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2004AMC10A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11:12

2004AMC10A真题

答案解析请参考文末

Problem 1

You and five friends need to raise $1500$ dollars in donations for a charity, dividing the fundraising equally. How many dollars will each of you need to raise?

$mathrm{(A) } 250qquad mathrm{(B) } 300 qquad mathrm{(C) } 1500 qquad mathrm{(D) } 7500 qquad mathrm{(E) } 9000$

Problem 2

For any three real numbers $a$ , $b$ , and $c$ , with $bneq c$ , the operation $otimes$ is defined by: $otimes(a,b,c)=frac{a}{b-c}.$ What is $otimes ( otimes (1,2,3), otimes (2,3,1), otimes (3,1,2))$ ?

$mathrm{(A) } -frac{1}{2}qquad mathrm{(B) } -frac{1}{4} qquad mathrm{(C) } 0 qquad mathrm{(D) } frac{1}{4} qquad mathrm{(E) } frac{1}{2}$

Problem 3

Alicia earns 20 dollars per hour, of which $1.45%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?

$mathrm{(A) } 0.0029 qquad mathrm{(B) } 0.029 qquad mathrm{(C) } 0.29 qquad mathrm{(D) } 2.9 qquad mathrm{(E) } 29$

Problem 4

What is the value of $x$ if $|x-1|=|x-2|$ ?

$mathrm{(A) } -frac12 qquad mathrm{(B) } frac12 qquad mathrm{(C) } 1 qquad mathrm{(D) } frac32 qquad mathrm{(E) } 2$

Problem 5

A set of three points is randomly chosen from the grid shown. Each three point set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

$[asy] unitsize(.5cm); defaultpen(linewidth(.8pt)); dotfactor=3; pair[] dotted={(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)}; dot(dotted); [/asy]$

$mathrm{(A) } frac{1}{21} qquad mathrm{(B) } frac{1}{14} qquad mathrm{(C) } frac{2}{21} qquad mathrm{(D) } frac17 qquad mathrm{(E) } frac27$

Problem 6

Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?

$mathrm{(A) } 22 qquad mathrm{(B) } 23 qquad mathrm{(C) } 24 qquad mathrm{(D) } 25 qquad mathrm{(E) } 26$

Problem 7

A grocer stacks oranges in a pyramid-like stack whose rectangular base is 5 oranges by 8 oranges. Each orange above the first level rests in a pocket formed by four oranges below. The stack is completed by a single row of oranges. How many oranges are in the stack?

$mathrm{(A) } 96 qquad mathrm{(B) } 98 qquad mathrm{(C) } 100 qquad mathrm{(D) } 101 qquad mathrm{(E) } 134$

Problem 8

A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ , $B$ , and $C$ start with 15, 14, and 13 tokens, respectively. How many rounds will there be in the game?

$mathrm{(A) } 36 qquad mathrm{(B) } 37 qquad mathrm{(C) } 38 qquad mathrm{(D) } 39 qquad mathrm{(E) } 40$

Problem 9

In the figure, $angle EAB$ and $angle ABC$ are right angles. $AB=4, BC=6, AE=8$ , and $AC$ and $BE$ intersect at $D$ . What is the difference between the areas of $triangle ADE$ and $triangle BDC$ ?

$[asy] unitsize(4mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A=(0,0), B=(4,0), C=(4,6), Ep=(0,8); pair D=extension(A,C,Ep,B); draw(A--C--B--A--Ep--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$E$",Ep,N); label("$D$",D,2.5*N); label("$4$",midpoint(A--B),S); label("$6$",midpoint(B--C),E); label("$8$",(0,3),W); [/asy]$

$mathrm{(A) } 2 qquad mathrm{(B) } 4 qquad mathrm{(C) } 5 qquad mathrm{(D) } 8 qquad mathrm{(E) } 9$

Problem 10

Coin $A$ is flipped three times and coin $B$ is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?

$mathrm{(A) } frac{29}{128} qquad mathrm{(B) } frac{23}{128} qquad mathrm{(C) } frac14 qquad mathrm{(D) } frac{35}{128} qquad mathrm{(E) } frac12$

Problem 11

A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25%$ without altering the volume, by what percent must the height be decreased?

$mathrm{(A) } 10 qquad mathrm{(B) } 25 qquad mathrm{(C) } 36 qquad mathrm{(D) } 50 qquad mathrm{(E) } 60$

Problem 12

Henry's Hamburger Heaven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two, or three meat patties, and any collection of condiments. How many different kinds of hamburgers can be ordered?

$mathrm{(A) } 24 qquad mathrm{(B) } 256 qquad mathrm{(C) } 768 qquad mathrm{(D) } 40,320 qquad mathrm{(E) } 120,960$

Problem 13

At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

$mathrm{(A) } 8 qquad mathrm{(B) } 12 qquad mathrm{(C) } 16 qquad mathrm{(D) } 18 qquad mathrm{(E) } 24$

Problem 14

The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average would be $21$ cents. How many dimes does she have in her purse?

