选择卷别Paper Set

2016

2016 AMC12 真题及解析2016 AMC12 Paper & Solutions 精选Pick A卷Paper A

25 题含解析，代数几何均衡，组合题出现容斥原理应用。整体难度适中。25 questions with solutions, balanced algebra and geometry, inclusion-exclusion in combinatorics. Moderate difficulty.

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

代数几何均衡，组合题涉及容斥原理。This exam emphasizes algebra, 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%)

89+

HR 荣誉奖Honor Roll

Honor Roll (Top 5%)

75+

Achievement Roll

十年级及以下Grade 10 and below

64+

Sample Problems

2016 AMC12 真题参考示例（共 12 题）2016 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(2) 的值。If f(x) = 2x² - 4x + 2, find f(2).

A) 0

B) 2

C) 4

D) 6

E) 8

解题步骤Steps

1f(2) = 2×2² - 4×2 + 2

2= 2×4 - 8 + 2 = 2

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=3，公差 d=3，求第 14 项。Arithmetic sequence: a₁=3, d=3, find the 14-th term.

A) 32

B) 37

C) 42

D) 47

E) 52

解题步骤Steps

1aₙ = a₁ + (n-1)d

2a_14 = 3 + 13×3 = 42

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 = 27，求 x 的值。If 3ˣ = 27, find x.

A) 1

B) 2

C) 4

D) 3

E) 5

解题步骤Steps

127 = 3^3

2故 x = 3

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 + 4 = 0 的两根之和是多少？Sum of roots of x² - 6x + 4 = 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 = 4

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 46 的正整数中，有多少个能被 4 整除？How many integers from 1 to 46 are divisible by 4?

A) 11

B) 7

C) 9

D) 13

E) 15

解题步骤Steps

1⌊46 / 4⌋ = 11

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 11 边形共有多少条对角线？How many diagonals does a regular 11-gon have?

A) 38

B) 44

C) 41

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：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 3^5 除以 5 的余数。Find the remainder of 3^5 divided by 5.

A) 1

B) 2

C) 4

D) 5

E) 3

解题步骤Steps

1计算 3^5 mod 5

2由模运算性质逐步化简

3=3

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 - 4x + 8 的最小值是多少？Find the minimum of f(x) = x² - 4x + 8.

A) 4

B) 0

C) 2

D) 6

E) 8

解题步骤Steps

1配方：f(x) = (x - 2)² + 4

2当 x = 2 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

第 1 题Q1Easy多项式求值Polynomial

若 f(x) = 2x² - 6x + 2，求 f(2) 的值。If f(x) = 2x² - 6x + 2, find f(2).

A) -6

B) -2

C) -4

D) 0

E) 2

解题步骤Steps

1f(2) = 2×2² - 6×2 + 2

2= 2×4 - 12 + 2 = -2

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=5，公差 d=3，求第 14 项。Arithmetic sequence: a₁=5, d=3, find the 14-th term.

A) 34

B) 39

C) 44

D) 49

E) 54

解题步骤Steps

1aₙ = a₁ + (n-1)d

2a_14 = 5 + 13×3 = 44

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 = 27，求 x 的值。If 3ˣ = 27, find x.

A) 1

B) 2

C) 4

D) 3

E) 5

解题步骤Steps

127 = 3^3

2故 x = 3

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 - 8x + 4 = 0 的两根之和是多少？Sum of roots of x² - 8x + 4 = 0?

A) 4

B) 6

C) 10

D) 12

E) 8

解题步骤Steps

1韦达定理：两根之和 = -(-8)/1 = 8

2两根之积 = 4

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 48 的正整数中，有多少个能被 4 整除？How many integers from 1 to 48 are divisible by 4?

A) 12

B) 8

C) 10

D) 14

E) 16

解题步骤Steps

1⌊48 / 4⌋ = 12

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 13 边形共有多少条对角线？How many diagonals does a regular 13-gon have?

A) 59

B) 65

C) 62

D) 68

E) 71

解题步骤Steps

1对角线数 = n(n-3)/2

2= 13×10/2 = 65

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(9, 3) 的值。Compute C(9, 3).

A) 66

B) 75

C) 84

D) 93

E) 102

解题步骤Steps

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

2= 9×8×7/6 = 84

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 8 和 15，求斜边长。Right triangle legs 8 and 15, find the hypotenuse.

A) 13

B) 15

C) 19

D) 17

E) 21

解题步骤Steps

1c² = 8² + 15² = 64 + 225 = 289

2c = √289 = 17

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 3^5 除以 5 的余数。Find the remainder of 3^5 divided by 5.

A) 1

B) 2

C) 4

D) 5

E) 3

解题步骤Steps

1计算 3^5 mod 5

2由模运算性质逐步化简

3=3

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 - 4x + 10 的最小值是多少？Find the minimum of f(x) = x² - 4x + 10.

A) 6

B) 2

C) 4

D) 8

E) 10

解题步骤Steps

1配方：f(x) = (x - 2)² + 6

2当 x = 2 时取最小值 6

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

A) 45

B) 51

C) 57

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：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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2016 AMC12 真题及解析2016 AMC12 Paper & Solutions 精选Pick A卷Paper A

扫码免费领取 2016 PDF版官方真题+答案Scan to get free 2016 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

2016 AMC12 真题参考示例（共 12 题）2016 AMC12 Sample Problems (12 questions)

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