AMC12 2007 A

The brick has volume $8000 cm^3$. The base of the aquarium has area $4000 cm^2$. For every cm the water rises, the volume increases by $4000 cm^3$; therefore, when the volume increases by $8000 cm^3$, the water level rises $2 cm \Rightarrow\fbox{D}$

Let $n$ be the smaller term. Then $n+2=3n \Longrightarrow 2n = 2 \Longrightarrow n=1$ - Thus, the answer is $1+(1+2)=4 \mathrm{(A)}$

\[16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14\] \[\frac{14}2=7\Rightarrow\boxed{A}\]

After paying his taxes, he has $0.8*0.9=0.72$ of his earnings left. Since $10500$ is $0.28$ of his income, he got a total of $\frac{10500}{0.28}=37500\ \mathrm{(D)}$.

Error creating thumbnail: Unable to save thumbnail to destination We angle chase and find out that: - $\angle DAC=\frac{180-140}{2} = 20$ - $\angle BAC=\frac{180-40}{2} = 70$ - $\angle BAD=\angle BAC- \angle DAC=50\ \mathrm{(D)}$

Let $f$ be the common difference between the terms. - $a=c-2f$ - $b=c-f$ - $c=c$ - $d=c+f$ - $e=c+2f$ $a+b+c+d+e=5c=30$, so $c=6$. But we can't find any more variables, because we don't know what $f$ is. So the answer is $\textrm{C}$.

Error creating thumbnail: Unable to save thumbnail to destination We look at the angle between 12, 5, and 10. It subtends $\frac 16$ of the circle, or $60$ degrees (or you can see that the arc is $\frac 23$ of the right angle). Thus, the angle at each vertex is an inscribed angle subtending $60$ degrees, making the answer $\frac 1260 = 30^{\circ} \Longrightarrow \mathrm{(C)}$

Let the distance from Yan's initial position to the stadium be $a$ and the distance from Yan's initial position to home be $b$. We are trying to find $b/a$, and we have the following identity given by the problem: \begin{align*}a &= b + \frac{a+b}{7}\\ \frac{6a}{7} &= \frac{8b}{7} \Longrightarrow 6a = 8b\end{align*} Thus $b/a = 6/8 = 3/4$ and the answer is $\mathrm{(B)}\ \frac{3}{4}$ Note: The identity can be given from an application of $t=\frac dr$.

Error creating thumbnail: Unable to save thumbnail to destination Since 3-4-5 is a Pythagorean triple, the triangle is a right triangle. Since the hypotenuse is a diameter of the circumcircle, the hypotenuse is $2r = 6$. Then the other legs are $\frac{24}5=4.8$ and $\frac{18}5=3.6$. The area is $\frac{4.8 \cdot 3.6}2 = 8.64\ \mathrm{(A)}$

A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S=111k=3 \cdot 37k$, so $S$ must be divisible by $37\ \mathrm{(D)}$. To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\boxed{(D)~~37}}$.

The only time when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity. The chance of $ad$ being odd is $\frac 12 \cdot \frac 12 = \frac 14$, since the only way to have $ad$ be odd is to have both $a$ and $d$ be odd. As a result, $ad$ has a $\frac 34$ probability of being even. $bc$ also has a $\frac 14$ chance of being odd and a $\frac34$ chance of being even. Therefore, the probability that $ad-bc$ will be even is $\left(\frac 14\right)^2+\left(\frac 34\right)^2=\boxed {\mathrm{(E )} \frac 58\ }$.

The point $(a,b)$ is the foot of the perpendicular from $(12,10)$ to the line $y=-5x+18$. The perpendicular has slope $\frac{1}{5}$, so its equation is $y=10+\frac{1}{5}(x-12)=\frac{1}{5}x+\frac{38}{5}$. The $x$-coordinate at the foot of the perpendicular satisfies the equation $\frac{1}{5}x+\frac{38}{5}=-5x+18$, so $x=2$ and $y=-5\cdot2+18=8$. Thus $(a,b) = (2,8)$, and $a+b = \boxed{10}$.

