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AMC8 2000

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AMC8 · 2000

Q1
Aunt Anna is 42 years old. Caitlin is 5 years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?
安娜阿姨42岁。凯特琳比布里安娜小5岁，布里安娜的年龄是安娜阿姨年龄的一半。凯特琳多大？
Correct Answer: B
If Brianna is half as old as Aunt Anna, then Brianna is $\frac{42}{2}$ years old, or $21$ years old. If Caitlin is $5$ years younger than Brianna, she is $21-5$ years old, or $16$. So, the answer is $\boxed{B}$
如果布里安娜的年龄是安娜阿姨年龄的一半，那么布里安娜是$\frac{42}{2}$岁，即$21$岁。 如果凯特琳比布里安娜小$5$岁，那么她是$21-5$岁，即$16$岁。 所以，答案是$\boxed{B}$
Q2
Which of these numbers is less than its reciprocal?
以下哪个数小于它的倒数？
Correct Answer: A
Answer (A): The number $0$ has no reciprocal, and $1$ and $-1$ are their own reciprocals. This leaves only $2$ and $-2$. The reciprocal of $2$ is $\frac{1}{2}$, but $2$ is not less than $\frac{1}{2}$. The reciprocal of $-2$ is $-\frac{1}{2}$, and $-2$ is less than $-\frac{1}{2}$.
答案（A）：数字 $0$ 没有倒数，而 $1$ 和 $-1$ 的倒数是它们本身。因此只剩下 $2$ 和 $-2$。$2$ 的倒数是 $\frac{1}{2}$，但 $2$ 并不小于 $\frac{1}{2}$。$-2$ 的倒数是 $-\frac{1}{2}$，并且 $-2$ 小于 $-\frac{1}{2}$。
Q3
How many whole numbers lie in the interval between $\frac{5}{3}$ and $2\pi$?
区间$\frac{5}{3}$和$2\pi$之间有多少个整数？
Correct Answer: D
The smallest whole number in the interval is $2$ because $5/3$ is more than $1$ but less than $2$. The largest whole number in the interval is $6$ because $2\pi$ is more than $6$ but less than $7$. There are five whole numbers in the interval. They are $2$, $3$, $4$, $5$, and $6$, so the answer is $\boxed{\text{(D)}\ 5}$.
区间最小整数是$2$，因为$\frac{5}{3} \approx 1.666$大于$1$小于$2$。最大整数是$6$，因为$2\pi \approx 6.28$大于$6$小于$7$。区间内有$2,3,4,5,6$五个整数，所以答案是$\boxed{\text{(D)}\ 5}$。
solution
Q4
In 1960 only 5% of the working adults in Carlin City worked at home. By 1970 the "at-home" work force had increased to 8%. In 1980 there were approximately 15% working at home, and in 1990 there were 30%. The graph that best illustrates this is:
1960年，卡林市仅有5%的在职成人居家工作。到1970年，居家劳动力增加到8%。1980年约为15%，1990年为30%。最能说明这一情况的图表是：
Correct Answer: E
The data are $1960 (5\%)$, $1970 (8\%)$, $1980 (15\%)$, and $1990 (30\%)$. Only one of these graphs has the answer and that is choice $\boxed{E}$.
数据为1960年(5%)，1970年(8%)，1980年(15%)，1990年(30%)。只有选择$\boxed{E}$的图表符合，所以答案是$\boxed{E}$。
Q5
Each principal of Lincoln High School serves exactly one 3-year term. What is the maximum number of principals this school could have during an 8-year period?
林肯高中的每个校长任期正好3年。8年内该校最多可能有多少位校长？
Correct Answer: C
Answer (C): If the first year of the 8-year period was the final year of a principal’s term, then in the next six years two more principals would serve, and the last year of the period would be the first year of the fourth principal’s term. Therefore, the maximum number of principals who can serve during an 8-year period is 4.
答案（C）：如果这 8 年期间的第一年是某位校长任期的最后一年，那么在接下来的 6 年里将有另外两位校长任职，而该期间的最后一年将是第四位校长任期的第一年。因此，在 8 年期间内最多可以有 4 位校长任职。
Q6
Figure ABCD is a square. Inside this square three smaller squares are drawn with side lengths as labeled. The area of the shaded L-shaped region is
图 ABCD 是一个正方形。在这个正方形内画了三个较小的正方形，边长如图所示。阴影 L 形区域的面积是
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Correct Answer: A
Answer (A): The L-shaped region is made up of two rectangles with area $3 \times 1 = 3$ plus the corner square with area $1 \times 1 = 1$, so the area of the L-shaped figure is $2 \times 3 + 1 = 7$.
