AMC12 2009 B

If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents. If Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\cdot1+50\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin at a total cost of $3.00$ dollars. Therefore, she bought $\boxed{\textbf{(B) } 2}$ bagels.

Losing three cans of paint corresponds to being able to paint five fewer rooms. So $\frac 35 \cdot 25 = \boxed{15}$. The answer is $\mathrm{(C)}$.

Twenty percent less than 60 is $\frac 45 \cdot 60 = 48$. One-third more than a number n is $\frac 43n$. Therefore $\frac 43n = 48$ and the number is $\boxed {36}$. The answer is $\mathrm{(D)}$.

Each triangle has leg length $\frac 12 \cdot (25 - 15) = 5$ meters and area $\frac 12 \cdot 5^2 = \frac {25}{2}$ square meters. Thus the flower beds have a total area of $25$ square meters. The entire yard has length $25$ m and width $5$ m, so its area is $125$ square meters. The fraction of the yard occupied by the flower beds is $\frac {25}{125} = \boxed{\frac15}$. The answer is $\mathrm{(C)}$.

The age of each person is a factor of $128 = 2^7$. So the twins could be $2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8$ years of age and, consequently Kiana could be $128$, $32$, $8$, or $2$ years old, respectively. Because Kiana is younger than her brothers, she must be $2$ years old. So the sum of their ages is $2 + 8 + 8 = \boxed{18}$. The answer is $\mathrm{(D)}$.

The three operations can be performed on any of $3! = 6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions \begin{align*} (2\times3) + (4\times5) &= 26,\\ (2\times3 + 4)\times5 &= 50,\\ 2\times(3 + 4\times5) &= 46,\\ 2\times(3 + 4)\times5 &= 70 \end{align*} are in fact all distinct. So the answer is $\boxed{4}$, which is choice $\mathrm{(C)}$.

Let $p$ be the price at the beginning of January. The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p.$ Because the price at the end of April was $p$, the price decreased by $0.2p$ during April, and the percent decrease was \[x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.\] So to the nearest integer $x$ is $\boxed{17}$. The answer is $\mathrm{(B)}$.

Let $x$ be the weight of the bucket and let $y$ be the weight of the water in a full bucket. Then we are given that $x + \frac 23y = a$ and $x + \frac 12y = b$. Hence $\frac 16y = a-b$, so $y = 6a-6b$. Thus $x = b - \frac 12 (6a-6b) = -3a + 4b$. Finally $x + y = \boxed {3a-2b}$. The answer is $\mathrm{(E)}$. Imagine that we take three buckets of the first type, to get rid of the fraction. We will have three buckets and two buckets' worth of water. On the other hand, if we take two buckets of the second type, we will have two buckets and enough water to fill one bucket. The difference between these is exactly one bucket full of water, hence the answer is $3a-2b$. We are looking for an expression of the form $xa + yb$. We must have $x+y=1$, as the desired result contains exactly one bucket. Also, we must have $\frac 23 x + \frac 12 y = 1$, as the desired result contains exactly one bucket of water. At this moment, it is easiest to check that only the options (A), (B), and (E) satisfy $x+y=1$, and out of these only (E) satisfies the second equation. Alternatively, we can directly solve the system, getting $x=3$ and $y=-2$. Since $a$ is one bucket plus two-thirds of the total amount of water, and $b$ is one bucket plus one-half of the total amount of water, $a-b$ would equal $\cancel{\text{bucket}}+\frac23\cdot\text{water}-(\cancel{\text{bucket}}+\frac12\cdot\text{water})=\frac16\cdot\text{water}$. Therefore, $a-b$ is one-sixth of the total mass of the water. Starting from $a$, we add two one-sixths of the total amount of the water to become full. Therefore, the full bucket of water is $a+2(a-b)=a+2a-2b=3a-2b\Rightarrow\textbf{(E)}$

