Doug can paint a room in $5$ hours. Dave can paint the same room in $7$ hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let $t$ be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by $t$?
Doug 刷完一间房间需要 $5$ 小时。Dave 刷完同一间房间需要 $7$ 小时。Doug 和 Dave 一起刷这间房间，并在中途午休 1 小时。设 $t$ 为他们完成工作所需的总时间（小时），包括午休。下列哪个方程由 $t$ 满足？
Correct Answer: D
Doug can paint $\frac{1}{5}$ of a room per hour, Dave can paint $\frac{1}{7}$ of a room per hour, and the time they spend working together is $t-1$.
Since rate multiplied by time gives output, $\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}$
If one person does a job in $a$ hours and another person does a job in $b$ hours, the time it takes to do the job together is $\frac{ab}{a+b}$ hours.
Since Doug paints a room in 5 hours and Dave paints a room in 7 hours, they both paint in $\frac{5\times7}{5+7} = \frac{35}{12}$ hours. They also take 1 hour for lunch, so the total time $t = \frac{35}{12} + 1$ hours.
Looking at the answer choices, $(D)$ is the only one satisfied by $t = \frac{35}{12} + 1$.
Doug 每小时能刷 $\frac{1}{5}$ 间房，Dave 每小时能刷 $\frac{1}{7}$ 间房，他们一起工作的时间为 $t-1$。
由于“效率 $\times$ 时间 = 完成量”，有 $\left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1 \Rightarrow \mathrm{(D)}$
若一人用 $a$ 小时完成工作，另一人用 $b$ 小时完成工作，则两人合作完成所需时间为 $\frac{ab}{a+b}$ 小时。
Doug 用 5 小时刷完，Dave 用 7 小时刷完，因此两人合作用时 $\frac{5\times7}{5+7} = \frac{35}{12}$ 小时。再加上 1 小时午休，总时间 $t = \frac{35}{12} + 1$ 小时。
查看选项，只有 $(D)$ 满足 $t = \frac{35}{12} + 1$。
Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum?
三个立方体每个都由所示图案制成。然后它们被一个叠一个地堆放在桌子上，使得 $13$ 个可见数字的和尽可能大。这个和是多少？
Correct Answer: C
To maximize the sum of the $13$ faces that are showing, we can minimize the sum of the numbers of the $5$ faces that are not showing.
The bottom $2$ cubes each have a pair of opposite faces that are covered up. When the cube is folded, $(1,32)$; $(2,16)$; and $(4,8)$ are opposite pairs. Clearly $4+8=12$ has the smallest sum.
The top cube has 1 number that is not showing. The smallest number on a face is $1$.
So, the minimum sum of the $5$ unexposed faces is $2\cdot12+1=25$. Since the sum of the numbers on all the cubes is $3(32+16+8+4+2+1)=189$, the maximum possible sum of $13$ visible numbers is $189-25=164 \Rightarrow C$.
为了使显示出来的 $13$ 个面的数字和最大，我们可以使未显示出来的 $5$ 个面的数字和最小。
下面的 $2$ 个立方体各有一对相对的面被遮住。将立方体折叠后，$(1,32)$、$(2,16)$ 和 $(4,8)$ 是相对面。显然 $4+8=12$ 的和最小。
最上面的立方体有 $1$ 个数字不显示。面上的最小数字是 $1$。
因此，$5$ 个未露出的面的最小和为 $2\cdot12+1=25$。由于所有立方体上数字总和为 $3(32+16+8+4+2+1)=189$，所以 $13$ 个可见数字的最大可能和为 $189-25=164 \Rightarrow C$。
A function $f$ has domain $[0,2]$ and range $[0,1]$. (The notation $[a,b]$ denotes $\{x:a \le x \le b \}$.) What are the domain and range, respectively, of the function $g$ defined by $g(x)=1-f(x+1)$?
