AMC12 2004 A

$20$ dollars is the same as $2000$ cents, and $1.45\%$ of $2000$ is $0.0145\times2000=29$ cents. $\Rightarrow\boxed{\mathrm{(E)}\ 29}$. Since there can't be decimal values of cents, the answer must be $\Rightarrow\boxed{\mathrm{(E)}\ 29}$.

She gets $8*2.5=20$ points for the problems she didn't answer. She must get $\left\lceil \frac{100-20}{6} \right\rceil =14 \Rightarrow \text {(C)}$ problems right to score at least 100.

Every integer value of $y$ leads to an integer solution for $x$ Since $y$ must be positive, $y\geq 1$ Also, $y = \frac{100-x}{2}$ Since $x$ must be positive, $y < 50$ $1 \leq y < 50$ This leaves $49$ values for y, which mean there are $49$ solutions to the equation $\Rightarrow \mathrm{(B)}$

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. $\boxed{\mathrm{(E)}\ 26}$

The line appears to have a slope of $-\dfrac{1}{2}$ and y-intercept of $\dfrac{4}{5}$ up. $\dfrac{4}{5}\cdot \dfrac{-1}{2}=\dfrac{-4}{10} \Rightarrow \mathrm {(B)}$

\begin{eqnarray*} U-V&=&2004*2004^{2004}\\ V-W&=&1*2004^{2004}\\ W-X&=&2001*2004^{2004}\\ X-Y&=&1*2004^{2004}\\ Y-Z&=&2003*2004^{2003} \end{eqnarray*} After comparison, $U-V$ is the largest. $\mathrm {(A)}$

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 them empty-handed, and thus the game will end. We then have there are $36+1=\boxed{\mathrm{(B)}\ 37}$ rounds until the game ends.

Looking, we see that the area of $[\triangle EBA]$ is 16 and the area of $[\triangle ABC]$ is 12. Set the area of $[\triangle ADB]$ to be x. We want to find $[\triangle ADE]$ - $[\triangle CDB]$. So, that would be $[\triangle EBA]-[\triangle ADB]=16-x$ and $[\triangle ABC]-[\triangle ADB]=12-x$. Therefore, $[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{\mathrm{(B)}\ 4}$

When the diameter is increased by $25\%$, it is increased by $\dfrac{5}{4}$, so the area of the base is increased by $\left(\dfrac54\right)^2=\dfrac{25}{16}$. 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. $\boxed{\mathrm{(C)}\ 36}$

The median of a sequence is the middle number of the sequence when the sequence is arranged in order. Since the integers are consecutive, the median is also the mean, so the median is $\frac{7^5}{49} = 7^3\ \mathrm{(C)}$.

Let the total value, in cents, of the coins Paula has originally be $v$, and the number of coins she has be $n$. Then $\frac{v}{n}=20\Longrightarrow v=20n$ and $\frac{v+25}{n+1}=21$. Substituting yields: $20n+25=21(n+1),$ so $n=4$, $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$ quarters and $1$ nickel in her purse. Thus, she has $\boxed{\mathrm{(A)}\ 0}$ dimes.

Error creating thumbnail: Unable to save thumbnail to destination 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 $2\sqrt{2}\ \mathrm{(B)}$.

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 $2(-14) + 29 = 1\Rightarrow\boxed{\mathrm{(A)}\ 1}$.

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 $x = 350\Longrightarrow\boxed{\mathrm{(C)}\ 350}$.

For all real numbers $a,b,$ and $c$ such that $b>1,$ note that: 1. $\log_b a$ is defined if and only if $a>0.$ 2. $\log_b a>c$ if and only if $a>b^c.$ Therefore, we have \begin{align*} \log_{2004}(\log_{2003}(\log_{2002}(\log_{2001}{x}))) \text{ is defined} &\implies \log_{2003}(\log_{2002}(\log_{2001}{x}))>0 \\ &\implies \log_{2002}(\log_{2001}{x})>1 \\ &\implies \log_{2001}{x}>2002 \\ &\implies x>2001^{2002}, \end{align*} from which $c=\boxed{\textbf {(B) }2001^{2002}}.$