答案（A）：这个 L 形区域由两个长方形组成，每个长方形的面积为 $3 \times 1 = 3$，再加上一个角上的正方形，其面积为 $1 \times 1 = 1$，因此该 L 形图形的面积为 $2 \times 3 + 1 = 7$。
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Q7
What is the minimum possible product of three different numbers of the set $\{-8, -6, -4, 0, 3, 5, 7\}$?
集合 $\{-8, -6, -4, 0, 3, 5, 7\}$ 中三个不同数的乘积的最小可能是多少？
Correct Answer: B
Answer (B): The only way to get a negative product using three numbers is to multiply one negative number and two positives or three negatives. Only two reasonable choices exist: $(-8)\times(-6)\times(-4)=(-8)\times(24)=-192$ and $(-8)\times5\times7=(-8)\times35=-280$. The latter is smaller.
答案（B）：用三个数得到负积的唯一方法是：一个负数与两个正数相乘，或三个负数相乘。只有两种合理选择：$(-8)\times(-6)\times(-4)=(-8)\times(24)=-192$，以及$(-8)\times5\times7=(-8)\times35=-280$。后者更小。
Q8
Three dice with faces numbered 1 through 6 are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is
三个骰子如图所示堆叠。十八个面中七个可见，留下十一个隐藏的面（背面、底面、之间）。在这个视图中不可见的点数总数是
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Correct Answer: D
Answer (D): The numbers on one die total $1+2+3+4+5+6=21$, so the numbers on the three dice total 63. Numbers 1, 1, 2, 3, 4, 5, 6 are visible, and these total 22. This leaves $63-22=41$ not seen.
答案（D）：一个骰子的点数总和为 $1+2+3+4+5+6=21$，所以三个骰子的点数总和为 63。可见的点数是 1、1、2、3、4、5、6，它们的和为 22。因此看不见的点数之和为 $63-22=41$。
Q9
Three-digit powers of 2 and 5 are used in this cross-number puzzle. What is the only possible digit for the outlined square? ACROSS 2. $2^m$ DOWN 1. $5^n$
这个填字谜中使用三位数的 2 和 5 的幂。虚线方框的唯一可能数字是？ ACROSS 2. $2^m$ DOWN 1. $5^n$
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Correct Answer: D
Answer (D): The 3-digit powers of 5 are 125 and 625, so space 2 is filled with a 2. The only 3-digit power of 2 beginning with 2 is 256, so the outlined block is filled with a 6.
答案（D）：5 的三位数幂是 125 和 625，所以第 2 个空格填 2。以 2 开头的唯一三位数的 2 的幂是 256，所以框出的方格填 6。
Q10
Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?
Ara 和 Shea 曾经身高相同。此后 Shea 长高了 20%，而 Ara 长高了 Shea 英寸数的一半。Shea 现在 60 英寸高。Ara 现在多高（英寸）？
Correct Answer: E
Answer (E): Shea is 60 inches tall. This is 1.2 times the common starting height, so the starting height was $\frac{60}{1.2}=50$ inches. Shea has grown $60-50=10$ inches. Therefore, Ara grew 5 inches and is now 55 inches tall.
答案（E）：Shea 现在身高 60 英寸。这是共同起始身高的 1.2 倍，因此起始身高为 $\frac{60}{1.2}=50$ 英寸。Shea 增长了 $60-50=10$ 英寸。因此，Ara 增长了 5 英寸，现在身高 55 英寸。
Q11
The number 64 has the property that it is divisible by its units digit. How many whole numbers between 10 and 50 have this property?
数字 64 具有能被其个位数整除的性质。在 10 到 50 之间，有多少个整数具有这种性质？
Correct Answer: C
Answer (C): Twelve numbers ending with 1, 2, or 5 have this property. They are 11, 12, 15, 21, 22, 25, 31, 32, 35, 41, 42, and 45. In addition, we have 33, 24, 44, 36, and 48, for a total of 17. (Note that 20, 30, and 40 are not divisible by 0, since division by 0 is not defined.)
答案（C）：有 12 个以 1、2 或 5 结尾的数具有这个性质。它们是 11、12、15、21、22、25、31、32、35、41、42 和 45。此外，还有 33、24、44、36 和 48，总共有 17 个。（注意 20、30 和 40 不能被 0 整除，因为除以 0 是没有定义的。）
Q12
A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wall?