If $45$ is expressed as a product of five distinct integer factors, the absolute value of the product of any four is at least $|(-3)(-1)(1)(3)|=9$, so no factor can have an absolute value greater than $5$. Thus the factors of the given expression are five of the integers $\pm 3, \pm 1, \pm 5$. The product of all six of these is $-225=(-5)(45)$, so the factors are $-3, -1, 1, 3,$ and $5.$ The corresponding values of $a, b, c, d,$ and $e$ are $9, 7, 5, 3,$ and $1,$ and their sum is $\fbox{25 (C)}$

The median must either be $6, 9,$ or $n$. Casework: - Median is $6$: Then $n \le 6$ and $\frac{3+6+9+10+n}{5} = 6 \Longrightarrow n = 2$. - Median is $9$: Then $n \ge 9$ and $\frac{3+6+9+10+n}{5} = 9 \Longrightarrow n = 17$. - Median is $n$: Then $6 < n < 9$ and $\frac{3+6+9+10+n}{5} = n \Longrightarrow n = 7$. All three cases are valid, so our solution is $2 + 7 + 17 = 26 \Longrightarrow \mathrm{(E)}$.

We can find the number of increasing arithmetic sequences of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences. This gives us a total of $2 + 4 + 6 + 8 = 20$ sequences. There are $3! = 6$ to permute these, for a total of $120$. However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are $2! \cdot 4 = 8$ numbers which start with zero, so our answer is $120 - 8 = 112 \Longrightarrow \mathrm{(C)}$.

We can make use the of the trigonometric Pythagorean identities: square both equations and add them up: This is just the cosine difference identity, which simplifies to $\cos (a - b) = \frac{1}{3} \Longrightarrow \mathrm{(B)}$

A fourth degree polynomial has four roots. Since the coefficients are real(meaning that complex roots come in conjugate pairs), the remaining two roots must be the complex conjugates of the two given roots. By the factor theorem, our roots are $2-i,-2i$. Now we work backwards for the polynomial: Thus our answer is $- 4 + 9 - 16 + 20 = 9\ \mathrm{(D)}$.

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Since the sides of a regular polygon are equal in length, we can call each side $x$. Examine one edge of the unit cube: each contains two slanted diagonal edges of an octagon and one straight edge. The diagonal edges form $45-45-90 \triangle$ right triangles, making the distance on the edge of the cube $\frac{x}{\sqrt{2}}$. Thus, $2 \cdot \frac{x}{\sqrt{2}} + x = 1$, and $x = \frac{1}{\sqrt{2} + 1} \cdot \left(\frac{\sqrt{2} - 1}{\sqrt{2} - 1}\right) = \sqrt{2} - 1$. Each of the cut off corners is a pyramid, whose volume can be calculated by $V = \frac 13 Bh$. Use the base as one of the three congruent isosceles triangles, with the height being one of the edges of the pyramid that sits on the edges of the cube. The height is $\frac x{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}}$. The base is a $45-45-90 \triangle$ with leg of length $1 - \frac{1}{\sqrt{2}}$, making its area $\frac 12\left(1 - \frac 1{\sqrt{2}}\right)^2 = \frac{3 - 2\sqrt{2}}4$. Plugging this in, we get that the volume of one of the tetrahedra is $\frac 13 \left(1 - \frac{1}{\sqrt{2}}\right)\left(\frac{3 - 2\sqrt{2}}4\right) = \frac{10 - 7\sqrt{2}}{24}$. Since there are 8 removed corners, we get an answer of $\frac{10 - 7\sqrt{2}}{3} \Longrightarrow \boxed{\mathrm{B}}$

By Vieta's Formulas, the sum of the roots of a quadratic equation is $\frac {-b}a$, the product of the zeros is $\frac ca$, and the sum of the coefficients is $a + b + c$. Setting equal the first two tells us that $\frac {-b}{a} = \frac ca \Rightarrow b = -c$. Thus, $a + b + c = a + b - b = a$, so the common value is also equal to the coefficient of $x^2 \Longrightarrow \fbox{\textrm{A}}$. To disprove the others, note that: - $\mathrm{B}$: then $b = \frac {-b}a$, which is not necessarily true. - $\mathrm{C}$: the y-intercept is $c$, so $c = \frac ca$, not necessarily true. - $\mathrm{D}$: an x-intercept of the graph is a root of the polynomial, but this excludes the other root. - $\mathrm{E}$: the mean of the x-intercepts will be the sum of the roots of the quadratic divided by 2.

