AMC12 2007 B

There are four walls in each bedroom (she can't paint floors or ceilings). Therefore, we calculate the number of square feet of walls there is in one bedroom: \[2\cdot(12\cdot8+10\cdot8)-60=2\cdot176-60=292\] We have three bedrooms, so she must paint $292\cdot3=\boxed{\textbf{(E) }876}$ square feet of walls.

The trip was $240$ miles long and took $\dfrac{120}{30}+\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\dfrac{240}{10}= \boxed{\textbf{(B) }24}$

Since triangles $ABO$ and $BOC$ are isosceles, $\angle ABO=20^o$ and $\angle OBC=30^o$. Therefore, $\angle ABC=50^o$, or $\mathrm{(D)}$.

$3$ bananas cost as much as $2$ apples, so $18$ bananas cost as much as $12$ apples. Since $6$ apples cost as much as $4$ oranges, $12$ apples cost as much as $8$ oranges, giving $(B)$.

She must get at least $100 - 4.5 = 95.5$ points, and that can only be possible by answering at least $\lceil \frac{95.5}{6}\rceil = 16 \Rightarrow \mathrm {(D)}$ questions correctly.

Error creating thumbnail: Unable to save thumbnail to destination One bug goes to $B$. The path that he takes is $\dfrac{5+6+7}{2}=9$ units long. The length of $BD$ is $9-AB=9-5=4 \Rightarrow \mathrm {(D)}$

Error creating thumbnail: Unable to save thumbnail to destination Since $A$ and $B$ are right angles, and $AE$ equals $BC$, and $AECB$ is a square. Since the length of $ED$ and $CD$ are also 5, triangle $CDE$ is equilateral. Angle $E$ is therefore $90+60=150 \Rightarrow \mathrm {(E)}$

Tom's age $N$ years ago was $T-N$. The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \begin{align*}T-N&=2(T-3N)\\ T-N&=2T-6N\\ T&=5N\\ T/N&=\boxed{\mathrm{(D) \ } 5}\end{align*} Note that actual values were not found.

$3x-1 =5 \implies x= 2$ $f(3(2)-1) = 2^2+2+1=7 \implies (A)$

If we let $p$ be the number of people initially in the group, then $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p$, but the number of girls is $0.4p-2$. Since only $30\%$ of the group are girls, \begin{align*} \frac{0.4p-2}{p}&=\frac{3}{10}\\ 4p-20&=3p\\ p&=20\end{align*} The number of girls initially in the group is $0.4p=0.4(20)=\boxed{\mathrm{(C) \ } 8}$

The sum of the interior angles of any quadrilateral is $360^\circ.$ \begin{align*} 360 &= \angle A + \angle B + \angle C + \angle D\\ &= \angle A + \frac{1}{2}A + \frac{1}{3}A + \frac{1}{4}A\\ &= \frac{12}{12}A + \frac{6}{12}A + \frac{4}{12}A + \frac{3}{12}A\\ &= \frac{25}{12}A \end{align*} \[\angle A = 360 \cdot \frac{12}{25} = 172.8 \approx \boxed{\mathrm{(D) \ } 173}\]

We can assume there are $10$ people in the class. Then there will be $1$ junior and $9$ seniors. The sum of everyone's scores is $10 \cdot 84 = 840$. Since the average score of the seniors was $83$, the sum of all the senior's scores is $9 \cdot 83 = 747$. The only score that has not been added to that is the junior's score, which is $840 - 747 = \boxed{\textbf{(C) } 93}$

The traffic light runs through a $63$ second cycle. Letting $t=0$ reference the moment it turns green, the light changes at three different times: $t=30$, $t=33$, and $t=63$ This means that the light will change if the beginning of Leah's interval lies in $[27,30]$, $[30,33]$ or $[60,63]$ This gives a total of $9$ seconds out of $63$ $\frac{9}{63} = \frac{1}{7} \Rightarrow \mathrm{(D)}$

Drawing $\overline{PA}$, $\overline{PB}$, and $\overline{PC}$, $\triangle ABC$ is split into three smaller triangles. The altitudes of these triangles are given in the problem as $PQ$, $PR$, and $PS$. Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get: \[\frac{s}{2} + \frac{2s}{2} + \frac{3s}{2} = \frac{s^2\sqrt{3}}{4}\] where $s$ is the length of a side of the equilateral triangle \[s = \boxed{\mathrm{(D) \ } 4\sqrt{3}}\]

