AMC12 2005 A

$2 = \frac {1}{10}x \Longrightarrow x = 20,\quad 2 = \frac{1}{5}y \Longrightarrow y = 10,\quad x-y = 20 - 10=10 \mathrm{(D)}$.

$2x + 7 = 3 \iff x = -2$, so we require $-2b-10 = -2 \iff -2b = 8 \iff b = \boxed{\textbf{(B) } -4}$.

Let $w$ be the width, so the length is $2w$. By the Pythagorean Theorem, $w^2 + 4w^2 = x^2 \Longrightarrow \frac{x}{\sqrt{5}} = w$. The area of the rectangle is $(w)(2w) = 2w^2 = 2\left(\frac{x}{\sqrt{5}}\right)^2 = \frac{2}{5}x^2 \ \mathrm{(B)}$.

For $n$ windows, the store offers a discount of $100 \cdot \left\lfloor\frac{n}{5}\right\rfloor$ (floor function). Dave receives a discount of $100 \cdot \left\lfloor \frac{7}{5}\right \rfloor = 100$ and Doug receives a discount of $100 \cdot \left\lfloor \frac{8}{5}\right\rfloor = 100$. These amount to $200$ dollars in discounts. Together, they receive a discount of $100 \cdot \left\lfloor \frac{15}{5} \right\rfloor = 300$, so they save $300-200=100\ \mathrm{(A)}$.

The sum of the first $20$ numbers is $20 \cdot 30$ and the sum of the other $30$ numbers is $30\cdot 20$. Hence the overall average is $\frac{20 \cdot 30 + 30 \cdot 20}{50} = 24 \ \mathrm{(B)}$. This is just the harmonic mean, so the answer is $\frac{2 \cdot 20 \cdot 30}{20+30}=24 \ \mathrm{(B)}$. Solution by franzliszt

Let $D_J, D_M$ be the distances traveled by Josh and Mike, respectively, and let $r,t$ be the time and rate of Mike. Using $d = rt$, we have that $D_M = rt$ and $D_J = \left(\frac{4}{5}r\right)\left(2t\right) = \frac 85rt$. Then $13 = D_M + D_J = rt + \frac{8}{5}rt = \frac{13}{5}rt$ $\Longrightarrow rt = D_M = 5\ \mathrm{(B)}$.

Error creating thumbnail: Unable to save thumbnail to destination Arguable the hardest part of this question is to visualize the diagram. Since each side of $EFGH$ can be extended to pass through a vertex of $ABCD$, we realize that $EFGH$ must be tilted in such a fashion. Let a side of $EFGH$ be $x$. Error creating thumbnail: Unable to save thumbnail to destination Notice the right triangle (in blue) with legs $1, x+1$ and hypotenuse $\sqrt{50}$. By the Pythagorean Theorem, we have $1^2 + (x+1)^2 = (\sqrt{50})^2 \Longrightarrow (x+1)^2 = 49 \Longrightarrow x = 6$. Thus, $[EFGH] = x^2 = 36\ \mathrm{(C)}$

Clearly the two quantities are both integers, so we check the prime factorization of $2005 = 5 \cdot 401$. It is easy to see now that $(A,M,C) = (4,0,1)$ works, so the answer is $\mathrm{(D)}$.

https://youtu.be/3dfbWzOfJAI?t=222

There are $6n^3$ sides total on the unit cubes, and $6n^2$ are painted red. $\dfrac{6n^2}{6n^3}=\dfrac{1}{4} \Rightarrow n=4 \rightarrow \mathrm {B}$

Let the digits be $A, B, C$ so that $B = \frac {A + C}{2}$. In order for this to be an integer, $A$ and $C$ have to have the same parity. There are $9$ possibilities for $A$, and $5$ for $C$. $B$ depends on the value of both $A$ and $C$ and is unique for each $(A,C)$. Thus our answer is $9 \cdot 5 \cdot 1 = 45 \implies E$.

