AMC12 2004 B

Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made $\frac{1}{2} \cdot 48 = 24$ free throws, on the third $12$, on the second $6$, and on the first $3 \Rightarrow \mathrm{(A)}$. Because there are five days, or four transformations between days (day 1 $\rightarrow$ day 2 $\rightarrow$ day 3 $\rightarrow$ day 4 $\rightarrow$ day 5), she makes $48 \cdot \frac{1}{2^4} = \boxed{\mathrm{(A)}\ 3}$

If $a=0$ or $c=0$, the expression evaluates to $-d<0$. If $b=0$, the expression evaluates to $c-d\leq 2$. Case $d=0$ remains. In that case, we want to maximize $c\cdot a^b$ where $\{a,b,c\}=\{1,2,3\}$. Trying out the six possibilities we get that the greatest is $(a,b,c)=(3,2,1)$, where $c\cdot a^b=1\cdot 3^2=\boxed{\mathrm{(D)}\ 9}$.

$1296 = 2^4 3^4$ and $4+4=\boxed{8} \Longrightarrow \mathrm{(A)}$.

The digit $7$ can be either the tens digit ($70, 71, \dots, 79$: $10$ possibilities), or the ones digit ($17, 27, \dots, 97$: $9$ possibilities), but we counted the number $77$ twice. This means that out of the $90$ two-digit numbers, $10+9-1=18$ have at least one digit equal to $7$. Therefore the probability is $\dfrac{18}{90} = \boxed{\dfrac{1}{5}} \Longrightarrow \mathrm{(B)}$.

Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=\frac{7}{10}\cdot(60+d)$, which solves to $d=140$, and the sum of digits of $d$ is $\boxed{\mathrm{(A)}\ 5}$.

The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs $8$ miles and $10$ miles long. The hypotenuse length is $\sqrt{8^2 + 10^2}\approx12.8$, and thus the answer is $\boxed{\mathrm{(A)}\ 13}$. Without a calculator one can note that $8^2+10^2=164<169=13^2\Rightarrow\boxed{\mathrm{(A)}\ 13}$.

The area of the circle is $S_{\bigcirc}=100\pi$; the area of the square is $S_{\square}=100$. Exactly $\frac{1}{4}$ of the circle lies inside the square. Thus the total area is $\dfrac34 S_{\bigcirc}+S_{\square}=\boxed{\mathrm{(B)\ }100+75\pi}$.

The sum of the first $n$ odd numbers is $n^2$. As in our case $n^2=100$, we have $n=\boxed{\mathrm{(D)\ }10}$.

The entire situation is in the picture below. The correct answer is $\boxed{\mathrm{(E)}\ (3,2)}$.

The area of the large circle is $\pi b^2$, the area of the small one is $\pi c^2$, hence the shaded area is $\pi(b^2-c^2)$. From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$, hence $b^2-c^2=a^2$ and thus the shaded area is $\boxed{\mathrm{(A)\ }\pi a^2}$.

Let the number of students be $n\geq 5$. Then the sum of their scores is at least $5\cdot 100 + (n-5)\cdot 60$. At the same time, we need to achieve the mean $76$, which is equivalent to achieving the sum $76n$. Hence we get a necessary condition on $n$: we must have $5\cdot 100 + (n-5)\cdot 60 \leq 76n$. This can be simplified to $200 \leq 16n$. The smallest integer $n$ for which this is true is $n=13$. To finish our solution, we now need to find one way how $13$ students could have scored on the test. We have $13\cdot 76 = 988$ points to divide among them. The five $100$s make $500$, hence we must divide the remaining $488$ points among the other $8$ students. This can be done e.g. by giving $61$ points to each of them. Hence the smallest possible number of students is $\boxed{\mathrm{(D)}\ 13}$.

We already know that $a_1=2001$, $a_2=2002$, $a_3=2003$, and $a_4=2000$. Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$, $a_6=2003+2000-2005=1998$, $a_7=2000+2005-1998=2007$, and so on. We can now discover the following pattern: $a_{2k+1} = 2001+2k$ and $a_{2k}=2004-2k$. This is easily proved by induction. It follows that $a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}$.

