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 |
|---|---|---|---|---|---|
6,801 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | 3 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $... | We evaluate $f(6)$ recursively: \begin{alignat*}{6} f(0)&=1, \\ f(1)&=3f(0)-1&&=2, \\ f(2)&=3f(1)-1&&=5, \\ f(3)&=3f(2)-1&&=14, \\ f(4)&=3f(3)-1&&=41, \\ f(5)&=3f(4)-1&&=122, \\ f(6)&=3f(5)-1&&=\boxed{365} ~MRENTHUSIASM | 365 |
6,802 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | 4 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $... | For all $n\geq1,$ we have \begin{align*} f(n) &= 3f(n-1)-1 \\ &= 3\left(3f(n-2)-1\right)-1 \\ &= 3^2f(n-2)-3-1 \\ &= 3^2\left(3f(n-3)-1\right)-3-1 \\ &= 3^3f(n-3)-3^2-3-1 \\ & \ \vdots \\ &= 3^nf(0)-3^{n-1}-3^{n-2}-3^{n-3}-\cdots-1 \\ &= 3^n-\left(3^{n-1}+3^{n-2}+3^{n-3}+\cdots+1\right) \\ &= 3^n-\frac{3^n-1}{2} \\ &= ... | 365 |
6,803 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | 5 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $... | We can perform casework based on the number of overlapping elements. If no elements overlap, there is $\binom60=1$ way to choose the overlapping elements, and $2^{6-0}$ ways to distribute the remaining elements--each element can go in one subset or the other. We must also divide by $2$ because the order of the subsets ... | 365 |
6,804 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_8 | 6 | Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $... | If we wanted to, we could use casework, being very careful not to double count cases. Let's first figure out what cases we need to look at. The notation I will be using is $x\rightarrow y,$ which implies that we pick $x$ numbers for the first set which then the second set can have $y$ numbers.
Clearly: \begin{align*} 0... | 365 |
6,805 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_9 | 1 | Two thousand points are given on a circle . Label one of the points $1$ . From this point, count $2$ points in the clockwise direction and label this point $2$ . From the point labeled $2$ , count $3$ points in the clockwise direction and label this point $3$ . (See figure.) Continue this process until the labels $1,2,... | The label $1993$ will occur on the $\frac12(1993)(1994) \pmod{2000}$ th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$
Simplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994... | 118 |
6,806 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_9 | 2 | Two thousand points are given on a circle . Label one of the points $1$ . From this point, count $2$ points in the clockwise direction and label this point $2$ . From the point labeled $2$ , count $3$ points in the clockwise direction and label this point $3$ . (See figure.) Continue this process until the labels $1,2,... | Two labels $a$ and $b$ occur on the same point if $\ a(a+1)/2\equiv \ b(b+1)/2\pmod{2000}$ . If we assume the final answer be $n$ , then we have $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$
Multiply $2$ on both side we have $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{4000}$ . As they hav... | 118 |
6,807 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_10 | 1 | Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$ | The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12$ equilateral pentagons) in which the $20$ vertices have all been truncated to form $20$ equilateral triangles with common vertices. The resulting solid has then $p=12$ smaller equilateral pentagons ... | 250 |
6,808 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_10 | 2 | Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$ | As seen above, $E=V+30$ . Every vertex $V$ , there is a triangle for every $T$ and a pentagon for every $P$ by the given. However, there are three times every triangle will be counted and five times every pentagon will be counted because of their numbers of vertices. From this observation, $\frac{VT}3+\frac{VP}5=32\imp... | 250 |
6,809 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_10 | 3 | Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$ | Notice that at each vertex, we must have the sum of the angles be less than $360$ degrees or we will not be able to fold the polyhedron. Therefore, we have $60T + 108P < 360.$ Now, let there be $t$ triangles and $p$ pentagons total such that $t+p = 32.$ From the given, we know that $E = V + 30.$ Lastly, we see that $E... | 250 |
6,810 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_10 | 4 | Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$ | We know that $V-E = -30 \implies V = E-30$ based off the problem condition. Furthermore, if we draw out a few pentagons as well as triangles on each of side of the pentagons, it's clear that each vertex has 4 edges connected to it, with two triangles and two pentagons for each vertex. However, each edge is used for two... | 250 |
6,811 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_12 | 1 | The vertices of $\triangle ABC$ are $A = (0,0)\,$ $B = (0,420)\,$ , and $C = (560,0)\,$ . The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's. Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ $P_3\,$ $P_4, \dots$ are generated by rolling the die ... | If we have points $(p,q)$ and $(r,s)$ and we want to find $(u,v)$ so $(r,s)$ is the midpoint of $(u,v)$ and $(p,q)$ , then $u=2r-p$ and $v=2s-q$ . So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have \[P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\bmod{560},... | 344 |
6,812 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_12 | 2 | The vertices of $\triangle ABC$ are $A = (0,0)\,$ $B = (0,420)\,$ , and $C = (560,0)\,$ . The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's. Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ $P_3\,$ $P_4, \dots$ are generated by rolling the die ... | Let $L_1$ be the $n^{th}$ roll that directly influences $P_{n + 1}$
Note that $P_7 = \cfrac{\cfrac{\cfrac{P_1 + L_1}2 + L_2}2 + \cdots}{2\ldots} = \frac {(k,m)}{64} + \frac {L_1}{64} + \frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}8 + \frac {L_5}4 + \frac {L_6}2 = (14,92)$
Then quickly checking each addend from the ri... | 344 |
6,813 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_13 | 1 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and K... | Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$ , they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are \begin{align}y&=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&=x... | 163 |
6,814 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_13 | 2 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and K... | Let $A$ and $B$ be Kenny's initial and final points respectively and define $C$ and $D$ similarly for Jenny. Let $O$ be the center of the building. Also, let $X$ be the intersection of $AC$ and $BD$ . Finaly, let $P$ and $Q$ be the points of tangency of circle $O$ to $AC$ and $BD$ respectively.
