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872e48f7d607d28a5ee88ab4bfc8852f | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_15 | The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$ | We could express this collection as integers $a\textsubscript{1}$ through $a\textsubscript{8}$ , with $a\textsubscript{1}$ being the smallest and $a\textsubscript{8}$ being the largest.
Since the mean is $8$ , we know that $a\textsubscript{4}$ and $a\textsubscript{5}$ must also be $8$ . If they were not, the other numbers, which are lesser and greater than $a\textsubscript{4}$ and $a\textsubscript{5}$ respectively, would not be able to satisfy the condition that $8$ is the mode.
There are $8$ terms and the mean is $8$ . This tells us that the sum of all the numbers is $64$
We want to maximize the value of $a\textsubscript{8}$ , so we set $a\textsubscript{6}$ and $a\textsubscript{7}$ to $8$ as well.
Knowing that we want to minimize numbers and that the range is $8$ , we set $a\textsubscript{1}$ $a\textsubscript{2}$ , and $a\textsubscript{3}$ equal to $a\textsubscript{8} - 8$
$a\textsubscript{1}$ $a\textsubscript{2}$ $a\textsubscript{3}$ $a\textsubscript{4}$ $a\textsubscript{5}$ $a\textsubscript{6}$ $a\textsubscript{7}$ $a\textsubscript{8}$ $=$ $a\textsubscript{8} - 8$ $a\textsubscript{8} - 8$ $a\textsubscript{8} - 8$ $8$ $8$ $8$ $8$ $a\textsubscript{8}$
Letting the sum of all the numbers be $64$ , we find that $32 + 4a_8 - 24 = 64$ , which simplifies to $4a_8 = 56$ . Solving, we get $\boxed{14}$ . ~By SK80, mod_x for minor edits | D | 14 |
e35e6ed6f840a81bd684327c7f898d61 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_17 | Several sets of prime numbers, such as $\{7,83,421,659\}$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?
$\text{(A) }193 \qquad \text{(B) }207 \qquad \text{(C) }225 \qquad \text{(D) }252 \qquad \text{(E) }447$ | 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 $207\implies \boxed{207}$ | B | 207 |
c79905bf1e067f606d15a45aef4df5bb | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_18 | Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$ respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$
$\text{(A) }15 \qquad \text{(B) }18 \qquad \text{(C) }20 \qquad \text{(D) }21 \qquad \text{(E) }24$ | 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: [asy] unitsize(0.3cm); defaultpen(0.8); path C1=circle((10,0),6); path C2=circle((-15,0),9); draw(C1); draw(C2); draw( (-25,0) -- (17,0) ); dot((10,0)); dot((-15,0)); pair[] p1 = intersectionpoints(C1, circle((5,0),5) ); pair[] p2 = intersectionpoints(C2, circle((-7.5,0),7.5) ); dot(p1[1]); dot(p2[0]); draw((10,0)--p1[1]--p2[0]--(-15,0)); label("$C_1$",(10,0) + 6*dir(-45), SE ); label("$C_2$",(-15,0) + 9*dir(225), SW ); label("$D_1$",(10,0), SE ); label("$D_2$",(-15,0), SW ); label("$Q$", p2[0], NE ); label("$P$", p1[1], SW ); label("$O$", (0,0), SW ); [/asy] 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 $D_1P$ and (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) | null | 20 |
c79905bf1e067f606d15a45aef4df5bb | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_18 | Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$ respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$
$\text{(A) }15 \qquad \text{(B) }18 \qquad \text{(C) }20 \qquad \text{(D) }21 \qquad \text{(E) }24$ | Using the above diagram, imagine that segment $\overline{QD_2}$ is shifted to the right to match up with $\overline{PD_1}$ . Then shift $\overline{QP}$ downwards to make a right triangle. We know $\overline{D_2D_1} = 25$ from the given information and the newly created leg has length $\overline{QD_2} + \overline{PD_1} = 9 + 6 = 15$ . Hence by Pythagorean theorem $15^2 + {\overline{QP}}^2 = 25^2$
$\overline{QP} = \boxed{20}$ , or C. | null | 20 |
cd9a557050bee03e7c49e591e9751027 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_19 | The graph of the function $f$ is shown below. How many solutions does the equation $f(f(x))=6$ have?
[asy] size(200); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6); real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; draw(P1--P2--P3--P4--P5); dot("(-7, -4)",P1); dot("(-2, 6)",P2,LeftSide); dot("(1, 6)",P4); dot("(5, -6)",P5); xaxis("$x$",-7.5,7,Ticks(xticks),EndArrow(6)); yaxis("$y$",-6.5,7,Ticks(yticks),EndArrow(6)); [/asy]
$\text{(A) }2 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }6 \qquad \text{(E) }7$ | 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 $2+4=\boxed{6}$ solutions. | D | 6 |
e37758662c5923d944fbedd06ddbbc3c | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_20 | Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?
$\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9$ | 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 $\boxed{5}$ possible denominators. | C | 5 |
e37758662c5923d944fbedd06ddbbc3c | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_20 | Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?
$\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9$ | 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. $\boxed{5}$ | C | 5 |
e37758662c5923d944fbedd06ddbbc3c | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_20 | Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?
$\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9$ | 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 $6-1=\boxed{5}$ \[\] ~AopsUser101 | null | 5 |
e37758662c5923d944fbedd06ddbbc3c | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_20 | Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?
$\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9$ | Since $0.\overline{ab} = \frac{ab}{99}$ , the denominator must be a factor of $99 = 3^2 \cdot 11$ . The factors of $99$ are $1,$ $3,$ $9,$ $11,$ $33,$ and $99$ . Since $a$ and $b$ are not both nine, the denominator cannot be $1$ . By choosing $a$ and $b$ appropriately, we can make fractions with each of the other denominators.
Thus, the answer is $\boxed{5}$ | null | 5 |
2917ea745fb0cb9dde5e13f4f3da1cd9 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_21 | Consider the sequence of numbers: $4,7,1,8,9,7,6,\dots$ For $n>2$ , the $n$ -th term of the sequence is the units digit of the sum of the two previous terms. Let $S_n$ denote the sum of the first $n$ terms of this sequence. The smallest value of $n$ for which $S_n>10,000$ is:
$\text{(A) }1992 \qquad \text{(B) }1999 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$ | 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 $\boxed{1999}$ | B | 1999 |
665aed5f79d526bc10f73328bd674f79 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_24 | Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$
$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$ | 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 $1+2003 = \boxed{2004}$ | null | 2004 |
665aed5f79d526bc10f73328bd674f79 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_24 | Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$
$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$ | As in the other solution, split the problem into when $s=0$ and when $s=1$ . When $s=1$ and $a+bi=\cos\theta+i\sin\theta$
$(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta)$ $=a-bi= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta)$
so we must have $2002\theta=-\theta+2\pi k$ and hence $\theta=\frac{2\pi k}{2003}$ . Since $\theta$ is restricted to $[0,2\pi)$ $k$ can range from $0$ to $2002$ inclusive, which is $2002-0+1=2003$ values. Thus the total is $1+2003 = \boxed{2004}$ | E | 2004 |
665aed5f79d526bc10f73328bd674f79 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_24 | Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$
$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$ | Let $r$ be the magnitude of $a+bi$ . Notice that $r$ must be either $0$ or $1$ for this to be true, as shown in the above solutions. We know this because we are taking magnitude to the $2003$ rd power, and if the magnitude of $a+bi$ is larger than $1$ , it will increase and if it is smaller than $1$ it will decrease. However, the magnitude on the RHS is still $r$ , so this is not possible. Again, only $r=0$ and $r=1$ satisfy the equation.
Now if $r=0$ , then $(a,b)$ must be $(0,0)$
However if $r=1$ , we then have:
$\cos(2002 \theta) = \cos(-\theta)$ . This has solution of $\theta = 0$ . This would represent the number $1+0i$ , with conjugate $1-0i$ . This works because the magnitude is the same and the angle is nothing anyways. We multiply angle by $2002$ through De Moivre's Theorem and also we do $-\theta$ because it is a reflection, angles therefore is negative.
We then write:
$\cos(2002 \theta) = \cos(360-\theta)$ which has solution of $\theta = \frac{360}{2003}$
We can also write:
$\cos(2002 \theta) = \cos(720-\theta)$ which has solution $\theta = \frac{720}{2003}$
We notice that it is simply headed upwards and the answer is of the form $\frac{720}{2003} n$ , where n is some integer from $0$ to infinity, inclusive.
Well wait, it repeats itself $n=2003$ , that is $360$ which is also $0$ ! Hence we only have $n=0$ to $2002$ as original solutions, or $2003$ solutions.
$1+2003 = \boxed{2004}$ | null | 2004 |
665aed5f79d526bc10f73328bd674f79 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12A_Problems/Problem_24 | Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$
$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$ | Let $z = a + bi = re^{i\theta}$ where $r = |z|$ and $0\leq \theta < 2\pi.$ We want to solve $z^{2002} = \overline{z}.$ Since $z\overline{z} = |z|^2,$ we multiply by $z$ on both sides to get \[z^{2003} = |z|^2 = r^{2003}e^{2023i\theta} = r^2,\] from which we get $r\in \{0, 1\}$ (since $r\in \mathbb{R}$ ), immediately yielding $z = 0$ as an answer. If $r=1$ then we have $z^{2003} = 1$ , which each of the $2003$ rd roots of unity satisfies. Altogether, there are $1 + 2003 = \boxed{2004}$ values of $z$ , each of which correspond to a unique ordered pair $(a, b)$ | E | 2004 |
fac31f66242c1aa4d81bc6e05334b315 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_1 | The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ doesn't contain the digit
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$ | We wish to find $\frac{9+99+\cdots +999999999}{9}$ , or $\frac{9(1+11+111+\cdots +111111111)}{9}=123456789$ . This doesn't have the digit 0, so the answer is $\boxed{0}$ | A | 0 |
fac31f66242c1aa4d81bc6e05334b315 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_1 | The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ doesn't contain the digit
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$ | Notice that the final number is guaranteed to have the digits $\{1, 3, 5, 7, 9\}$ and that each of these digits can be paired with an even number adding up to 9. $\boxed{0}$ can be taken out, with the other digits fulfilling divisibility by 9. | A | 0 |
fac31f66242c1aa4d81bc6e05334b315 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_1 | The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ doesn't contain the digit
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$ | The arithmetic mean is $\frac{(10^1-1)+(10^2-1)+\ldots+(10^9-1)}{9}=\frac{1111111101}{9}=123456789$ . So select $\boxed{0}$ .
