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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 $\{b, c, d, e, f\},\{a, c\}.$ | 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 $\{b, c, d, e, f\},\{a, c\}.$ | 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} \\ &= \frac{3^n+1}{2} \\ &= \frac{3^n-1}{2}+1, \end{align*} which resembles the result in Solutions 1 and 2. The answer is $f(6)=\boxed{365}.$ | 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 $\{b, c, d, e, f\},\{a, c\}.$ | 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 does not matter. Proceeding similarly for the other cases, our sum is \[\dbinom{6}0 \cdot 2^5+\dbinom{6}1 \cdot 2^4+\cdots +\dbinom{6}4 \cdot 2+\dbinom{6}5 \cdot 1+\dbinom{6}6 \cdot 1.\] (Note that we have to be especially careful with the last case, as it does not follow the pattern of the other cases.) Adding these up gives a total of $\boxed{365}.$ | 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 $\{b, c, d, e, f\},\{a, c\}.$ | 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&\rightarrow6 \\ 1&\rightarrow5\mid6 \\ 2&\rightarrow4\mid5\mid6 \\ 3&\rightarrow3\mid4\mid5\mid6 \\ 4&\rightarrow2\mid3\mid4\mid5\mid6 \\ 5&\rightarrow1\mid2\mid3\mid4\mid5\mid6 \\ 6&\rightarrow0\mid1\mid2\mid3\mid4\mid5\mid6 \end{align*} However notice that many of the cases are double counted as direction does not matter, e.g. $2\rightarrow4$ is the same as $4\rightarrow2.$ Get rid of those cases so we are just left with: \begin{align*} 0&\rightarrow6 \\ 1&\rightarrow5\mid6 \\ 2&\rightarrow4\mid5\mid6 \\ 3&\rightarrow3\mid4\mid5\mid6 \\ 4&\rightarrow4\mid5\mid6 \\ 5&\rightarrow5\mid6 \\ 6&\rightarrow6 \end{align*} Now we start computing, $0\rightarrow6$ is just $1$ case, $1\rightarrow5\mid6$ is just $\binom{6}{1}\cdot2$ cases, $2\rightarrow4\mid5\mid6$ is just $\binom{6}{2}\cdot2^2$ cases, and $3\rightarrow3\mid4\mid5\mid6$ is just $\binom{6}{3}\cdot2^3$ cases (If you have trouble understanding this, write down the six letters and then try to understand what $x\rightarrow y$ really means.). Now what you can do is continue on this same pattern like Solution 2 does, and then use simple symmetry to figure out the double counted cases. However, the purpose of this solution is to bash out the double counted cases, so we will do exactly that.
One quick thing though. We have a double counted case with the $3\rightarrow3,$ as choosing ABC and DEF is the same thing as choosing DEF and then ABC. There are $\frac{\binom{6}{3}}{2} = 10$ cases of this.
For computing $4\rightarrow4\mid5\mid6,$ we use the same process as before. We have $\binom{6}{4}\cdot(3+4+1)$ (Note, the $3$ comes from $\frac{\binom{4}{2}}{2}$ ), and for $5\rightarrow5\mid6$ we have $\binom{6}{5}\cdot\left(\frac{5}{2}+1\right),$ and then for $6\rightarrow6$ we just have $\binom{6}{6}$ (there is no double counted case since ABCDEF, ABCDEF is only counted once).
Summing case by case, we have $1+12+60+150+120+21+1 = \boxed{365}.$ | 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 $1,2,3\dots,1993\,$ are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as $1993$
AIME 1993 Problem 9.png | 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 $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{2000}$ . Therefore, one of $1993 - n$ or $1994 + n$ is odd, and each of them must be a multiple of $125$ or $16$
For $1993 - n$ to be a multiple of $125$ and $1994 + n$ to be a multiple of $16$ $n \equiv 118 \pmod {125}$ and $n\equiv 6 \pmod {16}$ . The smallest $n$ for this case is $118$
In order for $1993 - n$ to be a multiple of $16$ and $1994 + n$ to be a multiple of $125$ $n\equiv 9\pmod{16}$ and $n\equiv 6\pmod{125}$ . The smallest $n$ for this case is larger than $118$ , so $\boxed{118}$ is our answer. | 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 $1,2,3\dots,1993\,$ are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as $1993$
AIME 1993 Problem 9.png | 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 have different parities, the even one must be divisible by $32$ . As $(1993 - n)+(1994 + n)\equiv 2\pmod{5}$ , one of them is divisible by $5$ , which indicates it's divisible by $125$
Which leads to four different cases: $1993-n\equiv 0\pmod{4000}$ $1994+n\equiv 0\pmod{4000}$ $1993-n\equiv 0\pmod{32}$ and $1994+n\equiv 0\pmod{125}$ $1993-n\equiv 0\pmod{125}$ and $1994+n\equiv 0\pmod{32}$ . Which leads to $n\equiv 1993,2006,3881$ and $118\pmod{4000}$ respectively, and only $n=118$ satisfied.Therefore answer is $\boxed{118}$ .(by ZJY) | 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 and $t=20$ equilateral triangles yielding a total of $t+p=F=32$ faces. In each vertex, $T=2$ triangles and $P=2$ pentagons are concurrent. Now, the number of edges $E$ can be obtained if we count the number of sides that each triangle and pentagon contributes: $E=\frac{3t+5p}{2}$ , (the factor $2$ in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, $E=60$ . Finally, using Euler's formula we have $V=E-30=30$
In summary, the solution to the problem is $100P+10T+V=\boxed{250}$ | 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, $\frac{VT}3+\frac{VP}5=32\implies V(5T+3P)=480$ . Also, at every vertex $V$ , there are $T+P$ edges coming out from that vertex (one way to see this is to imagine the leftmost segment of each triangle and pentagon that is connected to the given vertex, and note that it includes every one of the edges exactly once), so $\frac{V(T+P)}2=E\implies V(T+P)=2E\implies V(5T+5P)=10E$ , and subtracting the other equation involving the vertices from this gives $2VP=10E-480\implies VP=5E-240=5(V+30)-240=5V-90$ $\implies V(5-P)=90$ .
Since $V|480$ from the first vertex-related observation and $P>0\implies5-P<5$ , and it quickly follows that $V=30\implies E=60\implies P=2\implies T=2\implies100P+10T+V=\boxed{250}$ | 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 $E = \frac{3t+5p}{2}$ and $V = \frac{3t}{T}=\frac{5p}{P}.$
Now, we do casework on what $P$ is.
Case 1: $P = 2$ Notice that we must have $t$ and $p$ integral. Trying $T = 1, 2$ yields a solution with $t=2.$ Trying other cases of $P$ and $T$ yields no solutions. Therefore, $T=2, P=2$ and after solving for $t, p$ we get $V=30.$ Finally, we have $100P+10T+V = \boxed{250}$ | 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 vertices and thus counted twice. Therefore, we have that $E = \frac{4V}{2} = 2V$ . Plugging this in, we have that $V=30$ and so our answer is $200+20+30 = \boxed{250}$ | 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 repeatedly and applying the rule: If the die shows label $L\,$ , where $L \in \{A, B, C\}$ , and $P_n\,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $\overline{P_n L}$ . Given that $P_7 = (14,92)\,$ , what is $k + m\,$ | 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 \[P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\bmod{560},\ 2y_n\bmod{420})\] Then $P_7=(14,92)$ , so $x_7=14$ and $y_7=92$ , and we get \[\begin{array}{c||ccccccc} n & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \hline\hline x_n & 14 & 28 & 56 & 112 & 224 & 448 & 336 \\ \hline y_n & 92 & 184 & 368 & 316 & 212 & 4 & 8 \end{array}\]
So the answer is $\boxed{344}$ | 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 repeatedly and applying the rule: If the die shows label $L\,$ , where $L \in \{A, B, C\}$ , and $P_n\,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $\overline{P_n L}$ . Given that $P_7 = (14,92)\,$ , what is $k + m\,$ | 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 right to the left, we have the following information (remembering that if a point must be $(0,0)$ , we can just ignore it!):
for $\frac {L_6}2,\frac {L_5}4$ , since all addends are nonnegative, a non- $(0,0)$ value will result in a $x$ or $y$ value greater than $14$ or $92$ , respectively, and we can ignore them,
for $\frac {L_4}8,\frac {L_3}{16},\frac {L_2}{32}$ in a similar way, $(0,0)$ and $(0,420)$ are the only possibilities,
and for $\frac {L_1}{64}$ , all three work.
Also, to be in the triangle, $0\le k\le560$ and $0\le m\le420$
Since $L_1$ is the only point that can possibly influence the $x$ coordinate other than $P_1$ , we look at that first.
