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baf2db916afa4f3027cb19ae92bcfe22 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_3 | Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number? | Let $x$ be the two-digit number, $y$ be the three-digit number. Putting together the given, we have $1000x+y=9xy \Longrightarrow 9xy-1000x-y=0$ . Using SFFT , this factorizes to $(9x-1)\left(y-\dfrac{1000}{9}\right)=\dfrac{1000}{9}$ , and $(9x-1)(9y-1000)=1000$
Since $89 < 9x-1 < 890$ , we can use trial and error on factors of 1000. If $9x - 1 = 100$ , we get a non-integer. If $9x - 1 = 125$ , we get $x=14$ and $y=112$ , which satisifies the conditions. Hence the answer is $112 + 14 = \boxed{126}$ | null | 126 |
baf2db916afa4f3027cb19ae92bcfe22 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_3 | Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number? | As shown above, we have $1000x+y=9xy$ , so $1000/y=9-1/x$ $1000/y$ must be just a little bit smaller than 9, so we find $y=112$ $x=14$ , and the solution is $\boxed{126}$ | null | 126 |
baf2db916afa4f3027cb19ae92bcfe22 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_3 | Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number? | To begin, we rewrite $(10a+b)*(100x+10y+z)*9 = 10000a + 1000b + 100x + 10y + z$
as
$(90a+9b-1)(100x+10y+z) = 10000a + 1000b$
and
$(90a+9b-1)(100x+10y+z) = 1000(10a + b)$
This is the most important part: Notice $(90a+9b-1)$ is $-1 \pmod{10a+b}$ and $1000(10a + b)$ is $0\pmod{10a+b}$ . That means $(100x+10y+z)$ is also $0\pmod{10+b}$ . Rewrite $(100x+10y+z)$ as $n\times(10a+b)$
$(90a+9b-1)\times n(10a+b)= 1000(10a + b)$
$(90a+9b-1)\times n= 1000$
Now we have to find a number that divides 1000 using prime factors 2 or 5 and is $8\pmod9$ . It is quick to find there is only one: 125. That gives 14 as $10a+b$ and 112 as $100x+10y+z$ . Therefore the answer is $112 + 14 = \boxed{126}$ | null | 126 |
e541c9483e476a60396f7feeada901f4 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_4 | Circles of radii $5, 5, 8,$ and $\frac mn$ are mutually externally tangent, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 1997 AIME-4.png
If (in the diagram above) we draw the line going through the centers of the circles with radii $8$ and $\frac mn = r$ , that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii $5$ . Then we form two right triangles , of lengths $5, x, 5+r$ and $5, 8+r+x, 13$ , wher $x$ is the distance between the center of the circle in question and the segment connecting the centers of the two circles of radii $5$ . By the Pythagorean Theorem , we now have two equations with two unknowns:
\begin{eqnarray*} 5^2 + x^2 &=& (5+r)^2 \\ x &=& \sqrt{10r + r^2} \\ && \\ (8 + r + \sqrt{10r+r^2})^2 + 5^2 &=& 13^2\\ 8 + r + \sqrt{10r+r^2} &=& 12\\ \sqrt{10r+r^2}&=& 4-r\\ 10r+r^2 &=& 16 - 8r + r^2\\ r &=& \frac{8}{9} \end{eqnarray*}
So $m+n = \boxed{17}$ | null | 17 |
e541c9483e476a60396f7feeada901f4 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_4 | Circles of radii $5, 5, 8,$ and $\frac mn$ are mutually externally tangent, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | We may also use Descartes' theorem, $k_4=k_1+k_2+k_3\pm 2\sqrt{k_1k_2+k_2k_3+k_3k_1}$ where each of $k_i$ is the curvature of a circle with radius $r_i$ , and the curvature is defined as $k_i=\frac{1}{r_i}$ . The larger solution for $k_4$ will give the curvature of the circle externally tangent to the other circles, while the smaller solution will give the curvature for the circle internally tangent to each of the other circles. Using Descartes' theorem, we get $k_4=\frac15+\frac15+\frac18+2\sqrt{\frac{1}{40}+\frac{1}{40}+\frac{1}{25}}=\frac{21}{40}+2\sqrt{\frac{45}{500}}=\frac{45}{40}$ . Thus, $r_4=\frac{1}{k_4}=\frac{40}{45}=\frac89$ , and the answer is $\boxed{017}$ | null | 017 |
e57a96ce3832bafcb7cf904152dcf96e | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_5 | The number $r$ can be expressed as a four-place decimal $0.abcd,$ where $a, b, c,$ and $d$ represent digits , any of which could be zero. It is desired to approximate $r$ by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to $r$ is $\frac 27.$ What is the number of possible values for $r$ | The nearest fractions to $\frac 27$ with numerator $1$ are $\frac 13, \frac 14$ ; and with numerator $2$ are $\frac 26, \frac 28 = \frac 13, \frac 14$ anyway. For $\frac 27$ to be the best approximation for $r$ , the decimal must be closer to $\frac 27 \approx .28571$ than to $\frac 13 \approx .33333$ or $\frac 14 \approx .25$
Thus $r$ can range between $\frac{\frac 14 + \frac{2}{7}}{2} \approx .267857$ and $\frac{\frac 13 + \frac{2}{7}}{2} \approx .309523$ . At $r = .2678, .3096$ , it becomes closer to the other fractions, so $.2679 \le r \le .3095$ and the number of values of $r$ is $3095 - 2679 + 1 = \boxed{417}$ | null | 417 |
00604caee60933feb5d21c26ea836b8d | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_6 | Point $B$ is in the exterior of the regular $n$ -sided polygon $A_1A_2\cdots A_n$ , and $A_1A_2B$ is an equilateral triangle . What is the largest value of $n$ for which $A_1$ $A_n$ , and $B$ are consecutive vertices of a regular polygon? | 1997 AIME-6.png
Let the other regular polygon have $m$ sides. Using the interior angle of a regular polygon formula, we have $\angle A_2A_1A_n = \frac{(n-2)180}{n}$ $\angle A_nA_1B = \frac{(m-2)180}{m}$ , and $\angle A_2A_1B = 60^{\circ}$ . Since those three angles add up to $360^{\circ}$
\begin{eqnarray*} \frac{(n-2)180}{n} + \frac{(m-2)180}{m} &=& 300\\ m(n-2)180 + n(m-2)180 &=& 300mn\\ 360mn - 360m - 360n &=& 300mn\\ mn - 6m - 6n &=& 0 \end{eqnarray*} Using SFFT
\begin{eqnarray*} (m-6)(n-6) &=& 36 \end{eqnarray*} Clearly $n$ is maximized when $m = 7, n = \boxed{042}$ | null | 042 |
00604caee60933feb5d21c26ea836b8d | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_6 | Point $B$ is in the exterior of the regular $n$ -sided polygon $A_1A_2\cdots A_n$ , and $A_1A_2B$ is an equilateral triangle . What is the largest value of $n$ for which $A_1$ $A_n$ , and $B$ are consecutive vertices of a regular polygon? | As above, find that $mn - 6m - 6n = 0$ using the formula for the interior angle of a polygon.
