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AMC10 2009 B

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AMC10 · 2009 (B)

Q1
Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?
Jane 在她五天工作周的每个早上，买了一个50美分的松饼或一个75美分的百吉饼。她一周的总花费是一个整数组美元。她买了多少个百吉饼？
Correct Answer: B
Answer (B): Make a table for the cost of the muffins and bagels: \[ \begin{array}{|c|c|c|} \hline \text{Cost of Muffins} & \text{Cost of Bagels} & \text{Total Cost} \\ \hline 0\cdot 0.50 = 0.00 & 5\cdot 0.75 = 3.75 & 3.75 \\ \hline 1\cdot 0.50 = 0.50 & 4\cdot 0.75 = 3.00 & 3.50 \\ \hline 2\cdot 0.50 = 1.00 & 3\cdot 0.75 = 2.25 & 3.25 \\ \hline 3\cdot 0.50 = 1.50 & 2\cdot 0.75 = 1.50 & 3.00 \\ \hline 4\cdot 0.50 = 2.00 & 1\cdot 0.75 = 0.75 & 2.75 \\ \hline 5\cdot 0.50 = 2.50 & 0\cdot 0.75 = 0.00 & 2.50 \\ \hline \end{array} \] The only combination which is a whole number of dollars is the cost of 3 muffins and 2 bagels.
答案（B）：制作一个松饼和百吉饼成本的表格： \[ \begin{array}{|c|c|c|} \hline \text{松饼的成本} & \text{百吉饼的成本} & \text{总成本} \\ \hline 0\cdot 0.50 = 0.00 & 5\cdot 0.75 = 3.75 & 3.75 \\ \hline 1\cdot 0.50 = 0.50 & 4\cdot 0.75 = 3.00 & 3.50 \\ \hline 2\cdot 0.50 = 1.00 & 3\cdot 0.75 = 2.25 & 3.25 \\ \hline 3\cdot 0.50 = 1.50 & 2\cdot 0.75 = 1.50 & 3.00 \\ \hline 4\cdot 0.50 = 2.00 & 1\cdot 0.75 = 0.75 & 2.75 \\ \hline 5\cdot 0.50 = 2.50 & 0\cdot 0.75 = 0.00 & 2.50 \\ \hline \end{array} \] 唯一使总价为整数美元的组合是：3 个松饼和 2 个百吉饼。
Q2
Which of the following is equal to （）
下列哪一项等于（）
stem
Correct Answer: C
Answer (C): The least common multiple of 2, 3, and 4 is 12, and $$\frac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}\cdot\frac{12}{12}=\frac{4-3}{6-4}=\frac{1}{2}.$$
答案（C）：2、3 和 4 的最小公倍数是 12，并且 $$\frac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}\cdot\frac{12}{12}=\frac{4-3}{6-4}=\frac{1}{2}.$$
solution
Q3
Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?
画家Paula原本有刚好足够的油漆刷30间大小相同的房间。不幸的是，在上班路上，三个油漆罐从她的卡车上掉了下来，所以她只够刷25间房间。她刷这25间房间用了多少罐油漆？
Correct Answer: C
Answer (C): The loss of 3 cans of paint resulted in 5 fewer rooms being painted, so the ratio of cans of paint to rooms painted is $3:5$. Hence for 25 rooms she would require $\frac{3}{5}\cdot 25 = 15$ cans of paint.
答案（C）：丢失了3罐油漆导致少刷了5个房间，因此油漆罐数与刷漆房间数之比为$3:5$。因此刷25个房间需要$\frac{3}{5}\cdot 25 = 15$罐油漆。
Q4
A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape. The parallel sides of the trapezoid have lengths 15 and 25 meters. What fraction of the yard is occupied by the flower beds?
一个矩形院子里有两个形状全等、直角等腰三角形的花坛。院子剩余部分呈梯形。梯形的平行边长分别为15米和25米。花坛占院子的几分之几？
stem
Correct Answer: C
Answer (C): Each triangle has leg length $\frac{1}{2}\cdot(25-15)=5$ meters and area $\frac{1}{2}\cdot 5^2=\frac{25}{2}$ square meters. Thus the flower beds have a total area of 25 square meters. The entire yard has length 25 and width 5, so its area is 125. The fraction of the yard occupied by the flower beds is $\frac{25}{125}=\frac{1}{5}$.