From the problem stateme... | 163 |
6,815 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_13 | 3 | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and K... | [asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,R,S; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); R=(100,106.667); S=(0,53.333); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(R); dot(S); draw(A--B--D--C--cycle); draw(P--O); draw(D--S); draw(O-... | 163 |
6,816 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_14 | 1 | A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter ... | Put the rectangle on the coordinate plane so its vertices are at $(\pm4,\pm3)$ , for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, $O$
Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is... | 448 |
6,817 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_14 | 2 | A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter ... | Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in l... | 448 |
6,818 | https://artofproblemsolving.com/wiki/index.php/1993_AIME_Problems/Problem_15 | 1 | Let $\overline{CH}$ be an altitude of $\triangle ABC$ . Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$ . If $AB = 1995\,$ $AC = 1994\,$ , and $BC = 1993\,$ , then $RS\,$ can be expressed as $m/n\,$ , where $m\,$ and $n\,$ are relativ... | [asy] unitsize(48); pair A,B,C,H; A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE); draw(circle((2,1),1)); pair [] x=intersectionpoints(C--H,circle((2,1),1)); dot(x[0]); label("$S$",x[0],SW); draw(circle((4.29843788128,1.298... | 997 |
6,819 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_1 | 1 | Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms | There are 8 fractions which fit the conditions between 0 and 1: $\frac{1}{30},\frac{7}{30},\frac{11}{30},\frac{13}{30},\frac{17}{30},\frac{19}{30},\frac{23}{30},\frac{29}{30}$
Their sum is 4. Note that there are also 8 terms between 1 and 2 which we can obtain by adding 1 to each of our first 8 terms. For example, $1+\... | 400 |
6,820 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_1 | 2 | Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms | Note that if $x$ is a solution, then $(300-x)$ is a solution. We know that $\phi(300) = 80.$ Therefore the answer is $\frac{80}{2} \cdot\frac{300}{30} = \boxed{400}.$ | 400 |
6,821 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_2 | 1 | positive integer is called ascending if, in its decimal representation , there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there? | Note that an ascending number is exactly determined by its digits : for any set of digits (not including 0, since the only position for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits.
So, there are nine digits that may be used: $1,2,3,4,5,6,7,8,9.$ Not... | 502 |
6,822 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_3 | 1 | A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly $.500$ . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater t... | Let $n$ be the number of matches won, so that $\frac{n}{2n}=\frac{1}{2}$ , and $\frac{n+3}{2n+4}>\frac{503}{1000}$
Cross multiplying $1000n+3000>1006n+2012$ , so $n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}$ . Thus, the answer is $\boxed{164}$ | 164 |
6,823 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_3 | 2 | A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly $.500$ . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater t... | Let $n$ be the number of matches she won before the weekend began. Since her win ratio started at exactly . $500 = \tfrac{1}{2},$ she must have played exactly $2n$ games total before the weekend began. After the weekend, she would have won $n+3$ games out of $2n+4$ total. Therefore, her win ratio would be $(n+3)/(2n+4)... | 164 |
6,824 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_5 | 1 | Let $S^{}_{}$ be the set of all rational numbers $r^{}_{}$ $0^{}_{}<r<1$ , that have a repeating decimal expansion in the form $0.abcabcabc\ldots=0.\overline{abc}$ , where the digits $a^{}_{}$ $b^{}_{}$ , and $c^{}_{}$ are not necessarily distinct. To write the elements of $S^{}_{}$ as fractions in lowest terms, how ma... | We consider the method in which repeating decimals are normally converted to fractions with an example:
$x=0.\overline{176}$
$\Rightarrow 1000x=176.\overline{176}$
$\Rightarrow 999x=1000x-x=176$
$\Rightarrow x=\frac{176}{999}$
Thus, let $x=0.\overline{abc}$
$\Rightarrow 1000x=abc.\overline{abc}$
$\Rightarrow 999x=1000x... | 660 |
6,825 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_6 | 1 | For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added? | For one such pair of consecutive integers, let the smaller integer be $\underline{1ABC},$ where $A,B,$ and $C$ are digits from $0$ through $9.$
We wish to count the ordered triples $(A,B,C).$ By casework, we consider all possible forms of the larger integer, as shown below. \[\begin{array}{c|c|c|c|c|c|c} & & & & & & \\... | 156 |
6,826 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_6 | 2 | For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added? | Consider what carrying means: If carrying is needed to add two numbers with digits $abcd$ and $efgh$ , then $h+d\ge 10$ or $c+g\ge 10$ or $b+f\ge 10$ . 6. Consider $c \in \{0, 1, 2, 3, 4\}$ $1abc + 1ab(c+1)$ has no carry if $a, b \in \{0, 1, 2, 3, 4\}$ . This gives $5^3=125$ possible solutions.
With $c \in \{5, 6, 7, ... | 156 |
6,827 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_6 | 3 | For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added? | Consider the ordered pair $(1abc , 1abc - 1)$ where $a,b$ and $c$ are digits. We are trying to find all ordered pairs where $(1abc) + (1abc - 1)$ does not require carrying. For the addition to require no carrying, $2a,2b < 10$ , so $a,b < 5$ unless $1abc$ ends in $00$ , which we will address later. Clearly, if $c \in ... | 156 |
6,828 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_7 | 1 | Faces $ABC^{}_{}$ and $BCD^{}_{}$ of tetrahedron $ABCD^{}_{}$ meet at an angle of $30^\circ$ . The area of face $ABC^{}_{}$ is $120^{}_{}$ , the area of face $BCD^{}_{}$ is $80^{}_{}$ , and $BC=10^{}_{}$ . Find the volume of the tetrahedron. | Since the area $BCD=80=\frac{1}{2}\cdot10\cdot16$ , the perpendicular from $D$ to $BC$ has length $16$
The perpendicular from $D$ to $ABC$ is $16 \cdot \sin 30^\circ=8$ . Therefore, the volume is $\frac{8\cdot120}{3}=\boxed{320}$ | 320 |
6,829 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | 1 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1... | Note that the $\Delta$ s are reminiscent of differentiation; from the condition $\Delta(\Delta{A}) = 1$ , we are led to consider the differential equation \[\frac{d^2 A}{dn^2} = 1\] This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; \[a_{n} = \frac{1}{2}(n-19)(n-92)\] as we must have ro... | 819 |
6,830 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | 2 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1... | Let $\Delta^1 A=\Delta A$ , and $\Delta^n A=\Delta(\Delta^{(n-1)}A)$
Note that in every sequence of $a_i$ $a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...$
Then $a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...$
Since $\Delta a_1... | 819 |
6,831 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | 3 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1... | The sequence $\Delta(\Delta A)$ is the second finite difference sequence, and the first $k-1$ terms of this sequence can be computed in terms of the original sequence as shown below.
