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2008 AMC12A 真题及答案详细解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月6日上午10:20

2008 AMC 12 A 真题

真题及解析

Problem 1

A bakery owner turns on his doughnut machine at $text{8:30} {smalltext{AM}}$ . At $text{11:10} {smalltext{AM}}$ the machine has completed one third of the day's job. At what time will the doughnut machine complete the job?

$mathrm{(A)} text{1:50} {smalltext{PM}}qquadmathrm{(B)} text{3:00} {smalltext{PM}}qquadmathrm{(C)} text{3:30} {smalltext{PM}}qquadmathrm{(D)} text{4:30} {smalltext{PM}}qquadmathrm{(E)} text{5:50} {smalltext{PM}}$

Problem 2

What is the reciprocal of $frac{1}{2}+frac{2}{3}$ ?

$mathrm{(A)} frac{6}{7}qquadmathrm{(B)} frac{7}{6}qquadmathrm{(C)} frac{5}{3}qquadmathrm{(D)} 3qquadmathrm{(E)} frac{7}{2}$

Problem 3

Suppose that $tfrac{2}{3}$ of $10$ bananas are worth as much as $8$ oranges. How many oranges are worth as much as $tfrac{1}{2}$ of $5$ bananas?

$mathrm{(A)} 2qquadmathrm{(B)} frac{5}{2}qquadmathrm{(C)} 3qquadmathrm{(D)} frac{7}{2}qquadmathrm{(E)} 4$

Problem 4

Which of the following is equal to the product $[frac{8}{4}cdotfrac{12}{8}cdotfrac{16}{12}cdotcdotscdotfrac{4n+4}{4n}cdotcdotscdotfrac{2008}{2004}?]$

$mathrm{(A)} 251qquadmathrm{(B)} 502qquadmathrm{(C)} 1004qquadmathrm{(D)} 2008qquadmathrm{(E)} 4016$

Problem 5

Suppose that $[frac{2x}{3}-frac{x}{6}]$ is an integer. Which of the following statements must be true about $x$ ?

$mathrm{(A)} text{It is negative.}qquadmathrm{(B)} text{It is even, but not necessarily a multiple of 3.}\qquadmathrm{(C)} text{It is a multiple of 3, but not necessarily even.}\qquadmathrm{(D)} text{It is a multiple of 6, but not necessarily a multiple of 12.}\qquadmathrm{(E)} text{It is a multiple of 12.}$

Problem 6

Heather compares the price of a new computer at two different stores. Store $A$ offers $15%$ off the sticker price followed by a $textdollar 90$ rebate, and store $B$ offers $25%$ off the same sticker price with no rebate. Heather saves $textdollar 15$ by buying the computer at store $A$ instead of store $B$ . What is the sticker price of the computer, in dollars?

$mathrm{(A)} 750qquadmathrm{(B)} 900qquadmathrm{(C)} 1000qquadmathrm{(D)} 1050qquadmathrm{(E)} 1500$

Problem 7

While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in more than 30 gallons of water. Steve starts rowing toward the shore at a constant rate of 4 miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?

$mathrm{(A)} 2qquadmathrm{(B)} 4qquadmathrm{(C)} 6qquadmathrm{(D)} 8qquadmathrm{(E)} 10$

Problem 8

What is the volume of a cube whose surface area is twice that of a cube with volume 1?

$mathrm{(A)} sqrt{2}qquadmathrm{(B)} 2qquadmathrm{(C)} 2sqrt{2}qquadmathrm{(D)} 4qquadmathrm{(E)} 8$

Problem 9

Older television screens have an aspect ratio of $4: 3$ . That is, the ratio of the width to the height is $4: 3$ . The aspect ratio of many movies is not $4: 3$ , so they are sometimes shown on a television screen by "letterboxing" - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of $2: 1$ and is shown on an older television screen with a $27$ -inch diagonal. What is the height, in inches, of each darkened strip?
$[asy]unitsize(1mm); filldraw((0,0)--(21.6,0)--(21.6,2.7)--(0,2.7)--cycle,grey,black); filldraw((0,13.5)--(21.6,13.5)--(21.6,16.2)--(0,16.2)--cycle,grey,black); draw((0,0)--(21.6,0)--(21.6,16.2)--(0,16.2)--cycle);[/asy]$
$mathrm{(A)} 2qquadmathrm{(B)} 2.25qquadmathrm{(C)} 2.5qquadmathrm{(D)} 2.7qquadmathrm{(E)} 3$

Problem 10

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 $t$ ?

$mathrm{(A)} left(frac{1}{5}+frac{1}{7}right)left(t+1right)=1qquadmathrm{(B)} left(frac{1}{5}+frac{1}{7}right)t+1=1qquadmathrm{(C)} left(frac{1}{5}+frac{1}{7}right)t=1\mathrm{(D)} left(frac{1}{5}+frac{1}{7}right)left(t-1right)=1qquadmathrm{(E)} left(5+7right)t=1$

Problem 11

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?

