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e6f44855c426eca8c3fb8a333f2a7a3e | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_11 | 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 $\boxed{945}$ | null | 945 |
62bee3e23133606a02b0e06d67cc42b5 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_12 | 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 $\binom{12}{5}=\boxed{792}$ | null | 792 |
62bee3e23133606a02b0e06d67cc42b5 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_12 | 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 $\binom{7+6-1}{6-1} = \boxed{792}$ | null | 792 |
03407b94835f1da087df5111f3eeda2c | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_13 | 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 $9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}$ | null | 820 |
03407b94835f1da087df5111f3eeda2c | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_13 | 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 $\dfrac{9\cdot \left(180+\frac{20}{9}\right)}{2}=\boxed{820}$ | null | 820 |
03407b94835f1da087df5111f3eeda2c | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_13 | 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 $\boxed{820}$ | null | 820 |
03407b94835f1da087df5111f3eeda2c | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_13 | 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 $\frac{9}{4} *\frac{9}{4}*\frac{3280}{9} = \boxed{820}$ | null | 820 |
702de842ecbddf4cf44e7b854f2a6915 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_14 | 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 $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$ | Let $K_A=[BOC], K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base, \[\frac{AO}{OA'}+1=\frac{AA'}{OA'}=\frac{[ABC]}{[BOC]}=\frac{K_A+K_B+K_C}{K_A}.\] Therefore, we have \[\frac{AO}{OA'}=\frac{K_B+K_C}{K_A}\] \[\frac{BO}{OB'}=\frac{K_A+K_C}{K_B}\] \[\frac{CO}{OC'}=\frac{K_A+K_B}{K_C}.\] Thus, we are given \[\frac{K_B+K_C}{K_A}+\frac{K_A+K_C}{K_B}+\frac{K_A+K_B}{K_C}=92.\] Combining and expanding gives \[\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}=92.\] We desire $\frac{(K_B+K_C)(K_C+K_A)(K_A+K_B)}{K_AK_BK_C}.$ Expanding this gives \[\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}+2=\boxed{094}.\] | null | 094 |
702de842ecbddf4cf44e7b854f2a6915 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_14 | 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 $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$ | Using mass points , let the weights of $A$ $B$ , and $C$ be $a$ $b$ , and $c$ respectively.
Then, the weights of $A'$ $B'$ , and $C'$ are $b+c$ $c+a$ , and $a+b$ respectively.
Thus, $\frac{AO^{}_{}}{OA'} = \frac{b+c}{a}$ $\frac{BO^{}_{}}{OB'} = \frac{c+a}{b}$ , and $\frac{CO^{}_{}}{OC'} = \frac{a+b}{c}$
Therefore: $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'} = \frac{b+c}{a} \cdot \frac{c+a}{b} \cdot \frac{a+b}{c}$ $= \frac{2abc+b^2c+bc^2+c^2a+ca^2+a^2b+ab^2}{abc} =$
$2+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}+\frac{ab(a+b)}{abc} = 2 + \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}$ $= 2 + \frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'} = 2+92 = \boxed{094}$ | null | 094 |
702de842ecbddf4cf44e7b854f2a6915 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_14 | 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} | null | 94 |
6726d01c779a643a59145bf58a592bd1 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_15 | Define a positive integer $n^{}_{}$ to be a factorial tail if there is some positive integer $m^{}_{}$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails? | Let the number of zeros at the end of $m!$ be $f(m)$ . We have $f(m) = \left\lfloor \frac{m}{5} \right\rfloor + \left\lfloor \frac{m}{25} \right\rfloor + \left\lfloor \frac{m}{125} \right\rfloor + \left\lfloor \frac{m}{625} \right\rfloor + \left\lfloor \frac{m}{3125} \right\rfloor + \cdots$
Note that if $m$ is a multiple of $5$ $f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)$
Since $f(m) \le \frac{m}{5} + \frac{m}{25} + \frac{m}{125} + \cdots = \frac{m}{4}$ , a value of $m$ such that $f(m) = 1991$ is greater than $7964$ . Testing values greater than this yields $f(7975)=1991$
There are $\frac{7975}{5} = 1595$ distinct positive integers, $f(m)$ , less than $1992$ . Thus, there are $1991-1595 = \boxed{396}$ positive integers less than $1992$ that are not factorial tails. | null | 396 |
6726d01c779a643a59145bf58a592bd1 | https://artofproblemsolving.com/wiki/index.php/1992_AIME_Problems/Problem_15 | Define a positive integer $n^{}_{}$ to be a factorial tail if there is some positive integer $m^{}_{}$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails? | After testing various values of $m$ in $f(m)$ of solution 1 to determine $m$ for which $f(m) = 1992$ , we find that $m \in \{7980, 7981, 7982, 7983, 7984\}$ . WLOG, we select $7980$ . Furthermore, note that every time $k$ reaches a multiple of $25$ $k!$ will gain two or more additional factors of $5$ and will thus skip one or more numbers.
With this logic, we realize that the desired quantity is simply $\left \lfloor \frac{7980}{25} \right \rfloor + \left \lfloor \frac{7980}{125} \right \rfloor \cdots$ , where the first term accounts for every time $1$ number is skipped, the second term accounts for each time $2$ numbers are skipped, and so on. Evaluating this gives us $319 + 63 + 12 + 2 = \boxed{396}$ . - Spacesam(edited by srisainandan6) | null | 396 |
1c75dc230f2bb83bdffb2646ac1cdc99 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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 $5^2 + 11^2 = \boxed{146}$ | null | 146 |
1c75dc230f2bb83bdffb2646ac1cdc99 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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*} | Let $a=x+y$ $b=xy$ then we get the equations \begin{align*} a+b&=71\\ ab&=880 \end{align*} After finding the prime factorization of $880=2^4\cdot5\cdot11$ , it's easy to obtain the solution $(a,b)=(16,55)$ . Thus \[x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}\] Note that if $(a,b)=(55,16)$ , the answer would exceed $999$ which is invalid for an AIME answer.