Solve for $n$ to find that $n = \frac{6m}{m-6}$ . Clearly, $m>6$ for $n$ to be positive.
With this restriction of $m>6$ , the larger $m$ gets, the smaller the fraction $\frac{6m}{m-6}$ becomes. This can be proven either by calculus, by noting that $n = \frac{6m}{m-6}$ is a transformed hyperbola, or by dividing out the rational function to get $n = 6 + \frac{36}{m - 6}.$
Either way, minimizng $m$ will maximize $n$ , and the smallest integer $m$ such that $n$ is positive is $m=7$ , giving $n = \boxed{042}$ | null | 042 |
00604caee60933feb5d21c26ea836b8d | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_6 | Point $B$ is in the exterior of the regular $n$ -sided polygon $A_1A_2\cdots A_n$ , and $A_1A_2B$ is an equilateral triangle . What is the largest value of $n$ for which $A_1$ $A_n$ , and $B$ are consecutive vertices of a regular polygon? | From the formula for the measure for an individual angle of a regular n-gon, $180 - \frac{360}{n}$ , the measure of $\angle A_2A_1A_n = 180 - \frac{360}{n}$ . Together with the fact that an equilateral triangle has angles measuring 60 degrees, the measure of $\angle A_nA_1B = 120 + \frac{360}{n}$ (Notice that this value decreases as $n$ increases; hence, we are looking for the least possible value of $\angle A_nA_1B$ ). For $A_n, A_1, B$ to be vertices of a regular polygon, $\angle A_nA_1B$ must be of the form $180 - \frac{360}{n}$ , where $n$ is a natural number greater than or equal to 3. It is obvious that $\angle A_nA_1B > 120$ . The least angle satisfying this condition is $180 - \frac{360}{7}$ . Equating this with $120 + \frac{360}{n}$ and solving yields $n = \boxed{042}$ | null | 042 |
d004542859146729787ba78f246da6a1 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_7 | A car travels due east at $\frac 23$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac 12\sqrt{2}$ mile per minute. At time $t=0$ , the center of the storm is $110$ miles due north of the car. At time $t=t_1$ minutes, the car enters the storm circle, and at time $t=t_2$ minutes, the car leaves the storm circle. Find $\frac 12(t_1+t_2)$ | We set up a coordinate system, with the starting point of the car at the origin . At time $t$ , the car is at $\left(\frac 23t,0\right)$ and the center of the storm is at $\left(\frac{t}{2}, 110 - \frac{t}{2}\right)$ . Using the distance formula,
\begin{eqnarray*} \sqrt{\left(\frac{2}{3}t - \frac 12t\right)^2 + \left(110-\frac{t}{2}\right)^2} &\le& 51\\ \frac{t^2}{36} + \frac{t^2}{4} - 110t + 110^2 &\le& 51^2\\ \frac{5}{18}t^2 - 110t + 110^2 - 51^2 &\le& 0\\ \end{eqnarray*}
Noting that $\frac 12 \left(t_1+t_2 \right)$ is at the maximum point of the parabola, we can use $-\frac{b}{2a} = \frac{110}{2 \cdot \frac{5}{18}} = \boxed{198}$ | null | 198 |
d004542859146729787ba78f246da6a1 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_7 | A car travels due east at $\frac 23$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac 12\sqrt{2}$ mile per minute. At time $t=0$ , the center of the storm is $110$ miles due north of the car. At time $t=t_1$ minutes, the car enters the storm circle, and at time $t=t_2$ minutes, the car leaves the storm circle. Find $\frac 12(t_1+t_2)$ | First do the same process for assigning coordinates to the car. The car moves $\frac{2}{3}$ miles per minute to the right, so the position starting from $(0,0)$ is $\left(\frac{2}{3}t, 0\right)$
Take the storm as circle. Given southeast movement, split the vector into component, getting position $\left(\frac{1}{2}t, 110 - \frac{1}{2}t\right)$ for the storm's center. This circle with radius 51 yields $\left(x - \frac{1}{2}t\right)^2 + \left(y -110 + \frac{1}{2}t\right)^2 = 51^2$
Now substitute the car's coordinates into the circle's:
$\left(\frac{2}{3}t - \frac{1}{2}t\right)^2 + \left(-110 + \frac{1}{2}t\right)^2 = 51^2$
Simplifying and then squaring:
$\left(\frac{1}{6}t\right)^2 + \left(-110 + \frac{1}{2}t\right)^2 = 51^2$
$\frac{1}{36}t^2 + \frac{1}{4}t^2 - 110t + 110^2$
Forming into a quadratic we get the following, then set equal to 0, since the first time the car hits the circumference of the storm is $t_{1}$ and the second is $t_{2}$
$\frac{5}{18}t^2 - 110t + 110^2 - 51^2 = 0$
The problem asks for sum of solutions divided by 2 so sum is equal to:
$-\frac{b}{a} = -\frac{-110}{\frac{5}{18}} = 110\cdot{\frac{18}{5}} = 396\cdot{\frac{1}{2}} = \boxed{198}$ | null | 198 |
d004542859146729787ba78f246da6a1 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_7 | A car travels due east at $\frac 23$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac 12\sqrt{2}$ mile per minute. At time $t=0$ , the center of the storm is $110$ miles due north of the car. At time $t=t_1$ minutes, the car enters the storm circle, and at time $t=t_2$ minutes, the car leaves the storm circle. Find $\frac 12(t_1+t_2)$ | We only need to know how the storm and car move relative to each other, so we can find this by subtracting the storm's movement vector from the car's. This gives the car's movement vector as $\left(\frac{1}{6}, \frac{1}{2}\right)$ . Labeling the car's starting position A, the storm center B, and the right triangle formed by AB with a right angle at B and the car's path, we get the following diagram, with AD as our desired length since D is the average of the points where the car enters and exits the storm.