$\begin{array}{rcl} a_3+a_1-2a_2&=&1\\ a_4+a_2-2a_3&=&1\\ &\vdots\\ a_k + a_{k-2} - 2a_{k-1} &= &1\\ a_{k+1} + a_{k-1} - 2a_k &=& 1.\\ \e... | 819 |
6,832 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | 5 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1... | Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that $f(19) = f(92) = 0$ so the quadratic is $f(x) = a(x-19)(x-92)$ for some constant $a.$ Now we use the conditions that the finite difference is $1$ to find $a.$ We know $f(19... | 819 |
6,833 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_8 | 6 | For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1... | Let $a_1=a,a_2=b.$ From the conditions, we have \[a_{n-1}+a_{n+1}=2a_n+1,\] for all $n>1.$ From this, we find that \begin{align*} a_3&=2b+1-a \\ a_4&=3b+3-2a\\ a_5&=4b+6-3a, \end{align*} or, in general, \[a_n=(n-1)b+\frac{(n-2)(n-1)}{2}-(n-2)a.\] Note: we can easily prove this by induction. Now, substituting $n=19,92,$... | 819 |
6,834 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_9 | 1 | Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime posit... | Let $AP=x$ so that $PB=92-x.$ Extend $AD, BC$ to meet at $X,$ and note that $XP$ bisects $\angle AXB;$ let it meet $CD$ at $E.$ Using the angle bisector theorem, we let $XB=y(92-x), XA=xy$ for some $y.$
Then $XD=xy-70, XC=y(92-x)-50,$ thus \[\frac{xy-70}{y(92-x)-50} = \frac{XD}{XC} = \frac{ED}{EC}=\frac{AP}{PB} = \frac... | 164 |
6,835 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_9 | 2 | Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime posit... | From $(1)$ above, $x = \frac{70r}{h}$ and $92-x = \frac{50r}{h}$ . Adding these equations yields $92 = \frac{120r}{h}$ . Thus, $x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}$ , and $m+n = \boxed{164}$ | 164 |
6,836 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_9 | 3 | Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ $BC=50^{}_{}$ $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime posit... | The area of the trapezoid is $\frac{(19+92)h}{2}$ , where $h$ is the height of the trapezoid.
Draw lines $CP$ and $BP$ . We can now find the area of the trapezoid as the sum of the areas of the three triangles $BPC$ $CPD$ , and $PBA$
$[BPC] = \frac{1}{2} \cdot 50 \cdot r$ (where $r$ is the radius of the tangent circle.... | 164 |
6,837 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_10 | 1 | Consider the region $A$ in the complex plane that consists of all points $z$ such that both $\frac{z}{40}$ and $\frac{40}{\overline{z}}$ have real and imaginary parts between $0$ and $1$ , inclusive. What is the integer that is nearest the area of $A$ | Let $z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i$ . Since $0\leq \frac{a}{40},\frac{b}{40}\leq 1$ we have the inequality \[0\leq a,b \leq 40\] which is a square of side length $40$
Also, $\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i$ so we have $0\leq a,b \leq \frac{a^2+b... | 572 |
6,838 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_11 | 1 | Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive x-axis. For any line $l^{}_{}$ , the transformation $R(l)^{}_{}$ produces another line as follows: $l^{}_{}$ is reflected in $l_1^{}$ , and the result... | Let $l$ be a line that makes an angle of $\theta$ with the positive $x$ -axis. Let $l'$ be the reflection of $l$ in $l_1$ , and let $l''$ be the reflection of $l'$ in $l_2$
The angle between $l$ and $l_1$ is $\theta - \frac{\pi}{70}$ , so the angle between $l_1$ and $l'$ must also be $\theta - \frac{\pi}{70}$ . Thus, $... | 945 |
6,839 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_12 | 1 | In a game of Chomp , two players alternately take bites from a 5-by-7 grid of unit squares . To take a bite, a player chooses one of the remaining squares , then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For exam... | By drawing possible examples of the subset, one can easily see that making one subset is the same as dividing the game board into two parts.
One can also see that it is the same as finding the shortest route from the upper left hand corner to the lower right hand corner; Such a route would require 5 lengths that go dow... | 792 |
6,840 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_12 | 2 | In a game of Chomp , two players alternately take bites from a 5-by-7 grid of unit squares . To take a bite, a player chooses one of the remaining squares , then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For exam... | If any square is eaten, the squares to the right of it must also be eaten. Thus, if $a_i$ is the number of squares remaining on row $i$ , there is exactly one way the row can be configured (the leftmost $a_i$ squares are uneaten and the ones to the right are eaten.) Additionally, the squares above an eaten square must ... | 792 |
6,841 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_13 | 1 | Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$ . What's the largest area that this triangle can have? | First, consider the triangle in a coordinate system with vertices at $(0,0)$ $(9,0)$ , and $(a,b)$ . Applying the distance formula , we see that $\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}$
We want to maximize $b$ , the height, with $9$ being the base.
Simplifying gives $-a^2 -\frac{3200}{9}a +... | 820 |
6,842 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_13 | 2 | Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$ . What's the largest area that this triangle can have? | Let $A, B$ be the endpoints of the side with length $9$ . Let $\Gamma$ be the Apollonian Circle of $AB$ with ratio $40:41$ ; let this intersect $AB$ at $P$ and $Q$ , where $P$ is inside $AB$ and $Q$ is outside. Then because $(A, B; P, Q)$ describes a harmonic set, $AP/AQ=BP/BQ\implies \dfrac{\frac{41}{9}}{BQ+9}=\dfrac{... | 820 |
6,843 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_13 | 3 | Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$ . What's the largest area that this triangle can have? | We can apply Heron's on this triangle after letting the two sides equal $40x$ and $41x$ . Heron's gives
$\sqrt{\left(\frac{81x+9}{2} \right) \left(\frac{81x-9}{2} \right) \left(\frac{x+9}{2} \right) \left(\frac{-x+9}{2} \right)}$
This can be simplified to
$\frac{9}{4} \cdot \sqrt{(81x^2-1)(81-x^2)}$
We can optimize the... | 820 |
6,844 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_13 | 4 | Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$ . What's the largest area that this triangle can have? | We can start how we did above in solution 4 to get $\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}$ .