For the sake of notation, let $T(n) = n + S(n) + S(S(n))$. Obviously $n<2007$. Then the maximum value of $S(n) + S(S(n))$ is when $n = 1999$, and the sum becomes $28 + 10 = 38$. So the minimum bound is $2007-38=1969$. We do casework upon the tens digit: Case 1: $196u \Longrightarrow u = 9$. Easy to directly disprove. Case 2: $197u$. $S(n) = 1 + 9 + 7 + u = 17 + u$, and $S(S(n)) = 8+u$ if $u \le 2$ and $S(S(n)) = 2 + (u-3) = u-1$ otherwise. Case 3: $198u$. $S(n) = 18 + u$, and $S(S(n)) = 9 + u$ if $u \le 1$ and $2 + (u-2) = u$ otherwise. Case 4: $199u$. But $S(n) > 19$, and $n + S(n)$ clearly sum to $> 2007$. Case 5: $200u$. So $S(n) = 2 + u$ and $S(S(n)) = 2 + u$ (recall that $n < 2007$), and $2000 + u + 2 + u + 2 + u = 2004 + 3u = 2007 \Longrightarrow u = 1$. Fourth solution. In total we have $4 \mathrm{(D)}$ solutions, which are $1977, 1980, 1983,$ and $2001$.

Let $x$ be the x-coordinate of $B$ and $C$, and $x_2$ be the x-coordinate of $A$ and $y$ be the y-coordinate of $A$ and $B$. Then $2\log_ax= y \Longrightarrow a^{y/2} = x$ and $\log_ax_2 = y \Longrightarrow x_2 = a^y = \left(a^{y/2}\right)^2 = x^2$. Since the distance between $A$ and $B$ is $6$, we have $x^2 - x - 6 = 0$, yielding $x = -2, 3$. However, we can discard the negative root (all three logarithmic equations are underneath the line $y = 3$ and above $y = 0$ when $x$ is negative, hence we can't squeeze in a square of side 6). Thus $x = 3$. Substituting back, $3\log_{a}x - 2\log_{a}x = 6 \Longrightarrow a^6 = x$, so $a = \sqrt[6]{3}\ \ \mathrm{(A)}$.

$F(2)=3$ By looking at various graphs, we obtain that, for most of the graphs $F(n) = n + 1$ Notice that the solutions are basically reflections across $x = \frac{\pi}{2}$. However, when $n \equiv 1 \pmod{4}$, the middle apex of the sine curve touches the sine curve at the top only one time (instead of two reflected points), so we get here $F(n) = n$. $3+4+5+5+7+8+9+9+\cdots+2008$ $= (1+2+3+4+5+\cdots+2008) - 3 - 501$ $= \frac{(2008)(2009)}{2} - 504 = 2016532$ $\mathrm{(D)}$

Let $S_{n}$ denote the number of spacy subsets of $\{ 1, 2, ... n \}$. We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$. The spacy subsets of $S_{n + 1}$ can be divided into two groups: - $A =$ those not containing $n + 1$. Clearly $|A|=S_{n}$. - $B =$ those containing $n + 1$. We have $|B|=S_{n - 2}$, since removing $n + 1$ from any set in $B$ produces a spacy set with all elements at most equal to $n - 2,$ and each such spacy set can be constructed from exactly one spacy set in $B$. Hence, From this recursion, we find that And so the answer is $\boxed{\textbf{(E)}129}$.