函数 $f$ 的定义域是 $[0,2]$，值域是 $[0,1]$。（记号 $[a,b]$ 表示 $\{x:a \le x \le b \}$。）由 $g(x)=1-f(x+1)$ 定义的函数 $g$ 的定义域和值域分别是什么？
Correct Answer: B
$g(x)$ is defined if $f(x + 1)$ is defined. Thus the domain is all $x| x + 1 \in [0,2] \rightarrow x \in [ - 1,1]$.
Since $f(x + 1) \in [0,1]$, $- f(x + 1) \in [ - 1,0]$. Thus $g(x) = 1 - f(x + 1) \in [0,1]$ is the range of $g(x)$.
Thus the answer is $[- 1,1],[0,1] \longrightarrow \boxed{B}$.
当且仅当 $f(x + 1)$ 有定义时，$g(x)$ 才有定义。因此定义域为所有满足 $x + 1 \in [0,2] \rightarrow x \in [ - 1,1]$ 的 $x$。
由于 $f(x + 1) \in [0,1]$，所以 $- f(x + 1) \in [ - 1,0]$。因此 $g(x) = 1 - f(x + 1) \in [0,1]$，这就是 $g(x)$ 的值域。
因此答案是 $[- 1,1],[0,1] \longrightarrow \boxed{B}$。
Points $A$ and $B$ lie on a circle centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?
点 $A$ 和 $B$ 位于以 $O$ 为圆心的圆上，且 $\angle AOB = 60^\circ$。第二个圆内切于第一个圆，并且与 $\overline{OA}$ 和 $\overline{OB}$ 都相切。较小圆的面积与较大圆面积的比是多少？
Correct Answer: B
Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$. Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$.
Then $PQO$ is a right triangle. Angle $POQ$ is $30$ degrees because line $OP$ bisects angle $AOB$ (this can be proved by dropping a perpendicular line from $P$ to line $OB$, letting their intersection be point $S$, and proving triangles $PQO$ and $PSO$ congruent), meaning that $PQO$ is a $30-60-90$ triangle. Therefore, $OP=2PQ$.
Since $OP=OC-PC=OC-r=R-r$, we have $R-r=2PQ$, or $R-r=2r$, or $\frac{1}{3}=\frac{r}{R}$.
Ratio of areas of circles is ratio of radii squared, so the answer is $\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}$
设 $P$ 为小圆圆心，半径为 $r$，设 $Q$ 为小圆与 $OA$ 相切的切点。再设 $C$ 为小圆与大圆（半径为 $R$）相切的切点。
则 $PQO$ 为直角三角形。由于直线 $OP$ 平分角 $AOB$（可通过从 $P$ 向直线 $OB$ 作垂线，垂足为 $S$，并证明三角形 $PQO$ 与 $PSO$ 全等来证明），所以 $\angle POQ$ 为 $30$ 度，这意味着 $PQO$ 是一个 $30-60-90$ 三角形。因此，$OP=2PQ$。
由于 $OP=OC-PC=OC-r=R-r$，我们有 $R-r=2PQ$，即 $R-r=2r$，从而 $\frac{1}{3}=\frac{r}{R}$。
圆面积之比等于半径平方之比，所以答案为 $\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}$。
Let $k={2008}^{2}+{2}^{2008}$. What is the units digit of $k^2+2^k$?
设 $k={2008}^{2}+{2}^{2008}$。$k^2+2^k$ 的个位数是多少？
Correct Answer: D
$k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$.
So, $k^2 \equiv 0 \pmod{10}$. Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$.
Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$. So the units digit is $6 \Rightarrow \boxed{D}$.
Another way to get $k \equiv 0 \pmod{10}$ is to find the cycles of the last digit.
For $2008^2$, we need only be concerned with the last digit $8$ since the other digits do not affect the last digit. Since $8^{2} = 64$, the last digit of $2008^2$ is $4$.
For $2^{2008}$, note that the last digit cycles through the pattern ${2, 4, 8, 6}$. (You can try to see this by calculating the first powers of $2$.)