一座长 100 英尺、高 7 英尺的砌块墙将使用高 1 英尺、长 2 英尺或 1 英尺的砌块建造（砌块不得切割）。砌块的垂直接缝必须如图所示错开，且墙两端必须平整。建造这座墙需要的最小砌块数是多少？
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Correct Answer: D
Answer (D): If the vertical joins were not staggered, the wall could be build with $\frac{1}{2}(100\times 7)=350$ of the two-foot blocks. To stagger the joins, we need only to replace, in every other row, one of the longer blocks by two shorter ones, placing one at each end. To minimize the number of blocks this should be done in rows 2, 4, and 6. This adds 3 blocks to the 350, making a total of 353.
答案（D）：如果竖直接缝不交错，这堵墙可以用 $\frac{1}{2}(100\times 7)=350$ 块两英尺的砖块砌成。为了让接缝交错，我们只需在每隔一行中，将其中一块较长的砖替换为两块较短的砖，并把它们分别放在两端。为使砖块数量最少，应在第 2、4、6 行这样做。这样在 350 的基础上增加 3 块，总数为 353。
Q13
In triangle CAT, we have $\angle ACT = \angle ATC$ and $\angle CAT = 36^\circ$. If $\overline{TR}$ bisects $\angle ATC$, then $\angle CRT =$
在三角形 CAT 中，$\\angle ACT = \\angle ATC$ 且 $\\angle CAT = 36^\circ$。如果 $\\overline{TR}$ 平分 $\\angle ATC$，则 $\\angle CRT =$
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Correct Answer: C
Answer (C): Since $\angle ACT=\angle ATC$ and $\angle CAT=36^\circ$, we have $2(\angle ATC)=180^\circ-36^\circ=144^\circ$ and $\angle ATC=\angle ACT=72^\circ$. Because $TR$ bisects $\angle ATC$, $\angle CTR=\frac12(72^\circ)=36^\circ$. In triangle $CRT$, $\angle CRT=180^\circ-36^\circ-72^\circ=72^\circ$. Note that some texts use $\angle ACT$ to define the angle and $m\angle ACT$ to indicate its measure.
答案（C）：由于 $\angle ACT=\angle ATC$ 且 $\angle CAT=36^\circ$，所以 $2(\angle ATC)=180^\circ-36^\circ=144^\circ$，并且 $\angle ATC=\angle ACT=72^\circ$。因为 $TR$ 平分 $\angle ATC$，所以 $\angle CTR=\frac12(72^\circ)=36^\circ$。在三角形 $CRT$ 中，$\angle CRT=180^\circ-36^\circ-72^\circ=72^\circ$。注意：有些教材用 $\angle ACT$ 表示角本身，用 $m\angle ACT$ 表示该角的度数。
Q14
What is the units digit of $19^{19} + 99^{99}$?
$19^{19} + 99^{99}$ 的个位数是多少？
Correct Answer: D
Answer (D): The units digit of a power of an integer is determined by the units digit of the integer; that is, the tens digit, hundreds digit, etc... of the integer have no effect on the units digit of the result. In this problem, the units digit of $19^{19}$ is the units digit of $9^{19}$. Note that $9^{1}=9$ ends in $9$, $9^{2}=81$ ends in $1$, $9^{3}=729$ ends in $9$, and, in general, the units digit of odd powers of $9$ is $9$, whereas the units digit of even powers of $9$ is $1$. Since both exponents are odd, the sum of their units digits is $9+9=18$, the units digit of which is $8$.
答案（D）：一个整数的幂的个位数由该整数的个位数决定；也就是说，这个整数的十位、百位等对结果的个位数没有影响。在本题中，$19^{19}$ 的个位数等同于 $9^{19}$ 的个位数。注意 $9^{1}=9$ 的个位是 $9$，$9^{2}=81$ 的个位是 $1$，$9^{3}=729$ 的个位是 $9$。一般而言，$9$ 的奇数次幂的个位是 $9$，而 $9$ 的偶数次幂的个位是 $1$。由于两个指数都是奇数，它们个位数之和为 $9+9=18$，其个位数为 $8$。
Q15
Triangle ABC, ADE, and EFG are all equilateral. Points D and G are midpoints of AC and AE, respectively. If AB = 4, what is the perimeter of figure ABCDEFG?