~ Nafer | null | 146 |
1c75dc230f2bb83bdffb2646ac1cdc99 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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*} | First, notice that you can factor $x^2y + xy^2$ as $xy(x + y)$ . From this, we notice that $xy$ and $x + y$ is a common occurrence, so that lends itself to a simple solution by substitution. Let $xy = b$ and $x + y = a$ . From this substitution, we get the following system: \[a + b = 71\] \[ab = 880\] Solving that system gives us the following two pairs $(a, b)$ $(16, 55)$ and $(55, 16)$ . The second one is obviously too big as $55^2$ is clearly out of bounds for an AIME problem (More on that later). Therefore, we proceed with substituting back the pair $(16, 55)$ . This means that $x + y = 16$ and $xy = 55$ Then, instead of solving the system, we can do a clever manipulation by squaring $x + y$ . Doing so, we get: \[(x + y)^2 = (x^2 + y^2) + 2xy\] We see that in this form, we can substitute everything in except for $(x^2 + y^2)$ , which is the desired answer. Substituting, we get: \[256 = (x^2 + y^2) + 110\] so $x^2 + y^2 = \boxed{146}$ . (If we were to go with the pair $(55, 16)$ , then the $(x + y)^2$ would be absurdly out of bounds) | null | 146 |
2625d118cccd9f014d7ed4fba3dffafb | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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. | 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. $2\sum\limits_{k=1}^{168} \frac{5(168-k)}{168}-5 =2\frac{(0+5)\cdot169}{2}-5 =168\cdot5 =\boxed{840}$ | null | 840 |
2625d118cccd9f014d7ed4fba3dffafb | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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, \[= \frac{835 \cdot 168}{168} +5 = 835+5 = \boxed{840}\] | null | 840 |
2625d118cccd9f014d7ed4fba3dffafb | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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. | 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 | null | 840 |
c842f8bc7ea4298cc4880cf6ff548174 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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? | 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 $k=\boxed{166}$ | null | 166 |
c842f8bc7ea4298cc4880cf6ff548174 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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 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 $\boxed{166}$ | null | 166 |
c842f8bc7ea4298cc4880cf6ff548174 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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 $\boxed{166}$ | null | 166 |
c842f8bc7ea4298cc4880cf6ff548174 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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? | 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 = $\boxed{166}$ | null | 166 |
c842f8bc7ea4298cc4880cf6ff548174 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_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? | 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 $k=m+1= \boxed{166}$ | null | 166 |
0cb901f2e5a6948f6d2ec672abdf05fd | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_4 | 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 $154 + 1 + 4 = \boxed{159}$ | null | 159 |
0cb901f2e5a6948f6d2ec672abdf05fd | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_4 | 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: $32 \cdot \frac{5}{2} \cdot 2 - 1 = \boxed{159}$ | null | 159 |
c503707e188b3ba6b67a5844a8f1c120 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_5 | 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 $\frac{2^8}{2} = \boxed{128}$ | null | 128 |
94a53012c117cd9385ccb0c2aa4462cf | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_6 | 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 $\lfloor 100r \rfloor = \boxed{743}$ | null | 743 |
94a53012c117cd9385ccb0c2aa4462cf | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_6 | 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 $\lfloor nx\rfloor = 546+19\cdot 7+8\cdot 8=\boxed{743}$ | null | 743 |
ee6ff675e62f12ad5901c08017608d01 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_7 | Find $A^2_{}$ , where $A^{}_{}$ is the sum of the absolute values of all roots of the following equation: | $x=\sqrt{19}+\underbrace{\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}}_{x}$
$x=\sqrt{19}+\frac{91}{x}$
$x^2=x\sqrt{19}+91$
$x^2-x\sqrt{19}-91 = 0$
$\left.\begin{array}{l}x_1=\frac{\sqrt{19}+\sqrt{383}}{2}\\\\x_2=\frac{\sqrt{19}-\sqrt{383}}{2}\end{array}\right\}A=|x_1|+|x_2|\Rightarrow\sqrt{383}$
$A^2=\boxed{383}$ | null | 383 |
ee6ff675e62f12ad5901c08017608d01 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_7 | 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 $A_{}^{2}=\boxed{383}$ | null | 383 |
44919e328d6a107da2be659f186a6678 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_8 | 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. | null | 10 |
44919e328d6a107da2be659f186a6678 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_8 | 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: $( - 36, - 1),( - 18, - 2),( - 12, - 3),( - 9, - 4),( - 6, - 6),(1,36),(2,18),(3,12),(4,9),(6,6)$ $\Rightarrow \boxed{10} a$ | null | 10 |
b8d4b7399f3d46dc184e0b147d21d311 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_9 | 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 $m+n^{}_{}.$ | Use the two trigonometric Pythagorean identities $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \csc^2 x$
If we square the given $\sec x = \frac{22}{7} - \tan x$ , we find that
\begin{align*} \sec^2 x &= \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x \\ 1 &= \left(\frac{22}7\right)^2 - \frac{44}7 \tan x \end{align*}
This yields $\tan x = \frac{435}{308}$
Let $y = \frac mn$ . Then squaring,
\[\csc^2 x = (y - \cot x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.\]
Substituting $\cot x = \frac{1}{\tan x} = \frac{308}{435}$ yields a quadratic equation $0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)$ . It turns out that only the positive root will work, so the value of $y = \frac{29}{15}$ and $m + n = \boxed{044}$ | null | 044 |
b8d4b7399f3d46dc184e0b147d21d311 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_9 | 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 $m+n^{}_{}.$ | We are given that $\frac{1+\sin x}{\cos x}=\frac{22}7\implies\frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}$ $=\frac{\cos x}{1-\sin x}$ , or equivalently, $\cos x=\frac{7+7\sin x}{22}=\frac{22-22\sin x}7\implies\sin x=\frac{22^2-7^2}{22^2+7^2}$ $\implies\cos x=\frac{2\cdot22\cdot7}{22^2+7^2}$ . Note that what we want is just $\frac{1+\cos x}{\sin x}=\frac{1+\frac{2\cdot22\cdot7}{22^2+7^2}}{\frac{22^2-7^2}{22^2+7^2}}=\frac{22^2+7^2+2\cdot22\cdot7}{22^2-7^2}=\frac{(22+7)^2}{(22-7)(22+7)}=\frac{22+7}{22-7}$ $=\frac{29}{15}\implies m+n=29+15=\boxed{044}$ | null | 044 |
b8d4b7399f3d46dc184e0b147d21d311 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_9 | 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 $m+n^{}_{}.$ | Assign a right triangle with angle $x$ , hypotenuse $c$ , adjacent side $a$ , and opposite side $b$ .
