AIME 2026 II · 第 2 题

Solution

Solution 1

The key observation is that the path is $\textbf{completely determined by the choice of horizontal segments}$. Once the horizontal segments are chosen, the vertical and diagonal segments are fixed.

For each of the $10$ squares, there are exactly $3$ choices for the horizontal segment:

$$\text{the upper horizontal edge, the lower horizontal edge, or no horizontal edge}.$$ Thus the total number of such paths is

$$N = 3^{10}.$$ Then

$$\sqrt{N} = \sqrt{3^{10}} = 3^{5} = \boxed{243}.$$ ~Steven Zheng

Solution 2

Consider each of the ten instances of

Let the number of ways to start at the top-left corner be $a$ and the number of ways to start at the bottom-left corner be $b$. Then, there are $a+b$ ways to get to the top-right corner without passing through the top-left corner, and there are $a$ ways to get to the bottom-right corner without passing through the top-right corner.

But on the right edge, it is also possible to swap the right-side corners. Therefore, there are an additional $b$ ways to get to the top-right corner AFTER passing through the bottom-right corner, and there are an additional $a+b$ ways to get to the bottom-right corner AFTER passing through the top-right corner. Combined, there are $a+2b$ ways to get to each of the right-side corners.

But we notice that both are the same, so it must also be true that $a=b$ by a simple induction. Therefore, there are $3a$ ways to get to each of the right-side corners.

At the very first vertical edge, there is one way to get to each corner on the edge. After 10 of these squares, the total number of ways is $3^{10}\implies \sqrt{3^{10}}=3^5=\boxed{243}$.

~ eevee9406

Solution 3

Label the 11 columns of this grid as column 0 through column 11, with the upper lattice points as row 0 and the lower lattice points as row 1. Let the path that first reaches column $i$ at row $j$ be denoted as $f_{i,j}$. Based on the initial condition, the first arrival at column 0 must be at the lower lattice point, so $f_{0,0}=0$, $f_{0,1}=1$.

Consider the moves from column $i$ to column $i+1$:

For each path that first reaches $(i,0)$, it can first reach $(i+1,0)$ in two ways: "down, then up-right" or "right"; and first reach $(i+1,1)$ in one way: "down, then right".

For each path that first reaches $(i,1)$, it can first reach $(i+1,0)$ in two ways: "up, then right" or "up-right"; and first reach $(i+1,1)$ in one way: "right".

Thus, we have

$$\begin{cases} f_{i+1,0}=2f_{i,0}+2f_{i,1} \\ f_{i+1,1}=f_{i,0}+f_{i,1} \end{cases}$$ Let $g_i = f_{i,0} + f_{i,1}$. When there are $n$ squares, since the path that ends at the bottom-right corner can reach the final point by taking one step upward, the number of ways to reach the top-right corner of the grid equals $g_n = f_{n,0} + f_{n,1}$. Moreover, since $g_{i+1} = 2f_{i,0} + 2f_{i,1} + f_{i,0} + f_{i,1} = 3g_i$, and $g_0 = f_{0,0} + f_{0,1} = 1$, we have $g_n = 3^n$. Therefore, when $n=10$, the total number of paths is $N = 3^{10}$, and the final answer is $\sqrt N = 3^5 = \fbox{243}$.

~Confringo