~hastapasta | A | 0 |
1a29198c9a6038d7c16c7c18fc5b1065 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_2 | What is the value of $(3x - 2)(4x + 1) - (3x - 2)4x + 1$ when $x=4$
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 11 \qquad\mathrm{(E)}\ 12$ | By the distributive property,
\[(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{11}\] | D | 11 |
1a29198c9a6038d7c16c7c18fc5b1065 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_2 | What is the value of $(3x - 2)(4x + 1) - (3x - 2)4x + 1$ when $x=4$
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 11 \qquad\mathrm{(E)}\ 12$ | Inputting 4 into $x$ in the original equation,
\[[3(4)-2][4(4)+1]-[3(4)-2][4(4)]+1 = 170-160+1 = \boxed{11}\] | D | 11 |
967b39271fdbabb35f006cd2e78c6802 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_4 | Let $n$ be a positive integer such that $\frac 12 + \frac 13 + \frac 17 + \frac 1n$ is an integer. Which of the following statements is not true:
$\mathrm{(A)}\ 2\ \text{divides\ }n \qquad\mathrm{(B)}\ 3\ \text{divides\ }n \qquad\mathrm{(C)}$ $\ 6\ \text{divides\ }n \qquad\mathrm{(D)}\ 7\ \text{divides\ }n \qquad\mathrm{(E)}\ n > 84$ | Since $\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}$ $0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2$
From which it follows that $\frac{41}{42} + \frac 1n = 1$ and $n = 42$ . The only answer choice that is not true is $\boxed{84}$ | E | 84 |
967b39271fdbabb35f006cd2e78c6802 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_4 | Let $n$ be a positive integer such that $\frac 12 + \frac 13 + \frac 17 + \frac 1n$ is an integer. Which of the following statements is not true:
$\mathrm{(A)}\ 2\ \text{divides\ }n \qquad\mathrm{(B)}\ 3\ \text{divides\ }n \qquad\mathrm{(C)}$ $\ 6\ \text{divides\ }n \qquad\mathrm{(D)}\ 7\ \text{divides\ }n \qquad\mathrm{(E)}\ n > 84$ | Since $\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}$ , it is very clear that $n=42$ makes the expression an integer. Because $n$ is a positive integer, $\frac{1}{n}$ must be less than or equal to $1$ . Thus, the only integer the expression can take is $1$ , which means the only value for $n$ is $42$ . Thus $\boxed{84}$ | E | 84 |
c0ab9b5b8f8255f6ec005a24c6bf7f5a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_7 | The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares
$\mathrm{(A)}\ 50 \qquad\mathrm{(B)}\ 77 \qquad\mathrm{(C)}\ 110 \qquad\mathrm{(D)}\ 149 \qquad\mathrm{(E)}\ 194$ | Let the three consecutive positive integers be $a-1$ $a$ , and $a+1$ . Since the mean is $a$ , the sum of the integers is $3a$ . So $8$ times the sum is just $24a$ . With this, we now know that $a(a-1)(a+1)=24a\Rightarrow(a-1)(a+1)=24$ $24=4\times6$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=\boxed{77}$ | B | 77 |
e3399a4e1cf78dc03b31ade795a71481 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_10 | How many different integers can be expressed as the sum of three distinct members of the set $\{1,4,7,10,13,16,19\}$
$\text{(A)}\ 13 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$ | Subtracting 10 from each number in the set, and dividing the results by 3, we obtain the set $\{-3, -2, -1, 0, 1, 2, 3\}$ . It is easy to see that we can get any integer between $-6$ and $6$ inclusive as the sum of three elements from this set, for the total of $\boxed{13}$ integers. | A | 13 |
e3399a4e1cf78dc03b31ade795a71481 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_10 | How many different integers can be expressed as the sum of three distinct members of the set $\{1,4,7,10,13,16,19\}$
$\text{(A)}\ 13 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$ | The set is an arithmetic sequence of numbers each $1$ more than a multiple of $3$ . Thus the sum of any three numbers will be a multiple of $3$ . All the multiples of $3$ from $1+4+7=12$ to $13+16+19=48$ are possible, totaling to $\boxed{13}$ integers. | A | 13 |
9ba1dec09d7a7ea393bd9f3ce6f88ff8 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_12 | For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | Let $x^2 = \frac{n}{20-n}$ , with $x \ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$ ). Then rewriting, $n = \frac{20x^2}{x^2 + 1}$ . Since $\text{gcd}(x^2, x^2 + 1) = 1$ , it follows that $x^2 + 1$ divides $20$ . Listing the factors of $20$ , we find that $x = 0, 1, 2 , 3$ are the only $\boxed{4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ). | D | 4 |
9ba1dec09d7a7ea393bd9f3ce6f88ff8 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_12 | For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | For $n<0$ and $n>20$ the fraction is negative, for $n=20$ it is not defined, and for $n\in\{1,\dots,9\}$ it is between 0 and 1.
Thus we only need to examine $n=0$ and $n\in\{10,\dots,19\}$
For $n=0$ and $n=10$ we obviously get the squares $0$ and $1$ respectively.
For prime $n$ the fraction will not be an integer, as the denominator will not contain the prime in the numerator.
This leaves $n\in\{12,14,15,16,18\}$ , and a quick substitution shows that out of these only $n=16$ and $n=18$ yield a square. Therefore, there are only $\boxed{4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ). | D | 4 |
9ba1dec09d7a7ea393bd9f3ce6f88ff8 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_12 | For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | If $\frac{n}{20-n} = k^2 \ge 0$ , then $n \ge 0$ and $20-n > 0$ , otherwise $\frac{n}{20-n}$ will be negative. Thus $0 \le n \le 19$ and \[0 = \frac{0}{20-(0)} \le \frac{n}{20-n} \le \frac{19}{20-(19)} = 19\] Checking all $k$ for which $0 \le k^2 \le 19$ , we have $0$ $1$ $2$ $3$ as the possibilities. $\boxed{4}$ | D | 4 |
9ba1dec09d7a7ea393bd9f3ce6f88ff8 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_12 | For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | For all integers x, $x^2$ is always a positive integer. So solve for $\frac{n}{20-n} = 0$ , getting $n=0$ and $\frac{n}{20-n} = 1$ , getting $n =10$ . For all values n less than 0 and greater than 20, the value $\frac{n}{20-n}$ is negative, so now try values of n between 10 and 20. Quick substitution finds $0$ $10$ $16$ , and $18$ which yields $x=0$ $x=1$ $x=2$ , and $x=3$ respectively. 4 values, or $\boxed{4}$ | D | 4 |
9ba1dec09d7a7ea393bd9f3ce6f88ff8 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_12 | For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | Simon's Favourite Factoring Trick
Since \frac{n}{20-n} is an integer k, w multiple both sides by 20-n. This gives us $n$ $20k^2$ $nk^2$ . We subtract $20k^2$ on both sides, then add $nk^2$ on both sides as a prerequisite for using Simon's Favorite Factoring Trick. We have (k^2+1)(n-20)=20. We then consider the different factors of 20 that k^2+1 can be. It could be $1$ $2$ $4$ $5$ $10$ , and $20$ . After checking case by case, we then are able to identify that there are 4 such k values that also yield an integer n value, meaning that there are 4 values, so the correct answer is $\boxed{4}$ ~CharmaineMa07292010 | D | 4 |
2ac41de2e0944529e28a1d9ea6d56754 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_14 | Four distinct circles are drawn in a plane . What is the maximum number of points where at least two of the circles intersect?
$\mathrm{(A)}\ 8 \qquad\mathrm{(B)}\ 9 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 16$ | For any given pair of circles, they can intersect at most $2$ times. Since there are ${4\choose 2} = 6$ pairs of circles, the maximum number of possible intersections is $6 \cdot 2 = 12$ . We can construct such a situation as below, so the answer is $\boxed{12}$ | D | 12 |
2ac41de2e0944529e28a1d9ea6d56754 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_14 | Four distinct circles are drawn in a plane . What is the maximum number of points where at least two of the circles intersect?
$\mathrm{(A)}\ 8 \qquad\mathrm{(B)}\ 9 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 16$ | Because a pair or circles can intersect at most $2$ times, the first circle can intersect the second at $2$ points, the third can intersect the first two at $4$ points, and the fourth can intersect the first three at $6$ points. This means that our answer is $2+4+6=\boxed{12}.$ | D | 12 |
2ac41de2e0944529e28a1d9ea6d56754 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_14 | Four distinct circles are drawn in a plane . What is the maximum number of points where at least two of the circles intersect?
$\mathrm{(A)}\ 8 \qquad\mathrm{(B)}\ 9 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 16$ | Pick a circle any circle- $4$ ways. Then, pick any other circle- $3$ ways. For each of these circles, there will be $2$ intersections for a total of $4*3*2$ $24$ intersections. However, we have counted each intersection twice, so we divide for overcounting. Therefore, we reach a total of $\frac{24}{2}=\boxed{12}$ | D | 12 |
451c9189930ace849cf6b6e4a06a3f22 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_15 | How many four-digit numbers $N$ have the property that the three-digit number obtained by removing the leftmost digit is one ninth of $N$
$\mathrm{(A)}\ 4 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 7 \qquad\mathrm{(E)}\ 8$ | Since N is a four digit number, assume WLOG that $N = 1000a + 100b + 10c + d$ , where a is the thousands digit, b is the hundreds digit, c is the tens digit, and d is the ones digit.
Then, $\frac{1}{9}N = 100b + 10c + d$ , so $N = 900b + 90c + 9d$ Set these equal to each other: \[1000a + 100b + 10c + d = 900b + 90c + 9d\] \[1000a = 800b + 80c + 8d\] \[1000a = 8(100b + 10c + d)\] Notice that $100b + 10c + d = N - 1000a$ , thus: \[1000a = 8(N - 1000a)\] \[1000a = 8N - 8000a\] \[9000a = 8N\] \[N = 1125a\]
Go back to our first equation, in which we set $N = 1000a + 100b + 10c + d$ ,
Then: \[1125a = 1000a + 100b + 10c + d\] \[125a = 100b + 10c + d\] The upper limit for the right hand side (RHS) is $999$ (when $b = 9$ $c = 9$ , and $d = 9$ ).
It's easy to prove that for an $a$ there is only one combination of $b, c,$ and $d$ that can make the equation equal. Just think about the RHS as a three digit number $bcd$ . There's one and only one way to create every three digit number with a certain combination of digits.
Thus, we test for how many as are in the domain set by the RHS. Since $125\cdot7 = 875$ which is the largest $a$ value, then $a$ can be $1$ through $7$ , giving us the answer of $\boxed{7}$ | D | 7 |
6cdcf2c9abd1afbe45388f383d2e6d62 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_19 | If $a,b,$ and $c$ are positive real numbers such that $a(b+c) = 152, b(c+a) = 162,$ and $c(a+b) = 170$ , then $abc$ is
$\mathrm{(A)}\ 672 \qquad\mathrm{(B)}\ 688 \qquad\mathrm{(C)}\ 704 \qquad\mathrm{(D)}\ 720 \qquad\mathrm{(E)}\ 750$ | Adding up the three equations gives $2(ab + bc + ca) = 152 + 162 + 170 = 484 \Longrightarrow ab + bc + ca = 242$ . Subtracting each of the above equations from this yields, respectively, $bc = 90, ca = 80, ab = 72$ . Taking their product, $ab \cdot bc \cdot ca = a^2b^2c^2 = 90 \cdot 80 \cdot 72 = 720^2 \Longrightarrow abc = \boxed{720}$ | D | 720 |
cece7c3eafed0d468927eafb84fb2a0c | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_20 | Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 28 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 32$ | 2002 12B AMC-20.png
Let $OM = x$ $ON = y$ . By the Pythagorean Theorem on $\triangle XON, MOY$ respectively, \begin{align*} (2x)^2 + y^2 &= 19^2\\ x^2 + (2y)^2 &= 22^2\end{align*}
Summing these gives $5x^2 + 5y^2 = 845 \Longrightarrow x^2 + y^2 = 169$
By the Pythagorean Theorem again, we have
\[(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{26}\] | B | 26 |
cece7c3eafed0d468927eafb84fb2a0c | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_20 | Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 28 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 32$ | Let $XO=x$ and $YO=y.$ Then, $XY=\sqrt{x^2+y^2}.$
Since $XN=19$ and $YM=22,$ \[XN^2=19^2=x^2+(\dfrac{y}{2})^2)=\dfrac{x^2}{4}+y^2\] \[YM^2=22^2=(\dfrac{x}{2})^2+y^2=\dfrac{x^2}{4}+y^2.\]
Adding these up: \[19^2+22^2=\dfrac{4x^2+y^2}{4}+\dfrac{x^2+4y^2}{4}\] \[845=\dfrac{5x^2+5y^2}{4}\] \[3380=5x^2+5y^2\] \[676=x^2+y^2.\]
Then, we substitute: $XY=\sqrt{x^2+y^2}=\sqrt{676}=\boxed{26}.$ | null | 26 |
eb1c63a56c8df590f0f4c7ff2e7d7271 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_21 | For all positive integers $n$ less than $2002$ , let
\begin{eqnarray*} a_n =\left\{ \begin{array}{lr} 11, & \text{if\ }n\ \text{is\ divisible\ by\ }13\ \text{and\ }14;\\ 13, & \text{if\ }n\ \text{is\ divisible\ by\ }14\ \text{and\ }11;\\ 14, & \text{if\ }n\ \text{is\ divisible\ by\ }11\ \text{and\ }13;\\ 0, & \text{otherwise}. \end{array} \right. \end{eqnarray*}
Calculate $\sum_{n=1}^{2001} a_n$
$\mathrm{(A)}\ 448 \qquad\mathrm{(B)}\ 486 \qquad\mathrm{(C)}\ 1560 \qquad\mathrm{(D)}\ 2001 \qquad\mathrm{(E)}\ 2002$ | Find the LCMs of the groups of the numbers.
