选择卷别Paper Set

2002

2002 AMC12 真题及解析2002 AMC12 Paper & Solutions 精选Pick A卷Paper A

25 题含解析，早期真题，适合入门练习。25 questions with solutions, early exam, good for beginners.

查看真题示例Sample Problems 下载完整PDF版真题Download PDF

扫码免费领取 2002 PDF版官方真题+答案Scan to get free 2002 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

早期真题适合入门。This exam emphasizes multiple math topics. Overall difficulty is moderate.

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%)

87+

HR 荣誉奖Honor Roll

Honor Roll (Top 5%)

77+

Achievement Roll

十年级及以下Grade 10 and below

62+

Sample Problems

2002 AMC12 真题参考示例（共 12 题）2002 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 - 5x + 3，求 f(4) 的值。If f(x) = 2x² - 5x + 3, find f(4).

A) 11

B) 13

C) 15

D) 17

E) 19

解题步骤Steps

1f(4) = 2×4² - 5×4 + 3

2= 2×16 - 20 + 3 = 15

正确答案：CAnswer: C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=5，公差 d=4，求第 12 项。Arithmetic sequence: a₁=5, d=4, find the 12-th term.

A) 39

B) 44

C) 54

D) 49

E) 59

解题步骤Steps

1aₙ = a₁ + (n-1)d

2a_12 = 5 + 11×4 = 49

正确答案：DAnswer: D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 4ˣ = 1024，求 x 的值。If 4ˣ = 1024, find x.

A) 3

B) 4

C) 6

D) 7

E) 5

解题步骤Steps

11024 = 4^5

2故 x = 5

正确答案：EAnswer: E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 - 7x + 6 = 0 的两根之和是多少？Sum of roots of x² - 7x + 6 = 0?

A) 7

B) 3

C) 5

D) 9

E) 11

解题步骤Steps

1韦达定理：两根之和 = -(-7)/1 = 7

2两根之积 = 6

正确答案：AAnswer: Ax²-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 32 的正整数中，有多少个能被 5 整除？How many integers from 1 to 32 are divisible by 5?

A) 2

B) 6

C) 4

D) 8

E) 10

解题步骤Steps

1⌊32 / 5⌋ = 6

正确答案：BAnswer: B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 9 边形共有多少条对角线？How many diagonals does a regular 9-gon have?

A) 21

B) 24

C) 27

D) 30

E) 33

解题步骤Steps

1对角线数 = n(n-3)/2

2= 9×6/2 = 27

正确答案：CAnswer: Cn 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(8, 3) 的值。Compute C(8, 3).

A) 44

B) 50

C) 62

D) 56

E) 68

解题步骤Steps

1C(n,3) = n(n-1)(n-2)/6

2= 8×7×6/6 = 56

正确答案：DAnswer: D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 6 和 8，求斜边长。Right triangle legs 6 and 8, find the hypotenuse.

A) 6

B) 8

C) 12

D) 14

E) 10

解题步骤Steps

1c² = 6² + 8² = 36 + 64 = 100

2c = √100 = 10

正确答案：EAnswer: E勾股定理 a²+b²=c²Pythagorean theorem a²+b²=c²

第 20 题Q20Hard模运算Modular Arith

求 5^6 除以 7 的余数。Find the remainder of 5^6 divided by 7.

A) 1

B) 0

C) 2

D) 3

E) 4

解题步骤Steps

1计算 5^6 mod 7

2由模运算性质逐步化简

3=1

正确答案：AAnswer: A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 - 8x + 21 的最小值是多少？Find the minimum of f(x) = x² - 8x + 21.

A) 1

B) 5

C) 3

D) 7

E) 9

解题步骤Steps

1配方：f(x) = (x - 4)² + 5

2当 x = 4 时取最小值 5

正确答案：BAnswer: B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 6 个不同球放入 3 个不同盒子，每盒至少 1 球，有多少种放法？6 distinct balls into 3 distinct boxes, each ≥1 ball, how many ways?

A) 430

B) 485

C) 540

D) 595

E) 650

解题步骤Steps

1总数 3^6 = 729

2减去有空盒: -C(3,1)×2^6 = -192

3加回两个空盒: +C(3,2)×1 = +3

4总计 729 - 192 + 3 = 540

正确答案：CAnswer: C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=7, b=8, cos C=1/2，求 c²。In △ABC, a=7, b=8, cos C=1/2, find c².

