AMC12 2008 B

If the basketball player makes $x$ three-point shots and $5-x$ two-point shots, he scores $3x+2(5-x)=10+x$ points. Clearly every value of $x$ yields a different number of total points. Since he can make any number of three-point shots between $0$ and $5$ inclusive, the number of different point totals is $6 \Rightarrow E$. Stars and bars can also be utilized to solve this problem. Since we need to decide what number of 2's and 3's are scored, and there are a total of 5 shots. It can be written like such: _ _ _ | _ _. Solving this, we get $6 \Rightarrow E$.

Answer (B): The two sums are $1+10+17+22=50$ and $4+9+16+25=54$, so the positive difference between the sums is $54-50=4$. Query: If a different $4\times 4$ block of dates had been chosen, the answer would be unchanged. Why? \[ \begin{array}{|c|c|c|c|} \hline 1 & 2 & 3 & 4\\ \hline 11 & 10 & 9 & 8\\ \hline 15 & 16 & 17 & 18\\ \hline 25 & 24 & 23 & 22\\ \hline \end{array} \]

We want to find the maximum any player could make, so assume that everyone else makes the minimum possible and that the combined salaries total the maximum of $700,000$ $700,000 = 20 * 15,000 + x$ $x = 400,000$ The maximum any player could make is $400,000$ dollars $\Rightarrow C$

$\angle COD = \angle AOB - \angle AOC - \angle BOD = 180^\circ - 30^\circ - 45^\circ = 105^\circ$. Since a circle has $360^\circ$, the desired ratio is $\frac{105^\circ}{360^\circ}=\frac{7}{24} \Rightarrow D$.

The class could send $25$ carnations and no roses, $22$ carnations and $2$ roses, $19$ carnations and $4$ roses, and so on, down to $1$ carnation and $16$ roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), $\Rightarrow \boxed{C}$

Every time the pedometer flips, Pete has walked $100,000$ steps. Therefore, Pete has walked a total of $100,000 \cdot 44 + 50,000 = 4,450,000$ steps, which is $4,450,000/1,800 = 2472.2$ miles, which is the closest to the answer choice $\boxed{A}$.

$\left[ (x-y)^2 - (y-x)^2 \right]^2$ $\left[ (x-y)^2 - (x-y)^2 \right]^2$ $[0]^2$ $0 \Rightarrow \textbf{(A)}$

Since $\overline{AB}=4\overline{BD}$ and $\overline{AB}+\overline{BD}=\overline{AD}$, $\overline{AB}=\frac{4}{5}\overline{AD}$. Since $\overline{AC}=9\overline{CD}$ and $\overline{AC}+\overline{CD}=\overline{AD}$, $\overline{AC}=\frac{9}{10}\overline{AD}$. Thus, $\overline{BC}=\overline{AC}-\overline{AB}=\left(\frac{9}{10}-\frac{4}{5}\right)\overline{AD} = \frac {1}{10}\overline{AD} \Rightarrow C$.

Let $\alpha$ be the angle that subtends the arc $AB$. By the law of cosines, $6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)$ implies $\cos(\alpha) = 7/25$. The half-angle formula says that $\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. The law of cosines tells us $AC = \sqrt{5^2+5^2-2\cdot 5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}$, which is answer choice $\boxed{\text{A}}$.

Let $h$ be the number of bricks in the chimney. Without talking, Brenda and Brandon lay $\frac{h}{9}$ and $\frac{h}{10}$ bricks per hour respectively, so together they lay $\frac{h}{9}+\frac{h}{10}-10$ per hour together. Since they finish the chimney in $5$ hours, $h=5\left( \frac{h}{9}+\frac{h}{10}-10 \right)$. Thus, $h=900 \Rightarrow B$.

In a cone, radius and height each vary inversely with increasing height (i.e. the radius of the cone formed by cutting off the mountain at $4,000$ feet is half that of the original mountain). Therefore, volume varies as the inverse cube of increasing height (expressed as a percentage of the total height of cone): $V_I\times \text{Height}^3 = V_N$ Plugging in our given condition, $\frac{1}{8} = \text{Height}^3 \Rightarrow \text{Height} = \frac{1}{2}$. $8000\cdot\frac{1}{2}=4000 \Rightarrow \boxed{\textbf{A}}$.