Then, through the given information above, we have that..
$\frac{c}{a}+\frac{b}{a}=\frac{22}{7}\implies \frac{c+b}{a}=\frac{22}{7}$
$\frac{c}{b}+\frac{a}{b}=\frac{m}{n}\implies \frac{a+c}{b}=\frac{m}{n}$
Hence, because similar right triangles can be scaled up by a factor, we can assume that this particular right triangle is indeed in simplest terms.
Hence, $a=7$ $b+c=22$
Furthermore, by the Pythagorean Theorem, we have that
$a^2+b^2=c^2\implies 49+b^2=c^2$
Solving for $c$ in the first equation and plugging in into the second equation...
$49+b^2=(22-b)^2\implies 49+b^2=484-44b+b^2\implies 44b=435\implies b=\frac{435}{44}$
Hence, $c=22-\frac{435}{44}=\frac{533}{44}$
Now, we want $\frac{a+c}{b}$
Plugging in, we find the answer is $\frac{\frac{7\cdot{44}}{44}+\frac{533}{44}}{\frac{435}{44}}=\frac{841}{435}=\frac{29}{15}$
Hence, the answer is $29+15=\boxed{044}$ | null | 044 |
b8d4b7399f3d46dc184e0b147d21d311 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_9 | 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 $m+n^{}_{}.$ | We know that $\sec(x) = \frac{h}{a}$ and that $\tan(x) = \frac{o}{a}$ where $h$ $a$ $o$ represent the hypotenuse, adjacent, and opposite (respectively) to angle $x$ in a right triangle. Thus we have that $\sec(x) + \tan(x) = \frac{h+o}{a}$ . We also have that $\csc(x) + \cot(x) = \frac{h}{o} + \frac{a}{o} = \frac{h+a}{o}$ . Set $\sec(x) + \tan(x) = \alpha$ and csc(x)+cot(x) = $\beta$ . Then, notice that $\alpha + \beta = \frac{h+o}{a} + \frac{h+a}{o} = \frac{oh+ah+o^2 + a^2}{oa} = \frac{h(o+a+h)}{oa}$ ( This is because of the Pythagorean Theorem, recall $o^2 +a^2 = h^2$ ). But then notice that $\alpha \cdot \beta = \frac{(o+h)(a+h)}{oa} = \frac{oa +oh +ha +h^2}{oa} = 1+ \frac{h(o+a+h)}{oa} = 1+ \alpha + \beta$ . From the information provided in the question, we can substitute $\alpha$ for $\frac{22}{7}$ . Thus, $\frac{22 \beta}{7}= \beta + \frac{29}{7} \Longrightarrow 22 \beta = 7 \beta + 29 \Longrightarrow 15 \beta = 29 \Longrightarrow \beta = \frac{29}{15}$ . Since, essentially we are asked to find the sum of the numerator and denominator of $\beta$ , we have $29 + 15 = \boxed{044}$ | null | 044 |
b8d4b7399f3d46dc184e0b147d21d311 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_9 | 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 $p+q=15+29=\boxed{44}$ | null | 44 |
b8d4b7399f3d46dc184e0b147d21d311 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_9 | 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 $m+n^{}_{}.$ | Rewriting $\sec{x}$ and $\tan{x}$ in terms of $\sin{x}$ and $\cos{x}$ , we know that $\frac{1+\sin{x}}{\cos{x}}=\frac{22}{7}.$
Clearing fractions, \[22\cos{x}=7+7\sin{x}.\]
Squaring to get an expression in terms of $\sin^2{x}$ and $\cos^2{x}$ \[484\cos^2{x}=49+49\sin^2{x}+98\sin{x}.\]
Substituting $\cos^2{x}=1-\sin^2{x},$
\[484(1-\sin^2{x})=49+49\sin^2{x}+98\sin{x}.\]
Expanding then collecting terms yields a quadratic in $\sin{x}:$
\[533\sin^2{x}+98\sin{x}-435=0.\]
To make calculations easier, let $y=\sin{x}.$
\[533y^2+98y-435=0.\]
Upon inspection, $y=-1$ is a root. Dividing by $y+1$
\[533y^2+98y-435=(533y-435)(y+1).\]
Substituting $y=\sin{x},$ we see that $\sin{x}=-1$ doesn't work, as $\cos{x}=0$ , leaving $\tan{x}$ undefined.
We conclude that $\sin{x}=\frac{435}{533}.$
Since $\sin^2{x}+\cos^2{x}=1,$
\[\cos{x}=\pm \sqrt{\frac{533^2-435^2}{533^2}}.\] \[=\pm \frac{308}{533}.\]
After checking via the given equation, we know that only the positive solution works.