Notice that the groups are relatively prime.
So $a_n=$
11 if $n$ is a multiple of 182.
13 if $n$ is a multiple of 154.
14 if $n$ is a multiple of 143.
When do we see ambiguities (for example: $n$ is a multiple of 11, 13, and 14)? This is only done when $n$ is a multiple of $\operatorname{lcm}(11,13,14)=2002$ . However, since $n<2002$ , this can never happen.
So we have 10 multiples of 182 we have to count (1 to 10 $*182$ ), and similarly, 12 multiples of 154, and 13 multiples of 143. The sum is $10*11+12*13+13*14=448$ . Select $\boxed{448}$ | A | 448 |
9ddf7fae51b84cf1738790ee87353839 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_22 | For all integers $n$ greater than $1$ , define $a_n = \frac{1}{\log_n 2002}$ . Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$ . Then $b- c$ equals
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ \frac{1}{2002} \qquad\mathrm{(D)}\ \frac{1}{1001} \qquad\mathrm{(E)}\ \frac 12$ | Note that $\frac{1}{\log_a b}=\log_b a$ . Thus $a_n=\log_{2002} n$ . Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following:
\[\log_{2002}\left(\frac{2*3*4*5}{10*11*12*13*14}=\frac{1}{11*13*14}=\frac{1}{2002}\right)=\boxed{1}\] | B | 1 |
9ddf7fae51b84cf1738790ee87353839 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_22 | For all integers $n$ greater than $1$ , define $a_n = \frac{1}{\log_n 2002}$ . Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$ . Then $b- c$ equals
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ \frac{1}{2002} \qquad\mathrm{(D)}\ \frac{1}{1001} \qquad\mathrm{(E)}\ \frac 12$ | Note that $a_2 = \frac{1}{\log_2 2002}$ $1$ is also equal to $\log_2 2$ . So $a_2 = \frac{\log_2 2}{\log_2 2002}$ . By the change of bases formula, $a_2 = \log_{2002} 2$ . Following the same reasoning, $a_3 = \log_{2002} 3$ $a_4 = \log_{2002} 4$ and so on.
\[b = \log_{2002} 2 + \log_{2002} 3 + .....+ \log_{2002} 5 = \log_{2002} 5! = \log_{2002} 120\] Now solving for $c$ , we see that it equals $\log_{2002} (10\cdot 11 \cdot 12 \cdot 13 \cdot 14)$ \[b-c = \log_{2002} 120 - \log_{2002} 240240 \rightarrow \log_{2002} \frac{1}{2002} = \boxed{1}\] | null | 1 |
39bca6811577736d9c9e3aed8c57c813 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_23 | In $\triangle ABC$ , we have $AB = 1$ and $AC = 2$ . Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$
$\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad\mathrm{(C)}\ \sqrt{2} \qquad\mathrm{(D)}\ \frac 32 \qquad\mathrm{(E)}\ \sqrt{3}$ | From Stewart's Theorem , we have $(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.$ Simplifying, we get $(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{2}.$ - awu2014 | null | 2 |
39bca6811577736d9c9e3aed8c57c813 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_23 | In $\triangle ABC$ , we have $AB = 1$ and $AC = 2$ . Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$
$\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad\mathrm{(C)}\ \sqrt{2} \qquad\mathrm{(D)}\ \frac 32 \qquad\mathrm{(E)}\ \sqrt{3}$ | There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have $\triangle{ABC}$ , and you draw a median from point $A$ to side $BC$ (label this as $M$ ), then: $(AM)^2 = \dfrac{2(b^2) + 2(c^2) - (a^2)}{4}$ . Note that $b$ is the length of side $\overline{AC}$ $c$ is the length of side $\overline{AB}$ , and $a$ is length of side $\overline{BC}$ . Let $MB = MC = x$ . Then $AM = 2x$ . Now, we can plug into the formula given above: $AM = 2x$ $b = 2$ $c = 1$ , and $a = 2x$ . After some simple algebra, we find $x = \dfrac{\sqrt{2}}{2}$ . Then, $BC = \boxed{2}$ | C | 2 |
9dd0f70aa16be1869f195fb50a64cb0f | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_25 | Let $f(x) = x^2 + 6x + 1$ , and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that \[f(x) + f(y) \le 0 \qquad \text{and} \qquad f(x)-f(y) \le 0\] The area of $R$ is closest to
$\textbf{(A) } 21 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 23 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 25$ | The first condition gives us that \[x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16\]
which is a circle centered at $(-3,-3)$ with radius $4$ . The second condition gives us that
\[x^2 + 6x + 1 - y^2 - 6y - 1 \le 0 \Longrightarrow (x^2 - y^2) + 6(x-y) \le 0 \Longrightarrow (x-y)(x+y+6) \le 0\]
Thus either
\[x - y \ge 0,\quad x+y+6 \le 0\]
or
\[x - y \le 0,\quad x+y+6 \ge 0\]
Each of those lines passes through $(-3,-3)$ and has slope $\pm 1$ , as shown above. Therefore, the area of $R$ is half of the area of the circle, which is $\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx \boxed{25}$ | E | 25 |
9dd0f70aa16be1869f195fb50a64cb0f | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_25 | Let $f(x) = x^2 + 6x + 1$ , and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that \[f(x) + f(y) \le 0 \qquad \text{and} \qquad f(x)-f(y) \le 0\] The area of $R$ is closest to
$\textbf{(A) } 21 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 23 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 25$ | Similar to Solution 1, we proceed to get the area of the circle satisfying $f(x)+f(y) \le 0$ , or $16 \pi$
Since $f(x)-f(y) \le 0 \implies f(x) \le f(y)$ , we have that by symmetry, if $(x,y)$ is in $R$ , then $(y,x)$ is not, and vice versa. Therefore, the shaded part of the circle above the line $y=x$ has the same area as the unshaded part below $y=x$ , and the unshaded part above $y=x$ has the same area as the shaded part below $y=x$ . This means that exactly half the circle is shaded, allowing us to divide by two to get $\frac{16 \pi }{2} = 8\pi \approx \boxed{25}$ . ~samrocksnature + ddot1 | E | 25 |
1a139e275f660b44dfb60a4c14d0c389 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_2 | The function $f$ is given by the table
\[\begin{tabular}{|c||c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 4 & 1 & 3 & 5 & 2 \\ \hline \end{tabular}\]
If $u_0=4$ and $u_{n+1} = f(u_n)$ for $n \ge 0$ , find $u_{2002}$
$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }3 \qquad \text{(D) }4 \qquad \text{(E) }5$ | We can guess that the series given by the problem is periodic in some way. Starting off, $u_0=4$ is given. $u_1=u_{0+1}=f(u_0)=f(4)=5,$ so $u_1=5.$ $u_2=u_{1+1}=f(u_1)=f(5)=2,$ so $u_2=2.$ $u_3=u_{2+1}=f(u_2)=f(2)=1,$ so $u_3=1.$ $u_4=u_{3+1}=f(u_3)=f(1)=4,$ so $u_4=4.$ Plugging in $4$ will give us $5$ as found before, and plugging in $5$ will give $2$ and so on. This means that our original guess of the series being periodic was correct. Summing up our findings in a nice table,
\[\begin{tabular}{|c||c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & 4 & ...\\ \hline un & 4 & 5 & 2 & 1 & 4 & ...\\ \hline \end{tabular}\]
in which the next $u_n$ is found by simply plugging in the number from the last box into $f(x).$ The function is periodic every $4$ terms. $2002 \equiv 2\pmod{4}$ , and counting $4$ starting from $u_1$ will give us our answer of $\boxed{2}$ | B | 2 |
128832d5549e454ad5ba0e91f1d246df | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_3 | The dimensions of a rectangular box in inches are all positive integers and the volume of the box is $2002$ in $^3$ . Find the minimum possible sum of the three dimensions.
$\text{(A) }36 \qquad \text{(B) }38 \qquad \text{(C) }42 \qquad \text{(D) }44 \qquad \text{(E) }92$ | Given an arbitrary product and an arbitrary amount of terms to multiply to get that product, to maximize the sum, make all of the terms $1$ with the last one being the number. To minimize the sum, make all of the terms equal to each other. (This is a corollary that follows from the $AM-GM$ proof.) Since $2002$ is not a perfect cube, we have to make the terms as close to each other as possible. A good rule of thumb is the memorize the prime factorization of the AMC year that you are doing, which is $2002= 2 \cdot 7 \cdot 11 \cdot 13$ . The three terms that are closest to each other that multiply to $2002$ are $11, 13,$ and $14$ , so our answer is $11+13+14=\boxed{38}$ | B | 38 |
a3d5b13c187626f189c9d83d73b23bd6 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_4 | Let $a$ and $b$ be distinct real numbers for which \[\frac{a}{b} + \frac{a+10b}{b+10a} = 2.\]
Find $\frac{a}{b}$
$\text{(A) }0.4 \qquad \text{(B) }0.5 \qquad \text{(C) }0.6 \qquad \text{(D) }0.7 \qquad \text{(E) }0.8$ | For sake of speed, WLOG, let $b=1$ . This means that the ratio $\frac{a}{b}$ will simply be $a$ because $\frac{a}{b}=\frac{a}{1}=a.$ Solving for $a$ with some very simple algebra gives us a quadratic which is $5a^2 -9a +4=0.$ Factoring the quadratic gives us $(5a-4)(a-1)=0$ . Therefore, $a=1$ or $a=\frac{4}{5}=0.8.$ However, since $a$ and $b$ must be distinct, $a$ cannot be $1$ so the latter option is correct, giving us our answer of $\boxed{0.8}.$ | E | 0.8 |
a3d5b13c187626f189c9d83d73b23bd6 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_4 | Let $a$ and $b$ be distinct real numbers for which \[\frac{a}{b} + \frac{a+10b}{b+10a} = 2.\]
Find $\frac{a}{b}$
$\text{(A) }0.4 \qquad \text{(B) }0.5 \qquad \text{(C) }0.6 \qquad \text{(D) }0.7 \qquad \text{(E) }0.8$ | The only tricky part about this equation is the fact that the left-hand side has fractions. Multiplying both sides by $b(b+10a)$ gives us $2ab+10a^2+10b^2=2b^2+20ab.$ Moving everything to the left-hand side and dividing by $2$ gives $5a^2-4b^2 -9ab,$ which factors into $(5a-4b)(a-b)=0.$ Because $a \neq b, 5a=4b \implies \frac{a}{b}=0.8$ giving us our answer of $\boxed{0.8}.$ | E | 0.8 |
acc9bab494ce332d55ff8f990b2adc4a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_7 | How many three-digit numbers have at least one $2$ and at least one $3$
$\text{(A) }52 \qquad \text{(B) }54 \qquad \text{(C) }56 \qquad \text{(D) }58 \qquad \text{(E) }60$ | We can do this problem with some simple case work.
Case 1: The hundreds place is not $2$ or $3.$ This means that the tens place and ones place must be $2$ and $3$ respectively or $3$ and $2$ respectively. This case covers $1, 4, 5, 6, 7, 8,$ and $9,$ so it gives us $2 \cdot 7 = 14$ cases.
Case 2: The hundreds place is $2.$ This means that $3$ must be in the tens place or ones place. Starting with cases in which the tens place is not $3$ , we get $203, 213, 223, 243, 253, 263, 273, 283,$ and $293.$ With cases in which the tens place is $3$ , we have $230-239$ , or $10$ more cases. This gives us $9 + 10=19$ cases.
Case 3: The hundreds place is $3.$ This case is almost identical to the second case, just swap the $2$ s with $3$ s and $3$ s with $2$ s in the reasoning and its the same, giving us an additional $19$ cases.
