AMC 8 2026 Problem 25

AMC8 2026 年第 25 题

四边形与多边形★★★★★
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 △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，所有内角均为 120°。下图展示了一个边长依次为 2、3、1、3、2、2 的例子，它内接于一个等边三角形 ABC 中。考虑所有边长为正整数、且六个顶点都在 △ABC 三边上的等角六边形。若两个六边形仅通过旋转或反射重合，则视为相同。这样的六边形共有多少个？
AMC 8 2026 Problem 25 figure
  1. A.4
  2. B.5
  3. C.6
  4. D.7
  5. E.8

正确答案 · Correct Answer

E

解析 · Solution

解法1

常规解法·倒过来想——从大三角形切三个角
第一步，看清结构。等角六边形每个内角 120∘120^\circ，六个顶点都在 △ABC\triangle ABC 的三条边上，所以三角形每个角落都被「切掉」一个小三角形：小三角形在 △ABC\triangle ABC 顶点处的角是 60∘60^\circ，在六边形顶点处的两个角都是 180∘−120∘=60∘180^\circ-120^\circ=60^\circ——三个角全是 60∘60^\circ，每个切下来的角都是等边三角形。设三个切角的边长为正整数 xx、yy、zz。
第二步，定大三角形的边长。沿 △ABC\triangle ABC 的一条边看：切角 xx＋六边形贴边的一条边＋切角 yy＝三角形边长。用题面例子（边长依次 2、3、1、3、2、2）验证：相邻三条边之和 2+3+1=6，所以 △ABC\triangle ABC 边长为 6。六边形六条边依次是
x,6−x−y, y,6−y−z, z,6−z−xx,6-x-y,\ y,6-y-z,\ z,6-z-x
（例子对应 (x,y,z)=(2,1,2)：代入得 2、3、1、3、2、2 ✓）
第三步，列规矩。六条边都要是正整数：x,y,z≥1x,y,z\ge1，且 6−x−y≥16-x-y\ge1 等三条——即任意两个切角之和 ≤5\le5。
第四步，按「最大的切角」分类枚举。旋转、翻转算同一个六边形，等于只看 {x,y,z}\{x,y,z\} 这组数、不管顺序：
最大 =1：(1,1,1)——1 种 最大 =2：(2,1,1),(2,2,1),(2,2,2)——3 种 最大 =3：另两个只能取 1 或 2（3+3=6>5 犯规）：(3,1,1),(3,2,1),(3,2,2)——3 种 最大 =4：另两个只能取 1（4+2=6>5）：(4,1,1)——1 种 最大 ≥5\ge5：连最小搭配 5+1=6 都犯规，无解
合计 1+3+3+1=8 种，选 E。
易错提醒：约束是两切角之和 ≤5\le5 而不是 ≤6\le6——等于 6 时贴边那条边长变 0，两个切口撞在一起，六边形退化；另外 (3,2,1) 和 (1,2,3) 转一转、翻一翻就是同一个，按从小到大只记一次。

Solution 1

*(Elementary — think backwards: cut three corners off the big triangle.)*
Step 1, see the structure. An equiangular hexagon has all interior angles 120∘120^\circ, and its six vertices lie on the three sides of △ABC\triangle ABC, so each corner of the triangle gets cut off by a small triangle. That small triangle's angles are 60∘60^\circ (at the triangle's vertex) and 180∘−120∘=60∘180^\circ-120^\circ=60^\circ at each hexagon vertex — all three angles are 60∘60^\circ, so every cut-off corner is an equilateral triangle. Let the three cut sizes be positive integers xx, yy, zz.
Step 2, find the big triangle's side. Along one side of △ABC\triangle ABC: cut xx + one hexagon side + cut yy = the triangle's side. Check with the example (side lengths 2, 3, 1, 3, 2, 2): three consecutive sides sum to 2+3+1=6, so △ABC\triangle ABC has side 6. The hexagon's sides, in order, are
x,6−x−y, y,6−y−z, z,6−z−xx,6-x-y,\ y,6-y-z,\ z,6-z-x
(the example is (x,y,z)=(2,1,2): substituting gives 2, 3, 1, 3, 2, 2 ✓)
Step 3, the rules: all six sides are positive integers, so x,y,z≥1x,y,z\ge1 and each of 6-x-y, 6-y-z, 6-z-x is ≥1\ge1 — that is, any two cuts sum to at most 5.
Step 4, classify by the largest cut. Rotations and reflections give the same hexagon, so only the multiset {x,y,z}\{x,y,z\} matters:
largest =1: (1,1,1) — 1 way largest =2: (2,1,1),(2,2,1),(2,2,2) — 3 ways largest =3: the others can only be 1 or 2 (since 3+3=6>5): (3,1,1),(3,2,1),(3,2,2) — 3 ways largest =4: the others must be 1 (4+2=6>5): (4,1,1) — 1 way largest ≥5\ge5: even the smallest pairing 5+1=6 breaks the rule Total 1+3+3+1=8, choice E.
Watch out: the constraint is "any two cuts sum to ≤5\le5," not ≤6\le6 — at 6 the side between them shrinks to 0 and the two cuts collide, degenerating the hexagon; also (3,2,1) and (1,2,3) become the same hexagon after a rotation or reflection — record each sorted triple once.

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