答案（C）：每个三角形的直角边长为 $\frac{1}{2}\cdot(25-15)=5$ 米，面积为 $\frac{1}{2}\cdot 5^2=\frac{25}{2}$ 平方米。因此花坛的总面积为 25 平方米。整个院子的长为 25、宽为 5，所以面积为 125。花坛占院子的比例为 $\frac{25}{125}=\frac{1}{5}$。
Q5
Twenty percent less than 60 is one-third more than what number?
60的百分之二十少于某个数的三分之一多于多少？
Correct Answer: D
Answer (D): Twenty percent less than 60 is $\frac{4}{5}\cdot 60 = 48$. One-third more than a number $n$ is $\frac{4}{3}n$. Therefore $\frac{4}{3}n = 48$, and the number is 36.
答案（D）：比 60 少 20% 等于 $\frac{4}{5}\cdot 60 = 48$。比一个数 $n$ 多三分之一是 $\frac{4}{3}n$。因此 $\frac{4}{3}n = 48$，这个数是 36。
Q6
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?
Kiana 有两个年龄相同的双胞胎哥哥。他们三人的年龄乘积是 128。他们三人的年龄和是多少？
Correct Answer: D
Answer (C): The three operations can be performed in any of $3! = 6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. Thus at most four distinct values can be obtained. It is easily checked that the values of the four expressions $$(2\times 3) + (4\times 5) = 26,$$ $$((2\times 3 + 4)\times 5) = 50,$$ $$2\times (3 + (4\times 5)) = 46,$$ $$2\times (3 + 4)\times 5 = 70$$ are in fact all distinct.
答案（C）：这三个运算可以按任意 $3! = 6$ 种顺序进行。然而，如果加法在最先或最后进行，那么按任意顺序做乘法都会得到相同结果。因此最多能得到四个不同的值。很容易检验，下面四个表达式的值分别为 $$(2\times 3) + (4\times 5) = 26,$$ $$((2\times 3 + 4)\times 5) = 50,$$ $$2\times (3 + (4\times 5)) = 46,$$ $$2\times (3 + 4)\times 5 = 70$$ 并且它们确实互不相同。
Q7
By inserting parentheses, it is possible to give the expression $2 \times 3 + 4 \times 5$ several values. How many different values can be obtained?
通过插入括号，可以使表达式 $2 \times 3 + 4 \times 5$ 有几种不同的值？
Correct Answer: C
Possible: (2×3)+(4×5)=26, ((2×3)+4)×5=50, 2×(3+(4×5))=46, 2×(3+4)×5=70. 4 distinct.
可能的值：(2×3)+(4×5)=26, ((2×3)+4)×5=50, 2×(3+(4×5))=46, 2×(3+4)×5=70。4 个不同值。
Q8
In a certain year the price of gasoline rose by 20% during January, fell by 20% during February, rose by 25% during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$?
在某一年，汽油价格在一月份上涨 20%，二月份下跌 20%，三月份上涨 25%，四月份下跌 $x\%$ 。四月底的汽油价格与一月初相同。四月份下跌的百分比 $x$ 取整到最近整数是多少？
Correct Answer: B
Answer (B): Let $p$ denote the price at the beginning of January. The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p$. Because the price at the end of April was $p$, the price decreased by $0.2p$ during April, and the percent decrease was $$ x = 100 \cdot \frac{0.2p}{1.2p} = \frac{100}{6} \approx 16.7. $$ To the nearest integer, $x$ is 17.
答案（B）：设 $p$ 表示一月初的价格。三月底的价格为 $(1.2)(0.8)(1.25)p = 1.2p$。由于四月底的价格是 $p$，因此四月份价格下降了 $0.2p$，降幅百分比为 $$ x = 100 \cdot \frac{0.2p}{1.2p} = \frac{100}{6} \approx 16.7. $$ 四舍五入到最接近的整数，$x$ 为 17。
Q9
Segment $BD$ and $AE$ intersect at $C$, as shown, $AB = BC = CD = CE$, and $\angle A = \frac{3}{2}\angle B$. What is the degree measure of $\angle D$?