Addition up all of these cases gives $14+19+19=52$ cases, or $\boxed{52}.$ | A | 52 |
f443dafe9823880f88da259be0c32d02 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_9 | Two walls and the ceiling of a room meet at right angles at point $P.$ A fly is in the air one meter from one wall, eight meters from the other wall, and nine meters from point $P$ . How many meters is the fly from the ceiling?
$\text{(A) }\sqrt{13} \qquad \text{(B) }\sqrt{14} \qquad \text{(C) }\sqrt{15} \qquad \text{(D) }4 \qquad \text{(E) }\sqrt{17}$ | We can use the formula for the diagonal of the rectangle, or $d=\sqrt{a^2+b^2+c^2}$ The problem gives us $a=1, b=8,$ and $c=9.$ Solving gives us $9=\sqrt{1^2 + 8^2 + c^2} \implies c^2=9^2-8^2-1^2 \implies c^2=16 \implies c=\boxed{4}.$ | D | 4 |
72109519a252cc009aaf595ce15f2ab6 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_10 | Let $f_n (x) = \text{sin}^n x + \text{cos}^n x.$ For how many $x$ in $[0,\pi]$ is it true that
\[6f_{4}(x)-4f_{6}(x)=2f_{2}(x)?\]
$\text{(A) }2 \qquad \text{(B) }4 \qquad \text{(C) }6 \qquad \text{(D) }8 \qquad \text{(E) more than }8$ | Divide by 2 on both sides to get \[3f_{4}(x)-2f_{6}(x)=f_{2}(x)\] Substituting the definitions of $f_{2}(x)$ $f_{4}(x)$ , and $f_{6}(x)$ , we may rewrite the expression as \[3(\text{sin}^4{x} + \text{cos}^4{x}) - 2(\text{sin}^6{x} + \text{cos}^6{x}) = 1\] We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.
We can rewrite $3(\text{sin}^4 x + \text{cos}^4 x)$ as $3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x$ , which is equivalent to $3 - 6\text{sin}^2 x \text{cos}^2 x$
As for $2(\text{sin}^6 x + \text{cos}^6 x)$ , we may factor it as $2(\text{sin}^2 x + \text{cos}^2 x)(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$ which can be rewritten as $2(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$ , and then as $2(\text{sin}^2 x + \text{cos}^2)^2) - 6\text{sin}^2 x \text{cos}^2 x$ , which is equivalent to $2 - 6\text{sin}^2 x \text{cos}^2 x$
Putting everything together, we have $(3 - 6\text{sin}^2 {x} \text{cos}^2 {x}) - (2 - 6\text{sin}^2 {x} \text{cos}^2 {x}) = 1$ or $1 = 1$ . Therefore, the given equation $3f_{4}(x)-2f_{6}(x)=f_{2}(x)$ is true for all real $x$ , meaning that there are more than 8 values of $x$ that satisfy the given equation and so the answer is $\boxed{8}$ | E | 8 |
c1a79ba195fae73d20ba12d90669b9c8 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_12 | For how many positive integers $n$ is $n^3 - 8n^2 + 20n - 13$ a prime number?
$\text{(A) 1} \qquad \text{(B) 2} \qquad \text{(C) 3} \qquad \text{(D) 4} \qquad \text{(E) more than 4}$ | Since this is a number theory question, it is clear that the main challenge here is factoring the given cubic. In general, the rational root theorem will be very useful for these situations.
The rational root theorem states that all rational roots of $n^3 - 8n^2 + 20n - 13$ will be among $1, 13, -1$ , and $-13$ . Evaluating the cubic at these values will give $n = 1$ as a root. Doing some synthetic division gives \[n^3 - 8n^2 + 20n - 13 = (n-1)(n^2 - 7n + 13)\]
Since $n > 0$ $n-1$ must be nonnegative. Since $(n-1)(n^2 - 7n + 13)$ evaluates to a prime, it is clear that exactly one of $n-1$ and $n^2 - 7n - 13$ is $1$ . We proceed by splitting the problem into 2 cases.
Case 1: $n-1 = 1$ It is clear that $n = 2$ . We have $2^2 - 7(2) + 13 = 3$ , so this case yields $n = 2$ as a solution.
Case 2: $n^2 - 7n + 13 = 1$ Solving for $n$ gives $n^2 - 7n + 12 = 0$ or $(n-3)(n-4) = 0$ . Therefore, $n = 3$ or $n = 4$ .
Since both $3-1 = 2$ and $4-1 = 3$ are prime, both $n = 3$ and $n = 4$ work, yielding 2 solutions.
Putting everything together, the answer is $1 + 2 = \boxed{3}$ | C | 3 |
381b8fd4038aed7a5f58456a30920567 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_13 | What is the maximum value of $n$ for which there is a set of distinct positive integers $k_1, k_2, ... k_n$ for which
\[k^2_1 + k^2_2 + ... + k^2_n = 2002?\]
$\text{(A) }14 \qquad \text{(B) }15 \qquad \text{(C) }16 \qquad \text{(D) }17 \qquad \text{(E) }18$ | Note that $k^2_1 + k^2_2 + ... + k^2_n = 2002 \leq \frac{n(n+1)(2n+1)}{6}$
When $n = 17$ $\frac{n(n+1)(2n+1)}{6} = \frac{(17)(18)(35)}{6} = 1785 < 2002$
When $n = 18$ $\frac{n(n+1)(2n+1)}{6} = 1785 + 18^2 = 2109 > 2002$
Therefore, we know $n \leq 17$
Now we must show that $n = 17$ works. We replace one of $1, 2, ... 17$ with an integer $a > 17$ to account for the amount under $2002$ , which is $2002-1785 = 217$
Essentially, this boils down to writing $217$ as a difference of squares. We know $217 = (7)(31)$ , so we assume there exist positive integers $a$ and $b$ where $a > 17$ and $b \leq 17$ such that $a^2 - b^2 = 217$
We can rewrite this as $(a+b)(a-b) = 217$ , so either $a+b = 217$ and $a-b = 1$ or $a+b = 31$ and $a-b = 7$ . We analyze each case separately.
Case 1: $a+b = 217$ and $a-b = 1$
Solving this system of equations gives $a = 109$ and $b = 108$ . However, $108 > 17$ , so this case does not yield a solution.
Case 2: $a+b = 31$ and $a-b = 7$
Solving this system of equations gives $a = 19$ and $b = 12$ . This satisfies all the requirements of the problem.
The list $1, 2 ... 11, 13, 14 ... 17, 19$ has $17$ terms whose sum of squares equals $2002$ . Therefore, the answer is $\boxed{17}$ | D | 17 |
8d46d720c449ecb415360c0081f595e0 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_16 | The altitudes of a triangle are $12, 15,$ and $20.$ The largest angle in this triangle is
$\text{(A) }72^\circ \qquad \text{(B) }75^\circ \qquad \text{(C) }90^\circ \qquad \text{(D) }108^\circ \qquad \text{(E) }120^\circ$ | Let $a, b,$ and $c$ denote the bases of altitudes $12, 15,$ and $20,$ respectively. Since they are all altitudes and bases of the same triangle, they have the same area, so $\frac{12a}{2}=\frac{15b}{2}=\frac{20c}{2}.$ Multiplying by $2$ , we get $12a=15b=20c.$ Notice that a simple solution to the equation is if all of them equal $12 \cdot 15 \cdot 20.$ That means $a=15 \cdot 20, b=12 \cdot 20,$ and $c=12 \cdot 15.$ Simplifying our solution to check for Pythagorean triples we see that this is just a Pythagorean triple, namely a $3-4-5$ triangle. Since the other two angles of a right triangle must be acute, the right angle must be the greatest angle. Therefore, our answer is $\boxed{90}.$ | C | 90 |
afaf9ead4ac53b3af04db3b4aa3e88f3 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_17 | Let $f(x) = \sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}}.$ An equivalent form of $f(x)$ is
$\text{(A) }1-\sqrt{2}\sin{x} \qquad \text{(B) }-1+\sqrt{2}\cos{x} \qquad \text{(C) }\cos{\frac{x}{2}} - \sin{\frac{x}{2}} \qquad \text{(D) }\cos{x} - \sin{x} \qquad \text{(E) }\cos{2x}$
Solution | By the Pythagorean identity we can rewrite the given expression as follows. \[\sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}} = \sqrt{\sin^4{x} + 4(1 - \sin^2{x})} - \sqrt{\cos^4{x} + 4(1 - \cos^2{x})}\]
Expanding each bracket gives \[\sqrt{\sin^4{x} - 4\sin^2{x} + 4} - \sqrt{\cos^4{x} - 4\cos^2{x} + 4}\]
The expressions under the square roots can be factored to get \[\sqrt{(\sin^2{x} - 2)^2} - \sqrt{(\cos^2{x} - 2)^2}\]
Since $\sin^2{x} < 2$ and $\cos^2{x} < 2$ for all real $x$ , the expression must evaluate to $(2 - \sin^2{x}) - (2 - \cos^2{x})$ , which simplifies to $\cos^2{x} - \sin^2{x} = \boxed{2}$ | E | 2 |
afaf9ead4ac53b3af04db3b4aa3e88f3 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_17 | Let $f(x) = \sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}}.$ An equivalent form of $f(x)$ is
$\text{(A) }1-\sqrt{2}\sin{x} \qquad \text{(B) }-1+\sqrt{2}\cos{x} \qquad \text{(C) }\cos{\frac{x}{2}} - \sin{\frac{x}{2}} \qquad \text{(D) }\cos{x} - \sin{x} \qquad \text{(E) }\cos{2x}$
Solution | We don't actually have to solve the question. Just let $x$ equal some easy value to calculate $\cos {x}, \cos {2x}, \sin {x}, \sin {\frac{x}{2}},$ and $\cos {\frac{x}{2}}.$ For this solution, let $x=60^\circ.$ This means that the expression in the problem will give $\sqrt{\sin^4{60^\circ} + 4 \cos^2{60^\circ}} - \sqrt{\cos^4{60^\circ} + 4 \sin^2{60^\circ}}=\sqrt{(\frac{\sqrt{3}}{2})^4 + 4(\frac{1}{2})^2} - \sqrt{(\frac{1}{2})^4 + 4(\frac{\sqrt{3}}{2})^2}=\sqrt{\frac{9}{16} +1} - \sqrt{\frac{1}{16} + 3} = \frac{-1}{2}.$ Plugging in $x=60^\circ$ for the rest of the expressions, we get
$\text{(A) }1-\sqrt{2}\sin{60^\circ}=1-\frac{\sqrt{6}}{2} \neq \frac{-1}{2}.$
$\text{(B) }1+\sqrt{2}\cos{60^\circ}=1+\frac{\sqrt{2}}{2} \neq \frac{-1}{2}.$
$\text{(C) }\cos{\frac{60^\circ}{2}} - \sin{\frac{60^\circ}{2}}=\frac{\sqrt{3}}{2}-1 \neq \frac{-1}{2}.$
$\text{(D) }\cos{60^\circ} - \sin{60^\circ}=\frac{1-\sqrt{3}}{2} \neq \frac{-1}{2}.$
$\text{(E) }\cos {120^\circ}=\frac{-1}{2}.$
Therefore, our answer is $\boxed{2}$ | E | 2 |
02baa6f66623afcab170619810485921 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_18 | If $a,b,c$ are real numbers such that $a^2 + 2b =7$ $b^2 + 4c= -7,$ and $c^2 + 6a= -14$ , find $a^2 + b^2 + c^2.$
$\text{(A) }14 \qquad \text{(B) }21 \qquad \text{(C) }28 \qquad \text{(D) }35 \qquad \text{(E) }49$ | Adding all of the equations gives $a^2 + b^2 +c^2 + 6a + 2b + 4c=-14.$ Adding 14 on both sides gives $a^2 + b^2 +c^2 + 6a + 2b + 4c+14=0.$ Notice that 14 can split into $9, 1,$ and $4,$ which coincidentally makes $a^2 +6a, b^2+2b,$ and $c^2+4c$ into perfect squares. Therefore, $(a+3)^2 + (b+1)^2 + (c+2) ^2 =0.$ An easy solution to this equation is $a=-3, b=-1,$ and $c=-2.$ Plugging in that solution, we get $a^2+b^2+c^2=-3^2+-1^2+-2^2=\boxed{14}.$ | A | 14 |
5b674e6c4ee28a67b12a325203767052 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_20 | Let $f$ be a real-valued function such that
\[f(x) + 2f(\frac{2002}{x}) = 3x\]
for all $x>0.$ Find $f(2).$
$\text{(A) }1000 \qquad \text{(B) }2000 \qquad \text{(C) }3000 \qquad \text{(D) }4000 \qquad \text{(E) }6000$ | Setting $x = 2$ gives $f(2) + 2f(1001) = 6$ .