Since $2008$ is a multiple of $4$, the last digit of $2^{2008}$ is evidently $6.$
Continue as follows.
Mathboy282,
That is actually what solution 1 is explaining in the first sentence but I think yours is a more detailed and easier to comprehend explanation.
$k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$。
所以，$k^2 \equiv 0 \pmod{10}$。由于 $k = 2008^2+2^{2008}$ 是 $4$ 的倍数，并且 $2$ 的幂的个位数以 $4$ 为周期循环，故 $2^k \equiv 2^4 \equiv 6 \pmod{10}$。
因此，$k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$。所以个位数是 $6 \Rightarrow \boxed{D}$。
另一种得到 $k \equiv 0 \pmod{10}$ 的方法是找个位数的循环。
对于 $2008^2$，我们只需关注个位数 $8$，因为其他数字不影响个位数。由于 $8^{2} = 64$，所以 $2008^2$ 的个位数是 $4$。
对于 $2^{2008}$，注意个位数按模式 ${2, 4, 8, 6}$ 循环。（你可以通过计算 $2$ 的前几次幂来观察。）
由于 $2008$ 是 $4$ 的倍数，$2^{2008}$ 的个位数显然是 $6$。
继续如下。
Mathboy282,
That is actually what solution 1 is explaining in the first sentence but I think yours is a more detailed and easier to comprehend explanation.
The numbers $\log(a^3b^7)$, $\log(a^5b^{12})$, and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence, and the $12^\text{th}$ term of the sequence is $\log{b^n}$. What is $n$?
数 $\log(a^3b^7)$，$\log(a^5b^{12})$ 和 $\log(a^8b^{15})$ 是等差数列的前三项，该数列的第 $12^\text{th}$ 项是 $\log{b^n}$。$n$ 是多少？
Correct Answer: D
Let $A = \log(a)$ and $B = \log(b)$.
The first three terms of the arithmetic sequence are $3A + 7B$, $5A + 12B$, and $8A + 15B$, and the $12^\text{th}$ term is $nB$.
Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$.
Since the first three terms in the sequence are $13B$, $22B$, and $31B$, the $k$th term is $(9k + 4)B$.
Thus the $12^\text{th}$ term is $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{D}$.
设 $A = \log(a)$，$B = \log(b)$。
等差数列的前三项分别为 $3A + 7B$，$5A + 12B$，$8A + 15B$，第 $12^\text{th}$ 项为 $nB$。
因此，$2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$。
由于前三项为 $13B$，$22B$，$31B$，第 $k$ 项为 $(9k + 4)B$。
所以第 $12^\text{th}$ 项为 $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{D}$。
Let $a_1,a_2,\ldots$ be a sequence determined by the rule $a_n=a_{n-1}/2$ if $a_{n-1}$ is even and $a_n=3a_{n-1}+1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \le 2008$ is it true that $a_1$ is less than each of $a_2$, $a_3$, and $a_4$?
设序列 $a_1,a_2,\ldots$ 由如下规则确定：若 $a_{n-1}$ 为偶数，则 $a_n=a_{n-1}/2$；若 $a_{n-1}$ 为奇数，则 $a_n=3a_{n-1}+1$。对于多少个满足 $a_1 \le 2008$ 的正整数 $a_1$，有 $a_1$ 小于 $a_2$、$a_3$ 和 $a_4$ 中的每一个？
Correct Answer: D
All positive integers can be expressed as $4n$, $4n+1$, $4n+2$, or $4n+3$, where $n$ is a nonnegative integer.
- If $a_1=4n$, then $a_2=\frac{4n}{2}=2n<a_1$.
- If $a_1=4n+1$, then $a_2=3(4n+1)+1=12n+4$, $a_3=\frac{12n+4}{2}=6n+2$, and $a_4=\frac{6n+2}{2}=3n+1<a_1$.
- If $a_1=4n+2$, then $a_2=2n+1<a_1$.