线段 $BD$ 和 $AE$ 相交于 $C$，如图所示，$AB = BC = CD = CE$，且 $\angle A = \frac{3}{2}\angle B$。$\angle D$ 的度数是多少？
stem
Correct Answer: A
Answer (A): Because $\triangle ABC$ is isosceles, $\angle A=\angle C$. Because $\angle A=\frac{5}{2}\angle B$, we have $\frac{5}{2}\angle B+\frac{5}{2}\angle B+\angle B=180^\circ$, so $\angle B=30^\circ$. Therefore $\angle ACB=\angle DCE=75^\circ$. Because $\triangle CDE$ is isosceles, $2\angle D+75^\circ=180^\circ$, so $\angle D=52.5^\circ$.
答案（A）：因为 $\triangle ABC$ 是等腰三角形，所以 $\angle A=\angle C$。又因为 $\angle A=\frac{5}{2}\angle B$，所以 $\frac{5}{2}\angle B+\frac{5}{2}\angle B+\angle B=180^\circ$，从而 $\angle B=30^\circ$。因此 $\angle ACB=\angle DCE=75^\circ$。因为 $\triangle CDE$ 是等腰三角形，所以 $2\angle D+75^\circ=180^\circ$，从而 $\angle D=52.5^\circ$。
Q10
A flagpole is originally 5 meters tall. A hurricane snaps the flagpole at a point $x$ meters above the ground so that the upper part, still attached to the stump, touches the ground 1 meter away from the base. What is $x$?
一根旗杆原来高 5 米。飓风从离地面 $x$ 米处将其折断，上部仍附着在残桩上，触地时距基部 1 米远。$x$ 是多少？
Correct Answer: E
Answer (E): Let $x$ be the height of the stump. Then $5-x$ is the height of the snapped part, now forming the hypotenuse of a right triangle. By the Pythagorean Theorem, $$ x^2+1^2=(5-x)^2=x^2-10x+25 $$ from which $x=2.4$.
答案（E）：设 $x$ 为树桩的高度，则 $5-x$ 为折断部分的长度，它现在构成一个直角三角形的斜边。由勾股定理， $$ x^2+1^2=(5-x)^2=x^2-10x+25 $$ 由此得 $x=2.4$。
solution
Q11
How many 7 digit palindromes (numbers that read the same backward as forward) can be formed using the digits 2, 2, 3, 3, 5, 5, 5?
使用数字 2, 2, 3, 3, 5, 5, 5 可以形成多少个7位回文数（正读反读都相同的数字）？
Correct Answer: A
Answer (A): Because the digit 5 appears three times, 5 must be the middle digit of any such palindrome. In the first three digits each of 2, 3, and 5 must appear once and the order in which they appear determines the last three digits. Since there are $3! = 6$ ways to order three distinct digits the number of palindromes is 6.
答案（A）：因为数字 5 出现了三次，所以在任何这样的回文数中，5 必须是中间一位。在前三位数字中，2、3、5 必须各出现一次，它们出现的顺序决定了后三位数字。由于三个不同数字的排列方式有 $3! = 6$ 种，因此回文数的个数是 6。
Q12
Distinct points $A, B, C, D$ lie on a line, with $AB = BC = CD = 1$. Points $E$ and $F$ lie on a second line, parallel to the first, with $EF = 1$. A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?
不同的点 $A, B, C, D$ 在一条直线上，且 $AB = BC = CD = 1$。点 $E$ 和 $F$ 在第二条与第一条平行的直线上，且 $EF = 1$。以六个点中的三个为顶点的三角形面积大于零。这样的三角形面积可能有几个不同的值？
Correct Answer: A
Answer (A): The base of the triangle can be 1, 2, or 3, and its altitude is the distance between the two parallel lines, so there are three possible values for the area.