If $L_1 = (0,0)$ , then $k = 2^6\cdot14 = 64\cdot14 > 40\cdot14 = 560$
so it can only be that $L_1 = (560,0)$ , and $k + 560 = 2^6\cdot14$
$\implies k = 64\cdot14 - 40\cdot14 = 24\cdot14 = 6\cdot56 = 336$
Now, considering the $y$ coordinate, note that if any of $L_2,L_3,L_4$ are $(0,0)$ $L_2$ would influence the least, so we test that),
then $\frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}8 < \frac {420}{16} + \frac {420}8 = 79\pm\epsilon < 80$
which would mean that $P_1 > 2^6\cdot(92 - 80) = 64\cdot12 > 42\cdot10 = 420\ge m$ , so $L_2,L_3,L_4 = (0,420)$
and now $\frac {P_1}{64} + \frac {420}{2^5} + \frac {420}{2^4} + \frac {420}{2^3} = 92$
$\implies P_1$
$= 64\cdot92 - 420(2 + 4 + 8)$
$= 64\cdot92 - 420\cdot14= 64(100 - 8) - 14^2\cdot30$
$= 6400 - 512 - (200 - 4)\cdot30$
$= 6400 - 512 - 6000 + 120$
$= - 112 + 120$
$= 8$
and finally, $k + m = 336 + 8 = \boxed{344}$ | 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 Kenny, they are 200 feet apart. Let $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | 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^2+y^2\end{align}. When they see each other again, the line connecting the two points will be tangent to the circle at the point $(x,y).$ Since the radius is perpendicular to the tangent we get \[-\frac{x}{y}=-\frac{100}{t}\] or $xt=100y.$ Now substitute \[y= \frac{xt}{100}\] into $(2)$ and get \[x=\frac{5000}{\sqrt{100^2+t^2}}.\] Now substitute this and \[y=\frac{xt}{100}\] into $(1)$ and solve for $t$ to get \[t=\frac{160}{3}.\] Finally, the sum of the numerator and denominator is $160+3=\boxed{163}.$ | 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 Kenny, they are 200 feet apart. Let $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | 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 statement, $AB=3t$ , and $CD=t$ . Since $\Delta ABX \sim \Delta CDX$ $CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100$
Since $PC=100$ $PX=200$ . So, $\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}$
Since circle $O$ is tangent to $BX$ and $AX$ $OX$ is the angle bisector of $\angle BXA$
Thus, $\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}$
Therefore, $t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}$ , and the answer is $\boxed{163}$ | 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 Kenny, they are 200 feet apart. Let $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | [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--Q--R--cycle); draw(Circle(O,50)); label("$A$",A,SW); label("$B$",B,NNW); label("$C$",(200,-205),S); label("$D$",D,NE); label("$P$",(100,-205),S); label("$Q$",Q,NE); label("$O$",O,SW); label("$R$",R,NE); label("$S$",S,W); [/asy]
Let $t$ be the time they walk. Then $CD=t$ and $AB=3t$
Draw a line from point $O$ to $Q$ such that $OQ$ is perpendicular to $BD$ . Further, draw a line passing through points $O$ and $P$ , so $OP$ is parallel to $AB$ and $CD$ and is midway between those two lines. Then $PR=\dfrac{AB+CD}{2}=\dfrac{3t+t}{2}=2t$ . Draw another line passing through point $D$ and parallel to $AC$ , and call the point of intersection of this line with $AB$ as $S$ . Then $SB=AB-AS=3t-t=2t$
We see that $m\angle SBD=m\angle ORQ$ since they are corresponding angles, and thus by angle-angle similarity, $\triangle QOR\sim\triangle SDB$
Then \begin{align*} \dfrac{OQ}{DS}=\dfrac{RO}{BD}&\implies\dfrac{50}{200}=\dfrac{RO}{\sqrt{200^2+4t^2}}\\ &\implies RO=\dfrac{1}{4}\left(\sqrt{200^2+4t^2}\right)\\ &\implies RO=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right) \end{align*}
And we obtain \begin{align*} PR-OP&=RO\\ 2t-50&=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)\\ 4t-100&=\sqrt{100^2+t^2}\\ (4t-100)^2&=\left(\sqrt{100^2+t^2}\right)^2\\ 16t^2-800t+100^2&=t^2+100^2\\ 15t^2&=800t\\ t&=\dfrac{800}{15} \end{align*}
so we have $t=\frac{160}{3}$ , and our answer is thus $160+3=\boxed{163}$ | 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 has the form $\sqrt{N}\,$ , for a positive integer $N\,$ . Find $N\,$ | 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 not the same rectangle as the big one. Let the four vertices of this rectangle be $A(4,y)$ $B(-x,3)$ $C(-4,-y)$ and $D(x,-3)$ for nonnegative $x,y$ . Then this is a rectangle, so $OA=OB$ , or $16+y^2=9+x^2$ , so $x^2=y^2+7$
Reflect $D$ across the side of the rectangle containing $C$ to $D'(-8-x,-3)$ . Then $BD'=\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10$ is constant, and the perimeter of the rectangle is equal to $2(BC+CD')$ . The midpoint of $\overline{BD'}$ is $(-4-x,0)$ , and since $-4>-4-x$ and $-y\le0$ $C$ always lies below $\overline{BD'}$
If $y$ is positive, it can be decreased to $y'<y$ . This causes $x$ to decrease as well, to $x'$ , where $x'^2=y'^2+7$ and $x'$ is still positive. If $B$ and $D'$ are held in place as everything else moves, then $C$ moves $(y-y')$ units up and $(x-x')$ units left to $C'$ , which must lie within $\triangle BCD'$ . Then we must have $BC'+C'D'<BC+CD'$ , and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with $y=0$ , so $x=\sqrt7$
By the distance formula, this minimum perimeter is \[2\left(\sqrt{(4-\sqrt7)^2+3^2}+\sqrt{(4+\sqrt7)^2+3^2}\right)=4\left(\sqrt{8-2\sqrt7}+\sqrt{8+2\sqrt7}\right)\] \[=4(\sqrt7-1+\sqrt7+1)=8\sqrt7=\sqrt{448}.\] Therefore $N$ would equal $\boxed{448}.$ | 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 has the form $\sqrt{N}\,$ , for a positive integer $N\,$ . Find $N\,$ | 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 length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let $a$ and $b$ be the sides of the rectangle. Then $ab = 3(8) = 24$ since both are twice the area of the same right triangle, and $a^2+b^2 = 64$ . So $(a+b)^2 = 64+2(24) = 112$ , so $2(a+b) = \sqrt{\boxed{448}$ | 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 relatively prime integers. Find $m + n\,$ | [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((4.29843788128,1.29843788128),1.29843788128)); pair [] y=intersectionpoints(C--H,circle((4.29843788128,1.29843788128),1.29843788128)); dot(y[0]); label("$R$",y[0],NE); label("$1993$",(1.5,2),NW); label("$1994$",(5.5,2),NE); label("$1995$",(4,0),S); [/asy]
From the Pythagorean Theorem $AH^2+CH^2=1994^2$ , and $(1995-AH)^2+CH^2=1993^2$
Subtracting those two equations yields $AH^2-(1995-AH)^2=3987$
After simplification, we see that $2*1995AH-1995^2=3987$ , or $AH=\frac{1995}{2}+\frac{3987}{2*1995}$
Note that $AH+BH=1995$
Therefore we have that $BH=\frac{1995}{2}-\frac{3987}{2*1995}$
Therefore $AH-BH=\frac{3987}{1995}$
Now note that $RS=|HR-HS|$ $RH=\frac{AH+CH-AC}{2}$ , and $HS=\frac{CH+BH-BC}{2}$
Therefore we have $RS=\left| \frac{AH+CH-AC-CH-BH+BC}{2} \right|=\frac{|AH-BH-1994+1993|}{2}$
Plugging in $AH-BH$ and simplifying, we have $RS=\frac{1992}{1995*2}=\frac{332}{665} \rightarrow 332+665=\boxed{997}$ | 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, $1+\frac{19}{30}=\frac{49}{30}.$ Following this pattern, our answer is $4(10)+8(1+2+3+\cdots+9)=\boxed{400}.$ | 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: $1,2,3,4,5,6,7,8,9.$ Note that each digit may be present or may not be present. Hence, there are $2^9=512$ potential ascending numbers, one for each subset of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$
However, we've counted one-digit numbers and the empty set , so we must subtract them off to get our answer, $512-10=\boxed{502}.$ | 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 than $.503$ . What's the largest number of matches she could've won before the weekend began? | 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 than $.503$ . What's the largest number of matches she could've won before the weekend began? | 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 $(n+3)/(2n+4).$ This means that \[\frac{n+3}{2n+4} > .503 = \frac{503}{1000}.\] Cross-multiplying, we get $1000(n+3) > 503(2n+4),$ which is equivalent to $n < \frac{988}{6} = 164.\overline{6}.$ Since $n$ must be an integer, the largest possible value for $n$ is $\boxed{164}.$ | 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 many different numerators are required? | 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 $x=0.\overline{abc}$