Since $f(f^{-1}(x))=x$, it follows that $a(bx+a)+b=x$, which implies $abx + a^2 +b = x$. This equation holds for all values of $x$ only if $ab=1$ and $a^2+b=0$. Then $b = -a^2$. Substituting into the equation $ab = 1$, we get $-a^3 = 1$. Then $a = -1$, so $b = -1$, and\[f(x)=-x-1.\]Likewise\[f^{-1}(x)=-x-1.\]These are inverses to one another since\[f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.\]\[f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.\]Therefore $a+b=\boxed{\mathrm{(A)}\ -2}$. note: you could just find that $a=-1$ and $b=-1$ and add, getting $-2$.

The triangle $ABC$ is clearly a right triangle, its area is $\frac{5\cdot 12}2 = 30$. If we knew the areas of triangles $AMJ$ and $BNK$, we could subtract them to get the area of the pentagon. Draw the height $CL$ from $C$ onto $AB$. As $AB=13$ and the area is $30$, we get $CL=\frac{60}{13}$. The situation is shown in the picture below: Now note that the triangles $ABC$, $AMJ$, $ACL$, $CBL$ and $NBK$ all have the same angles and therefore they are similar. We already know some of their sides, and we will use this information to compute their areas. Note that if two polygons are similar with ratio $k$, their areas have ratio $k^2$. We will use this fact repeatedly. Below we will use $[XYZ]$ to denote the area of the triangle $XYZ$. We have $\frac{CL}{BC} = \frac{60/13}{12} = \frac 5{13}$, hence $[ACL] = \frac{ 25[ABC] }{169} = \frac{750}{169}$. Also, $\frac{CL}{AC} = \frac{60/13}5 = \frac{12}{13}$, hence $[CBL] = \frac{ 144[ABC] }{169} = \frac{4320}{169}$. Now for the smaller triangles: We know that $\frac{AM}{AC}=\frac 15$, hence $[AMJ] = \frac{[ACL]}{25} = \frac{30}{169}$. Similarly, $\frac{BN}{BC}=\frac 8{12} = \frac 23$, hence $[NBK] = \frac{4[CBL]}9 = \frac{1920}{169}$. Finally, the area of the pentagon is $30 - \frac{30}{169} - \frac{1920}{169} = \boxed{\frac{240}{13}}$. Observe that all of the triangles in the problem are right triangles and similar with a ratio of 5-12-13. The largest triangle $\Delta ABC$ has area $\dfrac{5\cdot12}2=30$. $\Delta AMJ$ is linearly scaled down from $\Delta ABC$ by a factor of $\dfrac{5-4}{13}=\dfrac1{13}$ (as we can see from comparing the two hypotenuses), while $\Delta NBK$ is scaled by a factor of $\dfrac{12-4}{13}=\dfrac8{13}$. The area we desire is the combined area of $\Delta AMJ$ and $\Delta NBK$ subtracted from the area of $\Delta ABC$, which is \[30-30\left(\dfrac1{13}\right)^2-30\left(\dfrac8{13}\right)^2=30\left(1-\left(\dfrac1{13}\right)^2-\left(\dfrac8{13}\right)^2\right)=30\left(1-\dfrac{65}{169}\right)=30\left(\dfrac8{13}\right)=\boxed{\textbf{(D) }\dfrac{240}{13}}.\] Split the pentagon along a different diagonal as follows:

If Jack's current age is $\overline{ab}=10a+b$, then Bill's current age is $\overline{ba}=10b+a$. In five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$. We are given that $10a+b+5=2(10b+a+5)$. Thus $8a=19b+5 \Rightarrow a=\dfrac{19b+5}{8}$. For $b=1$ we get $a=3$. For $b=2$ and $b=3$ the value $\frac{19b+5}8$ is not an integer, and for $b\geq 4$, $a$ is more than $9$. Thus the only solution is $(a,b)=(3,1)$, and the difference in ages is $31-13=\boxed{\mathrm{(B)\ }18}$.

