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2004AMC10B真题与答案解析

Category: AMC美国数学竞赛, 国际竞赛 Date: 2019年12月5日上午11:22

2004AMC10B真题

答案解析请参考文末

Problem 1

Each row of the Misty Moon Amphitheater has $33$ seats. Rows $12$ through $22$ are reserved for a youth club. How many seats are reserved for this club?

$mathrm{(A) } 297 qquad mathrm{(B) } 330qquad mathrm{(C) } 363qquad mathrm{(D) } 396qquad mathrm{(E) } 726$

Problem 2

How many two-digit positive integers have at least one $7$ as a digit?

$mathrm{(A) } 10 qquad mathrm{(B) } 18qquad mathrm{(C) } 19 qquad mathrm{(D) } 20qquad mathrm{(E) } 30$

Problem 3

At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made $48$ free throws. How many free throws did she make at the first practice?

$mathrm{(A) } 3 qquad mathrm{(B) } 6 qquad mathrm{(C) } 9 qquad mathrm{(D) } 12qquad mathrm{(E) } 15$

Problem 4

A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$ ?

$mathrm{(A) } 6 qquad mathrm{(B) } 12 qquad mathrm{(C) } 24 qquad mathrm{(D) } 144qquad mathrm{(E) } 720$

Problem 5

In the expression $ccdot a^b-d$ , the values of $a$ , $b$ , $c$ , and $d$ are $0$ , $1$ , $2$ , and $3$ , although not necessarily in that order. What is the maximum possible value of the result?

$mathrm{(A) } 5 qquad mathrm{(B) } 6qquad mathrm{(C) } 8 qquad mathrm{(D) } 9qquad mathrm{(E) } 10$

Problem 6

Which of the following numbers is a perfect square?

$mathrm{(A) } 98! cdot 99! qquad mathrm{(B) } 98! cdot 100! qquad mathrm{(C) } 99! cdot 100! qquad mathrm{(D) } 99! cdot 101! qquad mathrm{(E) } 100! cdot 101!$

Problem 7

On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$ ?

$mathrm{(A) } 5 qquad mathrm{(B) } 6qquad mathrm{(C) } 7qquad mathrm{(D) } 8qquad mathrm{(E) } 9$

Problem 8

Minneapolis-St. Paul International Airport is $8$ miles southwest of downtown St. Paul and $10$ miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?

$mathrm{(A) } 13 qquad mathrm{(B) } 14qquad mathrm{(C) } 15qquad mathrm{(D) } 16qquad mathrm{(E) } 17$

Problem 9

A square has sides of length $10$ , and a circle centered at one of its vertices has radius $10$ . What is the area of the union of the regions enclosed by the square and the circle?

$mathrm{(A) } 200+25pi qquad mathrm{(B) } 100+75pi qquad mathrm{(C) } 75+100pi qquad mathrm{(D) } 100+100pi qquad mathrm{(E) } 100+125pi$

Problem 10

A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $100$ cans, how many rows does it contain?

$mathrm{(A) } 5 qquad mathrm{(B) } 8 qquad mathrm{(C) } 9 qquad mathrm{(D) } 10qquad mathrm{(E) } 411$

Problem 11

Two eight-sided dice each have faces numbered $1$ through $8$ . When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?

$mathrm{(A) } frac{1}{2} qquad mathrm{(B) } frac{47}{64} qquad mathrm{(C) } frac{3}{4} qquad mathrm{(D) } frac{55}{64} qquad mathrm{(E) } frac{7}{8}$

Problem 12

An annulus is the region between two concentric circles. The concentric circles in the ﬁgure have radii $b$ and $c$ , with $b>c$ . Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$ , and let $OY$ be the radius of the larger circle that contains $Z$ . Let $a=XZ$ , $d=YZ$ , and $e=XY$ . What is the area of the annulus?

$[asy] unitsize(1.5cm); defaultpen(0.8); real r1=1.5, r2=2.5; pair O=(0,0); path inner=Circle(O,r1), outer=Circle(O,r2); pair Y=(0,r2), Z=(0,r1), X=intersectionpoint( Z--(Z+(10,0)), outer ); filldraw(outer,lightgray,black); filldraw(inner,white,black); draw(X--O--Y); draw(Y--X--Z); label("$O$",O,SW); label("$X$",X,E); label("$Y$",Y,N); label("$Z$",Z,SW); label("$a$",X--Z,N); label("$b$",0.25*X,SE); label("$c$",O--Z,E); label("$d$",Y--Z,W); label("$e$",Y*0.65 + X*0.35,SW); defaultpen(0.5); dot(O); dot(X); dot(Z); dot(Y); [/asy]$

$mathrm{(A) } pi a^2 qquad mathrm{(B) } pi b^2 qquad mathrm{(C) } pi c^2 qquad mathrm{(D) } pi d^2 qquad mathrm{(E) } pi e^2$

Problem 13

In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack?

$mathrm{(A) } 7 qquad mathrm{(B) } 8 qquad mathrm{(C) } 9 qquad mathrm{(D) } 10 qquad mathrm{(E) } 11$

Problem 14

A bag initially contains red marbles and blue marbles only, with more blue than red. Red marbles are added to the bag until only $frac{1}{3}$ of the marbles in the bag are blue. Then yellow marbles are added to the bag until only $frac{1}{5}$ of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?