三角形 ABC、ADE 和 EFG 均为等边三角形。点 D 和 G 分别是 AC 和 AE 的中点。若 AB = 4，则图形 ABCDEFG 的周长是多少？
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Correct Answer: C
The large triangle $ABC$ has sides of length $4$. The medium triangle has sides of length $2$. The small triangle has sides of length $1$. There are $3$ segment sizes, and all segments depicted are one of these lengths. Starting at $A$ and going clockwise, the perimeter is: $AB + BC + CD + DE + EF + FG + GA$ $4 + 4 + 2 + 2 + 1 + 1 + 1$ $15$, thus the answer is $\boxed{C}$
大三角形 $ABC$ 边长为 $4$。中等三角形边长为 $2$。小三角形边长为 $1$。有 $3$ 种线段长度，图中所有线段均为这些长度之一。 从 A 顺时针开始，周长为： $AB + BC + CD + DE + EF + FG + GA$ $4 + 4 + 2 + 2 + 1 + 1 + 1$ $15$，答案 $\boxed{C}$。
Q16
In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?
为了让Mateen在他的矩形后院走一公里（1000米），他必须沿着长度走25次，或者沿着周长走10次。Mateen的后院面积有多少平方米？
Correct Answer: C
The length $L$ of the rectangle is $\frac{1000}{25}=40$ meters. The perimeter $P$ is $\frac{1000}{10}=100$ meters. Since $P_{rect} = 2L + 2W$, we plug values in to get: $100 = 2\cdot 40 + 2W$ $100 = 80 + 2W$ $2W = 20$ $W = 10$ meters Since $A_{rect} = LW$, the area is $40\cdot 10=400$ square meters or $\boxed{C}$.
矩形的长度$L=\frac{1000}{25}=40$米。周长$P=\frac{1000}{10}=100$米。因为$P_{rect}=2L+2W$，代入数值得： $100=2\cdot 40+2W$ $100=80+2W$ $2W=20$ $W=10$米 因为$A_{rect}=LW$，面积为$40\cdot 10=400$平方米，即\boxed{C}。
Q17
The operation $\otimes$ is defined for all nonzero numbers by $a \otimes b = \frac{a^2}{b}$. Determine $[[(1 \otimes 2) \otimes 3] - [1 \otimes (2 \otimes 3)]]$.
操作$\otimes$对于所有非零数定义为$a\otimes b=\frac{a^2}{b}$。求$[[(1\otimes 2)\otimes 3]-[1\otimes(2\otimes 3)]]$的值。
Correct Answer: A
Answer (A): We have $$(1\otimes 2)\otimes 3=\frac{1^2}{2}\otimes 3=\frac{1}{2}\otimes 3=\frac{\left(\frac{1}{2}\right)^2}{3}=\frac{\frac{1}{4}}{3}=\frac{1}{12},$$ and $$1\otimes(2\otimes 3)=1\otimes\left(\frac{2^2}{3}\right)=1\otimes\frac{4}{3}=\frac{1^2}{\frac{4}{3}}=\frac{3}{4}.$$ Therefore, $$(1\otimes 2)\otimes 3-1\otimes(2\otimes 3)=\frac{1}{12}-\frac{3}{4}=\frac{1}{12}-\frac{9}{12}=-\frac{8}{12}=-\frac{2}{3}.$$
答案（A）：我们有 $$(1\otimes 2)\otimes 3=\frac{1^2}{2}\otimes 3=\frac{1}{2}\otimes 3=\frac{\left(\frac{1}{2}\right)^2}{3}=\frac{\frac{1}{4}}{3}=\frac{1}{12},$$ 以及 $$1\otimes(2\otimes 3)=1\otimes\left(\frac{2^2}{3}\right)=1\otimes\frac{4}{3}=\frac{1^2}{\frac{4}{3}}=\frac{3}{4}.$$ 因此， $$(1\otimes 2)\otimes 3-1\otimes(2\otimes 3)=\frac{1}{12}-\frac{3}{4}=\frac{1}{12}-\frac{9}{12}=-\frac{8}{12}=-\frac{2}{3}.$$
Q18
Consider these two geoboard quadrilaterals. Which of the following statements is true?
考虑这两个地理板上的四边形。以下哪个陈述是正确的？
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Correct Answer: E
Answer (E): Divide each quadrilateral as shown. The resulting triangles each have base 1, altitude 1, and area $\frac{1}{2}$, so the quadrilaterals each have area 1. Three sides of quadrilateral I match those of quadrilateral II as indicated by matching marks. The fourth side of quadrilateral I is less than the fourth side of quadrilateral II, hence its perimeter is less, and choice (E) is correct.