Therefore,
\[\csc{x}+\cot{x}=\frac{1}{\sin{x}}+\frac{\cos{x}}{\sin{x}}\] \[=\frac{533}{435}+\frac{308}{435}\] \[=\frac{29}{15}=\frac{m}{n}.\]
Adding $m$ and $n$ , our answer is $\boxed{044}.$ | null | 044 |
2cc48ea6fc23c9cd1f94aa7e7cfa5380 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_10 | 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 $\boxed{532}$ | null | 532 |
2cc48ea6fc23c9cd1f94aa7e7cfa5380 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_10 | 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 $\boxed{532}$ | null | 532 |
2cc48ea6fc23c9cd1f94aa7e7cfa5380 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_10 | 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 $\boxed{532}$ | null | 532 |
2cc48ea6fc23c9cd1f94aa7e7cfa5380 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_10 | 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 $\boxed{532}$ | null | 532 |
e10df578cd4319b6293bcdd3a01b63e2 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_11 | 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 $84 + 48 + 3 = \boxed{135}$ | null | 135 |
f7bb8e637a3352dd7752f05ea7379ad1 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_12 | 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 $m + n = \boxed{677}$ | null | 677 |
f7bb8e637a3352dd7752f05ea7379ad1 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_12 | 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, $m+n = \frac{672}{5}\implies\boxed{677}$ | null | 677 |
f7bb8e637a3352dd7752f05ea7379ad1 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_12 | 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 $\frac{672}{5}\implies\boxed{677}$ | null | 677 |
702923fed8d309c573076f5574433330 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 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 $r=\boxed{990}$ | null | 990 |
702923fed8d309c573076f5574433330 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 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 $\frac{44^2+44}2=\frac12\cdot44(44+1)=22\cdot45=11\cdot90=\boxed{990}$ | null | 990 |
702923fed8d309c573076f5574433330 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 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 $r=\boxed{990}$ | null | 990 |
702923fed8d309c573076f5574433330 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 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, $r=\frac{t+\sqrt{t}}{2}=\frac{1936+44}{2}=\boxed{990}$ | null | 990 |
702923fed8d309c573076f5574433330 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 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 $r = \boxed{990}$ | null | 990 |
702923fed8d309c573076f5574433330 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_13 | 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 $\sqrt{8b+1}=87$
$\Rightarrow r=\frac{1980}{2}$
$\Rightarrow r=\boxed{990}$ | null | 990 |
3340ffa8f7c0603b196d3e1bca2b319c | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_14 | 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 $x+y+z=105+144+135=\boxed{384}$ | null | 384 |
1985980ebb1c6379c46a1d2a94ad9483 | https://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_15 | For positive integer $n_{}^{}$ , define $S_n^{}$ to be the minimum value of the sum $\sum_{k=1}^n \sqrt{(2k-1)^2+a_k^2},$ where $a_1,a_2,\ldots,a_n^{}$ are positive real numbers whose sum is 17. There is a unique positive integer $n^{}_{}$ for which $S_n^{}$ is also an integer. Find this $n^{}_{}$ | Consider $n$ right triangles joined at their vertices, with bases $a_1,a_2,\ldots,a_n$ and heights $1,3,\ldots, 2n - 1$ . The sum of their hypotenuses is the value of $S_n$ . The minimum value of $S_n$ , then, is the length of the straight line connecting the bottom vertex of the first right triangle and the top vertex of the last right triangle, so \[S_n \ge \sqrt {\left(\sum_{k = 1}^n (2k - 1)\right)^2 + \left(\sum_{k = 1}^n a_k\right)^2}.\] Since the sum of the first $n$ odd integers is $n^2$ and the sum of $a_1,a_2,\ldots,a_n$ is 17, we get \[S_n \ge \sqrt {17^2 + n^4}.\] If this is an integer, we can write $17^2 + n^4 = m^2$ , for an integer $m$ . Thus, $(m - n^2)(m + n^2) = 289\cdot 1 = 17\cdot 17 = 1\cdot 289.$ The only possible value, then, for $m$ is $145$ , in which case $n^2 = 144$ , and $n = \boxed{012}$ | null | 012 |
623d4bb6b99c9563038a10bd3c79d625 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_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 $\boxed{528}$ | null | 528 |
623d4bb6b99c9563038a10bd3c79d625 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_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. | 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 $529-1 = \boxed{528}$ | null | 528 |
c47ebb2a4cf1bf12bc6820fad35f1cb6 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_2 | Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$ | Suppose that $52+6\sqrt{43}$ is in the form of $(a + b\sqrt{43})^2$ FOILing yields that $52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}$ . This implies that $a$ and $b$ equal one of $\pm3, \pm1$ . The possible sets are $(3,1)$ and $(-3,-1)$ ; the latter can be discarded since the square root must be positive. This means that $52 + 6\sqrt{43} = (\sqrt{43} + 3)^2$ . Repeating this for $52-6\sqrt{43}$ , the only feasible possibility is $(\sqrt{43} - 3)^2$
Rewriting, we get $(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3$ . Using the difference of cubes , we get that $[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]$ $= (6)(3 \cdot 43 + 9) = \boxed{828}$ .
Note: You can also just use the formula $(a + b)^2 = a^2 + 2ab + b^2$ instead of foiling | null | 828 |
c47ebb2a4cf1bf12bc6820fad35f1cb6 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_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 $S=\boxed{828}$ | null | 828 |
c47ebb2a4cf1bf12bc6820fad35f1cb6 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_2 | 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 $(6)(138) = \boxed{828}$ | null | 828 |
c47ebb2a4cf1bf12bc6820fad35f1cb6 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_2 | 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 $S=\sqrt{685584}=\boxed{828}$ | null | 828 |
c47ebb2a4cf1bf12bc6820fad35f1cb6 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_2 | Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$ | (Similar to Solution 3, but with substitution)
Let $a=\sqrt{52+6\sqrt{43}}$ and $b=\sqrt{52-6\sqrt{43}}.$ We want to find $a^3-b^3=(a-b)(a^2+ab+b^2).$
We have \[a^2+b^2=102,\text{ and}\] \[ab=\sqrt{(52+6\sqrt{43})(52-6\sqrt{43})}=\sqrt{1156}=34.\] Then, $(a-b)^2=a^2+b^2-2ab=104-2\cdot 34= 36\implies a-b=6.$
Our answer is \[a^3-b^3=(a-b)(a^2+b^2+ab)=6\cdot 138=\boxed{828.}\] | null | 828. |
c47ebb2a4cf1bf12bc6820fad35f1cb6 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_2 | 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 $2 (27 + 9 \cdot 43) = 2 \cdot 414 = \boxed{828}.$ | null | 828 |
18c7d72010bc4020fd6d7ced6a423bc0 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_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 $s_{}^{}$ | 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 $s = \boxed{117}$ | null | 117 |
18c7d72010bc4020fd6d7ced6a423bc0 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_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 $s_{}^{}$ | 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.