The sum of an infinite geometric series is given by $\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio. In this series, $\frac{a}{1-r} = 7$ The series with odd powers of $r$ is given as \[ar + ar^3 + ar^5 ...\] Its sum can be given by $\frac{ar}{1-r^2} = 3$ Doing a little algebra $ar = 3(1-r)(1+r)$ $ar = 3\left(\frac{a}{7}\right)(1+r)$ $\frac{7}{3}r = 1 + r$ $r = \frac{3}{4}$ $a = 7(1-r) = \frac{7}{4}$ $a + r =\boxed{ \frac{5}{2}} \Rightarrow \mathrm{(E)}$

A tetrahedron has 4 sides. We can ignore the rotation part of the question completely and focus on the colors. The ratio of the number of faces with each color must be one of the following: $4:0:0$, $3:1:0$, $2:2:0$, or $2:1:1$ The first ratio yields $3$ appearances, one of each color. The second ratio yields $3\cdot 2 = 6$ appearances, three choices for the first color, and two choices for the second. The third ratio yields $\binom{3}{2} = 3$ appearances since the two colors are interchangeable. The fourth ratio yields $3$ appearances. There are three choices for the first color, and since the second two colors are interchangeable, there is only one distinguishable pair that fits them. The total is $3 + 6 + 3 + 3 = 15$ appearances $\Rightarrow \mathrm{(A)}$

Note that if $a$ is positive, then, the equation will have no solutions for $b$. This becomes more obvious by noting that at $a=1$, $ab^2 > \log_{10} b$. The LHS quadratic function will increase faster than the RHS logarithmic function, so they will never intersect. This puts $a$ as the smallest in the set since it must be negative. Checking the new equation: $-b^2 = \log_{10}b$ Near $b=0$, $-b^2 > \log_{10} b$ but at $b=1$, $-b^2 < \log_{10} b$ This implies that the solution occurs somewhere in between: $0 < b < 1$ This also implies that $\frac{1}{b} > 1$ This makes our set (ordered) $\{a,0,b,1,1/b\}$ The median is $b \Rightarrow \mathrm {(D)}$

The difference between $acb$ and $abc$ is given by $(100a + 10c + b) - (100a + 10b + c) = 9(c-b)$ The difference between the two squares is three times this amount or $27(c-b)$ The difference between two consecutive squares is always an odd number, therefore $c-b$ is odd. We will show that $c-b$ must be 1. Otherwise we would be looking for two consecutive squares that are at least 81 apart. But already the equation $(x+1)^2-x^2 = 27\cdot 3$ solves to $x=40$, and $40^2$ has more than three digits. The consecutive squares with common difference $27$ are $13^2=169$ and $14^2=196$. One third of the way between them is $178$ and two thirds of the way is $187$. This gives $a=1$, $b=7$, $c=8$. $a+b+c = 16 \Rightarrow \mathrm{(C)}$

$V_{\mathrm{Cylinder}} = \pi r^2 h$ Where $C = 2\pi r = 6$ and $h=6\sin\theta$ $r = \frac{3}{\pi}$ $V = \pi \left(\frac{3}{\pi}\right)^2\cdot 6\sin\theta$ $6 = \frac{9}{\pi} \cdot 6\sin\theta$ $\sin\theta = \frac{\pi}{9} \Rightarrow \mathrm{(A)}$

Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$. Because $72= 4\cdot 18$, we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$, which gives $c=3d$ (consider a homothety, or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by $4\times$, it follows that the stretch along the diagonal, or the ratio of side lengths, is $2\times$). The area of the triangular half of the parallelogram on the right side of the y-axis is given by $9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)$, so substituting $c = 3d$: \[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) = 9 \quad \Longrightarrow \quad 2d^2 = 9(a-b)\] Thus $3|d$, and we verify that $d = 3$, $a-b = 2 \Longrightarrow a = 3, b = 1$ will give us a minimum value for $a+b+c+d$. Then $a+b+c+d = 3 + 1 + 9 + 3 = \boxed{\mathbf{(D)} 16}$.

