📄 2006A AMC 12 真题

2006A AMC 12 — Official Competition Problems (February 2024)
📅 2024 A 年11月 📝 25题选择题 ⏱ 40分钟 🎯 满分25分 ✅ 含解题思路 👥 612 人已练习

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1
第 1 题
应用题
Sandwiches at Joe's Fast Food cost \3 each and sodas cost \2 each. How many dollars will it cost to purchase 5 sandwiches and 8 sodas?
💡 解题思路
The $5$ sandwiches cost $5\cdot 3=15$ dollars. The $8$ sodas cost $8\cdot 2=16$ dollars. In total, the purchase costs $15+16=\boxed{\textbf{(A) }31}$ dollars.
2
第 2 题
行程问题
Define x\otimes y=x^3-y . What is h\otimes (h\otimes h) ?
💡 解题思路
By the definition of $\otimes$ , we have $h\otimes h=h^{3}-h$ .
3
第 3 题
分数与比例
The ratio of Mary's age to Alice's age is 3:5 . Alice is 30 years old. How old is Mary?
💡 解题思路
Let $m$ be Mary's age. Then $\frac{m}{30}=\frac{3}{5}$ . Solving for $m$ , we obtain $m=\boxed{\textbf{(B) }18}.$
4
第 4 题
规律与数列
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
💡 解题思路
From the greedy algorithm , we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\boxed{\textbf{(E) }23}$
5
第 5 题
数论
Doug and Dave shared a pizza with 8 equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was 8 dollars, and there was an additional cost of 2 dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave pay than Doug?
💡 解题思路
Dave and Doug paid $8+2=10$ dollars in total. Doug paid for three slices of plain pizza, which cost $\frac{3}{8}\cdot 8=3$ . Dave paid $10-3=7$ dollars. Dave paid $7-3=\boxed{\textbf{(D) }4}$ more dol
6
第 6 题
几何·面积
The 8×18 rectangle ABCD is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is y ? [图] (A)\ 6 (B)\ 7 (C)\ 8 (D)\ 9 (E)\ 10
💡 解题思路
Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$ . This means the square will have four sides
7
第 7 题
规律与数列
Mary is 20\% older than Sally, and Sally is 40\% younger than Danielle. The sum of their ages is 23.2 years. How old will Mary be on her next birthday? (A) \ 7 (B) \ 8 (C) \ 9 (D) \ 10 (E) \ 11
💡 解题思路
Let $m$ be Mary's age, let $s$ be Sally's age, and let $d$ be Danielle's age.
8
第 8 题
规律与数列
How many sets of two or more consecutive positive integers have a sum of 15 ?
💡 解题思路
Notice that if the consecutive positive integers have a sum of $15$ , then their average (which could be a fraction) must be a divisor of $15$ . If the number of integers in the list is odd, then the
9
第 9 题
应用题
Oscar buys 13 pencils and 3 erasers for 1.00 . A pencil costs more than an eraser, and both items cost a whole number of cents. What is the total cost, in cents, of one pencil and one eraser? (A)\ 10 (B)\ 12 (C)\ 15 (D)\ 18 (E)\ 20
💡 解题思路
Let the price of a pencil be $p$ and an eraser $e$ . Then $13p + 3e = 100$ with $p > e > 0$ . Since $p$ and $e$ are positive integers , we must have $e \geq 1$ and $p \geq 2$ .
10
第 10 题
整数运算
For how many real values of x is √(120-\sqrt{x)} an integer?
💡 解题思路
For $\sqrt{120-\sqrt{x}}$ to be an integer, $120-\sqrt{x}$ must be a perfect square.
11
第 11 题
几何·面积
Which of the following describes the graph of the equation (x+y)^2=x^2+y^2 ? (A)\ the empty set (B)\ one point (C)\ two lines (D)\ a circle (E)\ the entire plane
💡 解题思路
\begin{align*}(x+y)^2&=x^2+y^2\\ x^2 + 2xy + y^2 &= x^2 + y^2\\ 2xy &= 0\end{align*} Either $x = 0$ or $y = 0$ . The union of them is 2 lines, and thus the answer is $\mathrm{(C)}$ .
12
第 12 题
行程问题
A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring? [图]
💡 解题思路
The inside diameters of the rings are the positive integers from $1$ to $18$ . The total distance needed is the sum of these values plus $2$ for the top of the first ring and the bottom of the last ri
13
第 13 题
几何·面积
The vertices of a 3-4-5 right triangle are the centers of three mutually externally tangent circles , as shown. What is the sum of the areas of the three circles? (A) \ 12π (B) \ \frac{25π}{2} (C) \ 13π (D) \ \frac{27π}{2} (E) \ 14π
💡 解题思路
Let the radius of the smallest circle be $r_A$ , the radius of the second largest circle be $r_B$ , and the radius of the largest circle be $r_C$ . \[r_A + r_B = 3\] \[r_A + r_C = 4\] \[r_ B + r_C = 5
14
第 14 题
方程
Two farmers agree that pigs are worth 300 dollars and that goats are worth 210 dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a 390 dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?
💡 解题思路
The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lo
15
第 15 题
方程
Suppose \cos x=0 and \cos (x+z)=1/2 . What is the smallest possible positive value of z ? (A) \ \frac{π}{6} (B) \ \frac{π}{3} (C) \ \frac{π}{2} (D) \ \frac{5π}{6} (E) \ \frac{7π}{6}
💡 解题思路
The smallest possible value of $z$ will be that of $\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow \mathrm{(A)}$ .
