AMC12 2003 B

\[\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}=\frac{2+(-4+6)+(-8+10)+(-12+14)}{3+(-6+9)+(-12+15)+(-18+21)}=\frac{2\cdot4}{3\cdot4}=\text {(C) } \frac{2}{3}\]

Because there are $14$ days in two weeks, Al spends $546/14 = 39$ dollars per day for the cost of a green pill and a pink pill. If the green pill costs $x$ dollars and the pink pill $x-1$ dollars, the sum of the two costs $2x-1$ should equal $39$ dollars. Then the cost of the green pill $x$ is $\boxed{\textbf{(D) }\$20}$.

The areas of the five regions from greatest to least are $21,20,15,6$ and $4$. If we want to minimize the cost, we want to maximize the area of the cheapest flower and minimize the area of the most expensive flower. Doing this, the cost is $1\cdot21+1.50\cdot20+2\cdot15+2.50\cdot6+3\cdot4$, which simplifies to $\$$108$. Therefore the answer is $\boxed{\textbf{(A) } 108}$.

If you divide the television screen into two right triangles, the legs are in the ratio of $4 : 3$, and we can let one leg be $4x$ and the other be $3x$. Then we can use the Pythagorean Theorem. \begin{align*}(4x)^2+(3x)^2&=27^2\\ 16x^2+9x^2&=729\\ 25x^2&=729\\ x^2&=\frac{729}{25}\\ x&=\frac{27}{5}\\ x&=5.4\end{align*} The horizontal length is $5.4\times4=21.6$, which is closest to $\boxed{\textbf{(D) \ } 21.5}$.

Let the first term be $a$ and the common ratio be $r$. Therefore, \[ar=2\ \ (1) \qquad \text{and} \qquad ar^3=6\ \ (2)\] Dividing $(2)$ by $(1)$ eliminates the $a$, yielding $r^2=3$, so $r=\pm\sqrt{3}$. Now, since $ar=2$, $a=\frac{2}{r}$, so $a=\frac{2}{\pm\sqrt{3}}=\pm\frac{2\sqrt{3}}{3}$. We therefore see that $\boxed{\textbf{(B)}\ -\frac{2\sqrt{3}}{3}}$ is a possible first term.

Where $a,b,c$ is the number of nickels, dimes, and quarters, respectively, we can set up two equations: \[(1)\ 5a+10b+25c=835\ \ \ \ (2)\ a+b+c=100\] Eliminate $a$ by subtracting $(2)$ from $(1)/5$ to get $b+4c=67$. Of the integer solutions $(b,c)$ to this equation, the number of dimes $b$ is least in $(3,16)$ and greatest in $(67,0)$, yielding a difference of $67-3=\boxed{\textbf{(D)}\ 64}$.

Let $y=\clubsuit (x)$. Since $x \leq 99$, we have $y \leq 18$. Thus if $\clubsuit (y)=3$, then $y=3$ or $y=12$. The 3 values of $x$ for which $\clubsuit (x)=3$ are 12, 21, and 30, and the 7 values of $x$ for which $\clubsuit (x)=12$ are 39, 48, 57, 66, 75, 84, and 93. There are $\boxed{E}$ = $\boxed{10}$ values in all.

Since $f$ is a linear function with slope $m$, \[m = \frac{f(6) - f(2)}{\Delta x} = \frac{12}{6 - 2} = 3\] \[f(12) - f(2) = m \Delta x = 3(12 - 2) = 30 \Rightarrow \text (D)\]

Place the first triangle. Now, we can place the second triangle either adjacent to the first, or with one side between them, for a total of $\boxed{\text{(B) }2}$

For every $60$ minutes that pass by in actual time, $57+\frac{36}{60}=57.6$ minutes pass by on Cassandra's watch. When her watch first reads, 10:00 pm, $10(60)=600$ minutes have passed by on her watch. Setting up a proportion, \[\frac{57.6}{60}=\frac{600}{x}\] where $x$ is the number of minutes that have passed by in actual time. Solve for $x$ to get $625$ minutes, or $10$ hours and $25$ minutes $\Rightarrow \boxed{\textbf{(C)}\ \text{10:25 PM}}$.

For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is $3 \cdot 5 = 15$, so ${\boxed{\textbf{(D)15}}}$ is the correct answer.

Let $r$ be the common radius of the sphere and the cone, and $h$ be the cone’s height. Then \[75\% \cdot \left(\frac 43 \pi r^3\right) = \frac 13 \pi r^2 h \Longrightarrow h = 3r\] Thus $h:r = 3:1$ and the answer is $\boxed{B}$.

$\triangle EFG \sim \triangle EAB$ because $FG \parallel AB.$ The ratio of $\triangle EFG$ to $\triangle EAB$ is $2:5$ since $AB=5$ and $FG=2$ from subtraction. If we let $h$ be the height of $\triangle EAB,$ \[\frac{2}{5} = \frac{h-3}{h}\] \[2h = 5h-15\] \[3h = 15\] \[h = 5\] The height is $5$ so the area of $\triangle EAB$ is $\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}$.