[asy] size(200,200); draw((0,0)--(0,110)); label("A",(0,0),S); dot((0,0)); dot((0,110)); label("B",(0,110),NE); draw(circle((0,110),51)); draw((0,0)--(161/3,161.0),EndArrow); draw((0,110)--(110/3,110.0)); label("C",(110/3,110.0),SE); dot((110/3,110.0)); label("D",(33,99),SE); dot((33,99)); draw((0,110)--(33,99)); markscalefactor=1; draw(rightanglemark((0,110),(33,99),(0,0))); [/asy]
$AB = 110$ , so $CB = \frac{110}{3}$ . The Pythagorean Theorem then gives $AC = \frac{110\sqrt{10}}{3}$ , and since $\bigtriangleup ABC \sim \bigtriangleup ADB$ $AD = (AB)\frac{AB}{AC} = 33\sqrt{10}$ . The Pythagorean Theorem now gives the car's speed as $\sqrt{\frac{5}{18}}$ , and finally $\frac{33\sqrt{10}}{\sqrt{\frac{5}{18}}} = \boxed{198}$ | null | 198 |
4ba106198c3fede8b1c08cd2f61fd48d | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_8 | How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0? | The problem is asking us for all configurations of $4\times 4$ grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:
Adding these cases up, we get $36 + 48 + 6 = \boxed{090}$ | null | 090 |
4ba106198c3fede8b1c08cd2f61fd48d | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_8 | How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0? | Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of $6$ ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange the one 1 and 2 -1's we have left to put. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is $6*3*(2+1+2)$ $\boxed{090}$ | null | 090 |
4ba106198c3fede8b1c08cd2f61fd48d | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_8 | How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0? | Notice that for every arrangement $A$ of the first rows of $-1$ s and $1$ s, we have the inverse of that row $A^{-1}$ so that the sum of the rows and columns of $A$ and $A^{-1}$ is $0$ . Therefore if we have another arrangement $B$ , we have $B^{-1}$ . For instance, if $A=(-1,1,1,-1)$ $A^{-1}=(1,-1,-1,1)$ . We then have that if we fix the first row $A$ , we have that first there are $\binom{4}{2}$ values of the fixed $A$ . We then have the following cases:
Case $1$ : ( $AA^{-1}BB^{-1}$ ). $\binom{4}{2}$
Case $2$ : ( $ABA^{-1}B^{-1}$ ). $\binom{4}{2}$
Case $3$ : ( $AAA^{-1}A^{-1}$ ). $\binom{4}{2}/2$ , where here we divided by $2$ because then we would overcount $AA$ and $A^{-1}A^{-1}$
Therefore the answer is $\binom{4}{2}(\binom{4}{2}+\binom{4}{2}+\binom{4}{2}/2)$ $6(6+6+3)$ $\boxed{090}$ | null | 090 |
a34d9685ee09a4c4c00ee44401e3bde9 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_9 | Given a nonnegative real number $x$ , let $\langle x\rangle$ denote the fractional part of $x$ ; that is, $\langle x\rangle=x-\lfloor x\rfloor$ , where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle$ , and $2<a^2<3$ . Find the value of $a^{12}-144a^{-1}$ | Looking at the properties of the number, it is immediately guess-able that $a = \phi = \frac{1+\sqrt{5}}2$ (the golden ratio ) is the answer. The following is the way to derive that:
Since $\sqrt{2} < a < \sqrt{3}$ $0 < \frac{1}{\sqrt{3}} < a^{-1} < \frac{1}{\sqrt{2}} < 1$ . Thus $\langle a^2 \rangle = a^{-1}$ , and it follows that $a^2 - 2 = a^{-1} \Longrightarrow a^3 - 2a - 1 = 0$ . Noting that $-1$ is a root, this factors to $(a+1)(a^2 - a - 1) = 0$ , so $a = \frac{1 + \sqrt{5}}{2}$ (we discard the negative root).
Our answer is $(a^2)^{6}-144a^{-1} = \left(\frac{3+\sqrt{5}}2\right)^6 - 144\left(\frac{2}{1 + \sqrt{5}}\right)$ Complex conjugates reduce the second term to $-72(\sqrt{5}-1)$ . The first term we can expand by the binomial theorem to get $\frac 1{2^6}\left(3^6 + 6\cdot 3^5\sqrt{5} + 15\cdot 3^4 \cdot 5 + 20\cdot 3^3 \cdot 5\sqrt{5} + 15 \cdot 3^2 \cdot 25 + 6 \cdot 3 \cdot 25\sqrt{5} + 5^3\right)$ $= \frac{1}{64}\left(10304 + 4608\sqrt{5}\right) = 161 + 72\sqrt{5}$ . The answer is $161 + 72\sqrt{5} - 72\sqrt{5} + 72 = \boxed{233}$ | null | 233 |
a34d9685ee09a4c4c00ee44401e3bde9 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_9 | Given a nonnegative real number $x$ , let $\langle x\rangle$ denote the fractional part of $x$ ; that is, $\langle x\rangle=x-\lfloor x\rfloor$ , where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle$ , and $2<a^2<3$ . Find the value of $a^{12}-144a^{-1}$ | Find $a$ as shown above. Note that, since $a$ is a root of the equation $a^3 - 2a - 1 = 0$ $a^3 = 2a + 1$ , and $a^{12} = (2a + 1)^4$ . Also note that, since $a$ is a root of $a^2 - a - 1 = 0$ $\frac{1}{a} = a - 1$ . The expression we wish to calculate then becomes $(2a + 1)^4 - 144(a - 1)$ . Plugging in $a = \frac{1 + \sqrt{5}}{2}$ , we plug in to get an answer of $(161 + 72\sqrt{5}) + 72 - 72\sqrt{5} = 161 + 72 = \boxed{233}$ | null | 233 |
a34d9685ee09a4c4c00ee44401e3bde9 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_9 | Given a nonnegative real number $x$ , let $\langle x\rangle$ denote the fractional part of $x$ ; that is, $\langle x\rangle=x-\lfloor x\rfloor$ , where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle$ , and $2<a^2<3$ . Find the value of $a^{12}-144a^{-1}$ | Find $a$ as shown above. Note that $a$ satisfies the equation $a^2 = a+1$ (this is the equation we solved to get it). Then, we can simplify $a^{12}$ as follows using the fibonacci numbers:
$a^{12} = a^{11}+a^{10}= 2a^{10} + a^{9} = 3a^9+ 2a^8 = ... = 144a^1+89a^0 = 144a+89$
So we want $144(a-\frac1a)+89 = 144(1)+89 = \boxed{233}$ since $a-\frac1a = 1$ is equivalent to $a^2 = a+1$ | null | 233 |
a34d9685ee09a4c4c00ee44401e3bde9 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_9 | Given a nonnegative real number $x$ , let $\langle x\rangle$ denote the fractional part of $x$ ; that is, $\langle x\rangle=x-\lfloor x\rfloor$ , where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle$ , and $2<a^2<3$ . Find the value of $a^{12}-144a^{-1}$ | As Solution 1 stated, $a^3 - 2a - 1 = 0$ $a^3 - 2a - 1 = a^3 - a^2 -a + a^2 -a -1 = (a+1)(a^2 - a - 1)$ . So, $a^2 - a - 1 = 0$ $1 = a^2 - a$ $\frac1a = a-1$ $a^3 = 2a+1$ $a^2 = a+1$
$a^6 = (a^3)^2 = (2a+1)^2= 4a^2 + 4a +1= 4(a+1) + 4a + 1= 8a+5$
$a^{12} = (a^6)^2 = (8a+5)^2 = 64a^2 + 80a + 25 = 64(a+1) + 80a + 25 = 144a + 89$
Therefore, $a^{12} - 144 a^{-1} = 144a + 89 - 144(a-1) = 89 + 144 = \boxed{233}$ | null | 233 |
de57ca06e10ba589dbc0c50a90b1e539 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_10 | Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:
i. Either each of the three cards has a different shape or all three of the cards have the same shape.
ii. Either each of the three cards has a different color or all three of the cards have the same color.
iii. Either each of the three cards has a different shade or all three of the cards have the same shade.
