id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
|---|---|---|---|---|---|
5,101 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_12 | 4 | 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... | 4 |
5,102 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_14 | 1 | 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 $\boxed{12}$ | 12 |
5,103 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_14 | 2 | 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 $2+4+6=\boxed{12}.$ | 12 |
5,104 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_14 | 3 | 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 $\frac{24}{2}=\boxed{12}$ | 12 |
5,105 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_15 | 1 | 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 + 9... | 7 |
5,106 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_19 | 1 | 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 ... | 720 |
5,107 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_20 | 1 | 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 \... | 26 |
5,108 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_20 | 2 | 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.... | 26 |
5,109 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_21 | 1 | 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{oth... | 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 $\... | 448 |
5,110 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_22 | 1 | 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)}\ \fr... | 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)=\b... | 1 |
5,111 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_22 | 2 | 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)}\ \fr... | 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 = \l... | 1 |
5,112 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_23 | 2 | 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 $\ove... | 2 |
5,113 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12B_Problems/Problem_25 | 1 | 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\... | 25 |
5,114 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_2 | 1 | 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 \te... | 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,... | 2 |
5,115 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_3 | 1 | 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... | 38 |
5,116 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_7 | 1 | 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 ... | 52 |
5,117 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_9 | 1 | 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{(... | 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 $9=\sqrt{1^2 + 8^2 + c^2} \implies c^2=9^2-8^2-1^2 \implies c^2=16 \implies c=\boxed{4}.$ | 4 |
5,118 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_10 | 1 | 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 manipu... | 8 |
5,119 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_12 | 1 | 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 $-13$ . Evalu... | 3 |
5,120 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_13 | 1 | 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... | 17 |
5,121 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_16 | 1 | 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 ... | 90 |
5,122 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_17 | 1 | 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}... | 2 |
5,123 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_17 | 2 | 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{... | 2 |
5,124 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_18 | 1 | 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... | 14 |
5,125 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_20 | 1 | 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, $f(2) = \frac{1003+2997}{2} = \boxed{2000}$ | 0 |
5,126 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_21 | 1 | 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} \... | 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} + \fr... | 2 |
5,127 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_12P_Problems/Problem_23 | 1 | 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 $\boxed{118}$ | 118 |
5,128 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_2 | 1 | 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 ... | 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 $\boxed{9}$ | 9 |
5,129 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_3 | 2 | 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 ... | 0 |
5,130 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_4 | 3 | 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 $3\cdot 10 = \boxed{30}$ | 30 |
5,131 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_4 | 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 - ... | 30 |
5,132 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_6 | 2 | 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 $1$ respe... | 8 |
5,133 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_7 | 4 | 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 ticket... | 782 |
5,134 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_7 | 5 | 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 $\implies 2Ax+(140-A)(x)=4002$
Sim... | 782 |
5,135 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_7 | 6 | 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 ( $f \cdot p$ ) pl... | 782 |
5,136 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_9 | 1 | 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 $f(600) = \boxed{52}$ | 52 |
5,137 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_9 | 2 | 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 $\frac{500 \cdot 3}{600} = \boxed{52}$ | 52 |
5,138 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_9 | 3 | 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 $f(y)=\boxed{52}$ | 52 |
5,139 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_10 | 2 | 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; dr... | 56 |
5,140 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_11 | 3 | 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 chip... | 35 |
5,141 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_11 | 4 | 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... | 35 |
5,142 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | 7 | 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 ... | 801 |
5,143 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | 8 | 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. $800+1=\boxed{801}$ | 801 |
5,144 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | 9 | 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 conditi... | 801 |
5,145 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | 10 | 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 $\boxed{801}$ | 801 |