$\Rightarrow 1000x=abc.\overline{abc}$
$\Rightarrow 999x=1000x-x=abc$
$\Rightarrow x=\frac{abc}{999}$
If $abc$ is not divisible by $3$ or $37$ , then this is in lowest terms. Let us consider the other multiples: $333$ multiples of $3$ $27$ of $37$ , and $9$ of both $3$ and $37$ , so $999-333-27+9 = 648$ , which is the amount that are neither. The $12$ numbers that are multiples of $81$ reduce to multiples of $3$ . We have to count these since it will reduce to a multiple of $3$ which we have removed from $999$ , but, this cannot be removed since the numerator cannot cancel the $3$ .There aren't any numbers which are multiples of $37^2$ , so we can't get numerators which are multiples of $37$ . Therefore $648 + 12 = \boxed{660}$ | 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. \[\begin{array}{c|c|c|c|c|c|c} & & & & & & \\ [-2.5ex] \textbf{Case} & \textbf{Thousands} & \textbf{Hundreds} & \textbf{Tens} & \textbf{Ones} & \textbf{Conditions for No Carrying} & \boldsymbol{\#}\textbf{ of Ordered Triples} \\ [0.5ex] \hline & & & & & & \\ [-2ex] 0\leq C\leq 8 & 1 & A & B & C+1 & 0\leq A,B,C\leq 4 & 5^3 \\ 0\leq B\leq 8; \ C=9 & 1 & A & B+1 & 0 & 0\leq A,B\leq 4; \ C=9 & 5^2 \\ 0\leq A\leq 8; \ B=C=9 & 1 & A+1 & 0 & 0 & 0\leq A\leq 4; \ B=C=9 & 5 \\ A=B=C=9 & 2 & 0 & 0 & 0 & A=B=C=9 & 1 \end{array}\] Together, the answer is $5^3+5^2+5+1=\boxed{156}.$ | 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 $c \in \{5, 6, 7, 8\}$ , there obviously must be a carry. Consider $c = 9$ $a, b \in \{0, 1, 2, 3, 4\}$ have no carry. This gives $5^2=25$ possible solutions. Considering $b = 9$ $a \in \{0, 1, 2, 3, 4, 9\}$ have no carry. Thus, the solution is $125 + 25 + 6=\boxed{156}$ | 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 $c \in \{0, 1, 2, 3, 4 ,5\}$ , then adding $(1abc) + (1abc - 1)$ will require no carrying. We have $5$ possibilities for the value of $a$ $5$ for $b$ , and $6$ for $c$ , giving a total of $(5)(5)(6) = 150$ , but we are not done yet.
We now have to consider the cases where $b,c = 0$ , specifically when $1abc \in \{1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000\}$ . We can see that $1100, 1200, 1300, 1400, 1500$ , and $2000$ all work, giving a grand total of $150 + 6 = \boxed{156}$ ordered pairs. | 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 roots at $n = 19$ and $n = 92$
Thus, $a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}$ | 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 $\Delta a_1 =a_2 -a_1$ $a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}$
$a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153$
$a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095$
Solving, $a_1=\boxed{819}$ | 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.
$\begin{array}{rcl} a_3+a_1-2a_2&=&1\\ a_4+a_2-2a_3&=&1\\ &\vdots\\ a_k + a_{k-2} - 2a_{k-1} &= &1\\ a_{k+1} + a_{k-1} - 2a_k &=& 1.\\ \end{array}$
Adding the above $k-1$ equations we find that
\[(a_{k+1} - a_k) = k-1 + (a_2-a_1).\tag{1}\]
We can sum equation $(1)$ from $k=1$ to $18$ , finding \[18(a_1-a_2) - a_1 = 153.\tag{2}\]
We can also sum equation $(1)$ from $k=1$ to $91$ , finding \[91(a_1-a_2) - a_1 = 4095.\tag{3}\] Finally, $18\cdot (3) - 91\cdot(2)$ gives $a_1=\boxed{819}$ | 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 $f(19) = 0$ and $f(20) = -72a$ and $f(18) = 74a.$ Therefore applying finite differences once yields the sequence $-74a,-72a$ and then applying finite differences one more time yields $2a$ so $a =\frac{1}{2}.$ Therefore $f(1) = 9 \cdot 91 = \boxed{819}.$ | 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 $n=19,92,$ we find that \begin{align*} 0=&18b+\frac{17\cdot18}{2}-17a\\ 0=&91b+\frac{90\cdot91}{2}-90a\\ b=&\frac{17a-\frac{17\cdot18}{2}}{18}=\frac{90a-\frac{90\cdot91}{2}}{91}. \end{align*} Now, cross multiplying, we find that \begin{align*} 91\left(17a-\frac{17\cdot18}{2}\right)&=18\left(90a-\frac{90\cdot91}{2}\right)\\ 1547a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\ 73a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\ a=\frac{18\cdot91}{2}=9\cdot91=\boxed{819} | 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 positive integers, find $m+n^{}_{}$ | 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 \[\frac{xy-70}{y(92-x)-50} = \frac{XD}{XC} = \frac{ED}{EC}=\frac{AP}{PB} = \frac{x}{92-x},\] which we can rearrange, expand and cancel to get $120x=70\cdot 92,$ hence $AP=x=\frac{161}{3}$ . This gives us a final answer of $161+3=\boxed{164}$ | 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 positive integers, find $m+n^{}_{}$ | 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 positive integers, find $m+n^{}_{}$ | 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.)
$[CPD] = \frac{1}{2} \cdot 19 \cdot h$
$[PBA] = \frac{1}{2} \cdot 70 \cdot r$
$[BPC] + [CPD] + [PBA] = 60r + \frac{19h}{2} = [ABCD] = \frac{(19+92)h}{2}$
$60r = 46h$
$r = \frac{23h}{30}$
From Solution 1 above, $\frac{h}{70} = \frac{r}{x}$
Substituting $r = \frac{23h}{30}$ , we find $x = \frac{161}{3}$ , hence the answer is $\boxed{164}$ | 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 $0\leq a,b \leq \frac{a^2+b^2}{40}$ , which leads to: \[(a-20)^2+b^2\geq 20^2\] \[a^2+(b-20)^2\geq 20^2\]
We graph them:
To find the area outside the two circles but inside the square, we want to find the unique area of the two circles. We can do this by adding the area of the two circles and then subtracting out their overlap. There are two methods of finding the area of overlap:
1. Consider that the area is just the quarter-circle with radius $20$ minus an isosceles right triangle with base length $20$ , and then doubled (to consider the entire overlapped area)
2. Consider that the circles can be converted into polar coordinates, and their equations are $r = 40sin\theta$ and $r = 40cos\theta$ . Using calculus with the appropriate bounds, we can compute the overlapped area.
Using either method, we compute the overlapped area to be $200\pi + 400$ , and so the area of the intersection of those three graphs is $40^2-(200\pi + 400) \Rightarrow 1200 - 200\pi \approx 571.68$
$\boxed{572}$ | 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 resulting line is reflected in $l_2^{}$ . Let $R^{(1)}(l)=R(l)^{}_{}$ and $R^{(n)}(l)^{}_{}=R\left(R^{(n-1)}(l)\right)$ . Given that $l^{}_{}$ is the line $y=\frac{19}{92}x^{}_{}$ , find the smallest positive integer $m^{}_{}$ for which $R^{(m)}(l)=l^{}_{}$ | 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, $l'$ makes an angle of $\frac{\pi}{70}-\left(\theta-\frac{\pi}{70}\right) = \frac{\pi}{35}-\theta$ with the positive $x$ -axis.
Similarly, since the angle between $l'$ and $l_2$ is $\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}$ , the angle between $l''$ and the positive $x$ -axis is $\frac{\pi}{54}-\left(\left(\frac{\pi}{35}-\theta\right)-\frac{\pi}{54}\right) = \frac{\pi}{27}-\frac{\pi}{35}+\theta = \frac{8\pi}{945} + \theta$
Thus, $R(l)$ makes an $\frac{8\pi}{945} + \theta$ angle with the positive $x$ -axis. So $R^{(n)}(l)$ makes an $\frac{8n\pi}{945} + \theta$ angle with the positive $x$ -axis.
Therefore, $R^{(m)}(l)=l$ iff $\frac{8m\pi}{945}$ is an integral multiple of $\pi$ . Thus, $8m \equiv 0\pmod{945}$ . Since $\gcd(8,945)=1$ $m \equiv 0 \pmod{945}$ , so the smallest positive integer $m$ is $\boxed{945}$ | 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 example, the bite determined by the shaded square in the diagram would remove the shaded square and the four squares marked by $\times.$ (The squares with two or more dotted edges have been removed form the original board in previous moves.)