$mathrm{(A) } 0 qquad mathrm{(B) } 1 qquad mathrm{(C) } 2 qquad mathrm{(D) } 3 qquad mathrm{(E) } 4$

Problem 15

Given that $-4leq xleq-2$ and $2leq yleq4$ , what is the largest possible value of $frac{x+y}{x}$ ?

$mathrm{(A) } -1 qquad mathrm{(B) } -frac12 qquad mathrm{(C) } 0 qquad mathrm{(D) } frac12 qquad mathrm{(E) } 1$

Problem 16

The $5times 5$ grid shown contains a collection of squares with sizes from $1times 1$ to $5times 5$ . How many of these squares contain the black center square?

$[asy] unitsize(6mm); defaultpen(linewidth(.8pt)); for(int i=0; i<=5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } fill((2,2)--(2,3)--(3,3)--(3,2)--cycle); [/asy]$

$mathrm{(A) } 12 qquad mathrm{(B) } 15 qquad mathrm{(C) } 17 qquad mathrm{(D) } 19qquad mathrm{(E) } 20$

Problem 17

Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?

$mathrm{(A) } 250 qquad mathrm{(B) } 300 qquad mathrm{(C) } 350 qquad mathrm{(D) } 400qquad mathrm{(E) } 500$

Problem 18

A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?

$mathrm{(A) } 1 qquad mathrm{(B) } 4 qquad mathrm{(C) } 36 qquad mathrm{(D) } 49 qquad mathrm{(E) } 81$

Problem 19

A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?

$[asy] size(250);defaultpen(linewidth(0.8)); draw(ellipse(origin, 3, 1)); fill((3,0)--(3,2)--(-3,2)--(-3,0)--cycle, white); draw((3,0)--(3,16)^^(-3,0)--(-3,16)); draw((0, 15)--(3, 12)^^(0, 16)--(3, 13)); filldraw(ellipse((0, 16), 3, 1), white, black); draw((-3,11)--(3, 5)^^(-3,10)--(3, 4)); draw((-3,2)--(0,-1)^^(-3,1)--(-1,-0.89)); draw((0,-1)--(0,15), dashed); draw((3,-2)--(3,-4)^^(-3,-2)--(-3,-4)); draw((-7,0)--(-5,0)^^(-7,16)--(-5,16)); draw((3,-3)--(-3,-3), Arrows(6)); draw((-6,0)--(-6,16), Arrows(6)); draw((-2,9)--(-1,9), Arrows(3)); label("$3$", (-1.375,9.05), dir(260), fontsize(7)); label("$A$", (0,15), N); label("$B$", (0,-1), NE); label("$30$", (0, -3), S); label("$80$", (-6, 8), W);[/asy]$

$mathrm{(A) } 120 qquad mathrm{(B) } 180 qquad mathrm{(C) } 240 qquad mathrm{(D) } 360 qquad mathrm{(E) } 480$

Problem 20

Points $E$ and $F$ are located on square $ABCD$ so that $triangle BEF$ is equilateral. What is the ratio of the area of $triangle DEF$ to that of $triangle ABE$ ?

$[asy] pair A=origin, B=(1,0), C=(1,1), D=(0,1), X=B+2*dir(165), E=intersectionpoint(B--X, A--D), Y=B+2*dir(105), F=intersectionpoint(B--Y, D--C); draw(B--C--D--A--B--F--E--B); pair point=(0.5,0.5); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); [/asy]$

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

Problem 21

Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is $frac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $pi$ radians is $180$ degrees.)

$[asy] defaultpen(linewidth(0.8)); pair O=origin; fill(O--Arc(O, 2, 20, 160)--cycle, mediumgray); fill(O--Arc(O, 1, 20, 160)--cycle, white); fill(O--Arc(O, 2, 200, 340)--cycle, mediumgray); fill(O--Arc(O, 1, 200, 340)--cycle, white); fill(O--Arc(O, 3, 160, 200)--cycle, mediumgray); fill(O--Arc(O, 2, 160, 200)--cycle, white); fill(O--Arc(O, 1, 160, 200)--cycle, mediumgray); fill(O--Arc(O, 3, -20, 20)--cycle, mediumgray); fill(O--Arc(O, 2, -20, 20)--cycle, white); fill(O--Arc(O, 1, -20, 20)--cycle, mediumgray); draw(Circle(origin, 1));draw(Circle(origin, 2));draw(Circle(origin, 3)); draw(5*dir(200)--5*dir(20)^^5*dir(160)--5*dir(-20)); [/asy]$

$mathrm{(A) } frac{pi}{8} qquad mathrm{(B) } frac{pi}{7} qquad mathrm{(C) } frac{pi}{6} qquad mathrm{(D) } frac{pi}{5} qquad mathrm{(E) } frac{pi}{4}$

Problem 22

Square $ABCD$ has side length $2$ . A semicircle with diameter $overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $overline{AD}$ at $E$ . What is the length of $overline{CE}$ ?