$mathrm{(A) } frac{1}{5} qquad mathrm{(B) } frac{1}{4} qquad mathrm{(C) } frac{1}{3} qquad mathrm{(D) } frac{2}{5} qquad mathrm{(E) } frac{1}{2}$

Problem 15

Patty has $20$ coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have $70$ cents more. How much are her coins worth?

$textbf{(A)} textdollar 1.15qquadtextbf{(B)} textdollar 1.20qquadtextbf{(C)} textdollar 1.25qquadtextbf{(D)} textdollar 1.30qquadtextbf{(E)} textdollar 1.35$

Problem 16

Three circles of radius $1$ are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?

$mathrm{(A) } frac{2 + sqrt{6}}{3} qquad mathrm{(B) } 2 qquad mathrm{(C) } frac{2 + 3sqrt{2}}{2} qquad mathrm{(D) } frac{3 + 2sqrt{3}}{3} qquad mathrm{(E) } frac{3 + sqrt{3}}{2}$

Problem 17

The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?

$mathrm{(A) } 9 qquad mathrm{(B) } 18 qquad mathrm{(C) } 27 qquad mathrm{(D) } 36qquad mathrm{(E) } 45$

Problem 18

In the right triangle $triangle ACE$ , we have $AC=12$ , $CE=16$ , and $EA=20$ . Points $B$ , $D$ , and $F$ are located on $AC$ , $CE$ , and $EA$ , respectively, so that $AB=3$ , $CD=4$ , and $EF=5$ . What is the ratio of the area of $triangle DBF$ to that of $triangle ACE$ ?

$mathrm{(A) } frac{1}{4} qquad mathrm{(B) } frac{9}{25} qquad mathrm{(C) } frac{3}{8} qquad mathrm{(D) } frac{11}{25} qquad mathrm{(E) } frac{7}{16}$

$[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label("$A$",A,N); label("$B$",B,W); label("$C$",C,SW); label("$D$",D,S); label("$E$",E,SE); label("$F$",F,NE); label("$3$",A--B,W); label("$9$",C--B,W); label("$4$",C--D,S); label("$12$",D--E,S); label("$5$",E--F,NE); label("$15$",F--A,NE); [/asy]$

Problem 19

In the sequence $2001$ , $2002$ , $2003$ , $ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$ . What is the $2004^textrm{th}$ term in this sequence?

$mathrm{(A) } -2004 qquad mathrm{(B) } -2 qquad mathrm{(C) } 0 qquad mathrm{(D) } 4003 qquad mathrm{(E) } 6007$

Problem 20

In $triangle ABC$ points $D$ and $E$ lie on $BC$ and $AC$ , respectively. If $AD$ and $BE$ intersect at $T$ so that $AT/DT=3$ and $BT/ET=4$ , what is $CD/BD$ ?

$mathrm{(A) } frac{1}{8} qquad mathrm{(B) } frac{2}{9} qquad mathrm{(C) } frac{3}{10} qquad mathrm{(D) } frac{4}{11} qquad mathrm{(E) } frac{5}{12}$

$[asy] unitsize(1cm); defaultpen(0.8); pair A=(0,0), B=5*dir(60), C=5*(1,0), D=B + (11/15)*(C-B), E = A + (11/16)*(C-A); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,S); label("$T$",T,2*WNW); [/asy]$

Problem 21

Let $1$ ; $4$ ; $ldots$ and $9$ ; $16$ ; $ldots$ be two arithmetic progressions. The set $S$ is the union of the first $2004$ terms of each sequence. How many distinct numbers are in $S$ ?

$mathrm{(A) } 3722 qquad mathrm{(B) } 3732 qquad mathrm{(C) } 3914 qquad mathrm{(D) } 3924 qquad mathrm{(E) } 4007$

Problem 22

A triangle with sides of $5, 12,$ and $13$ has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?

$mathrm{(A) } frac{3sqrt{5}}{2} qquad mathrm{(B) } frac{7}{2} qquad mathrm{(C) } sqrt{15} qquad mathrm{(D) } frac{sqrt{65}}{2} qquad mathrm{(E) } frac{9}{2}$

Problem 23

Each face of a cube is painted either red or blue, each with probability $1/2$ . The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

$mathrm{(A) } frac{1}{4} qquad mathrm{(B) } frac{5}{16} qquad mathrm{(C) } frac{3}{8} qquad mathrm{(D) } frac{7}{16} qquad mathrm{(E) } frac{1}{2}$

Problem 24

In $bigtriangleup ABC$ we have $AB = 7$ , $AC = 8$ , and $BC = 9$ . Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects $angle BAC$ . What is the value of $frac{AD}{CD}$ ?

$mathrm{(A) } frac{9}{8} qquad mathrm{(B) } frac{5}{3} qquad mathrm{(C) } 2 qquad mathrm{(D) } frac{17}{7} qquad mathrm{(E) } frac{5}{2}$

Problem 25

A circle of radius $1$ is internally tangent to two circles of radius $2$ at points $A$ and $B$ , where $AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?