Because the line $x + y = 7$ is parallel to $\overline {AB}$, the area of $\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$. In that case the triangle has base $AC = 4$ and altitude $3$, so its area is $\frac 12 \cdot 4 \cdot 3 = \boxed {6}$. The base of the triangle is $AB = \sqrt{3^2 + 3^2} = 3\sqrt 2$. Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$, which is $\frac 4{\sqrt 2} = 2\sqrt 2$. Therefore its area is $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$. The answer is $\mathrm{(A)}$. By Shoelace, our area is: \[\frac {1}{2} \cdot |(3\cdot 3 + 0 \cdot y+ 0 \cdot x)-(0*0+3(x+y))|.\] We know $x+y=7$ so we get: \[\frac {1}{2} \cdot |9-21|=\boxed 6\] WLOG, let the coordinates of $C$ be $(3,4)$ , or any coordinate, for that matter. Applying the shoelace formula, we get the area as $\boxed 6$.

The clock will display the incorrect time for the entire hours of $1, 10, 11$ and $12$. So the correct hour is displayed $\frac 23$ of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a $1$, so the minutes that will not display correctly are $10, 11, 12, \dots, 19$ and $01, 21, 31, 41,$ and $51$. This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the day that the clock shows the correct time is $\frac 23 \cdot \left(1 - \frac {15}{60}\right) = \frac 23 \cdot \frac 34 = \boxed{\frac 12}$. The answer is $\mathrm{(A)}$. The required fraction is the number of correct times divided by the total times. There are 60 minutes in an hour and 12 hours on a clock, so there are 720 total times. We count the correct times directly; let a correct time be $x:yz$, where $x$ is a number from 1 to 12 and $y$ and $z$ are digits, where $y<6$. There are 8 values of $x$ that will display the correct time: 2, 3, 4, 5, 6, 7, 8, and 9. There are five values of $y$ that will display the correct time: 0, 2, 3, 4, and 5. There are nine values of $z$ that will display the correct time: 0, 2, 3, 4, 5, 6, 7, 8, and 9. Therefore there are $8\cdot 5\cdot 9=40\cdot 9=360$ correct times. Therefore the required fraction is $\frac{360}{720}=\frac{1}{2}\Rightarrow \boxed{\mathrm{(A)}}$.

On Monday, day 1, the birds find $\frac 14$ quart of millet in the feeder. On Tuesday they find \[\frac 14 + \frac 34 \cdot \frac 14\] quarts of millet. On Wednesday, day 3, they find \[\frac 14 + \frac 34 \cdot \frac 14 + \left(\frac34\right)^2 \cdot \frac 14\] quarts of millet. The number of quarts of millet they find on day $n$ is \[\frac 14 + \frac 34 \cdot \frac 14 + \left(\frac34\right)^2 \cdot \frac 14 + \cdots + \left(\frac 34\right)^{n-1} \cdot \frac 14 = \frac {(\frac 14)(1 - (\frac 34)^n)}{1 - \frac 34} = 1 - \left(\frac 34\right)^n .\] The birds always find $\frac 34$ quart of other seeds, so more than half the seeds are millet if $1 - \left(\frac 34\right)^n > \frac 34$, that is, when $\left(\frac 34\right)^n < \frac 14$. Because $\left(\frac 34\right)^4 = \frac {81}{256} > \frac 14$ and $\left(\frac 34\right)^5 = \frac {243}{1024} < \frac 14$, this will first occur on day $5$ which is $\boxed {\text{Friday}}$. The answer is $\mathrm{(D)}$.

Let the $n$th term of the series be $ar^{n-1}$. Because \[\frac {8!}{7!} = \frac {ar^7}{ar^4} = r^3 = 8,\] it follows that $r = 2$ and the first term is $a = \frac {7!}{r^4} = \frac {7!}{16} = \boxed{315}$. The answer is $\mathrm{(E)}$.