$s * 116 = 118 * 116 - 116$
$s = 118 - 1$
$s = \boxed{117}$ | null | 117 |
18c7d72010bc4020fd6d7ced6a423bc0 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_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 $r.$ | null | 117 |
a6f42939d04dc8ae547156eecaa2679b | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_4 | Find the positive solution to | We could clear out the denominators by multiplying, though that would be unnecessarily tedious.
To simplify the equation, substitute $a = x^2 - 10x - 29$ (the denominator of the first fraction). We can rewrite the equation as $\frac{1}{a} + \frac{1}{a - 16} - \frac{2}{a - 40} = 0$ . Multiplying out the denominators now, we get:
\[(a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0\]
Simplifying, $-64a + 40 \times 16 = 0$ , so $a = 10$ . Re-substituting, $10 = x^2 - 10x - 29 \Longleftrightarrow 0 = (x - 13)(x + 3)$ . The positive root is $\boxed{013}$ | null | 013 |
7ec0df5049307c53bb63e0b146f5461b | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_5 | 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 $\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}$ | null | 432 |
f52d46b58b2387699590ed3e56467729 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_6 | 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, $\frac{3}{42} = \frac{60}{x} \Longrightarrow \boxed{840}$ | null | 840 |
f52d46b58b2387699590ed3e56467729 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_6 | 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 $n = \boxed{840}$ | null | 840 |
dec2030f51da889e181098ab99155f9d | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_7 | triangle has vertices $P_{}^{}=(-8,5)$ $Q_{}^{}=(-15,-19)$ , and $R_{}^{}=(1,-7)$ . The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0_{}^{}$ . Find $a+c_{}^{}$ | Use the angle bisector theorem to find that the angle bisector of $\angle P$ divides $QR$ into segments of length $\frac{25}{x} = \frac{15}{20 -x} \Longrightarrow x = \frac{25}{2},\ \frac{15}{2}$ . It follows that $\frac{QP'}{RP'} = \frac{5}{3}$ , and so $P' = \left(\frac{5x_R + 3x_Q}{8},\frac{5y_R + 3y_Q}{8}\right) = (-5,-23/2)$
The desired answer is the equation of the line $PP'$ $PP'$ has slope $\frac{-11}{2}$ , from which we find the equation to be $11x + 2y + 78 = 0$ . Therefore, $a+c = \boxed{089}$ | null | 089 |
dec2030f51da889e181098ab99155f9d | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_7 | triangle has vertices $P_{}^{}=(-8,5)$ $Q_{}^{}=(-15,-19)$ , and $R_{}^{}=(1,-7)$ . The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0_{}^{}$ . Find $a+c_{}^{}$ | Transform triangle $PQR$ so that $P$ is at the origin. Note that the slopes do not change when we transform the triangle.
We know that the slope of $PQ$ is $\frac{24}{7}$ and the slope of $PR$ is $-\frac{4}{3}$ . Thus, in the complex plane, they are equivalent to $\tan(\alpha)=\frac{24}{7}$ and $\tan(\beta)=-\frac{4}{3}$ , respectively. Here $\alpha$ is the angle formed by the $x$ -axis and $PQ$ , and $\beta$ is the angle formed by the $x$ -axis and $PR$ . The equation of the angle bisector is $\tan\left(\frac{\alpha+\beta}{2}\right)$
As the tangents are in very neat Pythagorean triples , we can easily calculate $\cos(\alpha)$ and $\cos(\beta)$
Angle $\alpha$ is in the third quadrant, so $\cos(\alpha)$ is negative. Thus $\cos(\alpha)=-\frac{7}{25}$
Angle $\beta$ is in the fourth quadrant, so $\cos(\beta)$ is positive. Thus $\cos(\beta)=\frac{3}{5}$
By the Half-Angle Identities $\tan\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1-\cos(\alpha)}{1+\cos(\alpha)}}=\pm\sqrt{\frac{\frac{32}{25}}{\frac{18}{25}}}=\pm\sqrt{\frac{16}{9}}=\pm\frac{4}{3}$ and $\tan\left(\frac{\beta}{2}\right)=\pm\sqrt{\frac{1-\cos(\beta)}{1+\cos(\beta)}}=\pm\sqrt{\frac{\frac{2}{5}}{\frac{8}{5}}}=\pm\sqrt{\frac{1}{4}}=\pm\frac{1}{2}$
Since $\frac{\alpha}{2}$ and $\frac{\beta}{2}$ must be in the second quadrant, their tangent values are both negative. Thus $\tan\left(\frac{\alpha}{2}\right)=-\frac{4}{3}$ and $\tan\left(\frac{\beta}{2}\right)=-\frac{1}{2}$
By the sum of tangents formula $\tan\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\frac{\tan\left(\frac{\alpha}{2}\right)+\tan\left(\frac{\beta}{2}\right)}{1-\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)}=\frac{-\frac{11}{6}}{\frac{1}{3}}=-\frac{11}{2}$ , which is the slope of the angle bisector.
Finally, the equation of the angle bisector is $y-5=-\frac{11}{2}(x+8)$ or $y=-\frac{11}{2}x-39$ . Rearranging, we get $11x+2y+78=0$ , so our sum is $a+c=11+78=\boxed{089}$ . ~eevee9406 | null | 089 |
15e5e4c56583d40c0b05c2cd5893a506 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_8 | 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: \[\frac{8!}{3!\cdot2!\cdot3!} = \boxed{560}.\] | null | 560 |
161edc1db5e74a0c2e2402ce15daae4b | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_9 | fair coin is to be tossed $10_{}^{}$ times. Let $\frac{i}{j}^{}_{}$ , in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$ | Clearly, at least $5$ tails must be flipped; any less, then by the Pigeonhole Principle there will be heads that appear on consecutive tosses.