- If $a_1=4n+3$, then $a_2=3(4n+3)+1=12n+10$, $a_3=\frac{12n+10}{2}=6n+5$, and $a_4=3(6n+5)+1=18n+16$.
Since $12n+10, 6n+5, 18n+16 > 4n+3$, every positive integer $a_1=4n+3$ will satisfy $a_1<a_2,a_3,a_4$.
Since one fourth of the positive integers $a_1 \le 2008$ can be expressed as $4n+3$, where $n$ is a nonnegative integer, the answer is $\frac{1}{4}\cdot 2008 = 502 \Rightarrow D$.
所有正整数都可表示为 $4n$、$4n+1$、$4n+2$ 或 $4n+3$，其中 $n$ 为非负整数。
- 若 $a_1=4n$，则 $a_2=\frac{4n}{2}=2n<a_1$。
- 若 $a_1=4n+1$，则 $a_2=3(4n+1)+1=12n+4$，$a_3=\frac{12n+4}{2}=6n+2$，$a_4=\frac{6n+2}{2}=3n+1<a_1$。
- 若 $a_1=4n+2$，则 $a_2=2n+1<a_1$。
- 若 $a_1=4n+3$，则 $a_2=3(4n+3)+1=12n+10$，$a_3=\frac{12n+10}{2}=6n+5$，$a_4=3(6n+5)+1=18n+16$。
由于 $12n+10, 6n+5, 18n+16 > 4n+3$，每个形如 $a_1=4n+3$ 的正整数都满足 $a_1<a_2,a_3,a_4$。
因为满足 $a_1 \le 2008$ 的正整数中有四分之一可表示为 $4n+3$（$n$ 为非负整数），所以答案为 $\frac{1}{4}\cdot 2008 = 502 \Rightarrow D$。
Triangle $ABC$, with sides of length $5$, $6$, and $7$, has one vertex on the positive $x$-axis, one on the positive $y$-axis, and one on the positive $z$-axis. Let $O$ be the origin. What is the volume of tetrahedron $OABC$?
三角形 $ABC$ 的边长分别为 $5$、$6$ 和 $7$，其中一个顶点在正 $x$ 轴上，一个在正 $y$ 轴上，一个在正 $z$ 轴上。设 $O$ 为原点。四面体 $OABC$ 的体积是多少？
Correct Answer: C
Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7. Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively. Then by the Pythagorean Theorem,
\begin{align*} a^2+b^2 &=5^2 , \\ b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*}
so $a^2 = (5^2+7^2-6^2)/2 = 19$; similarly, $b^2 = 6$ and $c^2 = 30$. Since $OA$, $OB$, and $OC$ are mutually perpendicular, the tetrahedron's volume is \[abc/6\] because we can consider the tetrahedron to be a right triangular pyramid.
\[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\]
which is answer choice $\boxed{\text{C}}$.
不妨设 $A$ 在 $x$ 轴上，$B$ 在 $y$ 轴上，$C$ 在 $z$ 轴上，并令 $AB, BC, CA$ 的长度分别为 $5, 6, 7$。设线段 $OA,OB,OC$ 的长度分别为 $a,b,c$。由勾股定理，
\begin{align*} a^2+b^2 &=5^2 , \\ b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*}
因此 $a^2 = (5^2+7^2-6^2)/2 = 19$；同理 $b^2 = 6$，$c^2 = 30$。由于 $OA$、$OB$、$OC$ 两两互相垂直，该四面体的体积为 \[abc/6\]，因为可将其视为直三棱锥。
\[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\]
对应选项为 $\boxed{\text{C}}$。
Triangle $ABC$ has $AC=3$, $BC=4$, and $AB=5$. Point $D$ is on $\overline{AB}$, and $\overline{CD}$ bisects the right angle. The inscribed circles of $\triangle ADC$ and $\triangle BCD$ have radii $r_a$ and $r_b$, respectively. What is $r_a/r_b$?