Let $D$ be the foot of the altitude to $\overline{BC}$. Then $BD = \sqrt {13^2 - 12^2} = 5$ and $DC = \sqrt {15^2 - 12^2} = 9$. Thus $BC = BD + BC = 5 + 9 = 14$. Otherwise, assume that the triangle is obtuse at angle $B$ then $BC = DC - BD = 9 -5 = 4$. The sum of the two possible values is $14 + 4 = \boxed{18}$. The answer is $\mathrm{(D)}$.

For $c\geq 1.5$ the shaded area is at most $1.5$, which is too little. Hence $c<1.5$, and therefore the point $(2,1)$ is indeed inside the shaded part, as shown in the picture. Then the area of the shaded part is one less than the area of the triangle with vertices $(c,0)$, $(3,0)$, and $(3,3)$. The area of the entire triangle is $\frac{3(3-c)}2$, therefore the area of the shaded part is $\frac{7-3c}{2}$. The entire figure has area $5$, hence we want the shaded part to have area $\frac 52$. Solving for $c$, we get $c=\boxed{\frac 23}$. The answer is $\mathrm{(C)}$ ( what a coincidence ;) ).

(B) Intuitively, $x$ will be largest for that option for which the value in the parentheses is smallest. Formally, first note that each of the values in parentheses is larger than $1$, which leads to the conclusion that $x$ will be the largest when the value of the parentheses is smallest. Now, each of the options is of the form $3f(r)^x = 7$. This can be rewritten as $x\log f(r) = \log\frac 73$. As $f(r)>1$, we have $\log f(r)>0$. Thus $x$ is the largest for the option for which $\log f(r)$ is smallest. And as $\log f(r)$ is an increasing function, this is the option for which $f(r)$ is smallest. We now get the following easier problem: Given that $0<r<3$, find the smallest value in the set $\{ 1+r, 1+r/10, 1+2r, 1+\sqrt r, 1+1/r\}$. Clearly $1+r/10$ is smaller than the first and the third option. We have $r^2 < 10$, dividing both sides by $10r$ we get $r/10 < 1/r$. And finally, $r/100 < 1$, therefore $r^2/100 < r$, and as both sides are positive, we can take the square root and get $r/10 < \sqrt r$. Thus the answer is $\boxed{\textbf{(B) } 3\left(1+\frac{r}{10}\right)^x = 7}$.

Extend $\overline {AB}$ and $\overline {DC}$ to meet at $E$. Then \begin{align*} \angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\ &= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}. \end{align*} Thus $\triangle BDE$ is isosceles with $DE = BD$. Because $\overline {AD} \parallel \overline {BC}$, it follows that the triangles $BCE$ and $ADE$ are similar. Therefore \[\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,\] so $CD = \boxed{\frac 45}.$