For convenience’s sake, we can transform $A$ to the origin and $B$ to $(99,999)$ (this does not change the problem). The line $AB$ has the equation $y = \frac{999}{99}x = \frac{111}{11}x$. The coordinates are integers if $11|x$, so the values of $x$ are $11, 22 \ldots 88$, with a total of $8\implies \boxed{\mathrm{(D)}}$ coordinates.

$(A+B) + (B+C) + (C+D) + (D+E) + (E+A) = 2(A+B+C+D+E)$ (i.e., each number is counted twice). The sum $A + B + C + D + E$ will always be $3 + 5 + 6 + 7 + 9 = 30$, so the arithmetic sequence has a sum of $2 \cdot 30 = 60$. The middle term must be the average of the five numbers, which is $\frac{60}{5} = 12 \Longrightarrow \mathrm{(D)}$.

There are $1 + 2 + 3 + 4 + 5 + 6 = 21$ dots total. Casework: - The dot is removed from an even face. There is a $\frac{2+4+6}{21} = \frac{4}{7}$ chance of this happening. Then there are 4 odd faces, giving us a probability of $\frac 47 \cdot \frac 46 = \frac{8}{21}$. - The dot is removed from an odd face. There is a $\frac{1+3+5}{21} = \frac{3}{7}$ chance of this happening. Then there are 2 odd faces, giving us a probability of $\frac 37 \cdot \frac 26 = \frac{1}{7}$. Thus the answer is $\frac 8{21} + \frac 17 = \frac{11}{21} \Longrightarrow \mathrm{(D)}$.

Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or $\frac{CF}{CD}$ ($F$ is the foot of the perpendicular from $C$ to $DE$). Call the radius $r$. Then $AC = \frac 13(2r) = \frac 23r$, $CO = \frac 13r$. Using the Pythagorean Theorem in $\triangle OCD$, we get $\frac{1}{9}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r$. Now we have to find $CF$. Notice $\triangle OCD \sim \triangle OFC$, so we can write the proportion: By the Pythagorean Theorem in $\triangle OFC$, we have $\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r$. Our answer is $\frac{CF}{CD} = \frac{\frac{2\sqrt{2}}{9}r}{\frac{2\sqrt{2}}{3}r} = \frac 13 \Longrightarrow \mathrm{(C)}$. Let the center of the circle be $O$. Note that $2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB$. $O$ is midpoint of $AB \Rightarrow \frac{3}{2}AC = AO \Rightarrow CO = \frac{1}{3}AO \Rightarrow CO = \frac{1}{6} AB$. $O$ is midpoint of $DE \Rightarrow$ Area of $\triangle DCE = 2 \cdot$ Area of $\triangle DCO = 2 \cdot (\frac{1}{6} \cdot$ Area of $\triangle ABD) = \frac{1}{3} \cdot$ Area of $\triangle ABD \Longrightarrow \mathrm{(C)}$. Let $r$ be the radius of the circle. Note that $AC+BC = 2r$ so $AC = \frac{2}{3}r$. By Power of a Point Theorem, $CD^2= AC \cdot BC = 2\cdot AC^2$, and thus $CD = \sqrt{2} \cdot AC = \frac{2\sqrt{2}}{3}r$ Then the area of $\triangle ABD$ is $\frac{1}{2} AB \cdot CD = \frac{2\sqrt{2}}{3}r^2$. Similarly, the area of $\triangle DCE$ is $\frac{1}{2}(r-AC) \cdot 2 \cdot CD = \frac{2\sqrt{2}}{9}r^2$, so the desired ratio is $\frac{\frac{2\sqrt{2}}{9}r^2}{\frac{2\sqrt{2}}{3}r^2} = \frac{1}{3} \Longrightarrow \mathrm{(C)}$ Let the center of the circle be $O$. Without loss of generality, let the radius of the circle be equal to $3$. Thus, $AO=3$ and $OB=3$. As a consequence of $2(AC)=BC$, $AC=2$ and $CO=1$. Also, we know that $DO$ and $OE$ are both equal to $3$ due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to $\sqrt{3^2-1^2}$ or $2\sqrt{2}$. Now we know that the area of $[ABD]$ is equal to $\frac{(3+2+1)(2\sqrt{2})}{2}$ or $6\sqrt{2}$. Know we need to find the area of $[DCE]$. By simple inspection $[COD]$ $\cong$ $[HOE]$ due to angles being equal and CPCTC. Thus $HE=2\sqrt{2}$ and $OH=1$. Know we know the area of $[CHE]=\frac{(1+1)(2\sqrt{2})}{2}$ or $2\sqrt{2}$. We also know that the area of $[OHE]=\frac{(1)(2\sqrt{2})}{2}$ or $\sqrt{2}$. Thus the area of $[COE]=2\sqrt{2}-\sqrt{2}$ or $\sqrt{2}$. We also can calculate the area of $[DOC]$ to be $\frac{(1)(2\sqrt{2})}{2}$ or $\sqrt{2}$. Thus $[DCE]$ is equal to $[COE]$ + $[DOC]$ or $\sqrt{2}+\sqrt{2}$ or $2\sqrt{2}$. The ratio between $[DCE]$ and $[ABD]$ is equal to $\frac{2\sqrt{2}}{6\sqrt{2}}$ or $\frac{1}{3}$ $\Longrightarrow \mathrm{(C)}$. We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for $D (x,y)$, and notice how $E$ is a 180 degree rotation of $D$, using the rotation matrix formula we get $E = (-x,-y)$. WLOG say that this circle has radius $3$. We can now find points $A$, $B$, and $C$ which are $(-3,0)$, $(3,0)$, and $(-1,0)$ respectively. By shoelace the area of $CED$ is $Y$, and the area of $ADB$ is $3Y$. Using division we get that the answer is $\mathrm{(C)}$. We set point $A$ as a mass of 2. This means that point $B$ has a mass of $1$ since $2\times{AC} = 1\times{BC}$. This implies that point $C$ has a mass of $2+1 = 3$ and the center of the circle has a mass of $3+1 = 4$. After this, we notice that points $D$ and $E$ both must have a mass of $2$ since $2+2 = 4$ and they are both radii of the circle. To find the ratio of the areas, we do the reciprocal of the multiplication of the mass points of DCE and the multiplication of ABD divided by each other. Which is simply $\frac{3\times2\times2}{2\times2\times1}$ which is $\boxed{\frac{1}{3}}$ (the reciprocal of 3)

Error creating thumbnail: Unable to save thumbnail to destination Set $s =1$ so that we only have to find $r$. Draw the segment between the center of the third circle and the large circle; this has length $r+1$. We then draw the radius of the large circle that is perpendicular to the x-axis, and draw the perpendicular from this radius to the center of the third circle. This gives us a right triangle with legs $r-3,r-1$ and hypotenuse $r+1$. The Pythagorean Theorem yields: Quite obviously $r > 1$, so $r = 9$.

It is a pyramid with height $1$ and base area $\frac{1}{4}$, so using the formula for the volume of a pyramid, $\frac{1}{3} \cdot \left(\frac{1}{4}\right) \cdot (1) = \frac {1}{12} \Rightarrow \boxed{(\mathrm {A})}$.

The given states that there are $168$ prime numbers less than $1000$, which is a fact we must somehow utilize. Since there seems to be no easy way to directly calculate the number of "prime-looking" numbers, we can apply complementary counting. We can split the numbers from $1$ to $1000$ into several groups: $\{1\},$ $\{\mathrm{numbers\ divisible\ by\ 2 = S_2}\},$ $\{\mathrm{numbers\ divisible\ by\ 3 = S_3}\},$ $\{\mathrm{numbers\ divisible\ by\ 5 = S_5}\}, \{\mathrm{primes\ not\ including\ 2,3,5}\},$ $\{\mathrm{prime-looking}\}$. Hence, the number of prime-looking numbers is $1000 - (168-3) - 1 - |S_2 \cup S_3 \cup S_5|$ (note that $2,3,5$ are primes). We can calculate $S_2 \cup S_3 \cup S_5$ using the Principle of Inclusion-Exclusion: (the values of $|S_2| \ldots$ and their intersections can be found quite easily) Substituting, we find that our answer is $1000 - 165 - 1 - 734 = 100 \Longrightarrow \mathrm{(A)}$.