答案（E）：如图将每个四边形分割。分割后得到的每个三角形底为 1，高为 1，面积为 $\frac{1}{2}$，所以每个四边形的面积都是 1。 四边形 I 的三条边与四边形 II 的对应三条边相等（由相同记号表示）。四边形 I 的第四条边小于四边形 II 的第四条边，因此它的周长更小，所以选项（E）正确。
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Q19
Three circular arcs of radius 5 units bound the region shown. Arcs AB and AD are quarter-circles, and arc BCD is a semi-circle. What is the area, in square units, of the region?
三个半径为5单位的圆弧包围了所示区域。弧AB和AD是四分之一圆，弧BCD是半圆。该区域面积有多少平方单位？
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Correct Answer: C
Answer (C): Divide the semicircle in half and rotate each half down to fill the space below the quarter-circles. The figure formed is a rectangle with dimensions 5 and 10. The area is 50.
答案（C）：将半圆分成两半，并将每一半向下旋转以填充四分之一圆下方的空间。形成的图形是一个长为 10、宽为 5 的矩形。面积为 50。
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Q20
You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02$, with at least one coin of each type. How many dimes must you have?
你有九枚硬币：便士、镍币、角币和25分币，总价值1.02美元，且至少有一种每种类型。必须有多少个角币？
Correct Answer: A
Answer (A): Since the total value is \$1.02, you must have either 2 or 7 pennies. It is impossible to have 7 pennies, since the two remaining coins cannot have a value of 95 cents. With 2 pennies the remaining 7 coins have a value of \$1.00. Either 2 or 3 of these must be quarters. If you have 2 quarters, the other 5 coins would be dimes, and you would have no nickels. The only possible solution is 3 quarters, 1 dime, 3 nickels and 2 pennies.
答案（A）：由于总价值为 \$1.02，你必须有 2 枚或 7 枚便士。拥有 7 枚便士是不可能的，因为剩下的两枚硬币不可能组成 95 美分。若有 2 枚便士，则其余 7 枚硬币的总值为 \$1.00。在这 7 枚中，必须有 2 枚或 3 枚是 25 美分硬币（quarter）。如果有 2 枚 quarter，那么另外 5 枚硬币就会都是 10 美分硬币（dime），从而不会有 5 美分硬币（nickel）。唯一可能的解是：3 枚 quarter、1 枚 dime、3 枚 nickel 和 2 枚 penny。
Q21
Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
惠子抛一枚硬币，以法莲抛两枚硬币。以法莲得到与惠子相同数量正面的概率是
Correct Answer: B
Answer (B): Make a complete list of equally likely outcomes: \[ \begin{array}{ccc} \text{Keiko} & \text{Ephraim} & \text{Same Number of Heads?} \\ H & HH & \text{No} \\ H & HT & \text{Yes} \\ H & TH & \text{Yes} \\ H & TT & \text{No} \\ T & HH & \text{No} \\ T & HT & \text{No} \\ T & TH & \text{No} \\ T & TT & \text{Yes} \end{array} \] The probability that they have the same number of heads is $\frac{3}{8}$.
答案（B）：列出所有等可能的结果： \[ \begin{array}{ccc} \text{Keiko} & \text{Ephraim} & \text{正面数量相同？} \\ H & HH & \text{否} \\ H & HT & \text{是} \\ H & TH & \text{是} \\ H & TT & \text{否} \\ T & HH & \text{否} \\ T & HT & \text{否} \\ T & TH & \text{否} \\ T & TT & \text{是} \end{array} \] 他们正面数量相同的概率是 $\frac{3}{8}$。
Q22
A cube has edge length 2. Suppose that we glue a cube of edge length 1 on top of the big cube so that one of its faces rests entirely on the top face of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to:
一个边长为2的立方体。我们将一个边长为1的立方体粘在大立方体的顶部，使其一个面完全贴在大立方体顶面上。从原立方体到新形成的固体的表面积（侧面、顶部和底部）的百分比增加最接近：
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Correct Answer: C
Answer (C): The area of each face of the larger cube is $2^2=4$. There are six faces of the cube, so its surface area is $6(4)=24$. When we add the smaller cube, we decrease the original surface area by $1$, but we add $5(1^2)=5$ units of area (1 unit for each of the five unglued faces of the smaller cube). This is a net increase of $4$ from the original surface area, and $4$ is $\frac{4}{24}=\frac{1}{6}\approx16.7\%$ of $24$. The closest value given is $17$.