$[asy] unitsize(.8cm); pen p = linewidth(1); draw(shift(-2,0)*unitsquare,p); label("1",(-1.5,0.5)); draw(shift(-1,0)*unitsquare,p); label("2",(-0.5,0.5)); draw(unitsquare,p); label("32",(0.5,0.5)); draw(shift(1,0)*unitsquare,p); label("16",(1.5,0.5)); draw(shift(0,1)*unitsquare,p); label("4",(0.5,1.5)); draw(shift(0,-1)*unitsquare,p); label("8",(0.5,-0.5)); [/asy]$

$mathrm{(A)} 154qquadmathrm{(B)} 159qquadmathrm{(C)} 164qquadmathrm{(D)} 167qquadmathrm{(E)} 189$

Problem 12

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 $g(x)=1-f(x+1)$ ?

$mathrm{(A)} [-1,1],[-1,0]qquadmathrm{(B)} [-1,1],[0,1]qquadtextbf{(C)} [0,2],[-1,0]qquadmathrm{(D)} [1,3],[-1,0]qquadmathrm{(E)} [1,3],[0,1]$

Problem 13

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?

$mathrm{(A)} frac{1}{16}qquadmathrm{(B)} frac{1}{9}qquadmathrm{(C)} frac{1}{8}qquadmathrm{(D)} frac{1}{6}qquadmathrm{(E)} frac{1}{4}$

Problem 14

What is the area of the region defined by the inequality $|3x-18|+|2y+7|le3$ ?

$mathrm{(A)} 3qquadmathrm{(B)} frac {7}{2}qquadmathrm{(C)} 4qquadmathrm{(D)} frac{9}{2}qquadmathrm{(E)} 5$

Problem 15

Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$ ?

$mathrm{(A)} 0qquadmathrm{(B)} 2qquadmathrm{(C)} 4qquadmathrm{(D)} 6qquadmathrm{(E)} 8$

Problem 16

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 $n$ ?

$mathrm{(A)} 40qquadmathrm{(B)} 56qquadmathrm{(C)} 76qquadmathrm{(D)} 112qquadmathrm{(E)} 143$

Problem 17

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 $a_4$ ?

$mathrm{(A)} 250qquadmathrm{(B)} 251qquadmathrm{(C)} 501qquadmathrm{(D)} 502qquadmathrm{(E)} 1004$

Problem 18

A triangle $triangle ABC$ with sides $5$ , $6$ , $7$ is placed in the three-dimensional plane with 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 $OABC$ ?

$mathrm{(A)} sqrt{85}qquadmathrm{(B)} sqrt{90}qquadmathrm{(C)} sqrt{95}qquadmathrm{(D)} 10qquadmathrm{(E)} sqrt{105}$

Problem 19

In the expansion of $[left(1 + x + x^2 + cdots + x^{27}right)left(1 + x + x^2 + cdots + x^{14}right)^2,]$ what is the coefficient of $x^{28}$ ?

$mathrm{(A)} 195qquadmathrm{(B)} 196qquadmathrm{(C)} 224qquadmathrm{(D)} 378qquadmathrm{(E)} 405$

Problem 20

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 $r_a/r_b$ ?

$mathrm{(A)} frac{1}{28}left(10-sqrt{2}right)qquadmathrm{(B)} frac{3}{56}left(10-sqrt{2}right)qquadmathrm{(C)} frac{1}{14}left(10-sqrt{2}right)qquadmathrm{(D)} frac{5}{56}left(10-sqrt{2}right)\mathrm{(E)} frac{3}{28}left(10-sqrt{2}right)$

Problem 21

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?

$mathrm{(A)} 36qquadmathrm{(B)} 40qquadtextbf{(C)} 44qquadmathrm{(D)} 48qquadmathrm{(E)} 52$

Problem 22

A round table has radius $4$ . Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$ . Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$ ?

$[asy]unitsize(4mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("(x)",(-1.55,2.1),E); label("(1)",(-0.5,3.8),S);[/asy]$

$mathrm{(A)} 2sqrt{5}-sqrt{3}qquadmathrm{(B)} 3qquadmathrm{(C)} frac{3sqrt{7}-sqrt{3}}{2}qquadmathrm{(D)} 2sqrt{3}qquadmathrm{(E)} frac{5+2sqrt{3}}{2}$

Problem 23

The solutions of the equation $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon?

$mathrm{(A)} 2^{frac{5}{8}}qquadmathrm{(B)} 2^{frac{3}{4}}qquadmathrm{(C)} 2qquadmathrm{(D)} 2^{frac{5}{4}}qquadmathrm{(E)} 2^{frac{3}{2}}$

Problem 24

Triangle $ABC$ has $angle C = 60^{circ}$ and $BC = 4$ . Point $D$ is the midpoint of $BC$ . What is the largest possible value of $tan{angle BAD}$ ?

$mathrm{(A)} frac{sqrt{3}}{6}qquadmathrm{(B)} frac{sqrt{3}}{3}qquadmathrm{(C)} frac{sqrt{3}}{2sqrt{2}}qquadmathrm{(D)} frac{sqrt{3}}{4sqrt{2}-3}qquadmathrm{(E)} 1$

Problem 25

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 $a_1 + b_1$ ?

