mathbf{b}_{j}}$ for $j=1, \ldots, 2 n$. The polygonal line $\mathcal{Q}=B_{1} B_{2} \ldots B_{2 n}$ is closed, since $\sum_{j=1}^{2 n} \overrightarrow{\mathbf{b}_{j}}=\overrightarrow{0}$. Moreover, $\mathcal{Q}$ is a convex $(2 n)$-gon due to the arrangement of the vectors $\overrightarrow{\mathbf{b}_{j}}$, possibly with $180^{\circ}$-angles. The side vectors of $\mathcal{Q}$ are $\pm \overrightarrow{A_{i} A_{i+1}}$, $1 \leq i \leq n$. So in particular $\mathcal{Q}$ is centrally symmetric, because it contains as side vectors $\overrightarrow{A_{i} A_{i+1}}$ and $-\overrightarrow{A_{i} A_{i+1}}$ for each $i=1, \ldots, n$. Note that $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ are opposite sides of $\mathcal{Q}$, $1 \leq j \leq n$. We call $\mathcal{Q}$ the associate of $\mathcal{P}$.

Let $S_{i}$ be the maximum area of a triangle with side $A_{i} A_{i+1}$ in $\mathcal{P}, 1 \leq i \leq n$. We prove that

$$
\left[B_{1} B_{2} \ldots B_{2 n}\right]=2 \sum_{i=1}^{n} S_{i} \tag{1}
$$

and

$$
\left[B_{1} B_{2} \ldots B_{2 n}\right] \geq 4\left[A_{1} A_{2} \ldots A_{n}\right] \tag{2}
$$

It is clear that (1) and (2) imply the conclusion of the original problem.

Lemma. For a side $A_{i} A_{i+1}$ of $\mathcal{P}$, let $h_{i}$ be the maximum distance from a point of $\mathcal{P}$ to line $A_{i} A_{i+1}$, $i=1, \ldots, n$. Denote by $B_{j} B_{j+1}$ the side of $\mathcal{Q}$ such that $\overrightarrow{A_{i} A_{i+1}}=\overrightarrow{B_{j} B_{j+1}}$. Then the distance between $B_{j} B_{j+1}$ and its opposite side in $\mathcal{Q}$ is equal to $2 h_{i}$.

Proof. Choose a vertex $A_{k}$ of $\mathcal{P}$ at distance $h_{i}$ from line $A_{i} A_{i+1}$. Let $\mathbf{u}$ be the unit vector perpendicular to $A_{i} A_{i+1}$ and pointing inside $\mathcal{P}$. Denoting by $\mathbf{x} \cdot \mathbf{y}$ the dot product of vectors $\mathbf{x}$ and $\mathbf{y}$, we have

$$
h=\mathbf{u} \cdot \overrightarrow{A_{i} A_{k}}=\mathbf{u} \cdot\left(\overrightarrow{A_{i} A_{i+1}}+\cdots+\overrightarrow{A_{k-1} A_{k}}\right)=\mathbf{u} \cdot\left(\overrightarrow{A_{i} A_{i-1}}+\cdots+\overrightarrow{A_{k+1} A_{k}}\right)
$$

In $\mathcal{Q}$, the distance $H_{i}$ between the opposite sides $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ is given by

$$
H_{i}=\mathbf{u} \cdot\left(\overrightarrow{B_{j} B_{j+1}}+\cdots+\overrightarrow{B_{j+n-1} B_{j+n}}\right)=\mathbf{u} \cdot\left(\overrightarrow{\mathbf{b}_{j}}+\overrightarrow{\mathbf{b}_{j+1}}+\cdots+\overrightarrow{\mathbf{b}_{j+n-1}}\right) .
$$

The choice of vertex $A_{k}$ implies that the $n$ consecutive vectors $\overrightarrow{\mathbf{b}_{j}}, \overrightarrow{\mathbf{b}_{j+1}}, \ldots, \overrightarrow{\mathbf{b}_{j+n-1}}$ are precisely $\overrightarrow{A_{i} A_{i+1}}, \ldots, \overrightarrow{A_{k-1} A_{k}}$ and $\overrightarrow{A_{i} A_{i-1}}, \ldots, \overrightarrow{A_{k+1} A_{k}}$, taken in some order. This implies $H_{i}=2 h_{i}$.

For a proof of (1), apply the lemma to each side of $\mathcal{P}$. If $O$ the centre of $\mathcal{Q}$ then, using the notation of the lemma,

$$
\left[B_{j} B_{j+1} O\right]=\left[B_{j+n} B_{j+n+1} O\right]=\left[A_{i} A_{i+1} A_{k}\right]=S_{i}
$$

Summation over all sides of $\mathcal{P}$ yields (1).

