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2001

2001 AMC12 真题及解析2001 AMC12 Paper & Solutions 精选Pick

25 题含解析,AMC12 早期真题,题型经典。25 questions with solutions, early AMC12 exam, classic problem types.

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扫码免费领取 2001 PDF版官方真题+答案Scan to get free 2001 PDF + solutions

25 道选择题25 Questions
75 分钟限时75 Minutes
满分 150 分Max Score 150
不可用计算器No Calculator
Exam Overview

考试概览Exam Overview

题型经典适合基础训练。This exam emphasizes multiple math topics. Overall difficulty is relatively basic.

D难度分布Difficulty

  • Easy 基础Easy第 1-10 题Q1-10
  • Medium 中等Medium第 11-20 题Q11-20
  • Hard 较难Hard第 21-25 题Q21-25

T考点分布Topics

  • 代数Algebra35%
  • 几何Geometry30%
  • 数论Number Theory15%
  • 组合Combinatorics20%

A奖项分数线Awards

DHR 卓越荣誉奖AIME Qualification
AIME Qualification (Top 2.5%)
86+
HR 荣誉奖Honor Roll
Honor Roll (Top 5%)
76+
Achievement Roll
十年级及以下Grade 10 and below
61+
Sample Problems

2001 AMC12 真题参考示例(共 12 题)2001 AMC12 Sample Problems (12 questions)

以下为 Easy / Medium / Hard 难度参考示例题目,仅供练习参考,点击选项查看答案Sample reference problems by difficulty — click an option to check your answer

第 1 题Q1Easy多项式求值Polynomial
若 f(x) = 2x² - 4x + 2,求 f(3) 的值。If f(x) = 2x² - 4x + 2, find f(3).
A) 4
B) 8
C) 6
D) 10
E) 12

解题步骤Steps

1f(3) = 2×3² - 4×3 + 2
2= 2×9 - 12 + 2 = 8
正确答案:BAnswer: B代入后先乘方再乘除最后加减Substitute, then exponentiate, multiply/divide, add/subtract
第 3 题Q3Easy等差数列Arithmetic Seq
等差数列首项 a₁=4,公差 d=3,求第 11 项。Arithmetic sequence: a₁=4, d=3, find the 11-th term.
A) 26
B) 30
C) 34
D) 38
E) 42

解题步骤Steps

1aₙ = a₁ + (n-1)d
2a_11 = 4 + 10×3 = 34
正确答案:CAnswer: C通项公式 aₙ = a₁ + (n-1)dGeneral term: aₙ=a₁+(n-1)d
第 5 题Q5Easy指数运算Exponent
若 3ˣ = 81,求 x 的值。If 3ˣ = 81, find x.
A) 2
B) 3
C) 5
D) 4
E) 6

解题步骤Steps

181 = 3^4
2故 x = 4
正确答案:DAnswer: D化为同底数幂,比较指数Convert to same base, compare exponents
第 7 题Q7Easy韦达定理Vieta's
方程 x² - 6x + 5 = 0 的两根之和是多少?Sum of roots of x² - 6x + 5 = 0?
A) 2
B) 4
C) 8
D) 10
E) 6

解题步骤Steps

1韦达定理:两根之和 = -(-6)/1 = 6
2两根之积 = 5
正确答案:EAnswer: Ex²-px+q=0 两根之和为 pFor x²-px+q=0, sum of roots = p
第 10 题Q10Medium整除计数Divisibility
从 1 到 31 的正整数中,有多少个能被 4 整除?How many integers from 1 to 31 are divisible by 4?
A) 7
B) 3
C) 5
D) 9
E) 11

解题步骤Steps

1⌊31 / 4⌋ = 7
正确答案:AAnswer: A能被 k 整除的个数 = ⌊n/k⌋Count divisible by k: ⌊n/k⌋
第 12 题Q12Medium多边形对角线Diagonals
正 8 边形共有多少条对角线?How many diagonals does a regular 8-gon have?
A) 14
B) 20
C) 17
D) 23
E) 26

解题步骤Steps

1对角线数 = n(n-3)/2
2= 8×5/2 = 20
正确答案:BAnswer: Bn 边形对角线公式 n(n-3)/2n-gon diagonals: n(n-3)/2
第 15 题Q15Medium组合数Combination
计算 C(7, 3) 的值。Compute C(7, 3).
A) 27
B) 31
C) 35
D) 39
E) 43

解题步骤Steps

1C(n,3) = n(n-1)(n-2)/6
2= 7×6×5/6 = 35
正确答案:CAnswer: C组合数公式 C(n,k)=n!/(k!(n-k)!)Combination: C(n,k)=n!/(k!(n-k)!)
第 17 题Q17Medium勾股定理Pythagorean
直角三角形两直角边长为 5 和 12,求斜边长。Right triangle legs 5 and 12, find the hypotenuse.
A) 9
B) 11
C) 15
D) 13
E) 17

解题步骤Steps

1c² = 5² + 12² = 25 + 144 = 169
2c = √169 = 13
正确答案:DAnswer: D勾股定理 a²+b²=c²Pythagorean theorem a²+b²=c²
第 20 题Q20Hard模运算Modular Arith
求 4^5 除以 6 的余数。Find the remainder of 4^5 divided by 6.
A) 2
B) 3
C) 5
D) 6
E) 4

解题步骤Steps

1计算 4^5 mod 6
2由模运算性质逐步化简
3=4
正确答案:EAnswer: E模运算可逐步取余简化大指数Modular arithmetic simplifies large exponents
第 22 题Q22Hard配方法Completing Sq
函数 f(x) = x² - 6x + 13 的最小值是多少?Find the minimum of f(x) = x² - 6x + 13.
A) 4
B) 0
C) 2
D) 6
E) 8

解题步骤Steps

1配方:f(x) = (x - 3)² + 4
2当 x = 3 时取最小值 4
正确答案:AAnswer: A配方法是求二次函数最值的标准方法Completing the square for quadratic extrema
第 24 题Q24Hard容斥原理Inclusion-Excl
将 5 个不同球放入 3 个不同盒子,每盒至少 1 球,有多少种放法?5 distinct balls into 3 distinct boxes, each ≥1 ball, how many ways?
A) 118
B) 150
C) 134
D) 166
E) 182

解题步骤Steps

1总数 3^5 = 243
2减去有空盒: -C(3,1)×2^5 = -96
3加回两个空盒: +C(3,2)×1 = +3
4总计 243 - 96 + 3 = 150
正确答案:BAnswer: B容斥原理处理"至少"条件Inclusion-exclusion handles 'at least'
第 25 题Q25Hard余弦定理Law of Cosines
△ABC 中 a=6, b=7, cos C=1/2,求 c²。In △ABC, a=6, b=7, cos C=1/2, find c².
A) 33
B) 38
C) 43
D) 48
E) 53

解题步骤Steps

1余弦定理 c² = a² + b² - 2ab·cos C
2= 36 + 49 - 42 = 43
正确答案:CAnswer: C余弦定理是解三角形的核心Law of cosines is key for solving triangles

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