Letting $S_n$ be the nth partial sum of the sequence: $\frac{S_n}{n} = n$ $S_n = n^2$ The only possible sequence with this result is the sequence of odd integers. $a_n = 2n - 1$ $a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}$

The region is the shaded area: We can find the area of the shaded region by subtracting the pentagon from the middle third of the square. The area of the middle third of the square is $\left(\frac13\right)(1)=\frac13$. The pentagon can be split into a rectangle and an equilateral triangle. The base of the equilateral triangle is $\frac13$ and the height is $\left(\frac13\right)\left(\frac12\right)(\sqrt{3})=\frac{\sqrt{3}}{6}$. Thus, the area is $\left(\frac{\sqrt3}{6}\right)\left(\frac13\right)\left(\frac12\right)=\frac{\sqrt3}{36}$. The base of the rectangle is $\frac13$ and the height is the height of the equilateral triangle minus the height of the smaller equilateral triangle. This is: $\frac{\sqrt3}{2}-\frac{\sqrt3}{6}=\frac{\sqrt3}{3}$ Therefore, the area of the shaded region is $\frac13-\frac{\sqrt3}{9}-\frac{\sqrt3}{36}=\boxed{\text{(B) }\frac{12-5\sqrt3}{36}}.$

Let $C$ be the circumference of the circle, and let $r$ be the radius of the circle. Using log properties, $C=\log_{10}{(b^4)}=4\log_{10}{(b)}$ and $r=\log_{10}{(a^2)}=2\log_{10}{(a)}$. Since $C=2\pi r$, $4\log_{10}{(b)}=2\pi\cdot2\log_{10}{(a)} \Rightarrow \log_{a}{b} = \frac{\log_{10}{(b)}}{\log_{10}{(a)}}=\pi \Rightarrow C$.

The equilateral triangles form trapezoids with side lengths $1, 1, 1, 2$ (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengths $1$ and an angle of $30^{\circ}$ in between them, so the total area of these triangles (which is the area of $S - R$) is, $4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1$ which makes the answer $\boxed{C}$.

$A_{outer}=ab$ $A_{inner}=(a-2)(b-2)$ $A_{outer}=2A_{inner}$ $ab=2(a-2)(b-2)=2ab-4a-4b+8$ $0=ab-4a-4b+8$ By Simon's Favorite Factoring Trick: $8=ab-4a-4b+16=(a-4)(b-4)$ Since $8=1\times8$ and $8=2\times4$ are the only positive factorings of $8$. $(a,b)=(5,12)$ or $(a,b)=(6,8)$ yielding $\Rightarrow\textbf{(B)}$ $2$ solutions. Notice that because $b>a$, the reversed pairs are invalid.

Supposing $\angle A=90^\circ$, $AC$ is perpendicular to $AB$ and, it follows, to the $x$-axis, making $AC$ a segment of the line $x=m$. But that would mean that the coordinates of $C$ are $(m, m^2)$, contradicting the given that points $A$ and $C$ are distinct. So $\angle A$ is not $90^\circ$. By a similar logic, neither is $\angle B$. This means that $\angle C=90^\circ$ and $AC$ is perpendicular to $BC$. Let C be the point $(n, n^2)$. So the slope of $BC$ is the negative reciprocal of the slope of $AC$, yielding $m+n=\frac{1}{m-n}$ $\Rightarrow$ $m^2-n^2=1$. Because $m^2-n^2$ is the length of the altitude of triangle $ABC$ from $AB$, and $2m$ is the length of $AB$, the area of $\triangle ABC=m(m^2-n^2)=2008$. Since $m^2-n^2=1$, $m=2008$. Substituting, $2008^2-n^2=1$ $\Rightarrow$ $n^2=2008^2-1=(2000+8)^2-1=4000000+32000+64-1=4032063$, whose digits sum to $18 \Rightarrow \textbf{(C)}$.

Let $h$ be the height of the pyramid and $a$ be the distance from $h$ to $CD$. The side length of the base is $14$. The heights of $\triangle ABE$ and $\triangle CDE$ are $2\cdot105\div14=15$ and $2\cdot91\div14=13$, respectively. Consider a side view of the pyramid from $\triangle BCE$. We have a systems of equations through the Pythagorean Theorem: $13^2-(14-a)^2=h^2 \\ 15^2-a^2=h^2$ Setting them equal to each other and simplifying gives $-27+28a=225 \implies a=9$. Therefore, $h=\sqrt{15^2-9^2}=12$, and the volume of the pyramid is $\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}$.

