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2023

2023 AMC12 真题及解析2023 AMC12 Paper & Solutions 精选Pick A卷Paper A

25 题逐步拆解解析，含考点归纳与解法对比。组合计数题难度突出，几何涉及相似三角形。25 questions with step-by-step solutions, topic summaries and method comparisons. Hard combinatorics, similar triangles in geometry.

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

组合计数题难度突出，几何涉及相似三角形。This exam emphasizes geometry, combinatorics. 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%)

96+

HR 荣誉奖Honor Roll

Honor Roll (Top 5%)

82+

Achievement Roll

十年级及以下Grade 10 and below

65+

Sample Problems

2023 AMC12 真题参考示例（共 12 题）2023 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 - 6x + 3，求 f(5) 的值。If f(x) = 2x² - 6x + 3, find f(5).

A) 17

B) 20

C) 26

D) 23

E) 29

解题步骤Steps

1f(5) = 2×5² - 6×5 + 3

2= 2×25 - 30 + 3 = 23

正确答案：DAnswer: D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=5，求第 15 项。Arithmetic sequence: a₁=6, d=5, find the 15-th term.

A) 60

B) 68

C) 84

D) 92

E) 76

解题步骤Steps

1aₙ = a₁ + (n-1)d

2a_15 = 6 + 14×5 = 76

正确答案：EAnswer: E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 = 4096，求 x 的值。If 4ˣ = 4096, find x.

A) 6

B) 4

C) 5

D) 7

E) 8

解题步骤Steps

14096 = 4^6

2故 x = 6

正确答案：AAnswer: A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 - 8x + 7 = 0 的两根之和是多少？Sum of roots of x² - 8x + 7 = 0?

A) 4

B) 8

C) 6

D) 10

E) 12

解题步骤Steps

1韦达定理：两根之和 = -(-8)/1 = 8

2两根之积 = 7

正确答案：BAnswer: Bx²-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 33 的正整数中，有多少个能被 5 整除？How many integers from 1 to 33 are divisible by 5?

A) 2

B) 4

C) 6

D) 8

E) 10

解题步骤Steps

1⌊33 / 5⌋ = 6

正确答案：CAnswer: C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 12 边形共有多少条对角线？How many diagonals does a regular 12-gon have?

A) 48

B) 51

C) 57

D) 54

E) 60

解题步骤Steps

1对角线数 = n(n-3)/2

2= 12×9/2 = 54

正确答案：DAnswer: Dn 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(9, 3) 的值。Compute C(9, 3).

A) 66

B) 75

C) 93

D) 102

E) 84

解题步骤Steps

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

2= 9×8×7/6 = 84

正确答案：EAnswer: E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 8 和 15，求斜边长。Right triangle legs 8 and 15, find the hypotenuse.

A) 17

B) 13

C) 15

D) 19

E) 21

解题步骤Steps

1c² = 8² + 15² = 64 + 225 = 289

2c = √289 = 17

正确答案：AAnswer: A勾股定理 a²+b²=c²Pythagorean theorem a²+b²=c²

第 20 题Q20Hard模运算Modular Arith

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

A) 1

B) 0

C) 2

D) 3

E) 4

解题步骤Steps

1计算 6^6 mod 8

2由模运算性质逐步化简

3=0

正确答案：BAnswer: B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 + 30 的最小值是多少？Find the minimum of f(x) = x² - 10x + 30.

A) 1

B) 3

C) 5

D) 7

E) 9

解题步骤Steps

1配方：f(x) = (x - 5)² + 5

2当 x = 5 时取最小值 5

正确答案：CAnswer: C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) 595

D) 540

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：DAnswer: D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) 69

E) 57

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：EAnswer: E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 - 8x + 3，求 f(5) 的值。If f(x) = 2x² - 8x + 3, find f(5).

A) 9

B) 11

C) 15

D) 13

E) 17

解题步骤Steps

1f(5) = 2×5² - 8×5 + 3

2= 2×25 - 40 + 3 = 13

正确答案：DAnswer: D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=5，求第 15 项。Arithmetic sequence: a₁=8, d=5, find the 15-th term.

A) 62

B) 70

C) 86

D) 94

E) 78

解题步骤Steps

1aₙ = a₁ + (n-1)d

2a_15 = 8 + 14×5 = 78

正确答案：EAnswer: E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 = 4096，求 x 的值。If 4ˣ = 4096, find x.

A) 6

B) 4

C) 5

D) 7

E) 8

解题步骤Steps

14096 = 4^6

2故 x = 6

正确答案：AAnswer: A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 - 10x + 7 = 0 的两根之和是多少？Sum of roots of x² - 10x + 7 = 0?

A) 6

B) 10

C) 8

D) 12

E) 14

解题步骤Steps

1韦达定理：两根之和 = -(-10)/1 = 10

2两根之积 = 7

正确答案：BAnswer: Bx²-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 35 的正整数中，有多少个能被 5 整除？How many integers from 1 to 35 are divisible by 5?

A) 3

B) 5

C) 7

D) 9

E) 11

解题步骤Steps

1⌊35 / 5⌋ = 7

正确答案：CAnswer: C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 14 边形共有多少条对角线？How many diagonals does a regular 14-gon have?

A) 71

B) 74

C) 80

D) 77

E) 83

解题步骤Steps

1对角线数 = n(n-3)/2

2= 14×11/2 = 77

正确答案：DAnswer: Dn 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(11, 3) 的值。Compute C(11, 3).

A) 131

B) 148

C) 182

D) 199

E) 165

解题步骤Steps

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

2= 11×10×9/6 = 165

正确答案：EAnswer: E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 3 和 4，求斜边长。Right triangle legs 3 and 4, find the hypotenuse.

A) 5

B) 1

C) 3

D) 7

E) 9

解题步骤Steps

1c² = 3² + 4² = 9 + 16 = 25

2c = √25 = 5

正确答案：AAnswer: A勾股定理 a²+b²=c²Pythagorean theorem a²+b²=c²

第 20 题Q20Hard模运算Modular Arith

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

A) 1

B) 0

C) 2

D) 3

E) 4

解题步骤Steps

1计算 6^6 mod 8

2由模运算性质逐步化简

3=0

正确答案：BAnswer: B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 + 32 的最小值是多少？Find the minimum of f(x) = x² - 10x + 32.

A) 3

B) 5

C) 7

D) 9

E) 11

解题步骤Steps

1配方：f(x) = (x - 5)² + 7

2当 x = 5 时取最小值 7

正确答案：CAnswer: C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) 595

D) 540

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：DAnswer: D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) 89

E) 73

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：EAnswer: E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

2023 AMC12 真题及解析2023 AMC12 Paper & Solutions 精选Pick A卷Paper A

扫码免费领取 2023 PDF版官方真题+答案Scan to get free 2023 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

2023 AMC12 真题参考示例（共 12 题）2023 AMC12 Sample Problems (12 questions)

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