答案（A）：三角形的底边可以是 1、2 或 3，而它的高是两条平行线之间的距离，因此面积有三种可能的取值。
Q13
As shown below, convex pentagon ABCDE has sides AB = 3, BC = 4, CD = 6, DE = 3, and EA = 7. The pentagon is originally positioned in the plane with vertex A at the origin and vertex B on the positive x-axis. The pentagon is then rolled clockwise to the right along the x-axis. Which side will touch the point x = 2009 on the x-axis?
如图所示，凸五边形ABCDE的边长AB = 3, BC = 4, CD = 6, DE = 3, EA = 7。五边形最初位于平面中，顶点A在原点，顶点B在正x轴上。然后五边形沿x轴向右顺时针滚动。哪条边会接触x轴上的点x = 2009？
stem
Correct Answer: C
Answer (C): Define a rotation of the pentagon to be a sequence that starts with $AB$ on the $x$-axis and ends when $AB$ is on the $x$-axis the first time thereafter. Because the pentagon has perimeter $23$ and $2009 = 23 \cdot 87 + 8$, it follows that after $87$ rotations, point $A$ will be at $x = 23 \cdot 87 = 2001$ and point $B$ will be at $x = 2001 + 3 = 2004$. Points $C$ and $D$ will next touch the $x$-axis at $x = 2004 + 4 = 2008$ and $x = 2008 + 6 = 2014$, respectively. Therefore a point on $CD$ will touch $x = 2009$.
答案（C）：将五边形的一次“旋转”定义为：从线段 $AB$ 位于 $x$ 轴开始，到之后第一次再次出现 $AB$ 位于 $x$ 轴时结束的一段过程。由于五边形的周长为 $23$，且 $2009 = 23 \cdot 87 + 8$，可知经过 $87$ 次旋转后，点 $A$ 将位于 $x = 23 \cdot 87 = 2001$，点 $B$ 将位于 $x = 2001 + 3 = 2004$。接着，点 $C$ 和点 $D$ 将分别在 $x = 2004 + 4 = 2008$ 与 $x = 2008 + 6 = 2014$ 处再次接触 $x$ 轴。因此，线段 $CD$ 上会有一点接触到 $x = 2009$。
Q14
On Monday, Millie puts a quart of seeds, 25% of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only 25% of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?
周一，Millie向鸟食器中放入一夸脱种子，其中25%是小米。此后每天她再添加一夸脱相同混合的种子，而不移除剩余的种子。每天鸟儿只吃掉食器中25%的小米，但吃掉所有其他种子。在Millie放入种子后哪一天，鸟儿会发现食器中超过一半的种子是小米？
Correct Answer: D
Answer (D): On Monday, day 1, the birds find $\frac{1}{4}$ quart of millet in the feeder. On Tuesday they find \[ \frac{1}{4}+\frac{3}{4}\cdot\frac{1}{4} \] quarts of millet. On Wednesday, day 3, they find \[ \frac{1}{4}+\frac{3}{4}\cdot\frac{1}{4}+\left(\frac{3}{4}\right)^2\cdot\frac{1}{4} \] quarts of millet. The number of quarts of millet they find on day $n$ is \[ \frac{1}{4}+\frac{3}{4}\cdot\frac{1}{4}+\left(\frac{3}{4}\right)^2\cdot\frac{1}{4}+\cdots+\left(\frac{3}{4}\right)^{n-1}\cdot\frac{1}{4} =\frac{\left(\frac{1}{4}\right)\left(1-\left(\frac{3}{4}\right)^n\right)}{1-\frac{3}{4}} =1-\left(\frac{3}{4}\right)^n. \] The birds always find $\frac{3}{4}$ quart of other seeds, so more than half the seeds are millet if $1-\left(\frac{3}{4}\right)^n>\frac{3}{4}$, that is, when $\left(\frac{3}{4}\right)^n<\frac{1}{4}$. Because $\left(\frac{3}{4}\right)^4=\frac{81}{256}>\frac{1}{4}$ and $\left(\frac{3}{4}\right)^5=\frac{243}{1024}<\frac{1}{4}$, this will first occur on day 5 which is Friday.