Here is an easy way to look at this, where $p$ is the perimeter, and $a$ is the apothem: Area of Octagon: $\frac{ap}{2}=1$. Area of Rectangle: $\frac{p}{8}\times 2a=\dfrac{2ap}{8}=\frac{ap}{4}$. You can see from this that the octagon's area is twice as large as the rectangle's area is $\boxed{\textbf{(D)}\ \frac{1}{2}}$.

By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into $\frac{5}{6}$ of a circle with radius $1$ and two equilateral triangles with side length $1$. This gives the area of the white region as $\frac{5}{6}\pi+\frac{2\cdot\sqrt3}{4}=\frac{5}{6}\pi+\frac{\sqrt3}{2}$. The area of the shaded region is the area of the white region subtracted from the area of the large semicircle. This is equivalent to $2\pi-\left(\frac{5}{6}\pi+\frac{\sqrt3}{2}\right)=\frac{7}{6}\pi-\frac{\sqrt3}{2}$. Thus the answer is $\boxed{\textbf{(E)}\ \frac{7}{6}\pi-\frac{\sqrt3}{2}}$. The reason why it is $\frac{5}{6}$ of a circle and why the triangles are equilateral are because, first, the radii are the same and they make up the equilateral triangles. Secondly, the reason it is $\frac{5}{6}$ of a circle is because the middle sector has a degree of $180-2 \cdot 60 = 60$ and thus $\frac{60}{360}=\frac{1}{6}$ of a circle. The other two have areas of $\frac{180-60}{360}=\frac{1}{3}$ of a triangle each. Therefore, the total fraction of the circle(since they have the same radii) is $\frac{1}{6} + 2 \cdot \frac{1}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}.$

Since \begin{align*} &\log(xy) +2\log y = 1 \\ \log(xy) + \log x = 1 \quad \Longrightarrow \quad &2\log(xy) + 2\log x = 2 \end{align*} Summing gives \[3\log(xy) + 2\log y + 2\log x = 3 \Longrightarrow 5\log(xy) = 3\] Hence $\log (xy) = \frac 35 \Rightarrow \mathrm{(D)}$. It is not difficult to find $x = 10^{\frac{2}{5}}, y = 10^{\frac{1}{5}}$.

Substitute $a^cb^d$ into $x$. We then have $7(a^{5c}b^{5d}) = 11y^{13}$. Divide both sides by $7$, and it follows that: \[(a^{5c}b^{5d}) = \frac{11y^{13}}{7}.\] Note that because $11$ and $7$ are prime, the minimum value of $x$ must involve factors of $7$ and $11$ only. Thus, we try to look for the lowest power $p$ of $11$ such that $13p + 1 \equiv 0 \pmod{5}$, so that we can take $11^{13p + 1}$ to the fifth root. Similarly, we want to look for the lowest power $n$ of $7$ such that $13n - 1 \equiv 0 \pmod{5}$. Again, this allows us to take the fifth root of $7^{13n - 1}$. Obviously, we want to add $1$ to $13p$ and subtract $1$ from $13n$ because $11^{13p}$ and $7^{13n}$ are multiplied by $11$ and divided by $7$, respectively. With these conditions satisfied, we can simply multiply $11^{p}$ and $7^{n}$ and substitute this quantity into $y$ to attain our answer. We can simply look for suitable values for $p$ and $n$. We find that the lowest $p$, in this case, would be $3$ because $13(3) + 1 \equiv 0 \pmod{5}$. Moreover, the lowest $q$ should be $2$ because $13(2) - 1 \equiv 0 \pmod{5}$. Hence, we can substitute the quantity $11^{3} \cdot 7^{2}$ into $y$. Doing so gets us: \[(a^{5c}b^{5d}) = \frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}.\] Taking the fifth root of both sides, we are left with $a^cb^d = 11^{8} \cdot 7^{5}$. $a + b + c + d = 11 + 7 + 8 + 5 = \boxed{\textbf{(B)}\ 31}$

There are $4$ choices for the first element of $S$, and for each of these choices there are $4!$ ways to arrange the remaining elements. If the second element must be $2$, then there are only $3$ choices for the first element and $3!$ ways to arrange the remaining elements. Hence the answer is $\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}$, and $a+b=19 \Rightarrow \mathrm{(E)}$.