How many different complementary three-card sets are there? | Adding the cases up, we get $27 + 54 + 36 = \boxed{117}$ | null | 117 |
de57ca06e10ba589dbc0c50a90b1e539 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_10 | Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:
i. Either each of the three cards has a different shape or all three of the cards have the same shape.
ii. Either each of the three cards has a different color or all three of the cards have the same color.
iii. Either each of the three cards has a different shade or all three of the cards have the same shade.
How many different complementary three-card sets are there? | Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different there is only one option- choose the only value that is remaining. In this way, every two card pick corresponds to exactly one set, for a total of $\binom{27}{2} = 27*13 = 351$ possibilities. Note, however, that each set is generated by ${3\choose 2} = 3$ pairs, so we've overcounted by a multiple of 3 and the answer is $\frac{351}{3} = \boxed{117}$ | null | 117 |
de57ca06e10ba589dbc0c50a90b1e539 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_10 | Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:
i. Either each of the three cards has a different shape or all three of the cards have the same shape.
ii. Either each of the three cards has a different color or all three of the cards have the same color.
iii. Either each of the three cards has a different shade or all three of the cards have the same shade.
How many different complementary three-card sets are there? | Treat the sets as ordered. Then for each of the three criterion, there are $3!=6$ choices if the attribute is different and there are $3$ choices is the attribute is the same. Thus all three attributes combine to a total of $(6+3)^3=729$ possibilities. However if all three attributes are the same then the set must be composed of three cards that are the same, which is impossible. This takes out $3^3=27$ possibilities. Notice that we have counted every set $3!=6$ times by treating the set as ordered. The final solution is then $\frac{729-27}{6}=\boxed{117}$ | null | 117 |
dba966c1900ece63a6bc6b5de4a47b60 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | Note that $\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=1}^{44} \cos n}{\sum_{n=46}^{89} \cos n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}$ by the cofunction identities.(We could have also written it as $\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=46}^{89} \sin n}{\sum_{n=1}^{44} \sin n} = \frac {\sin 89 + \sin 88 + \dots + \sin 46}{\sin 1 + \sin 2 + \dots + \sin 44}$ .)
Now use the sum-product formula $\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$ We want to pair up $[1, 44]$ $[2, 43]$ $[3, 42]$ , etc. from the numerator and $[46, 89]$ $[47, 88]$ $[48, 87]$ etc. from the denominator. Then we get: \[\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=1}^{44} \cos n}{\sum_{n=46}^{89} \cos n} = \frac{2\cos(\frac{45}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})]}{2\cos(\frac{135}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})]} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}\]
To calculate this number, use the half angle formula. Since $\cos\left(\frac{x}{2}\right) = \pm \sqrt{\frac{\cos x + 1}{2}}$ , then our number becomes: \[\frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}\] in which we drop the negative roots (as it is clear cosine of $22.5$ and $67.5$ are positive). We can easily simplify this:
\begin{eqnarray*} \frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}} &=& \sqrt{\frac{\frac{2+\sqrt{2}}{4}}{\frac{2-\sqrt{2}}{4}}} \\ &=& \sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} \cdot \sqrt{\frac{2+\sqrt{2}}{2+\sqrt{2}}} \\ &=& \sqrt{\frac{(2+\sqrt{2})^2}{2}} \\ &=& \frac{2+\sqrt{2}}{\sqrt{2}} \cdot \sqrt{2} \\ &=& \sqrt{2}+1 \end{eqnarray*}
And hence our answer is $\lfloor 100x \rfloor = \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}$ | null | 241 |
dba966c1900ece63a6bc6b5de4a47b60 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | \begin{eqnarray*} x &=& \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\ &=& \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44} \end{eqnarray*}
Using the identity $\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}$ $\Longrightarrow \sin x + \cos x$ $= \sin x + \sin (90-x)$ $= 2 \sin 45 \cos (45-x)$ $= \sqrt{2} \cos (45-x)$ , that summation reduces to
\begin{eqnarray*}x &=& \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\ &=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right) \end{eqnarray*}
This fraction is equivalent to $x$ . Therefore, \begin{eqnarray*} x &=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\ \frac {1}{\sqrt {2}} &=& x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\ x &=& \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\ \lfloor 100x \rfloor &=& \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241} | null | 241 |
dba966c1900ece63a6bc6b5de4a47b60 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | A slight variant of the above solution, note that
\begin{eqnarray*} \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &=& \sum_{n=1}^{44} \sin n + \sin(90-n)\\ &=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\ \sum_{n=1}^{44} \sin n &=& (\sqrt{2}-1)\sum_{n=1}^{44} \cos n \end{eqnarray*}
This is the ratio we are looking for. $x$ reduces to $\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$ , and $\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}$ | null | 241 |
dba966c1900ece63a6bc6b5de4a47b60 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | Consider the sum $\sum_{n = 1}^{44} \text{cis } n^\circ$ . The fraction is given by the real part divided by the imaginary part.
The sum can be written $- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}$ (by De Moivre's Theorem with geometric series)
$= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}$ (after multiplying by complex conjugate
$= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}$
$= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}$
Using the tangent half-angle formula , this becomes $\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)$
Dividing the two parts and multiplying each part by 4, the fraction is $\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}$
Although an exact value for $\cot (1/2^\circ)$ in terms of radicals will be difficult, this is easily known: it is really large!
So treat it as though it were $\infty$ . The fraction is approximated by $\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}$ | null | 241 |
dba966c1900ece63a6bc6b5de4a47b60 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | Consider the sum $\sum_{n = 1}^{44} \text{cis } n^\circ$ . The fraction is given by the real part divided by the imaginary part.
The sum can be written as $\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)$ .