There are two possible stripe orientations for each of the six faces of the cube, so there are $2^6 = 64$ possible stripe combinations. There are three pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that does not contribute uniquely determines the stripe orientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in either of two different ways. So a total of $3 \cdot 2 \cdot 2 = 12$ stripe combinations on the cube result in a continuous stripe around the cube. The required probability is $\frac {12}{64} = \boxed{\frac {3}{16}}$. Without loss of generality, orient the cube so that the stripe on the top face goes from front to back. There are two mutually exclusive ways for there to be an encircling stripe: either the front, bottom and back faces are painted to complete an encircling stripe with the top face's stripe or the front, right, back and left faces are painted to form an encircling stripe. The probability of the first case is $\left(\frac 12\right)^3 = \frac 18$, and the probability of the second case is $\left(\frac 12\right)^4 = \frac {1}{16}$. The cases are disjoint, so the probabilities sum to $\frac 18 + \frac {1}{16} = \boxed {\frac {3}{16}}$. There are three possible orientations of an encircling stripe (as each encircling would leave out 1 pair of opposite faces, and there are 3 pairs in total). For any one of these to appear, the stripes on the four faces through which the continuous stripe is to pass must be properly aligned. The probability of each such stripe alignment is $\left(\frac 12\right)^4 = \frac {1}{16}$. Since there are three such possibilities and they are disjoint, the total probability is $3 \cdot \frac {1}{16} = \boxed{\frac {3}{16}}$. The answer is $\mathrm{(B)}$. Consider a vertex on the cube and the three faces that are adjacent to that vertex. If no two stripes on those three faces are aligned, then there is no stripe encircling the cube. The probability that the stripes aren't aligned is $\frac{1}{4}$, since for each alignment of one stripe, there is one and only one way to align the other two stripes out of four total possibilities. therefore there is a probability of $\frac{3}{4}$ that two stripes are aligned. Now consider the opposing vertex and the three sides adjacent to it. Given the two connected stripes next to our first vertex, we have two more that must be connected to make a continuous stripe. There is a probability of $\left(\frac{1}{2}\right)^2=\frac{1}{4}$ that they are aligned, so there is a probability of $\frac{3}{4}\cdot \frac{1}{4}=\boxed{\frac {3}{16}}$ that there is a continuous stripe. Each face has two possible stripes, so there are $2^6=64$ ways in total. We need to choose $2$ opposite faces to leave out. Since the second is fixed once we pick the first face, we only need to calculate $\binom{3}{1}=3$ orientations. Next, the two faces also each have 2 orientations, since they don't matter to the stripe. Therefore there are $3\cdot2\cdot2=12$ ways. The probability is $\frac{12}{64}=\frac{3}{16} \Longrightarrow \boxed{\textbf{(B) } \frac{3}{16}}$.

After $10$ minutes $(600$ seconds$),$ Rachel will have completed $6$ laps and be $30$ seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in $22.5$ seconds, she will be in the picture between $18.75$ seconds and $41.25$ seconds of the tenth minute. After 10 minutes, Robert will have completed $7$ laps and be $40$ seconds from completing his eighth lap. Because Robert runs one-fourth of a lap in $20$ seconds, he will be in the picture between $30$ seconds and $50$ seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between $30$ seconds and $41.25$ seconds of the tenth minute. So the probability that both runners are in the picture is $\frac {41.25 - 30} {60} = \boxed{\frac {3}{16}}$. The answer is $\mathrm{(C)}$.

To find the answer it was enough to play around with $f$. One can easily find that $f(1)=41$ is a prime, then $f$ becomes negative for $n$ between $2$ and $18$, and then $f(19)=761$ is again a prime number. And as $41 + 761 = 802$ is already the largest option, the answer must be $\boxed{802}$.

Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Three edges of $R$ meet at each vertex, so $R$ has a total of $\frac 12 \cdot 3 \cdot 200 = \boxed {300}$ edges.

Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is \[\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \boxed{89}.\] Recall that the number of ways to arrange 1x1 and 2x1 blocks to form a 1xn rectangle results in Fibonacci numbers. Clearly, $\boxed{89}$ is the only fibonacci number, so no calculation is needed for this problem.

The area of any parallelogram $ABCD$ can be computed as the size of the vector product of $\overrightarrow{AB}$ and $\overrightarrow{AD}$. In our setting where $A=(0,0)$, $B=(s,s)$, and $D=(t,kt)$ this is simply $s\cdot kt - s\cdot t = (k-1)st$. In other words, we need to count the triples of integers $(k,s,t)$ where $k>1$, $s,t>0$ and $(k-1)st = 1,\!000,\!000 = 2^6 5^6$. These can be counted as follows: We have $6$ identical red balls (representing powers of $2$), $6$ blue balls (representing powers of $5$), and three labeled urns (representing the factors $k-1$, $s$, and $t$). The red balls can be distributed in ${8\choose 2} = 28$ ways, and for each of these ways, the blue balls can then also be distributed in $28$ ways. (See Distinguishability for a more detailed explanation.) Thus there are exactly $28^2 = 784$ ways how to break $1,\!000,\!000$ into three positive integer factors, and for each of them we get a single parallelogram. Hence the number of valid parallelograms is $784 \longrightarrow \boxed{C}$. Without the vector product the area of $ABCD$ can be computed for example as follows: If $B=(s,s)$ and $D=(t,kt)$, then clearly $C=(s+t,s+kt)$. Let $B'=(s,0)$, $C'=(s+t,0)$ and $D'=(t,0)$ be the orthogonal projections of $B$, $C$, and $D$ onto the $x$ axis. Let $[P]$ denote the area of the polygon $P$. We can then compute: \begin{align*} [ABCD] & = [ADD'] + [DD'C'C] - [BB'C'C] - [ABB'] \\ & = \frac{kt^2}2 + \frac{s(s+2kt)}2 - \frac{t(2s+kt)}2 - \frac{s^2}2 \\ & = kst - st \\ & = (k-1)st. \end{align*} The remainder of the solution is the same as the above.