A) 45

B) 51

C) 63

D) 57

E) 69

解题步骤Steps

1余弦定理 c² = a² + b² - 2ab·cos C

2= 49 + 64 - 56 = 57

正确答案：DAnswer: D余弦定理是解三角形的核心Law of cosines is key for solving triangles

第 1 题Q1Easy多项式求值Polynomial

若 f(x) = 2x² - 7x + 3，求 f(4) 的值。If f(x) = 2x² - 7x + 3, find f(4).

A) 3

B) 5

C) 7

D) 9

E) 11

解题步骤Steps

1f(4) = 2×4² - 7×4 + 3

2= 2×16 - 28 + 3 = 7

正确答案：CAnswer: C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=7，公差 d=4，求第 12 项。Arithmetic sequence: a₁=7, d=4, find the 12-th term.

A) 39

B) 45

C) 57

D) 51

E) 63

解题步骤Steps

1aₙ = a₁ + (n-1)d

2a_12 = 7 + 11×4 = 51

正确答案：DAnswer: D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 4ˣ = 1024，求 x 的值。If 4ˣ = 1024, find x.

A) 3

B) 4

C) 6

D) 7

E) 5

解题步骤Steps

11024 = 4^5

2故 x = 5

正确答案：EAnswer: E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 - 9x + 6 = 0 的两根之和是多少？Sum of roots of x² - 9x + 6 = 0?

A) 9

B) 5

C) 7

D) 11

E) 13

解题步骤Steps

1韦达定理：两根之和 = -(-9)/1 = 9

2两根之积 = 6

正确答案：AAnswer: Ax²-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 34 的正整数中，有多少个能被 5 整除？How many integers from 1 to 34 are divisible by 5?

A) 2

B) 6

C) 4

D) 8

E) 10

解题步骤Steps

1⌊34 / 5⌋ = 6

正确答案：BAnswer: B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 11 边形共有多少条对角线？How many diagonals does a regular 11-gon have?

A) 38

B) 41

C) 44

D) 47

E) 50

解题步骤Steps

1对角线数 = n(n-3)/2

2= 11×8/2 = 44

正确答案：CAnswer: Cn 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(10, 3) 的值。Compute C(10, 3).

A) 94

B) 107

C) 133

D) 120

E) 146

解题步骤Steps

1C(n,3) = n(n-1)(n-2)/6

2= 10×9×8/6 = 120

正确答案：DAnswer: D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 7 和 24，求斜边长。Right triangle legs 7 and 24, find the hypotenuse.

A) 21

B) 23

C) 27

D) 29

E) 25

解题步骤Steps

1c² = 7² + 24² = 49 + 576 = 625

2c = √625 = 25

正确答案：EAnswer: E勾股定理 a²+b²=c²Pythagorean theorem a²+b²=c²

第 20 题Q20Hard模运算Modular Arith

求 5^6 除以 7 的余数。Find the remainder of 5^6 divided by 7.

A) 1

B) 0

C) 2

D) 3

E) 4

解题步骤Steps

1计算 5^6 mod 7

2由模运算性质逐步化简

3=1

正确答案：AAnswer: A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 - 8x + 23 的最小值是多少？Find the minimum of f(x) = x² - 8x + 23.

A) 3

B) 7

C) 5

D) 9

E) 11

解题步骤Steps

1配方：f(x) = (x - 4)² + 7

2当 x = 4 时取最小值 7

正确答案：BAnswer: B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 6 个不同球放入 3 个不同盒子，每盒至少 1 球，有多少种放法？6 distinct balls into 3 distinct boxes, each ≥1 ball, how many ways?

A) 430

B) 485

C) 540

D) 595

E) 650

解题步骤Steps

1总数 3^6 = 729

2减去有空盒: -C(3,1)×2^6 = -192

3加回两个空盒: +C(3,2)×1 = +3

4总计 729 - 192 + 3 = 540

正确答案：CAnswer: C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=9, b=8, cos C=1/2，求 c²。In △ABC, a=9, b=8, cos C=1/2, find c².

A) 57

B) 65

C) 81

D) 73

E) 89

解题步骤Steps

1余弦定理 c² = a² + b² - 2ab·cos C

2= 81 + 64 - 72 = 73

正确答案：DAnswer: D余弦定理是解三角形的核心Law of cosines is key for solving triangles

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2000-2025 全部 AMC12 真题年份，点击即可进入对应年份页面All AMC12 exam years from 2000-2025, click to view

2002 AMC12 真题及解析2002 AMC12 Paper & Solutions 精选Pick A卷Paper A

扫码免费领取 2002 PDF版官方真题+答案Scan to get free 2002 PDF + solutions

考试概览Exam Overview

D难度分布Difficulty

T考点分布Topics

A奖项分数线Awards

2002 AMC12 真题参考示例（共 12 题）2002 AMC12 Sample Problems (12 questions)

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