AMC8 2026

AMC8 2026 · Q25

AMC8 2026 · Q25. It mainly tests Polygons, Geometry misc.

In an equiangular hexagon, all interior angles measure 120°. An example of such a hexagon with side lengths 2, 3, 1, 3, 2, and 2 is shown below, inscribed in equilateral triangle ABC. Consider all equiangular hexagons with positive integer side lengths that can be inscribed in triangle ABC, with all six vertices on the sides of the triangle. What is the total number of such hexagons? Hexagons that differ only by a rotation or a reflection are considered the same.

在一个等角六边形中，所有内角都为120°。下面展示了一个这样的六边形的例子，其边长依次为2、3、1、3、2和2，且内切于正三角形ABC。考虑所有边长为正整数且可以内切于三角形ABC的等角六边形，六个顶点均在三角形的边上。这类六边形共有多少个？仅通过旋转或反射而不同的六边形视为相同。

(A) 4 4

(B) 5 5

(D) 7 7

(E) 8 8

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let the side lengths of the hexagon be $a, b, c, d, e, f$. Obviously, each one of these is to be in between 1, 2, or 3 . Every 2 can be followed by a 2 or a 3. Every 3 must be followed by a 1 or 2, and every 1 must be followed by a 3. The question is now how many strings of length 6 satisfy these conditions, in which the first and last numbers also satisfy these conditions. To count that, we use recursion. Let $a_n$ be a string of length $n$ ending in 1, $b_n$ is a string of length $n$ ending in 2, and $c_n$ is a string that ends in 3. For $a_{n+1}$, we have that it must come from $c_n$, so $a_{n+1} = c_n$. Additionally, $b_{n+1} = b_n + c_n$, and $c_{n+1} = a_n$. Then, $a_1 = 1$, $b_1 = 1$, and $c_1 = 1$. Thus, $a_2 = 1$, $b_2 = 2$, and $c_2 = 1$; $a_3 = 1$, $b_3 = 3$, and $c_3 = 1$; $a_4 = 1$, $b_4 = 4$, $c_4 = 1$; $a_5 = 1$, $b_5 = 5$, $c_5 = 1$; and finally $a_6 = 1$, $b_6 = 6$, and $c_6 = 1$. Our answer is $a_6 + b_6 + c_6$, or $1+6+1 = \boxed{8}$.

设六边形的边长依次为 $a, b, c, d, e, f$。显然，这些边长必须是1、2或3中的数）。每个2后面可跟2或3；每个3后面必须跟1或2；每个1后面必须跟3。现在的问题是，有多少长度为6的数列满足这些条件，并且首尾数字也满足这些条件。为计数，我们用递推。设 $a_n$ 是以1结尾的长度为 $n$ 的数列个数，$b_n$ 是以2结尾的数列个数，$c_n$ 是以3结尾的数列个数。根据规则，$a_{n+1}$ 必须由 $c_n$ 继承，即 $a_{n+1} = c_n$；同时，$b_{n+1} = b_n + c_n$，$c_{n+1} = a_n$。初始值为 $a_1=1$，$b_1=1$，$c_1=1$。由此得出：$a_2=1$，$b_2=2$，$c_2=1$；$a_3=1$，$b_3=3$，$c_3=1$；$a_4=1$，$b_4=4$，$c_4=1$；$a_5=1$，$b_5=5$，$c_5=1$；最后，$a_6=1$，$b_6=6$，$c_6=1$。答案即为 $a_6 + b_6 + c_6 = 1 + 6 + 1 = \boxed{8}$。

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