From (ii), note that \begin{alignat*}{8} f(2) &= 1\cdot f(1) &&= 1, \\ f\left(2^2\right) &= 2\cdot f(2) &&= 2, \\ f\left(2^3\right) &= 2^2\cdot f\left(2^2\right) &&= 2^{2+1}, \\ f\left(2^4\right) &= 2^3\cdot f\left(2^3\right) &&= 2^{3+2+1}, \end{alignat*} and so on. In general, we have \[f\left(2^n\right)=2^{(n-1)+(n-2)+(n-3)+\cdots+3+2+1}\] for any positive integer $n.$ Therefore, the answer is \begin{align*} f\left(2^{100}\right)&=2^{99+98+97+\cdots+3+2+1} \\ &=2^{99\cdot100/2} \\ &= \boxed{\textbf {(D)}\ 2^{4950}}. \end{align*}

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 $CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.

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 $O_{D}O_{B}O_{E}$ 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 $O_{A}O_{B}O_{E}$ 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+2\sqrt{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 $\boxed{\mathrm{(D)}\ \frac{8}{9}}$.

Casework: 1. $0 + 0 = 0$. The probability that $a < \frac{1}{2}$ and $b < \frac{1}{2}$ is $\left(\frac 12\right)^2 = \frac{1}{4}$. Notice that the sum $a+b$ ranges from $0$ to $1$ with a symmetric distribution across $a+b=c=\frac 12$, and we want $c < \frac 12$. Thus the chance is $\frac{\frac{1}{4}}2 = \frac 18$. 2. $0 + 1 = 1$. The probability that $a < \frac 12$ and $b > \frac 12$ is $\frac 14$, but now $\frac{1}{2} < a+b = c < \frac 32$, which makes $C = 1$ automatically. Hence the chance is $\frac 14$. 3. $1 + 0 = 1$. This is the same as the previous case. 4. $1 + 1 = 2$. We recognize that this is equivalent to the first case. Our answer is $2\left(\frac 18 + \frac 14 \right) = \frac 34 \Rightarrow \mathrm{(E)}$. Use areas to deal with this continuous probability problem. Set up a unit square with values of $a$ on x-axis and $b$ on y-axis. If $a + b < 1/2$ then this will work because $A = B = C = 0$. Similarly if $a + b > 3/2$ then this will work because in order for this to happen, $a$ and $b$ are each greater than $1/2$ making $A = B = 1$, and $C = 2$. Each of these triangles in the unit square has area of 1/8. The only case left is when $C = 1$. Then each of $A$ and $B$ must be 1 and 0, in any order. These cut off squares of area 1/2 from the upper left and lower right corners of the unit square. Then the area producing the desired result is 3/4. Since the area of the unit square is 1, the probability is $\frac 34$.

This is an infinite geometric series, which sums to $\frac{\cos^0 \theta}{1 - \cos^2 \theta} = 5 \Longrightarrow 1 = 5 - 5\cos^2 \theta \Longrightarrow \cos^2 \theta = \frac{4}{5}$. Using the formula $\cos 2\theta = 2\cos^2 \theta - 1 = 2\left(\frac 45\right) - 1 = \frac 35 \Rightarrow \mathrm{(D)}$. We can more directly solve this with superficially less work. Again, applying the formula for an infinite geometric series, \[\sum_{i=0}^{\infty}\cos^{2i}\theta=\dfrac1{1-\cos^2\theta}=\dfrac1{\sin^2\theta}=5.\] Thus, $\sin^2\theta=\dfrac15$, so $\cos(2\theta)=1-2\sin^2\theta=1-\dfrac25=\dfrac35.$

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 $2$. Error creating thumbnail: Unable to save thumbnail to destination We now need the vertical height of the centers. If we connect centers, we get a rectangular pyramid with an equilateral triangle base. The distance from the vertex of the equilateral triangle to its centroid can be found by $30\text-60\text-90 \triangle$s to be $\frac{2\sqrt{3}}{3}$. Error creating thumbnail: Unable to save thumbnail to destination 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 $\frac{\sqrt{69}}{3} + 1 + 2 \Rightarrow 3+\frac{\sqrt{69}}{3}$, or $\boxed{B}$

