AMC12 2002 A

We expand to get $2x^2-8x+3x-12+2x^2-12x+3x-18=0$ which is $4x^2-14x-30=0$ after combining like terms. Using the quadratic part of Vieta's Formulas, we find the sum of the roots is $\frac{14}4 = \boxed{\textbf{(A) }7/2}$.

We work backwards; the number that Cindy started with is $3(43)+9=138$. Now, the correct result is $\frac{138-3}{9}=\frac{135}{9}=15$. Our answer is $\boxed{\textbf{(A) }15}$.

The best way to solve this problem is by simple brute force. It is convenient to drop the usual way how exponentiation is denoted, and to write the formula as $2\uparrow 2\uparrow 2\uparrow 2$, where $\uparrow$ denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so: 1. $2\uparrow (2\uparrow (2\uparrow 2))$ 2. $2\uparrow ((2\uparrow 2)\uparrow 2)$ 3. $((2\uparrow 2)\uparrow 2)\uparrow 2$ 4. $(2\uparrow (2\uparrow 2))\uparrow 2$ 5. $(2\uparrow 2)\uparrow (2\uparrow 2)$ We can note that $2\uparrow (2\uparrow 2) = (2\uparrow 2)\uparrow 2 =16$. Therefore options 1 and 2 are equal, and options 3 and 4 are equal. Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5. $((2\uparrow 2)\uparrow 2)\uparrow 2 = 16\uparrow 2 = 256$ $(2\uparrow 2)\uparrow (2\uparrow 2) = 4 \uparrow 4 = 256$ Thus the only other result is $256$, and our answer is $\boxed{\textbf{(B) } 1}$.

We can create an equation for the question, $4(90-x)=(180-x)$ $360-4x=180-x$ $3x=180$ After simplifying, we get $x=60 \Rightarrow \mathrm {(B)}$ Given that the complementary angle is $\frac{1}{4}$ of the supplementary angle. Subtracting the complementary angle from the supplementary angle, we have $90^{\circ}$ as $\frac{3}{4}$ of the supplementary angle. Thus the degree measure of the supplementary angle is $120^{\circ}$, and the degree measure of the desired angle is $180^{\circ} - 120^{\circ} = 60^{\circ}$. $\mathrm {(B)}$

The outer circle has radius $1+1+1=3$, and thus area $9\pi$. The little circles have area $\pi$ each; since there are 7, their total area is $7\pi$. Thus, our answer is $9\pi-7\pi=\boxed{2\pi\Rightarrow \textbf{(C)}}$.

For any $m$ we can pick $n=1$, we get $m \cdot 1 \le m + 1$, therefore the answer is $\boxed{\textbf{(E) } \text{infinitely many}}$.

Let $r_1$ and $r_2$ be the radii of circles $A$ and$B$, respectively. It is well known that in a circle with radius $r$, a subtended arc opposite an angle of $\theta$ degrees has length $\frac{\theta}{360} \cdot 2\pi r$. Using that here, the arc of circle A has length $\frac{45}{360}\cdot2\pi{r_1}=\frac{r_1\pi}{4}$. The arc of circle B has length $\frac{30}{360} \cdot 2\pi{r_2}=\frac{r_2\pi}{6}$. We know that they are equal, so $\frac{r_1\pi}{4}=\frac{r_2\pi}{6}$, so we multiply through and simplify to get $\frac{r_1}{r_2}=\frac{2}{3}$. As all circles are similar to one another, the ratio of the areas is just the square of the ratios of the radii, so our answer is $\boxed{\textbf{(A) } 4/9}$.

The blue that's touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is 1 square and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have $\boxed{B=W\Rightarrow \text{(A)}}$.

A $0.8$ MB file can either be on its own disk, or share it with a $0.4$ MB. Clearly it is better to pick the second possibility. Thus we will have $3$ disks, each with one $0.8$ MB file and one $0.4$ MB file. We are left with $12$ files of $0.7$ MB each, and $12$ files of $0.4$ MB each. Their total size is $12\cdot (0.7 + 0.4) = 13.2$ MB. The total capacity of $9$ disks is $9\cdot 1.44 = 12.96$ MB, hence we need at least $10$ more disks. And we can easily verify that $10$ disks are indeed enough: six of them will carry two $0.7$ MB files each, and four will carry three $0.4$ MB files each. Thus our answer is $3+10 = \boxed{\textbf{(B) }13 }$.