三角形 $ABC$ 满足 $AC=3$，$BC=4$，$AB=5$。点 $D$ 在 $\overline{AB}$ 上，且 $\overline{CD}$ 平分直角。$\triangle ADC$ 与 $\triangle BCD$ 的内切圆半径分别为 $r_a$ 与 $r_b$。求 $r_a/r_b$。
Correct Answer: E
By the Angle Bisector Theorem,
\[\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7\]
By Law of Sines on $\triangle BCD$,
\[\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}\]
Since the area of a triangle satisfies $[\triangle]=rs$, where $r =$ the inradius and $s =$ the semiperimeter, we have
\[\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}\]
Since $\triangle ACD$ and $\triangle BCD$ share the altitude (to $\overline{AB}$), their areas are the ratio of their bases, or \[\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}\]
The semiperimeters are $s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}$ and $s_B = \frac{24+ 6\sqrt{2}}{7}$. Thus,
\begin{align*} \frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\ &= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}
由角平分定理，
\[\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7\]
在 $\triangle BCD$ 中用正弦定理，
\[\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}\]
由于三角形面积满足 $[\triangle]=rs$，其中 $r$ 为内切圆半径，$s$ 为半周长，有
\[\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}\]
因为 $\triangle ACD$ 与 $\triangle BCD$ 共享同一高（到底边 $\overline{AB}$），它们面积之比等于底边之比，即
\[\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}\]
半周长为 $s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}$，$s_B = \frac{24+ 6\sqrt{2}}{7}$。因此，
\begin{align*} \frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\ &= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}
A permutation $(a_1,a_2,a_3,a_4,a_5)$ of $(1,2,3,4,5)$ is heavy-tailed if $a_1 + a_2 < a_4 + a_5$. What is the number of heavy-tailed permutations?
$(1,2,3,4,5)$ 的一个排列 $(a_1,a_2,a_3,a_4,a_5)$ 若满足 $a_1 + a_2 < a_4 + a_5$，则称为重尾排列。重尾排列的个数是多少？
Correct Answer: D
There are $5!=120$ total permutations.
For every permutation $(a_1,a_2,a_3,a_4,a_5)$ such that $a_1 + a_2 < a_4 + a_5$, there is exactly one permutation such that $a_1 + a_2 > a_4 + a_5$. Thus it suffices to count the permutations such that $a_1 + a_2 = a_4 + a_5$.
$1+4=2+3$, $1+5=2+4$, and $2+5=3+4$ are the only combinations of numbers that can satisfy $a_1 + a_2 = a_4 + a_5$.
There are $3$ combinations of numbers, $2$ possibilities of which side of the equation is $a_1+a_2$ and which side is $a_4+a_5$, and $2^2=4$ possibilities for rearranging $a_1,a_2$ and $a_4,a_5$. Thus, there are $3\cdot2\cdot4=24$ permutations such that $a_1 + a_2 = a_4 + a_5$.
Thus, the number of heavy-tailed permutations is $\frac{120-24}{2}=48 \Rightarrow D$.
共有 $5!=120$ 个排列。
对于每一个满足 $a_1 + a_2 < a_4 + a_5$ 的排列 $(a_1,a_2,a_3,a_4,a_5)$，恰好存在一个排列满足 $a_1 + a_2 > a_4 + a_5$。因此只需统计满足 $a_1 + a_2 = a_4 + a_5$ 的排列个数。
只有 $1+4=2+3$、$1+5=2+4$、$2+5=3+4$ 这三种数组合能满足 $a_1 + a_2 = a_4 + a_5$。
共有 $3$ 种数组合；等式两边哪一边是 $a_1+a_2$、哪一边是 $a_4+a_5$ 有 $2$ 种；并且 $a_1,a_2$ 与 $a_4,a_5$ 各自内部交换共有 $2^2=4$ 种。因此满足 $a_1 + a_2 = a_4 + a_5$ 的排列共有 $3\cdot2\cdot4=24$ 个。
所以重尾排列的个数为 $\frac{120-24}{2}=48 \Rightarrow D$。
A sequence $(a_1,b_1)$, $(a_2,b_2)$, $(a_3,b_3)$, $\ldots$ of points in the coordinate plane satisfies
$(a_{n + 1}, b_{n + 1}) = (\sqrt {3}a_n - b_n, \sqrt {3}b_n + a_n)$ for $n = 1,2,3,\ldots$.
