Question

设 $A = (0, 0)$ 且 $B = (b, 2)$ 是坐标平面上的点。设 $ABCDEF$ 是一个凸等边六边形，满足 $\angle FAB = 120^\circ$ ， $\overline{AB} \parallel \overline{DE}$ ， $\overline{BC} \parallel \overline{EF}$ ， $\overline{CD} \parallel \overline{FA}$ ，并且其顶点的 $y$ -坐标是集合 $\{0, 2, 4, 6, 8, 10\}$ 中互不相同的元素。该六边形的面积可写成 $m\sqrt{n}$ ，其中 $m$ 和 $n$ 是正整数，且 $n$ 不被任何质数的平方整除。求 $m + n$ 。

Let $A = (0, 0)$ and $B = (b, 2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB} \parallel \overline{DE},$ $\overline{BC} \parallel \overline{EF},$ $\overline{CD} \parallel \overline{FA},$ and the $y$ -coordinates of its vertices are distinct elements of the set $\{0, 2, 4, 6, 8, 10\}.$ The area of the hexagon can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$

答案

答案：51

知识点：坐标几何向量对称性

难度评级：3060

解答：

对边平行、长度相等，且按相反方向经过，所以 $\overrightarrow{AB} = \overrightarrow{ED}$ ， $\overrightarrow{BC} = \overrightarrow{FE}$ ， $\overrightarrow{CD} = \overrightarrow{AF}$ ：这个六边形中心对称，相对顶点的 $y$ -坐标有共同的和，即 $\frac{0 + 2 + \cdots + 10}{3} = 10$ 。由 $y_A = 0$ 和 $y_B = 2$ 得到 $y_D = 10$ ， $y_E = 8$ ，凸性又给出 $y_C = 6$ ， $y_F = 4$ 。写 $\overrightarrow{AB} = (b, 2)$ ， $\overrightarrow{BC} = (p, 4)$ ， $\overrightarrow{CD} = (q, 4)$ 边长相等给出 $s^2 = b^2 + 4$ $= p^2 + 16$ $= q^2 + 16$ ，所以 $p = \pm q$ ；若 $p = q$ ，则 $B$ ， $C$ ， $D$ 共线，因此 $p = -q$ 。

因为 $\overrightarrow{AF} = \overrightarrow{CD}$ ，所以 $F = (q, 4)$ ，且 $\angle FAB = 120^\circ$ 给出 $\begin{aligned} \overrightarrow{AB} \cdot \overrightarrow{AF} &= bq + 8 \\ &= -\frac{s^2}{2} = -\frac{b^2 + 4}{2}. \end{aligned}$ 取 $b \gt 0$ 会迫使 $q \lt 0$ ，所以 $q = -\sqrt{b^2 - 12}$ ，方程变为 $b\sqrt{b^2 - 12} = \frac{b^2 + 20}{2}$ 。平方得到 $3b^4 - 88b^2 - 400 = 0$ ，所以 $b^2 = \frac{100}{3}$ ，进而 $b = \frac{10}{\sqrt{3}}$ ， $q = -\frac{8}{\sqrt{3}}$ ， $p = \frac{8}{\sqrt{3}}$ 。

各顶点为 $A = (0, 0)$ ， $B = \left(\frac{10}{\sqrt{3}}, 2\right)$ ， $C = (6\sqrt{3}, 6)$ ， $D = \left(\frac{10}{\sqrt{3}}, 10\right)$ ， $E = (0, 8)$ ， $F = \left(-\frac{8}{\sqrt{3}}, 4\right)$ 。该六边形可分成平行四边形 $ABDE$ ，和两个全等三角形 $BCD$ 与 $EFA$ ，平行四边形的竖直边 $AE = 8$ ，水平偏移为 $b$ 面积为 $8b$ ；每个三角形的竖直底为 $8$ ，水平高为 $\frac{8}{\sqrt{3}}$ 。总面积为 $\begin{aligned} &8 \cdot \frac{10}{\sqrt{3}} + 2 \cdot \frac{1}{2} \cdot 8 \cdot \frac{8}{\sqrt{3}} \\ &= \frac{144}{\sqrt{3}} = 48\sqrt{3}, \end{aligned}$ 所以 $m + n = 48 + 3 = 51$ 。

Opposite sides are parallel, equal in length, and traversed in opposite directions, so $\overrightarrow{AB} = \overrightarrow{ED},$ $\overrightarrow{BC} = \overrightarrow{FE},$ $\overrightarrow{CD} = \overrightarrow{AF}:$ the hexagon is centrally symmetric, and opposite vertices' $y$ -coordinates share a common sum, namely $\frac{0 + 2 + \cdots + 10}{3} = 10.$ From $y_A = 0$ and $y_B = 2$ we get $y_D = 10,$ $y_E = 8,$ and convexity puts $y_C = 6,$ $y_F = 4.$ Write $\overrightarrow{AB} = (b, 2),$ $\overrightarrow{BC} = (p, 4),$ $\overrightarrow{CD} = (q, 4).$ Equal side lengths give $s^2 = b^2 + 4$ $= p^2 + 16$ $= q^2 + 16,$ so $p = \pm q;$ since $p = q$ would make $B,$ $C,$ $D$ collinear, $p = -q.$

Since $\overrightarrow{AF} = \overrightarrow{CD},$ we have $F = (q, 4),$ and $\angle FAB = 120^\circ$ gives $\begin{aligned} \overrightarrow{AB} \cdot \overrightarrow{AF} &= bq + 8 \\ &= -\frac{s^2}{2} = -\frac{b^2 + 4}{2}. \end{aligned}$ Taking $b \gt 0$ forces $q \lt 0,$ so $q = -\sqrt{b^2 - 12},$ and the equation becomes $b\sqrt{b^2 - 12} = \frac{b^2 + 20}{2}.$ Squaring yields $3b^4 - 88b^2 - 400 = 0,$ so $b^2 = \frac{100}{3},$ giving $b = \frac{10}{\sqrt{3}},$ $q = -\frac{8}{\sqrt{3}},$ $p = \frac{8}{\sqrt{3}}.$

The vertices are $A = (0, 0),$ $B = \left(\frac{10}{\sqrt{3}}, 2\right),$ $C = (6\sqrt{3}, 6),$ $D = \left(\frac{10}{\sqrt{3}}, 10\right),$ $E = (0, 8),$ $F = \left(-\frac{8}{\sqrt{3}}, 4\right).$ The hexagon splits into the parallelogram $ABDE,$ with vertical side $AE = 8$ and horizontal offset $b$ (area $8b$ ), plus the two congruent triangles $BCD$ and $EFA,$ each with vertical base $8$ and horizontal height $\frac{8}{\sqrt{3}}.$ The total area is $\begin{aligned} &8 \cdot \frac{10}{\sqrt{3}} + 2 \cdot \frac{1}{2} \cdot 8 \cdot \frac{8}{\sqrt{3}} \\ &= \frac{144}{\sqrt{3}} = 48\sqrt{3}, \end{aligned}$ so $m + n = 48 + 3 = 51.$

2003 AIME II 第 14 题

答案：51

难度评级：3060

解答：

其他年份的第 14 题