We will simulate the process in steps. In the beginning, we have: - $4$ ounces of coffee in cup $1$ - $4$ ounces of cream in cup $2$ In the first step we pour $4/2=2$ ounces of coffee from cup $1$ to cup $2$, getting: - $2$ ounces of coffee in cup $1$ - $2$ ounces of coffee and $4$ ounces of cream in cup $2$ In the second step we pour $2/2=1$ ounce of coffee and $4/2=2$ ounces of cream from cup $2$ to cup $1$, getting: - $2+1=3$ ounces of coffee and $0+2=2$ ounces of cream in cup $1$ - the rest in cup $2$ Hence at the end we have $3+2=5$ ounces of liquid in cup $1$, and out of these $2$ ounces is cream. Thus the answer is $\boxed{\text{(D) } \frac 25}$. Let's consider this in steps. We have 4 ounces of coffee in the first cup. We hace 4 ounces of cream in the second cup. We take half of the coffee in the first cup(2 ounces), and add it to the second cup, yielding 6 ounces in total in the second cup(in a 1:2 ratio between coffee and cream, respecitvely). We then take half of the second cup and pour it into the first cup. $6/2=3$, so there is now 5 ounces in the first cup, 2 coffee and 3 the mixture. Remember that the mixture is in a 1:2 ratio between coffee and cream. So, coffee has one ounce and cream has 2 ounces. In total, there is 2 ounces in the 5 ounce first cup. Putting 2 over 5, we get the answer. Therefore, the answer is $\boxed{\text{(D) } \frac 25}$.

Let the time he needs to get there in be $t$ and the distance he travels be $d$. From the given equations, we know that $d=\left(t+\frac{1}{20}\right)40$ and $d=\left(t-\frac{1}{20}\right)60$. Setting the two equal, we have $40t+2=60t-3$ and we find $t=\frac{1}{4}$ of an hour. Substituting t back in, we find $d=12$. From $d=rt$, we find that $r$, our answer, is $\boxed{\textbf{(B) }48 }$. Since either time he arrives at is $3$ minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. Substituting $t=\frac ds$ and dividing both sides by $d$, we get $\frac 2s = \frac 1{40} + \frac 1{60}$ hence $s=\boxed{\textbf{(B) }48}$. (Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighted sum in step two, and hence obtain a weighted harmonic mean in step three.) Let x be equal to the total amount of distance he needs to cover. Let y be equal to the amount of time he would travel correctly. Setting up a system of equations, $\frac x{40} -3 = y$ and $\frac x{60} +3 = y$ Solving, we get x = 720 and y = 15. We divide x by y to get the average speed, $\frac {720}{15} = 48$. Therefore, the answer is $\boxed{\textbf{(B) }48}$. Let $v$ be Mr Bird's speed in miles per hour and $t$ be the desired time in hours. No matter what, the product of Mr Bird's speed and time must always be constant.

Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$. It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$. Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to Vieta's Formulas). We now have that the sum of the two roots is $63$ while the product is $k$. Since both roots are primes, one must be $2$, otherwise, the sum would be even. That means the other root is $61$ and the product must be $122$. Hence, our answer is $\boxed{\text{(B)}\ 1 }$.

Each of the numbers $a$ and $b$ is a solution to $\left| x - \frac 1x \right| = 1$. Hence it is either a solution to $x - \frac 1x = 1$, or to $\frac 1x - x = 1$. Then it must be a solution either to $x^2 - x - 1 = 0$, or to $x^2 + x - 1 = 0$. There are in total four such values of $x$, namely $\frac{\pm 1 \pm \sqrt 5}2$. Out of these, two are positive: $\frac{-1+\sqrt 5}2$ and $\frac{1+\sqrt 5}2$. We can easily check that both of them indeed have the required property, and their sum is $\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{(C) \sqrt 5}$.

First, note that $2002 = 11 \cdot 13 \cdot 14$. Using the fact that for any base we have $\log a + \log b = \log ab$, we get that $N = \log_{2002} (11^2 \cdot 13^2 \cdot 14^2) = \log_{2002} 2002^2 = \boxed{(D) N=2}$.

As the unique mode is $8$, there are at least two $8$s. As the range is $8$ and one of the numbers is $8$, the largest one can be at most $16$. If the largest one is $16$, then the smallest one is $8$, and thus the mean is strictly larger than $8$, which is a contradiction. If we have 2 8's we can find the numbers 4, 6, 7, 8, 8, 9, 10, 12. This is a possible solution but has not reached the maximum. If we have 4 8's we can find the numbers 6, 6, 6, 8, 8, 8, 8, 14. We can also see that they satisfy the need for the mode, median, and range to be 8. This means that the answer will be $\boxed{\text{(D)}\ 14 }$. ~By QWERTYUIOPASDFGHJKLZXCVBNM