$[asy] defaultpen(linewidth(0.8)); pair A=origin, B=(1,0), C=(1,1), D=(0,1), X=tangent(C, (0.5,0), 0.5, 1), F=C+2*dir(C--X), E=intersectionpoint(C--F, A--D); draw(C--D--A--B--C--E); draw(Arc((0.5,0), 0.5, 0, 180)); pair point=(0.5,0.5); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); [/asy]$

$mathrm{(A) } frac{2+sqrt{5}}{2} qquad mathrm{(B) } sqrt{5} qquad mathrm{(C) } sqrt{6} qquad mathrm{(D) } frac{5}{2} qquad mathrm{(E) } 5-sqrt{5}$

Problem 23

Circles $A$ , $B$ , and $C$ are externally tangent to each other and internally tangent to circle $D$ . Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$ . What is the radius of circle $B$ ?

$[asy] size(6cm); defaultpen(linewidth(0.8));defaultpen(fontsize(8)); draw(Circle(origin, 2)); draw(Circle((-1,0), 1)); draw(Circle((6/9, 8/9), 8/9)); draw(Circle((6/9, -8/9), 8/9)); label("$A$", (-1.2, -0.2), NE); label("$B$", (6/9, 7/9), N); label("$C$", (6/9, -7/9), S); label("$D$", 2*dir(110), dir(110)); [/asy]$

$mathrm{(A) } frac{2}{3} qquad mathrm{(B) } frac{sqrt{3}}{2} qquad mathrm{(C) } frac{7}{8} qquad mathrm{(D) } frac{8}{9} qquad mathrm{(E) } frac{1+sqrt{3}}{3}$

Problem 24

Let $a_1,a_2,cdots$ , be a sequence with the following properties.

(i) $a_1=1$ , and

(ii) $a_{2n}=ncdot a_n$ for any positive integer $n$ .

What is the value of $a_{2^{100}}$ ?

$mathrm{(A) } 1 qquad mathrm{(B) } 2^{99} qquad mathrm{(C) } 2^{100} qquad mathrm{(D) } 2^{4950} qquad mathrm{(E) } 2^{9999}$

Problem 25

Three pairwise-tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere?

$mathrm{(A) } 3+dfrac{sqrt{30}}{2} qquad mathrm{(B) } 3+dfrac{sqrt{69}}{3} qquad mathrm{(C) } 3+dfrac{sqrt{123}}{4} qquad mathrm{(D) } dfrac{52}{9} qquad mathrm{(E) } 3+2sqrt{2}$

2004AMC10A详细解析

There are $6$ people to split the $1500$ dollars among, so each person must raise $frac{1500}6=250$ dollars. $Rightarrowboxed{mathrm{(A)} 250}$

$otimes(frac{1}{2-3},frac{2}{3-1},frac{3}{1-2})=otimes(-1,1,-3)=frac{-1}{1+3}=-frac{1}{4}Longrightarrowboxed{mathrm{(B)} -frac{1}{4}}$

$20$ dollars is the same as $2000$ cents, and $1.45%$ of $2000$ is $0.0145times2000=29$ cents. $Rightarrowboxed{mathrm{(E)} 29}$ .

$|x-1|$ is the distance between $x$ and $1$ ; $|x-2|$ is the distance between $x$ and $2$ .Therefore, the given equation says $x$ is equidistant from $1$ and $2$ , so $x=frac{1+2}2=frac{3}{2}Rightarrowboxed{mathrm{(D)} frac{3}{2}}$ .Alternatively, we can solve by casework (a method which should work for any similar problem involving absolute values of real numbers). If $x leq 1$ , then $|x - 1| = 1-x$ and $|x - 2| = 2 - x$ , so we must solve $1 - x = 2 - x$ , which has no solutions. Similarly, if $x geq 2$ , then $|x - 1| = x - 1$ and $|x - 2| = x - 2$ , so we must solve $x - 1 = x- 2$ , which also has no solutions. Finally, if $1 leq x leq 2$ , then $|x - 1| = x - 1$ and $|x - 2| = 2-x$ , so we must solve $x - 1 = 2 - x$ , which has the unique solution $x = frac32$ .We know that either $x-1=x-2$ or $x-1=-(x-2)$ . The first equation simplifies to $-1=-2$ , which is false, so $x-1=-(x-2)$ . Solving, we get $x=frac{3}{2}Rightarrowboxed{mathrm{(D)} frac{3}{2}}$ .

There are $binom{9}{3}$ ways to choose three points out of the 9 there. There are 8 combinations of dots such that they lie in a straight line: three vertical, three horizontal, and the diagonals. $dfrac{8}{binom{9}{3}}=dfrac{8}{84}=dfrac{2}{21} Rightarrowboxed{mathrm{(C)} frac{2}{21}}$

Since Bertha has $6$ daughters, she has $30-6=24$ granddaughters, of which none have daughters. Of Bertha's daughters, $frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters. Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters. $boxed{mathrm{(E)} 26}$

- Alcumus**

Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = boxed{26}$ .

Draw a tree diagram and see that the answer can be found in the sum of $6+6$ granddaughters, $5+5$ daughters, and $4$ more daughters. Adding them together gives the answer of $boxed{mathrm{(E)} 26}$ .