Then, we can notice the inside is a quadratic in terms of $x^2$ , which is $-81(x^2)^2+6562x^2-81$ . This is maximized when $x^2 = \frac{3281}{81}$ .If we plug it into the equation, we get $\frac{9}{4} *\frac{9}{4}*\frac{3280}... | 820 |
6,845 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_14 | 3 | In triangle $ABC^{}_{}$ $A'$ $B'$ , and $C'$ are on the sides $BC$ $AC^{}_{}$ , and $AB^{}_{}$ , respectively. Given that $AA'$ $BB'$ , and $CC'$ are concurrent at the point $O^{}_{}$ , and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$ , find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$ | A consequence of Ceva's theorem sometimes attributed to Gergonne is that $\frac{AO}{OA'}=\frac{AC'}{C'B}+\frac{AB'}{B'C}$ , and similarly for cevians $BB'$ and $CC'$ . Now we apply Gergonne several times and do algebra:
\begin{align*} \frac{AO}{OA'}\frac{BO}{OB'}\frac{CO}{OC'} &= \left(\frac{AB'}{B'C}+\frac{AC'}{C'B}\... | 94 |
6,846 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_1 | 1 | Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that \begin{align*} xy+x+y&=71, \\ x^2y+xy^2&=880. \end{align*} | Define $a = x + y$ and $b = xy$ . Then $a + b = 71$ and $ab = 880$ . Solving these two equations yields a quadratic $a^2 - 71a + 880 = 0$ , which factors to $(a - 16)(a - 55) = 0$ . Either $a = 16$ and $b = 55$ or $a = 55$ and $b = 16$ . For the first case, it is easy to see that $(x,y)$ can be $(5,11)$ (or vice versa)... | 146 |
6,847 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_2 | 1 | Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$ , and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$ . For $1_{}^{}... | The length of the diagonal is $\sqrt{3^2 + 4^2} = 5$ (a 3-4-5 right triangle ). For each $k$ $\overline{P_kQ_k}$ is the hypotenuse of a $3-4-5$ right triangle with sides of $3 \cdot \frac{168-k}{168}, 4 \cdot \frac{168-k}{168}$ . Thus, its length is $5 \cdot \frac{168-k}{168}$ . Let $a_k=\frac{5(168-k)}{168}$ . We want... | 840 |
6,848 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_2 | 2 | Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$ , and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$ . For $1_{}^{}... | Using the above diagram, we have that $\Delta ABC \sim \Delta P_k B Q_k$ and each one of these is a dilated 3-4-5 right triangle (This is true since $\Delta ABC$ is a 3-4-5 right triangle). Now, for all $k$ , we have that $\overline{P_k Q_k}$ is the hypotenuse for the triangle $P_k B Q_k$ . Therefore we want to know th... | 840 |
6,849 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_2 | 3 | Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$ , and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$ . For $1_{}^{}... | First, count the diagonal which has length $5$ . For the rest of the segments, think about pairing them up so that each pair makes $5$ . For example, the parallel lines closest to the diagonal would have length $\frac{167}{168}\cdot{5}$ while the parallel line closest to the corner of the rectangle would have length $\... | 840 |
6,850 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_3 | 1 | Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives
${1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \cdots + A_{1000},$ where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000$ . For ... | Let $0<x_{}^{}<1$ . Then we may write $A_{k}^{}={N\choose k}x^{k}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}$ . Taking logarithms in both sides of this last equation and using the well-known fact $\log(a_{}^{}b)=\log a + \log b$ (valid if $a_{}^{},b_{}^{}>0$ ), we have
$\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x... | 166 |
6,851 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_3 | 2 | Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives
${1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \cdots + A_{1000},$ where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000$ . For ... | We know that once we have found the largest value of $k$ , all values after $A_k$ are less than $A_k$ . Therefore, we are looking for the smallest possible value such that:
${\frac{1}{5}}^k\cdot {{1000} \choose {k}}>{\frac{1}{5}}^{k+1}\cdot {{1000} \choose {k+1}}$
Dividing by ${\frac{1}{5}}^k$ gives:
${1000\choose k}>{... | 166 |
6,852 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_3 | 3 | Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives
${1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \cdots + A_{1000},$ where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000$ . For ... | We know that $A_k$ will increase as $k$ increases until certain $k=m$ , where $A_0 < A_1 < A_2 < \dots < A_{m-2} < A_{m-1} < A_m$ and
$A_m > A_{m+1} > A_{m+2} > \dots > A_{1000}.$
Next, to change $A_{k-1}$ to $A_k$ , we multiply $A_{k-1}$ by $\frac{1000-k+1}{5k}$ . It follows that the numerator must be greater than the... | 166 |
6,853 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_3 | 4 | Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives
${1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \cdots + A_{1000},$ where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000$ . For ... | Notice that the expansion is the largest the moment BEFORE the $nC_p < 5$ (this reasoning can probably be found in the other solutions; basically, if we have a number k and then k+1, the value is the largest when k+1 is larger than k, or in other words $nC_p*\frac{1}{5} > 1$
Say we have ${1000 \choose 5}$ ... this equa... | 166 |
6,854 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_3 | 5 | Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives
${1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \cdots + A_{1000},$ where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000$ . For ... | Notice the relation between $A_m$ and $A_{m+1}$ . We have that: $A_{m+1} = A_m \cdot \frac{1}{5} \cdot \frac{1000-m}{m+1}$ . This is true because from $A_m$ to $A_{m+1}$ we have to multiply by $\frac{1}{5}=0.2$ once,and then we must resolve the factorial issue. To do this, we must realize that
\[{1000 \choose m+1} = \... | 166 |