$mathrm{(A) } frac{5}{3} pi - 3sqrt 2 qquad mathrm{(B) } frac{5}{3} pi - 2sqrt 3 qquad mathrm{(C) } frac{8}{3} pi - 3sqrt 3 qquad mathrm{(D) } frac{8}{3} pi - 3sqrt 2 qquad mathrm{(E) } frac{8}{3} pi - 2sqrt 3$

$[asy] unitsize(1cm); defaultpen(0.8); pair O=(0,0), A=(0,1), B=(0,-1); path bigc1 = Circle(A,2), bigc2 = Circle(B,2), smallc = Circle(O,1); pair[] P = intersectionpoints(bigc1, bigc2); filldraw( arc(A,P[0],P[1])--arc(B,P[1],P[0])--cycle, lightgray, black ); draw(bigc1); draw(bigc2); unfill(smallc); draw(smallc); dot(O); dot(A); dot(B); label("$A$",A,N); label("$B$",B,S); draw( O--dir(30) ); draw( A--(A+2*dir(30)) ); draw( B--(B+2*dir(210)) ); label("$1$", O--dir(30), N ); label("$2$", A--(A+2*dir(30)), N ); label("$2$", B--(B+2*dir(210)), S ); [/asy]$

2004AMC10B详细解析

There are $22-12+1=11$ rows of $33$ seats, giving $11times 33=boxed{mathrm{(C)} 363}$ seats.
Ten numbers $(70,71,dots,79)$ have $7$ as the tens digit. Nine numbers $(17,27,dots,97)$ have it as the ones digit. Number $77$ is in both sets.Thus the result is $10+9-1=18 Rightarrow$ $boxed{mathrm{(B)} 18}$ .We use complementary counting. The complement of having at least one $7$ as a digit is having no $7$ s as a digit.We have $9$ digits to choose from for the first digit and $10$ digits for the second. This gives a total of $9 times 10 = 90$ two-digit numbers.But since we cannot have $7$ as a digit, we have $8$ first digits and $9$ second digits to choose from.Thus there are $8 times 9 = 72$ two-digit numbers without a $7$ as a digit.
$90$ (The total number of two-digit numbers) $- 72$ (The number of two-digit numbers without a $7$ ) $= 18 Rightarrow$ $boxed{mathrm{(B)} 18}$
At the fourth practice she made $48/2=24$ throws, at the third one it was $24/2=12$ , then we get $12/2=6$ throws for the second practice, and finally $6/2=3Rightarrowboxed{mathrm{(A)} 3}$ throws at the first one.
The product of all six numbers is $6!=720$ . The products of numbers that can be visible are $720/1$ , $720/2$ , ..., $720/6$ . The answer to this problem is their greatest common divisor -- which is $720/L$ , where $L$ is the least common multiple of ${1,2,3,4,5,6}$ . Clearly $L=60$ and the answer is $720/60=boxed{mathrm{(B)} 12}$ .Clearly, $P$ cannot have a prime factor other than $2$ , $3$ and $5$ .We can not guarantee that the product will be divisible by $5$ , as the number $5$ can end on the bottom.We can guarantee that the product will be divisible by $3$ (one of $3$ and $6$ will always be visible), but not by $3^2$ .Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by $2^2$ . This is the most we can guarantee, as when the $4$ is on the bottom side, the two visible even numbers are $2$ and $6$ , and their product is not divisible by $2^3$ .Hence $P=3cdot2^2=boxed{mathrm{(B)} 12}$ .
If $a=0$ or $c=0$ , the expression evaluates to $-d<0$ .
If $b=0$ , the expression evaluates to $c-dleq 2$ .
Case $d=0$ remains. In that case, we want to maximize $ccdot a^b$ where ${a,b,c}={1,2,3}$ . Trying out the six possibilities we get that the best one is $(a,b,c)=(3,2,1)$ , where $ccdot a^b=1cdot 3^2=boxed{mathrm{(D)} 9}$ .
Using the fact that $n! = ncdot (n-1)!$ , we can write: $begin{align} A&=98! cdot (99cdot 98!) = 99 cdot (98!)^2 = 11cdot3^2cdot(98!)^2 \ B&=100 cdot 99 cdot (98!)^2 = 11cdot10^2cdot3^2cdot( 98!)^2 \ C&=100cdot (99!)^2 = 10^2cdot (99!)^2\ D&=101cdot 100cdot (99!)^2 = 101 cdot 10^2 cdot (99!)^2\ E& =101cdot (100!)^2 end{align}$ We see that $boxed{mathrm{(C) } 99! cdot 100!}$ is a square, and because $11$ , and $101$ are primes, none of the other four choices are squares.
Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=frac{7}{10}cdot(60+d)$ , which solves to $d=140$ , and the sum of digits of $d$ is $boxed{mathrm{(A)} 5}$ .Each time Isabella exchanges $7$ U.S. dollars, she gets $7$ Canadian dollars and $3$ Canadian dollars extra. Isabella received a total of $60$ Canadian dollars extra, therefore she exchanged $7$ U.S. dollars $frac{60}{3}=20$ times. Thus $d=7cdot20=140$ , and the sum of the digits is $boxed{mathrm{(A)} 5}$ .
The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs $8$ miles and $10$ miles long. The hypotenuse length is $sqrt{8^2 + 10^2}approx12.8$ , and thus the answer is $boxed{mathrm{(A)} 13}$ .Without a calculator one can note that $8^2+10^2=164<169=13^2Rightarrowboxed{mathrm{(A)} 13}$ .
The area of the circle is $S_{bigcirc}=100pi$ ; the area of the square is $S_{square}=100$ .Exactly $frac{1}{4}$ of the circle lies inside the square. Thus the total area is $dfrac34 S_{bigcirc}+S_{square}=boxed{mathrm{(B) }100+75pi}$ . $[asy] Draw(Circle((0,0),10)); Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0)); label("$10$",(5,0),S); label("$10$",(0,5),W); dot((0,0)); [/asy]$
The sum of the first $n$ odd numbers is $n^2$ . As in our case $n^2=100$ , we have $n=boxed{mathrm{(D) }10}$ .
We have $1times n = n < 1 + n$ , hence if at least one of the numbers is $1$ , the sum is larger. There $15$ such possibilities.We have $2times 2 = 2+2$ .For $n>2$ we already have $2times n = n + n > 2 + n$ , hence all other cases are good.Out of the $8times 8$ possible cases, we find that in $15+1=16$ the sum is greater than or equal to the product, hence in $64-16=48$ cases the sum is smaller, satisfying the condition. Therefore the answer is $frac{48}{64}=boxed{mathrm{(C) }frac{3}{4}}$ .Let the two rolls be $m$ , and $n$ .From the restriction: $mn > m + n$
$mn - m - n > 0$