We have 4 cases, if Tina chooses $1, 2, 3,$ or $4$ and always chooses numbers greater than the first number she chose. The number of ways of choosing 2 numbers from $5$ are $\binom{5}{2}$. Case 1: Tina chooses $1$. In this case, since the numbers are distinct, Tina can choose $(1, 2), (1, 3), (1, 4),$ or $(1, 5).$ If Tina chooses $1$ and $2$ which sum to $3$, Sergio only has $10-3=7$ choices. Since the sum of the combined numbers increases by $1$ every time for this specific case, Sergio has $1$ less choice every time. Therefore, the probability of this is $\frac{7+6+5+4}{10 \cdot \binom{5}{2}}$. Case 2: Tina chooses $2$. In this case, Tina can choose $(2, 3), (2, 4),$ or $(2, 5).$ If Tina chooses $2$ and $3$ which sum to $5$, Sergio only has $10-5=5$ choices. Since the sum of the combined numbers increases by $1$ every time for this specific case, Sergio has $1$ less choice every time. Therefore, the probability of this is $\frac{5+4+3}{10 \cdot \binom{5}{2}}$. Case 3: Tina chooses $3$. In this case, Tina can choose $(3, 4),$ or $(3, 5).$ If Tina chooses $3$ and $4$ which sum to $7$, Sergio only has $10-7=3$ choices. Since the sum of the combined numbers increases by $1$ every time for this specific case, Sergio has $1$ less choice every time. Therefore, the probability of this is $\frac{3+2}{10 \cdot \binom{5}{2}}$. Case 4: Tina chooses $4$. In this case, Tina can only choose $(4,5).$ If Tina chooses $4$ and $5$ which sum to $9$, Sergio only has $10-9=1$ choice. Therefore, the probability of this is $\frac{1}{10 \cdot \binom{5}{2}}$. Once you add these probabilities up, you will have $\frac{(7+6+5+4)+(5+4+3)+(3+2)+(1)}{10 \cdot\binom{5}{2}} = \frac{40}{100} = \frac{2}{5}$ probability. Thus our answer is $\frac{2}{5}$. Assume Sergio chooses from ${2,3,\ldots,10}$. The probability of Tina getting a sum of $6+x$ and a sum of $6-x$, where $x \leq 4$, are equal due to symmetry. The probability of Sergio choosing numbers higher/lower than $6+x$ is equal to him choosing numbers lower/higher than $6-x$. Therefore over all of Tina's sums, the probability of Sergio choosing a number higher is equal to the probability of choosing a number lower.

Neither of the digits $4$, $6$, and $8$ can be a units digit of a prime. Therefore the sum of the set is at least $40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207$. We can indeed create a set of primes with this sum, for example the following sets work: $\{ 41, 67, 89, 2, 3, 5 \}$ or $\{ 43, 61, 89, 2, 5, 7 \}$. Thus the answer is $207\implies \boxed{\mathrm{(B)}}$.

First examine the formula $(x-10)^2+y^2=36$, for the circle $C_1$. Its center, $D_1$, is located at (10,0) and it has a radius of $\sqrt{36}$ = 6. The next circle, using the same pattern, has its center, $D_2$, at (-15,0) and has a radius of $\sqrt{81}$ = 9. So we can construct this diagram: Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles $D_2QO$ and $D_1PO$ similar by AA, with a scale factor of $6:9$, or $2:3$. Next, we must subdivide the line $D_2D_1$ in a 2:3 ratio to get the length of the segments $D_2O$ and $D_1O$. The total length is $10 - (-15)$, or $25$, so applying the ratio, $D_2O$ = 15 and $D_1O$ = 10. These are the hypotenuses of the triangles. We already know the length of $D_2Q$ and $D_1P$, 9 and 6 (they're radii). So in order to find $PQ$, we must find the length of the longer legs of the two triangles and add them. $15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144$ $\sqrt{144} = 12$ $10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64$ $\sqrt{64} = 8$ Finally, the length of PQ is $12+8=\boxed{20}$, or (C).

First of all, note that the equation $f(t)=6$ has two solutions: $t=-2$ and $t=1$. Given an $x$, let $f(x)=t$. Obviously, to have $f(f(x))=6$, we need to have $f(t)=6$, and we already know when that happens. In other words, the solutions to $f(f(x))=6$ are precisely the solutions to ($f(x)=-2$ or $f(x)=1$). Without actually computing the exact values, it is obvious from the graph that the equation $f(x)=-2$ has two and $f(x)=1$ has four different solutions, giving us a total of $2+4=\boxed{(D)6}$ solutions.