答案（C）：较大立方体每个面的面积为 $2^2=4$。立方体有 6 个面，所以其表面积为 $6(4)=24$。当我们加上较小的立方体时，原来的表面积减少 $1$，但会新增 $5(1^2)=5$ 个面积单位（小立方体有 5 个未粘合的面，每个贡献 1 个单位面积）。因此相对于原表面积净增加 $4$，而 $4$ 占 $24$ 的比例为 $\frac{4}{24}=\frac{1}{6}\approx16.7\%$。所给选项中最接近的是 $17$。
Q23
There is a list of seven numbers. The average of the first four numbers is 5, and the average of the last four numbers is 8. If the average of all seven numbers is $6\frac{4}{7}$, then the number common to both sets of four numbers is
有一列七个数。前四个数的平均数是5，后四个数的平均数是8。如果全部七个数的平均数是$6\frac{4}{7}$，则两个四数集合共有的那个数是
Correct Answer: B
Answer (B): Since the average of all seven numbers is $6\frac{4}{7}=\frac{46}{7}$, the sum of the seven numbers is $7\times\frac{46}{7}=46$. The sum of the first four numbers is $4\times5=20$ and the sum of the last four numbers is $4\times8=32$. Since the fourth number is used in each of these two sums, the fourth number must be $(20+32)-46=6$.
答案（B）：由于这七个数的平均数是 $6\frac{4}{7}=\frac{46}{7}$，因此这七个数的和为 $7\times\frac{46}{7}=46$。前四个数的和为 $4\times5=20$，后四个数的和为 $4\times8=32$。由于第四个数在这两个和中都被计算了一次，所以第四个数应为 $(20+32)-46=6$。
Q24
If $\angle A = 20^\circ$ and $\angle AFG = \angle AGF$, then $\angle B + \angle D =$
若$\angle A = 20^\circ$且$\angle AFG = \angle AGF$，则$\angle B + \angle D =$
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Correct Answer: D
Answer (D): Since $\angle AFG = \angle AGF$ and $\angle GAF + \angle AFG + \angle AGF = 180^\circ$, we have $20^\circ + 2(\angle AFG) = 180^\circ$. So $\angle AFG = 80^\circ$. Also, $\angle AFG + \angle BFD = 190^\circ$, so $\angle BFD = 100^\circ$. The sum of the angles of $\triangle BFD$ is $180^\circ$, so $\angle B + \angle D = 80^\circ$. Note: In $\triangle AFG$, $\angle AFG = \angle B + \angle D$. In general, an exterior angle of a triangle equals the sum of its remote interior angles. For example, in $\triangle GAF$, $\angle x = \angle GAF + \angle AGF$. Note that, as in Problem 13, some texts use different symbols to represent an angle and its degree measure.
答案（D）：由于 $\angle AFG = \angle AGF$ 且 $\angle GAF + \angle AFG + \angle AGF = 180^\circ$，我们有 $20^\circ + 2(\angle AFG) = 180^\circ$。所以 $\angle AFG = 80^\circ$。另外，$\angle AFG + \angle BFD = 190^\circ$，因此 $\angle BFD = 100^\circ$。$\triangle BFD$ 的内角和为 $180^\circ$，所以 $\angle B + \angle D = 80^\circ$。 注：在 $\triangle AFG$ 中，$\angle AFG = \angle B + \angle D$。一般地，三角形的一个外角等于与之不相邻的两个内角之和。例如，在 $\triangle GAF$ 中，$\angle x = \angle GAF + \angle AGF$。 还要注意：如同第 13 题所示，有些教材会用不同的符号来表示“角”及其“度数”。
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Q25
The area of rectangle ABCD is 72. If point A and the mid-points of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is
矩形ABCD的面积是72。若将点A与BC和CD的中点连接形成一个三角形，则该三角形的面积是
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Correct Answer: B
Answer (B): Three right triangles lie outside $\triangle AMN$. Their areas are $\frac{1}{4}$, $\frac{1}{4}$, and $\frac{1}{8}$ for a total of $\frac{5}{8}$ of the rectangle. The area of $\triangle AMN$ is $\frac{3}{8}(72)=27$.
答案（B）：有三个直角三角形位于 $\triangle AMN$ 的外部。它们的面积分别是 $\frac{1}{4}$、$\frac{1}{4}$ 和 $\frac{1}{8}$，总共占矩形面积的 $\frac{5}{8}$。因此 $\triangle AMN$ 的面积为 $\frac{3}{8}(72)=27$。