$mathrm{(A)} -frac{1}{2^{97}}qquadmathrm{(B)} -frac{1}{2^{99}}qquadmathrm{(C)} 0qquadmathrm{(D)} frac{1}{2^{98}}qquadmathrm{(E)} frac{1}{2^{96}}$

2008 AMC12 A 真题答案详细解析

The machine completes one-third of the job in $text{11:10}-text{8:30}=text{2:40}$ hours. Thus, the entire job is completed in $3cdot(text{2:40})=text{8:00}$ hours.
Since the machine was started at $text{8:30 AM}$ , the job will be finished $8$ hours later, at $text{4:30 PM}$ . The answer is $mathrm{(D)}$ .
Solution 1

Here's a cheapshot: Obviously, $frac{1}{2}+frac{2}{3}$ is greater than $1$ . Therefore, its reciprocal is less than $1$ , and the answer must be $boxed{frac{6}{7}}$ .

Solution 2

$left(frac{1}{2}+frac{2}{3}right)^{-1}=left(frac{3}{6}+frac{4}{6}right)^{-1}=left(frac{7}{6}right)^{-1}=frac{6}{7}Rightarrow A$ .
If $frac{2}{3}cdot10 text{bananas}=8 text{oranges}$ , then $frac{1}{2}cdot5 text{bananas}=left(frac{1}{2}cdot 5 text{bananas}right)cdotleft(frac{8 text{oranges}}{frac{2}{3}cdot10 text{bananas}}right)=3 text{oranges}Longrightarrowmathrm{(C)}$ .
Solution 1

$frac {8}{4}cdotfrac {12}{8}cdotfrac {16}{12}cdotsfrac {4n + 4}{4n}cdotsfrac {2008}{2004} = frac {1}{4}cdotleft(frac {8}{8}cdotfrac {12}{12}cdotsfrac {4n}{4n}cdotsfrac {2004}{2004}right)cdot 2008 = frac{2008}{4} =$ $502 Rightarrow B$ .

Solution 2

Notice that everything cancels out except for $2008$ in the numerator and $4$ in the denominator.

Thus, the product is $frac{2008}{4}=502$ , and the answer is $textbf{(B)}$ .
$[frac{2x}{3}-frac{x}{6}quadLongrightarrowquadfrac{4x}{6}-frac{x}{6}quadLongrightarrowquadfrac{3x}{6}quadLongrightarrowquadfrac{x}{2}]$
For $frac{x}{2}$ to be an integer, $x$ must be even, but not necessarily divisible by $3$ . Thus, the answer is $mathrm{(B)}$ .
Solution 1

Let the sticker price be $x$ .

The price of the computer is $0.85x-90$ at store $A$ , and $0.75x$ at store $B$ .

Heather saves $$$$ $15$ at store $A$ , so $0.85x-90+15=0.75x$ .

Solving, we find $x=750$ , and the thus answer is $mathrm{(A)}$ .

Solution 2

The $textdollar 90$ in store $A$ is $textdollar 15$ better than the additional $10%$ off at store $B$ .

Thus the $10%$ off is equal to $textdollar 90$ - $textdollar 15$ $=$ $textdollar 75$ , and therefore the sticker price is $textdollar 750$ .
It will take $frac{1}{4}$ of an hour or $15$ minutes to get to shore.Since only $30$ gallons of water can enter the boat, only $frac{30}{15}=2$ net gallons can enter the boat per minute.Since $10$ gallons of water enter the boat each minute, LeRoy must bail $10-2=8$ gallons per minute $Rightarrowmathrm{(D)}$ .
A cube with volume $1$ has a side of length $sqrt[3]{1}=1$ and thus a surface area of $6 cdot 1^2=6$ .A cube whose surface area is $6cdot2=12$ has a side of length $sqrt{frac{12}{6}}=sqrt{2}$ and a volume of $(sqrt{2})^3=2sqrt{2}Rightarrowmathrm{(C)}$ .
Let the width and height of the screen be $4x$ and $3x$ respectively, and let the width and height of the movie be $2y$ and $y$ respectively.By the Pythagorean Theorem, the diagonal is $sqrt{(3x)^2+(4x)^2}=5x = 27$ . So $x=frac{27}{5}$ .Since the movie and the screen have the same width, $2y=4xRightarrow y=2x$ .Thus, the height of each strip is $frac{3x-y}{2}=frac{3x-2x}{2}=frac{x}{2}=frac{27}{10}=2.7Longrightarrowmathrm{(D)}$ .
Solution 1

Doug can paint $frac{1}{5}$ of a room per hour, Dave can paint $frac{1}{7}$ of a room in an hour, and the time they spend working together is $t-1$ .

Since rate times time gives output, $left(frac{1}{5}+frac{1}{7}right)left(t-1right)=1 Rightarrow mathrm{(D)}$