$2007_{10} = 2202100_{3}$ All numbers of six or less digits in base 3 have been written. The form of each palindrome is as follows 1 digit - $a$ 2 digits - $aa$ 3 digits - $aba$ 4 digits - $abba$ 5 digits - $abcba$ 6 digits - $abccba$ Where $a,b,c$ are base 3 digits Since $a \neq 0$, this gives a total of $2 + 2 + 2\cdot 3 + 2\cdot 3 + 2\cdot 3^2 + 2\cdot 3^2 = 52$ palindromes so far. 7 digits - $abcdcba$, but not all of the numbers are less than $2202100_{3}$ Case: $a=1$ All of these numbers are less than $2202100_3$ giving $3^3$ more palindromes Case: $a=2$, $b\neq 2$ All of these numbers are also small enough, giving $2\cdot 3^2$ more palindromes Case: $a=2$, $b=2$ It follows that $c=0$, since any other $c$ would make the value too large. This leaves the number as $220d022_3$. Checking each value of d, all of the three are small enough, so that gives $3$ more palindromes. Summing our cases there are $52 + 3^3 + 2\cdot 3^2 + 3 = 100 \Rightarrow \mathrm{(A)}$

First, notice that each of the midpoints of $AB$,$BC$, and $CA$ are on the locus. Suppose after some time the particles have each been displaced by a short distance $x$, to new positions $A'$ and $M'$ respectively. Consider $\triangle ABM$ and drop a perpendicular from $M'$ to hit $AB$ at $Y$. Then, $BA'=1-x$ and $BM'=1/2+x$. From here, we can use properties of a $30-60-90$ triangle to determine the lengths $YA'$ and $YM'$ as monomials in $x$. Thus, the locus of the midpoint will be linear between each of the three special points mentioned above. It follows that the locus consists of the only triangle with those three points as vertices. (A cheaper way to find the shape of the region is to look at the answer choices: if it were any sort of conic section then the ratio would not generally be rational.) Comparing inradii between this "midpoint" triangle and the original triangle, the area contained by $R$ must be $\textbf{(A)} \frac{1}{16}$ of the total area.

Let $a$ and $b$ be the two legs of the triangle. We have $\frac{1}{2}ab = 3(a+b+c)$. Then $ab=6 \left(a+b+\sqrt {a^2 + b^2}\right)$. We can complete the square under the root, and we get, $ab=6 \left(a+b+\sqrt {(a+b)^2 - 2ab}\right)$. Let $ab=p$ and $a+b=s$, we have $p=6 \left(s+ \sqrt {s^2 - 2p}\right)$. After rearranging, squaring both sides, and simplifying, we have $p=12s-72$. Putting back $a$ and $b$, and after factoring using Simon's Favorite Factoring Trick, we've got $(a-12)(b-12)=72$. Factoring 72, we get 6 pairs of $a$ and $b$ $(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).$ And this gives us $6$ solutions $\Rightarrow \mathrm{(A)}$. Alternatively, note that $72 = 2^3 \cdot 3^2$. Then 72 has $(3+1)(2+1) = (4)(3) = 12$ factors. However, half of these are repeats, so we have $\frac{12}{2} = 6$ solutions.

Combining the fraction, $\frac{9a^2 + 14b^2}{9ab}$ must be an integer. Since the denominator contains a factor of $9$, $9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b$ Since $b = 3n$ for some positive integer $n$, we can rewrite the fraction(divide by $9$ on both top and bottom) as $\frac{a^2 + 14n^2}{3an}$ Since the denominator now contains a factor of $n$, we get $n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2$. But since $1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)$, we must have $n=1$, and thus $b=3$. For $b=3$ the original fraction simplifies to $\frac{a^2 + 14}{3a}$. For that to be an integer, $a$ must be a factor of $14$, and therefore we must have $a\in\{1,2,7,14\}$. Each of these values does indeed yield an integer. Thus there are four solutions: $(1,3)$, $(2,3)$, $(7,3)$, $(14,3)$ and the answer is $\mathrm{(A)}$

Let $A=(0,0,0)$, and $B=(2,0,0)$. Since $EA=2$, we could let $C=(2,0,2)$, $D=(2,2,2)$, and $E=(2,2,0)$. Now to get back to $A$ we need another vertex $F=(0,2,0)$. Now if we look at this configuration as if it was two dimensions, we would see a square missing a side if we don't draw $FA$. Now we can bend these three sides into an equilateral triangle, and the coordinates change: $A=(0,0,0)$, $B=(2,0,0)$, $C=(2,0,2)$, $D=(1,\sqrt{3},2)$, and $E=(1,\sqrt{3},0)$. Checking for all the requirements, they are all satisfied. Now we find the area of triangle $BDE$. The side lengths of this triangle are $2, 2, 2\sqrt{2}$, which is an isosceles right triangle. Thus the area of it is $\frac{2\cdot2}{2}=2\Rightarrow \mathrm{(C)}$.