16
第 16 题
几何·面积
Circles with centers A and B have radius 3 and 8, respectively. A common internal tangent intersects the circles at C and D , respectively. Lines AB and CD intersect at E , and AE=5 . What is CD ? [图]
💡 解题思路
$\triangle ADE \sim \triangle BCE$ because \[\angle AED \cong \angle BEC \text{ (vertical angles)}\] \[\angle ADE \cong \angle BCE \text{ (tangent perpendicularity)}\]
17
第 17 题
几何·面积
Square ABCD has side length s , a circle centered at E has radius r , and r and s are both rational. The circle passes through D , and D lies on \overline{BE} . Point F lies on the circle, on the same side of \overline{BE} as A . Segment AF is tangent to the circle, and AF=√(9+5\sqrt{2)} . What is r/s ? [图] (A) \ \frac{1}{2} (B) \ \frac{5}{9} (C) \ \frac{3}{5} (D) \ \frac{5}{3} (E) \ \frac{9}{5}
💡 解题思路
One possibility is to use the coordinate plane , setting $B$ at the origin. Point $A$ will be $(0,s)$ and $E$ will be $\left(s + \frac{r}{\sqrt{2}},\ s + \frac{r}{\sqrt{2}}\right)$ since $B, D$ , and
18
第 18 题
函数
The function f has the property that for each real number x in its domain, 1/x is also in its domain and f(x)+f(\frac{1}{x})=x What is the largest set of real numbers that can be in the domain of f ? (A) \ \{x|x\ne 0\} (B) \ \{x|x<0\} (C) \ \{x|x>0\} (D) \ \{x|x\ne -1\;\rm{and}\; x\ne 0\;\rm{and}\; x\ne 1\} (E) \ \{-1,1\}
💡 解题思路
Quickly verifying by plugging in values verifies that $-1$ and $1$ are in the domain.
19
第 19 题
几何·面积
Circles with centers (2,4) and (14,9) have radii 4 and 9 , respectively. The equation of a common external tangent to the circles can be written in the form y=mx+b with m>0 . What is b ? [图] (A) \ \frac{908}{119} (B) \ \frac{909}{119} (C) \ \frac{130}{17} (D) \ \frac{911}{119} (E) \ \frac{912}{119}
💡 解题思路
Let $L_1$ be the line that goes through $(2,4)$ and $(14,9)$ , and let $L_2$ be the line $y=mx+b$ . If we let $\theta$ be the measure of the acute angle formed by $L_1$ and the x-axis, then $\tan\thet
20
第 20 题
概率
A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?
💡 解题思路
Call this cube $ABCDEFGH$ , with $A$ being the starting point.
21
第 21 题
几何·面积
Let S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\} and S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\} . What is the ratio of the area of S_2 to the area of S_1 ? (A) \ 98 (B) \ 99 (C) \ 100 (D) \ 101 (E) \ 102
💡 解题思路
Looking at the constraints of $S_1$ :
22
第 22 题
几何·面积
A circle of radius r is concentric with and outside a regular hexagon of side length 2 . The probability that three entire sides of hexagon are visible from a randomly chosen point on the circle is 1/2 . What is r ? (A) \ 2√(2)+2√(3) (B) \ 3√(3)+√(2) (C) \ 2√(6)+√(3) \rm{(D) \ } 3√(2)+√(6) (E) \ 6√(2)-√(3)
💡 解题思路
Project any two non-adjacent and non-opposite sides of the hexagon to the circle ; the arc between the two points formed is the location where all three sides of the hexagon can be fully viewed. Since
23
第 23 题
规律与数列
Given a finite sequence S=(a_1,a_2,\ldots ,a_n) of n real numbers, let A(S) be the sequence (\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}) of n-1 real numbers. Define A^1(S)=A(S) and, for each integer m , 2\le m\le n-1 , define A^m(S)=A(A^{m-1}(S)) . Suppose x>0 , and let S=(1,x,x^2,\ldots ,x^{100}) . If A^{100}(S)=(1/2^{50}) , then what is x ? (A) \ 1-\frac{√(2)}{2} (B) \ √(2)-1 (C) \ \frac{1}{2} (D) \ 2-√(2) (E) \ \frac{√(2)}{2}
💡 解题思路
\[A^1(S)=\left(\frac{1+x}{2},\frac{x+x^2}{2},...,\frac{x^{99}+x^{100}}{2}\right)\] \[A^2(S)=\left(\frac{1+2x+x^2}{2^2},\frac{x+2x^2+x^3}{2^2},...,\frac{x^{98}+2x^{99}+x^{100}}{2^2}\right)\] \[\Rightar
24
第 24 题
规律与数列
The expression \[(x+y+z)^{2006}+(x-y-z)^{2006}\] is simplified by expanding it and combining like terms. How many terms are in the simplified expression? (A) \ 6018 (B) \ 671,676 (C) \ 1,007,514 (D) \ 1,008,016 (E) \ 2,015,028
💡 解题思路
By the Multinomial Theorem , the summands can be written as
25
第 25 题
逻辑推理
How many non- empty subsets S of \{1,2,3,\ldots ,15\} have the following two properties? (1) No two consecutive integers belong to S . (2) If S contains k elements , then S contains no number less than k . (A) \ 277 (B) \ 311 (C) \ 376 (D) \ 377 (E) \ 405
💡 解题思路
This question can be solved fairly directly by casework and pattern-finding. We give a somewhat more general attack, based on the solution to the following problem:
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