$mn - m - n + 1 > 1$

$(m-1)(n-1) > 1$

Since $m-1$ and $n-1$ are non-negative integers between $1$ and $8$ , either $(m-1)(n-1) = 0$ , $(m-1)(n-1) = 1$ , or $(m-1)(n-1) > 1$

$(m-1)(n-1) = 0$ if and only if $m=1$ or $n=1$ .

There are $8$ ordered pairs $(m,n)$ with $m=1$ , $8$ ordered pairs with $n=1$ , and $1$ ordered pair with $m=1$ and $n=1$ . So, there are $8+8-1 = 15$ ordered pairs $(m,n)$ such that $(m-1)(n-1) = 0$ .

$(m-1)(n-1) = 1$ if and only if $m-1=1$ and $n-1=1$ or equivalently $m=2$ and $n=2$ . This gives $1$ ordered pair $(m,n) = (2,2)$ .

So, there are a total of $15+1=16$ ordered pairs $(m,n)$ with $(m-1)(n-1) < 1$ .

Since there are a total of $8cdot8 = 64$ ordered pairs $(m,n)$ , there are $64-16 = 48$ ordered pairs $(m,n)$ with $(m-1)(n-1) > 1$ .

Thus, the desired probability is $frac{48}{64}=boxed{mathrm{(C) }frac{3}{4}}$ .
The area of the large circle is $pi b^2$ , the area of the small one is $pi c^2$ , hence the shaded area is $pi(b^2-c^2)$ .From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$ , hence $b^2-c^2=a^2$ and thus the shaded area is $boxed{mathrm{(A) }pi a^2}$ .
All numbers in this solution will be in hundredths of a millimeter.The thinnest coin is the dime, with thickness $135$ . A stack of $n$ dimes has height $135n$ .The other three coin types have thicknesses $135+20$ , $135+40$ , and $135+60$ . By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set ${135n, 135n+20, 135n+40, dots, 195n}$ .If we take an odd $n$ , then all the possible heights will be odd, and thus none of them will be $1400$ . Hence $n$ is even.If $n<8$ the stack will be too low and if $n>10$ it will be too high. Thus we are left with cases $n=8$ and $n=10$ .If $n=10$ the possible stack heights are $1350,1370,1390,dots$ , with the remaining ones exceeding $1400$ .
Therefore there are $boxed{mathrm{(B) }8}$ coins in the stack.

Using the above observation we can easily construct such a stack. A stack of $8$ dimes would have height $8cdot 135=1080$ , thus we need to add $320$ . This can be done for example by replacing five dimes by nickels (for $+60cdot 5 = +300$ ), and one dime by a penny (for $+20$ ).

Let $p,n,d$ , and $q$ be the number of pennies, nickels, dimes, and quarters used in the stack.

From the conditions above, we get the following equation:

$[155p+195n+135d+175q=1400.]$

Then we divide each side by five to get

$[31p+39n+27d+35q=280.]$

Writing both sides in terms of mod 4, we have $-p-n-d-q equiv 0 pmod 4$ .