The repeating decimal $0.\overline{ab}$ is equal to \[\frac{10a+b}{100} + \frac{10a+b}{10000} + \cdots = (10a+b)\cdot\left(\frac 1{10^2} + \frac 1{10^4} + \cdots \right) = (10a+b) \cdot \frac 1{99} = \frac{10a+b}{99}\] When expressed in the lowest terms, the denominator of this fraction will always be a divisor of the number $99 = 3\cdot 3\cdot 11$. This gives us the possibilities $\{1,3,9,11,33,99\}$. As $a$ and $b$ are not both nine and not both zero, the denominator $1$ can not be achieved, leaving us with $\boxed{\mathrm{(C) }5}$ possible denominators. (The other ones are achieved e.g. for $ab$ equal to $33$, $11$, $9$, $3$, and $1$, respectively.) Another way to convert the decimal into a fraction (simplifying, I guess?). We have \[100(0.\overline{ab}) = ab.\overline{ab}\] \[99(0.\overline{ab}) = 100(0.\overline{ab}) - 0.\overline{ab} = ab.\overline{ab} - 0.\overline{ab} = ab\] \[0.\overline{ab} = \frac{ab}{99}\] where $a, b$ are digits. Continuing in the same way by looking at the factors of 99, we have 5 different possibilities for the denominator. $\boxed{(C)}$ Since $\frac{1}{99}=0.\overline{01}$, we know that $0.\overline{ab} = \frac{ab}{99}$. From here, we wish to find the number of factors of $99$, which is $6$. However, notice that $1$ is not a possible denominator, so our answer is $6-1=\boxed{5}$. \[\]

The sequence is infinite. As there are only $100$ pairs of digits, sooner or later a pair of consecutive digits will occur for the second time. As each next digit only depends on the previous two, from this point on the sequence will be periodic. (Additionally, as every two consecutive digits uniquely determine the previous one as well, the first pair of digits that will occur twice must be the first pair $4,7$.) Hence it is a good idea to find the period. Writing down more terms of the sequence, we get: \[4,7,1,8,9,7,6,3,9,2,1,3,4,7,\dots\] and we found the period. The length of the period is $12$, and its sum is $4+7+\cdots+1+3 = 60$. Hence for each $k$ we have $S_{12k} = 60k$. We have $\lfloor 10000/60 \rfloor = 166$ and $166\cdot 12 = 1992$, therefore $S_{1992} = 60\cdot 166 = 9960$. The rest can now be computed by hand, we get $S_{1998} = 9960+4+7+1+8+9+7= 9996$, and $S_{1999}=9996 + 6 = 10002$, thus the answer is $\boxed{\text{(B) }1999}$.

Clearly $BC=5$ and $AC=5\sqrt{3}$. Choose a $P'$ and get a corresponding $D'$ such that $BD'= 5\sqrt{2}$ and $CD'=5$. For $BD > 5\sqrt2$ we need $CD>5$, creating an isosceles right triangle with hypotenuse $5\sqrt {2}$ . Thus the point $P$ may only lie in the triangle $ABD'$. The probability of it doing so is the ratio of areas of $ABD'$ to $ABC$, or equivalently, the ratio of $AD'$ to $AC$ because the triangles have identical altitudes when taking $AD'$ and $AC$ as bases. This ratio is equal to $\frac{AC-CD'}{AC}=1-\frac{CD'}{AC}=1-\frac{5}{5\sqrt{3}}=1-\frac{\sqrt{3}}{3}= \frac{3-\sqrt{3}}{3}$. Thus the answer is $\boxed{C}$.

Looking at the triangle $BCD$, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let $x = \angle C$, so that $B=2x$ from given and the previous deducted. Then $\angle ABD=x, \angle ADB=2x$ because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means $\triangle ABD$ and $\triangle ACB$ are similar, so $\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12$. Then by using Heron's Formula on $ABD$ (with sides $12,7,9$), we have $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$.

Let $s=\sqrt{a^2+b^2}$ be the magnitude of $a+bi$. Then the magnitude of $(a+bi)^{2002}$ is $s^{2002}$, while the magnitude of $a-bi$ is $s$. We get that $s^{2002}=s$, hence either $s=0$ or $s=1$. For $s=0$ we get a single solution $(a,b)=(0,0)$. Let's now assume that $s=1$. Multiply both sides by $a+bi$. The left hand side becomes $(a+bi)^{2003}$, the right hand side becomes $(a-bi)(a+bi)=a^2 + b^2 = 1$. Hence the solutions for this case are precisely all the $2003$rd complex roots of unity, and there are $2003$ of those. The total number of solutions is therefore $1+2003 = \boxed{2004}$.

The sum of the coefficients of $P$ and of $Q$ will be equal, so $P(1) = Q(1)$. The only answer choice with an intersection between the two graphs at $x = 1$ is (B). (The polynomials in the graph are $P(x) = 2x^4-3x^2-3x-4$ and $Q(x) = -2x^4-2x^2-2x-2$.)