AIME 1992 Problem 12.png
The object of the game is to make one's opponent take the last bite. The diagram shows one of the many subsets of the set of 35 unit squares that can occur during the game of Chomp. How many different subsets are there in all? Include the full board and empty board in your count. | 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 down, and 7 that go across, with the shape on the right "carved" out by the path a possible subset.
Therefore, the total number of such paths is $\binom{12}{5}=\boxed{792}$ | 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 example, the bite determined by the shaded square in the diagram would remove the shaded square and the four squares marked by $\times.$ (The squares with two or more dotted edges have been removed form the original board in previous moves.)
AIME 1992 Problem 12.png
The object of the game is to make one's opponent take the last bite. The diagram shows one of the many subsets of the set of 35 unit squares that can occur during the game of Chomp. How many different subsets are there in all? Include the full board and empty board in your count. | 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 be eaten, so $\{a_i\}$ is nondecreasing. We can thus write $0 \leq a_1 \leq a_2 \leq \dots \leq a_5 \leq 7$ ; the number of sequences $\{a_i\}$ satisfying this inequality may be found by Stars and Bars to be $\binom{7+6-1}{6-1} = \boxed{792}$ | 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 $-a^2 -\frac{3200}{9}a +1600 = b^2$
To maximize $b$ , we want to maximize $b^2$ . So if we can write: $b^2=-(a+n)^2+m$ , then $m$ is the maximum value of $b^2$ (this follows directly from the trivial inequality , because if ${x^2 \ge 0}$ then plugging in $a+n$ for $x$ gives us ${(a+n)^2 \ge 0}$ ).
$b^2=-a^2 -\frac{3200}{9}a +1600=-\left(a +\frac{1600}{9}\right)^2 +1600+\left(\frac{1600}{9}\right)^2$
$\Rightarrow b\le\sqrt{1600+\left(\frac{1600}{9}\right)^2}=40\sqrt{1+\frac{1600}{81}}=\frac{40}{9}\sqrt{1681}=\frac{40\cdot 41}{9}$
Then the area is $9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}$ | 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, $AP/AQ=BP/BQ\implies \dfrac{\frac{41}{9}}{BQ+9}=\dfrac{\frac{40}{9}}{BQ}\implies BQ=360$ . Finally, this means that the radius of $\Gamma$ is $\dfrac{360+\frac{40}{9}}{2}=180+\dfrac{20}{9}$
Since the area is maximized when the altitude to $AB$ is maximized, clearly we want the last vertex to be the highest point of $\Gamma$ , which just makes the altitude have length $180+\dfrac{20}{9}$ . Thus, the area of the triangle is $\dfrac{9\cdot \left(180+\frac{20}{9}\right)}{2}=\boxed{820}$ | 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 area of the triangle by finding when the derivative of the expression inside the square root equals 0.
We have that $-324x^3+13124x=0$ , so $x=\frac{\sqrt{3281}}{9}$
Plugging this into the expression, we have that the area is $\boxed{820}$ | 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 $\frac{9}{4} *\frac{9}{4}*\frac{3280}{9} = \boxed{820}$ | 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}\right) \left(\frac{BC'}{C'A}+\frac{BA'}{A'C}\right) \left(\frac{CB'}{B'A}+\frac{CA'}{A'B}\right)\\ &=\underbrace{\frac{AB'\cdot CA'\cdot BC'}{B'C\cdot A'B\cdot C'A}}_{\text{Ceva}} + \underbrace{\frac{AC'\cdot BA'\cdot CB'}{C'B\cdot A'C\cdot B'A}}_{\text{Ceva}} + \underbrace{\frac{AB'}{B'C} + \frac{AC'}{C'B}}_{\text{Gergonne}} + \underbrace{\frac{BA'}{A'C} + \frac{BC'}{C'A}}_{\text{Gergonne}} + \underbrace{\frac{CA'}{A'B} + \frac{CB'}{B'A}}_{\text{Gergonne}}\\ &= 1 + 1 + \underbrace{\frac{AO}{OA'} + \frac{BO}{OB'} + \frac{CO}{OC'}}_{92} = \boxed{94} | 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). In the second case, since all factors of $16$ must be $\le 16$ , no two factors of $16$ can sum greater than $32$ , and so there are no integral solutions for $(x,y)$ . The solution is $5^2 + 11^2 = \boxed{146}$ | 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 $1_{}^{} \le k \le 167$ , draw the segments $\overline {P_kQ_k}$ . Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$ , and then draw the diagonal $\overline {AC}$ . Find the sum of the lengths of the 335 parallel segments drawn. | 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 to find $2\sum\limits_{k=1}^{168} a_k-5$ since we are over counting the diagonal. $2\sum\limits_{k=1}^{168} \frac{5(168-k)}{168}-5 =2\frac{(0+5)\cdot169}{2}-5 =168\cdot5 =\boxed{840}$ | 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 $1_{}^{} \le k \le 167$ , draw the segments $\overline {P_kQ_k}$ . Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$ , and then draw the diagonal $\overline {AC}$ . Find the sum of the lengths of the 335 parallel segments drawn. | 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 the sum of the lengths of all $\overline{P_k Q_k}$ .This is given by the following: \[2 \cdot(\sum_{k=1}^{168} P_kQ_k) + 5\]
\[= 2 \cdot \frac{ 0+5+10+...+835}{168} +5\] Then by the summation formula for the sum of the terms of an arithmetic series, \[= \frac{835 \cdot 168}{168} +5 = 835+5 = \boxed{840}\] | 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 $1_{}^{} \le k \le 167$ , draw the segments $\overline {P_kQ_k}$ . Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$ , and then draw the diagonal $\overline {AC}$ . Find the sum of the lengths of the 335 parallel segments drawn. | 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 $\frac{1}{168}\cdot{5}$ by similar triangles. If you add the two lengths together, it is $\frac{167}{168}\cdot{5} + \frac{1}{168}\cdot{5} = 5.$ There are $\frac{335-1}{2}$ pairs of these segments, for a total of $5+(167)(5)=168(5)=\boxed{840}.$ ~justlearningmathog | 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 which $k_{}^{}$ is $A_k^{}$ the largest? | 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
$\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\log\left[\prod_{j=1}^{k}\frac{(N-j+1)x}{j}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .$
Now, $\log(A_{k}^{})$ keeps increasing with $k_{}^{}$ as long as the arguments $\frac{(N-j+1)x}{j}>1$ in each of the $\log\big[\;\big]$ terms (recall that $\log y_{}^{} <0$ if $0<y_{}^{}<1$ ). Therefore, the integer $k_{}^{}$ that we are looking for must satisfy $k=\Big\lfloor\frac{(N+1)x}{1+x}\Big\rfloor$ , where $\lfloor z_{}^{}\rfloor$ denotes the greatest integer less than or equal to $z_{}^{}$
In summary, substituting $N_{}^{}=1000$ and $x_{}^{}=0.2$ we finally find that $k=\boxed{166}$ | 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 which $k_{}^{}$ is $A_k^{}$ the largest? | 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:
${1000\choose k}>{\frac{1}{5}}\cdot{1000\choose {k+1}}$
We can express these binomial coefficients as factorials.