Solution 2

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 $frac{5*7}{5+7} = frac{35}{12}$ 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 $t = frac{35}{12} + 1$ .
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 $2cdot12+1=25$ . 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 $189-25=164 Rightarrow C$ .
$g(x)$ 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 $[- 1,1],[0,1] longrightarrow boxed{B}$ .
$[asy]size(200); defaultpen(fontsize(10)); pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5); picture p = new picture; draw(p,Circle(O,0.2)); clip(p,O--C--A--cycle); add(p); draw(Circle(O,3)); dot(A); dot(B); dot(C); dot(O); draw(A--O--B); draw(O--C--D); draw(C--F); draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2)); draw(Circle(C,1)); label("(30^{circ})",(0.65,0.15),O); label("(r)",(C+D)/2,E); label("(2r)",(O+C)/2,NNE); label("(O)",O,SW); label("(r)",(C+F)/2,SE); label("(R)",(O+A)/2-(0,0.3),S); label("(P)",C,NW); label("(Q)",D,SE);[/asy]$ 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 $2008 AMC12A 真题及答案详细解析$ . 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, and a $30-60-90$ triangle at that. Therefore, $OP=2PQ$ .Since $OP=OC-PC=OC-r=R-r$ , we have $R-r=2PQ$ , or $R-r=2r$ , or $frac{1}{3}=frac{r}{R}$ .Then the ratio of areas will be $frac{1}{3}$ squared, or $frac{1}{9}Rightarrow boxed{text{B}}$ .
Area is invariant under translation, so after translating left $6$ and up $7/2$ units, we have the inequality $[|3x| + |2y|leq 3]$ which forms a diamond centered at the origin and vertices at $(pm 1, 0), (0, pm 1.5)$ . Thus the diagonals are of length $2$ and $3$ . Using the formula $A = frac 12 d_1 d_2$ , the answer is $frac{1}{2}(2)(3) = 3 Rightarrow mathrm{(A)}$ .
$k equiv 2008^2 + 2^{2008} equiv 8^2 + 2^4 equiv 4+6 equiv 0 pmod{10}$ .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 $6 Rightarrow boxed{D}$ .
Solution 1

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 $(9cdot12 + 4)B = 112B = nB Rightarrow n = 112Rightarrow boxed{D}$ .

Solution 2

If $log(a^3b^7)$ , $log(a^5b^{12})$ , and $log(a^8b^{15})$ are in arithmetic progression, then $a^3b^7$ , $a^5b^{12}$ , and $a^8b^{15}$ are in geometric progression. Therefore,

$[a^2b^5=a^3b^3 Rightarrow a=b^2]$

Therefore, $a^3b^7=b^{13}$ , $a^5b^{12}=b^{22}$ , therefore the 12th term in the sequence is $b^{13+9*11}=b^{112} Rightarrow boxed{D}$

Solution 3

$text{If a, b, and c are in a arithmetic progression then } b = frac{a+c}{2} text{ which means}$ $log(a^5b^{12}) = frac{log(a^3b^7) + log(a^8b^{15})}{2}= frac{log(a^{11}b^{22})}{2} text{ therefore}$ $2log(a^5b^{12}) = log(a^{10}b^{24}) = log(a^{11}b^{22}) Rightarrow a=b^2$ $text{This means that the Kth term of the series would be } log(b^{13+9(k-1)})$ $text{The 12th term would be } log(b^{112}) Rightarrow n=112 Rightarrow D$
All positive integers can be expressed as , , , or , where 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 $frac{1}{4}cdot 2008 = 502 Rightarrow D$ .

Alternate Solution

After checking the first few $a_n$ such as $1$ , $2$ through $7$ , we can see that the only $a_1$ that satisfy the conditions are odd numbers that when tripled and added 1 to, are double an odd number. For example, for $a_n=3$ , we notice the sequence yields $10$ , $5$ , and $16$ , a valid sequence.

So we can set up an equation, $3x + 1 = 2(2k - 1)$ where x is equal to $a_1$ . Rearranging the equation yields $(3x + 3)/4 = k$ . Experimenting yields that every 4th $x$ after 3 creates an integer, and thus satisfies the sequence condition. So the number of valid solutions is equal to $2008/4 = 502 Rightarrow D$ .
$[asy] defaultpen(fontsize(8)); draw((0,10)--(0,0)--(8,0));draw((-3,-4)--(0,0));draw((0,10)--(-3,-4)--(8,0)--cycle); label("A",(8,0),(1,0));label("B",(0,10),(0,1));label("C",(-3,-4),(-1,-1));label("O",(0,0),(1,1)); label("$a$",(4,0),(0,1));label("$b$",(0,5),(1,0));label("$c$",(-1.5,-2),(1,0)); label("$5$",(4,5),(1,1));label("$6$",(-1.5,3),(-1,0));label("$7$",(2.5,-2),(1,-1)); [/asy]$ 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 $boxed{text{C}}$ .
Solution 1

Let $A = left(1 + x + x^2 + cdots + x^{14}right)$ and $B = left(1 + x + x^2 + cdots + x^{27}right)$ . We are expanding $A cdot A cdot B$ .

Since there are $15$ terms in $A$ , there are $15^2 = 225$ ways to choose one term from each $A$ . The product of the selected terms is $x^n$ for some integer $n$ between $0$ and $28$ inclusive. For each $n neq 0$ , there is one and only one $x^{28 - n}$ in $B$ . For example, if I choose $x^2$ from $A$ , then there is exactly one power of $x$ in $B$ that I can choose; in this case, it would be $x^{24}$ . Since there is only one way to choose one term from each $A$ to get a product of $x^0$ , there are $225 - 1 = 224$ ways to choose one term from each $A$ and one term from $B$ to get a product of $x^{28}$ . Thus the coefficient of the $x^{28}$ term is $224 Rightarrow boxed{C}$ .