Setting $x = 1001$ gives $2f(2) + f(1001) = 3003$
Adding these 2 equations and dividing by 3 gives $f(2) + f(1001) = \frac{6+3003}{3} = 1003$
Subtracting these 2 equations gives $f(2) - f(1001) = 2997$
Therefore, $f(2) = \frac{1003+2997}{2} = \boxed{2000}$ | B | 2000 |
acffc92b3e970ea35f961a3258ba2264 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_21 | Let $a$ and $b$ be real numbers greater than $1$ for which there exists a positive real number $c,$ different from $1$ , such that
\[2(\log_a{c} + \log_b{c}) = 9\log_{ab}{c}.\]
Find the largest possible value of $\log_a b.$
$\text{(A) }\sqrt{2} \qquad \text{(B) }\sqrt{3} \qquad \text{(C) }2 \qquad \text{(D) }\sqrt{6} \qquad \text{(E) }3$ | We may rewrite the given equation as \[2(\frac {\log c}{\log a} + \frac {\log c}{\log b}) = \frac {9\log c}{\log a + \log b}\] Since $c \neq 1$ , we have $\log c \neq 0$ , so we may divide by $\log c$ on both sides. After making the substitutions $x = \log a$ and $y = \log b$ , our equation becomes \[\frac {2}{x} + \frac {2}{y} = \frac {9}{x+y}\]
Rewriting the left-hand side gives \[\frac {2(x+y)}{xy} = \frac {9}{x+y}\]
Cross-multiplying gives $2(x+y)^2 = 9xy$ or \[2x^2 - 5xy + 2y^2 = 0\]
Factoring gives $(2x-y)(x-2y) = 0$ or $\frac {x}{y} = 2, \frac {1}{2}$
Recall that $\frac {x}{y} = \frac {\log a}{\log b} = \log_{a} b$ . Therefore, the maximum value of $\log_{a} b$ is $\boxed{2}$ | C | 2 |
166befbcb69b0e3d2ed096f385a10201 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_22 | Under the new AMC $10, 12$ scoring method, $6$ points are given for each correct answer, $2.5$ points are given for each unanswered question, and no points are given for an incorrect answer. Some of the possible scores between $0$ and $150$ can be obtained in only one way, for example, a score of $104.5$ can be obtained with $17$ correct answers, $1$ unanswered question, and $7$ incorrect answers, and also with $12$ correct answers and $13$ unanswered questions. There are scores that can be obtained in exactly three ways. What is their sum?
$\text{(A) }175 \qquad \text{(B) }179.5 \qquad \text{(C) }182 \qquad \text{(D) }188.5 \qquad \text{(E) }201$ | $\boxed{188.5}$ | D | 188.5 |
f6733f480506d48c17bc725ceb0aa577 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_23 | The equation $z(z+i)(z+3i)=2002i$ has a zero of the form $a+bi$ , where $a$ and $b$ are positive real numbers. Find $a.$
$\text{(A) }\sqrt{118} \qquad \text{(B) }\sqrt{210} \qquad \text{(C) }2 \sqrt{210} \qquad \text{(D) }\sqrt{2002} \qquad \text{(E) }100 \sqrt{2}$ | According to Wolfram-Alpha, the answer is $\boxed{118}$ | A | 118 |
cea5126fa4c42a5f46672a0e63785daf | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_2 | Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits
of the integer $n$ . For example, $P(23) = 6$ and $S(23) = 5$ . Suppose $N$ is a
two-digit number such that $N = P(N)+S(N)$ . What is the units digit of $N$
$\text{(A)}\ 2\qquad \text{(B)}\ 3\qquad \text{(C)}\ 6\qquad \text{(D)}\ 8\qquad \text{(E)}\ 9$ | Denote $a$ and $b$ as the tens and units digit of $N$ , respectively. Then $N = 10a+b$ . It follows that $10a+b=ab+a+b$ , which implies that $9a=ab$ . Since $a\neq0$ $b=9$ . So the units digit of $N$ is $\boxed{9}$ | E | 9 |
227cbd9a094a48b5f95a026123843b8c | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_3 | problem_id
227cbd9a094a48b5f95a026123843b8c The state income tax where Kristin lives is le...
227cbd9a094a48b5f95a026123843b8c The state income tax where Kristin lives is le...
Name: Text, dtype: object | Let $A$ $T$ be Kristin's annual income and the income tax total, respectively. Notice that \begin{align*} T &= p\%\cdot28000 + (p + 2)\%\cdot(A - 28000) \\ &= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000) \\ &= p\%\cdot A + 2\%\cdot(A - 28000) \end{align*} We are also given that \[T = (p + 0.25)\%\cdot A = p\%\cdot A + 0.25\%\cdot A\] Thus, \[p\%\cdot A + 2\%\cdot(A - 28000) = p\%\cdot A + 0.25\%\cdot A\] \[2\%\cdot(A - 28000) = 0.25\%\cdot A\] Solve for $A$ to obtain $A = \boxed{32000}$ | B | 32000 |
ae79010feec50f73241383732e6c476e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_4 | problem_id
ae79010feec50f73241383732e6c476e The mean of three numbers is $10$ more than th...
ae79010feec50f73241383732e6c476e The mean of three numbers is $10$ more than th...
Name: Text, dtype: object | Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$ . The middle of the three numbers is the median, $5$ . So $\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$ , which can be solved to get $m=10$ .
Hence, the sum of the three numbers is $3\cdot 10 = \boxed{30}$ | D | 30 |
ae79010feec50f73241383732e6c476e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_4 | problem_id
ae79010feec50f73241383732e6c476e The mean of three numbers is $10$ more than th...
ae79010feec50f73241383732e6c476e The mean of three numbers is $10$ more than th...
Name: Text, dtype: object | Say the three numbers are $x$ $y$ and $z$ . When we arrange them in ascending order then we can assume $y$ is in the middle therefore $y = 5$
We can also assume that the smallest number is $x$ and the largest number of the three is $y$ .
Therefore,
\[\frac{x+y+z}{3} = x + 10 = z - 15\] \[\frac{x+5+z}{3} = x + 10 = z - 15\]
Taking up the first equation $\frac{x+5+z}{3} = x + 10$ and simplifying we obtain $z - 2x - 25 = 0$ doing so for the equation $\frac{x+5+z}{3} = z - 15$ we obtain the equation $x - 2z + 50 = 0$
\[x - 2z + 50 = 2x - 4z + 100\]
when solve the above obtained equation and $z - 2x - 25 = 0$ we obtain the values of $z = 25$ and $x = 0$
Therefore the sum of the three numbers is $25 + 5 + 0 = \boxed{30}$ | D | 30 |
74b973e4f94621e9337c1a9c0077ccfc | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_6 | problem_id
74b973e4f94621e9337c1a9c0077ccfc A telephone number has the form $\text{ABC-DEF...
74b973e4f94621e9337c1a9c0077ccfc A telephone number has the form $\text{ABC-DEF...
Name: Text, dtype: object | We start by noting that there are $10$ letters, meaning there are $10$ digits in total. Listing them all out, we have $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ . Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that.
Case 1: $G$ $H$ $I$ , and $J$ are $7$ $5$ $3$ , and $1$ respectively.
A cursory glance allows us to deduce that the smallest possible sum of $A + B + C$ is $11$ when $D$ $E$ , and $F$ are $8$ $6$ , and $4$ respectively, so this is out of the question.
Case 2: $G$ $H$ $I$ , and $J$ are $3$ $5$ $7$ , and $9$ respectively.
A cursory glance allows us to deduce the answer. Clearly, when $D$ $E$ , and $F$ are $6$ $4$ , and $2$ respectively, $A + B + C$ is $9$ when $A$ $B$ , and $C$ are $8$ $1$ , and $0$ respectively, giving us a final answer of $\boxed{8}$ | E | 8 |
afa106734f55c02711ecd5e8bbf4e8d3 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_7 | problem_id
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
Name: Text, dtype: object | Let's multiply ticket costs by $2$ , then the half price becomes an integer, and the charity sold $140$ tickets worth a total of $4002$ dollars.
Let $h$ be the number of half price tickets, we then have $140-h$ full price tickets. The cost of $140-h$ full price tickets is equal to the cost of $280-2h$ half price tickets.
Hence we know that $h+(280-2h) = 280-h$ half price tickets cost $4002$ dollars. Then a single full price ticket costs $\frac{4002}{280-h}$ dollars, and this must be an integer. Thus $280-h$ must be a divisor of $4002$ . Keeping in mind that $0\leq h\leq 140$ , we are looking for a divisor between $140$ and $280$ , inclusive.
The prime factorization of $4002$ is $4002=2\cdot 3\cdot 23\cdot 29$ . We can easily find out that the only divisor of $4002$ within the given range is $2\cdot 3\cdot 29 = 174$
This gives us $280-h=174$ , hence there were $h=106$ half price tickets and $140-h = 34$ full price tickets.
In our modified setting (with prices multiplied by $2$ ) the price of a half price ticket is $\frac{4002}{174} = 23$ . In the original setting this is the price of a full price ticket. Hence $23\cdot 34 = \boxed{782}$ dollars are raised by the full price tickets. | A | 782 |
afa106734f55c02711ecd5e8bbf4e8d3 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_7 | problem_id
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
Name: Text, dtype: object | Let the cost of the full price ticket be $x$ , the number of full-price tickets be $A$ , and the number of half-price tickets be $B$
Let's multiply both sides of the equation that naturally follows by 2. We have
\[2Ax+Bx=4002\]
And we have $A+B=140\implies B=140-A$
Plugging in, we get $\implies 2Ax+(140-A)(x)=4002$
Simplifying, we get $Ax+140x=4002$
Factoring out the $x$ , we get $x(A+140)=4002\implies x=\frac{4002}{A+140}$
We see that the fraction has to simplify to an integer (the full price is a whole dollar amount)
Thus, $A+140$ must be a factor of 4002.
Consider the prime factorization of $4002$ $2\times3\times23\times29$
$A$ must be a positive integer. So, we seek a factor of $4002$ to set equal to $A+140$ so that we get an integer solution for $A$ that is less than $140$ . By guess-and-check OR inspection, the appropriate factor is $174$ $2\times3\times29$ ), meaning that $A$ has a value of $34$ . Plug this into the above equation for $x$ to get $x = 23$
Therefore, the price of full tickets out of $2001$ is $23\times34=\boxed{782}$ | A | 782 |
afa106734f55c02711ecd5e8bbf4e8d3 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_7 | problem_id
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
afa106734f55c02711ecd5e8bbf4e8d3 A charity sells $140$ benefit tickets for a to...
