Question

三个不重叠的正多边形，至少两个全等，边长都为 $1$ 。这些多边形在点 $A$ 处相接，使得在 $A$ 处的三个内角之和为 $360^\circ$ 。于是这三个多边形形成一个新的多边形，且 $A$ 为内部点。这个新多边形的最大可能周长是多少？

Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1.$ The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^\circ.$ Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?

$12$

$14$

$18$

$21$

$24$

答案：D

知识点：正多边形丢番图方程周长

难度评级：2090

解答：

设两个全等的多边形为正 $a$ 边形，另一个为正 $b$ 边形，在点 $A$ 处相接。内角条件为化简得 $(a - 4)(b - 2) = 8$ 。 $\begin{aligned} &2 \cdot 180\left(1 - \tfrac2a\right) \\ &\quad {}+ 180\left(1 - \tfrac2b\right) = 360, \end{aligned}$

解为 $(a, b)$ 等于 $(5, 10), (6, 6), (8, 4)$ 或 $(12, 3)$ 。新多边形周长为 $2a + b - 6$ 分别为 $14, 12, 14$ 和 $21$ 。最大值为 $21$ 。

所以正确答案是 D。

Let two congruent $a$ -gons and one $b$ -gon meet at $A.$ Their interior angles satisfy $\begin{aligned} &2 \cdot 180\left(1 - \tfrac2a\right) \\ &\quad {}+ 180\left(1 - \tfrac2b\right) = 360, \end{aligned}$ which reduces to $(a - 4)(b - 2) = 8.$

The solutions $(a, b)$ are $(5, 10), (6, 6), (8, 4),$ and $(12, 3).$ The new polygon's perimeter is $2a + b - 6,$ giving $14, 12, 14,$ and $21.$ The largest is $21.$

Thus, the correct answer is D.

1999 AMC 12 第 26 题

答案：D

难度评级：2090

解答：