We need only concern ourselves with the imaginary portions of $f(1)$ and $f(i)$ (both of which must be 0). These are: \begin{align*} \text{Im}(f(1)) & = i+i\text{Im}(\alpha)+i\text{Im}(\gamma) \\ \text{Im}(f(i)) & = -i+i\text{Re}(\alpha)+i\text{Im}(\gamma) \end{align*} Let $p=\text{Im}(\gamma)$ and $q=\text{Re}{(\gamma)},$ then we know $\text{Im}(\alpha)=-p-1$ and $\text{Re}(\alpha)=1-p.$ Therefore \[|\alpha|+|\gamma|=\sqrt{(1-p)^2+(-1-p)^2}+\sqrt{q^2+p^2}=\sqrt{2p^2+2}+\sqrt{p^2+q^2},\] which reaches its minimum $\sqrt 2$ when $p=q=0$ by the Trivial Inequality. Thus, the answer is $\boxed B.$

Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$. Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds. Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds. Meetings occur whenever $D(t)=0$. We have $D(0)=200$. The truck always moves for $20$ seconds, then stands still for $30$. During the first $20$ seconds of the cycle the truck moves by $200$ feet and Michael by $100$, hence during the first $20$ seconds of the cycle $D(t)$ increases by $100$. During the remaining $30$ seconds $D(t)$ decreases by $150$. From this observation it is obvious that after four full cycles, i.e. at $t=200$, we will have $D(t)=0$ for the first time. During the fifth cycle, $D(t)$ will first grow from $0$ to $100$, then fall from $100$ to $-50$. Hence Michael overtakes the truck while it is standing at the pail. During the sixth cycle, $D(t)$ will first grow from $-50$ to $50$, then fall from $50$ to $-100$. Hence the truck starts moving, overtakes Michael on their way to the next pail, and then Michael overtakes the truck while it is standing at the pail. During the seventh cycle, $D(t)$ will first grow from $-100$ to $0$, then fall from $0$ to $-150$. Hence the truck meets Michael at the moment when it arrives to the next pail. Obviously, from this point on $D(t)$ will always be negative, meaning that Michael is already too far ahead. Hence we found all $\boxed{5 \Longrightarrow B}$ meetings. The movement of Michael and the truck is plotted below: Michael in blue, the truck in red.

Circles centered at $A$ and $B$ will overlap if $A$ and $B$ are closer to each other than if the circles were tangent. The circles are tangent when the distance between their centers is equal to the sum of their radii. Thus, the distance from $A$ to $B$ will be $2$. Since $A$ and $B$ are separated by $1$ vertically, they must be separated by $\sqrt{3}$ horizontally. Thus, if $|A_x-B_x|<\sqrt{3}$, the circles intersect. Now, plot the two random variables $A_x$ and $B_x$ on the coordinate plane. Each variable ranges from $0$ to $2$. The circles intersect if the variables are within $\sqrt{3}$ of each other. Thus, the area in which the circles don't intersect is equal to the total area of two small triangles on opposite corners, each of area $\frac{(2-\sqrt{3})^2}{2}$. So, the total area of the 2 triangles sums to $(2-\sqrt{3})^2$. Since the total $2 \times 2$ square has an area of $4$, the probability of the circles not intersecting is $\frac{(2-\sqrt{3})^2}{4}$. But remember, we want the probability that they do intersect. We conclude the probability the circles intersect is:\[1-\frac{(2-\sqrt{3})^2}{4}=\boxed{\textbf{(E)}\frac{4\sqrt{3}-3}{4}}.\]

Auntie Em won't be able to park only when none of the four available spots touch. We can form a bijection between all such cases and the number of ways to pick four spots out of 13: since none of the spots touch, remove a spot from between each of the cars. From the other direction, given four spots out of 13, simply add a spot between each. So the probability she can park is \[1-\frac{{13 \choose 4}}{{16 \choose 4}}=1-\frac{13\cdot12\cdot11\cdot10}{16\cdot15\cdot14\cdot13}=1-\frac{11}{28}={\textbf{(E)}\frac{17}{28}}.\] (Bijection: When elements of two sets are perfectly paired with each other, i.e. each and every element from both sets has exactly one match in the other, and no elements are left out. In the context of this problem, this means the number of distinct ways to order the cars such that no two spaces are adjacent is exactly the number of ways to pick 4 spots out of 13.)

