2009 AMC 10A — Official Competition Problems (February 2009)
📅 2009 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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题目涉及图形的部分，原题以文字描述代替（图形题建议配合原版试卷使用）
1
第 1 题
综合
One can holds 12 ounces of soda, what is the minimum number of cans needed to provide a gallon ( 128 ounces) of soda?
💡 解题思路
$10$ cans would hold $120$ ounces, but $128>120$ , so $11$ cans are required. Thus, the answer is $\mathrm{\boxed{{(E)}11}}$ .
2
第 2 题
概率
Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could not be the total value of the four coins, in cents?
💡 解题思路
Pre-Note: This solution is kinda just guessing, idk you decide.
3
第 3 题
综合
Which of the following is equal to 1 + \frac {1}{1 + \frac {1}{1 + 1}} ?
💡 解题思路
We compute:
4
第 4 题
统计
Eric plans to compete in a triathlon. He can average 2 miles per hour in the \frac{1}{4} -mile swim and 6 miles per hour in the 3 -mile run. His goal is to finish the triathlon in 2 hours. To accomplish his goal what must his average speed in miles per hour, be for the 15 -mile bicycle ride? (A)\ \frac{120}{11} (B)\ 11 (C)\ \frac{56}{5} (D)\ \frac{45}{4} (E)\ 12
💡 解题思路
Since $d=rt$ , Eric takes $\frac{\frac{1}{4}}{2}=\frac{1}{8}$ hours for the swim. Then, he takes $\frac{3}{6}=\frac{1}{2}$ hours for the run.
5
第 5 题
几何·面积
What is the sum of the digits of the square of \text 111111111 ? (A)\ 18 (B)\ 27 (C)\ 45 (D)\ 63 (E)\ 81
💡 解题思路
Using the standard multiplication algorithm, $111,111,111^2=12,345,678,987,654,321,$ whose digit sum is $\boxed{(E)\text{ }81.}$
6
第 6 题
几何·面积
A circle of radius 2 is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded? [图] (A)\ \frac{1}{2} (B)\ \frac{π}{6} (C)\ \frac{2}{π} (D)\ \frac{2}{3} (E)\ \frac{3}{π}
💡 解题思路
Area of the circle inscribed inside the semicircle $= \pi r^2 \Rightarrow \pi(2^2) = 4 \pi .$ Area of the larger circle (semicircle's area x 2) $= \pi r^2 \Rightarrow \pi(4^2)= 16 \pi$ (4, or the diam
7
第 7 题
分数与比例
A carton contains milk that is 2 % fat, an amount that is 40 % less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk? (A)\ \frac{12}{5} (B)\ 3 (C)\ \frac{10}{3} (D)\ 38 (E)\ 42
💡 解题思路
Rewording the question, we are being asked " $2$ is $40\%$ less than what number?" If $x$ represents the number we are looking for, then $40\%$ less than the number would be represented by $x-0.4x$ or
8
第 8 题
分数与比例
Three generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a 50 % discount as children. The two members of the oldest generation receive a 25\% discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs \6.00 , is paying for everyone. How many dollars must he pay? (A)\ 34 (B)\ 36 (C)\ 42 (D)\ 46 (E)\ 48$
💡 解题思路
A senior ticket costs $\$6.00$ , so a regular ticket costs $6 \cdot \frac{1}{\frac{3}{4}}\:=\:6\cdot\frac{4}{3}\:=\:8$ dollars. Therefore children's tickets cost half that, or $\$4.00$ , so we have:
9
第 9 题
分数与比例
Positive integers a , b , and 2009 , with a<b<2009 , form a geometric sequence with an integer ratio. What is a ? (A)\ 7 (B)\ 41 (C)\ 49 (D)\ 289 (E)\ 2009
💡 解题思路
The prime factorization of $2009$ is $2009 = 7\cdot 7\cdot 41$ . As $a<b<2009$ , the ratio must be positive and larger than $1$ , hence there is only one possibility: the ratio must be $7$ , and then
10
第 10 题
几何·面积
Triangle ABC has a right angle at B . Point D is the foot of the altitude from B , AD=3 , and DC=4 . What is the area of \triangle ABC ? [图] (A)\ 4\sqrt3 (B)\ 7\sqrt3 (C)\ 21 (D)\ 14\sqrt3 (E)\ 42
💡 解题思路
It is a well-known fact that in any right triangle $ABC$ with the right angle at $B$ and $D$ the foot of the altitude from $B$ onto $AC$ we have $BD^2 = AD\cdot CD$ . (See below for a proof.) Then $BD
11
第 11 题
立体几何
One dimension of a cube is increased by 1 , another is decreased by 1 , and the third is left unchanged. The volume of the new rectangular solid is 5 less than that of the cube. What was the volume of the cube?
💡 解题思路
Let the original cube have edge length $a$ . Then its volume is $a^3$ . The new box has dimensions $a-1$ , $a$ , and $a+1$ , hence its volume is $(a-1)a(a+1) = a^3-a$ . The difference between the two
12
第 12 题
整数运算
In quadrilateral ABCD , AB = 5 , BC = 17 , CD = 5 , DA = 9 , and BD is an integer. What is BD ?
💡 解题思路
By the triangle inequality we have $BD BC$ , hence $BD > BC - CD = 17 - 5 = 12$ .