6,855 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_4 | 1 | How many real numbers $x^{}_{}$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$ | The range of the sine function is $-1 \le y \le 1$ . It is periodic (in this problem) with a period of $\frac{2}{5}$
Thus, $-1 \le \frac{1}{5} \log_2 x \le 1$ , and $-5 \le \log_2 x \le 5$ . The solutions for $x$ occur in the domain of $\frac{1}{32} \le x \le 32$ . When $x > 1$ the logarithm function returns a positive... | 159 |
6,856 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_4 | 2 | How many real numbers $x^{}_{}$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$ | Notice that the equation is satisfied twice for every sine period (which is $\frac{2}{5}$ ), except in the sole case when the two equations equate to $0$ . In that case, the equation is satisfied twice but only at the one instance when $y=0$ . Hence, it is double-counted in our final solution, so we have to subtract it... | 159 |
6,857 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_5 | 1 | Given a rational number , write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator . For how many rational numbers between $0$ and $1$ will $20_{}^{}!$ be the resulting product | If the fraction is in the form $\frac{a}{b}$ , then $a < b$ and $\gcd(a,b) = 1$ . There are 8 prime numbers less than 20 ( $2, 3, 5, 7, 11, 13, 17, 19$ ), and each can only be a factor of one of $a$ or $b$ . There are $2^8$ ways of selecting some combination of numbers for $a$ ; however, since $a<b$ , only half of them... | 128 |
6,858 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_6 | 1 | Suppose $r^{}_{}$ is a real number for which
Find $\lfloor 100r \rfloor$ . (For real $x^{}_{}$ $\lfloor x \rfloor$ is the greatest integer less than or equal to $x^{}_{}$ .) | There are $91 - 19 + 1 = 73$ numbers in the sequence . Since the terms of the sequence can be at most $1$ apart, all of the numbers in the sequence can take one of two possible values. Since $\frac{546}{73} = 7 R 35$ , the values of each of the terms of the sequence must be either $7$ or $8$ . As the remainder is $35$ ... | 743 |
6,859 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_6 | 2 | Suppose $r^{}_{}$ is a real number for which
Find $\lfloor 100r \rfloor$ . (For real $x^{}_{}$ $\lfloor x \rfloor$ is the greatest integer less than or equal to $x^{}_{}$ .) | Recall by Hermite's Identity that $\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +...+\lfloor x+\frac{n-1}{n}\rfloor = \lfloor nx\rfloor$ for positive integers $n$ , and real $x$ . Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, $\lfloor r\rflo... | 743 |
6,860 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_7 | 1 | Find $A^2_{}$ , where $A^{}_{}$ is the sum of the absolute values of all roots of the following equation: | $x=\sqrt{19}+\underbrace{\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}}_{x}$
$x=\sqrt{19}+\frac{91}{x}$
$x^2=x\sqrt{19}+91$
$x^2-x\sqrt{19}-91 = 0$
$\left.\begin{array}{l}x_1=\frac{\sqrt{19}+\sqrt{383}}{2}\\\\x_2=\frac{\sqrt{19}-\sqrt{383}}{2}\end{array}\right\}A=|x_1|... | 383 |
6,861 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_7 | 2 | Find $A^2_{}$ , where $A^{}_{}$ is the sum of the absolute values of all roots of the following equation: | Let $f(x) = \sqrt{19} + \frac{91}{x}$ . Then $x = f(f(f(f(f(x)))))$ , from which we realize that $f(x) = x$ . This is because if we expand the entire expression, we will get a fraction of the form $\frac{ax + b}{cx + d}$ on the right hand side, which makes the equation simplify to a quadratic. As this quadratic will ha... | 383 |
6,862 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_8 | 1 | For how many real numbers $a$ does the quadratic equation $x^2 + ax + 6a=0$ have only integer roots for $x$ | By Vieta's formulas $x_1 + x_2 = -a$ where $x_1, x_2$ are the roots of the quadratic, and since $x_1,x_2$ are integers, $a$ must be an integer. Applying the quadratic formula
\[x = \frac{-a \pm \sqrt{a^2 - 24a}}{2}\]
Since $-a$ is an integer, we need $\sqrt{a^2-24a}$ to be an integer (let this be $b$ ): $b^2 = a^2 - 24... | 10 |
6,863 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_8 | 2 | For how many real numbers $a$ does the quadratic equation $x^2 + ax + 6a=0$ have only integer roots for $x$ | Let $x^2 + ax + 6a = (x - s)(x - r)$ . Vieta's yields $s + r = - a, sr = 6a$ \begin{eqnarray*}sr + 6s + 6r &=& 0\\ sr + 6s + 6r + 36 &=& 36\\ (s + 6)(r + 6) &=& 36 \end{eqnarray*}
Without loss of generality let $r \le s$
The possible values of $(r + 6,s + 6)$ are: $( - 36, - 1),( - 18, - 2),( - 12, - 3),( - 9, - 4),( -... | 10 |
6,864 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_9 | 5 | Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ where $\frac mn$ is in lowest terms. Find $m+n^{}_{}.$ | Firstly, we write $\sec x+\tan x=a/b$ where $a=22$ and $b=7$ . This will allow us to spot factorable expressions later. Now, since $\sec^2x-\tan^2x=1$ , this gives us \[\sec x-\tan x=\frac{b}{a}\] Adding this to our original expressions gives us \[2\sec x=\frac{a^2+b^2}{ab}\] or \[\cos x=\frac{2ab}{a^2+b^2}\] Now since... | 44 |
6,865 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_10 | 1 | Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$ , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a^{}_{}$ when it should have been a $b^{}_{}$ , or as a $b^{}_{}$ when it should be an $a^... | Let us make a chart of values in alphabetical order, where $P_a,\ P_b$ are the probabilities that each string comes from $aaa$ and $bbb$ multiplied by $27$ , and $S_b$ denotes the partial sums of $P_b$ (in other words, $S_b = \sum_{n=1}^{b} P_b$ ): \[\begin{array}{|r||r|r|r|} \hline \text{String}&P_a&P_b&S_b\\ \hline a... | 532 |
6,866 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_10 | 2 | Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$ , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a^{}_{}$ when it should have been a $b^{}_{}$ , or as a $b^{}_{}$ when it should be an $a^... | Let $S(a,n)$ be the $n$ th letter of string $S(a)$ .
Compare the first letter of the string $S(a)$ to the first letter of the string $S(b)$ .
There is a $(2/3)^2=4/9$ chance that $S(a,1)$ comes before $S(b,1)$ .