Every factor of $10^n$ will be of the form $2^a \times 5^b , a\leq n , b\leq n$. Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property $\log(a \times b) = \log(a)+\log(b)$. For any factor $2^a \times 5^b$, there will be another factor $2^{n-a} \times 5^{n-b}$. Note this is not true if $10^n$ is a perfect square. When these are multiplied, they equal $2^{a+n-a} \times 5^{b+n-b} = 10^n$. $\log 10^n=n$ so the number of factors divided by 2 times n equals the sum of all the factors, 792. There are $n+1$ choices for the exponent of 5 in each factor, and for each of those choices, there are $n+1$ factors (each corresponding to a different exponent of 2), yielding $(n+1)^2$ total factors. $\frac{(n+1)^2}{2}*n = 792$. We then plug in answer choices and arrive at the answer $\boxed {11}$

Let $a_n=|A_{n-1}A_n|$. We need to rewrite the recursion into something manageable. The two strange conditions, $B$'s lie on the graph of $y=\sqrt{x}$ and $A_{n-1}B_nA_n$ is an equilateral triangle, can be compacted as follows: \[\left(a_n\frac{\sqrt{3}}{2}\right)^2=\frac{a_n}{2}+a_{n-1}+a_{n-2}+\cdots+a_1\] which uses $y^2=x$, where $x$ is the height of the equilateral triangle and therefore $\frac{\sqrt{3}}{2}$ times its base. The relation above holds for $n=k$ and for $n=k-1$ $(k>1)$, so \[\left(a_k\frac{\sqrt{3}}{2}\right)^2-\left(a_{k-1}\frac{\sqrt{3}}{2}\right)^2=\] \[=\left(\frac{a_k}{2}+a_{k-1}+a_{k-2}+\cdots+a_1\right)-\left(\frac{a_{k-1}}{2}+a_{k-2}+a_{k-3}+\cdots+a_1\right)\] Or, \[a_k-a_{k-1}=\frac23\] This implies that each segment of a successive triangle is $\frac23$ more than the last triangle. To find $a_{1}$, we merely have to plug in $k=1$ into the aforementioned recursion and we have $a_{1} - a_{0} = \frac23$. Knowing that $a_{0}$ is $0$, we can deduce that $a_{1} = 2/3$.Thus, $a_n=\frac{2n}{3}$, so $A_0A_n=a_n+a_{n-1}+\cdots+a_1=\frac{2}{3} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{3}$. We want to find $n$ so that $n^2<300<(n+1)^2$. $n=\boxed{17}$ is our answer.

Error creating thumbnail: Unable to save thumbnail to destination Note: In the image AB and CD have been swapped from their given lengths in the problem. However, this doesn't affect any of the solving. Drop perpendiculars to $CD$ from $A$ and $B$, and call the intersections $X,Y$ respectively. Now, $DA^2-BC^2=(7-5)(7+5)=DX^2-CY^2$ and $DX+CY=19-11=8$. Thus, $DX-CY=3$. We conclude $DX=\frac{11}{2}$ and $CY=\frac{5}{2}$. To simplify things even more, notice that $90^{\circ}=\frac{\angle D+\angle A}{2}=180^{\circ}-\angle APD$, so $\angle P=\angle Q=90^{\circ}$. Also, \[\sin(\angle PAD)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}\] So the area of $\triangle APD$ is: \[R\cdot c\sin a\sin b =\frac{7\cdot7}{2}\sqrt{\frac{3}{28}}\sqrt{1-\frac{3}{28}}=\frac{35}{8}\sqrt{3}\] Over to the other side: $\triangle BCY$ is $30-60-90$, and is therefore congruent to $\triangle BCQ$. So $[BCQ]=\frac{5\cdot5\sqrt{3}}{8}$. The area of the hexagon is clearly \begin{align*} [ABCD]-([BCQ]+[APD]) &=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}\\ &=30\sqrt{3}\implies\boxed{\mathrm{B}} \end{align*} Note: Once $DY$ is found, there is no need to do the trig. Notice that the hexagon consists of two trapezoids, $ABPQ$ and $CDPQ$. $PQ = \frac{19-7-5 +11}{2} = 9$. The height is one half of $BY$ which is $\frac{5\sqrt{3}}{4}$. So the area is \[\frac{1}{2} \cdot \frac{5\sqrt{3}}{4}(19+9+11+9)=30\sqrt{3}\]