Name: Text, dtype: object | Let $f$ equal the number of full-price tickets, and let $h$ equal the number of half-price tickets. Additionally, suppose that the price of $f$ is $p$ . We are trying to solve for $f \cdot p$
Since the total number of tickets sold is $140$ , we know that \[f+h=140.\] The sales from full-price tickets ( $f \cdot p$ ) plus the sales from half-price tickets $\Big(\frac{h \cdot p}{2}$ , because each hall-price ticket costs $\frac{p}{2}$ dollars $\Big)$ equals $2001.$ Then we can write \[fx + \frac{hx}{2}=2001.\]
Substituting $h=140-f$ into the second equation, we get \[f \cdot p +\frac{(140-f)p}{2}=f \cdot p+\frac{140p-f\cdot p}{2}=\frac{f\cdot p+140p}{2}=2001.\]
Multiplying by $2$ and subtracting $140p$ gives us \[f\cdot p=4002-140p.\]
Since the problem states that $x$ is a whole number, $140p$ will be some integer multiple of $140$ that ends in a $0$ . Thus, $4002-140p$ will end in a $2$ . Looking at the answer choices, only $\boxed{782}$ satisfies that condition. | A | 782 |
e693d56583fdeed51d6927382415ee98 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_9 | Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all positive real numbers $x$ and $y$ . If $f(500) =3$ , what is the value of $f(600)$
$(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5$ | Letting $x = 500$ and $y = \dfrac65$ in the given equation, we get $f(500\cdot\frac65) = \frac3{\frac65} = \frac52$ , or $f(600) = \boxed{52}$ | C | 52 |
e693d56583fdeed51d6927382415ee98 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_9 | Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all positive real numbers $x$ and $y$ . If $f(500) =3$ , what is the value of $f(600)$
$(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5$ | The only function that satisfies the given condition is $y = \frac{k}{x}$ , for some constant $k$ . Thus, the answer is $\frac{500 \cdot 3}{600} = \boxed{52}$ | C | 52 |
e693d56583fdeed51d6927382415ee98 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_9 | Let $f$ be a function satisfying $f(xy) = \frac{f(x)}y$ for all positive real numbers $x$ and $y$ . If $f(500) =3$ , what is the value of $f(600)$
$(\mathrm{A})\ 1 \qquad (\mathrm{B})\ 2 \qquad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5$ | Note that the equation given above is symmetric, so we have $x \cdot f(x)=y \cdot f(y)$ . Plugging in $x=500$ and $y=600$ gives $f(y)=\boxed{52}$ | C | 52 |
c1c2900151c908ac390988a490c7e35c | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_10 | problem_id
c1c2900151c908ac390988a490c7e35c The plane is tiled by congruent squares and co...
c1c2900151c908ac390988a490c7e35c The plane is tiled by congruent squares and co...
Name: Text, dtype: object | Consider any single tile:
[asy] unitsize(1cm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; draw(p); [/asy]
If the side of the small square is $a$ , then the area of the tile is $9a^2$ , with $4a^2$ covered by squares and $5a^2$ by pentagons.
Hence exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane. When expressed as a percentage, this is $55.\overline{5}\%$ , and the closest integer to this value is $\boxed{56}$ | D | 56 |
44dac98b900fb2d03612e3e20d26762f | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_11 | problem_id
44dac98b900fb2d03612e3e20d26762f A box contains exactly five chips, three red a...
44dac98b900fb2d03612e3e20d26762f A box contains exactly five chips, three red a...
Name: Text, dtype: object | Let's assume we don't stop picking until all of the chips are picked. To satisfy this condition, we have to arrange the letters: $W, W, R, R, R$ such that both $W$ 's appear in the first $4$ . We find the number of ways to arrange the white chips in the first $4$ and divide that by the total ways to choose all the chips. The probability of this occurring is $\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{35}$ | D | 35 |
44dac98b900fb2d03612e3e20d26762f | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_11 | problem_id
44dac98b900fb2d03612e3e20d26762f A box contains exactly five chips, three red a...
44dac98b900fb2d03612e3e20d26762f A box contains exactly five chips, three red a...
Name: Text, dtype: object | The amount of ways to end with a white chip is by having $RRWW, RWW,$ and $WW$ . The amount of arrangements for $RRWW$ with $W$ at the end is $3$ , the number of arrangements of $RWW$ with $W$ at the end is $2$ , and the number of arrangements with $WW$ is just $1$ . This gives us $6$ total ways to end with white. Next, the cases to end with a red are $RWRR$ , and $RRR$ $RWRR$ gives us $3$ ways and $RRR$ gives us $1$ way. So the number of ways to end with a red is $4$ . Thus, our answer is simply $\frac{6}{4+6}$ $\boxed{35}$ | D | 35 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | We can solve this problem by finding the cases where the number is divisible by $3$ or $4$ , then subtract from the cases where none of those cases divide $5$ . To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor function to every case to get $\left\lfloor \frac{2001}{3} \right\rfloor$ $\left\lfloor \frac{2001}{4} \right\rfloor$ , and $\left\lfloor \frac{2001}{12} \right\rfloor$ . The first two floor functions were for calculating the number of individual cases for $3$ and $4$ . The third case was to find any overlapping numbers. The numbers were $667$ $500$ , and $166$ , respectively. We add the first two terms and subtract the third to get $1001$ . The first case is finished.
The second case is more or less the same, except we are applying $3$ and $4$ to $5$ . We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions $\left\lfloor \frac{2001}{3\cdot5} \right\rfloor$ $\left\lfloor \frac{2001}{4\cdot5} \right\rfloor$ , and $\left\lfloor \frac{2001}{3\cdot4\cdot5} \right\rfloor$ yields the numbers $133$ $100$ , and $33$ . The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$ . The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$ . Subtracting this number from the original $1001$ numbers procures $\boxed{801}$ | B | 801 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | First find the number of such integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of $3$
There are $\frac45\cdot2000=1600$ numbers that are not multiples of $5$ $\frac23\cdot\frac34\cdot1600=800$ are not multiples of $3$ or $4$ , so $800$ numbers are. $800+1=\boxed{801}$ | B | 801 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | Take a good-sized sample of consecutive integers; for example, the first $25$ positive integers. Determine that the numbers $3, 4, 6, 8, 9, 12, 16, 18, 21,$ and $24$ exhibit the properties given in the question. $25$ is a divisor of $2000$ , so there are $\frac{10}{25}\cdot2000=800$ numbers satisfying the given conditions between $1$ and $2000$ . Since $2001$ is a multiple of $3$ , add $1$ to $800$ to get $800+1=\boxed{801}$ | B | 801 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | By PIE, there are $1001$ numbers that are multiples of $3$ or $4$ and less than or equal to $2001$ $80\%$ of them will not be divisible by $5$ , and by far the closest number to $80\%$ of $1001$ is $\boxed{801}$ | B | 801 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | Similar to some of the above solutions.
We can divide $2001$ by $3$ and $4$ to find the number of integers divisible by $3$ and $4$ . Hence, we find that there are $667$ numbers less than $2001$ that are divisible by $3$ , and $500$ numbers that are divisible by $4$ . However, we will need to subtract the number of multiples of $15$ from 667 and that of $20$ from $500$ , since they're also divisible by 5 which we don't want. There are $133$ $100$ $233$ such numbers. Note that during this process, we've subtracted the multiples of $60$ twice because they're divisible by both $15$ and $20$ , so we have to add $33$ back to the tally (there are $33$ multiples of $60$ that does not exceed $2001$ ). Lastly, we have to subtract multiples of both $3$ AND $4$ since we only want multiples of either $3$ or $4$ . This is tantamount to subtracting the number of multiples of $12$ . And there are $166$ such numbers. Let's now collect our numbers and compute the total: $667$ $500$ $133$ $100$ $33$ $166$ $\boxed{801}$ | B | 801 |
21b33a597f802b0a4ad56201b9aeba1e | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | problem_id
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e How many positive integers not exceeding $2001...
Name: Text, dtype: object | Similar to @above:
Let the function $M_{2001}(n)$ return how many multiples of $n$ are there not exceeding $2001$ . Then we have that the desired number is: \[M_{2001}(3)+M_{2001}(4)-M_{2001}(3\cdot 4)-M_{2001}(3 \cdot 5) - M_{2001}(4 \cdot 5)+M_{2001}(3 \cdot 4 \cdot 5)\]
Evaluating each of these we get: \[667+500-166-133-100+33 = 1100-299 = 801.\]
Thus, the answer is $\boxed{801}.$ | B | 801 |
e256412f33446d81ec5cb1420a8ecd98 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_14 | Given the nine-sided regular polygon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$ , how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set $\{A_1,A_2,\dots,A_9\}$
$\text{(A) }30 \qquad \text{(B) }36 \qquad \text{(C) }63 \qquad \text{(D) }66 \qquad \text{(E) }72$ | Each of the $\binom{9}{2} = 36$ pairs of vertices determines two equilateral triangles, for a total of 72 triangles. However, the three triangles $A_1A_4A_7$ $A_2A_5A_8$ , and $A_3A_6A_9$ are each counted 3 times, resulting in an overcount of 6. Thus, there are $\boxed{66}$ distinct equilateral triangles. | null | 66 |
046784104ca77029d564b057b0cd05a0 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_15 | An insect lives on the surface of a regular tetrahedron with edges of length 1. It wishes to travel on the surface of the tetrahedron from the midpoint of one edge to the midpoint of the opposite edge. What is the length of the shortest such trip? (Note: Two edges of a tetrahedron are opposite if they have no common endpoint.)
$\text{(A) }\frac {1}{2} \sqrt {3} \qquad \text{(B) }1 \qquad \text{(C) }\sqrt {2} \qquad \text{(D) }\frac {3}{2} \qquad \text{(E) }2$ | Given any path on the surface, we can unfold the surface into a plane to get a path of the same length in the plane. Consider the net of a tetrahedron in the picture below. A pair of opposite points is marked by dots. It is obvious that in the plane the shortest path is just a segment that connects these two points. Its length is the same as the length of the tetrahedron's edge, i.e. $\boxed{1}$ | null | 1 |
b0081ef57e024b17892d43b8c2f3b9d6 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_18 | A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle?
[asy] unitsize(0.75cm); pair A=(0,1), B=(4,4); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( A -- (A+(-1,0)) ); label("$1$", A -- (A+(-1,0)), N ); draw( B -- (B+(4,0)) ); label("$4$", B -- (B+(4,0)), N ); label("$A$",A,E); label("$B$",B,W); filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black ); dot( (12/9,4/9) ); [/asy]
$\text{(A) }\frac {1}{3} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {5}{12} \qquad \text{(D) }\frac {4}{9} \qquad \text{(E) }\frac {1}{2}$ | [asy] unitsize(1cm); pair A=(0,1), B=(4,4), C=(4,1), S=(12/9,4/9); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( (A+(4,0)) -- A -- (A+(0,-1)) ); draw( A -- B -- (B+(0,-4)) ); label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$S$",S,N); filldraw( circle(S,4/9), lightgray, black ); dot(S); draw( rightanglemark(A,C,B) ); draw( S -- A ); draw( S -- B ); [/asy]
In the triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$ , thus by the Pythagorean theorem we have $AC=4$
Let $r$ be the radius of the small circle, and let $s$ be the perpendicular distance from $S$ to $\overline{AC}$ . Moreover, the small circle is tangent to both other circles, hence we have $SA=1+r$ and $SB=4+r$
We have $SA = \sqrt{s^2 + (1-r)^2}$ and $SB=\sqrt{(4-s)^2 + (4-r)^2}$ . Hence we get the following two equations:
\begin{align*} s^2 + (1-r)^2 & = (1+r)^2 \\ (4-s)^2 + (4-r)^2 & = (4+r)^2 \end{align*}
Simplifying both, we get
\begin{align*} s^2 & = 4r \\ (4-s)^2 & = 16r \end{align*}
As in our case both $r$ and $s$ are positive, we can divide the second one by the first one to get $\left( \frac{4-s}s \right)^2 = 4$
Now there are two possibilities: either $\frac{4-s}s=-2$ , or $\frac{4-s}s=2$
In the first case clearly $s<0$ , which puts the center on the wrong side of $A$ , so this is not the correct case.
(Note: This case corresponds to the other circle that is tangent to both given circles and the common tangent line. By coincidence, due to the $4:1$ ratio between radii of $A$ and $B$ , this circle turns out to have the same radius as circle $B$ , with center directly left of center $B$ , and tangent to $B$ directly above center $A$ .)