答案（D）：在星期一（第 1 天），鸟在喂食器里找到 $\frac{1}{4}$ 夸脱的小米。星期二它们找到 \[ \frac{1}{4}+\frac{3}{4}\cdot\frac{1}{4} \] 夸脱的小米。星期三（第 3 天），它们找到 \[ \frac{1}{4}+\frac{3}{4}\cdot\frac{1}{4}+\left(\frac{3}{4}\right)^2\cdot\frac{1}{4} \] 夸脱的小米。第 $n$ 天它们找到的小米数量（单位：夸脱）为 \[ \frac{1}{4}+\frac{3}{4}\cdot\frac{1}{4}+\left(\frac{3}{4}\right)^2\cdot\frac{1}{4}+\cdots+\left(\frac{3}{4}\right)^{n-1}\cdot\frac{1}{4} =\frac{\left(\frac{1}{4}\right)\left(1-\left(\frac{3}{4}\right)^n\right)}{1-\frac{3}{4}} =1-\left(\frac{3}{4}\right)^n. \] 鸟总能找到 $\frac{3}{4}$ 夸脱的其他种子，所以当 $1-\left(\frac{3}{4}\right)^n>\frac{3}{4}$（即 $\left(\frac{3}{4}\right)^n<\frac{1}{4}$）时，小米会超过种子总量的一半。因为 $\left(\frac{3}{4}\right)^4=\frac{81}{256}>\frac{1}{4}$ 且 $\left(\frac{3}{4}\right)^5=\frac{243}{1024}<\frac{1}{4}$，这种情况首次发生在第 5 天，也就是星期五。
Q15
When a bucket is two-thirds full of water, the bucket and water weigh a kilograms. When the bucket is one-half full of water the total weight is b kilograms. In terms of a and b, what is the total weight in kilograms when the bucket is full of water?
当水桶装满水的三分之二时，水桶和水的总重量为a千克。当水桶装满水的一半时，总重量为b千克。用a和b表示水桶满水时的总重量（千克）？
Correct Answer: E
Answer (E): Let $x$ be the weight of the bucket and let $y$ be the weight of the water in a full bucket. Then we are given that $x+\frac{2}{3}y=a$ and $x+\frac{1}{2}y=b$. Hence $\frac{1}{6}y=a-b$, so $y=6a-6b$. Thus $x=b-\frac{1}{2}(6a-6b)=-3a+4b$. Finally, $x+y=3a-2b$.
答案（E）：设 $x$ 为桶的重量，$y$ 为一满桶水的重量。已知 $x+\frac{2}{3}y=a$ 且 $x+\frac{1}{2}y=b$。因此 $\frac{1}{6}y=a-b$，所以 $y=6a-6b$。于是 $x=b-\frac{1}{2}(6a-6b)=-3a+4b$。最后，$x+y=3a-2b$。
Q16
Points A and C lie on a circle centered at O, each of \(\overline{BA}\) and \(\overline{BC}\) are tangent to the circle, and \(\triangle ABC\) is equilateral. The circle intersects BO at D. What is BD/BO?
点 A 和 C 位于以 O 为圆心的圆上，\(\overline{BA}\) 和 \(\overline{BC}\) 各切于该圆，且 \(\triangle ABC\) 是等边三角形。该圆与 BO 相交于 D。求 BD/BO？
Correct Answer: B
Answer (B): Let the radius of the circle be $r$. Because $\triangle BCO$ is a right triangle with a $30^\circ$ angle at $B$, the hypotenuse $BO$ is twice as long as $OC$, so $BO=2r$. It follows that $BD=2r-r=r$, and \[ \frac{BD}{BO}=\frac{r}{2r}=\frac{1}{2}. \]
答案（B）：设圆的半径为$r$。因为$\triangle BCO$是一直角三角形，且在$B$处的角为$30^\circ$，所以斜边$BO$是$OC$的两倍，因此$BO=2r$。由此可得$BD=2r-r=r$，并且 \[ \frac{BD}{BO}=\frac{r}{2r}=\frac{1}{2}. \]
solution
Q17
Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(a, 0)$ to $(3, 3)$, divides the entire region into two regions of equal area. What is $a$?