$\frac{1000!}{(1000-k)!\cdot(k)!}>{\frac{1}{5}}\cdot\frac{1000!}{(1000-k-1)!\cdot{(k+1)!}}$
We note that the $1000!$ can cancel. Also, $(1000-k)!=(1000-k)(1000-k-1)!$ . Similarly, $(k+1)!=(k+1)k!$
Canceling these terms yields,
$\frac{1}{(1000-k)}>{\frac{1}{5}}\cdot\frac{1}{(k+1)}$
Cross multiplying gives:
$5k+5>1000-k \Rightarrow k>165.8$
Therefore, since this identity holds for all values of $k>165.8$ , the largest possible value of $k$ is $\boxed{166}$ | 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 which $k_{}^{}$ is $A_k^{}$ the largest? | 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
denominator or
\[1000-k+1>5k\] \[k<166.8\]
We conclude that the answer is $\boxed{166}$ | 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 which $k_{}^{}$ is $A_k^{}$ the largest? | 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 equals $\frac{1000*999*998*997*996}{5*4*3*2*1}$ , and if we have k+1, then it's just going to be equivalent to multiplying this fraction by $\frac{995}{6}$ . Notice that this fraction's numerator plus denominator is equal to $1001$ . Calling the numerator x and the denominator y, we get that \[x+y = 1001\] \[\frac{x}{y} > 5\] \[x>5y\] \[6y<1001\] \[y<166.83333\] and since the denominator is what we are specifically choosing, or in other words what k is, we conclude k = $\boxed{166}$ | 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 which $k_{}^{}$ is $A_k^{}$ the largest? | 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
\[{1000 \choose m+1} = \frac{1000!}{(m+1)!(1000-m-1)!} = \frac{1000 \cdots (1001-m)(1000-m)}{(m+1)!}\] and that \[{1000 \choose m} = \frac{1000!}{m!(1000-m)!} = \frac{1000 \cdots (1001-m)}{m!}\]
\[\Longrightarrow {1000 \choose m} \cdot \frac{1000-m}{m+1}\]
So now, we must find out for which $m$ is $1000-m< 5 \cdot (m+1)$ (when this happens the numerator is less than the denominator for the fraction $\frac{1000-m}{m+1}$ so then we will have $A_m > A_{m+1}$ ) Then we find that for all $m > 165.833333$ , the above inequality ( $1000-m< 5 \cdot (m+1)$ ) holds, so then $m=165$ so $k=m+1= \boxed{166}$ | 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 value; up to $x = 32$ it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of $x$ ) of the sine curve and another curve that is $< 1$ , so there are $\frac{32}{2} \cdot 10 - 6 = 160 - 6 = 154$ values (the subtraction of 6 since all the “intersections” when $x < 1$ must be disregarded). When $y = 0$ , there is exactly $1$ touching point between the two functions: $\left(\frac{1}{5},0\right)$ . When $y < 0$ or $x < 1$ , we can count $4$ more solutions. The solution is $154 + 1 + 4 = \boxed{159}$ | 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 out. We then compute: $32 \cdot \frac{5}{2} \cdot 2 - 1 = \boxed{159}$ | 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 will be between $0 < \frac{a}{b} < 1$ . Therefore, the solution is $\frac{2^8}{2} = \boxed{128}$ | 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 $35$ $8$ must take on $35$ of the values, with $7$ being the value of the remaining $73 - 35 = 38$ numbers. The 39th number is $\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor$ , which is also the first term of this sequence with a value of $8$ , so $8 \le r + \frac{57}{100} < 8.01$ . Solving shows that $\frac{743}{100} \le r < \frac{744}{100}$ , so $743\le 100r < 744$ , and $\lfloor 100r \rfloor = \boxed{743}$ | 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, $\lfloor r\rfloor \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7$ and $\lfloor r+\frac{92}{100}\rfloor \ge ...\ge \lfloor r+1\rfloor \ge 8$ . We can see that $\lfloor r\rfloor +1=\lfloor r+1\rfloor$ . Because $\lfloor r\rfloor$ is at most 7, and $\lfloor r+1\rfloor$ is at least 8, we can clearly see their values are $7$ and $8$ respectively.
So, $\lfloor r\rfloor = ... = \lfloor r+\frac{18}{100}\rfloor = 7$ , and $\lfloor r+\frac{92}{100}\rfloor = ...= \lfloor r+1\rfloor = 8$ . Since there are 19 terms in the former equation and 8 terms in the latter, our answer is $\lfloor nx\rfloor = 546+19\cdot 7+8\cdot 8=\boxed{743}$ | 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: | $x=\sqrt{19}+\underbrace{\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}}_{x}$
$x=\sqrt{19}+\frac{91}{x}$
$x^2=x\sqrt{19}+91$
$x^2-x\sqrt{19}-91 = 0$
$\left.\begin{array}{l}x_1=\frac{\sqrt{19}+\sqrt{383}}{2}\\\\x_2=\frac{\sqrt{19}-\sqrt{383}}{2}\end{array}\right\}A=|x_1|+|x_2|\Rightarrow\sqrt{383}$
$A^2=\boxed{383}$ | 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 have two roots, they must be the same roots as the quadratic $f(x)=x$
The given finite expansion can then be easily seen to reduce to the quadratic equation $x_{}^{2}-\sqrt{19}x-91=0$ . The solutions are $x_{\pm}^{}=$ $\frac{\sqrt{19}\pm\sqrt{383}}{2}$ . Therefore, $A_{}^{}=\vert x_{+}\vert+\vert x_{-}\vert=\sqrt{383}$ . We conclude that $A_{}^{2}=\boxed{383}$ | 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 $b$ ): $b^2 = a^2 - 24a$ Completing the square , we get
\[(a - 12)^2 = b^2 + 144\]
Which implies that $b^2 + 144$ is a perfect square also (let this be $c^2$ ). Then
\[c^2 - b^2 = 144 \Longrightarrow (c+b)(c-b) = 144\]
The pairs of factors of $144$ are $(\pm1,\pm144),( \pm 2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)$ ; since $c$ is the average of each respective pair and is also an integer, the pairs that work must have the same parity . Thus we get $\boxed{10}$ pairs (counting positive and negative) of factors that work, and substituting them backwards show that they all work. | 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: $( - 36, - 1),( - 18, - 2),( - 12, - 3),( - 9, - 4),( - 6, - 6),(1,36),(2,18),(3,12),(4,9),(6,6)$ $\Rightarrow \boxed{10} a$ | 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 $\sin^2x+\cos^2x=1$ $\sin x=\sqrt{1-\cos^2x}$ So we can write \[\sin x=\sqrt{1-\frac{4a^2b^2}{(a^2+b^2)^2}}\] Upon simplification, we get \[\sin x=\frac{a^2-b^2}{a^2+b^2}\] We are asked to find $1/\sin x+\cos x/\sin x$ so we can write that as \[\csc x+\cot x=\frac{1}{\sin x}+\frac{\cos x}{\sin x}\] \[\csc x+\cot x=\frac{a^2+b^2}{a^2-b^2}+\frac{2ab}{a^2+b^2}\frac{a^2+b^2}{a^2-b^2}\] \[\csc x+\cot x=\frac{a^2+b^2+2ab}{a^2-b^2}\] \[\csc x+\cot x=\frac{(a+b)^2}{(a-b)(a+b)}\] \[\csc x+\cot x=\frac{a+b}{a-b}\] Now using the fact that $a=22$ and $b=7$ yields, \[\csc x+\cot x=\frac{29}{15}=\frac{p}{q}\] so $p+q=15+29=\boxed{44}$ | 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 $a^{}_{}$ . However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let $S_a^{}$ be the three-letter string received when $aaa^{}_{}$ is transmitted and let $S_b^{}$ be the three-letter string received when $bbb^{}_{}$ is transmitted. Let $p$ be the probability that $S_a^{}$ comes before $S_b^{}$ in alphabetical order. When $p$ is written as a fraction in lowest terms , what is its numerator | 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, $S_b = \sum_{n=1}^{b} P_b$ ): \[\begin{array}{|r||r|r|r|} \hline \text{String}&P_a&P_b&S_b\\ \hline aaa & 8 & 1 & 1 \\ aab & 4 & 2 & 3 \\ aba & 4 & 2 & 5 \\ abb & 2 & 4 & 9 \\ baa & 4 & 2 & 11 \\ bab & 2 & 4 & 15 \\ bba & 2 & 4 & 19 \\ bbb & 1 & 8 & 27 \\ \hline \end{array}\]
The probability is $p=\sum P_a \cdot (27 - S_b)$ , so the answer turns out to be $\frac{8\cdot 26 + 4 \cdot 24 \ldots 2 \cdot 8 + 1 \cdot 0}{27^2} = \frac{532}{729}$ , and the solution is $\boxed{532}$ | 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 $a^{}_{}$ . However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let $S_a^{}$ be the three-letter string received when $aaa^{}_{}$ is transmitted and let $S_b^{}$ be the three-letter string received when $bbb^{}_{}$ is transmitted. Let $p$ be the probability that $S_a^{}$ comes before $S_b^{}$ in alphabetical order. When $p$ is written as a fraction in lowest terms , what is its numerator | 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 for the second letters of the strings. But you have to multiply the $4/9$ chance that $S(a,2)$ comes before $S(b,2)$ as there is a $4/9$ chance we will get to this step.