First, turn $\frac34 + \frac34i$ into polar form as $\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}$. Restated using geometric probabilities, we are trying to find the portion of a square enlarged by a factor of $\frac{3\sqrt{2}}{4}$ and rotated $45$ degrees that lies within the original square. This skips all the absolute values required before. Finish with the symmetry method stated above.

First of all, we have to agree on the range of $\sin^{-1}$ and $\cos^{-1}$. This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: $\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2$ and $\forall x: 0\leq \cos^{-1}(x) \leq \pi$. Hence we get that $\forall x\in[0,\pi]: \cos^{ - 1}(\cos x) = x$, thus our equation simplifies to $\sin^{ - 1}(\sin 6x) = x$. Consider the function $f(x) = \sin^{ - 1}(\sin 6x) - x$. We are looking for roots of $f$ on $[0,\pi]$. By analyzing properties of $\sin$ and $\sin^{-1}$ (or by computing the derivative of $f$) one can discover the following properties of $f$: - $f(0)=0$. - $f$ is increasing and then decreasing on $[0,\pi/6]$. - $f$ is decreasing and then increasing on $[\pi/6,2\pi/6]$. - $f$ is increasing and then decreasing on $[2\pi/6,3\pi/6]$. For $x=\pi/6$ we have $f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0$. Hence $f$ has exactly one root on $(0,\pi/6)$. For $x=2\pi/6$ we have $f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0$. Hence $f$ is negative on the entire interval $[\pi/6,2\pi/6]$. Now note that $\forall t: \sin^{-1}(t) \leq \pi/2$. Hence for $x > 3\pi/6$ we have $f(x) < 0$, and we can easily check that $f(3\pi/6)<0$ as well. Thus the only unknown part of $f$ is the interval $(2\pi/6,3\pi/6)$. On this interval, $f$ is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval. To prove that there are two roots, it is enough to find any $x$ from this interval such that $f(x)>0$. A good guess is its midpoint, $x=5\pi/12$, where the function $\sin^{-1}(\sin 6x)$ has its local maximum. We can evaluate: $f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0$. Summary: The function $f$ has $\boxed{\textbf{(B) }4}$ roots on $[0,\pi]$: the first one is $0$, the second one is in $(0,\pi/6)$, and the last two are in $(2\pi/6,3\pi/6)$. Actual solutions are $x=0$, $x=\pi/7$, $x=2\pi/5$, and $x=3\pi/7$.

We need to find a reasonably easy way to count the squares. First, obviously the maximum distance between two points in the same quadrant is $4\sqrt 2 < 6$, hence each square has exactly one vertex in each quadrant. Given any square, we can circumscribe another axes-parallel square around it. In the picture below, the original square is red and the circumscribed one is blue. Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it? Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations. The size of the blue square can range from $6\times 6$ to $14\times 14$, and for the intermediate sizes, there is more than one valid placement. We will now examine the cases one after another. Also, we can use symmetry to reduce the number of cases. Summing the last column, we get that the answer is $\boxed{225}$.