Problem 1165

First, we prove a lemma.
Lemma: For each convex $2n$-gon with area $S$, the area of the triangles formed by its sides and vertices is not less than $\frac{S}{2}$.
Proof of the lemma: The main diagonal of a $2n$-gon is a diagonal that divides the $2n$-gon into two polygons, each containing the same side. For any side $b$ of the $2n$-gon, $\triangle_{b}$ denotes the triangle $ABP$, where $A, B$ are the endpoints of $b$, and $P$ is the intersection of the main diagonals $AA'$ and $BB'$. We will prove that the union of the triangles $\triangle_{b}$ taken over all sides covers the entire polygon.

To this end, select any side $AB$, and consider the main diagonal $AA'$ as a directed segment. Let $X$ be any point in the polygon that is not on any main diagonal, and assume without loss of generality that $X$ is to the left of the ray $AA'$. Consider the sequence of main diagonals $AA', BB', CC', \cdots$, where $A, B, C, \cdots$ are consecutive vertices to the right of $AA'$.

The $n$-th term in this sequence is the diagonal $A'A$, and $X$ is to its right. Therefore, before $A'$, there exist two consecutive vertices $K, L$ in the sequence $A, B, C, \cdots$ such that $X$ is to the left of $KK'$ and to the right of $LL'$. This implies that $X$ is in the triangle $\triangle l'$, where $l' = K'L'$. A similar argument can be made for points $X$ to the right of $AA'$ (points on the main diagonals can be ignored). Thus, the union of the triangles $\triangle_{b}$ covers the entire polygon.

Their total area is not less than $S$, so we can find two opposite sides, such as $b = AB$ and $b' = A'B'$ (where $AA'$ and $BB'$ are main diagonals), such that $[\triangle_{b}] + [\triangle_{b'}] \geq \frac{S}{n}$, where $[\cdots]$ denotes the area of a region. Let $AA'$ and $BB'$ intersect at $P$. Without loss of generality, assume $PB \geq PB'$, then $[ABA'] = [ABP] + [PBA'] \geq [ABP] + [PA'B'] = [\triangle_{b}] + [\triangle_{b'}] \geq \frac{S}{n}$. The lemma is proved.

Now, assume the convex polygon $P$ has area $S$ and $m$ sides $a_1, a_2, \cdots, a_m$. Let $S_i$ be the largest triangle in $P$ with side $a_i$. If the conclusion does not hold, then
$$\sum_{i=1}^{m} \frac{S_i}{S} < 1.$$
Let $n$ be the common denominator of the $m$ fractions $q_1, q_2, \cdots, q_m$. Let $q_i = \frac{k_i}{n}$, then
$$\sum k_i = 2n.$$
Divide each side $a_i$ of $P$ into $k_i$ equal parts, resulting in a convex $2n$-gon with area $S$ (some angles may be $180^\circ$). Apply the lemma to this $2n$-gon. According to the lemma, there is a side $b$ and a vertex $H$ such that the triangle $T$ formed by $b$ and $H$ has area $[T] \geq \frac{S}{n}$. If $b$ is a part of the side $a_i$ of $P$, then the triangle $W$ with base $a_i$ and the highest vertex $H$ has area
$$[W] = k_i \cdot [T] \geq k_i \cdot \frac{S}{n} = q_i \cdot S > S_i,$$
which contradicts the definition of $S_i$. The proof is complete.