The second case solves to $s=\frac 43$ . We then have $4r = s^2 = \frac {16}9$ , hence $r = \boxed{49}$ | null | 49 |
b0081ef57e024b17892d43b8c2f3b9d6 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_18 | A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle?
[asy] unitsize(0.75cm); pair A=(0,1), B=(4,4); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( A -- (A+(-1,0)) ); label("$1$", A -- (A+(-1,0)), N ); draw( B -- (B+(4,0)) ); label("$4$", B -- (B+(4,0)), N ); label("$A$",A,E); label("$B$",B,W); filldraw( circle( (12/9,4/9), 4/9 ), lightgray, black ); dot( (12/9,4/9) ); [/asy]
$\text{(A) }\frac {1}{3} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {5}{12} \qquad \text{(D) }\frac {4}{9} \qquad \text{(E) }\frac {1}{2}$ | The horizontal line is the equivalent of a circle of curvature $0$ , thus we can apply Descartes' Circle Formula
The four circles have curvatures $0, 1, \frac 14$ , and $\frac 1r$
We have $2\left(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2}\right)=\left(0+1+\frac 14+\frac 1r\right)^2$
Simplifying, we get $\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}$
\[\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0\] \[\frac{16}{r^2}-\frac{40}{r}+9=0\] \[\left(\frac{4}{r}-9\right)\left(\frac{4}{r}-1\right)=0\]
Obviously $r$ cannot equal $4$ , therefore $r = \boxed{49}$ | null | 49 |
0cacc648619069ed75df2430e817d0b5 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_20 | Points $A = (3,9)$ $B = (1,1)$ $C = (5,3)$ , and $D=(a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ABCD$ . The quadrilateral formed by joining the midpoints of $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ , and $\overline{DA}$ is a square. What is the sum of the coordinates of point $D$
$\text{(A) }7 \qquad \text{(B) }9 \qquad \text{(C) }10 \qquad \text{(D) }12 \qquad \text{(E) }16$ | [asy] pair A=(3,9), B=(1,1), C=(5,3), D=(7,3); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,E); pair AB = (A + B)/2, BC = (B + C)/2, CD = (C + D)/2, DA = (D + A)/2; draw(AB--BC--CD--DA--cycle); [/asy]
We already know two vertices of the square: $(A+B)/2 = (2,5)$ and $(B+C)/2 = (3,2)$
There are only two possibilities for the other vertices of the square: either they are $(6,3)$ and $(5,6)$ , or they are $(0,1)$ and $(-1,4)$ . The second case would give us $D$ outside the first quadrant, hence the first case is the correct one. As $(6,3)$ is the midpoint of $CD$ , we can compute $D=(7,3)$ , and $7+3=\boxed{10}$ | null | 10 |
152e427e5124490adafab1a5789da027 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_21 | Four positive integers $a$ $b$ $c$ , and $d$ have a product of $8!$ and satisfy:
\[\begin{array}{rl} ab + a + b & = 524 \\ bc + b + c & = 146 \\ cd + c + d & = 104 \end{array}\]
What is $a-d$
$\text{(A) }4 \qquad \text{(B) }6 \qquad \text{(C) }8 \qquad \text{(D) }10 \qquad \text{(E) }12$ | Using Simon's Favorite Factoring Trick, we can rewrite the three equations as follows:
\begin{align*} (a+1)(b+1) & = 525 \\ (b+1)(c+1) & = 147 \\ (c+1)(d+1) & = 105 \end{align*}
Let $(e,f,g,h)=(a+1,b+1,c+1,d+1)$ . We get:
\begin{align*} ef & = 3\cdot 5\cdot 5\cdot 7 \\ fg & = 3\cdot 7\cdot 7 \\ gh & = 3\cdot 5\cdot 7 \end{align*}
Clearly $7^2$ divides $fg$ . On the other hand, $7^2$ can not divide $f$ , as it then would divide $ef$ . Similarly, $7^2$ can not divide $g$ . Hence $7$ divides both $f$ and $g$ . This leaves us with only two cases: $(f,g)=(7,21)$ and $(f,g)=(21,7)$
The first case solves to $(e,f,g,h)=(75,7,21,5)$ , which gives us $(a,b,c,d)=(74,6,20,4)$ , but then $abcd \not= 8!$ . We do not need to multiply, it is enough to note e.g. that the left hand side is not divisible by $7$ . (Also, a - d equals $70$ in this case, which is way too large to fit the answer choices.)
The second case solves to $(e,f,g,h)=(25,21,7,15)$ , which gives us a valid quadruple $(a,b,c,d)=(24,20,6,14)$ , and we have $a-d=24-14 =\boxed{10}$ | null | 10 |
5d12ff87ac01f7c268ce4e7d6a9b62a5 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$ | [asy] unitsize(0.5cm); defaultpen(0.8); pair A=(0,0), B=(10,0), C=(10,7), D=(0,7), E=(C+D)/2, F=(2*A+B)/3, G=(A+2*B)/3; pair H = intersectionpoint(A--C,E--F); pair J = intersectionpoint(A--C,E--G); draw(A--B--C--D--cycle); draw(G--E--F); draw(A--C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,N); label("$F$",F,S); label("$G$",G,S); label("$H$",H,SE); label("$J$",J,ESE); filldraw(E--H--J--cycle,lightgray,black); draw(H--D, dashed); [/asy] Note that the triangles $AFH$ and $CEH$ are similar, as they have the same angles. Hence $\frac {AH}{HC} = \frac{AF}{EC} = \frac 23$
Also, triangles $AGJ$ and $CEJ$ are similar, hence $\frac {AJ}{JC} = \frac {AG}{EC} = \frac 43$
We can now compute $[EHJ]$ as $[ACD]-[AHD]-[DEH]-[EJC]$ . We have:
Therefore $[EHJ]=[ACD]-[AHD]-[DEH]-[EJC]=35-14-\frac {21}2-\frac{15}2 = \boxed{3}$ | null | 3 |
5d12ff87ac01f7c268ce4e7d6a9b62a5 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$ | As in the previous solution, we note the similar triangles and prove that $H$ is in $2/5$ and $J$ in $4/7$ of $AC$
We can then compute that $HJ = AC \cdot \left( \frac 47 - \frac 25 \right) = AC \cdot \frac{6}{35}$
As $E$ is the midpoint of $CD$ , the height from $E$ onto $AC$ is $1/2$ of the height from $D$ onto $AC$ . Therefore we have $[EHJ] = \frac{6}{35} \cdot \frac 12 \cdot [ACD] = \frac 3{35} \cdot 35 = \boxed{3}$ | null | 3 |
5d12ff87ac01f7c268ce4e7d6a9b62a5 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$ | Because we see that there are only lines and there is a rectangle, we can coordbash (place this figure on coordinates). Because this is a general figure, we can assume the sides are $7$ and $10$ (or any other two positive real numbers that multiply to 70). We can find $H$ and $J$ by intersecting lines, and then we calculate the area of $EHJ$ using shoelace formula. This yields $\boxed{3}$ | null | 3 |
5d12ff87ac01f7c268ce4e7d6a9b62a5 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$ | Note that triangle $AFH$ is similar to triangle $CEH$ with ratio $\frac{2}{3}$ . Similarly, triangle $AGJ$ is similar to triangle $ECJ$ with ratio $\frac{4}{3}$ . Thus, if $AC = a$ then we know that $AH = \frac{2}{5}a$ and $JC = \frac{3}{7}a$ meaning $HJ = \frac{6}{35}a$ and thus the ratio of $HJ$ to $JC$ is $\frac{\frac{6}{35}}{\frac{3}{7}} = \frac{2}{5}$ which equals the ratio of the areas of $HJE$ to $JEC$ . If $y = AD, x = DC$ , then we know that $JEC = \text{(altitude from J to EC)} \cdot EC = \frac{3}{7}y \cdot \frac{1}{2}x \cdot \frac{1}{2}$ and since $xy = 70$ and we want to find $\frac{2}{5}$ of this, we get our answer is $\frac{2}{5} \cdot \frac{3}{7} \cdot \frac{1}{2} \cdot 70 \cdot \frac{1}{2} = \boxed{3}$ . -SuperJJ | null | 3 |
5d12ff87ac01f7c268ce4e7d6a9b62a5 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$ | $[CEF] = \frac{[ABCD]}{4} = \frac{35}{2}$
$\triangle CEH \sim \triangle AFH$ $\frac{HE}{HF} = \frac{CE}{AF} = \frac{3}{2}$ $\frac{HE}{EF} = \frac{3}{5}$
$[CEH] = \frac{HE}{EF} \cdot [CEF] = \frac{3}{5} \cdot \frac{35}{2} = \frac{21}{2}$
$\triangle CEH \sim \triangle AFH$ $\frac{AH}{HC} = \frac{AF}{CE} = \frac{2}{3}$ $\frac{AH}{AC} = \frac{2}{5}$ $\frac{CH}{AC} = \frac{3}{5}$
$\triangle CEJ \sim \triangle AGJ$ $\frac{AJ}{JC} = \frac{AG}{CE} = \frac{4}{3}$ $\frac{AJ}{AC} = \frac{4}{7}$
$\frac{HJ}{AC} = \frac{AJ}{AC} - \frac{AH}{AC} = \frac{4}{7} - \frac{2}{5} = \frac{6}{35}$
$\frac{HJ}{CH} = \frac{HJ}{AC} \cdot \frac{AC}{CH} = \frac{6}{35} \cdot \frac{5}{3} = \frac{2}{7}$
$[EHJ] = \frac{HJ}{CH} \cdot [CEH] = \frac{2}{7} \cdot \frac{21}{2} = \boxed{3}$ | C | 3 |
5d12ff87ac01f7c268ce4e7d6a9b62a5 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }\frac {35}{12} \qquad \text{(C) }3 \qquad \text{(D) }\frac {7}{2} \qquad \text{(E) }\frac {35}{8}$ | We need one more pair of ratios to fully define our mass point system. Let's use $\triangle AFH \sim \triangle CEH\implies EH:HF = 3:2$ and now do mass points on $\triangle AEG$
[asy] size(250); pair A = (0,0), B = (10,0), C = (10,7), D = (0,7); pair F = (10/3,0), G = (20/3,0), E = (5,7); pair H = intersectionpoint(A--C, E--F); pair J = intersectionpoint(A--C, E--G); filldraw(A--E--G--cycle, rgb(1,1,1)+opacity(0.3), red+2bp); draw(A--B--C--D--cycle); draw(A--C); draw(E--F); draw(E--G); draw(A--E, dashed); draw(E--B, dashed); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, N); dot("$F$", F, S); dot("$G$", G, S); dot("$H$", H, ESE); dot("$J$", J, W); // mass point labels pair mass = A + SW; label(scale(0.8)*"$3$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = F + S; label(scale(0.8)*"$6$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = G + SE; label(scale(0.8)*"$3$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = J + .7*ESE; label(scale(0.8)*"$7$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = E + N; label(scale(0.8)*"$4$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = H + .7*WNW; label(scale(0.8)*"$10$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); [/asy]
Now it's just a standard "Area Reduction by Ratios"™ problem going from:
\[[ABCD]\xrightarrow[]{\frac{1}{2}}[AEB]\xrightarrow[]{\frac{2}{3}}[AEG]\xrightarrow[]{\frac{3}{7}}[AEJ]\xrightarrow[]{\frac{3}{10}}[HEJ]\]
or,
\[70 \cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{7}\cdot \frac{3}{10} = \boxed{3}\] | C | 3 |
602643cde4ac92dce65c137bfaa97d06 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | In $\triangle ABC$ $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB.$