Consider the rhombus $OABC$ on the complex plane such that $O$ is the origin, $A$ represents $\text{cis } n^\circ$ $B$ represents $\text{cis } n^\circ + \text{cis } 45-n^\circ$ and $C$ represents $\text{cis } n^\circ$ . Simple geometry shows that $\angle BOA = 22.5-k^\circ$ , so the angle that $\text{cis } n^\circ + \text{cis } 45-n^\circ$ makes with the real axis is simply $22.5^\circ$ . So $\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)$ is the sum of collinear complex numbers, so the angle the sum makes with the real axis is $22.5^\circ$ . So our answer is $\lfloor 100 \cot(22.5^\circ) \rfloor = \boxed{241}$ | null | 241 |
dba966c1900ece63a6bc6b5de4a47b60 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | We write $x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}$ since $\cos x = \sin (90^{\circ}-x).$ Now we by the sine angle sum we know that $\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).$ So the expression simplifies to $\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} (\sin n^{\circ}+\cos n^{\circ})}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).$ Therefore we have the equation $x = \sin 45^{\circ}(1+x) \implies x = \sqrt{2}+1.$ Finishing, we have $\lfloor 100x \rfloor = \boxed{241}.$ | null | 241 |
dba966c1900ece63a6bc6b5de4a47b60 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_11 | Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ | We can pair the terms of the summations as below.
\[\dfrac{(\cos{1} + \cos{44}) + (\cos{2} + \cos{43}) + (\cos{3} + \cos{42}) + \cdots + (\cos{22} + \cos{23})}{(\sin{1} + \sin{44}) + (\sin{2} + \sin{43}) + (\sin{3} + \sin{42}) + \cdots + (\sin{22} + \sin{23})}.\]
From here, we use the cosine and sine subtraction formulas as shown.
\begin{align*}
&\dfrac{(\cos(45-44) + \cos(45-1)) + (\cos(45-43) + \cos(45-2))+ \cdots (\cos(45-23) + \cos(45-22))}{(\sin(45-44) + \sin(45-1)) + (\sin(45-43) + \sin(45-2))+ \cdots (\sin(45-23) + \sin(45-22))} \\
&= \dfrac{(\cos{45}\cos{44} +\sin{45}\sin{44} + \cos{45}\cos{1} + \sin{45}\sin{1})+ \cdots + (\cos{45}\cos{23} + \sin{45}\sin{23} + \cos{45}\cos{22} + \sin{45}\sin{22})}{(\sin{45}\cos{44}-\cos{45}\sin{44} + \sin{45}\cos{1} - \cos{45}\sin{1}) + \cdots + (\sin{45}\cos{23} -\cos{45}\sin{23} + \sin{45}\cos{22} - \cos{45}\sin{22})} \\
&= \dfrac{\sqrt{2}/2(\cos{44} + \sin{44} + \cos{1}+\sin{1} +\cos{43} + \sin{43} + \cos{2} + \sin{2} + \cdots + \cos{23} + \sin{23} + \cos{22} + \sin{22})}{\sqrt{2}/2(\cos{44} -\sin{44} +\cos{1} - \sin{1} + \cos{43}-\sin{43} + \cos{2} -\sin{2} +\cdots + \cos{23}-\sin{23} + \cos{22} - \sin{22})} \\
&=\dfrac{\sum\limits_{n=1}^{44} \cos n^\circ + \sum\limits_{n=1}^{44} \sin n^\circ}{\sum\limits_{n=1}^{44} \cos n^\circ - \sum\limits_{n=1}^{44} \sin n^\circ}.
\end{align*}
Since all of these operations have been performed to ensure equality, we can set the final line of the above mess equal to our original expression.
\[\dfrac{\sum\limits_{n=1}^{44} \cos n^\circ + \sum\limits_{n=1}^{44} \sin n^\circ}{\sum\limits_{n=1}^{44} \cos n^\circ - \sum\limits_{n=1}^{44} \sin n^\circ} = \frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}.\]
For the sake of clarity, let $\sum\limits_{n=1}^{44} \cos n^\circ = C$ and $\sum\limits_{n=1}^{44} \sin n^\circ = S$ . Then, we have
\[\dfrac{C+S}{C-S} = \dfrac{C}{S} \implies CS+S^2 = C^2-CS.\]
Finishing, we have $S^2+2CS=C^2$ . Adding $C^2$ to both sides gives $(C+S)^2 = 2C^2$ , or $C+S = \pm C\sqrt{2}$ . Taking the positive case gives $S= C(\sqrt{2}-1)$ . Finally,
\[x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ} = \dfrac{C}{S} = \dfrac{1}{\sqrt{2}-1} = \sqrt{2} +1 \implies \lfloor{100x\rfloor} = \boxed{241}.\] | null | 241 |
843fbe2770f01a95f16ccc7df9a9cfa4 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_12 | The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ $b$ $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ | From $f(f(x))=x$ , it is obvious that $\frac{-d}{c}$ is the value not in the range. First notice that since $f(0)=\frac{b}{d}$ $f(\frac{b}{d})=0$ which means $a(\frac{b}{d})+b=0$ so $a=-d$ . Using $f(19)=19$ , we have that $b=361c+38d$ ; on $f(97)=97$ we obtain $b=9409c+194d$ . Solving for $d$ in terms of $c$ leads us to $d=-58c$ , so the answer is $\boxed{058}$ | null | 058 |
843fbe2770f01a95f16ccc7df9a9cfa4 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_12 | The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ $b$ $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ | Begin by finding the inverse function of $f(x)$ , which turns out to be $f^{-1}(x)=\frac{19d-b}{a-19c}$ . Since $f(f(x))=x$ $f(x)=f^{-1}(x)$ , so substituting 19 and 97 yields the system, $\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}$ , and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get $116c=a-d$ . Coincidentally, then $116c+d=a$ , which is familiar because $f(116)=\frac{116a+b}{116c+d}$ , and since $116c+d=a$ $f(116)=\frac{116a+b}{a}$ . Also, $f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116$ , due to $f(f(x))=x$ . This simplifies to $\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116$ $116a+2b=116(c(\frac{116a+b}{a})+d)$ $116a+2b=116(c(116+\frac{b}{a})+d)$ $116a+2b=116c(116+\frac{b}{a})+116d$ , and substituting $116c+d=a$ and simplifying, you get $2b=116c(\frac{b}{a})$ , then $\frac{a}{c}=58$ . Looking at $116c=a-d$ one more time, we get $116=\frac{a}{c}+\frac{-d}{c}$ , and substituting, we get $\frac{-d}{c}=\boxed{58}$ , and we are done. | null | 58 |
843fbe2770f01a95f16ccc7df9a9cfa4 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_12 | The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ $b$ $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ | Because there are no other special numbers other than $19$ and $97$ , take the average to get $\boxed{58}$ . (Note I solved this problem the solution one way but noticed this and this probably generalizes to all $f(x)=x, f(y)=y$ questions like these) | null | 58 |
843fbe2770f01a95f16ccc7df9a9cfa4 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_12 | The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ $b$ $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ | By the function definition, $f(f(x))$ $f$ is its own inverse, so the only value not in the range of $f$ is the value not in the domain of $f$ (which is $-d/c$ ).