Solution 2

Let $P(x) = left(1 + x + x^2 + cdots + x^{14}right)^2 = a_0 + a_1x + a_2x^2 + cdots + a_{28}x^{28}$ . Then the $x^{28}$ term from the product in question $left(1 + x + x^2 + cdots + x^{27}right)(a_0 + a_1x + a_2x^2 + cdots + a_{28}x^{28})$ is

$1a_{28}x^{28} + xa_{27}x^{27} + x^2a_{26}x^{26} + cdots + x^{27}a_1x = left(a_1 + a_2 + cdots a_{28}right)x^{28}$

So we are trying to find the sum of the coefficients of $P(x)$ minus $a_0$ . Since the constant term $a_0$ in $P(x)$ (when expanded) is $1$ , and the sum of the coefficients of $P(x)$ is $P(1)$ , we find the answer to be $P(1) - a_0 = left(1 + 1 + 1^2 + cdots 1^{14}right)^2 - 1 = 15^2 - 1 = 224 Rightarrow boxed{C}$ .

Solution 3

We expand $(1 + x + x^2 + x^3 + cdots + x^{14})^2$ to $(1 + x + x^2 + x^3 + cdots + x^{14}) * (1 + x + x^2 + x^3 + cdots + x^{14})$ and use FOIL to multiply. It expands out to:

$1 + x + x^2 + x^3 + x^4 + cdots + x^{14} +$

$qquad x + x^2 + x^3 + x^4 + cdots + x^{14} + x^{15} +$

$qquad qquad x^2 + x^3 + x^4 + cdots + x^{14} + x^{15} + x^{16} + cdots$

It becomes apparent that

$(1 + x + x^2 + x^3 + cdots + x^{14})^2 = 1 + 2x + 3x^2 + 4x^3 + cdots + 15x^{14} + 14x^{15} + 13x^{16} + cdots + x^{28}$ .

Now we have to find the coefficient of $x^{28}$ in the product:

$(1 + 2x + 3x^2 + 4x^3 + cdots + 15x^{14} + 14x^{15} + 13x^{16} + cdots + x^{28}) cdot (1 + x + x^2 + x^3 + cdots + x^{27})$ .

We quickly see that the we get $x^{28}$ terms from $x^{27} cdot 2x$ , $x^{26} cdot 3x^2$ , $x^{25} cdot 4x^3$ , ... $15x^{14} cdot x^{14}$ , ... $x^{28} cdot 1$ . The coefficient of $x^{28}$ is just the sum of the coefficients of all these terms. $1 + 2 + 3 + 4 + cdots + 15 + 14 + 13 + cdots + 4 + 3 + 2 = 224$ , so the answer is $boxed{C}$ .

Solution 4

Rewrite the product as $frac{(x^{28} - 1)(x^{15} - 1)(x^{15} - 1)}{(x - 1)^3}$ . It is known that

$[frac{1}{(1 - x)^n} = binom{n - 1}{n - 1} + binom{n}{n - 1}x + binom{n + 1}{n - 1}x^2 + binom{n + 2}{n - 1}x^3 + cdots + binom{n - 1 + k}{n - 1}x^k + cdots .]$

Thus, our product becomes

$[-left( x^{28} - 1 right) left( x^{15} - 1 right) left( x^{15} - 1 right) left( binom{2}{2} + binom{3}{2}x + binom{4}{2}x^2 + cdots right).]$

$[= -left( x^{28} - 1 right) left( x^{15} - 1 right) left( x^{15} - 1 right) left( 1 + 3x + 6x^2 + cdots right).]$

We determine the $x^{28}$ coefficient by doing casework on the first three terms in our product. We can obtain an $x^{28}$ term by choosing $x^{28}$ in the first term, $-1$ in the second and third terms, and $1$ in the fourth term. We can get two $x^{28}$ terms by choosing $x^{15}$ in either the second or third term, $-1$ in the first term, $-1$ in the second or third term from which $x^{15}$ has not been chosen, and the $binom{15}{2}x^{13}$ in the fourth term. We get $binom{15}{2} * 2 = 210$ $x^{28}$ terms this way. (We multiply by $2$ because the $x^{15}$ term could have been chosen from the second term or the third term). Lastly, we can get an $x^{28}$ term by choosing $-1$ in the first three terms and a $binom{30}{2}x^{28}$ from the fourth term. We have a total of $1 + 210 - 435 = -224$ for the $x^{28}$ coefficient, but we recall that we have a negative sign in front of our product, so we obtain an answer of $224 Rightarrow boxed{(C)}$ .
Solution 1

$[asy] import olympiad; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); picture p = new picture; draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2)); clip(p,B--C--D--cycle); add(p); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); dot(O);dot(P); label("(A)",A,W); label("(B)",B,E); label("(C)",C,W); label("(D)",D,NE); label("(O_A)",O,W); label("(O_B)",P,W); label("(3)",(A+C)/2,W); label("(4)",(B+C)/2,S); label("(frac{15}{7})",(A+D)/2,NE); label("(frac{20}{7})",(B+D)/2,NE); label("(45^{circ})",(.2,.1),E); label("(sin theta = frac{3}{5})",B-(.2,-.1),W); [/asy]$ By the Angle Bisector Theorem, $[frac{BD}{4} = frac{5-BD}{3} Longrightarrow BD = frac{20}7]$ By Law of Sines on $triangle BCD$ , $[frac{BD}{sin 45^{circ}} = frac{CD}{sin angle B} Longrightarrow frac{20/7}{sqrt{2}/2} = frac{CD}{3/5} Longrightarrow CD=frac{12sqrt{2}}{7}]$ 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 $s_A = left(3 + frac{15}{7} + frac{12sqrt{2}}{7}right)left/right.2 = frac{18+6sqrt{2}}{7}$ and $s_B = frac{24+ 6sqrt{2}}{7}$ . Thus, $begin{align*} frac{r_A}{r_B} &= frac{[ACD] cdot s_B}{[BCD] cdot s_A} = frac{3}{4} cdot frac{(24+ 6sqrt{2})/7}{(18+6sqrt{2})/7} \ &= frac{3(4+sqrt{2})}{4(3+sqrt{2})} cdot left(frac{3-sqrt{2}}{3-sqrt{2}}right) = frac{3}{28}(10-sqrt{2}) Rightarrow mathrm{(E)}qquad blacksquare end{align*}$