五个单位正方形如图在坐标平面中排列，左下角位于原点。从 $(a, 0)$ 到 $(3, 3)$ 的斜线将整个区域分为两个面积相等的区域。求 $a$？
stem
Correct Answer: C
Answer (C): The area of the entire region is 5. The shaded region consists of a triangle with base $3-a$ and altitude $3$, with one unit square removed. Therefore $$ \frac{3(3-a)}{2}-1=\frac{5}{2}. $$ Solving this equation yields $a=\frac{2}{3}$.
答案（C）：整个区域的面积为 $5$。阴影区域由一个底为 $3-a$、高为 $3$ 的三角形组成，并去掉了一个单位正方形。因此 $$ \frac{3(3-a)}{2}-1=\frac{5}{2}. $$ 解此方程得到 $a=\frac{2}{3}$。
Q18
Rectangle $ABCD$ has $AB = 8$ and $BC = 6$. Point $M$ is the midpoint of diagonal $\overline{AC}$, and $E$ is on $\overline{AB}$ with $ME \perp \overline{AC}$. What is the area of $\triangle AME$?
矩形 $ABCD$ 有 $AB = 8$ 和 $BC = 6$。点 $M$ 是对角线 $\overline{AC}$ 的中点，$E$ 在 $\overline{AB}$ 上且 $ME \perp \overline{AC}$。求 \(\triangle AME\) 的面积？
Correct Answer: D
Answer (D): By the Pythagorean Theorem, $AC=10$, so $AM=5$. Triangles $AME$ and $ABC$ are similar, so $\frac{ME}{AM}=\frac{6}{8}$ and $ME=\frac{15}{4}$. The area of $\triangle AME$ is $\frac{1}{2}\cdot 5 \cdot \frac{15}{4}=\frac{75}{8}$.
答案（D）：由勾股定理，$AC=10$，所以 $AM=5$。三角形 $AME$ 与 $ABC$ 相似，因此 $\frac{ME}{AM}=\frac{6}{8}$，并且 $ME=\frac{15}{4}$。$\triangle AME$ 的面积为 $\frac{1}{2}\cdot 5 \cdot \frac{15}{4}=\frac{75}{8}$。
solution
Q19
A particular 12-hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a 1, it mistakenly displays a 9. For example, when it is 1:16 PM the clock incorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?
一个特殊的 12 小时数字时钟显示一天的小时和分钟。不幸的是，每当它应该显示 1 时，它错误地显示 9。例如，当是下午 1:16 时，时钟错误显示 9:96 PM。这一天中，时钟显示正确时间的几分之几？
Correct Answer: A
Answer (A): The clock will display the incorrect time for the entire hours of 1, 10, 11, and 12. So the correct hour is displayed correctly $\frac{2}{3}$ of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a 1, so the minutes that will not display correctly are 10, 11, 12, . . . , 19, and 01, 21, 31, 41, and 51. This is 15 of the 60 possible minutes for a given hour. Hence the fraction of the day that the clock shows the correct time is $\frac{2}{3}\cdot\left(1-\frac{15}{60}\right)=\frac{2}{3}\cdot\frac{3}{4}=\frac{1}{2}$.
答案（A）：这个时钟在整点为 1 点、10 点、11 点和 12 点时会显示错误的时间。因此，小时显示正确的比例是 $\frac{2}{3}$。当分钟的十位或个位数字为 1 时，分钟将无法正确显示，所以无法正确显示的分钟是 10、11、12、…、19，以及 01、21、31、41、51。这在每小时的 60 个可能分钟中共有 15 个。因此，这个时钟在一天中显示正确时间的比例为 $\frac{2}{3}\cdot\left(1-\frac{15}{60}\right)=\frac{2}{3}\cdot\frac{3}{4}=\frac{1}{2}$。
Q20
Triangle $ABC$ has a right angle at $B$, $AB = 1$, and $BC = 2$. The bisector of $\angle BAC$ meets $\overline{BC}$ at $D$. What is $BD$?