Since \begin{align*} -f(-1) = a - b + c - d = 0 = f(1) = a + b + c + d \end{align*} It follows that $b + d = 0$. Also, $d = f(0) = 2$, so $b = -2 \Rightarrow \mathrm{(B)}$. Two of the roots of $f(x) = 0$ are $\pm 1$, and we let the third one be $n$. Then \[a(x-1)(x+1)(x-n) = ax^3-anx^2-ax+an = ax^3 + bx^2 + cx + d = 0\] Notice that $f(0) = d = an = 2$, so $b = -an = -2 \Rightarrow \mathrm{(B)}$. Notice that if $g(x) = 2 - 2x^2$, then $f - g$ vanishes at $x = -1, 0, 1$ and so \[f(x) - g(x) = ax(x-1)(x+1) = ax^3 - ax\] implies by $x^2$ coefficient, $b + 2 = 0, b = -2 \Rightarrow \mathrm{(B)}$. The roots of this equation are $-1, 1, \text{ and } x$, letting $x$ be the root not shown in the graph. By Vieta, we know that $-1+1+x=x=-\frac{b}{a}$ and $-1\cdot 1\cdot x=-x=-\frac{d}{a}$. Therefore, $x=\frac{d}{a}$. Setting the two equations for $x$ equal to each other, $\frac{d}{a}=-\frac{b}{a}$. We know that the y-intercept of the polynomial is $d$, so $d=2$. Plugging in for $d$, $\frac{2}{a}=-\frac{b}{a}$. Therefore, $b=-2 \Rightarrow \boxed{B}$ From the graph, we have $f(0)=2$ so $d=2$. Also from the graph, we have $f(1)=a+b+c+2=0$. But we also have from the graph $f(-1)=-a+b-c+2=0$. Summing $f(1)+f(2)$ we get $2b+4=0$ so $b = -2 \Rightarrow \mathrm{(B)}$. Solution by franzliszt

By the Law of Cosines, \begin{align*} AB^2 + BC^2 - 2 AB \cdot BC \cos \alpha = 89 - 80 \cos \alpha = AC^2 &< 49\\ \cos \alpha &> \frac 12\\ \end{align*} It follows that $0 < \alpha < \frac {\pi}3$, and the probability is $\frac{\pi/3}{\pi} = \boxed{\textbf{(D) } \frac13 }$.

Let $\overline{AC}$ and $\overline{BD}$ intersect at $O$. Since $ABCD$ is a rhombus, then $\overline{AC}$ and $\overline{BD}$ are perpendicular bisectors. Thus $\angle POQ = 90^{\circ}$, so $OPNQ$ is a rectangle. Since the diagonals of a rectangle are of equal length, $PQ = ON$, so we want to minimize $ON$. It follows that we want $ON \perp AB$. Finding the area in two different ways, \[\frac{1}{2} AO \cdot BO = 60 = \frac{1}{2} ON \cdot AB = \frac{\sqrt{8^2 + 15^2}}{2} \cdot ON \Longrightarrow ON = \frac{120}{17} \approx 7.06 \Rightarrow \mathrm{(C)}\]

The function $f(x) = \sin x$ has roots in the form of $\pi n$ for all integers $n$. Therefore, we want $\frac{1}{x} = \pi n$ on $\frac{1}{10000} \le x \le \frac{1}{1000}$, so $1000 \le \frac 1x = \pi n \le 10000$. There are $\frac{10000-1000}{\pi} \approx \boxed{2900} \Rightarrow \mathrm{(A)}$ solutions for $n$ on this interval.

Consider the graph of $f(x)=|x-a|+|x-b|+|x-c|$. When $x<a$, the slope is $-3$. When $a<x<b$, the slope is $-1$. When $b<x<c$, the slope is $1$. When $c<x$, the slope is $3$. Setting $x=b$ gives $y=|b-a|+|b-b|+|b-c|=c-a$, so $(b,c-a)$ is a point on $f(x)$. In fact, it is the minimum of $f(x)$ considering the slope of lines to the left and right of $(b,c-a)$. Thus, graphing this will produce a figure that looks like a cup: From the graph, it is clear that $f(x)$ and $2x+y=2003$ have one intersection point if and only if they intersect at $x=a$. Since the line where $a<x<b$ has slope $-1$, the positive difference in $y$-coordinates from $x=a$ to $x=b$ must be $b-a$. Together with the fact that $(b,c-a)$ is on $f(x)$, we see that $P=(a,c-a+b-a)$. Since this point is on $x=a$, the only intersection point with $2x+y=2003$, we have $2 \cdot a+(b+c-2a)=2003 \implies b+c=2003$. As $c>b$, the smallest possible value of $c$ occurs when $b=1001$ and $c=1002$. This is indeed a solution as $a=1000$ puts $P$ on $y=2003-2x$, and thus the answer is $\boxed{\mathrm{(C)}\ 1002}$.

The first point is placed anywhere on the circle, because it doesn't matter where it is chosen. The next point must lie within $60$ degrees of arc on either side, a total of $120$ degrees possible, giving a total $\frac{1}{3}$ chance. The last point must lie within $60$ degrees of both points. The minimum area of freedom we have to place the third point is a $60$ degrees arc(if the first two are $60$ degrees apart), with a $\frac{1}{6}$ probability. The maximum amount of freedom we have to place the third point is a $120$ degree arc(if the first two are the same point), with a $\frac{1}{3}$ probability. As the second point moves farther away from the first point, up to a maximum of $60$ degrees, the probability changes linearly (every degree it moves, adds one degree to where the third could be). Therefore, we can average probabilities at each end to find $\frac{1}{4}$, the average probability we can place the third point based on a varying second point. Therefore the total probability is $1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$ or $\boxed{\text{(D)}}$