Solution 2

$[asy] import olympiad; import geometry; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); add(p); add(q); add(r); add(s); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); label("(A)",A,W); label("(B)",B,E); label("(C)",C,W); label("(D)",D,NE); label("(O_a)",O,W); label("(O_b)",P,E); label("(3)",(A+C)/2,W); label("(4)",(B+C)/2,S); label("(frac{15}{7})",(A+D)/2,NE); label("(frac{20}{7})",(B+D)/2,NE); label("(M)", inter1, 2W); label("(N)", inter2, 2E); [/asy]$ We start by finding the length of $AD$ and $BD$ as in solution 1. Using the angle bisector theorem, we see that $AD = frac{15}{7}$ and $BD = frac{20}{7}$ . Using Stewart's Theorem gives us the equation $5d^2 + frac{1500}{49} = frac{240}{7} + frac{180}{7}$ , where $d$ is the length of $CD$ . Solving gives us $d = frac{12sqrt{2}}{7}$ , so $CD = frac{12sqrt{2}}{7}$ .

Call the incenters of triangles $ACD$ and $BCD$ $O_a$ and $O_b$ respectively. Since $O_a$ is an incenter, it lies on the angle bisector of $angle ACD$ . Similarly, $O_b$ lies on the angle bisector of $angle BCD$ . Call the point on $CD$ tangent to $O_a$ $M$ , and the point tangent to $O_b$ $N$ . Since $triangle CO_aM$ and $triangle CO_bN$ are right, and $angle O_aCM = angle O_bCN$ , $triangle CO_aM sim triangle CO_bN$ . Then, $frac{r_a}{r_b} = frac{CM}{CN}$ .

We now use common tangents to find the length of $CM$ and $CN$ . Let $CM = m$ , and the length of the other tangents be $n$ and $p$ . Since common tangents are equal, we can write that $m + n = frac{12sqrt{2}}{7}$ , $n + p = frac{15}{7}$ and $m + p = 3$ . Solving gives us that $CM = m = frac{6sqrt{2} + 3}{7}$ . Similarly, $CN = frac{6sqrt{2} + 4}{7}$ .

We see now that $frac{r_a}{r_b} = frac{frac{6sqrt{2} + 3}{7}}{frac{6sqrt{2} + 4}{7}} = frac{6sqrt{2} + 3}{6sqrt{2} + 4} = frac{60-6sqrt{2}}{56} = frac{3}{28}(10 - sqrt{2}) Rightarrow boxed{E}$
Solution 1

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 $3cdot2cdot4=24$ permutations such that $a_1 + a_2 = a_4 + a_5$ .

Thus, the number of heavy-tailed permutations is $frac{120-24}{2}=48 Rightarrow D$ .

Solution 2 (Casework)

We use case work on the value of $a_3$ .

Case 1: $a_3 = 1$ . Since $a_1 + a_2 < a_4 + a_5$ , $(a_1, a_2)$ can only be a permutation of $(2, 3)$ or $(2, 4)$ . The values of $a_1$ and $a_2$ , as well as the values of $a_4$ and $a_5$ , are interchangeable, so this case produces a total of $2(2 cdot 2) = 8$ solutions.

Case 2: $a_3 = 2$ . Similarly, we have $(a_1, a_2)$ is a permutation of $(1, 3)$ , $(1, 4)$ , or $(1, 5)$ , which gives a total of $3(2 cdot 2) = 12$ solutions.

Case 3: $a_3 = 3$ . $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 4)$ , which gives a total of $2(2 cdot 2) = 8$ solutions.

Case 4: $a_3 = 4$ . $(a_1, a_2)$ is a permutation of $(1, 2)$ , $(1, 3)$ , or $(2, 3)$ , which gives a total of $3(2 cdot 2) = 12$ solutions.

Case 5: $a_3 = 5$ . $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 3)$ , which gives a total of $2(2 cdot 2) = 8$ solutions.

Therefore, our answer is $8 + 12 + 8 + 12 + 8 = 48 Rightarrow boxed{D}$ .
Solution 1 (trigonometry)

Let one of the mats be $ABCD$ , and the center be $O$ as shown:

$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("(x)",(-1.55,2.1),E); label("(x)",(0.03,1.5),E); label("(A)",(-3.6,2.5513),E); label("(B)",(-3.15,1.35),E); label("(C)",(0.05,3.20),E); label("(D)",(-0.75,4.15),E); label("(O)",(0.00,-0.10),E); label("(1)",(-0.1,3.8),S); label("(4)",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686));[/asy]$

Since there are $6$ mats, $Delta BOC$ is equilateral. So, $BC=CO=x$ . Also, $angle OCD = angle OCB + angle BCD = 60^circ+90^circ=150^circ$ .

