选择卷别Paper Set

2006

2006 AMC10 真题及解析2006 AMC10 Paper & Solutions 精选Pick A卷Paper A

25 题含解析，几何侧重圆和三角形，应用题贴近实际。25 questions with solutions, geometry focuses on circles and triangles, practical word problems.

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

扫码免费领取 2006 PDF版官方真题+答案Scan to get free 2006 PDF + solutions

25 道选择题25 Questions

75 分钟限时40 Minutes

满分 150 分Max Score 25

不可用计算器No Calculator

Exam Overview

考试概览Exam Overview

几何侧重圆和三角形。This exam emphasizes geometry. 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%)

109+

HR 荣誉奖Honor Roll

Honor Roll (Top 5%)

99+

Achievement Roll

六年级及以下 · 15分以上Grade 6 and below · 15+ points

88+

Sample Problems

2006 AMC10 真题参考示例（共 12 题）2006 AMC10 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 + 1，求 f(4) 的值。If f(x) = 2x² - 4x + 1, find f(4).

A) 11

B) 17

C) 14

D) 20

E) 23

解题步骤Steps

1f(4) = 2×4² - 4×4 + 1

2= 2×16 - 16 + 1 = 17

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 10 项。Arithmetic sequence: a₁=5, d=3, find the 10-th term.

A) 24

B) 28

C) 32

D) 36

E) 40

解题步骤Steps

1aₙ = a₁ + (n-1)d

2a_10 = 5 + 9×3 = 32

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 2ˣ = 32，求 x 的值。If 2ˣ = 32, find x.

A) 3

B) 4

C) 6

D) 5

E) 7

解题步骤Steps

132 = 2^5

2故 x = 5

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

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 36 的正整数中，有多少个能被 3 整除？How many integers from 1 to 36 are divisible by 3?

A) 12

B) 8

C) 10

D) 14

E) 16

解题步骤Steps

1⌊36 / 3⌋ = 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 7 边形共有多少条对角线？How many diagonals does a regular 7-gon have?

A) 8

B) 14

C) 11

D) 17

E) 20

解题步骤Steps

1对角线数 = n(n-3)/2

2= 7×4/2 = 14

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 5^4 除以 7 的余数。Find the remainder of 5^4 divided by 7.

A) 0

B) 1

C) 3

D) 4

E) 2

解题步骤Steps

1计算 5^4 mod 7

2由模运算性质逐步化简

3=2

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 - 8x + 19 的最小值是多少？Find the minimum of f(x) = x² - 8x + 19.

A) 3

B) 1

C) 5

D) 7

E) 9

解题步骤Steps

1配方：f(x) = (x - 4)² + 3

2当 x = 4 时取最小值 3

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 4 个不同球放入 3 个不同盒子，每盒至少 1 球，有多少种放法？4 distinct balls into 3 distinct boxes, each ≥1 ball, how many ways?

A) 26

B) 36

C) 31

D) 41

E) 46

解题步骤Steps

1总数 3^4 = 81

2减去有空盒: -C(3,1)×2^4 = -48

3加回两个空盒: +C(3,2)×1 = +3

4总计 81 - 48 + 3 = 36

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=5, b=6, cos C=1/2，求 c²。In △ABC, a=5, b=6, cos C=1/2, find c².

A) 23

B) 27

C) 31

D) 35

E) 39

解题步骤Steps

1余弦定理 c² = a² + b² - 2ab·cos C

2= 25 + 36 - 30 = 31

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 + 1，求 f(4) 的值。If f(x) = 2x² - 6x + 1, find f(4).

A) 5

B) 9

C) 7

D) 11

E) 13

解题步骤Steps

1f(4) = 2×4² - 6×4 + 1

2= 2×16 - 24 + 1 = 9

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=7，公差 d=3，求第 10 项。Arithmetic sequence: a₁=7, d=3, find the 10-th term.

A) 26

B) 30

C) 34

D) 38

E) 42

解题步骤Steps

1aₙ = a₁ + (n-1)d

2a_10 = 7 + 9×3 = 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 2ˣ = 32，求 x 的值。If 2ˣ = 32, find x.

A) 3

B) 4

C) 6

D) 5

E) 7

解题步骤Steps

132 = 2^5

2故 x = 5

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 + 6 = 0 的两根之和是多少？Sum of roots of x² - 8x + 6 = 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 = 6

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 38 的正整数中，有多少个能被 3 整除？How many integers from 1 to 38 are divisible by 3?

A) 12

B) 8

C) 10

D) 14

E) 16

解题步骤Steps

1⌊38 / 3⌋ = 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 9 边形共有多少条对角线？How many diagonals does a regular 9-gon have?

A) 21

B) 27

C) 24

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：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 5^4 除以 7 的余数。Find the remainder of 5^4 divided by 7.

A) 0

B) 1

C) 3

D) 4

E) 2

解题步骤Steps

1计算 5^4 mod 7

2由模运算性质逐步化简

3=2

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 - 8x + 21 的最小值是多少？Find the minimum of f(x) = x² - 8x + 21.

A) 5

B) 1

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：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 4 个不同球放入 3 个不同盒子，每盒至少 1 球，有多少种放法？4 distinct balls into 3 distinct boxes, each ≥1 ball, how many ways?

A) 26

B) 36

C) 31

D) 41

E) 46

解题步骤Steps

1总数 3^4 = 81

2减去有空盒: -C(3,1)×2^4 = -48

3加回两个空盒: +C(3,2)×1 = +3

4总计 81 - 48 + 3 = 36

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=7, b=6, cos C=1/2，求 c²。In △ABC, a=7, b=6, 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= 49 + 36 - 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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2006 AMC10 真题及解析2006 AMC10 Paper & Solutions 精选Pick A卷Paper A

扫码免费领取 2006 PDF版官方真题+答案Scan to get free 2006 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

2006 AMC10 真题参考示例（共 12 题）2006 AMC10 Sample Problems (12 questions)

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