Similarly, if $S(a,2)=S(b,2)$ , then there is a $(4/9)^3$ chance that we will get to comparing the third letters and that $S(a)$ comes before $S(b)$
So we have $p=(4/9)+(4/9)^2+(4/9)^3=4/9+16/81+64/729=532/729$ . Therefore, the answer is $\boxed{532}$ | 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 $a^{}_{}$ . However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let $S_a^{}$ be the three-letter string received when $aaa^{}_{}$ is transmitted and let $S_b^{}$ be the three-letter string received when $bbb^{}_{}$ is transmitted. Let $p$ be the probability that $S_a^{}$ comes before $S_b^{}$ in alphabetical order. When $p$ is written as a fraction in lowest terms , what is its numerator | 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: $p_n=\frac23\cdot\frac23+2\cdot\frac23\cdot\frac13p_{n-1}=\frac49+\frac49p_{n-1}$ . Clearly, $p_0=0\implies p_1=\frac49\implies p_2=\frac{52}{81}\implies p_3=\frac{532}{729}$ . Therefore, the answer is $\boxed{532}$ | 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 $a^{}_{}$ . However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let $S_a^{}$ be the three-letter string received when $aaa^{}_{}$ is transmitted and let $S_b^{}$ be the three-letter string received when $bbb^{}_{}$ is transmitted. Let $p$ be the probability that $S_a^{}$ comes before $S_b^{}$ in alphabetical order. When $p$ is written as a fraction in lowest terms , what is its numerator | 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 sequence are the same, then we must now consider these same probabilities for the second letter of each sequence. The probability that when the first two letters of both sequences are the same, that the second letter of $S_a$ is $a$ and that the second letter of $S_b$ is $b$ is $\frac{2}{3}\cdot\frac{2}{3}=\frac{4}{9}$ . Similarly, the probability that when the first two letters of both sequences are the same, that the second set of letters in both sets of sequences are the same is $2\cdot\frac{2}{3}\cdot\frac{1}{3} = \frac{4}{9}$ . Now, if the last case is true then the probability that $S_a$ precedes $S_b$ is $\frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}$ . Therefore the total probability would be: $\frac{4}{9} + \frac{4}{9}\left(\frac{4}{9} + \frac{4}{9}\left(\frac{4}{9}\right)\right) = \frac{4}{9}+\frac{4}{9}\left(\frac{52}{81}\right) = \frac{4}{9} + \frac{208}{729} = \frac{532}{729}$ . Therefore the answer is $\boxed{532}$ | 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 twelve disks can be written in the from $\pi(a-b\sqrt{c})$ , where $a,b,c^{}_{}$ are positive integers and $c^{}_{}$ is not divisible by the square of any prime. Find $a+b+c^{}_{}$
[asy] unitsize(100); draw(Circle((0,0),1)); dot((0,0)); draw((0,0)--(1,0)); label("$1$", (0.5,0), S); for (int i=0; i<12; ++i) { dot((cos(i*pi/6), sin(i*pi/6))); } for (int a=1; a<24; a+=2) { dot(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))); draw(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))--((1/cos(pi/12))*cos((a+2)*pi/12), (1/cos(pi/12))*sin((a+2)*pi/12))); draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12))); }[/asy]
_Diagram by 1-1 is 3_ | 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 to the larger circle at the midpoint of the two centers. Thus, we have essentially have a regular dodecagon whose vertices are the centers of the smaller triangles circumscribed about a circle of radius $1$
We thus know that the apothem of the dodecagon is equal to $1$ . To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote $A, M,$ and $O$ respectively. Notice that $OM=1$ , and that $\triangle OMA$ is a right triangle with hypotenuse $OA$ and $m \angle MOA = 15^\circ$ . Thus $AM = (1) \tan{15^\circ} = 2 - \sqrt {3}$ , which is the radius of one of the circles. The area of one circle is thus $\pi(2 - \sqrt {3})^{2} = \pi (7 - 4 \sqrt {3})$ , so the area of all $12$ circles is $\pi (84 - 48 \sqrt {3})$ , giving an answer of $84 + 48 + 3 = \boxed{135}$ | 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 $QS^{}_{}=40$ . Let $\frac{m}{n}$ , in lowest terms, denote the perimeter of $ABCD^{}_{}$ . Find $m+n^{}_{}$ | 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 for a side of the rhombus; $PQ = \sqrt{15^2 + 20^2} = 25$ . Since the diagonals of a rhombus are perpendicular bisectors , we have that $OP = 15, OQ = 20$ . Also, $\angle POQ = 90^{\circ}$ , so quadrilateral $BPOQ$ is cyclic . By Ptolemy's Theorem $25 \cdot OB = 20 \cdot 15 + 15 \cdot 20 = 600$
By similar logic, we have $APOS$ is a cyclic quadrilateral. Let $AP = x$ $AS = y$ . The Pythagorean Theorem gives us $x^2 + y^2 = 625\quad \mathrm{(1)}$ . Ptolemy’s Theorem gives us $25 \cdot OA = 20x + 15y$ . Since the diagonals of a rectangle are equal, $OA = \frac{1}{2}d = OB$ , and $20x + 15y = 600\quad \mathrm{(2)}$ . Solving for $y$ , we get $y = 40 - \frac 43x$ . Substituting into $\mathrm{(1)}$
\begin{eqnarray*}x^2 + \left(40-\frac 43x\right)^2 &=& 625\\ 5x^2 - 192x + 1755 &=& 0\\ x = \frac{192 \pm \sqrt{192^2 - 4 \cdot 5 \cdot 1755}}{10} &=& 15, \frac{117}{5}\end{eqnarray*}
We reject $15$ because then everything degenerates into squares, but the condition that $PR \neq QS$ gives us a contradiction . Thus $x = \frac{117}{5}$ , and backwards solving gives $y = \frac{44}5$ . The perimeter of $ABCD$ is $2\left(20 + 15 + \frac{117}{5} + \frac{44}5\right) = \frac{672}{5}$ , and $m + n = \boxed{677}$ | 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 $QS^{}_{}=40$ . Let $\frac{m}{n}$ , in lowest terms, denote the perimeter of $ABCD^{}_{}$ . Find $m+n^{}_{}$ | 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 straightforward to find the answer, $2(a+b+35) \Rightarrow \frac{672}{5}$ . Thus, $m+n = \frac{672}{5}\implies\boxed{677}$ | 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 $QS^{}_{}=40$ . Let $\frac{m}{n}$ , in lowest terms, denote the perimeter of $ABCD^{}_{}$ . Find $m+n^{}_{}$ | 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, $\sin\angle APS=\sin\angle SPB=\sin(\angle SPQ+\angle QPB)=\sin\angle SPQ\cos\angle QPB+\sin\angle QPB\cos\angle SPQ=\sin\angle PSR\cos\angle QPB-\sin\angle QPB\cos\angle PSR=\frac{24}{25}\cdot\frac{15}{25}-\frac{20}{25}\cdot\frac{7}{25}=\frac{44}{125}$ . Thus, since $PS=25$ , we get that $AS=\frac{44}{5}$ . Now, by the Pythagorean Theorem, $AP=\frac{117}{5}$
Using the same idea, $\cos\angle RSD=-\cos\angle RSA=-\cos(\angle RSP+\angle PSA)=\sin\angle RSP\sin\angle PSA-\cos\angle RSP\cos\angle PSA=\frac{24}{25}\cdot\frac{117}{125}-\frac{7}{25}\cdot\frac{44}{125}=\frac{4}{5}$ . Thus, since $SR=20$
Now, we can finish. We know $AB=\frac{117}{5}+15=\frac{192}{5}$ . We also know $AD=\frac{44}{5}+20=\frac{144}{5}$ . Thus, our perimeter is $\frac{672}{5}\implies\boxed{677}$ | 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 consistent with this data? | 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
\[\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}.\]
Solving the resulting quadratic equation $r^{2}-rt+t(t-1)/4=0$ , for $r$ in terms of $t$ , one obtains that
\[r=\frac{t\pm\sqrt{t}}{2}\, .\]
Now, since $r$ and $t$ are positive integers, it must be the case that $t=n^{2}$ , with $n\in\mathbb{N}$ . Hence, $r=n(n\pm 1)/2$ would correspond to the general solution. For the present case $t\leq 1991$ , and so one easily finds that $n=44$ is the largest possible integer satisfying the problem conditions.
In summary, the solution is that the maximum number of red socks is $r=\boxed{990}$ | 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 consistent with this data? | 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 $1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)$ $=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2$ $=(r-b)^2=r+b$ , so $r+b$ must be a perfect square $k^2$ . Clearly, $r=\frac{k^2+k}2$ , so the larger $k$ , the larger $r$ $k^2=44^2$ is the largest perfect square below $1991$ , and our answer is $\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}$ | 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 consistent with this data? | 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 $t$ for $r + b$ and cross multiplying, we get $2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b$
Combining terms, we get $b^2 - 2br + r^2 - b - r = 0$
To make this expression factorable, we add $2r$ to both sides, resulting in $(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r$
From this equation, we can test values for the expression $(b - r - 1)(b - r)$ , which is the multiplication of two consecutive integers, until we find the highest value of $b$ or $r$ such that $b + r \leq 1991$
By testing $(b - r - 1) = 43$ and $(b - r) = 44$ , we get that $r = 43(22) = 946$ and $b = 990$ . Testing values one integer higher, we get that $r = 990$ and $b = 1035$ . Since $990 + 1035 = 2025$ is greater than $1991$ , we conclude that $(946, 990)$ is our answer.