We have to evaluate the answer choices and use process of elimination: - $\mathrm{(A)}$: We are given that $a_1 = b_1 = 0$, so $z_1 = 0$. If one of the roots is zero, then $P(0) = c_0 = 0$. - $\mathrm{(B)}$: By Vieta's formulas, we know that $-\frac{c_{2003}}{c_{2004}}$ is the sum of all of the roots of $P(x)$. Since that is real, $\sum_{k = 1}^{2004}{b_k}=0=\sum_{k = 1}^{2004}{a_k}$, and $\frac{c_{2003}}{c_{2004}}=0$, so $c_{2003}=0$. - $\mathrm{(C)}$: All of the coefficients are real. For sake of contradiction suppose none of $b_{2\ldots 2004}$ are zero. Then for each complex root $z_k$, its complex conjugate $\overline{z_k} = a_k - b_k i$ is also a root. So the roots should pair up, but we have an odd number of imaginary roots! (Remember that $b_1 = 0$.) This gives us the contradiction, and therefore the product is equal to zero. - $\mathrm{(D)}$: We are given that $\sum_{k = 1}^{2004}{a_k} = \sum_{k = 1}^{2004}{b_k}$. Since the coefficients are real, it follows that if a root is complex, its conjugate is also a root; and the sum of the imaginary parts of complex conjugates is zero. Hence the RHS is zero. There is, however, no reason to believe that $\boxed{\mathrm{E}}$ should be zero (in fact, that quantity is $P(1)$, and there is no evidence that $1$ is a root of $P(x)$).

As the red circles move about segment $AB$, they cover the area we are looking for. On the left side, the circle must move around pivoted on $B$. On the right side, the circle must move pivoted on $A$ However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds. This egg-like shape is $S$. The area of the region can be found by dividing it into several sectors, namely \begin{align*} A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\ A &= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi (1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\ A &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\ A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}

This is an infinite geometric series with common ratio $\frac{1}{x^3}$ and initial term $x^{-1} + 3x^{-2} + 3x^{-3}$, so $a_x = \left(\frac{1}{x} + \frac{3}{x^2} + \frac{3}{x^3}\right)\left(\frac{1}{1-\frac{1}{x^3}}\right)$ $= \frac{x^2 + 3x + 3}{x^3} \cdot \frac{x^3}{x^3 - 1}$ $= \frac{x^2 + 3x + 3}{x^3 - 1}$ $= \frac{(x+1)^3 - 1}{x(x^3 - 1)}$. Alternatively, we could have used the algebraic manipulation for repeating decimals, \begin{align*} a_x &= \frac{1}{x}+\frac{3}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}+\frac{3}{x^5}+\frac{3}{x^6}+\cdots \\ a_x \cdot x^3 &= x^2+3x+3+a_x\\ a_x(x^3-1) &= x^2+3x+3\\ a_x &= \frac{x^2+3x+3}{x^3-1}=\frac{(x+1)^3-1}{x(x^3-1)} \end{align*} Telescoping, \begin{align*} a_4a_5\cdots a_{99}&= \frac{(5^3-1)(6^3-1)\cdots (100^3-1)}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot (4^3-1)(5^3-1)\cdots(99^3-1)}\\ a_4a_5\cdots a_{99}&= \frac{999999}{4 \cdot 5 \cdot 6 \cdot \cdots \cdot 99 \cdot 63}=\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}\end{align*} Some factors cancel, (after all, $13 \cdot 37 \cdot 33 \cdot 6$ isn't one of the answer choices) \[\frac{13 \cdot 37 \cdot 33 \cdot 6}{99!}=\frac{13 \cdot 37 \cdot 2}{98!}\] Since the only factor in the numerator that goes into $98$ is $2$, $n$ is minimized. Therefore the answer is $13 \cdot 37 \cdot 2=962 \Rightarrow \text {(E)}$.