We look at a set of three rounds, where the players begin with $x+1$ , $x$ , and $x-1$ tokens. After three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$ , $2$ , and $1$ tokens, respectively. After $1$ more round, player $A$ will give away $3$ tokens, leaving it empty-handed, and thus the game will end. We then have there are $36+1=boxed{mathrm{(B)} 37}$ rounds until the game ends.

Since $AE perp AB$ and $BC perp AB$ , $AE parallel BC$ . By alternate interior angles and $AAsim$ , we find that $triangle ADE sim triangle CDB$ , with side length ratio $frac{4}{3}$ . Their heights also have the same ratio, and since the two heights add up to $4$ , we have that $h_{ADE} = 4 cdot frac{4}{7} = frac{16}{7}$ and $h_{CDB} = 3 cdot frac 47 = frac {12}7$ . Subtracting the areas, $frac{1}{2} cdot 8 cdot frac {16}7 - frac 12 cdot 6 cdot frac{12}7 = 4$ $Rightarrow$ $boxed{mathrm{(B)} 4}$ .Let $[X]$ represent the area of figure $X$ . Note that $[triangle BEA]=[triangle ABD]+[triangle ADE]$ and $[triangle BCA]=[triangle ABD]+[triangle BDC]$ . $[triangle ADE]-[triangle BDC]=[triangle BEA]-[triangle BCA]=frac{1}{2}times8times4-frac{1}{2}times6times4= 16-12=4Rightarrowboxed{mathrm{(B)} 4}$ Put figure $ABCDE$ on a graph. $overline{AC}$ goes from (0, 0) to (4, 6) and $overline{BE}$ goes from (4, 0) to (0, 8). $overline{AC}$ is on line $y = 1.5x$ . $overline{BE}$ is on line $y = -2x + 8$ . Finding intersection between these points, $1.5x = -2x + 8$ . $3.5x = 8$

$x = 8 times frac{2}{7}$

$= frac{16}{7}$

This gives us the x-coordinate of D. So, $frac{16}{7}$ is the height of $triangle ADE$ , then area of $triangle ADE$ is $frac{16}{7} times 8 times frac{1}{2}$ $= frac{64}{7}$

Now, the height of $triangle BDC$ is $4-frac{16}{7} = frac{12}{7}$ And the area of $triangle BDC$ is $6 times frac{12}{7} times frac{1}{2} = frac{36}{7}$

This gives us $frac{64}{7} - frac{36}{7} = 4$

There are $4$ ways that the same number of heads will be obtained; $0$ , $1$ , $2$ , or $3$ heads.The probability of both getting $0$ heads is $left(frac12right)^3{3choose0}left(frac12right)^4{4choose0}=frac1{128}$ The probability of both getting $1$ head is $left(frac12right)^3{3choose1}left(frac12right)^4{4choose1}=frac{12}{128}$ The probability of both getting $2$ heads is $left(frac12right)^3{3choose2}left(frac12right)^4{4choose2}=frac{18}{128}$ The probability of both getting $3$ heads is $left(frac12right)^3{3choose3}left(frac12right)^4{4choose3}=frac{4}{128}$ Therefore, the probabiliy of flipping the same number of heads is: $frac{1+12+18+4}{128}=frac{35}{128}Rightarrowboxed{mathrm{(D)} frac{35}{128}}$

When the diameter is increased by $25%$ , it is increased by $dfrac{5}{4}$ , so the area of the base is increased by $left(dfrac54right)^2=dfrac{25}{16}$ .To keep the volume the same, the height must be $dfrac{1}{frac{25}{16}}=dfrac{16}{25}$ of the original height, which is a $36%$ reduction. $boxed{mathrm{(C)} 36}$

2004 AMC 12A-12.png The equation of $overline{AA'}$ can be found using points $A, C$ to be $y - 9 = left(frac{9-8}{0-2}right)(x - 0) Longrightarrow y = -frac{1}{2}x + 9$ . Similarily, $overline{BB'}$ has the equation $y - 12 = left(frac{12-8}{0-2}right)(x-0) Longrightarrow y = -2x + 12$ . These two equations intersect the line $y=x$ at $(6,6)$ and $(4,4)$ . Using the distance formula or $45-45-90$ right triangles, the answer is $2sqrt{2} mathrm{(B)}$ .

Let's count them by cases:

Case 1: The line is horizontal or vertical, clearly $3 cdot 2 = 6$ .
Case 2: The line has slope $pm 1$ , with $2$ through $(0,0)$ and $4$ additional ones one unit above or below those. These total $6$ .
Case 3: The only remaining lines pass through two points, a vertex and a non-vertex point on the opposite side. Thus we have each vertex pairing up with two points on the two opposites sides, giving $4 cdot 2 = 8$ lines.

These add up to $6+6+8=20 mathrm{(B)}$ .