Let $z = a+bi$, so $\overline{z} = a-bi$. By definition, $z = a+bi = f(z) = i(a-bi) = b+ai$, which implies that all solutions to $f(z) = z$ lie on the line $y=x$ on the complex plane. The graph of $|z| = 5$ is a circle centered at the origin, and there are $2 \Rightarrow \mathrm{(C)}$ intersections.

Let the three roots be $x_1,x_2,x_3$. \[\log_2 x_1 + \log_2 x_2 + \log_2 x_3 = \log_2 x_1x_2x_3= 5 \Longrightarrow x_1x_2x_3 = 32\] By Vieta’s formulas, \[8(x-x_1)(x-x_2)(x-x_3) = 8x^3 + 4ax^2 + 2bx + a\] gives us that $a = -8x_1x_2x_3 = -256 \Rightarrow \mathrm{(A)}$.

Let the coordinates of $A$ be $(x_A,y_A)$. As $A$ lies on the parabola, we have $y_A=4x_A^2+7x_A-1$. As the origin is the midpoint of $AB$, the coordinates of $B$ are $(-x_A,-y_A)$. We need to choose $x_A$ so that $B$ will lie on the parabola as well. In other words, we need $-y_A = 4(-x_A)^2 + 7(-x_A) - 1$. Substituting for $y_A$, we get: $-4x_A^2 - 7x_A + 1 = 4(-x_A)^2 + 7(-x_A) - 1$. This simplifies to $8x_A^2 - 2 = 0$, which solves to $x_A = \pm 1/2$. Both roots lead to the same pair of points: $(1/2,7/2)$ and $(-1/2,-7/2)$. Their distance is $\sqrt{ 1^2 + 7^2 } = \sqrt{50} = \boxed{5\sqrt2}$.

Consider a trapezoid (label it $ABCD$ as follows) cross-section of the truncate cone along a diameter of the bases: Above, $E,F,$ and $G$ are points of tangency. By the Two Tangent Theorem, $BF = BE = 18$ and $CF = CG = 2$, so $BC = 20$. We draw $H$ such that it is the foot of the altitude $\overline{HD}$ to $\overline{AB}$: By the Pythagorean Theorem, \[r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12\] Therefore, the answer is $\boxed{{A (6)}}.$

There are $2^6$ possible colorings of the cube. Consider the color that appears with greater frequency. The property obviously holds true if $5$ or $6$ of the faces are colored the same, which for each color can happen in $6 + 1 = 7$ ways. If $4$ of the faces are colored the same, there are $3$ possible cubes (corresponding to the $3$ possible ways to pick pairs of opposite faces for the other color). If $3$ of the faces are colored the same, the property obviously cannot be satisfied. Thus, there are a total of $2(7 + 3) = 20$ ways for this to occur, and the desired probability is $\frac{20}{2^6} = \frac{5}{16}\ \mathbf{(B)}$.

Error creating thumbnail: Unable to save thumbnail to destination $\frac yx$ represents the slope of a line passing through the origin. It follows that since a line $y = mx$ intersects the ellipse at either $0, 1,$ or $2$ points, the minimum and maximum are given when the line $y = mx$ is a tangent, with only one point of intersection. Substituting, \[2x^2 + x(mx) + 3(mx)^2 - 11x - 20(mx) + 40 = 0\] Rearranging by the degree of $x$, \[(3m^2 + m + 2)x^2 - (20m + 11)x + 40 = 0\] Since the line $y=mx$ is tangent to the ellipse, we want the discriminant, \[(20m+11)^2 - 4\cdot 40 \cdot (3m^2 + m + 2) = -80m^2 + 280m - 199\] to be equal to $0$. We want $a+b$, which is the sum of the roots of the above quadratic. By Vieta’s formulas, that is $\frac{280}{80} = \frac{7}{2} \Rightarrow \mathrm{(C)}$.