Consider the case when $5$ tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled $(H)$
There are six slots for the heads to be placed, but only $5$ heads remaining. Thus, using stars-and-bars there are ${6\choose5}$ possible combinations of 6 heads. Continuing this pattern, we find that there are $\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144$ . There are a total of $2^{10}$ possible flips of $10$ coins, making the probability $\frac{144}{1024} = \frac{9}{64}$ . Thus, our solution is $9 + 64 = \boxed{073}$ | null | 073 |
161edc1db5e74a0c2e2402ce15daae4b | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_9 | fair coin is to be tossed $10_{}^{}$ times. Let $\frac{i}{j}^{}_{}$ , in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$ | Call the number of ways of flipping $n$ coins and not receiving any consecutive heads $S_n$ . Notice that tails must be received in at least one of the first two flips.
If the first coin flipped is a T, then the remaining $n-1$ flips must fall under one of the configurations of $S_{n-1}$
If the first coin flipped is a H, then the second coin must be a T. There are then $S_{n-2}$ configurations.
Thus, $S_n = S_{n-1} + S_{n-2}$ . By counting, we can establish that $S_1 = 2$ and $S_2 = 3$ . Therefore, $S_3 = 5,\ S_4 = 8$ , forming the Fibonacci sequence . Listing them out, we get $2,3,5,8,13,21,34,55,89,144$ , and the 10th number is $144$ . Putting this over $2^{10}$ to find the probability, we get $\frac{9}{64}$ . Our solution is $9+64=\boxed{073}$ | null | 073 |
161edc1db5e74a0c2e2402ce15daae4b | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_9 | fair coin is to be tossed $10_{}^{}$ times. Let $\frac{i}{j}^{}_{}$ , in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$ | We can also split the problem into casework.
Case 1: 0 Heads
There is only one possibility.
Case 2: 1 Head
There are 10 possibilities.
Case 3: 2 Heads
There are 36 possibilities.
Case 4: 3 Heads
There are 56 possibilities.
Case 5: 4 Heads
There are 35 possibilities.
Case 6: 5 Heads
There are 6 possibilities.
We have $1+10+36+56+35+6=144$ , and there are $1024$ possible outcomes, so the probability is $\frac{144}{1024}=\frac{9}{64}$ , and the answer is $\boxed{073}$ | null | 073 |
3e60b58fcf8ab8cf5140ef3b6c8a3d17 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_10 | 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 $\boxed{144}$ | null | 144 |
3e60b58fcf8ab8cf5140ef3b6c8a3d17 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_10 | 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. | null | 144 |
3e60b58fcf8ab8cf5140ef3b6c8a3d17 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_10 | 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 $3960$ $(\text{lcm}(55,360)) \implies \frac{3960 \cdot 2}{55}=\boxed{144}$ | null | 144 |
e58105445a683b9ce02c41fb6a90e62d | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_12 | 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 $144 \cdot 5 = \boxed{720}$ | null | 720 |
e58105445a683b9ce02c41fb6a90e62d | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_12 | 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 $144\cdot 5 = \boxed{720}$ | null | 720 |
e58105445a683b9ce02c41fb6a90e62d | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_12 | 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 $\boxed{720}$ | null | 720 |
3173ec80d3ca533b29e2b2f9b97d2732 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_13 | Let $T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}$ . Given that $9^{4000}_{}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T_{}^{}$ have 9 as their leftmost digit? | Lemma : For all positive integers n, there's exactly one n-digit power of 9 that does not have a left-most digit 9
(Not-so-rigorous) Proof: One can prove by contradiction that there must be at least either one or two n-digit power of 9 for all n. If there is exactly 1 n-digit power of 9, then such a number $m$ cannot begin with 9 since that would result in $\frac{m}{9}$ also being an n-digits, hence a contradiction. Therefore, this single n-digit power of 9 must not begin with 9. In the case that there are 2 n-digit powers of 9, we have already discovered that one of them must begin with 9, let it be $9^k = m$ . The integer $9^{k - 1} = \frac{m}{9}$ must then be an n-digit number that begins with $1$
Using this lemma, out of the 4001 powers of 9 in T, exactly 3816 must not begin with 9 (one for each digit). Therefore, the answer is $4000 - 3816 = \boxed{184}$ numbers have 9 as their leftmost digits. | null | 184 |
3173ec80d3ca533b29e2b2f9b97d2732 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_13 | Let $T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}$ . Given that $9^{4000}_{}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T_{}^{}$ have 9 as their leftmost digit? | Let's divide all elements of $T$ into sections. Each section ranges from $10^{i}$ to $10^{i+1}$ And, each section must have 1 or 2 elements. So, let's consider both cases.
If a section has 1 element, we claim that the number doesn't have 9 as the leftmost digit. Let this element be $9^{k}$ and the section ranges from $10^{i}$ to $10^{i+1}$ . To the contrary, let's assume the number ( $9^{k}$ ) does have 9 as the leftmost digit. Thus, $9 \cdot 10^i \leq 9^k$ . But, if you divide both sides by 9, you get $10^i \leq 9^{k-1}$ , and because $9^{k-1} < 9^{k} < 10^{i+1}$ , so we have another number ( $9^{k-1}$ ) in the same section ( $10^i \leq 9^{k-1} < 10^{i+1}$ ). Which is a contradiction to our assumption that the section only has 1 element. So in this case, the number doesn't have 9 as the leftmost digit.
If a section has 2 elements, we claim one has to have a 9 as the leftmost digit, one doesn't. Let the elements be $9^{k}$ and $9^{k+1}$ , and the section ranges from $10^{i}$ to $10^{i+1}$ .
So We know $10^{i} \leq 9^{k} < 9^{k+1} < 10^{i+1}$ . From $10^{i} \leq 9^{k}$ , we know $9 \cdot 10^{i} \leq 9^{k+1}$ , and since $9^{k+1} < 10^{i+1}$ . The number( $9^{k+1}$ )'s leftmost digit must be 9, and the other number( $9^{k}$ )'s leftmost digit is 1.