Set $d(\mathcal{P})=[\mathcal{Q}]-4[\mathcal{P}]$ for a convex polygon $\mathcal{P}$ with associate $\mathcal{Q}$. Inequality $(2)$ means that $d(\mathcal{P}) \geq 0$ for each convex polygon $\mathcal{P}$. The last inequality will be proved by induction on the



number $\ell$ of side directions of $\mathcal{P}$, i. e. the number of pairwise nonparallel lines each containing a side of $\mathcal{P}$.

We choose to start the induction with $\ell=1$ as a base case, meaning that certain degenerate polygons are allowed. More exactly, we regard as degenerate convex polygons all closed polygonal lines of the form $X_{1} X_{2} \ldots X_{k} Y_{1} Y_{2} \ldots Y_{m} X_{1}$, where $X_{1}, X_{2}, \ldots, X_{k}$ are points in this order on a line segment $X_{1} Y_{1}$, and so are $Y_{m}, Y_{m-1}, \ldots, Y_{1}$. The initial construction applies to degenerate polygons; their associates are also degenerate, and the value of $d$ is zero. For the inductive step, consider a convex polygon $\mathcal{P}$ which determines $\ell$ side directions, assuming that $d(\mathcal{P}) \geq 0$ for polygons with smaller values of $\ell$.

Suppose first that $\mathcal{P}$ has a pair of parallel sides, i. e. sides on distinct parallel lines. Let $A_{i} A_{i+1}$ and $A_{j} A_{j+1}$ be such a pair, and let $A_{i} A_{i+1} \leq A_{j} A_{j+1}$. Remove from $\mathcal{P}$ the parallelogram $R$ determined by vectors $\overrightarrow{A_{i} A_{i+1}}$ and $\overrightarrow{A_{i} A_{j+1}}$. Two polygons are obtained in this way. Translating one of them by vector $\overrightarrow{A_{i} A_{i+1}}$ yields a new convex polygon $\mathcal{P}^{\prime}$, of area $[\mathcal{P}]-[R]$ and with value of $\ell$ not exceeding the one of $\mathcal{P}$. The construction just described will be called operation A.
<img_3326>

The associate of $\mathcal{P}^{\prime}$ is obtained from $\mathcal{Q}$ upon decreasing the lengths of two opposite sides by an amount of $2 A_{i} A_{i+1}$. By the lemma, the distance between these opposite sides is twice the distance between $A_{i} A_{i+1}$ and $A_{j} A_{j+1}$. Thus operation $\mathbf{A}$ decreases $[\mathcal{Q}]$ by the area of a parallelogram with base and respective altitude twice the ones of $R$, i. e. by $4[R]$. Hence $\mathbf{A}$ leaves the difference $d(\mathcal{P})=[\mathcal{Q}]-4[\mathcal{P}]$ unchanged.

Now, if $\mathcal{P}^{\prime}$ also has a pair of parallel sides, apply operation $\mathbf{A}$ to it. Keep doing so with the subsequent polygons obtained for as long as possible. Now, A decreases the number $p$ of pairs of parallel sides in $\mathcal{P}$. Hence its repeated applications gradually reduce $p$ to 0 , and further applications of $\mathbf{A}$ will be impossible after several steps. For clarity, let us denote by $\mathcal{P}$ again the polygon obtained at that stage.

The inductive step is complete if $\mathcal{P}$ is degenerate. Otherwise $\ell>1$ and $p=0$, i. e. there are no parallel sides in $\mathcal{P}$. Observe that then $\ell \geq 3$. Indeed, $\ell=2$ means that the vertices of $\mathcal{P}$ all lie on the boundary of a parallelogram, implying $p>0$.

Furthermore, since $\mathcal{P}$ has no parallel sides, consecutive collinear vectors in the sequence $\left(\overrightarrow{\mathbf{b}_{k}}\right)$ (if any) correspond to consecutive $180^{\circ}$-angles in $\mathcal{P}$. Removing the vertices of such angles, we obtain a convex polygon with the same value of $d(\mathcal{P})$.

In summary, if operation $\mathbf{A}$ is impossible for a nondegenerate polygon $\mathcal{P}$, then $\ell \geq 3$. In addition, one may assume that $\mathcal{P}$ has no angles of size $180^{\circ}$.