There is a $2(1/3)(2/3)=4/9$ that $S(a,1)$ is the same as $S(b,1)$
If $S(a,1)=S(b,1)$ , then you do the same... | 532 |
6,867 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_10 | 3 | Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$ , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a^{}_{}$ when it should have been a $b^{}_{}$ , or as a $b^{}_{}$ when it should be an $a^... | Consider $n$ letter strings instead. If the first letters all get transmitted correctly, then the $a$ string will be first. Otherwise, the only way is for both of the first letters to be the same, and then we consider the next $n-1$ letter string following the first letter. This easily leads to a recursion: $p_n=\frac2... | 532 |
6,868 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_10 | 4 | Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$ , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a^{}_{}$ when it should have been a $b^{}_{}$ , or as a $b^{}_{}$ when it should be an $a^... | The probability that $S_a$ will take the form $a$ _ _ and that $S_b$ will take the form $b$ _ _ is $\frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}$ . Then, the probability that both $S_a$ and $S_b$ will share the same first digit is $2\cdot\frac{2}{3}\cdot\frac{1}{3} = \frac{4}{9}$ . Now if the first digits of either sequen... | 532 |
6,869 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_11 | 1 | Twelve congruent disks are placed on a circle $C^{}_{}$ of radius 1 in such a way that the twelve disks cover $C^{}_{}$ , no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelv... | We wish to find the radius of one circle, so that we can find the total area.
Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length $2r$ that is tangent... | 135 |
6,870 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_12 | 1 | Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$ $Q^{}_{}$ $R^{}_{}$ , and $S^{}_{}$ are interior points on sides $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ , and $\overline{DA}$ , respectively. It is given that $PB^{}_{}=15$ $BQ^{}_{}=20$ $PR^{}_{}=30$ , and $QS^{}_{}=40$ . L... | Let $O$ be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent $\triangle BPQ \cong \triangle DRS$ $\triangle APS \cong \triangle CRQ$ ). Quickly we realize that $O$ is also the center of the rectangle.
By the Pythagorean Theorem , we can solve f... | 677 |
6,871 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_12 | 2 | Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$ $Q^{}_{}$ $R^{}_{}$ , and $S^{}_{}$ are interior points on sides $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ , and $\overline{DA}$ , respectively. It is given that $PB^{}_{}=15$ $BQ^{}_{}=20$ $PR^{}_{}=30$ , and $QS^{}_{}=40$ . L... | We can just use areas. Let $AP = b$ and $AS = a$ $a^2 + b^2 = 625$ . Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, $(a+20)(b+15)$ . This gives $3a + 4b = 120$ . Solving this system of equation gives $\frac{44}{5} = a$ $\frac{117}{5} = b$ , from which it is st... | 677 |
6,872 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_12 | 3 | Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$ $Q^{}_{}$ $R^{}_{}$ , and $S^{}_{}$ are interior points on sides $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ , and $\overline{DA}$ , respectively. It is given that $PB^{}_{}=15$ $BQ^{}_{}=20$ $PR^{}_{}=30$ , and $QS^{}_{}=40$ . L... | We will bash with trigonometry.
Firstly, by Pythagoras Theorem, $PQ=QR=RS=SP=25$ . We observe that $[PQRS]=\frac{1}{2}\cdot30\cdot40=600$ . Thus, if we drop an altitude from $P$ to $\overline{SR}$ to point $E$ , it will have length $\frac{600}{25}=24$ . In particular, $SE=7$ since we form a 7-24-25 triangle.
Now, $\sin... | 677 |
6,873 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 1 | A drawer contains a mixture of red socks and blue socks, at most $1991$ in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consist... | Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$ . The probability $P$ that when two socks are drawn randomly, without replacement, both are red or both are blue is given by
\[\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.\... | 990 |
6,874 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 2 | A drawer contains a mixture of red socks and blue socks, at most $1991$ in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consist... | Let $r$ and $b$ denote the number of red and blue socks such that $r+b\le1991$ . Then by complementary counting, the number of ways to get a red and a blue sock must be equal to $1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)$ $=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2$ $=(r-b)^... | 990 |
6,875 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 3 | A drawer contains a mixture of red socks and blue socks, at most $1991$ in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consist... | Let $r$ and $b$ denote the number of red and blue socks, respectively. In addition, let $t = r + b$ , the total number of socks in the drawer.
From the problem, it is clear that $\frac{r(r-1)}{t(t-1)} + \frac{b(b-1)}{t(t-1)} = \frac{1}{2}$
Expanding, we get $\frac{r^2 + b^2 - r - b}{t^2 - t} = \frac{1}{2}$
Substituting... | 990 |
6,876 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 4 | A drawer contains a mixture of red socks and blue socks, at most $1991$ in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consist... | As above, let $r$ $b$ , and $t$ denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that $\frac{r(r-1)}{t(t-1)}+\frac{b(b-1)}{t(t-1)}=\frac{1}{2}$ , so $r^2+b^2-r-b=\frac{t(t-1)}{2}=r^2+b^2-t=\frac{t^2}{2}-\frac{t}{2}$
Seeing that we can rewrite $r^2+b^2$ as $(r... | 990 |
6,877 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 5 | A drawer contains a mixture of red socks and blue socks, at most $1991$ in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consist... | Let $r$ be the number of socks that are red, and $t$ be the total number of socks. We get:
$2(r(r-1)+(t-r)(t-r-1))=t(t-1)$ Expanding the left hand side and the right hand side, we get: $4r^2-4rt+2t^2-2t = t^2-t$
And, moving terms, we will get that: $4r^2-4rt+t^2 = t$
We notice that the left side is a perfect square. $(... | 990 |
6,878 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 6 | A drawer contains a mixture of red socks and blue socks, at most $1991$ in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consist... | Let $r$ and $b$ denote the red socks and blue socks, respectively. Thus the equation in question is:
$\frac{r(r-1)+b(b-1)}{(r+b)(r+b-1)}=\frac{1}{2}$
$\Rightarrow 2r^2-2r+2b^2-2b=r^2+2rb+b^2-r-b$
$\Rightarrow r^2+b^2-r-b-2rb=0$
$\Rightarrow (r-b)^2=r+b\le 1991$
Because we wish to maximize $r$ , we have $r\ge b$ and thu... | 990 |
6,879 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_14 | 1 | hexagon is inscribed in a circle . Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$ , has length $31$ . Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$ | Let $x=AC=BF$ $y=AD=BE$ , and $z=AE=BD$
Ptolemy's Theorem on $ABCD$ gives $81y+31\cdot 81=xz$ , and Ptolemy on $ACDF$ gives $x\cdot z+81^2=y^2$ .