By the Law of Cosines: $4^2=1^2+x^2-2cdot1cdot x cdot cos(150^circ) Rightarrow x^2 + xsqrt{3} - 15 = 0 Rightarrow x = frac{-sqrt{3}pm 3sqrt{7}}{2}$ .

Since $x$ must be positive, $x = frac{3sqrt{7}-sqrt{3}}{2} Rightarrow C$ .

Solution 2 (without trigonometry)

Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$

$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); pair D = rotate(300)*(-3.687,1.5513); pair C = rotate(300)*(-2.687,1.5513); pair EE = foot((0.00,0.00),D,C); draw(D--EE--(0,0)); label("(x)",(-1.55,2.1),E); label("(x)",(0.03,1.5),E); label("(A)",(-3.6,2.5513),E); label("(B)",(-3.15,1.35),E); label("(C)",(0.05,3.20),E); label("(D)",(-0.75,4.15),E); label("(O)",(0.00,-0.10),E); label("(1)",(-0.1,3.8),S); label("(4)",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); label("(E)", EE,SE); [/asy]$

As proved in the first solution, $angle OCD = 150^circ$ . That makes $Delta OCE$ a $30-60-90$ triangle, so $OE = frac{x}{2}$ and $CE= frac{xsqrt 3}{2}$

Since $Delta OEC$ is a right triangle, $left({frac{x}{2}}right)^2 + left({frac{xsqrt 3}{2} +1}right)^2 = 4^2 Rightarrow x^2+xsqrt3-15 = 0$

Solving for $x$ gives $x =frac{3sqrt{7}-sqrt{3}}{2}Rightarrow C$

Solution 3

$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("(x)",(-1.95,3),E); label("(A)",(-3.6,2.5513),E); label("(C)",(0.05,3.20),E); label("(E)",(0.40,-3.60),E); label("(B)",(-0.75,4.15),E); label("(D)",(-2.62,1.5),E); label("(F)",(-2.64,-1.43),E); label("(G)",(-0.2,-2.8),E); label("( sqrt{3}x)",(-1.5,-0.5),E); label("(M)",(-2,-0.9),E); label("(O)",(0.00,-0.10),E); label("(1)",(-2.7,2.3),S); label("(1)",(0.1,-3.4),S); label("(8)",(-0.3,0),S); draw((0,-3.103)--(-2.687,1.5513)); draw((0.5,-3.9686)--(-0.5,3.9686));[/asy]$

Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^circ$ . $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$ , notice that if we draw a line from $F$ to $M$ , the midpoint of line $DG$ , it creates two $30$ - $60$ - $90$ triangles. Therefore, $MD = MG = frac{sqrt{3}x}{2} Rightarrow DG = sqrt{3}x$ . $AE = 2 + sqrt{3}x$

Use the Pythagorean theorem on triangle $ABE$ , we get $[(2+sqrt{3}x)^2 + x^2 = 8^2 Rightarrow 4 + 3x^2 + 4sqrt{3}x + x^2 = 64 Rightarrow x^2 + sqrt{3}x - 15 = 0]$ Using the pythagorean theorem to solve, we get $[x = frac{-sqrt{3} pm sqrt{3 -4(1)(-15)}}{2} = frac{pm 3sqrt{7} - sqrt{3}}{2}]$ $x$ must be positive, therefore $[x = frac{3sqrt{7} - sqrt{3}}{2} Rightarrow C]$
Looking at the coefficients, we are immediately reminded of the binomial expansion of ${left(x+1right)}^{4}$ .Modifying this slightly, we can write the given equation as: $[{left(z+iright)}^{4}=1+i=2^{frac{1}{2}}cdot i cdot cos frac {pi}{4} + 2^{frac{1}{2}}cdot sin frac {pi}{4}]$ We can apply a translation of $-i$ and a rotation of $-frac{pi}{4}$ (both operations preserve area) to simplify the problem: $[z^{4}=2^{frac{1}{2}}]$ Because the roots of this equation are created by rotating $frac{pi}{2}$ radians successively about the origin, the quadrilateral is a square.We know that half the diagonal length of the square is ${left(2^{frac{1}{2}}right)}^{frac{1}{4}}=2^{frac{1}{8}}$ Therefore, the area of the square is $frac{{left( 2 cdot 2^{frac{1}{8}}right)}^2}{2}=frac{2^{frac{9}{4}}}{2}=2^{frac{5}{4}} Rightarrow D.$
Solution 1

$[asy]unitsize(12mm); pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60)); pair E=(1,0), F=(2,0); draw(C--B--A--C); draw(A--D);draw(D--E);draw(B--F); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("(C)",C,SW); label("(B)",B,N); label("(A)",A,SE); label("(D)",D,NW); label("(E)",E,S); label("(F)",F,S); label("(60^circ)",C+(.1,.1),ENE); label("(2)",1*dir(60),NW); label("(2)",3*dir(60),NW); label("(theta)",(7,.4)); label("(1)",(.5,0),S); label("(1)",(1.5,0),S); label("(x-2)",(5,0),S);[/asy]$