There are total 4001 elements in $T$ and 3817 sections that have 1 or 2 elements. And, no matter how many elements a section has, each section contains exactly one element that doesn't begin with 9. We can take 4001 elements, subtract 3817 elements that don't have 9 as the leftmost digit, and get $\boxed{184}$ numbers that have 9 as the leftmost digit. | null | 184 |
3173ec80d3ca533b29e2b2f9b97d2732 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_13 | Let $T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}$ . Given that $9^{4000}_{}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T_{}^{}$ have 9 as their leftmost digit? | We know that $9$ is very close to $10$ . But we also know that for all $k \in \mathbb{Z}^+$ $10^k$ has $k+1$ digits. Hence $10^{4000}$ has $4001$ digits. Now, notice that if a power of $9^n$ has $9$ as the leftmost digit, then $9^{n-1}$ has $1$ as the leftmost digit and will preserve the same number of digits (just due to division). Hence we find that consecutive powers of $9$ are "supposed" to increase by one digit, however, when consecutive powers of $9$ have leading digit $1$ then leading digit $9$ they "lose" this one digit. Therefore, it suffices to find the number of times that $9^k$ has "lost" a digit since the number of times that this occurs shows us the number of pairs where the leading digit of $9^k$ is $1$ and the leading digit of $9^{k+1}$ is $9$ . Hence, we find that $9^{4000}$ is "supposed" to have $4001$ digits, but only has $3817$ digits. Thus, the number of times $9^k$ has lost a digit is $4001-3817 = 184$ . Thus, $\boxed{184}$ elements of $T$ have $9$ as their leading digit. | null | 184 |
c6f43c0a401c0a50fd7566fba848f8ef | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_14 | 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 $V = 6\sqrt{133}\left(\frac{99}{\sqrt{133}}\right) = \boxed{594}$ | null | 594 |
c6f43c0a401c0a50fd7566fba848f8ef | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_14 | 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 $\frac13 \cdot 9\cdot13\cdot \frac{66\cdot3}{13}=\boxed{594}$ | null | 594 |
4ff7758eeea95adc1120449999fce657 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_15 | Find $ax^5 + by^5$ if the real numbers $a,b,x,$ and $y$ satisfy the equations \begin{align*} ax + by &= 3, \\ ax^2 + by^2 &= 7, \\ ax^3 + by^3 &= 16, \\ ax^4 + by^4 &= 42. \end{align*} | Set $S = (x + y)$ and $P = xy$ . Then the relationship
\[(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})\]
can be exploited:
\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\ (ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}
Therefore:
\begin{eqnarray*}7S & = & 16 + 3P \\ 16S & = & 42 + 7P\end{eqnarray*}
Consequently, $S = - 14$ and $P = - 38$ . Finally:
\begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \\ (42)(S) & = & (ax^5 + by^5) + (P)(16) \\ (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\ ax^5 + by^5 & = & \boxed{020} | null | 020 |
4ff7758eeea95adc1120449999fce657 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_15 | Find $ax^5 + by^5$ if the real numbers $a,b,x,$ and $y$ satisfy the equations \begin{align*} ax + by &= 3, \\ ax^2 + by^2 &= 7, \\ ax^3 + by^3 &= 16, \\ ax^4 + by^4 &= 42. \end{align*} | recurrence of the form $T_n=AT_{n-1}+BT_{n-2}$ will have the closed form $T_n=ax^n+by^n$ , where $x,y$ are the values of the starting term that make the sequence geometric, and $a,b$ are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.
Suppose we have such a recurrence with $T_1=3$ and $T_2=7$ . Then $T_3=ax^3+by^3=16=7A+3B$ , and $T_4=ax^4+by^4=42=16A+7B$
Solving these simultaneous equations for $A$ and $B$ , we see that $A=-14$ and $B=38$ . So, $ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}$ | null | 020 |
4ff7758eeea95adc1120449999fce657 | https://artofproblemsolving.com/wiki/index.php/1990_AIME_Problems/Problem_15 | Find $ax^5 + by^5$ if the real numbers $a,b,x,$ and $y$ satisfy the equations \begin{align*} ax + by &= 3, \\ ax^2 + by^2 &= 7, \\ ax^3 + by^3 &= 16, \\ ax^4 + by^4 &= 42. \end{align*} | We first let the answer to this problem be $k.$ Multiplying the first equation by $x$ gives $ax^2 + bxy=3x$
Subtracting this equation from the second equation gives $by^2-bxy=7-3x$ . Similarly, doing the same for the other equations, we obtain: $by^2-bxy=7-3x$ $by^3-bxy^2=16-7x$ $by^4-bxy^3=42-16x$ , and $by^5-bxy^4=k-42x$
Now lets take the first equation. Multiplying this by $y$ and subtracting this from the second gives us $by^3-bxy^2=(7-3x)y$ . We can also obtain $by^4-bxy^3=(16-7x)y$
Now we can solve for $x$ and $y$ $(7-3x)y = 16-7x$ and $(16-7x)y=42-16x$ . Solving for $x$ and $y$ gives us $(-7+\sqrt{87},-7-\sqrt{87})$ (It can be switched, but since the given equations are symmetric, it doesn't matter). $k-42x= (42-16x)y$ , and solving for $k$ gives us $k= \boxed{020}$ | null | 020 |
96193976a32b3c29ba6565e615be0f4b | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | Note that the four numbers to multiply are symmetric with the center at $29.5$ .