Since $f(f(x))$ $f(f(0)=0$ (0 is a convenient value to use). $f(f(0))=f(f(\tfrac{b}{d})=\dfrac{a\cdot\tfrac{b}{d}+b}{c\cdot\tfrac{b}{d}+d}=\dfrac{ab+bd}{bc+d^2}=0 \Rightarrow ab+bd=0$
Then $ab+bd=b(a+d)=0$ and since $b$ is nonzero, $a=-d$
The answer we are searching for, $\dfrac{-d}{c}$ (the only value not in the range of $f$ ), can now be expressed as $\dfrac{a}{c}$
We are given $f(19)=19$ and $f(97)$ , and they satisfy the equation $f(x)=x$ , which simplifies to $\dfrac{ax+b}{cx+d}=x\Rightarrow x(cx+d)=ax+b\Rightarrow cx^2+(d-a)x+b=0$ . We have written this quadratic with roots $19$ and $97$
By Vieta, $\dfrac{-(d-a)}{c}=\dfrac{-(-a-a)}{c}=\dfrac{2a}{c}=19+97$
So our answer is $\dfrac{116}{2}=\boxed{058}$ | null | 058 |
843fbe2770f01a95f16ccc7df9a9cfa4 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_12 | The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ $b$ $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ | Notice that the function is just an involution on the real number line. Since the involution has two fixed points, namely $19$ and $97$ , we know that the involution is an inversion with respect to a circle with a diameter from $19$ to $97$ . The only point that is undefined under an inversion is the center of the circle, which we know is $\frac{19+97}{2}=\boxed{58}$ in both $x$ and $y$ dimensions. | null | 58 |
8c8e542a115aa4a20e4f922070bf2ade | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_13 | Let $S$ be the set of points in the Cartesian plane that satisfy
If a model of $S$ were built from wire of negligible thickness, then the total length of wire required would be $a\sqrt{b}$ , where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime number. Find $a+b$ | Let $f(x) = \Big|\big||x|-2\big|-1\Big|$ $f(x) \ge 0$ . Then $f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4$ . We only have a $4\times 4$ area, so guessing points and graphing won't be too bad of an idea. Since $f(x) = f(-x)$ , there's a symmetry about all four quadrants , so just consider the first quadrant. We now gather some points:
We can now graph the pairs of coordinates which add up to $1$ . Just using the first column of information gives us an interesting lattice pattern:
1997 AIME-13a.png
Plotting the remaining points and connecting lines, the graph looks like:
1997 AIME-13b.png
Calculating the lengths is now easy; each rectangle has sides of $\sqrt{2}, 3\sqrt{2}$ , so the answer is $4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}$ . For all four quadrants, this is $64\sqrt{2}$ , and $a+b=\boxed{066}$ | null | 066 |
8c8e542a115aa4a20e4f922070bf2ade | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_13 | Let $S$ be the set of points in the Cartesian plane that satisfy
If a model of $S$ were built from wire of negligible thickness, then the total length of wire required would be $a\sqrt{b}$ , where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime number. Find $a+b$ | Since $0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1$ and $0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1$ $- 1 \le \big||x| - 2\big| - 1 \le 1$ $0 \le \big||x| - 2\big| \le 2$ $- 2 \le |x| - 2 \le 2$ $- 4 \le x \le 4$ Also $- 4 \le y \le 4$
Define $f(a) = \Big|\big||a| - 2\big| - 1\Big|$
[this is also true for horizontal reflection, with $3 \le y \le 4$ , etc]
So it is only necessary to find the length of the function at $3 \le x \le 4$ and $3 \le y \le 4$ $\Big|\big||x| - 2\big| - 1\Big| + \Big|\big||y| - 2\big| - 1\Big| = 1$ $x - 3 + y - 3 = 1$ $y = - x + 7$ (Length = $\sqrt {2}$
This graph is reflected over the line y=3, the quantity of which is reflected over y=2,
So a total of $6$ doublings = $2^6$ $64$ , the total length = $64 \cdot \sqrt {2} = a\sqrt {b}$ , and $a + b = 64 + 2 = \boxed{066}$ | null | 066 |
8c8e542a115aa4a20e4f922070bf2ade | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_13 | Let $S$ be the set of points in the Cartesian plane that satisfy
If a model of $S$ were built from wire of negligible thickness, then the total length of wire required would be $a\sqrt{b}$ , where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime number. Find $a+b$ | We make use of several consecutive substitutions.
Let $||x| - 2|= x_1$ and similarly with $y$ .
Therefore, our graph is $|x_1 - 1| + |y_1 - 1| = 1$ . This is a rhombus with perimeter $4\sqrt{2}$ . Now, we make use of the following fact for a function of two variables $x$ and $y$ : Suppose we have $f(x, y) = c$ . Then $f(|x|, |y|)$ is equal to the graph of $f(x, y)$ reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of of $f(|x|, |y|)$ is 4 times the perimeter of $f(x, y)$ . Now, we continue making substitutions at each absolute value sign ( $|x| - 2 = x_2$ and finally $x = x_3$ , similarly for y as well. ), noting that the constants don't matter as they just translate the graph and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is $4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}$ , and $a+b = \boxed{066}$ | null | 066 |
8c8e542a115aa4a20e4f922070bf2ade | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_13 | Let $S$ be the set of points in the Cartesian plane that satisfy
If a model of $S$ were built from wire of negligible thickness, then the total length of wire required would be $a\sqrt{b}$ , where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime number. Find $a+b$ | In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It's possible to go right for the smallest possible piece that'll help us by letting $a = \big||x|-2\big|-1$ and $b = \big||y|-2\big|-1$ . We then get that $b = 1 - a$ (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where $b = 1 - x - 2 = -x - 1$ . Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes; the length of this piece also won't be affected by the constants that are hidden in $b$ . Using the Pythagorean Theorem / distance formula, that tiny length comes out to be $\sqrt{2}$ .
Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function's graph at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are $2^6$ of these pieces of length $\sqrt{2}$ , making our answer $2 + 64 = \boxed{066}$ | null | 066 |
1d06f7c8317efdb3f1d48d6ce677356d | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_14 | Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers . Find $m+n$ | Define $\theta = 2\pi/1997$ . By De Moivre's Theorem the roots are given by
Now, let $v$ be the root corresponding to $m\theta=2m\pi/1997$ , and let $w$ be the root corresponding to $n\theta=2n\pi/ 1997$ . Then \begin{align*} |v+w|^2 &= \left(\cos(m\theta) + \cos(n\theta)\right)^2 + \left(\sin(m\theta) + \sin(n\theta)\right)^2 \\ &= 2 + 2\cos\left(m\theta\right)\cos\left(n\theta\right) + 2\sin\left(m\theta\right)\sin\left(n\theta\right) \end{align*} The cosine difference identity simplifies that to \[|v+w|^2 = 2+2\cos((m-n)\theta)\]
We need $|v+w|^2 \ge 2+\sqrt{3}$ , which simplifies to \[\cos((m-n)\theta) \ge \frac{\sqrt{3}}{2}\] Thus, \[|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \left\lfloor \frac{1997}{12} \right\rfloor =166\]
Therefore, $m$ and $n$ cannot be more than $166$ away from each other. This means that for a given value of $m$ , there are $332$ values for $n$ that satisfy the inequality; $166$ of them $> m$ , and $166$ of them $< m$ . Since $m$ and $n$ must be distinct, $n$ can have $1996$ possible values. Therefore, the probability is $\frac{332}{1996}=\frac{83}{499}$ . The answer is then $499+83=\boxed{582}$ | null | 582 |
1d06f7c8317efdb3f1d48d6ce677356d | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_14 | Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers . Find $m+n$ | The solutions of the equation $z^{1997} = 1$ are the $1997$ th roots of unity and are equal to $\text{cis}(\theta_k)$ , where $\theta_k = \tfrac {2\pi k}{1997}$ for $k = 0,1,\ldots,1996.$ Thus, they are located at uniform intervals on the unit circle in the complex plane.
