The 92nd question: Divide a regular 2017-gon into 2015 triangular regions using 2014 non-intersecting internal diagonals. Find the maximum possible number of isosceles triangles among these 2015 triangles.
Question 92:
Solution: Label the 2017 vertices of the regular 2017-gon as $\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{~A}_{2017}$.
For any side $A_{i} A_{j}(1 \leq i<j \leq 2017)$, denote its span as $\min \{j-i, 2017-(j-i)\}$. Clearly, the span of any side does not exceed $\left[\frac{2017}{2}\right]=1008$.

For any isosceles triangle $\mathrm{T}$, denote $\mathrm{d}(\mathrm{T})$ as the span of one of its legs. Let the number of isosceles triangles satisfying $\mathrm{d}(\mathrm{T})=\mathrm{k}(\mathrm{k}=1, 2, \ldots, 1008)$ be $s_{k}$, then the total number of isosceles triangles is $\sum_{\mathrm{k}=1}^{1008} s_{\mathrm{k}}$.

For any natural number $\mathrm{i}$, consider the isosceles triangles with $\mathrm{d}(\mathrm{T})=2^{i}, 2^{i}+1, \ldots, 2^{i+1}-1$. The sides covered by these isosceles triangles are distinct, and each isosceles triangle covers at least $2 \cdot 2^{i}=2^{i+1}$ sides, thus:
$$
\begin{array}{l}
2^{i+1}\left(\mathrm{~s}_{2^{i}}+\mathrm{s}_{2^{\mathrm{i}}+1}+\ldots+\mathrm{s}_{2^{i+1}-1}\right) \leq 2017 \\
\Rightarrow \mathrm{s}_{2^{i}}+\mathrm{s}_{2^{\mathrm{i}}+1}+\ldots+\mathrm{s}_{2^{i+1}-1} \leq\left[\frac{2017}{2^{i+1}}\right]
\end{array}
$$

Thus:
$$
\sum_{\mathrm{k}=1}^{1008} \mathrm{~s}_{\mathrm{k}} \leq\left[\frac{2017}{2^{1}}\right]+\left[\frac{2017}{2^{2}}\right]+\ldots+\left[\frac{2017}{2^{10}}\right]=2010
$$

On the other hand, first connect all points with indices congruent to 1 modulo $2^{1}$ in ascending order; then connect all points with indices congruent to 1 modulo $2^{2}$ in ascending order; $\cdots$; then connect all points with indices congruent to 1 modulo $2^{10}$ in ascending order. Finally, connect the remaining diagonals arbitrarily according to the conditions. It is easy to see that these segments do not intersect internally, and the number of isosceles triangles formed is $\left[\frac{2017}{2^{1}}\right]+\left[\frac{2017}{2^{2}}\right]+\ldots+\left[\frac{2017}{2^{10}}\right]=2010$.
In summary, the maximum number of isosceles triangles that can be formed is 2010.
Note 1: Generally, for any regular $\mathrm{n}(\mathrm{n} \geq 3)$-gon, let the maximum number of isosceles triangles that can be formed be $\mathrm{f}(\mathrm{n})$. Then $f(n)=\left\{\begin{array}{c}n-S_{2}(n), \text { when } n \text { is not a power of } 2 \\ n-S_{2}(n)-1, \text { when } n \text { is a power of } 2\end{array}\right.$, where $S_{2}(n)$ represents the sum of the digits of the positive integer $n$ in its binary representation.

In fact, as analyzed in the solution, for any isosceles triangle $\mathrm{T}$ inscribed in a regular $\mathrm{n}$-gon, the span of its legs $\mathrm{d}(\mathrm{T}) \leq\left\lceil\frac{\mathrm{n}}{2}\right]-1$. And $\mathrm{s}_{2^{i}}+\mathrm{s}_{2^{i}+1}+\ldots+\mathrm{s}_{2^{i+1}-1} \leq\left[\frac{\mathrm{n}}{2^{i+1}}\right]$, hence
$$
\begin{array}{l}
\leq \sum_{\left.0 \leq i \leq \log _{2}\left(\left\lvert\, \frac{n}{2}\right.\right]-1\right)}\left[\frac{n}{2^{i+1}}\right]=\left\{\begin{array}{c}
n-S_{2}(n), \text { when } n \text { is not a power of } 2 \\
n-S_{2}(n)-1, \text { when } n \text { is a power of } 2
\end{array}\right. \\
\end{array}
$$

By the construction in the solution, this value can be achieved.
Note 2: The 2008 USAMO has the following problem: Let $\mathrm{P}$ be a convex $\mathrm{n}$-gon. Any $\mathrm{n}-3$ diagonals that do not intersect inside $\mathrm{P}$ divide $\mathrm{P}$ into $\mathrm{n}-2$ triangles. If $\mathrm{P}$ is a regular $\mathrm{n}$-gon and there exists a triangulation such that all triangles are isosceles, find all possible values of $\mathrm{n}$.

According to the conclusion in Note 1, we need to find all positive integers $n \geq 3$ such that $f(n)=n-2$. Clearly, the values of $n$ that satisfy this are $2^{m}(m \geq 2)$ or $2^{m}+2^{t}\left(m \in Z^{+}, t \in N\right)$.