5,146 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | 11 | 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 mu... | 801 |
5,147 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_12 | 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... | 801 |
5,148 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_14 | 1 | 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. | 66 |
5,149 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_15 | 1 | 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 en... | 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. It... | 1 |
5,150 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_18 | 1 | 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); dra... | [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), li... | 49 |
5,151 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_18 | 2 | 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); dra... | 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... | 49 |
5,152 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_20 | 1 | 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 $D$
$\te... | [asy] pair A=(3,9), B=(1,1), C=(5,3), D=(7,3); draw(A--B--C--D--cycle); label("$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+... | 10 |
5,153 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_21 | 1 | 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... | 10 |
5,154 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | 1 | 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 $EHJ$
$\text{(A) }\frac {5}{2} \qquad \text{(B) }... | [asy] unitsize(0.5cm); defaultpen(0.8); pair A=(0,0), B=(10,0), C=(10,7), D=(0,7), E=(C+D)/2, F=(2*A+B)/3, G=(A+2*B)/3; pair H = intersectionpoint(A--C,E--F); pair J = intersectionpoint(A--C,E--G); draw(A--B--C--D--cycle); draw(G--E--F); draw(A--C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D... | 3 |
5,155 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | 2 | 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) }... | 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 $AC$ ... | 3 |
5,156 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | 3 | 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) }... | 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 calc... | 3 |
5,157 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | 4 | 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) }... | 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{\fr... | 3 |
5,158 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | 5 | 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) }... | $[CEF] = \frac{[ABCD]}{4} = \frac{35}{2}$
$\triangle CEH \sim \triangle AFH$ $\frac{HE}{HF} = \frac{CE}{AF} = \frac{3}{2}$ $\frac{HE}{EF} = \frac{3}{5}$
$[CEH] = \frac{HE}{EF} \cdot [CEF] = \frac{3}{5} \cdot \frac{35}{2} = \frac{21}{2}$
$\triangle CEH \sim \triangle AFH$ $\frac{AH}{HC} = \frac{AF}{CE} = \frac{2}{3}$ $\... | 3 |
5,159 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_22 | 6 | 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) }... | 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-... | 3 |
5,160 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | 1 | 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... | 75 |
5,161 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | 2 | 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 ... | 75 |
5,162 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | 3 | 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... | 75 |
5,163 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_12_Problems/Problem_24 | 4 | 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 trigo... | 75 |
5,164 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_1 | 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 \qq... | 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 $18$ ... | 671 |
5,165 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_1 | 2 | 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 \qq... | The sum is the highest if two factors are the lowest.
So, $1 \cdot 3 \cdot 667 = 2001$ and $1+3+667=671 \Longrightarrow \boxed{671}$ | 671 |
5,166 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_1 | 3 | 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 \qq... | 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. | 671 |
5,167 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_3 | 1 | Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?
$\textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50 \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75$ | We can begin by labeling the number of initial jellybeans $x$ . If she ate $20\%$ of the jellybeans, then $80\%$ is remaining. Hence, after day 1, there are: $0.8 * x$
After day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$ , so $x = \boxed{50}$ | 50 |
5,168 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_3 | 2 | Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?
$\textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50 \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75$ | Testing the answers choices out, we see that the answer is $\boxed{50}$ | 50 |
5,169 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_4 | 1 | The Fibonacci sequence $1,1,2,3,5,8,13,21,\ldots$ starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?
$\textbf{(A)} \ 0 \qquad \textbf{(B)} \ 4 \qquad \textbf{(C)} \ 6 \qquad \t... | Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $\bmod{10}$
$1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$
The last digit to appear in the units position of a number in the Fibonacci sequence is $6 \Longrightarr... | 6 |
5,170 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_6 | 1 | Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$ | Any two prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate A, B, and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$ . Thus, we can eliminate... | 119 |
5,171 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_6 | 2 | Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$ | Let the two primes be $p$ and $q$ . We wish to obtain the value of $pq-(p+q)$ , or $pq-p-q$ . Using Simon's Favorite Factoring Trick , we can rewrite this expression as $(1-p)(1-q) -1$ or $(p-1)(q-1) -1$ . Noticing that $(13-1)(11-1) - 1 = 120-1 = 119$ , we see that the answer is $\boxed{119}$ | 119 |
5,172 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_7 | 1 | How many positive integers $b$ have the property that $\log_{b} 729$ is a positive integer?
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 }$ | If $\log_{b} 729 = n$ , then $b^n = 729$ . Since $729 = 3^6$ $b$ must be $3$ to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of $b \Longrightarrow \boxed{4}$ | 4 |
5,173 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_8 | 1 | Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)-... | Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$ . Thus, the original sequence can be generated from a quadratic function.