Subtracting these equations give $y^2-81y-112\cdot 81=0$ , and from this $y=144$ . Ptolemy on $ADEF$ gives $81y+81^2=z^2$ , and from this $z=135$ . Finally, plugging back int... | 384 |
6,880 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_1 | 1 | The increasing sequence $2,3,5,6,7,10,11,\ldots$ consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence. | Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than $500$ . This happens to be $23^2=529$ . Notice that there are $23$ squares and $8$ cubes less than or equal to $529$ , but $1$ and $2^6$ are both squares and cubes. Thus, there are $529-23-8+2=500$ numbers i... | 528 |
6,881 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_1 | 2 | The increasing sequence $2,3,5,6,7,10,11,\ldots$ consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence. | This solution is similar as Solution 1, but to get the intuition why we chose to consider $23^2 = 529$ , consider this:
We need $n - T = 500$ , where $n$ is an integer greater than 500 and $T$ is the set of numbers which contains all $k^2,k^3\le 500$
Firstly, we clearly need $n > 500$ , so we substitute n for the small... | 528 |
6,882 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_2 | 2 | Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$ | The $3/2$ power is quite irritating to work with so we look for a way to eliminate that. Notice that squaring the expression will accomplish that.
Let $S$ be the sum of the given expression. \[S^2= ((52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2})^2\] \[S^2 = (52+6\sqrt{43})^{3} + (52-6\sqrt{43})^{3} - 2((52+6\sqrt{43})(5... | 828 |
6,883 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_2 | 3 | Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$ | Factor as a difference of cubes. \[\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[\left(\left(\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\right)^2+\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}}+\left(\left(52-6\sqrt{43}\right)^{\frac{1}... | 828 |
6,884 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_2 | 4 | Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$ | Let $x=52+6\sqrt{43}$ $y=52-6\sqrt{43}$ . Similarly to solution 2, we let \[S=x^{\frac{3}{2}}+y^{\frac{3}{2}}\] \begin{align*} S^2&=(x^{\frac{3}{2}}+y^{\frac{3}{2}})^2\\ &=x^3+y^3+2x^{\frac{3}{2}}y^{\frac{3}{2}} \end{align*} The expression can be simplified as follow \begin{align*} S^2&=x^3+y^3+2x^{\frac{3}{2}}y^{\frac... | 828 |
6,885 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_2 | 6 | Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$ | (Similar to Solution 1, but expanding the cubes instead)
Like in Solution 1, we have $\sqrt{52 + 6\sqrt{43}} = \sqrt{43} + 3$ and $\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3.$
Therefore we have that $(52 + 6\sqrt{43})^{3/2} - (52 + 6\sqrt{43})^{3/2}$ $= \sqrt{52 + 6\sqrt{43}}^3 - \sqrt{52 - 6\sqrt{43}}^3$ $= (\sqrt{43} + 3... | 828 |
6,886 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_3 | 1 | Let $P_1^{}$ be a regular $r~\mbox{gon}$ and $P_2^{}$ be a regular $s~\mbox{gon}$ $(r\geq s\geq 3)$ such that each interior angle of $P_1^{}$ is $\frac{59}{58}$ as large as each interior angle of $P_2^{}$ . What's the largest possible value of $s_{}^{}$ | The formula for the interior angle of a regular sided polygon is $\frac{(n-2)180}{n}$
Thus, $\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}$ . Cross multiplying and simplifying, we get $\frac{58(r-2)}{r} = \frac{59(s-2)}{s}$ . Cross multiply and combine like terms again to yield $58rs - 58 \cdot 2s = 59r... | 117 |
6,887 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_3 | 2 | Let $P_1^{}$ be a regular $r~\mbox{gon}$ and $P_2^{}$ be a regular $s~\mbox{gon}$ $(r\geq s\geq 3)$ such that each interior angle of $P_1^{}$ is $\frac{59}{58}$ as large as each interior angle of $P_2^{}$ . What's the largest possible value of $s_{}^{}$ | Like above, use the formula for the interior angles of a regular sided polygon
$\frac{(r-2)180}{r} = \frac{59}{58} * \frac{(s-2)180}{s}$
$59 * 180 * (s-2) * r = 58 * 180 * (r-2) * s$
$59 * (rs - 2r) = 58 * (rs - 2s)$
$rs - 118r = -116s$
$rs = 118r-116s$
This equation tells us $s$ divides $118r$ . If $s$ specifically di... | 117 |
6,888 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_3 | 3 | Let $P_1^{}$ be a regular $r~\mbox{gon}$ and $P_2^{}$ be a regular $s~\mbox{gon}$ $(r\geq s\geq 3)$ such that each interior angle of $P_1^{}$ is $\frac{59}{58}$ as large as each interior angle of $P_2^{}$ . What's the largest possible value of $s_{}^{}$ | As in above, we have $rs = 118r - 116s.$ This means that $rs + 116s - 118r = 0.$ Using SFFT we obtain $s(r+116) - 118(r+116) = -118 \cdot 116 \implies (s-118)(r+116) = -118 \cdot 116.$ Since $r+116$ is always positive, we know thta $s-118$ must be negative. Therefore the maximum value of $s$ must be $\boxed{117}$ which... | 117 |
6,889 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_5 | 1 | Let $n^{}_{}$ be the smallest positive integer that is a multiple of $75_{}^{}$ and has exactly $75_{}^{}$ positive integral divisors, including $1_{}^{}$ and itself. Find $\frac{n}{75}$ | The prime factorization of $75 = 3^15^2 = (2+1)(4+1)(4+1)$ . For $n$ to have exactly $75$ integral divisors, we need to have $n = p_1^{e_1-1}p_2^{e_2-1}\cdots$ such that $e_1e_2 \cdots = 75$ . Since $75|n$ , two of the prime factors must be $3$ and $5$ . To minimize $n$ , we can introduce a third prime factor, $2$ . Al... | 432 |
6,890 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_6 | 1 | A biologist wants to calculate the number of fish in a lake. On May 1 she catches a random sample of 60 fish, tags them, and releases them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate the number of fish in the lake on May 1, she assumes that 25% of these fish ... | Of the $70$ fish caught in September, $40\%$ were not there in May, so $42$ fish were there in May. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, $\frac{3}{42} = \frac{60}{x} \Longrightarrow \boxed{840}$ | 840 |
6,891 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_6 | 2 | A biologist wants to calculate the number of fish in a lake. On May 1 she catches a random sample of 60 fish, tags them, and releases them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate the number of fish in the lake on May 1, she assumes that 25% of these fish ... | First, we notice that there are 45 tags left, after 25% of the original fish have went away/died. Then, some $x$ percent of fish have been added such that $\frac{x}{x+75} = 40 \%$ , or $\frac{2}{5}$ . Solving for $x$ , we get that $x = 50$ , so the total number of fish in September is $125 \%$ , or $\frac{5}{4}$ times ... | 840 |
6,892 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_8 | 1 | In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:
1) The marksman first chooses a column from which a target is to be broken.