There are ${9 choose 2} = 36$ ways to pick two points, but we've clearly overcounted all of the lines which pass through three points. In fact, each line which passes through three points will have been counted ${3 choose 2} = 3$ times, so we have to subtract $2$ for each of these lines. Quick counting yields $3$ horizontal, $3$ vertical, and $2$ diagonal lines, so the answer is $36 - 2(3+3+2) = 20$ distinct lines.

First consider how many lines go through $(-1, -1)$ and hit two points in $S$ . You can see that there are $5$ such lines. Now, we cross out $(-1, -1)$ and make sure to never consider consider lines that go through it anymore (As doing so would be double counting). Repeat for $(-1, 0)$ , making sure not to count the vertical line as it goes through the crossed out $(-1, -1)$ . Then cross out $(-1, 0)$ . Repeat for the rest, and count $20$ lines in total.

Let $d$ be the common difference. Then $9$ , $9+d+2=11+d$ , $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and the third term is $2(-14) + 29 = 1Rightarrowboxed{mathrm{(A)} 1}$ .Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ , $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ , $11+d$ , and $29+2d$ . Thus, we get the following equations: $9r=11+dRightarrow d=9r-11$ $9r^2=29+2d$ Plugging in the first equation into the second, our equation becomes $9r^2=29+18r-22Longrightarrow9r^2-18r-7=0$ . By the quadratic formula, $r$ can either be $-frac{1}{3}$ or $frac{7}{3}$ . If $r$ is $-frac{1}{3}$ , the third term (of the geometric sequence) would be $1$ , and if $r$ is $frac{7}{3}$ , the third term would be $49$ . Clearly the minimum possible value for the third term of the geometric sequence is $boxed{mathrm{(A)} 1}$ .Let the three numbers be, in increasing order, $z,y,9$

Hence, we have that $9-y=y-zimplies 9+z=2y$ .

Also, from the second part of information given, we get that

$frac{9}{y+2}=frac{y+2}{z+20}implies 9(z+20)=(y+2)^2implies y=3(sqrt{z+20})-2$

Plugging back in..

$9+z=6(sqrt{z+20})-4implies (9+z)^2=36(z+20)$

Simplifying, we get that $z^2-10z-551=0$

Applying the quadratic formula, we get that $z=frac{10pm sqrt{2304}}{2}implies frac{10pm48}{2}$

Obviously, in order to minimize the value of $z$ , we have to subtract. Hence, $z=-19$

However, the problem asks for the minimum value of the third term in a geometric progression.

Hence, the answer is $-19+20=boxed{1} implies boxed{A}$

Rewrite $frac{(x+y)}x$ as $frac{x}x+frac{y}x=1+frac{y}x$ .We also know that $frac{y}x<0$ because $x$ and $y$ are of opposite sign.Therefore, $1+frac{y}x$ is maximized when $|frac{y}x|$ is minimized, which occurs when $|x|$ is the largest and $|y|$ is the smallest.This occurs at $(-4,2)$ , so $frac{x+y}x=1-frac12=frac12Rightarrowboxed{mathrm{(D)} frac{1}{2}}$ .

Since there are five types of squares: $1 times 1, 2 times 2, 3 times 3, 4 times 4,$ and $5 times 5.$ We must find how many of each square contain the black shaded square in the center.If we list them, we get that

There is $1$ of all $1times 1$ squares, containing the black square
There are $4$ of all $2times 2$ squares, containing the black square
There are $9$ of all $3times 3$ squares, containing the black square
There are $4$ of all $4times 4$ squares, containing the black square
There is $1$ of all $5times 5$ squares, containing the black square

Thus, the answer is $1+4+9+4+1=19Rightarrowboxed{mathrm{(D)} 19}$ .

We use complementary counting. There are only $2times2$ and $1times1$ squares that do not contain the black square. Counting, there are $12$ $2times2$ , and $25-1 = 24$ $1times1$ squares that do not contain the black square. That gives $12+24=36$ squares that don't contain it. There are a total of $25+16+9+4+1 = 55$ squares possible, therefore there are $55-36 = 19$ squares that contains the black square, which is $boxed{mathrm{(D)} 19}$

Call the length of the race track $x$ . When they meet at the first meeting point, Brenda has run $100$ meters, while Sally has run $frac{x}{2} - 100$ meters. By the second meeting point, Sally has run $150$ meters, while Brenda has run $x - 150$ meters. Since they run at a constant speed, we can set up a proportion: $frac{100}{x- 150} = frac{frac{x}{2} - 100}{150}$ . Cross-multiplying, we get that $x = 350Longrightarrowboxed{mathrm{(C)} 350}$ .Sidenote by carlos8:Since they run at constant speeds, Brenda must've ran 200 meters to get to the second meeting point, therefore we can make an equation $200 = x-150$ , solving for $x$ , gives us our answer $350$ .The total distance the girls run between the start and the first meeting is one half of the track length. The total distance they run between the two meetings is the track length. As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting. Thus between the meetings Brenda will run $2times100=200$ meters. Therefore the length of the track is $150 + 200 = 350$ meters $Rightarrowboxed{mathrm{(C)} 350}$

Hence, we have that $9-y=y-zimplies 9+z=2y$ .