We find the number of numbers with a $4$ and subtract from $2005$. Quick counting tells us that there are $200$ numbers with a 4 in the hundreds place, $200$ numbers with a 4 in the tens place, and $201$ numbers with a 4 in the units place (counting $2004$). Now we apply the Principle of Inclusion-Exclusion. There are $20$ numbers with a 4 in the hundreds and in the tens, and $20$ for both the other two intersections. The intersection of all three sets is just $2$. So we get: Alternatively, consider that counting without the number $4$ is almost equivalent to counting in base $9$; only, in base $9$, the number $9$ is not counted. Since $4$ is skipped, the symbol $5$ represents $4$ miles of travel, and we have traveled $2004_9$ miles. By base conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}$ Since any numbers containing one or more $4$s were skipped, we need only to find the numbers that don't contain a $4$ at all. First we consider $1$ - $1999$. Single digits are simply four digit numbers with a zero in all but the ones place (this concept applies to double and triple digits numbers as well). From $1$ - $1999$, we have $2$ possibilities for the thousands place, and $9$ possibilities for the hundreds, tens, and ones places. This is $2 \cdot 9 \cdot 9 \cdot 9-1$ possibilities (because $0000$ doesn't count) or $1457$ numbers. From $2000$ - $2005$ there are $6$ numbers, $5$ of which don't contain a $4$. Therefore the total is $1457 + 5$, or $1462$ $\Rightarrow$ $\boxed{\text{B}}$. We seek to find the amount of numbers that contain at least one $4,$ and subtract this number from $2005.$ We can simply apply casework to this problem. The amount of numbers with at least one $4$ that are one or two digit numbers are $4,14,24,34,40-49,54,\cdots,94$ which gives $19$ numbers. The amount of three digit numbers with at least one $4$ is $8*19+100=252.$ The amount of four digit numbers with at least one $4$ is $252+1+19=272$ This, our answer is $2005-19-252-272=1462,$ or $\boxed{B}.$ This is very analogous to base $9$. But, in base $9$, we don't have a $9$. So, this means that these are equal except for that base 9 will be one more than the operation here. $2005_9 = 5+0+0+1458 = 1463$. $1463 - 1 = 1462$

For the two functions $f(x)=2x,0\le x\le \frac{1}{2}$ and $f(x)=2-2x,\frac{1}{2}\le x\le 1$,as long as $f(x)$ is between $0$ and $1$, $x$ will be in the right domain, so we don't need to worry about the domain of $x$. Also, every time we change $f(x)$, the expression for the final answer in terms of $x$ will be in a different form (although they'll all satisfy the final equation), so we get a different starting value of $x$. Every time we have two choices for $f(x$) and altogether we have to choose $2005$ times. Thus, $2^{2005}\Rightarrow\boxed{E}$. Note: the values of x that satisfy $f^{[n]}(x) = \frac {1}{2}$ are $\frac{1}{2^{n+1}}$, $\frac{3}{2^{n+1}}$, $\frac{5}{2^{n+1}}$, $\cdots$ ,$\frac{2^{n+1}-1}{2^{n+1}}$.

$a^{c^{2005}} = b$ Casework upon $c$: - $c = 0$: Then $a^0 = b \Longrightarrow b = 1$. Thus we get $(2004,1,0)$. - $c = 1$: Then $a^1 = b \Longrightarrow a = b$. Thus we get $(1002,1002,1)$. - $c \ge 2$: Then the exponent of $a$ becomes huge, and since $a \ge 2$ there is no way we can satisfy the second condition. Hence we have two ordered triples $\mathrm{(C)}$.