13
第 13 题
整数运算
Suppose that P = 2^m and Q = 3^n . Which of the following is equal to 12^{mn} for every pair of integers (m,n) ?
💡 解题思路
We have $12^{mn} = (2\cdot 2\cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = \boxed{\bold{E)} P^{2n} Q^m}$ .
14
第 14 题
几何·面积
Four congruent rectangles are placed as shown. The area of the outer square is 4 times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?
💡 解题思路
The area of the outer square is $4$ times that of the inner square. Therefore the side of the outer square is $\sqrt 4 = 2$ times that of the inner square.
15
第 15 题
几何·面积
The figures F_1 , F_2 , F_3 , and F_4 shown are the first in a sequence of figures. For n\ge3 , F_n is constructed from F_{n - 1} by surrounding it with a square and placing one more diamond on each side of the new square than F_{n - 1} had on each side of its outside square. For example, figure F_3 has 13 diamonds. How many diamonds are there in figure F_{20} ?
💡 解题思路
Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$ th triangular numb
16
第 16 题
规律与数列
Let a , b , c , and d be real numbers with |a-b|=2 , |b-c|=3 , and |c-d|=4 . What is the sum of all possible values of |a-d| ? (A)\ 9 (B)\ 12 (C)\ 15 (D)\ 18 (E)\ 24
💡 解题思路
From $|a-b|=2$ we get that $a=b\pm 2$
17
第 17 题
几何·面积
Rectangle ABCD has AB=4 and BC=3 . Segment EF is constructed through B so that EF is perpendicular to DB , and A and C lie on DE and DF , respectively. What is EF ? (A)\ 9 (B)\ 10 (C)\ \frac {125}{12} (D)\ \frac {103}{9} (E)\ 12
💡 解题思路
The situation is shown in the picture below.
18
第 18 题
分数与比例
At Jefferson Summer Camp, 60\% of the children play soccer, 30\% of the children swim, and 40\% of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer? (A)\ 30\% (B)\ 40\% (C)\ 49\% (D)\ 51\% (E)\ 70\%
💡 解题思路
Out of the soccer players, $40\%$ swim. As the soccer players are $60\%$ of the whole, the swimming soccer players are $0.4 \cdot 0.6 = 0.24 = 24\%$ of all children.
19
第 19 题
几何·面积
Circle A has radius 100 . Circle B has an integer radius r<100 and remains internally tangent to circle A as it rolls once around the circumference of circle A . The two circles have the same points of tangency at the beginning and end of circle B 's trip. How many possible values can r have? (A)\ 4\ (B)\ 8\ (C)\ 9\ (D)\ 50\ (E)\ 90\
💡 解题思路
The circumference of circle $A$ is $200\pi$ , and the circumference of circle $B$ with radius $r$ is $2r\pi$ . Since circle $B$ makes a complete revolution and ends up on the same point , the circumfe
20
第 20 题
行程问题
Andrea and Lauren are 20 kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of 1 kilometer per minute. After 5 minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea? (A)\ 20 (B)\ 30 (C)\ 55 (D)\ 65 (E)\ 80
💡 解题思路
Let their speeds in kilometers per hour be $v_A$ and $v_L$ . We know that $v_A=3v_L$ and that $v_A+v_L=60$ . (The second equation follows from the fact that $1\mathrm km/min = 60\mathrm km/h$ .) This
21
第 21 题
几何·面积
Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle? [图] (A)\ 3-2\sqrt2 (B)\ 2-\sqrt2 (C)\ 4(3-2\sqrt2) (D)\ \frac12(3-\sqrt2) (E)\ 2\sqrt2-2
💡 解题思路
Draw some of the radii of the small circles as in the picture below.
22
第 22 题
概率
Two cubical dice each have removable numbers 1 through 6 . The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is 7 ? (A)\ \frac{1}{9} (B)\ \frac{1}{8} (C)\ \frac{1}{6} (D)\ \frac{2}{11} (E)\ \frac{1}{5}
💡 解题思路
At the moment when the numbers are in the bag, imagine that each of them has a different color. Clearly the situation is symmetric at this moment. Hence after we draw them, attach them and throw the d
23
第 23 题
几何·面积
Convex quadrilateral ABCD has AB = 9 and CD = 12 . Diagonals AC and BD intersect at E , AC = 14 , and \triangle AED and \triangle BEC have equal areas. What is AE ?
💡 解题思路
Let $[ABC]$ denote the area of triangle $ABC$ . $[AED] = [BEC]$ , so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$ . Since triangles $ABD$ and $ABC$ share a base, they also have the same height and
24
第 24 题
概率
Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube? (A)\ \frac{1}{4} (B)\ \frac{3}{8} (C)\ \frac{4}{7} (D)\ \frac{5}{7} (E)\ \frac{3}{4}
💡 解题思路
First of all, the number of planes determined by any three vertices of a cube is $20$ ( $6$ faces, $6$ planes through two opposite edges, $8$ planes through three vertices forming a tetrahedron). Amon
25
第 25 题
数论
For k > 0 , let I_k = 10\ldots 064 , where there are k zeros between the 1 and the 6 . Let N(k) be the number of factors of 2 in the prime factorization of I_k . What is the maximum value of N(k) ?
💡 解题思路
The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$ .