This means that the sum $p+n+d+q$ is divisible by 4. Therefore, the answer must be $boxed{(B),, 8}.$
We can ignore most of the problem statement. The only important information is that immediately before the last step blue marbles formed $frac{1}{5}$ of the marbles in the bag. This means that there were $x$ blue and $4x$ other marbles, for some $x$ . When we double the number of blue marbles, there will be $2x$ blue and $4x$ other marbles, hence blue marbles now form $boxed{mathrm{(C) }frac{1}{3}}$ of all marbles in the bag.
She has $n$ nickels and $d=20-n$ dimes. Their total cost is $5n+10d=5n+10(20-n)=200-5n$ cents. If the dimes were nickels and vice versa, she would have $10n+5d=10n+5(20-n)=100+5n$ cents. This value should be $70$ cents more than the previous one. We get $200-5n+70=100+5n$ , which solves to $n=17$ . Her coins are worth $200-5n=boxed{mathrm{(A) }textdollar1.15}$ .Changing a nickel into a dime increases the sum by $5$ cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by $70$ cents, there are $70/5=14$ more nickels than dimes. As the total count is $20$ , this means that there are $17$ nickels and $3$ dimes, which is equal to $boxed{mathrm{(A) }textdollar1.15}$ .
The situation is shown in the picture below. The radius we seek is $SD = AD + AS$ . Clearly $AD=1$ . The point $S$ is clearly the center of the equilateral triangle $ABC$ , thus $AS$ is $2/3$ of the altitude of this triangle. We get that $AS = frac23 cdot sqrt 3$ . Therefore the radius we seek is $1 + frac23 cdot sqrt 3 = boxed{mathrm{(D) }frac{3+2sqrt{3}}3}$ . $[asy] unitsize(2cm); pair A=(0,0), B=dir(0)*2, C=dir(60)*2; draw(circle(A,1)); draw(circle(B,1)); draw(circle(C,1)); dot(A); dot(B); dot(C); draw(A--B--C--cycle); pair D=A+dir(210), E=B+dir(-30), F=C+dir(90); draw(circumcircle(D,E,F)); dot(D); dot(E); dot(F); pair S=(A+B+C)/3; dot(S); draw(S--D); draw(S--E); draw(S--F); label("$S$",S,S); label("$A$",A,SE); label("$D$",D,SW); label("$B$",B,NE); label("$C$",C,ENE); [/asy]$

Solution 2

Using Descartes' Circle Formula, we can assign curvatures to all the circles: $1$ , $1$ , $1$ , and $-frac{1}{r}$ (b/c it is the circle internally tangent to all the other circles, the radius of the bigger circle is negative). Then, we can solve:

$2(1^2+1^2+1^2+(-frac{1}{r})^2) = (1+1+1-frac{1}{r})^2$

$r = boxed{mathrm{(D) }frac{3+2sqrt{3}}3}$
If Jack's current age is $overline{ab}=10a+b$ , then Bill's current age is $overline{ba}=10b+a$ .In five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$ .We are given that $10a+b+5=2(10b+a+5)$ . Thus $8a=19b+5$ .For $b=1$ we get $a=3$ . For $b=2$ and $b=3$ the value $frac{19b+5}8$ is not an integer, and for $bgeq 4$ it is more than $9$ . Thus the only solution is $(a,b)=(3,1)$ , and the difference in ages is $31-13=boxed{mathrm{(B) }18}$ .Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.The age difference is $(10a+b)-(10b+a)=9(a-b)$ , hence it is a multiple of $9$ . Thus Bill's current age modulo $9$ must be $4$ .
Thus Bill's age is in the set ${13,22,31,40,49,58,67,76,85,94}$ .

As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options ${13,49,58,67}$ .

Checking each of them, we see that only $13$ works, and gives the solution $31-13=boxed{mathrm{(B) }18}$ .
Let $x = [DBF]$ . Because $triangle ACE$ is divided into four triangles, $[ACE] = [BCD] + [ABF] + [DEF] + x$ .Because of $SAS$ triangle area, $frac12 cdot 12 cdot 16 = frac12 cdot 9 cdot 4 + frac12 cdot 3 cdot 15 cdot sin(angle A) + frac12 cdot 5 cdot 12 cdot sin(angle E) + x$ . $sin(angle A) = frac{16}{20}$ and $sin(angle E) = frac{12}{20}$ , so $96 = 18 + 18 + 18 + x$ . $x = 42$ , so $frac{[DBF]}{[ACE]} = frac{42}{96} = boxed{textbf{(E)}frac 7{16}}$ .First of all, note that $frac{AB}{AC} = frac{CD}{CE} = frac{EF}{EA} = frac 14$ , and therefore $frac{BC}{AC} = frac{DE}{CE} = frac{FA}{EA} = frac 34$ .Draw the height from $F$ onto $AB$ as in the picture below:
$[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label("$A$",A,N); label("$B$",B,W); label("$C$",C,SW); label("$D$",D,S); label("$E$",E,SE); label("$F$",F,NE); label("$3$",A--B,W); label("$9$",0.5*C + 0.5*B,4*W); label("$4$",C--D,S); label("$12$",D--E,S); label("$5$",E--F,NE); label("$15$",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label("$G$",G,W); [/asy]$

Now consider the area of $triangle ABF$ . Clearly the triangles $triangle AFG$ and $triangle AEC$ are similar, as they have all angles equal. Their ratio is $frac {AF}{AE} = frac 34$ , hence $FG = frac 34 cdot CE$ . Now the area $S_{ABF}$ of $triangle ABF$ can be computed as $S_{ABF} = frac 12 cdot AB cdot FG$ = $frac 12 cdot left( frac 14 cdot AC right) cdot left( frac 34 cdot EC right) = frac 14 cdot frac 34 cdot S_{ACE}$ .

