2022 AMC 12B 第 25 题

先试着解答 2022 AMC 12B 第 25 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2022 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

25.

四个正六边形围绕一个边长为 11 的正方形,每个六边形都与该正方形共用一条边,如下图所示。所得外侧 1212 边非凸多边形的面积可写成 mn+pm\sqrt n + p,其中 mm、nn 和 pp 都是整数,且 nn 不被任何质数的平方整除。求 m+n+pm + n + p。
Four regular hexagons surround a square with a side length 1,1, each one sharing an edge with the square, as shown in the figure below. The area of the resulting 1212-sided outer nonconvex polygon can be written as mn+p,m\sqrt n + p, where m,m, n,n, and pp are integers and nn is not divisible by the square of any prime. What is m+n+p?m + n + p?
−12-12
−4-4
44
2424
3232
答案:B
知识点:正多边形坐标几何鞋带公式
小提示:
每个六边形与正方形共用一条边并延伸到对侧;把正方形中心设为原点建立坐标。
Each hexagon shares an edge with the square and reaches across to the far side; set up coordinates with the square centered at the origin
大提示:
外边界有距离中心 3−12\sqrt3 - \tfrac12 的平边,以及相邻六边形边相交形成的凹点 (52−3, 52−3)\left(\tfrac52 - \sqrt3,\ \tfrac52 - \sqrt3\right)。
The outer boundary has flat edges at distance 3−12\sqrt3 - \tfrac12 and reflex notches where adjacent hexagons’ edges cross, at (52−3, 52−3)\left(\tfrac52 - \sqrt3,\ \tfrac52 - \sqrt3\right)
解答:
将正方形中心置于原点,顶点为 (±12,±12)\left(\pm\tfrac12, \pm\tfrac12\right)。每个六边形与正方形共用一条边,并向正方形的对侧延伸。例如底边上的六边形,其远端上边从 (−12,3−12)\left(-\tfrac12, \sqrt3 - \tfrac12\right) 到 (12,3−12)\left(\tfrac12, \sqrt3 - \tfrac12\right)。
外边界是一个 1212 边形,其平边到中心的距离为 3−12\sqrt3 - \tfrac12,凸顶点包括 (3−12,12)\left(\sqrt3 - \tfrac12, \tfrac12\right),而相邻六边形的斜边在 (52−3,52−3)\left(\tfrac52 - \sqrt3, \tfrac52 - \sqrt3\right) 及其对称点相交,形成四个凹入的顶点。
对这 1212 个顶点使用鞋带公式,面积为 163−2316\sqrt3 - 23,所以 m=16m = 16,n=3n = 3,p=−23p = -23,从而 m+n+p=−4m + n + p = -4。
所以正确答案是 B。
Center the square at the origin with vertices (±12,±12).\left(\pm\tfrac12, \pm\tfrac12\right). Each hexagon shares one edge with the square and extends across to the opposite side; the hexagon on the bottom edge, for instance, has its far (top) edge from (−12,3−12)\left(-\tfrac12, \sqrt3 - \tfrac12\right) to (12,3−12).\left(\tfrac12, \sqrt3 - \tfrac12\right).
The outer boundary is a 1212-gon with flat edges at distance 3−12\sqrt3 - \tfrac12 from the center, convex vertices such as (3−12,12),\left(\sqrt3 - \tfrac12, \tfrac12\right), and four reflex notches where adjacent hexagons’ slanted edges meet, at (52−3,52−3)\left(\tfrac52 - \sqrt3, \tfrac52 - \sqrt3\right) and its symmetric images.
Applying the shoelace formula to these 1212 vertices gives area 163−23,16\sqrt3 - 23, so m=16,m = 16, n=3,n = 3, p=−23,p = -23, and m+n+p=−4.m + n + p = -4.
Thus, the correct answer is B.
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