Multiply the symmetric pairs to get $31\cdot 28=868$ and $30\cdot 29=870$ $\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}$ | null | 869 |
96193976a32b3c29ba6565e615be0f4b | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | Notice that $(a+1)^2 = a \cdot (a+2) +1$ . Then we can notice that $30 \cdot 29 =870$ and that $31 \cdot 28 = 868$ . Therefore, $\sqrt{(31)(30)(29)(28) +1} = \sqrt{(870)(868) +1} = \sqrt{(868 +1)^2} = \boxed{869}$ . This is because we have that $a=868$ as per the equation $(a+1)^2 = a \cdot (a+2) +1$ | null | 869 |
96193976a32b3c29ba6565e615be0f4b | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | More generally, we can prove that one more than the product of four consecutive integers must be a perfect square: \begin{align*} (a+3)(a+2)(a+1)(a)+1 &= [(a+3)(a)][(a+2)(a+1)]+1 \\ &= [a^2+3a][a^2+3a+2]+1 \\ &= [a^2+3a]^2+2[a^2+3a]+1 \\ &= [a^2+3a+1]^2. \end{align*} At $a=28,$ we have \[\sqrt{(a+3)(a+2)(a+1)(a)+1}=a^2+3a+1=\boxed{869}.\] ~Novus677 (Fundamental Logic) | null | 869 |
96193976a32b3c29ba6565e615be0f4b | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | Similar to Solution 1 above, call the consecutive integers $\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)$ to make use of symmetry. Note that $n$ itself is not an integer - in this case, $n = 29.5$ . The expression becomes $\sqrt{\left(n-\frac{3}{2}\right)\left(n + \frac{3}{2}\right)\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) + 1}$ . Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives $\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}$ . The inside is a perfect square trinomial, since $b^2 = 4ac$ . It's equal to $\sqrt{\left(n^2 - \frac{5}{4}\right)^2}$ , which simplifies to $n^2 - \frac{5}{4}$ . You can plug in the value of $n$ from there, or further simplify to $\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) - 1$ , which is easier to compute. Either way, plugging in $n=29.5$ gives $\boxed{869}$ | null | 869 |
96193976a32b3c29ba6565e615be0f4b | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | We have $(31)(30)(29)(28)+1=755161.$ Since the alternating sum of the digits $7-5+5-1+6-1=11$ is divisible by $11,$ we conclude that $755161$ is divisible by $11.$
We evaluate the original expression by prime factorization: \begin{align*} \sqrt{(31)(30)(29)(28)+1}&=\sqrt{755161} \\ &=\sqrt{11\cdot68651} \\ &=\sqrt{11^2\cdot6241} \\ &=\sqrt{11^2\cdot79^2} \\ &=11\cdot79 \\ &=\boxed{869} ~Vrjmath (Fundamental Logic) | null | 869 |
96193976a32b3c29ba6565e615be0f4b | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_1 | Compute $\sqrt{(31)(30)(29)(28)+1}$ | The last digit under the radical is $1$ , so the square root must either end in $1$ or $9$ , since $x^2 = 1\pmod {10}$ means $x = \pm 1$ . Additionally, the number must be near $29 \cdot 30 = 870$ , narrowing the reasonable choices to $869$ and $871$
Continuing the logic, the next-to-last digit under the radical is the same as the last digit of $28 \cdot 29 \cdot 3 \cdot 31$ , which is $6$ . Quick computation shows that $869^2$ ends in $61$ , while $871^2$ ends in $41$ . Thus, the answer is $\boxed{869}$ | null | 869 |
249dd81c9ec98c831a0ec35a2bc37b4f | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_2 | Ten points are marked on a circle . How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices | Any subset of the ten points with three or more members can be made into exactly one such polygon. Thus, we need to count the number of such subsets. There are $2^{10} = 1024$ total subsets of a ten-member set , but of these ${10 \choose 0} = 1$ have 0 members, ${10 \choose 1} = 10$ have 1 member and ${10 \choose 2} = 45$ have 2 members. Thus the answer is $1024 - 1 - 10 - 45 = \boxed{968}$ | null | 968 |
2302e884c1ddecbf272bd2b005101643 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3 | Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if | Repeating decimals represent rational numbers . To figure out which rational number, we sum an infinite geometric series $0.d25d25d25\ldots = \sum_{n = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}$ . Thus $\frac{n}{810} = \frac{100d + 25}{999}$ so $n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}$ . Since 750 and 37 are relatively prime $4d + 1$ must be divisible by 37, and the only digit for which this is possible is $d = 9$ . Thus $4d + 1 = 37$ and $n = \boxed{750}$ | null | 750 |
2302e884c1ddecbf272bd2b005101643 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3 | Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if | To get rid of repeating decimals, we multiply the equation by 1000. We get $\frac{1000n}{810} = d25.d25d25...$ We subtract the original equation from the second to get $\frac{999n}{810}=d25$ We simplify to $\frac{37n}{30} = d25$ Since $\frac{37n}{30}$ is an integer, $n=(30)(5)(2k+1)$ because $37$ is relatively prime to $30$ , and d25 is divisible by $5$ but not $10$ . The only odd number that yields a single digit $d$ and 25 at the end of the three digit number is $k=2$ , so the answer is $\boxed{750}$ | null | 750 |
2302e884c1ddecbf272bd2b005101643 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3 | Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if | Similar to Solution 2, we start off by writing that $\frac{1000n}{810} = d25.d25d25 \dots$ .Then we subtract this from the original equation to get:
\[\frac{999n}{810} =d25 \Longrightarrow \frac{37n}{30} = d25 \Longrightarrow 37n = d25 \cdot 30\]
Since n is an integer, we have that $37 \mid d25 \cdot 30$
Since $37$ is prime, we can apply Euclid's Lemma (which states that if $p$ is a prime and if $a$ and $b$ are integers and if $p \mid ab$ , then $p \mid a$ or $p \mid b$ ) to realize that $37 \mid d25$ , since $37 \nmid 30$ . Then we can expand $d25$ as $25 \cdot (4d +1)$ . Since $37 \nmid 25$ , by Euclid, we can arrive at $37 \mid 4d+1 \Longrightarrow d=9$ . From this we know that $n= 25 \cdot 30 = \boxed{750}$ . (This is true because $37n = 925 \cdot 30 \rightarrow n= 25 \cdot 30 = 750$ | null | 750 |
2302e884c1ddecbf272bd2b005101643 | https://artofproblemsolving.com/wiki/index.php/1989_AIME_Problems/Problem_3 | Suppose $n$ is a positive integer and $d$ is a single digit in base 10 . Find $n$ if | Write out these equations:
$\frac{n}{810} = \frac{d25}{999}$
$\frac{n}{30} = \frac{d25}{37}$
$37n = 30(d25)$
Thus $n$ divides 25 and 30. The only solution for this under 1000 is $\boxed{750}$ | null | 750 |