Similarly we can find that $S_{BCD} = S_{DEF} = frac 3{16}cdot S_{ACE}$ as well.

Hence $S_{BDF} = S_{ACE} - 3cdotleft( frac 3{16} cdot S_{ACE} right) = frac 7{16} cdot S_{ACE}$ , and the answer is $frac{S_{BDF}}{S_{ACE}} = boxed{frac 7{16}}$ .

The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9), and let F be (12, 3). You can then use the shoelace theorem to find the area of DBF, which is 42. $frac {42}{96} = boxed{frac 7{16}}$
We already know that $a_1=2001$ , $a_2=2002$ , $a_3=2003$ , and $a_4=2000$ . Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$ , $a_6=2003+2000-2005=1998$ , $a_7=2000+2005-1998=2007$ , and so on.We can now discover the following pattern: $a_{2k+1} = 1999+2k$ and $a_{2k}=2004-2k$ . This is easily proved by induction. It follows that $a_{2004}=a_{2cdot 1002} = 2004 - 2cdot 1002 = boxed{0}$ .Note that the recurrence $a_n+a_{n+1}-a_{n+2}~=~a_{n+3}$ can be rewritten as $a_n+a_{n+1} ~=~ a_{n+2}+a_{n+3}$ .Hence we get that $a_1+a_2 ~=~ a_3+a_4 ~=~ a_5+a_6 ~= cdots$ and also $a_2+a_3 ~=~ a_4+a_5 ~=~ a_6+a_7 ~= cdots$ From the values given in the problem statement we see that $a_3=a_1+2$ .From $a_1+a_2 = a_3+a_4$ we get that $a_4=a_2-2$ .
From $a_2+a_3 = a_4+a_5$ we get that $a_5=a_3+2$ .

Following this pattern, we get $a_{2004} = a_{2002} - 2 = a_{2000} - 4 = cdots = a_2 - 2002 = boxed{0}$ .
We use the square bracket notation $[cdot]$ to denote area.WLOG, we can assume $[triangle BTD] = 1$ . Then $[triangle BTA] = 3$ , and $[triangle ATE] = 3/4$ . We have $CD/BD = [triangle ACD]/[triangle ABD]$ , so we need to find the area of quadrilateral $TDCE$ .Draw the line segment $TC$ to form the two triangles $triangle TDC$ and $triangle TEC$ . Let $x = [triangle TDC]$ , and $y = [triangle TEC]$ . By considering triangles $triangle BTC$ and $triangle ETC$ , we obtain $(1+x)/y=4$ , and by considering triangles $triangle ATC$ and $triangle DTC$ , we obtain $(3/4+y)/x=3$ . Solving, we get $x=4/11$ , $y=15/44$ , so the area of quadrilateral $TDEC$ is $x+y=31/44$ .Therefore $frac{CD}{BD}=frac{frac{3}{4}+frac{31}{44}}{3+1}=boxed{textbf{(D)} frac{4}{11}}$ The presence of only ratios in the problem essentially cries out for mass points.As per the problem, we assign a mass of $1$ to point $A$ , and a mass of $3$ to $D$ . Then, to balance $A$ and $D$ on $T$ , $T$ has a mass of $4$ .
Now, were we to assign a mass of $1$ to $B$ and a mass of $4$ to $E$ , we'd have $5T$ . Scaling this down by $4/5$ (to get $4T$ , which puts $B$ and $E$ in terms of the masses of $A$ and $D$ ), we assign a mass of $frac{4}{5}$ to $B$ and a mass of $frac{16}{5}$ to $E$ .

Now, to balance $A$ and $C$ on $E$ , we must give $C$ a mass of $frac{16}{5}-1=frac{11}{5}$ .

Finally, the ratio of $CD$ to $BD$ is given by the ratio of the mass of $B$ to the mass of $C$ , which is $frac{4}{5}cdotfrac{5}{11}=boxed{textbf{(D)} frac{4}{11}}$ .

Affine transformations preserve ratios of distances, and for any pair of triangles, there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any $triangle ABC$ , and we just need to compute it for any single triangle.

We can choose the points $A=(-3,0)$ , $B=(0,4)$ , and $D=(1,0)$ . This way we will have $T=(0,0)$ , and $E=(0,-1)$ . The situation is shown in the picture below:

$[asy] unitsize(1cm); defaultpen(0.8); pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,S); label("$T$",T,NW); label("$3$",A--T,N); label("$4$",B--T,W); label("$1$",D--T,N); label("$1$",E--T,W); [/asy]$

The point $C$ is the intersection of the lines $BD$ and $AE$ . The points on the first line have the form $(t,4-4t)$ , the points on the second line have the form $(t,-1-t/3)$ . Solving for $t$ we get $t=15/11$ , hence $C=(15/11,-16/11)$ .