The last two conditions then also hold for the associate $\mathcal{Q}$ of $\mathcal{P}$, and we perform the following construction. Since $\ell \geq 3$, there is a side $B_{j} B_{j+1}$ of $\mathcal{Q}$ such that the sum of the angles at $B_{j}$ and $B_{j+1}$ is greater than $180^{\circ}$. (Such a side exists in each convex $k$-gon for $k>4$.) Naturally, $B_{j+n} B_{j+n+1}$ is a side with the same property. Extend the pairs of sides $B_{j-1} B_{j}, B_{j+1} B_{j+2}$



and $B_{j+n-1} B_{j+n}, B_{j+n+1} B_{j+n+2}$ to meet at $U$ and $V$, respectively. Let $\mathcal{Q}^{\prime}$ be the centrally symmetric convex 2(n+1)-gon obtained from $\mathcal{Q}$ by inserting $U$ and $V$ into the sequence $B_{1}, \ldots, B_{2 n}$ as new vertices between $B_{j}, B_{j+1}$ and $B_{j+n}, B_{j+n+1}$, respectively. Informally, we adjoin to $\mathcal{Q}$ the congruent triangles $B_{j} B_{j+1} U$ and $B_{j+n} B_{j+n+1} V$. Note that $B_{j}, B_{j+1}, B_{j+n}$ and $B_{j+n+1}$ are kept as vertices of $\mathcal{Q}^{\prime}$, although $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ are no longer its sides.

Let $A_{i} A_{i+1}$ be the side of $\mathcal{P}$ such that $\overrightarrow{A_{i} A_{i+1}}=\overrightarrow{B_{j} B_{j+1}}=\overrightarrow{\mathbf{b}_{j}}$. Consider the point $W$ such that triangle $A_{i} A_{i+1} W$ is congruent to triangle $B_{j} B_{j+1} U$ and exterior to $\mathcal{P}$. Insert $W$ into the sequence $A_{1}, A_{2}, \ldots, A_{n}$ as a new vertex between $A_{i}$ and $A_{i+1}$ to obtain an $(n+1)$-gon $\mathcal{P}^{\prime}$. We claim that $\mathcal{P}^{\prime}$ is convex and its associate is $\mathcal{Q}^{\prime}$.
<img_3469>

Vectors $\overrightarrow{A_{i} W}$ and $\overrightarrow{\mathbf{b}_{j-1}}$ are collinear and have the same direction, as well as vectors $\overrightarrow{W A_{i+1}}$ and $\overrightarrow{\mathbf{b}_{j+1}}$. Since $\overrightarrow{\mathbf{b}_{j-1}}, \overrightarrow{\mathbf{b}_{j}}, \overrightarrow{\mathbf{b}_{j+1}}$ are consecutive terms in the sequence $\left(\overrightarrow{\mathbf{b}_{k}}\right)$, the angle inequalities $\angle\left(\overrightarrow{\mathbf{b}_{j-1}}, \overrightarrow{\mathbf{b}_{j}}\right) \leq \angle\left(\overrightarrow{A_{i-1} A_{i}}, \overrightarrow{\mathbf{b}_{j}}\right)$ and $\angle\left(\overrightarrow{\mathbf{b}_{j}}, \overrightarrow{\mathbf{b}_{j+1}}\right) \leq \angle\left(\overrightarrow{\mathbf{b}_{j}}, \overrightarrow{A_{i+1} A_{i+2}}\right)$ hold true. They show that $\mathcal{P}^{\prime}$ is a convex polygon. To construct its associate, vectors $\pm \overrightarrow{A_{i} A_{i+1}}= \pm \overrightarrow{\mathbf{b}_{j}}$ must be deleted from the defining sequence $\left(\overrightarrow{\mathbf{b}_{k}}\right)$ of $\mathcal{Q}$, and the vectors $\pm \overrightarrow{A_{i} W}, \pm \overrightarrow{W A_{i+1}}$ must be inserted appropriately into it. The latter can be done as follows:

$$
\ldots, \overrightarrow{\mathbf{b}_{j-1}}, \overrightarrow{A_{i} W}, \overrightarrow{W A_{i+1}}, \overrightarrow{\mathbf{b}_{j+1}}, \ldots,-\overrightarrow{\mathbf{b}_{j-1}},-\overrightarrow{A_{i} W},-\overrightarrow{W A_{i+1}},-\overrightarrow{\mathbf{b}_{j+1}}, \ldots
$$

This updated sequence produces $\mathcal{Q}^{\prime}$ as the associate of $\mathcal{P}^{\prime}$.

It follows from the construction that $\left[\mathcal{P}^{\prime}\right]=[\mathcal{P}]+\left[A_{i} A_{i+1} W\right]$ and $\left[\mathcal{Q}^{\prime}\right]=[\mathcal{Q}]+2\left[A_{i} A_{i+1} W\right]$. Therefore $d\left(\mathcal{P}^{\prime}\right)=d(\mathcal{P})-2\left[A_{i} A_{i+1} W\right]<d(\mathcal{P})$.

To finish the induction, it remains to notice that the value of $\ell$ for $\mathcal{P}^{\prime}$ is less than the one for $\mathcal{P}$. This is because side $A_{i} A_{i+1}$ was removed. The newly added sides $A_{i} W$ and $W A_{i+1}$ do not introduce new side directions. Each one of them is either parallel to a side of $\mathcal{P}$ or lies on the line determined by such a side. The proof is complete.