The quantity $|v+w|$ is unchanged upon rotation around the origin, so, WLOG, we can assume $v=1$ after rotating the axis till $v$ lies on the real axis. Let $w=\text{cis}(\theta_k)$ . Since $w\cdot \overline{w}=|w|^2=1$ and $w+\overline{w}=2\text{Re}(w) = 2\cos\theta_k$ , we have \[|v + w|^2 = (1+w)(1+\overline{w}) = 2+2\cos\theta_k\] We want $|v + w|^2\ge 2 + \sqrt {3}.$ From what we just obtained, this is equivalent to \[\cos\theta_k\ge \frac {\sqrt {3}}2 \qquad \Leftrightarrow \qquad -\frac {\pi}6\le \theta_k \le \frac {\pi}6\] which is satisfied by $k = 166,165,\ldots, - 165, - 166$ (we don't include 0 because that corresponds to $v$ ). So out of the $1996$ possible $k$ $332$ work. Thus, $m/n = 332/1996 = 83/499.$ So our answer is $83 + 499 = \boxed{582}.$ | null | 582 |
1d06f7c8317efdb3f1d48d6ce677356d | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_14 | Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers . Find $m+n$ | We can solve a geometrical interpretation of this problem.
Without loss of generality, let $u = 1$ . We are now looking for a point exactly one unit away from $u$ such that the point is at least $\sqrt{2 + \sqrt{3}}$ units away from the origin. Note that the "boundary" condition is when the point will be exactly $\sqrt{2+\sqrt{3}}$ units away from the origin; these points will be the intersections of the circle centered at $(1,0)$ with radius $1$ and the circle centered at $(0,0)$ with radius $\sqrt{2+\sqrt{3}}$ . The equations of these circles are $(x-1)^2 = 1$ and $x^2 + y^2 = 2 + \sqrt{3}$ . Solving for $x$ yields $x = \frac{\sqrt{3}}{2}$ . Clearly, this means that the real part of $v$ is greater than $\frac{\sqrt{3}}{2}$ . Solving, we note that $332$ possible $v$ s exist, meaning that $\frac{m}{n} = \frac{332}{1996} = \frac{83}{499}$ . Therefore, the answer is $83 + 499 = \boxed{582}$ | null | 582 |
1d06f7c8317efdb3f1d48d6ce677356d | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_14 | Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers . Find $m+n$ | Since $z^{1997}=1$ , the roots will have magnitude $1$ . Thus, the roots can be written as $\cos(\theta)+i\sin(\theta)$ and $\cos(\omega)+i\sin(\omega)$ for some angles $\theta$ and $\omega$ . We rewrite the requirement as $\sqrt{2+\sqrt3}\le|\cos(\theta)+\cos(\omega)+i\sin(\theta)+i\sin(\omega)|$ , which can now be easily manipulated to $2+\sqrt{3}\le(\cos(\theta)+\cos(\omega))^2+(\sin(\theta)+\sin(\omega))^2$
WLOG, let $\theta = 0$ . Thus, our inequality becomes $2+\sqrt{3}\le(1+\cos(\omega))^2+(\sin(\omega))^2$ $2+\sqrt{3}\le2+2\cos(\omega)$ , and finally $\cos(\omega)\ge\frac{\sqrt{3}}{2}$ . Obviously, $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$ , and thus it follows that, on either side of a given point, $\frac{1997}{12}\approx166$ points will work. The probability is $\frac{166\times2}{1996} = \frac{83}{499}$ , and thus our requested sum is $\boxed{582}$ ~SigmaPiE | null | 582 |
eca8cd8b0a5cf1c7c43c4ebdf2f6caf1 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_15 | The sides of rectangle $ABCD$ have lengths $10$ and $11$ . An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$ . The maximum possible area of such a triangle can be written in the form $p\sqrt{q}-r$ , where $p$ $q$ , and $r$ are positive integers, and $q$ is not divisible by the square of any prime number. Find $p+q+r$ | Consider points on the complex plane $A (0,0),\ B (11,0),\ C (11,10),\ D (0,10)$ . Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at $A$ , and the other two points $E$ and $F$ on $BC$ and $CD$ , respectively. Let $E (11,a)$ and $F (b, 10)$ . Since it's equilateral, then $E\cdot\text{cis}60^{\circ} = F$ , so $(11 + ai)\left(\frac {1}{2} + \frac {\sqrt {3}}{2}i\right) = b + 10i$ , and expanding we get $\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i$
We can then set the real and imaginary parts equal, and solve for $(a,b) = (20 - 11\sqrt {3},22 - 10\sqrt {3})$ . Hence a side $s$ of the equilateral triangle can be found by $s^2 = AE^2 = a^2 + AB^2 = 884 - 440\sqrt{3}$ . Using the area formula $\frac{s^2\sqrt{3}}{4}$ , the area of the equilateral triangle is $\frac{(884-440\sqrt{3})\sqrt{3}}{4} = 221\sqrt{3} - 330$ . Thus $p + q + r = 221 + 3 + 330 = \boxed{554}$ | null | 554 |
eca8cd8b0a5cf1c7c43c4ebdf2f6caf1 | https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_15 | The sides of rectangle $ABCD$ have lengths $10$ and $11$ . An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$ . The maximum possible area of such a triangle can be written in the form $p\sqrt{q}-r$ , where $p$ $q$ , and $r$ are positive integers, and $q$ is not divisible by the square of any prime number. Find $p+q+r$ | Since $\angle{BAD}=90$ and $\angle{EAF}=60$ , it follows that $\angle{DAF}+\angle{BAE}=90-60=30$ . Rotate triangle $ADF$ $60$ degrees clockwise. Note that the image of $AF$ is $AE$ . Let the image of $D$ be $D'$ . Since angles are preserved under rotation, $\angle{DAF}=\angle{D'AE}$ . It follows that $\angle{D'AE}+\angle{BAE}=\angle{D'AB}=30$ . Since $\angle{ADF}=\angle{ABE}=90$ , it follows that quadrilateral $ABED'$ is cyclic with circumdiameter $AE=s$ and thus circumradius $\frac{s}{2}$ . Let $O$ be its circumcenter. By Inscribed Angles, $\angle{BOD'}=2\angle{BAD'}=60$ . By the definition of circle, $OB=OD'$ . It follows that triangle $OBD'$ is equilateral. Therefore, $BD'=r=\frac{s}{2}$ . Applying the Law of Cosines to triangle $ABD'$ $\frac{s}{2}=\sqrt{10^2+11^2-(2)(10)(11)(\cos{30})}$ . Squaring and multiplying by $\sqrt{3}$ yields $\frac{s^2\sqrt{3}}{4}=221\sqrt{3}-330\implies{p+q+r=221+3+330=\boxed{554}$ | null | 554 |
ef6e63e14b72fd5e296daaa721a33f3f | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_1 | In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$ | Let's make a table.
