LIVE by Po-Shen Loh标识

LIVE

中文

2021 AMC 12A Fall 第 24 题

先试着解答 2021 AMC 12A Fall 第 24 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 2021 AMC 12A Fall 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。

24.

凸四边形 ABCDABCD 满足 AB=18AB = 18、∠A=60∘\angle A = 60^\circ，且 AB‾∥CD‾\overline{AB} \parallel \overline{CD}。四条边的长度按某种顺序构成一个等差数列，并且边 ABAB 是最大长度的边。另一条边的长度为 aa。所有可能的 aa 值之和是多少？

Convex quadrilateral ABCDABCD has AB=18,AB = 18, ∠A=60∘,\angle A = 60^\circ, and AB‾∥CD‾.\overline{AB} \parallel \overline{CD}. In some order, the lengths of the four sides form an arithmetic progression, and side ABAB is a side of maximum length. The length of another side is a.a. What is the sum of all possible values of a?a?

2424

4242

6060

6666

8484

答案：E
知识点：梯形等差数列分类讨论
难度评级：2520
解答：

因为 AB=18AB = 18 是最大边，四条边为 18,18−d,18−2d,18−3d.18, 18 - d, 18 - 2d, 18 - 3d. 取 A=(0,0),A = (0,0), B=(18,0),B = (18,0), D=(m2,m32)D = \left(\tfrac{m}{2}, \tfrac{m\sqrt3}{2}\right)，其中 m=DA,m=DA,令 n=CDn=CD 且 ℓ=BC.\ell=BC. 则 C=(m2+n,m32),C=(\tfrac m2+n,\tfrac{m\sqrt3}{2}),所以 ℓ2=m2+(18−n)2−m(18−n). \begin{aligned} \ell^2&=m^2+(18-n)^2 \\ &\quad {}-m(18-n). \end{aligned}

令 uj=18−jdu_j=18-jd，其中 j=1,2,3.j=1,2,3. 将 (u1,u2,u3)(u_1,u_2,u_3) 的排列 123,132,213,231,312,321123,132,213,231,312,321 依次代入 (m,n,ℓ)(m,n,\ell)，得到非零候选值 d=18,2,9,5,6,6,d=18,2,9,5,6,6,因为 u3u_3 为正要求 d<6,d<6,所以只留下 d=2d=2 和 d=5.d=5. 情形 d=0d=0 也给出一个有效的菱形。

因此边长集合为 {18,16,14,12},\{18,16,14,12\}, {18,13,8,3},\{18,13,8,3\}, 和 {18,18,18,18}.\{18,18,18,18\}.

非 ABAB 边长的可能值为 {3,8,12,13,14,16,18},\{3, 8, 12, 13, 14, 16, 18\},其和为 84.84.

因此，正确答案是 E。

Since AB=18AB = 18 is the largest, the four sides are 18,18−d,18−2d,18−3d.18, 18 - d, 18 - 2d, 18 - 3d. Placing A=(0,0),A = (0,0), B=(18,0),B = (18,0), and D=(m2,m32)D = \left(\tfrac{m}{2}, \tfrac{m\sqrt3}{2}\right) with m=DA,m=DA, let n=CDn=CD and ℓ=BC.\ell=BC. Then C=(m2+n,m32),C=(\tfrac m2+n,\tfrac{m\sqrt3}{2}), so ℓ2=m2+(18−n)2−m(18−n). \begin{aligned} \ell^2&=m^2+(18-n)^2 \\ &\quad {}-m(18-n). \end{aligned}

Write uj=18−jdu_j=18-jd for j=1,2,3.j=1,2,3. Substituting the permutations 123,132,213,231,312,321123,132,213,231,312,321 of (u1,u2,u3)(u_1,u_2,u_3) for (m,n,ℓ)(m,n,\ell) gives the nonzero candidates d=18,2,9,5,6,6,d=18,2,9,5,6,6, respectively. Positivity of u3u_3 requires d<6,d<6, leaving only d=2d=2 and d=5.d=5. The case d=0d=0 also gives a valid rhombus.

Thus the side sets are {18,16,14,12},\{18,16,14,12\}, {18,13,8,3},\{18,13,8,3\}, and {18,18,18,18}.\{18,18,18,18\}.

The possible values of a non-ABAB side length are {3,8,12,13,14,16,18},\{3, 8, 12, 13, 14, 16, 18\}, whose sum is 84.84.

Thus, the correct answer is E.

← 第 23 题#23
完整试卷
第 25 题#25 →

其他年份的第 24 题

1999 AMC 12 · 2000 AMC 12 · 2001 AMC 12 · 2002 AMC 12A · 2002 AMC 12B · 2003 AMC 12A · 2003 AMC 12B · 2004 AMC 12A · 2004 AMC 12B · 2005 AMC 12A · 2005 AMC 12B · 2006 AMC 12A · 2006 AMC 12B · 2007 AMC 12A · 2007 AMC 12B · 2008 AMC 12A · 2008 AMC 12B · 2009 AMC 12A · 2009 AMC 12B · 2010 AMC 12A · 2010 AMC 12B · 2011 AMC 12A · 2011 AMC 12B · 2012 AMC 12A · 2012 AMC 12B · 2013 AMC 12A · 2013 AMC 12B · 2014 AMC 12A · 2014 AMC 12B · 2015 AMC 12A · 2015 AMC 12B · 2016 AMC 12A · 2016 AMC 12B · 2017 AMC 12A · 2017 AMC 12B · 2018 AMC 12A · 2018 AMC 12B · 2019 AMC 12A · 2019 AMC 12B · 2020 AMC 12A · 2020 AMC 12B · 2021 AMC 12A Spring · 2021 AMC 12B Spring · 2021 AMC 12B Fall · 2022 AMC 12A · 2022 AMC 12B · 2023 AMC 12A · 2023 AMC 12B · 2024 AMC 12A · 2024 AMC 12B · 2025 AMC 12A · 2025 AMC 12B