Also, from the second part of information given, we get that

$frac{9}{y+2}=frac{y+2}{z+20}implies 9(z+20)=(y+2)^2implies y=3(sqrt{z+20})-2$

Plugging back in..

$9+z=6(sqrt{z+20})-4implies (9+z)^2=36(z+20)$

Simplifying, we get that $z^2-10z-551=0$

Applying the quadratic formula, we get that $z=frac{10pm sqrt{2304}}{2}implies frac{10pm48}{2}$

Obviously, in order to minimize the value of $z$ , we have to subtract. Hence, $z=-19$

However, the problem asks for the minimum value of the third term in a geometric progression.

Hence, the answer is $-19+20=boxed{1} implies boxed{A}$

The cylinder can be "unwrapped" into a rectangle, and we see that the stripe is a parallelogram with base $3$ and height $80$ . Thus, we get $3times80=240Rightarrowboxed{mathrm{(C)} 240}$

Since triangle $BEF$ is equilateral, $EA=FC$ , and $EAB$ and $FCB$ are $SAS$ congruent. Thus, triangle $DEF$ is an isosceles right triangle. So we let $DE=x$ . Thus $EF=EB=FB=xsqrt{2}$ . If we go angle chasing, we find out that $angle AEB=75^{circ}$ , thus $angle ABE=15^{circ}$ . $frac{AE}{EB}=sin{15^{circ}}=frac{sqrt{6}-sqrt{2}}{4}$ . Thus $frac{AE}{xsqrt{2}}=frac{sqrt{6}-sqrt{2}}{4}$ , or $AE=frac{x(sqrt{3}-1)}{2}$ . Thus $AB=frac{x(sqrt{3}+1)}{2}$ , and $[ABE]=frac{x^2}{4}$ , and $[DEF]=frac{x^2}{2}$ . Thus the ratio of the areas is $boxed{mathrm{(D)} 2}$ WLOG, let the side length of $ABCD$ be 1. Let $DE = x$ . It suffices that $AE = 1 - x$ . Then triangles $ABE$ and $CBF$ are congruent by HL, so $CF = AE$ and $DE = DF$ . We find that $BE = EF = x sqrt{2}$ , and so, by the Pythagorean Theorem, we have $(1 - x)^2 + 1 = 2x^2.$ This yields $x^2 + 2x = 2$ , so $x^2 = 2 - 2x$ . Thus, the desired ratio of areas is $[frac{frac{x^2}{2}}{frac{1-x}{2}} = frac{x^2}{1 - x} = boxed{text{(D) }2}.]$ $bigtriangleup BEF$ is equilateral, so $angle EBF = 60^{circ}$ , and $angle EBA = angle FBC$ so they must each be $15^{circ}$ . Then let $BE=EF=FB=1$ , which gives $EA=sin{15^{circ}}$ and $AB=cos{15^{circ}}$ . The area of $bigtriangleup ABE$ is then $frac{1}{2}sin{15^{circ}}cos{15^{circ}}=frac{1}{4}sin{30^{circ}}=frac{1}{8}$ . $bigtriangleup DEF$ is an isosceles right triangle with hypotenuse 1, so $DE=DF=frac{1}{sqrt{2}}$ and therefore its area is $frac{1}{2}left(frac{1}{sqrt{2}}cdotfrac{1}{sqrt{2}}right)=frac{1}{4}$ . The ratio of areas is then $frac{frac{1}{4}}{frac{1}{8}}=framebox{(D) 2}$

Let the area of the shaded region be $S$ , the area of the unshaded region be $U$ , and the acute angle that is formed by the two lines be $theta$ . We can set up two equations between $S$ and $U$ : $S+U=9pi$ $S=dfrac{8}{13}U$ Thus $dfrac{21}{13}U=9pi$ , and $U=dfrac{39pi}{7}$ , and thus $S=dfrac{8}{13}cdot dfrac{39pi}{7}=dfrac{24pi}{7}$ .Now we can make a formula for the area of the shaded region in terms of $theta$ : $dfrac{2theta}{2pi} cdot pi +dfrac{2(pi-theta)}{2pi} cdot (4pi-pi)+dfrac{2theta}{2pi}(9pi-4pi)=theta +3pi-3theta+5theta=3theta+3pi=dfrac{24pi}{7}$

Thus $3theta=dfrac{3pi}{7}Rightarrow theta=dfrac{pi}{7}Rightarrowboxed{mathrm{(B)} frac{pi}{7}}$

As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of $theta$ .

$impliesdfrac{2theta}{2pi} cdot pi +dfrac{2(pi-theta)}{2pi} cdot (4pi-pi)+dfrac{2theta}{2pi}(9pi-4pi)=theta +3pi-3theta+5theta=3theta+3pi$ .

So, the shaded region is $3theta+3pi$ . This means that the unshaded region is $9pi-(3theta+3pi)$ .

Also, the shaded region is $frac{8}{13}$ of the unshaded region. Hence, we can now make an equation and solve for theta!