The ratio $CD/BD$ can now be computed simply by observing the $x$ coordinates of $B$ , $C$ , and $D$ :

$[frac{CD}{BD} = frac{15/11 - 1}{1 - 0} = boxed{textbf{(D)}frac 4{11}}]$
The two sets of terms are $A={ 3k+1 : 0leq k < 2004 }$ and $B={ 7l+9 : 0leq l<2004}$ .Now $S=Acup B$ . We can compute $|S|=|Acup B|=|A|+|B|-|Acap B|=4008-|Acap B|$ . We will now find $|Acap B|$ .Consider the numbers in $B$ . We want to find out how many of them lie in $A$ . In other words, we need to find out the number of valid values of $l$ for which $7l+9in A$ .The fact " $7l+9in A$ " can be rewritten as " $1leq 7l+9 leq 3cdot 2003 + 1$ , and $7l+9equiv 1pmod 3$ ".The first condition gives $0leq lleq 857$ , the second one gives $lequiv 1pmod 3$ .Thus the good values of $l$ are ${1,4,7,dots,856}$ , and their count is $858/3 = 286$ .
Therefore $|Acap B|=286$ , and thus $|S|=4008-|Acap B|=boxed{3722}$ .

Shift down the first sequence by $1$ and the second by $9$ so that the two sequences become $0,3,6,cdots,6009$ and $0,7,14,cdots,14028$ . The first becomes multiples of $3$ and the second becomes multiples of $7$ . Their intersection is the multiples of $21$ up to $6009$ . There are $lfloor frac{6009}{21} rfloor=286$ multiples of $21$ . There are $4008-286=boxed{textbf{(A)} 3722}$ distinct numbers in $S$ .
$[asy] import geometry; unitsize(0.6 cm); pair A, B, C, D, E, F, I, O; A = (5^2/13,5*12/13); B = (0,0); C = (13,0); I = incenter(A,B,C); D = (I + reflect(B,C)*(I))/2; E = (I + reflect(C,A)*(I))/2; F = (I + reflect(A,B)*(I))/2; O = (B + C)/2; draw(A--B--C--cycle); draw(incircle(A,B,C)); draw(I--D); draw(I--E); draw(I--F); draw(I--O); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); dot("$D$", D, S); dot("$E$", E, NE); dot("$F$", F, NW); dot("$I$", I, N); dot("$O$", O, S); [/asy]$ This is obviously a right triangle. Pick a coordinate system so that the right angle is at $(0,0)$ and the other two vertices are at $(12,0)$ and $(0,5)$ .As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at $(6,2.5)$ .The radius $r$ of the inscribed circle can be computed using the well-known identity $frac{rP}2=S$ , where $S$ is the area of the triangle and $P$ its perimeter. In our case, $S=frac{5cdot 12}{2}=30$ and $P=5+12+13=30$ . Thus, $r=2$ . As the inscribed circle touches both legs, its center must be at $(r,r)=(2,2)$ .The distance of these two points is then $sqrt{ (6-2)^2 + (2.5-2)^2 } = sqrt{16.25} = sqrt{frac{65}4} = boxed{frac{sqrt{65}}2}$ .We directly apply Euler’s Theorem, which states that if the circumcenter is $O$ and the incenter $I$ , and the inradius is $r$ and the circumradius is $R$ , then $[OI^2=R(R-2r)]$ We notice that this is a right triangle, and hence has area $30$ . We then find the inradius with the formula $A=rs$ , where $s$ denotes semiperimeter. We easily see that $s=15$ , so $r=2$ .
We now find the circumradius with the formula $A=frac{abc}{4R}$ . Solving for $R$ gives $R=frac{13}{2}$ .

Substituting all of this back into our formula gives: $[OI^2= frac{65}{4}]$ So, $OI=frac{sqrt{65}}{2}implies boxed{D}$
If the orientation of the cube is fixed, there are $2^6=64$ possible arrangements of colors on the faces. There are $[2dbinom{6}{6}=2]$ arrangements in which all six faces are the same color and $[2dbinom{6}{5}=12]$ arrangements in which exactly five faces have the same color. In each of these cases the cube can be placed so that the four vertical faces have the same color. The only other suitable arrangements have four faces of one color, with the other color on a pair of opposing faces. Since there are three pairs of opposing faces, there are $2(3)=6$ such arrangements. The total number of suitable arrangements is therefore $2+12+6=20$ , and the probability is $dfrac{20}{64}=dfrac{5}{16}.$ Label the six sides of the cube by numbers $1$ to $6$ as on a classic dice. Then the "four vertical faces" can be: ${1,2,5,6}$ , ${1,3,4,6}$ , or ${2,3,4,5}$ .Let $A$ be the set of colorings where $1,2,5,6$ are all of the same color, similarly let $B$ and $C$ be the sets of good colorings for the other two sets of faces.There are $2^6=64$ possible colorings, and there are $|Acup Bcup C|$ good colorings. Thus the result is $frac{|Acup Bcup C|}{64}$ . We need to compute $|Acup Bcup C|$ .Using the Principle of Inclusion-Exclusion we can write $[|Acup Bcup C| = |A|+|B|+|C| - |Acap B| - |Acap C| - |Bcap C| + |Acap Bcap C|]$ Clearly $|A|=|B|=|C|=2^3=8$ , as we have two possibilities for the common color of the four vertical faces, and two possibilities for each of the horizontal faces.
What is $Acap B$ ? The faces $1,2,5,6$ must have the same color, and at the same time faces $1,3,4,6$ must have the same color. It turns out that $Acap B=Acap C=Bcap C= Acap Bcap C =$ the set containing just the two cubes where all six faces have the same color.