Since it doesn't matter whether the number of blue or red socks is $990$ , we take the lower value for $r$ , thus the maximum number of red socks is $r=\boxed{990}$ | 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 consistent with this data? | 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 $(r+b)^2-2rb$ , and remembering that $r+b=t$ , we have $\frac{t^2}{2}-\frac{t}{2}=t^2-2rb-t$ , so $2rb=\frac{t^2}{2}-\frac{t}{2}$ , which equals $r^2+b^2-t$
We now have $r^2+b^2-t=2rb$ , so $r^2-2rb+b^2=t$ and $r-b=\pm\sqrt{t}$ . Adding this to $r+b=t$ , we have $2r=t\pm\sqrt{t}$ . To maximize $r$ , we must use the positive square root and maximize $t$ . The largest possible value of $t$ is the largest perfect square less than 1991, which is $1936=44^2$ . Therefore, $r=\frac{t+\sqrt{t}}{2}=\frac{1936+44}{2}=\boxed{990}$ | 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 consistent with this data? | 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. $(2r-t)^2 = t$
Thus $t$ is a perfect square. And, the higher $t$ is, the higher $r$ will be. So, we should set $t = 44^2 = 1936$
And, we see, $2r-1936 = \pm44$ We will use the positive root, to get that $2r-1936 = 44$ $2r = 1980$ , and $r = \boxed{990}$ | 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 consistent with this data? | 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 thus $r-b\le 44$ as both $r$ and $b$ must be integers and ${44}^2=1936$ is the largest square less than or equal to $1991$ . We now know that the maximum difference between the number of socks is $44$ . Now we return to an earlier equation:
$r^2+b^2-r-b-2rb=0$
$\Rightarrow r^2-(2b+1)r+(b^2-b)=0$
Solving by the Quadratic formula , we have:
$r=\frac{2b+1\pm\sqrt{4b^2+4b+1-4b^2+4b}}{2}$
$\Rightarrow r=\frac{2b+1\pm\sqrt{8b+1}}{2}$
$\Rightarrow r-b=\frac{1\pm\sqrt{8b+1}}{2}$
$\Rightarrow 44\ge r-b=\frac{1\pm\sqrt{8b+1}}{2}$
$\Rightarrow 44\ge\frac{1\pm\sqrt{8b+1}}{2}$
$\Rightarrow \pm\sqrt{8b+1}\le 87$
$\Rightarrow b\le 946$ (Here we use the positive sign to maximize $r$ .)
Thus the optimal case would be $b=946$ as $r$ increases only when $b$ increases. In addition, $\sqrt{8b+1}=87$ as we are using the equality case. We then plug back in:
$r=\frac{2b+1\pm\sqrt{8b+1}}{2}$
$\Rightarrow r=\frac{1893+87}{2}$ (using the established $\sqrt{8b+1}=87$
$\Rightarrow r=\frac{1980}{2}$
$\Rightarrow r=\boxed{990}$ | 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 into the first equation gives $x=105$ , so $x+y+z=105+144+135=\boxed{384}$ | 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 in our sequence less than $529$ . Magically, we want the $500th$ term, so our answer is the biggest non-square and non-cube less than $529$ , which is $\boxed{528}$ | 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 smallest square or cube greater than $500$ . However, if we use $n=8^3=512$ , the number of terms in $T$ will exceed $n-500$ . Therefore, $n=23^2=529$ , and the number of terms in $T$ is $23+8-2=29$ by the Principle of Inclusion-Exclusion , fulfilling our original requirement of $n-T=500$ .
As a result, our answer is $529-1 = \boxed{528}$ | 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. \[S^2= ((52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2})^2\] \[S^2 = (52+6\sqrt{43})^{3} + (52-6\sqrt{43})^{3} - 2((52+6\sqrt{43})(52-6\sqrt{43}))^{3/2}\] After doing the arithmetic (note that the first two terms will have some cancellation and that the last term will simplify quickly using difference of squares), we arrive at $S^2 = 685584$ which gives $S=\boxed{828}$ | 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. \[\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[\left(\left(\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\right)^2+\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}}+\left(\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right)^2\right)\right] =\] \[\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+\left(52^2-\left(36\right)\left(43\right)\right)^{\frac{1}{2}}\right] =\] \[\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+34\right].\] We can simplify the left factor as follows. \[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x\] \[104-2\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x^2\] \[104-68 = x^2\] \[36 = x^2.\] Since $\left(52+6\sqrt{43}\right)^{\frac{1}{2}} > \left(52-6\sqrt{43}\right)^{\frac{1}{2}}$ , we know that $x=6$ , so our final answer is $(6)(138) = \boxed{828}$ | 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{3}{2}}\\ &=(x+y)(x^2-xy+y^2)+2(xy)^{\frac{3}{2}}\\ &=(x+y)((x+y)^2-xy)+2\sqrt{xy}^3\\ &=(x+y)((x+y)^2-\sqrt{xy}^2)+\sqrt{xy}^3\\ &=(x+y)(x+y+\sqrt{xy})(x+y-\sqrt{xy})+2\sqrt{xy}^3\\ &=104((104+34)(104-34)+2\cdot34^3\\ &=685584 \end{align*} Thus $S=\sqrt{685584}=\boxed{828}$ | 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 $(52 + 6\sqrt{43})^{3/2} - (52 + 6\sqrt{43})^{3/2}$ $= \sqrt{52 + 6\sqrt{43}}^3 - \sqrt{52 - 6\sqrt{43}}^3$ $= (\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3.$
From here, we use the formula $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ . Applying them to our problem we get that $(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3 = (27 + 27\sqrt{43} + 9 \cdot 43 + 43\sqrt{43}) - (-27 + 27\sqrt{43} - 9*43 + 43\sqrt{43}).$ We see that all the terms with square roots cancel, leaving us with $2 (27 + 9 \cdot 43) = 2 \cdot 414 = \boxed{828}.$ | 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 $58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs$ . Solving for $r$ , we get $r = \frac{116s}{118 - s}$
$r \ge 0$ and $s \ge 0$ , making the numerator of the fraction positive. To make the denominator positive $s < 118$ ; the largest possible value of $s$ is $117$
This is achievable because the denominator is $1$ , making $r$ a positive number $116 \cdot 117$ and $s = \boxed{117}$ | 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 divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is $s=59$ , which does give a solution: $s=59, r=116$ . Although, the problem asks for $s$ , not $r$ . The only conceivable reasoning behind this is that $r$ is greater than 1000. This prompts us to look into the second case, where $s$ divides $r$ . Make $r = s * k$ . Rewrite the equation using this new information.
$s * s * k = 118 * s * k - 116 * s$
$s * k = 118 * k - 116$
Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum: 116.
$s * 116 = 118 * 116 - 116$
$s = 118 - 1$
$s = \boxed{117}$ | 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 indeed yields an integral value of $r.$ | 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, $2$ . Also to minimize $n$ , we want $5$ , the greatest of all the factors, to be raised to the least power. Therefore, $n = 2^43^45^2$ and $\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}$ | 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 are no longer in the lake on September 1 (because of death and emigrations), that 40% of the fish were not in the lake May 1 (because of births and immigrations), and that the number of untagged fish and tagged fish in the September 1 sample are representative of the total population. What does the biologist calculate for the number of fish in the lake on May 1? | 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 are no longer in the lake on September 1 (because of death and emigrations), that 40% of the fish were not in the lake May 1 (because of births and immigrations), and that the number of untagged fish and tagged fish in the September 1 sample are representative of the total population. What does the biologist calculate for the number of fish in the lake on May 1? | 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 the total number of fish in May.
Since $\frac{3}{70}$ of the fish in September were tagged, $\frac{45}{5n/4} = \frac{3}{70}$ , where $n$ is the number of fish in May. Solving for $n$ , we see that $n = \boxed{840}$ | 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 remaining target in the chosen column.
If the rules are followed, in how many different orders can the eight targets be broken? | 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 one-to-one correspondence, we count the letter arrangements: \[\frac{8!}{3!\cdot2!\cdot3!} = \boxed{560}.\] | 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 $n^{8k_1 + 3k_2}$
$8$ and $3$ are relatively prime, and by the Chicken McNugget Theorem , for two relatively prime integers $a,b$ , the largest number that cannot be expressed as the sum of multiples of $a,b$ is $a \cdot b - a - b$ . For $3,8$ , this is $13$ ; however, we can easily see that the numbers $145$ to $157$ can be written in terms of $3,8$ . Since the exponents are of roots of unities, they reduce $\mod{144}$ , so all numbers in the range are covered. Thus the answer is $\boxed{144}$ | 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 $47$ , respectively.
$zw = \text{cis}\,\left(\frac{\pi k_1}{9} + \frac{\pi k_2}{24}\right) = \text{cis}\,\left(\frac{8\pi k_1 + 3 \pi k_2}{72}\right)$ . Since the trigonometric functions are periodic every $2\pi$ , there are at most $72 \cdot 2 = \boxed{144}$ distinct elements in $C$ . As above, all of these will work. | 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.