Suppose that $(a_{100},b_{100}) = (2,4)$. What is $a_1 + b_1$?
坐标平面上的点序列 $(a_1,b_1)$、$(a_2,b_2)$、$(a_3,b_3)$、$\ldots$ 满足
$(a_{n + 1}, b_{n + 1}) = (\sqrt {3}a_n - b_n, \sqrt {3}b_n + a_n)$，其中 $n = 1,2,3,\ldots$。
已知 $(a_{100},b_{100}) = (2,4)$。求 $a_1 + b_1$。
Correct Answer: D
This sequence can also be expressed using matrix multiplication as follows:
$\left[ \begin{array}{c} a_{n+1} \\ b_{n+1} \end{array} \right] = \left[ \begin{array}{cc} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{array} \right] \left[ \begin{array}{c} a_{n} \\ b_{n} \end{array} \right] = 2 \left[ \begin{array}{cc} \cos 30^\circ & -\sin 30^\circ \\ \sin 30^\circ & \ \cos 30^\circ \end{array} \right] \left[ \begin{array}{c} a_{n} \\ b_{n} \end{array} \right]$.
Thus, $(a_{n+1} , b_{n+1})$ is formed by rotating $(a_n , b_n)$ counter-clockwise about the origin by $30^\circ$ and dilating the point's position with respect to the origin by a factor of $2$.
So, starting with $(a_{100},b_{100})$ and performing the above operations $99$ times in reverse yields $(a_1,b_1)$.
Rotating $(2,4)$ clockwise by $99 \cdot 30^\circ \equiv 90^\circ$ yields $(4,-2)$. A dilation by a factor of $\frac{1}{2^{99}}$ yields the point $(a_1,b_1) = \left(\frac{4}{2^{99}}, -\frac{2}{2^{99}} \right) = \left(\frac{1}{2^{97}}, -\frac{1}{2^{98}} \right)$.
Therefore, $a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \frac{1}{2^{98}} \Rightarrow D$.
该序列也可用矩阵乘法表示如下：
$\left[ \begin{array}{c} a_{n+1} \\ b_{n+1} \end{array} \right] = \left[ \begin{array}{cc} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{array} \right] \left[ \begin{array}{c} a_{n} \\ b_{n} \end{array} \right] = 2 \left[ \begin{array}{cc} \cos 30^\circ & -\sin 30^\circ \\ \sin 30^\circ & \ \cos 30^\circ \end{array} \right] \left[ \begin{array}{c} a_{n} \\ b_{n} \end{array} \right]$。
因此，$(a_{n+1} , b_{n+1})$ 是将 $(a_n , b_n)$ 绕原点逆时针旋转 $30^\circ$，并以原点为中心将其位置按比例因子 $2$ 放缩得到的。
所以，从 $(a_{100},b_{100})$ 出发，反向进行上述操作 $99$ 次可得到 $(a_1,b_1)$。
将 $(2,4)$ 顺时针旋转 $99 \cdot 30^\circ \equiv 90^\circ$ 得到 $(4,-2)$。再按比例因子 $\frac{1}{2^{99}}$ 缩放，得到点 $(a_1,b_1) = \left(\frac{4}{2^{99}}, -\frac{2}{2^{99}} \right) = \left(\frac{1}{2^{97}}, -\frac{1}{2^{98}} \right)$。
因此，$a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \frac{1}{2^{98}} \Rightarrow D$。