Therefore $|Acup Bcup C| = 8+8+8-2-2-2+2 = 20$ , and the result is $frac{20}{64}=boxed{frac{5}{16}}$ .

Suppose we break the situation into cases that contain four vertical faces of the same color:

I. Two opposite sides of same color: There are 3 ways to choose the two sides, and then two colors possible, so $3 cdot 2=6$ .

II. One face different from all the others: There are 6 ways to choose this face, and 2 colors, so $6 cdot 2=12$ .

III. All faces are the same: There are 2 colors, and so two ways for all faces to be the same.

Adding them up, we have a total of $20$ ways to have four vertical faces the same color. The are $2^6$ ways to color the cube, so the answer is $frac{20}{64}=boxed{frac{5}{16}}$ .
Set $overline{BD}$ 's length as $x$ . $overline{CD}$ 's length must also be $x$ since $angle BAD$ and $angle DAC$ intercept arcs of equal length (because $angle BAD=angle DAC$ ). Using Ptolemy's Theorem, $7x+8x=9(AD)$ . The ratio is $frac{5}{3}impliesboxed{text{(B)}}$ $[asy] import graph; import geometry; import markers; unitsize(0.5 cm); pair A, B, C, D, E, I; A = (11/3,8*sqrt(5)/3); B = (0,0); C = (9,0); I = incenter(A,B,C); D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C)); E = extension(A,D,B,C); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(D--A); draw(D--B); draw(D--C); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, NE); markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true))); markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true))); markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true))); markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true))); markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true))); [/asy]$ Let $E = overline{BC}cap overline{AD}$ . Observe that $angle ABC cong angle ADC$ because they subtend the same arc, $overarc{AC}$ . Furthermore, $angle BAE cong angle EAC$ because $overline{AE}$ is an angle bisector, so $triangle ABE sim triangle ADC$ by $text{AA}$ similarity. Then $dfrac{AD}{AB} = dfrac{CD}{BE}$ . By the Angle Bisector Theorem, $dfrac{7}{BE} = dfrac{8}{CE}$ , so $dfrac{7}{BE} = dfrac{8}{9-BE}$ . This in turn gives $BE = frac{21}{5}$ . Plugging this into the similarity proportion gives: $dfrac{AD}{7} = dfrac{CD}{dfrac{21}{5}} implies dfrac{AD}{CD} = {dfrac{5}{3}} = boxed{text{(B)}}$ .
The area of the small circle is $pi$ . We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.Let $C$ and $D$ be the intersections of the two large circles. Connect them to $A$ and $B$ to get the picture below: $[asy] unitsize(1.5cm); defaultpen(0.8); pair O=(0,0), A=(0,1), B=(0,-1); path bigc1 = Circle(A,2), bigc2 = Circle(B,2), smallc = Circle(O,1); pair[] P = intersectionpoints(bigc1, bigc2); filldraw( arc(A,P[0],P[1])--arc(B,P[1],P[0])--cycle, lightgray, black ); draw(bigc1); draw(bigc2); dot(A); dot(B); label("$A$",A,N); label("$B$",B,S); /* dot(O); unfill(smallc); draw(smallc); draw( O--dir(30) ); draw( A--(A+2*dir(30)) ); draw( B--(B+2*dir(210)) ); label("$1$", O--dir(30), N ); label("$2$", A--(A+2*dir(30)), N ); label("$2$", B--(B+2*dir(210)), S ); */ label("$C$",P[0],W); label("$D$",P[1],E); draw( P[0]--A--P[1] ); draw( P[0]--B--P[1] ); draw( A--B ); [/asy]$ We can see that the triangles $triangle ABC$ and $triangle ABD$ are both equilateral with side $2$ .Take a look at the lower circle. The angle $ABC$ is $60^circ$ , thus sector $ABC$ is $1/6$ of the circle. The same is true for sector $ABD$ of the lower circle, and sectors $CAB$ and $BAD$ of the upper circle.If we now sum the areas of these four sectors, we will almost get the area of the new shaded region - except that each of the two equilateral triangles will be counted twice. The area of an equilateral triangle given its side, $s,$ is $frac{sqrt3}4s^2.$
Therefore, the area of the new shaded region is $4cdot left( frac 16 cdot picdot 2^2 right) - 2 cdot left( 4 cdot frac{sqrt3}4 right) = frac 83 pi - 2sqrt 3$ . Lastly, we must subtract the area of the circle that we added earlier, $pi$ , and we get

$left( frac 83 pi - 2sqrt 3 right) - pi = boxed{mathrm{(B) } frac 53 pi - 2sqrt 3 }$ .