选择卷别Paper Set

2019

2019 AMC12 真题及解析2019 AMC12 Paper & Solutions 精选Pick A卷Paper A

25 题完整解析，代数侧重方程应用，几何涉及圆的基本性质。组合出现递推思想。25 complete solutions, algebra focuses on equation applications, geometry covers circle properties. Combinatorics introduces recursion.

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

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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

代数侧重方程应用，组合出现递推思想。This exam emphasizes algebra, combinatorics, equations. 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%)

92+

HR 荣誉奖Honor Roll

Honor Roll (Top 5%)

78+

Achievement Roll

十年级及以下Grade 10 and below

61+

Sample Problems

2019 AMC12 真题参考示例（共 12 题）2019 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 - 7x + 2，求 f(5) 的值。If f(x) = 2x² - 7x + 2, find f(5).

A) 11

B) 14

C) 20

D) 23

E) 17

解题步骤Steps

1f(5) = 2×5² - 7×5 + 2

2= 2×25 - 35 + 2 = 17

正确答案：EAnswer: E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=6，公差 d=6，求第 11 项。Arithmetic sequence: a₁=6, d=6, find the 11-th term.

A) 66

B) 52

C) 59

D) 73

E) 80

解题步骤Steps

1aₙ = a₁ + (n-1)d

2a_11 = 6 + 10×6 = 66

正确答案：AAnswer: A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 = 729，求 x 的值。If 3ˣ = 729, find x.

A) 4

B) 6

C) 5

D) 7

E) 8

解题步骤Steps

1729 = 3^6

2故 x = 6

正确答案：BAnswer: B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 + 7 = 0 的两根之和是多少？Sum of roots of x² - 9x + 7 = 0?

A) 5

B) 7

C) 9

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 = 7

正确答案：CAnswer: Cx²-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 49 的正整数中，有多少个能被 4 整除？How many integers from 1 to 49 are divisible by 4?

A) 8

B) 10

C) 14

D) 12

E) 16

解题步骤Steps

1⌊49 / 4⌋ = 12

正确答案：DAnswer: D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) 17

C) 23

D) 26

E) 20

解题步骤Steps

1对角线数 = n(n-3)/2

2= 8×5/2 = 20

正确答案：EAnswer: En 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) 120

B) 94

C) 107

D) 133

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：AAnswer: A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) 25

C) 23

D) 27

E) 29

解题步骤Steps

1c² = 7² + 24² = 49 + 576 = 625

2c = √625 = 25

正确答案：BAnswer: B勾股定理 a²+b²=c²Pythagorean theorem a²+b²=c²

第 20 题Q20Hard模运算Modular Arith

求 6^5 除以 8 的余数。Find the remainder of 6^5 divided by 8.

A) 1

B) 2

C) 0

D) 3

E) 4

解题步骤Steps

1计算 6^5 mod 8

2由模运算性质逐步化简

3=0

正确答案：CAnswer: C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 - 10x + 29 的最小值是多少？Find the minimum of f(x) = x² - 10x + 29.

A) 0

B) 2

C) 6

D) 4

E) 8

解题步骤Steps

1配方：f(x) = (x - 5)² + 4

2当 x = 5 时取最小值 4

正确答案：DAnswer: D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) 134

C) 166

D) 182

E) 150

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：EAnswer: E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) 43

B) 33

C) 38

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：AAnswer: A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 - 9x + 2，求 f(5) 的值。If f(x) = 2x² - 9x + 2, find f(5).

A) 3

B) 5

C) 9

D) 11

E) 7

解题步骤Steps

1f(5) = 2×5² - 9×5 + 2

2= 2×25 - 45 + 2 = 7

正确答案：EAnswer: E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=8，公差 d=6，求第 11 项。Arithmetic sequence: a₁=8, d=6, find the 11-th term.

A) 68

B) 54

C) 61

D) 75

E) 82

解题步骤Steps

1aₙ = a₁ + (n-1)d

2a_11 = 8 + 10×6 = 68

正确答案：AAnswer: A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 = 729，求 x 的值。If 3ˣ = 729, find x.

A) 4

B) 6

C) 5

D) 7

E) 8

解题步骤Steps

1729 = 3^6

2故 x = 6

正确答案：BAnswer: B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 - 11x + 7 = 0 的两根之和是多少？Sum of roots of x² - 11x + 7 = 0?

A) 7

B) 9

C) 11

D) 13

E) 15

解题步骤Steps

1韦达定理：两根之和 = -(-11)/1 = 11

2两根之积 = 7

正确答案：CAnswer: Cx²-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 51 的正整数中，有多少个能被 4 整除？How many integers from 1 to 51 are divisible by 4?

A) 8

B) 10

C) 14

D) 12

E) 16

解题步骤Steps

1⌊51 / 4⌋ = 12

正确答案：DAnswer: D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 10 边形共有多少条对角线？How many diagonals does a regular 10-gon have?

A) 29

B) 32

C) 38

D) 41

E) 35

解题步骤Steps

1对角线数 = n(n-3)/2

2= 10×7/2 = 35

正确答案：EAnswer: En 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(12, 3) 的值。Compute C(12, 3).

A) 220

B) 174

C) 197

D) 243

E) 266

解题步骤Steps

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

2= 12×11×10/6 = 220

正确答案：AAnswer: A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) 13

C) 11

D) 15

E) 17

解题步骤Steps

1c² = 5² + 12² = 25 + 144 = 169

2c = √169 = 13

正确答案：BAnswer: B勾股定理 a²+b²=c²Pythagorean theorem a²+b²=c²

第 20 题Q20Hard模运算Modular Arith

求 6^5 除以 8 的余数。Find the remainder of 6^5 divided by 8.

A) 1

B) 2

C) 0

D) 3

E) 4

解题步骤Steps

1计算 6^5 mod 8

2由模运算性质逐步化简

3=0

正确答案：CAnswer: C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 - 10x + 31 的最小值是多少？Find the minimum of f(x) = x² - 10x + 31.

A) 2

B) 4

C) 8

D) 6

E) 10

解题步骤Steps

1配方：f(x) = (x - 5)² + 6

2当 x = 5 时取最小值 6

正确答案：DAnswer: D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) 134

C) 166

D) 182

E) 150

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：EAnswer: E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=8, b=7, cos C=1/2，求 c²。In △ABC, a=8, b=7, cos C=1/2, find c².

A) 57

B) 45

C) 51

D) 63

E) 69

解题步骤Steps

1余弦定理 c² = a² + b² - 2ab·cos C

2= 64 + 49 - 56 = 57

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

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2019 AMC12 真题及解析2019 AMC12 Paper & Solutions 精选Pick A卷Paper A

扫码免费领取 2019 PDF版官方真题+答案Scan to get free 2019 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

2019 AMC12 真题参考示例（共 12 题）2019 AMC12 Sample Problems (12 questions)

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