$3theta+3pi=frac{8}{13}(9pi-(3theta+3pi)implies 39theta+39pi=8(6pi-3theta)implies 39theta+39pi=48pi-24theta$ .

Simplifying, we get $63theta=9piimplies theta=boxed{mathrm{(B)} frac{pi}{7}}$

$[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2; draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("$A$",A,(-1,-1)); label("$B$",B,( 1,-1)); label("$C$",C,( 1, 1)); label("$D$",D,(-1, 1)); label("$E$",E,(-1, 0)); label("$F$",F,( 0, 1)); label("$x$",(A+E)/2,(-1, 0)); label("$x$",(E+F)/2,( 0, 1)); label("$2$",(F+C)/2,( 0, 1)); label("$2$",(D+C)/2,( 0, 1)); label("$2$",(B+C)/2,( 1, 0)); label("$2-x$",(D+E)/2,(-1, 0)); [/asy]$ Let the point of tangency be $F$ . By the Two Tangent Theorem $BC = FC = 2$ and $AE = EF = x$ . Thus $DE = 2-x$ . The Pythagorean Theorem on $triangle CDE$ yields $begin{align*} DE^2 + CD^2 &= CE^2\ (2-x)^2 + 2^2 &= (2+x)^2\ x^2 - 4x + 8 &= x^2 + 4x + 4\ x &= frac{1}{2}end{align*}$ Hence $CE = FC + x = frac{5}{2} Rightarrowboxed{mathrm{(D)} frac{5}{2}}$ .Call the point of tangency point $F$ and the midpoint of $AB$ as $G$ . $CF=2$ by Tangent Theorem. Notice that $angle EGF=frac{180-2cdotangle CGF}{2}=90-angle CGF$ . Thus, $angle EGF=angle FCG$ and $tanEGF=tanFCG=frac{1}{2}$ . Solving $EF=frac{1}{2}$ . Adding, the answer is $frac{5}{2}$ . 2004 AMC12A-18.png Clearly, $EA = EF = BG$ . Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$ , $CE = 5/2$ .

$[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("$A$",A,(-1,-1)); label("$B$",B,( 1,-1)); label("$C$",C,( 1, 1)); label("$D$",D,(-1, 1)); label("$E$",E,(-1, 0)); label("$F$",F,( 0, 1)); label("$x$",(A+E)/2,(-1, 0)); label("$x$",(E+F)/2,( 0, 1)); label("$2$",(F+C)/2,( 0, 1)); label("$2$",(D+C)/2,( 0, 1)); label("$2$",(B+C)/2,( 1, 0)); label("$2-x$",(D+E)/2,(-1, 0)); label("$G$",G,(0,-1)); dot(G); draw(G--C); label("$sqrt{5}$",(G+C)/2,(-1,0)); [/asy]$

Let us call the midpoint of side $AB$ , point $G$ . Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$ . We get $GC=sqrt{5}$ . We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$ , we get similar triangles $EFG$ and $GFC$ . Solving the ratios, we get $x=frac{1}{2}$ , so the answer is $frac{5}{2} Rightarrowboxed{mathrm{(D)} frac{5}{2}}$ .

Using the diagram as drawn in Solution 5, let the total area of square $ABCD$ be divided into the triangles $DCE$ , $EAG$ , $CGB$ , and $EGC$ . Let x be the length of AE. Thus, the area of each triangle can be determined as follows:

$[DCE = frac{DCcdot{DE}}{2} = frac{2cdot(2-x)}{2} = 1-x]$

$[EAG= frac{AEcdot{AG}}{2} = frac{1cdot{x}}{2} = frac{x}{2}]$

$[CGB = frac{GBcdot{CB}}{2} = frac{1cdot(2)}{2} = 1]$

$[EGC= frac{EGcdot{GC}}{2} = frac{sqrt{4+(2-x)^2}}{2}]$ (the length of CE is calculated with the Pythagorean Theorem, lines GE and CE are perpendicular by definition of tangent)

Adding up the areas and equating to the area of the total square (2*2=4), we get

$[1-x+frac{x}{2}+1+ frac{sqrt{4+(2-x)^2}}{2} = 4]$

Solving for x:

$[2-2x+x+1+sqrt{x^2-4x+8}=8]$ $[sqrt{x^2-4x+8}=x+2]$ $[x^2-4x+8=x^2+4x+4]$ $[8-4x=4x+4 rightarrow x=frac{1}{2}]$

Solving for length of CE with the value we have for x: $[sqrt{4+(2-x)^2} = sqrt{4+(3/2)^2} = sqrt{25/4} = boxed{frac{5}{2}}]$