\(\triangle ABC\) 在 B 处直角，$AB = 1$，$BC = 2$。\(\angle BAC\) 的平分线与 $\overline{BC}$ 相交于 D。求 $BD$？
stem
Correct Answer: B
Answer (B): By the Pythagorean Theorem, $AC=\sqrt{5}$. By the Angle Bisector Theorem, $\frac{BD}{AB}=\frac{CD}{AC}$. Therefore $CD=\sqrt{5}\cdot BD$ and $BD+CD=2$, from which $$ BD=\frac{2}{1+\sqrt{5}}=\frac{\sqrt{5}-1}{2}. $$
答案（B）：由勾股定理，$AC=\sqrt{5}$。由角平分线定理，$\frac{BD}{AB}=\frac{CD}{AC}$。因此 $CD=\sqrt{5}\cdot BD$ 且 $BD+CD=2$，从而 $$ BD=\frac{2}{1+\sqrt{5}}=\frac{\sqrt{5}-1}{2}. $$
Q21
What is the remainder when $3^0 + 3^1 + 3^2 + \dots + 3^{2009}$ is divided by 8?
将 $3^0 + 3^1 + 3^2 + \dots + 3^{2009}$ 除以 8 所得的余数是多少？
Correct Answer: D
Answer (D): The sum of any four consecutive powers of 3 is divisible by $3^0+3^1+3^2+3^3=40$ and hence is divisible by 8. Therefore $$(3^2+3^3+3^4+3^5)+\cdots+(3^{2006}+3^{2007}+3^{2008}+3^{2009})$$ is divisible by 8. So the required remainder is $3^0+3^1=4$.
答案（D）：任意连续四个 3 的幂之和可被 $3^0+3^1+3^2+3^3=40$ 整除，因此也可被 8 整除。于是 $$(3^2+3^3+3^4+3^5)+\cdots+(3^{2006}+3^{2007}+3^{2008}+3^{2009})$$ 可被 8 整除。因此所求余数为 $3^0+3^1=4$。
Q22
A cubical cake with edge length 2 inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where M is the midpoint of a top edge. The piece whose top is triangle B contains c cubic inches of cake and s square inches of icing. What is $c + s$?
一个边长 2 英寸的立方体蛋糕在侧面和顶面涂了糖霜。如顶视图所示，它被垂直切成三块，其中 M 是顶边中点。三角形 B 的顶部那块蛋糕包含 $c$ 立方英寸的蛋糕和 $s$ 平方英寸的糖霜。求 $c + s$。
stem
Correct Answer: B
Answer (B): The area of triangle $A$ is $1$, and its hypotenuse has length $\sqrt{5}$. Triangle $B$ is similar to triangle $A$ and has a hypotenuse of $2$, so its area is $(\frac{2}{\sqrt{5}})^2=\frac{4}{5}$. The volume of the required piece is $c=\frac{4}{5}\cdot 2=\frac{8}{5}$ cubic inches. The icing on this piece has an area of $s=\frac{4}{5}+2^2=\frac{24}{5}$ square inches. Therefore $c+s=\frac{8}{5}+\frac{24}{5}=\frac{32}{5}$.
答案（B）：三角形$A$的面积是$1$，其斜边长为$\sqrt{5}$。三角形$B$与三角形$A$相似，且斜边为$2$，所以其面积为$(\frac{2}{\sqrt{5}})^2=\frac{4}{5}$。所需那块的体积为$c=\frac{4}{5}\cdot 2=\frac{8}{5}$立方英寸。这块上面的糖霜面积为$s=\frac{4}{5}+2^2=\frac{24}{5}$平方英寸。因此$c+s=\frac{8}{5}+\frac{24}{5}=\frac{32}{5}$。
Q23
Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the start line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?