If $f(n) = an^2 + bn + c$ , and $f(0) = 1$ $... | 201 |
5,174 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_8 | 2 | Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)-... | We can see that each figure $n$ has a central box and 4 columns of $n$ boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are $\sum_{... | 201 |
5,175 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_8 | 4 | Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)-... | Let $f_n$ denote the number of unit cubes in a figure. We have \[f_0=1\] \[f_1=5\] \[f_2=13\] \[f_3=25\] \[f_4=41\] \[...\]
Computing the difference between the number of cubes in each figure yields \[4,8,12,16,...\] It is easy to notice that this is an arithmetic sequence, with the first term being $4$ and the differe... | 201 |
5,176 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_12 | 1 | Let [mathjax]A, M,[/mathjax] and [mathjax]C[/mathjax] be nonnegative integers such that [mathjax]A + M + C=12[/mathjax]. What is the maximum value of [mathjax]A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C[/mathjax]?
[katex] \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D... | It is not hard to see that \[(A+1)(M+1)(C+1)=\] \[AMC+AM+AC+MC+A+M+C+1\] Since $A+M+C=12$ , we can rewrite this as \[(A+1)(M+1)(C+1)=\] \[AMC+AM+AC+MC+13\] So we wish to maximize \[(A+1)(M+1)(C+1)-13\] Which is largest when all the factors are equal (consequence of AM-GM). Since $A+M+C=12$ , we set $A=M=C=4$ Which giv... | 112 |
5,177 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_12 | 2 | Let [mathjax]A, M,[/mathjax] and [mathjax]C[/mathjax] be nonnegative integers such that [mathjax]A + M + C=12[/mathjax]. What is the maximum value of [mathjax]A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C[/mathjax]?
[katex] \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D... | Assume $A$ $M$ , and $C$ are equal to $4$ . Since the resulting value of $AMC+AM+AC+MC$ will be $112$ and this is the largest answer choice, our answer is $\boxed{112}$ | 112 |
5,178 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_12 | 3 | Let [mathjax]A, M,[/mathjax] and [mathjax]C[/mathjax] be nonnegative integers such that [mathjax]A + M + C=12[/mathjax]. What is the maximum value of [mathjax]A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C[/mathjax]?
[katex] \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D... | We start off the same way as Solution 4, using AM-GM to observe that $AMC \leq 64$ . We then observe that
$(A + M + C)^2 = A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144$ , since $A + M + C = 12$
We can use the AM-GM inequality again, this time observing that $\frac{A^2 + M^2 + C^2}{3} \geq \sqrt[3]{{(AMC)}^2}$
Since $AMC \le... | 112 |
5,179 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_12 | 4 | Let [mathjax]A, M,[/mathjax] and [mathjax]C[/mathjax] be nonnegative integers such that [mathjax]A + M + C=12[/mathjax]. What is the maximum value of [mathjax]A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C[/mathjax]?
[katex] \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D... | The largest number for our value would be $A = M = C.$ So $3A = 12$ and $A = M = C = 4.$ $4\times4\times4 + 4\times4 + 4\times4 = 112$ or $\boxed{112}$ | 112 |
5,180 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_13 | 1 | One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$\text {(A)}\ 3 \qquad \te... | If there were 4 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be too much coffee. If there were 6 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be not enough milk. Thus, it has to be $\boxed... | 5 |
5,181 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_15 | 1 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | Let $y = \frac{x}{3}$ ; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$ . Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$ , and $z = -\frac{1}{3}, \frac{2}{9}$ . These sum up to $\boxed{19}$ | 19 |
5,182 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_15 | 2 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | This is quite trivially solved, as $3x = \dfrac{9x}{3}$ , so $P(3x) = P(9x/3) = 81x^2 + 9x + 1 = 7$ $81x^2+9x-6 = 0$ has solutions $-\frac{1}{3}$ and $\frac{2}{9}$ . Adding these yields a solution of $\boxed{19}$ | 19 |
5,183 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_15 | 3 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | Similar to Solution 1, we have $=81z^2+9z-6=0.$ The answer is the sum of the roots, which by Vieta's Formulas is $-\frac{b}{a}=-\frac{9}{81}=\boxed{19}$ | 19 |
5,184 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_15 | 4 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | Set $f\left(\frac{x}{3} \right) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots (-3 and 2), or using Vieta's formulas, we find the sum of these roots to be $-1.$ Each root of this equation is $9$ times greater than a corresponding root of $f(3z) = 7$ (because $\frac{x}{3} = 3z$ gives $x = 9z$ ), thus the... | 19 |