All the unknown entries can be expressed in terms of $b$. Since $100e = beh = ceg = def$, it follows that $h = \frac{100}{b}, g = \frac{100}{c}$, and $f = \frac{100}{d}$. Comparing rows $1$ and $3$ then gives $50bc = 2 \cdot \frac{100}{b} \cdot \frac{100}{c}$, from which $c = \frac{20}{b}$. Comparing columns $1$ and $3$ gives $50d \cdot \frac{100}{c}= 2c \cdot \frac{100}{d}$, from which $d = \frac{c}{5} = \frac{4}{b}$. Finally, $f = 25b, g = 5b$, and $e = 10$. All the entries are positive integers if and only if $b = 1, 2,$ or $4$. The corresponding values for $g$ are $5, 10,$ and $20$, and their sum is $\boxed{\mathbf{(C)}35}$.

Let the roots be $r,s,r + s$, and let $t = rs$. Then $(x - r)(x - s)(x - (r + s))$ $= x^3 - (r + s + r + s) x^2 + (rs + r(r + s) + s(r + s))x - rs(r + s) = 0$ and by matching coefficients, $2(r + s) = 2004 \Longrightarrow r + s = 1002$. Then our polynomial looks like \[x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0\] and we need the number of possible products $t = rs = r(1002 - r)$. Because $m=t+1002^2$ is an integer, we also note that $t$ must be an integer. Since $r > 0$ and $t > 0$, it follows that $0 < t = r(1002-r) < 501^2 = 251001$, with the endpoints not achievable because the roots must be distinct and positive. Because neither $r$ nor $1002-r$ can be an integer, there are $251,000 - 500 = \boxed{\textbf{(C) } 250,500}$ possible values of $n = -1002t$.

Let $\alpha = DBC$. Then the first condition tells us that \[\tan^2 DBE = \tan(DBE - \alpha)\tan(DBE + \alpha) = \frac {\tan^2 DBE - \tan^2 \alpha}{1 - \tan ^2 DBE \tan^2 \alpha},\] and multiplying out gives us $(\tan^4 DBE - 1) \tan^2 \alpha = 0$. Since $\tan\alpha \neq 0$, we have $\tan^4 DBE = 1 \Longrightarrow \angle DBE = 45^{\circ}$. The second condition tells us that $2\cot (45 - \alpha) = 1 + \cot \alpha$. Expanding, we have $1 + \cot \alpha = 2\left[\frac {\cot \alpha + 1}{\cot \alpha - 1}\right] \Longrightarrow (\cot \alpha - 3)(\cot \alpha + 1) = 0$. Evidently $\cot \alpha \neq - 1$, so we get $\cot \alpha = 3$. Now $BD = 5\sqrt {2}$ and $AC = \frac {2BD} {\cot \alpha} = \frac {10\sqrt {2}}{3}$. Thus, $[ABC] = \frac {1}{2} \cdot 5\sqrt {2} \cdot \frac {10\sqrt {2}}{3} = \frac {50}{3}\ \mathrm{(B)}$.

Given $n$ digits, there must be exactly one power of $2$ with $n$ digits such that the first digit is $1$. Thus $S$ contains $603$ elements with a first digit of $1$. For each number in the form of $2^k$ such that its first digit is $1$, then $2^{k+1}$ must either have a first digit of $2$ or $3$, and $2^{k+2}$ must have a first digit of $4,5,6,7$. Thus there are also $603$ numbers with first digit $\{2,3\}$ and $603$ numbers with first digit $\{4,5,6,7\}$. By using complementary counting, there are $2004 - 3 \times 603 = 195$ elements of $S$ with a first digit of $\{8,9\}$. Now, $2^k$ has a first digit of $\{8,9\}$ if and only if the first digit of $2^{k-1}$ is $4$, so there are $\boxed{195} \Rightarrow \mathrm{(B)}$ elements of $S$ with a first digit of $4$.