Rachel 和 Robert 在一个圆形跑道上跑步。Rachel 逆时针跑，每 90 秒完成一圈，Robert 顺时针跑，每 80 秒完成一圈。他们同时从起点开始。在他们开始跑后 10 分钟到 11 分钟之间的某个随机时刻，站在跑道内侧的摄影师拍了一张照片，照片显示以起点线为中心的一刻跑道。Rachel 和 Robert 同时出现在照片中的概率是多少？
Correct Answer: C
Answer (C): After 10 min. $=600$ sec., Rachel will have completed 6 laps and be 30 seconds from the finish line. Because Rachel runs one-fourth of a lap in 22.5 seconds, she will be in the picture taking region between $$ 30-\frac{22.5}{2}=18.75 \quad \text{and} \quad 30+\frac{22.5}{2}=41.25 $$ seconds of the 10th minute. After 10 minutes Robert will have completed 7 laps and will be 40 seconds from the starting line. Because Robert runs one-fourth of a lap in 20 seconds, he will be in the picture taking region between 30 and 50 seconds of the 10th minute. Hence both Rachel and Robert will be in the picture if it is taken between 30 and 41.25 seconds of the 10th minute. The probability that the picture is snapped during this time is $$ \frac{41.25-30}{60}=\frac{3}{16}. $$
答案（C）：10 分钟（$=600$ 秒）后，Rachel 将完成 6 圈，并距离终点线还有 30 秒。由于 Rachel 跑四分之一圈需要 22.5 秒，所以在第 10 分钟内，她出现在拍照区域的时间在 $$ 30-\frac{22.5}{2}=18.75 \quad \text{和} \quad 30+\frac{22.5}{2}=41.25 $$ 秒之间。10 分钟后，Robert 将完成 7 圈，并距离起点线还有 40 秒。由于 Robert 跑四分之一圈需要 20 秒，所以在第 10 分钟内，他出现在拍照区域的时间在 30 到 50 秒之间。因此，如果照片在第 10 分钟的 30 到 41.25 秒之间拍摄，Rachel 和 Robert 都会出现在照片中。照片在这段时间内拍下的概率为 $$ \frac{41.25-30}{60}=\frac{3}{16}. $$
Q24
The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with 9 trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$?
拱顶是古老的建筑特征。它由沿非平行边拼合的全等等腰梯形组成，如图所示。两端梯形的底边是水平的。用 9 个梯形构成的拱顶中，设 $x$ 为梯形较大内角的度数。求 $x$。
stem
Correct Answer: A
Answer (A): Add a symmetric arch to the given arch to create a closed loop of trapezoids. Consider the regular 18-sided polygon created by the interior of the completed loop. Each interior angle of a regular 18-gon measures $$(18-2)\cdot 180^\circ/18 = 160^\circ.$$ Then $x + x + 160^\circ = 360^\circ$, so $x = 100^\circ$.
答案（A）：在给定的拱形旁添加一个对称的拱形，形成由梯形组成的闭合环。考虑由该闭合环内部形成的正十八边形。正十八边形的每个内角为 $$(18-2)\cdot 180^\circ/18 = 160^\circ.$$ 因此 $x + x + 160^\circ = 360^\circ$，所以 $x = 100^\circ$。
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Q25
Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of its opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?
立方体每个面上从一条边的中心画一条窄条纹到对边中心的中心。每面边对选择随机独立。存在一条环绕立方体的连续条纹的概率是多少？
Correct Answer: B
Answer (B): The stripe on each face of the cube will be oriented in one of two possible directions, so there are $2^6=64$ possible stripe combinations on the cube. There are 3 pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripe orientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in 2 different ways. Thus a total of $3\cdot2\cdot2=12$ stripe combinations on the cube result in a continuous stripe around the cube, and the requested probability is $\frac{12}{64}=\frac{3}{16}$.
答案（B）：立方体每个面的条纹方向有两种可能，因此立方体上一共有 $2^6=64$ 种条纹组合。立方体有 3 对互相平行的面，所以如果存在一条环绕的条纹，那么那一对“不起唯一贡献”的面会唯一确定其余各面的条纹方向。另外，这两个“不起贡献”的面上的条纹各自仍可有 2 种取向。因此，使立方体出现连续环绕条纹的组合总数为 $3\cdot2\cdot2=12$，所求概率为 $\frac{12}{64}=\frac{3}{16}$。