Let $x = CA$ . Then $tantheta = tan(angle BAF - angle DAE)$ , and since $tanangle BAF = frac{2sqrt{3}}{x-2}$ and $tanangle DAE = frac{sqrt{3}}{x-1}$ , we have

$[tantheta = frac{frac{2sqrt{3}}{x-2} - frac{sqrt{3}}{x-1}}{1 + frac{2sqrt{3}}{x-2}cdotfrac{sqrt{3}}{x-1}}= frac{xsqrt{3}}{x^2-3x+8}]$

With calculus, taking the derivative and setting equal to zero will give the maximum value of $tan theta$ . Otherwise, we can apply AM-GM:

$begin{align*} frac{x^2 - 3x + 8}{x} = left(x + frac{8}{x}right) -3 &geq 2sqrt{x cdot frac 8x} - 3 = 4sqrt{2} - 3\ frac{x}{x^2 - 3x + 8} &leq frac{1}{4sqrt{2}-3}\ frac{xsqrt{3}}{x^2 - 3x + 8} = tan theta &leq frac{sqrt{3}}{4sqrt{2}-3}end{align*}$

Thus, the maximum is at $frac{sqrt{3}}{4sqrt{2}-3} Rightarrow mathbf{(D)}$ .

Solution 2

We notice that $tan(x)$ is strictly increasing on the interval $[0, frac{pi}{2})$ (if $angle BADge 90^circ$ , then it is impossible for $angle C=60^circ$ ), so we want to maximize $angle BAD$ .

Consider the circumcircle of $BAD$ and let it meet $AC$ again at $F$ . Any point $P$ between $A$ and $F$ on line $AC$ is inside this circle, so it follows that $angle BPD>angle BAD$ . Therefore to maximize $angle BAD$ , the circumcircle of $BAD$ must be tangent to $AC$ at $A$ . By PoP we find that $CA^2=CDcdot CB Rightarrow AC = 2sqrt{2}$ .

Now our computations are straightforward: $[tanangle BAD = frac{sin angle BAD}{cos angle BAD} = frac{frac{2sinangle ABD}{AD}}{frac{AB^2+AD^2-BD^2}{2ABcdot AD}}]$ $[=frac{4sin angle ABDcdot AB}{AB^2+AD^2-4} = frac{4 AC sin angle ACB}{AB^2 + AD^2 - 4}]$ $[=frac{4sqrt{6}}{(4^2+(2sqrt{2})^2-4cdot 2sqrt{2}) + (2^2+(2sqrt{2})^2 - 2cdot 2sqrt{2}) - 4} = frac{4sqrt{6}}{32-12sqrt{2}}]$ $[=boxed{frac{sqrt{3}}{4sqrt{2}-3}}]$
Solution 1

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, $a_1 + b_1 = frac{1}{2^{97}} - frac{1}{2^{98}} = frac{1}{2^{98}} Rightarrow D$ .

[s]Shortcut: no answer has $3$ in the numerator. So the point cannot have orientation $(2,4)$ or $(-2,-4)$ . Also there are no negative answers. Any other non-multiple of $90^circ$ rotation of $30n^circ$ would result in the need of radicals. So either it has orientation $(4,-2)$ or $(-4,2)$ . Both answers add up to $2$ . Thus, $2/2^{99}=boxed{textbf{(D) }frac{1}{2^{98}}}$ .[/s] Does not work as there are negative answers.

Solution 2 (algebra)

Let $(x,y)=(a_1,b_1)$ . Then, we can begin to list out terms as follows:

$(a_2,b_2)=(xsqrt{3}-y,ysqrt{3}+x)$

$(a_3,b_3)=(2x-2ysqrt{3},2y+2xsqrt{3})$

$(a_4,b_4)=(-8y,8x)$

We notice that the sequence follows the rule $(a_{n+3},b_{n+3})=(-2^3b_n,2^3a_n)$

We can now start listing out every third point, getting:

$(a_1,b_1)=(x,y)$

$(a_4,b_4)=(-2^3y,2^3x)$

$(a_7,b_7)=(-2^6x,-2^6y)$

$(a_{10},b_{10})=(2^9y,-2^9x)$

$(a_{13},b_{13})=(2^{12}x,2^{12}y)$

We can make two observations from this:

(1) In $a_n$ , the coefficient of $x$ and $y$ is $2^{n-1}$

(2) The positioning of $x$ and $y$ , and their signs, cycle with every $12$ terms.

We know then that from (1), the coefficients of $x$ and $y$ in $(a_{100},b_{100})$ are both $2^{99}$

We can apply (2), finding $100 text{(mod }12)=4$ , so the positions and signs of $x$ and $y$ are the same in $(a_{100},b_{100})$ as they are in $(a_{4},b_{4})$ .

From this, we can get $(a_{100},b_{100})=(-2^{99}y,2^{99}x)$ . We know that $(a_{100},b_{100})=(2,4)$ , so we get the following:

$-2^{99}y=2 Rightarrow y=-frac{1}{2^{98}}$

$2^{99}x=4 Rightarrow x=frac{1}{2^{97}}$

The answer is $x+y=frac{1}{2^{97}}-frac{1}{2^{98}}=boxed{textbf{(D) }frac{1}{2^{98}}}$