5,185 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_15 | 5 | Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\] | Since we have $f(x/3)$ $f(3z)$ occurs at $x=9z.$ Thus, $f(9z/3) = f(3z) = (9z)^2 + 9z + 1$ . We set this equal to 7:
$81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0$ . For any quadratic $ax^2 + bx +c = 0$ , the sum of the roots is $-\frac{b}{a}$ . Thus, the sum of the roots of this equation is $-\frac{9}{81} = \b... | 19 |
5,186 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_16 | 1 | A checkerboard of $13$ rows and $17$ columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered $1,2,\ldots,17$ , the second row $18,19,\ldots,34$ , and so on down the board. If the board is renumbered so that the left column, top to bottom, is $1,2,\ldots,13,$ , ... | Index the rows with $i = 1, 2, 3, ..., 13$ Index the columns with $j = 1, 2, 3, ..., 17$
For the first row number the cells $1, 2, 3, ..., 17$ For the second, $18, 19, ..., 34$ and so on
So the number in row = $i$ and column = $j$ is $f(i, j) = 17(i-1) + j = 17i + j - 17$
Similarly, numbering the same cells columnwise ... | 555 |
5,187 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_19 | 1 | In triangle $ABC$ $AB = 13$ $BC = 14$ $AC = 15$ . Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC$ . Which of the following is closest to the area of the triangle $ADE$
$\text {(A)}\ 2 \qquad \text {(B)}\ 2.5 \qquad \text {(C)}\ 3 \qq... | [asy] pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$E$",E,NW); label("$D$",D,NE); label("$13$",A--B,NW); label("$15$",A--C,NE); label("$14$",B--C,S);... | 21 |
5,188 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_24 | 1 | If circular arcs $AC$ and $BC$ have centers at $B$ and $A$ , respectively, then there exists a circle tangent to both $\overarc {AC}$ and $\overarc{BC}$ , and to $\overline{AB}$ . If the length of $\overarc{BC}$ is $12$ , then the circumference of the circle is
[asy] label("A", (0,0), W); label("B", (64,0), E); label("... | First, note the triangle $ABC$ is equilateral. Next, notice that since the arc $BC$ has length 12, it follows that we can find the radius of the sector centered at $A$ $\frac {1}{6}({2}{\pi})AB=12 \implies AB=36/{\pi}$ . Next, connect the center of the circle to side $AB$ , and call this length $r$ , and call the foot ... | 27 |
5,189 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_25 | 1 | Eight congruent equilateral triangles , each of a different color, are used to construct a regular octahedron . How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)
$\textbf {(A)}\ 210 \qquad \textbf {(B... | This problem can be approached by Graph Theory . Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge.
Now the problem becomes how many distinguishable ways to colo... | 680 |
5,190 | https://artofproblemsolving.com/wiki/index.php/2000_AMC_12_Problems/Problem_25 | 2 | Eight congruent equilateral triangles , each of a different color, are used to construct a regular octahedron . How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)
$\textbf {(A)}\ 210 \qquad \textbf {(B... | Let the colors be $1$ to $8$ inclusive, then rotate the octahedron such that color $1$ is on top. You have $7$ choices of what color is on the bottom, WLOG $2$ . Then, there's two rings of each $3$ colors on the top and bottom. For the top ring, you can choose any $3$ out of the $6$ remaining colors, and there's two wa... | 680 |
5,191 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_3 | 1 | Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for whi... | Let's first try some experimentation. Alice obviously wins if there is one coin. She will just take it and win. If there are 2 remaining, then Alice will take one and then Bob will take one, so Bob wins. If there are $3$ , Alice will take $1$ , Bob will take one, and Alice will take the final one. If there are $4$ , Al... | 809 |
5,192 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_3 | 2 | Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for whi... | We will use winning and losing positions, where a $W$ marks when Alice wins and an $L$ marks when Bob wins.
$1$ coin: $W$
$2$ coins: $L$
$3$ coins: $W$
$4$ coins: $W$
$5$ coins: $L$
$6$ coin: $W$
$7$ coins: $L$
$8$ coins: $W$
$9$ coins: $W$
$10$ coins: $L$
$11$ coin: $W$
$12$ coins: $L$
$13$ coins: $W$
$14$ coins: $W$
... | 809 |
5,193 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_3 | 3 | Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for whi... | Denote by $A_i$ and $B_i$ Alice's or Bob's $i$ th moves, respectively.
Case 1: $n \equiv 0 \pmod{5}$
Bob can always take the strategy that $B_i = 5 - A_i$ .
This guarantees him to win.