$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$ | Draw a good diagram! Now, let's call $BD=t$ , so $DC=2t$ . Given the rather nice angles of $\angle ABD = 45^\circ$ and $\angle ADC = 60^\circ$ as you can see, let's do trig. Drop an altitude from $A$ to $BC$ ; call this point $H$ . We realize that there is no specific factor of $t$ we can call this just yet, so let $AH=kt$ . Notice that in $\triangle{ABH}$ we get $BH=kt$ . Using the 60-degree angle in $\triangle{ADH}$ , we obtain $DH=\frac{\sqrt{3}}{3}kt$ . The comparable ratio is that $BH-DH=t$ . If we involve our $k$ , we get:
$kt(\frac{3}{3}-\frac{\sqrt{3}}{3})=t$ . Eliminating $t$ and removing radicals from the denominator, we get $k=\frac{3+\sqrt{3}}{2}$ . From there, one can easily obtain $HC=3t-kt=\frac{3-\sqrt{3}}{2}t$ . Now we finally have a desired ratio. Since $\tan\angle ACH = 2+\sqrt{3}$ upon calculation, we know that $\angle ACH$ can be simplified. Indeed, if you know that $\tan(75)=2+\sqrt{3}$ or even take a minute or two to work out the sine and cosine using $\sin(x)^2+\cos(x)^2=1$ , and perhaps the half- or double-angle formulas, you get $\boxed{75}$ | null | 75 |
602643cde4ac92dce65c137bfaa97d06 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | In $\triangle ABC$ $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB.$
$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$ | Without loss of generality, we can assume that $BD = 1$ and $CD = 2$ . As above, we are able to find that $\angle ADC = 60^\circ$ and $\angle ADB = 120^\circ$
Using Law of Sines on triangle $ADB$ , we find that \[\frac{1}{\sin15^\circ} = \frac{AD}{\sin 45^\circ} = \frac{AB}{\sin 120^\circ}.\] Since we know that \[\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4},\] \[\sin 45^\circ = \frac{\sqrt{2}}{2},\] \[\sin 120^\circ = \frac{\sqrt{3}}{2},\] we can compute $AD$ to equal $1+\sqrt{3}$ and $AB$ to be $\frac{3\sqrt{2}+\sqrt{6}}{2}$
Next, we apply Law of Cosines to triangle $ADC$ to see that \[AC^2 = (1+\sqrt{3})^2 + 2^2 - (2)(1+\sqrt{3})(2)(\cos 60^\circ).\] Simplifying the right side, we get $AC^2 = 6$ , so $AC = \sqrt{6}$
Now, we apply Law of Sines to triangle $ABC$ to see that \[\frac{\sqrt{6}}{\sin 45^\circ} = \frac{\frac{3\sqrt{2}+\sqrt{6}}{2}}{\sin ACB}.\] After rearranging and noting that $\sin 45^\circ = \frac{\sqrt{2}}{2}$ , we get \[\sin ACB = \frac{\sqrt{6}+3\sqrt{2}}{4\sqrt{3}}.\]
Dividing the right side by $\sqrt{3}$ , we see that \[\sin ACB = \frac{\sqrt{6}+\sqrt{2}}{4},\] so $\angle ACB$ is either $75^\circ$ or $105^\circ$ . Since $105^\circ$ is not a choice, we know $\angle ACB = \boxed{75}$ | null | 75 |
602643cde4ac92dce65c137bfaa97d06 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | In $\triangle ABC$ $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB.$
$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$ | $\angle ADB = 120^\circ$ $\angle ADC = 60^\circ$ $\angle DAB = 15^\circ$ , let $\angle ACB = \theta$ $\angle DAC = 120^\circ - \theta$
By the Law of Sines , we have $\frac{CD}{\sin(120^\circ - \theta)} = \frac{AD}{\sin \theta}$
$\space$ $\frac{BD}{\sin15^\circ} = \frac{AD}{\sin45^\circ}$
$\frac{BD}{CD} \cdot \frac{\sin(120^\circ - \theta)}{\sin15^\circ} = \frac{\sin \theta}{\sin45^\circ}$
$\frac{1}{2} \cdot \frac{\sin(120^\circ - \theta)}{\sin15^\circ} = \frac{\sin \theta}{\sin45^\circ}$
By the Triple-angle Identities $\sin 45^\circ = 3\sin15^\circ -4\sin^3 15^\circ$
$\frac{\sin(120^\circ - \theta)}{2} = \frac{\sin \theta}{3\ -4\sin^2 15^\circ}$
$\sin^2 15^\circ = \frac{1 - \cos 30^\circ}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}$
$\frac{\sin(120^\circ - \theta)}{2} = \frac{\sin \theta}{3 - 4 \cdot \frac{2 - \sqrt{3}}{4}} = \frac{\sin \theta}{1+\sqrt{3}}$
$\frac{\sin \theta}{\sin(120^\circ - \theta)} = \frac{1+\sqrt{3}}{2}$
$\sin(120^\circ - \theta) = \sin120^\circ \cos\theta - \sin\theta \cos120^\circ = \frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{1}{2} \cdot \sin\theta$ ，so
$\frac{\sin \theta}{\frac{\sqrt{3}}{2} \cdot \cos\theta + \frac{1}{2} \cdot \sin\theta} = \frac{1+\sqrt{3}}{2}$
$\frac{\sqrt{3} \cdot \cos\theta + \sin\theta}{2 \sin \theta} = \frac{2}{1+\sqrt{3}}$
$\frac{\sqrt{3}}{2} \cdot \cot \theta = \frac{2}{1+\sqrt{3}} - \frac{1}{2} = \frac{3 - \sqrt{3}}{2 + 2 \sqrt{3}}$
$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
Suppose $\cos \theta = k(\sqrt{3} - 1)$ , and $\sin \theta = k(\sqrt{3} + 1)$
$\sin^2 \theta + \cos^2 \theta = 1$ $k^2(\sqrt{3} + 1)^2 + k^2(\sqrt{3} - 1)^2 = 8k^2 = 1$ $k = \frac{1}{2 \sqrt{2}}$
$\sin \theta = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$ $\cos \theta = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$
$\sin 2 \theta = 2 \cdot \frac{\sqrt{3} + 1}{2 \sqrt{2}} \cdot \frac{\sqrt{3} - 1}{2 \sqrt{2}} = \frac{1}{2}$
Two possible values of $2 \theta$ are $150^\circ$ and $30^\circ$ . However we can rule out $30^\circ$ because $\cos 15^\circ$ is positive, while $\cos \theta$ is negative.
Therefore $2 \theta = 150^\circ$ $\angle ACB = \boxed{75}$ | D | 75 |
602643cde4ac92dce65c137bfaa97d06 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | In $\triangle ABC$ $\angle ABC=45^\circ$ . Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$ . Find $\angle ACB.$
$\text{(A) }54^\circ \qquad \text{(B) }60^\circ \qquad \text{(C) }72^\circ \qquad \text{(D) }75^\circ \qquad \text{(E) }90^\circ$ | For starters, we have $\angle ABD=120^\circ.$ Dropping perpendiculars $\overline{DX}$ and $\overline{CY}$ from $D$ and $C$ to $\overline{AB}$ gives $\angle ADX=120^\circ-45^\circ=75^\circ,$ since $\angle BDX=45^\circ.$
Without loss of generality, let $BD=1$ and $CD=2.$ This tells us that $BX=DX=\sqrt{2}/2.$ Using trigonometric identities, we find that
\[\tan \angle ADX=\tan 75^\circ=\dfrac{\sqrt 6+\sqrt 2}{\sqrt 6-\sqrt 2}.\] Thus, $AX/DX=\dfrac{\sqrt 6+\sqrt 2}{\sqrt 6-\sqrt 2},$ which gives $AX=\dfrac{2\sqrt 2+\sqrt 6}{2}.$ Thus, $AB=AX+BX=\dfrac{3\sqrt 2+\sqrt 6}{2}.$
Now, note that $BCY$ is a $45-45-90$ triangle, so $CY=BY=\dfrac{3\sqrt 2}{2}.$ Thus, we have
\[[ABC]=\dfrac{1}{2}\cdot AB\cdot CY=\dfrac{1}{2}\cdot \dfrac{3\sqrt 2+\sqrt 6}{2}\cdot \dfrac{3\sqrt 2}{2}=\dfrac{9+3\sqrt 3}{4}.\] Additionally, note that $AY=AB-BY=\sqrt{6}/2.$ Applying the Pythagorean Theorem to triangle $AYC$ then tells us that $AC=\sqrt{6}.$ By the trigonometric formula for area,
\[[ABC]=\dfrac{1}{2}\cdot BC\cdot AC\cdot \sin \angle ACB=\dfrac{3\sqrt 6}{2}\sin \angle ACB.\] Setting this equal to our other area and solving gives $\sin \angle ACB=\dfrac{\sqrt 6+\sqrt 2}{4},$ so $\angle ACB=\boxed{75}.$ ~vaporwave | null | 75 |
e30be0f17c74bf3ec3bf34880ba054ad | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_25 | Consider sequences of positive real numbers of the form $x, 2000, y, \dots$ in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of $x$ does the term $2001$ appear somewhere in the sequence?
$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }3 \qquad \text{(D) }4 \qquad \text{(E) more than }4$ | It never hurts to compute a few terms of the sequence in order to get a feel how it looks like. In our case, the definition is that $\forall$ (for all) $n>1:~ a_n = a_{n-1}a_{n+1} - 1$ . This can be rewritten as $a_{n+1} = \frac{a_n +1}{a_{n-1}}$ . We have $a_1=x$ and $a_2=2000$ , and we compute:
\begin{align*} a_3 & = \frac{a_2+1}{a_1} = \frac{2001}x \\ a_4 & = \frac{a_3+1}{a_2} = \frac{ \dfrac{2001}x + 1 }{ 2000 } = \frac{2001 + x}{2000x} \\ a_5 & = \frac{a_4+1}{a_3} = \frac{ \frac{2001 + x}{2000x} + 1 }{ \frac{2001}x } = \frac{ \frac{2001 + 2001x}{2000x} }{ \frac{2001}x } = \frac{1+x}{2000} \\ a_6 & = \frac{a_5+1}{a_4} = \frac{ \frac{1+x}{2000} + 1 }{ \frac{2001 + x}{2000x} } = \frac{ \frac{2001+x}{2000} }{ \frac{2001 + x}{2000x} } = x \\ a_7 & = \frac{a_6+1}{a_5} = \frac{ x+1 }{ \frac{1+x}{2000} } = 2000 \end{align*}
At this point we see that the sequence will become periodic: we have $a_6=a_1$ $a_7=a_2$ , and each subsequent term is uniquely determined by the previous two.
Hence if $2001$ appears, it has to be one of $a_1$ to $a_5$ . As $a_2=2000$ , we only have four possibilities left. Clearly $a_1=2001$ for $x=2001$ , and $a_3=2001$ for $x=1$ . The equation $a_4=2001$ solves to $x = \frac{2001}{2000\cdot 2001 - 1}$ , and the equation $a_5=2001$ to $x=2000\cdot 2001 - 1$
No two values of $x$ we just computed are equal, and therefore there are $\boxed{4}$ different values of $x$ for which the sequence contains the value $2001$ | null | 4 |
217fbd4845427e43723d361afc3b0bfc | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_1 | In the year $2001$ , the United States will host the International Mathematical Olympiad . Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$ | First, we need to recognize that a number is going to be lowest only if, of the $3$ factors , two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like $30$ . It becomes much more clear that this is true, and in this situation, the value of $I + M + O$ would be $18$ . Now, we use this process on $2001$ to get $667 * 3 * 1$ as our $3$ factors.
Hence, we have $667 + 3 + 1 = \boxed{671}$ | E | 671 |
217fbd4845427e43723d361afc3b0bfc | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_1 | In the year $2001$ , the United States will host the International Mathematical Olympiad . Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$ | The sum is the highest if two factors are the lowest.
So, $1 \cdot 3 \cdot 667 = 2001$ and $1+3+667=671 \Longrightarrow \boxed{671}$ | E | 671 |
217fbd4845427e43723d361afc3b0bfc | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_1 | In the year $2001$ , the United States will host the International Mathematical Olympiad . Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ . What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$ | We see since $2 + 0 + 0 + 1$ is divisible by $3$ , we can eliminate all of the first $4$ answer choices because they are way too small and get $\boxed{671}$ as our final answer. | E | 671 |