Box P has dimensions $l$, $w$, and $h$. Its surface area is \[2lw+2lh+2wh=384,\] and the sum of all its edges is \[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.\] The diameter of the sphere is the space diagonal of the prism, which is \[\sqrt{l^2 + w^2 +h^2}.\] Notice that \begin{align*}(l + w + h)^2 - (2lw + 2lh + 2wh) &= l^2 + w^2 + h^2 \\ &= 784 - 384 \\ &= 400,\end{align*} so the diameter is \[\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.\] The radius is half of the diameter, so \[r=\frac{20}{2} = \boxed{\textbf{(B) } 10}.\]

Let $\log_{a}b = z$, so $a^z = b$. Define $a = 2^x$, $b = 2^y$; then $\left(2^x\right)^z = 2^{xz}= 2^y$, so $x \mid y$. Here we can just make a table and count the number of values of $y$ per value of $x$. The largest possible value of $x$ is $12$, and so we get $\sum_{x=1}^{12} \left\lfloor\frac{25}{x}-1\right\rfloor = 24 + 11 + 7 + 5 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 = 62$. The total number of ways to pick two distinct numbers (where the order matters) is $\frac{25!}{(25-2)!}= 25 \cdot 24 = 600$, so we get a probability of $\frac{62}{600} = \boxed{\textbf{(B) }\frac{31}{300}}$.

We can write the problem as Since $\deg P(x) = 3$ and $\deg R(x) = 3$, $\deg P(x)\cdot R(x) = 6$. Thus, $\deg P(Q(x)) = 6$, so $\deg Q(x) = 2$. Hence, we conclude $Q(1)$, $Q(2)$, and $Q(3)$ must each be $1$, $2$, or $3$. Since a quadratic is uniquely determined by three points, there can be $3*3*3 = 27$ different quadratics $Q(x)$ after each of the values of $Q(1)$, $Q(2)$, and $Q(3)$ are chosen. However, we have included polynomials $Q(x)$ which are linear, rather than quadratic. Namely, Clearly, we could not have included any other constant functions. For any linear function, we have $2\cdot Q(2) = Q(1) + Q(3)$ because $Q(2)$ is the $y$-coordinate of the midpoint of $(1, Q(1))$ and $(3, Q(3))$. So we have not included any other linear functions. Therefore, the desired answer is $27 - 5 = \boxed{\textbf{(B) }22}$.

For this solution, we will just find as many equilateral triangles as possible, until it becomes intuitive that there are no more size of triangles left. First, we observe that we can form an equilateral triangle with vertices in $S$ by taking any point in $S$ and connecting it to the $2$ adjacent points. This triangle will have a side length of $\sqrt{2}$; a quick further examination of this cube will show us that this is the only possible side length (the red triangle in the diagram below). Each of these triangles is determined by one vertex of the cube, so in one cube we have $8$ equilateral triangles. We have $8$ unit cubes, as well as the entire $2 \times 2 \times 2$ cube (giving the green triangle in the diagram), for a total of $8+1 = 9$ cubes, and thus $9 \cdot 8 = 72$ equilateral triangles. (Note that connecting the centers of the faces will actually give an octahedron, not a cube, because it only has $6$ vertices.) Now we look for any further equilateral triangles. Connecting the midpoints of $3$ non-adjacent, non-parallel edges indeed gives us more equilateral triangles (e.g. the blue triangle in the diagram below). Notice that picking these $3$ edges leaves $2$ vertices alone (labelled A and B in the diagram), and that picking any $2$ opposite vertices determines $2$ equilateral triangles. Hence there are $\frac{8 \cdot 2}{2} = 8$ of these equilateral triangles, so adding these to the triangles already found above gives a total of $72+8 = \boxed{\textbf{(C) }80}$.