In this case, the number of $n$ is $\left\lfloor \frac{2024}{5} \right\rfloor = 404$
Case 2: $n \equiv 1 \pmod{5}$
In this case, consid... | 809 |
5,194 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_3 | 4 | Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for whi... | Since the game Alice and Bob play is impartial (the only difference between player 1 and player 2 is that player 1 goes first (note that games like chess are not impartial because each player can only move their own pieces)), we can use the Sprague-Grundy Theorem to solve this problem. We will use induction to calculat... | 809 |
5,195 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_4 | 1 | Jen enters a lottery by picking $4$ distinct numbers from $S=\{1,2,3,\cdots,9,10\}.$ $4$ numbers are randomly chosen from $S.$ She wins a prize if at least two of her numbers were $2$ of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of... | This is a conditional probability problem. Bayes' Theorem states that \[P(A|B)=\dfrac{P(B|A)\cdot P(A)}{P(B)}\]
Let us calculate the probability of winning a prize. We do this through casework: how many of Jen's drawn numbers match the lottery's drawn numbers?
To win a prize, Jen must draw at least $2$ numbers identica... | 116 |
5,196 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_4 | 2 | Jen enters a lottery by picking $4$ distinct numbers from $S=\{1,2,3,\cdots,9,10\}.$ $4$ numbers are randomly chosen from $S.$ She wins a prize if at least two of her numbers were $2$ of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of... | For getting all $4$ right, there is only $1$ way.
For getting $3$ right, there is $\dbinom43$ multiplied by $\dbinom61$ $24$ ways.
For getting $2$ right, there is $\dbinom42$ multiplied by $\dbinom62$ $90$ ways.
$\frac{1}{1+24+90}$ $\frac{1}{115}$
Therefore, the answer is $1+115 = \boxed{116}$ | 116 |
5,197 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_6 | 1 | Consider the paths of length $16$ that follow the lines from the lower left corner to the upper right corner on an $8\times 8$ grid. Find the number of such paths that change direction exactly four times, as in the examples shown below.
[asy] size(10cm); usepackage("tikz");label("\begin{tikzpicture}[scale=.5]\draw(0,0)... | We divide the path into eight “ $R$ ” movements and eight “ $U$ ” movements. Five sections of alternative $RURUR$ or $URURU$ are necessary in order to make four “turns.” We use the first case and multiply by $2$
For $U$ , we have seven ordered pairs of positive integers $(a,b)$ such that $a+b=8$
For $R$ , we subtract $... | 294 |
5,198 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_6 | 2 | Consider the paths of length $16$ that follow the lines from the lower left corner to the upper right corner on an $8\times 8$ grid. Find the number of such paths that change direction exactly four times, as in the examples shown below.
[asy] size(10cm); usepackage("tikz");label("\begin{tikzpicture}[scale=.5]\draw(0,0)... | Notice that the $RURUR$ case and the $URURU$ case is symmetrical. WLOG, let's consider the RURUR case.
Now notice that there is a one-to-one correspondence between this problem and the number of ways to distribute 8 balls into 3 boxes and also 8 other balls into 2 other boxes, such that each box has a nonzero amount of... | 294 |
5,199 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_6 | 3 | Consider the paths of length $16$ that follow the lines from the lower left corner to the upper right corner on an $8\times 8$ grid. Find the number of such paths that change direction exactly four times, as in the examples shown below.
[asy] size(10cm); usepackage("tikz");label("\begin{tikzpicture}[scale=.5]\draw(0,0)... | Starting at the origin, you can either first go up or to the right. If you go up first, you will end on the side opposite to it (the right side) and if you go right first, you will end up on the top. It can then be observed that if you choose the turning points in the middle $7 \times 7$ grid, that will automatically d... | 294 |
5,200 | https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_6 | 4 | Consider the paths of length $16$ that follow the lines from the lower left corner to the upper right corner on an $8\times 8$ grid. Find the number of such paths that change direction exactly four times, as in the examples shown below.
[asy] size(10cm); usepackage("tikz");label("\begin{tikzpicture}[scale=.5]\draw(0,0)... | As in Solution 1, there are two cases: $RURUR$ or $URURU$ . We will work with the first case and multiply by $2$ at the end. We use stars and bars; we can treat the $R$ s as the stars and the $U$ s as the bars. However, we must also use stars and bars on the $U$ s to see how many different patterns of bars we can creat... | 294 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.