$[asy] import graph; size(400); defaultpen(fontsize(10)); pair OA=(-1,0),OB=(2/3,8/9),OC=(2/3,-8/9),OD=(0,0),E=(2/3,0); real t = 2.5; pair OA1=(-2+2*t,0),OB1=(4/3+2*t,16/9),OC1=(4/3+2*t,-16/9),OD1=(0+2*t,0),E1=(4/3+2*t,0); draw(Circle(OD,2)); draw(Circle(OA,1)); draw(Circle(OB,8/9)); draw(Circle(OC,8/9)); draw(OA--OB--OC--cycle); draw(OD--OB--OB+(OB-OD)*4/5); draw(OA--E); label("$O_{A}$",OA,(-1,0)); label("$O_{B}$",OB,(-1,1)); label("$O_{C}$",OC,(-1,-1)); label("$O_{D}$",OD,(-1,-1)); label("$E$",E,(0.5,-1)); label("$r$",OB+(OB-OD)*2/5,(-0.5,1)); label("$r$",(1*OA+3*OB)/4,(-0.5,1)); dot(OA^^OB^^OC^^OD^^E); draw(OA1--OB1--OC1--cycle); draw(OD1--OB1); draw(OA1--E1); label("$O_{A}$",OA1,(-1,0)); label("$O_{B}$",OB1,(1,1)); label("$O_{C}$",OC1,(1,-1)); label("$O_{D}$",OD1,(0,-1)); label("$E$",E1,(1,0)); label("$1+r$",(OA1+OB1)/2,(-0.5,1)); label("$r$",(E1+OB1)/2,(1,0)); label("$r$",(E1+OC1)/2,(1,0)); label("$2-r$",(OB1+OD1)/2,(-1,0)); label("$1$",(OA1+OD1)/2,(0,-1)); label("$x$",(E1+OD1)/2,(0,-1)); dot(OA1^^OB1^^OC1^^OD1^^E1); [/asy]$ Let $O_{i}$ be the center of circle $i$ for all $i in {A,B,C,D}$ and let $E$ be the tangent point of $B,C$ . Since the radius of $D$ is the diameter of $A$ , the radius of $D$ is $2$ . Let the radius of $B,C$ be $r$ and let $O_{D}E = x$ . If we connect $O_{A},O_{B},O_{C}$ , we get an isosceles triangle with lengths $1 + r, 2r$ . Then right triangle $O_{D}O_{B}E$ has legs $r, x$ and hypotenuse $2-r$ . Solving for $x$ , we get $x^2 = (2-r)^2 - r^2 Longrightarrow x = sqrt{4-4r}$ .Also, right triangle $O_{A}O_{B}E$ has legs $r, 1+x$ , and hypotenuse $1+r$ . Solving, $begin{eqnarray*} r^2 + (1+sqrt{4-4r})^2 &=& (1+r)^2\ 1+4-4r+2sqrt{4-4r}&=& 2r + 1\ 1-r &=& left(frac{6r-4}{4}right)^2\ frac{9}{4}r^2-2r&=& 0\ r &=& frac 89 end{eqnarray*}$ So the answer is $boxed{mathrm{(D)} frac{8}{9}}$ .Using Descartes' Circle Formula, $left(1-frac{1}{2}+frac{1}{r}+frac{1}{r}right)^2=2left(frac{1}{4}+1+frac{1}{r^2}+frac{1}{r^2}right)$ . Solving this gives us linear equation with $r=boxed{mathrm{(D)} frac{8}{9}}$ .

$f(2^{100}) = f(2 times 2^{99}) = 2^{99} times f(2^{99})$ $= 2^{99} cdot 2^{98} times f(2^{98}) = ldots$ $= 2^{99}2^{98}cdots 2^{1} cdot 1 cdot f(1)$ $= 2^{99 + 98 + ldots + 2 + 1}$ $= 2^{frac{99(100)}{2}} = 2^{4950}$ $Rightarrow mathrm{(D)}$ .

We draw the three spheres of radius $1$ : 2004AMC12A 22 1 rerender.png And then add the sphere of radius $2$ : 2004AMC12A 22 2 rerender.png The height from the center of the bottom sphere to the plane is $1$ , and from the center of the top sphere to the tip is $2$ . 2004 AMC12A-22a.png

We now need the vertical height of the centers. If we connect the centers, we get a triangular pyramid with an equilateral triangle base. The distance from the vertex of the equilateral triangle to its centroid can be found by $30-60-90 triangle$ s to be $frac{2sqrt{3}}{3}$ .

2004 AMC12A-22b.png

By the Pythagorean Theorem, we have $left(frac{2}{sqrt{3}}right)^2 + h^2 = 3^2 Longrightarrow h = frac{sqrt{69}}{3}$ . Adding the heights up, we get $frac{sqrt{69}}{3} + 1 + 2 Rightarrowboxed{mathrm{(B)} 3+frac{sqrt{69}}{3}}$

Connect the centers of the spheres. Note that the resulting prism is a tetrahedron with base lengths of 2 and side lengths of 3. Drop a height from the top of the tetrahedron to the centroid of its equilateral triangle base. Using the Pythagorean Theorem, it is easy to see that the circumradius of the base is $dfrac { 2 sqrt {3 } } {3}$ . We can use PT again to find the height of the tetrahedron given its base's circumradius and it's leg lengths. Finally, we add the distance from the top of the tetrahedron to the top of the sphere of radius 2 and the distance from the bottom of the prism to the ground to get an answer of $3 + dfrac{sqrt{69}}{3}$ .