AMC12 2009 A

AMC12 2009 A · Q11

AMC12 2009 A · Q11. It mainly tests Sequences & recursion (algebra), Arithmetic misc.

The figures $F_1$, $F_2$, $F_3$, and $F_4$ shown are the first in a sequence of figures. For $n\ge3$, $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$?

所示的图形 $F_1$、$F_2$、$F_3$ 和 $F_4$ 是一个图形序列中的前几个。对于 $n\ge3$，$F_n$ 由 $F_{n - 1}$ 构造：用一个正方形将其围住，并且在新正方形的每一边上放置的菱形数比 $F_{n - 1}$ 的外层正方形每边上的菱形数多 $1$ 个。例如，图形 $F_3$ 有 $13$ 个菱形。图形 $F_{20}$ 中有多少个菱形？

(A) 401 401

(B) 485 485

(D) 626 626

(E) 761 761

Answer

Correct choice: (E)

正确答案：（E）

Solution

Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$th triangular number (i.e. the diamonds within the triangles or between the diagonals) and $4(n-1)+1$ (the diamonds on sides of the triangles or on the diagonals). The $n$th triangular number is $\frac{n(n+1)}{2}$. Putting this together for $F_{20}$ this gives: $\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}$

用两条对角线将 $F_n$ 分成 $4$ 个全等三角形（如题图所示）。这表明其中菱形的总数等于 $4$ 倍的第 $(n-2)$ 个三角数（即三角形内部或两条对角线之间的菱形数）再加上 $4(n-1)+1$（即在三角形边上或对角线上的菱形数）。第 $n$ 个三角数为 $\frac{n(n+1)}{2}$。将这些用于 $F_{20}$，得到： $\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}$

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