2023 AMC 10A Problem 4

In any polygon, each side must be strictly shorter than the sum of all the other sides; otherwise the other sides cannot reach around to close the figure. If the longest side has length $L$, the other three sides sum to $26 - L$, so
$$L < 26 - L \quad\Longrightarrow\quad L < 13.$$
Since $L$ is an integer, $L \le 12$.

Now check that $12$ works with a side of length $4$: the remaining two sides must total $26 - 12 - 4 = 10$, for example $5$ and $5$. The sides $4, 5, 5, 12$ satisfy $12 < 4 + 5 + 5 = 14$, so a genuine quadrilateral exists.

The answer is $\boxed{\textbf{(D)}\ 12}$.

The triangle inequality generalizes: a closed path of segments can never have one segment as long as all the rest combined. Maximum problems of this type always have two parts: bound the quantity (here $L < 13$) and exhibit a configuration that reaches the bound. The given side of $4$ is a distraction that only matters for the construction step.

• Answering 13 by allowing the longest side to equal the sum of the other three, which gives a degenerate flat figure.
• Assuming the side of length $4$ must be the shortest side or must be adjacent to the longest side; neither constraint exists.