Adding all of these up, we get $12[6(\sqrt{6} - \sqrt{2}) + 12 + 12\sqrt{2} + 12\sqrt{3} + 6(\sqrt{6}+\sqrt{2})] + 6 \cdot 24$
$= 12(12 + 12\sqrt{2} + 12\sqrt{3} + 12\sqrt{6}) + 144 = 288 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6}$ . Thus, the answer is $144 \cdot 5 = \boxed{720}$ | 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:
$a^2 = 12^2 + 12^2 - 2\cdot 12\cdot 12\cdot \cos \theta$
$a^2 = 2\cdot 12^2 - 2\cdot 12^2 \cos \theta$
$a^2 = 2\cdot 12^2 (1 - \cos \theta)$
$a = 12\sqrt{2} \cdot \sqrt{1 - \cos \theta}$
There are six lengths of sides/diagonals, corresponding to $\theta = {30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}, 180^{\circ}}$
Call these lengths $a_1, a_2, a_3, a_4, a_5, a_6$ from shortest to longest. The total length $l$ that is asked for is
$l = 12(a_1 + a_2 + a_3 + a_4 + a_5) + 6a_6$ , noting that $a_6$ as written gives the diameter of the circle, which is the longest diagonal.
$l = 12[12\sqrt{2} (\sqrt{1 - \cos 30^{\circ}}+\sqrt{1-\cos 60^{\circ}}+\sqrt{1-\cos 90^{\circ}}+\sqrt{1-\cos 120^{\circ}}+\sqrt{1 - \cos 150^{\circ}})] + 6 \cdot 24$
$l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 - \frac{1}{2}} + \sqrt{1-0} + \sqrt{1 + \frac{1}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}}\right) + 144$
$l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{6}}{2} + \sqrt{1 + \frac{\sqrt{3}}{2}}\right) + 144$
To simplify the two nested radicals, add them, and call the sum $x$
$x = \sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}}$
Squaring both sides, the F and L part of FOIL causes the radicals to cancel, leaving:
$x^2 = 2 + 2\sqrt{\left(1 - \frac{\sqrt{3}}{2}\right)\left(1 + \frac{\sqrt{3}}{2}\right)}$
$x^2 = 2 + 2\sqrt{1 - \frac{3}{4}}$
$x^2 = 2 + 2\sqrt{\frac{1}{4}}$
$x^2 = 2 + 2\cdot\frac{1}{2}$
$x = \sqrt{3}.$
Plugging that sum $x$ back into the equation for $l$ , we find
$l = 144\sqrt{2} \left(\frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{6}}{2} + \sqrt{3}\right) + 144$
$l = 144 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6} + 144$
Thus, the desired quantity is $144\cdot 5 = \boxed{720}$ | 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 points on the 12-gon. Now, $d=\sqrt{288-288 \cos \theta}$ . Instead of factoring out $288$ as in solution 2, factor out $\sqrt{576}$ instead-the motivation for this is to make the expression look like the half angle identity, and the fact that $\sqrt{576}$ is an integer doesn't hurt. Now, we have that $d=24 \sin \frac{\theta}{2}$ , which simplifies things quite nicely. Continue as in solution 2, computing half-angle sines instead of nested radicals, to obtain $\boxed{720}$ | 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 segments $\overline{CP}$ and $\overline{DP}$ , we obtain a triangular pyramid , all four of whose faces are isosceles triangles . Find the volume of this pyramid.
[asy] pair D=origin, A=(13,0), B=(13,12), C=(0,12), P=(6.5, 6); draw(B--C--P--D--C^^D--A); filldraw(A--P--B--cycle, gray, black); label("$A$", A, SE); label("$B$", B, NE); label("$C$", C, NW); label("$D$", D, SW); label("$P$", P, N); label("$13\sqrt{3}$", A--D, S); label("$12\sqrt{3}$", A--B, E);[/asy] | [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.x, Ci.y, (99/133)^.5); Pa=(P.x,P.y,0); draw(P--Pa--A); draw(C--Pa--D); draw((C+D)/2--A--C--D--P--C--P--A--D); label("A", A, NE); label("P", P, N); label("C", C); label("D", D, S); label("$13\sqrt{3}$", (A+D)/2, E, small); label("$13\sqrt{3}$", (A+C)/2, N, small); label("$12\sqrt{3}$", (C+D)/2, SW, small); [/asy]
Our triangular pyramid has base $12\sqrt{3} - 13\sqrt{3} - 13\sqrt{3} \triangle$ . The area of this isosceles triangle is easy to find by $[ACD] = \frac{1}{2}bh$ , where we can find $h_{ACD}$ to be $\sqrt{399}$ by the Pythagorean Theorem . Thus $A = \frac 12(12\sqrt{3})\sqrt{399} = 18\sqrt{133}$
[asy] size(280); import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0,399^.5,0); triple D=(108^.5,0,0); triple C=(-108^.5,0,0); pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, 99/133^.5); triple Pa=(P.x,P.y,0); draw(A--C--D--P--C--P--A--D); draw(P--Pa--A); draw(C--Pa--D); draw(circle(Pa, h)); label("A", A, NE); label("C", C, NW); label("D", D, S); label("P",P , N); label("P$'$", Pa, SW); label("$13\sqrt{3}$", (A+D)/2, E, small); label("$13\sqrt{3}$", (A+C)/2, NW, small); label("$12\sqrt{3}$", (C+D)/2, SW, small); label("h", (P + Pa)/2, W); label("$\frac{\sqrt{939}}{2}$", (C+P)/2 ,NW); [/asy]
To find the volume, we want to use the equation $\frac 13Bh = 6\sqrt{133}h$ , so we need to find the height of the tetrahedron . By the Pythagorean Theorem, $AP = CP = DP = \frac{\sqrt{939}}{2}$ . If we let $P$ be the center of a sphere with radius $\frac{\sqrt{939}}{2}$ , then $A,C,D$ lie on the sphere. The cross section of the sphere that contains $A,C,D$ is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of $\triangle ACD$
From here we just need to perform some brutish calculations. Using the formula $A = 18\sqrt{133} = \frac{abc}{4R}$ (where $R$ is the circumradius ), we find $R = \frac{12\sqrt{3} \cdot (13\sqrt{3})^2}{4\cdot 18\sqrt{133}} = \frac{13^2\sqrt{3}}{2\sqrt{133}}$ (there are slightly simpler ways to calculate $R$ since we have an isosceles triangle). By the Pythagorean Theorem,
\begin{align*}h^2 &= PA^2 - R^2 \\ &= \left(\frac{\sqrt{939}}{2}\right)^2 - \left(\frac{13^2\sqrt{3}}{2\sqrt{133}}\right)^2\\ &= \frac{939 \cdot 133 - 13^4 \cdot 3}{4 \cdot 133} = \frac{13068 \cdot 3}{4 \cdot 133} = \frac{99^2}{133}\\ h &= \frac{99}{\sqrt{133}} \end{align*}
Finally, we substitute $h$ into the volume equation to find $V = 6\sqrt{133}\left(\frac{99}{\sqrt{133}}\right) = \boxed{594}$ | 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 segments $\overline{CP}$ and $\overline{DP}$ , we obtain a triangular pyramid , all four of whose faces are isosceles triangles . Find the volume of this pyramid.
[asy] pair D=origin, A=(13,0), B=(13,12), C=(0,12), P=(6.5, 6); draw(B--C--P--D--C^^D--A); filldraw(A--P--B--cycle, gray, black); label("$A$", A, SE); label("$B$", B, NE); label("$C$", C, NW); label("$D$", D, SW); label("$P$", P, N); label("$13\sqrt{3}$", A--D, S); label("$12\sqrt{3}$", A--B, E);[/asy] | 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 $MD=6\sqrt3$ , we have $MX=\sqrt{3\cdot13^2-3\cdot6^2} = \sqrt{399}$
Let $\theta=\angle XPM$ . For calculating trig, let us double all sides of $\triangle XMP$ . By Law of Cosines, $\cos\theta = \frac{3\cdot13^2+939-4\cdot399}{2\cdot13\cdot3\sqrt{313}}=\frac{-150}{6\cdot13\sqrt{313}}=-\frac{25}{13\sqrt{313}}$
Hence, \[\sin\theta = \sqrt{1-\cos^2\theta}\] \[=\sqrt{1-\frac{625}{13^2\cdot313}}\] \[=\frac{\sqrt{13^2\cdot313-625}}{13\sqrt{313}}\] \[=\frac{\sqrt{3\cdot132^2}}{13\sqrt{313}}\] \[=\frac{132\sqrt3}{13\sqrt{313}}\] Thus, the height of the pyramid is $PX\sin\theta=\frac{\sqrt{939}}{2}\cdot\frac{132\sqrt3}{13\sqrt{313}}=\frac{66\cdot3}{13}$ . Since $[CPD]=3\sqrt{3}\cdot13\sqrt{3}=9\cdot13$ , the volume of the pyramid is $\frac13 \cdot 9\cdot13\cdot \frac{66\cdot3}{13}=\boxed{594}$ | 594 |
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