2) The marksman must then break the lowest... | Clearly, the marksman must shoot the left column three times, the middle column two times, and the right column three times.
From left to right, suppose that the columns are labeled $L,M,$ and $R,$ respectively. We consider the string $LLLMMRRR:$
Since the letter arrangements of $LLLMMRRR$ and the shooting orders have ... | 560 |
6,893 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_10 | 1 | The sets $A = \{z : z^{18} = 1\}$ and $B = \{w : w^{48} = 1\}$ are both sets of complex roots of unity . The set $C = \{zw : z \in A ~ \mbox{and} ~ w \in B\}$ is also a set of complex roots of unity. How many distinct elements are in $C_{}^{}$ | The least common multiple of $18$ and $48$ is $144$ , so define $n = e^{2\pi i/144}$ . We can write the numbers of set $A$ as $\{n^8, n^{16}, \ldots n^{144}\}$ and of set $B$ as $\{n^3, n^6, \ldots n^{144}\}$ $n^x$ can yield at most $144$ different values. All solutions for $zw$ will be in the form of $n^{8k_1 + 3k_2}$... | 144 |
6,894 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_10 | 2 | The sets $A = \{z : z^{18} = 1\}$ and $B = \{w : w^{48} = 1\}$ are both sets of complex roots of unity . The set $C = \{zw : z \in A ~ \mbox{and} ~ w \in B\}$ is also a set of complex roots of unity. How many distinct elements are in $C_{}^{}$ | The 18 and 48th roots of $1$ can be found by De Moivre's Theorem . They are $\text{cis}\,\left(\frac{2\pi k_1}{18}\right)$ and $\text{cis}\,\left(\frac{2\pi k_2}{48}\right)$ respectively, where $\text{cis}\,\theta = \cos \theta + i \sin \theta$ and $k_1$ and $k_2$ are integers from $0$ to $17$ and $0$ to $47$ , respect... | 144 |
6,895 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_10 | 3 | The sets $A = \{z : z^{18} = 1\}$ and $B = \{w : w^{48} = 1\}$ are both sets of complex roots of unity . The set $C = \{zw : z \in A ~ \mbox{and} ~ w \in B\}$ is also a set of complex roots of unity. How many distinct elements are in $C_{}^{}$ | The values in polar form will be $(1, 20x)$ and $(1, 7.5x)$ . Multiplying these gives $(1, 27.5x)$ . Then, we get $27.5$ $55$ $82.5$ $110$ $...$ up to $3960$ $(\text{lcm}(55,360)) \implies \frac{3960 \cdot 2}{55}=\boxed{144}$ | 144 |
6,896 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_12 | 1 | regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},$ where $a^{}_{}$ $b^{}_{}$ $c^{}_{}$ , and $d^{}_{}$ are positive integers. Find $a + b + c + d^{}_{}$ | 1990 AIME-12.png
The easiest way to do this seems to be to find the length of each of the sides and diagonals. To do such, draw the radii that meet the endpoints of the sides/diagonals; this will form isosceles triangles . Drawing the altitude of those triangles and then solving will yield the respective lengths.
Addin... | 720 |
6,897 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_12 | 2 | regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},$ where $a^{}_{}$ $b^{}_{}$ $c^{}_{}$ , and $d^{}_{}$ are positive integers. Find $a + b + c + d^{}_{}$ | A second method involves drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon. Since the distance from the center to a vertex of the 12-gon is $12$ , the Law of Cosines can be applied to this isosceles triangle, to give:
$a^2 = 12^2 + 12^2 - 2\cdot 12\cdot 12\cdot \cos \theta$
$a^2 = 2\... | 720 |
6,898 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_12 | 3 | regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},$ where $a^{}_{}$ $b^{}_{}$ $c^{}_{}$ , and $d^{}_{}$ are positive integers. Find $a + b + c + d^{}_{}$ | Begin as in solution 2, drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon. Apply law of cosines on $\theta = {30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}, 180^{\circ}}$ to get $d^2 = 288 - 288 \cos \theta$ where $d$ is the diagonal or sidelength distance between two p... | 720 |
6,899 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_14 | 1 | The rectangle $ABCD^{}_{}$ below has dimensions $AB^{}_{} = 12 \sqrt{3}$ and $BC^{}_{} = 13 \sqrt{3}$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P^{}_{}$ . If triangle $ABP^{}_{}$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segment... | [asy] import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0, 399^(0.5), 0); triple D=(108^(0.5), 0, 0); triple C=(-108^(0.5), 0, 0); triple Pa; pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci... | 594 |
6,900 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_14 | 2 | The rectangle $ABCD^{}_{}$ below has dimensions $AB^{}_{} = 12 \sqrt{3}$ and $BC^{}_{} = 13 \sqrt{3}$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P^{}_{}$ . If triangle $ABP^{}_{}$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segment... | Let $X$ be the apex of the pyramid and $M$ be the midpoint of $\overline{CD}$ . We find the side lengths of $\triangle XMP$
$MP = \frac{13\sqrt3}{2}$ $PX$ is half of $AC$ , which is $\frac{\sqrt{3\cdot13^2+3\cdot12^2}}{2} = \frac{\sqrt{939}}{2}$ . To find $MX$ , consider right triangle $XMD$ ; since $XD=13\sqrt3$ and $... | 594 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.