\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&d&e\\\hline \end{array}\]
\begin{eqnarray*} x+19+96=x+1+c\Rightarrow c=19+96-1=114,\\ 114+96+a=x+1+114\Rightarrow a=x-95 \end{eqnarray*}
\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table in progress}}\\\hline x&19&96\\\hline 1&x-95&b\\\hline 114&d&e\\\hline \end{array}\]
\begin{eqnarray*}19+x-95+d=x+d-76=115+x\Rightarrow d=191,\\ 114+191+e=x+115\Rightarrow e=x-190 \end{eqnarray*} \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table in progress}}\\\hline x&19&96\\\hline 1&x-95&b\\\hline 114&191&x-190\\\hline \end{array}\]
\[3x-285=x+115\Rightarrow 2x=400\Rightarrow x=\boxed{200}\] | null | 200 |
ef6e63e14b72fd5e296daaa721a33f3f | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_1 | In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$ | Use the table from above. Obviously $c = 114$ . Hence $a+e = 115$ . Similarly, $1+a = 96 + e \Rightarrow a = 95+e$
Substitute that into the first to get $2e = 20 \Rightarrow e=10$ , so $a=105$ , and so the value of $x$ is just $115+x = 210 + 105 \Rightarrow x = \boxed{200}$ | null | 200 |
ef6e63e14b72fd5e296daaa721a33f3f | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_1 | In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$ | \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&d&e\\\hline \end{array}\] The formula \[e=\frac{1+19}{2}\] can be used. Therefore, $e=10$ . Similarly, \[96=\frac{1+d}{2}\] So $d=191$
Now we have this table: \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&191&10\\\hline \end{array}\] By property of magic squares, observe that \[x+a+10=19+a+191\] The $a$ 's cancel! We now have \[x+10=19+191\] Thus $x=\boxed{200}.$ | null | 200 |
585269501a67abed6d44067e70131867 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_2 | For each real number $x$ , let $\lfloor x \rfloor$ denote the greatest integer that does not exceed x. For how many positive integers $n$ is it true that $n<1000$ and that $\lfloor \log_{2} n \rfloor$ is a positive even integer? | For integers $k$ , we want $\lfloor \log_2 n\rfloor = 2k$ , or $2k \le \log_2 n < 2k+1 \Longrightarrow 2^{2k} \le n < 2^{2k+1}$ . Thus, $n$ must satisfy these inequalities (since $n < 1000$ ):
There are $4$ for the first inequality, $16$ for the second, $64$ for the third, and $256$ for the fourth, so the answer is $4+16+64+256=\boxed{340}$ | null | 340 |
13b351a7ada20018581a10e040b61542 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_3 | Find the smallest positive integer $n$ for which the expansion of $(xy-3x+7y-21)^n$ , after like terms have been collected, has at least 1996 terms. | Using Simon's Favorite Factoring Trick , we rewrite as $[(x+7)(y-3)]^n = (x+7)^n(y-3)^n$ . Both binomial expansions will contain $n+1$ non-like terms; their product will contain $(n+1)^2$ terms, as each term will have an unique power of $x$ or $y$ and so none of the terms will need to be collected. Hence $(n+1)^2 \ge 1996$ , the smallest square after $1996$ is $2025 = 45^2$ , so our answer is $45 - 1 = \boxed{044}$ | null | 044 |
13b351a7ada20018581a10e040b61542 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_3 | Find the smallest positive integer $n$ for which the expansion of $(xy-3x+7y-21)^n$ , after like terms have been collected, has at least 1996 terms. | Notice that the coefficients in the problem statement have no effect on how many unique terms there will be in the expansion. Therefore this problem is synonymous with finding the amount of terms in the expansion of $(xy+x+y+1)^n$ (we do this to simplify the problem).
If we expand the exponent the expression becomes $\underbrace{(xy+x+y+1)\cdot(xy+x+y+1)\cdots(xy+x+y+1)}_{n\;times}$
This is equivalent to starting off with a $x^0y^0$ terms and choosing between $4$ options $n$ different times:
$\bullet$ Adding nothing to either exponent (choosing $1$ ).
$\bullet$ Adding $1$ to the $x$ exponent (choosing $x$ ).
$\bullet$ Adding $1$ to the $y$ exponent (choosing $y$ ).
$\bullet$ Adding $1$ to the $x$ exponent and adding $1$ to the $y$ exponent (choosing $xy$ ).
Doing this $n$ times, you can see that you end up with a term in the form $k\cdot x^i\cdot y^j$ where $k$ is some coefficient (which we don't care about) and $0\le i,j\le n$ and $i,j \in \mathbb{N}$
Repeating this for all possible combinations of choices yields $n+1$ options for each of $i$ and $j$ which means there are a total of $(n+1)^2$ possible terms in the form $x^i\cdot y^j$ . Therefore $(xy+x+y+1)^n$ has $(n+1)^2$ terms.
$(n+1)^2 \ge 1996$ which yields $n=44$ $\boxed{044}$ | null | 044 |
3ffafc792fc2ea601d085a1dc3e8bba3 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_4 | A wooden cube , whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is $x$ centimeters directly above an upper vertex , the cube casts a shadow on the horizontal surface. The area of the shadow, which does not include the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed $1000x$ | [asy] import three; size(250);defaultpen(0.7+fontsize(9)); real unit = 0.5; real r = 2.8; triple O=(0,0,0), P=(0,0,unit+unit/(r-1)); dot(P); draw(O--P); draw(O--(unit,0,0)--(unit,0,unit)--(0,0,unit)); draw(O--(0,unit,0)--(0,unit,unit)--(0,0,unit)); draw((unit,0,0)--(unit,unit,0)--(unit,unit,unit)--(unit,0,unit)); draw((0,unit,0)--(unit,unit,0)--(unit,unit,unit)--(0,unit,unit)); draw(P--(r*unit,0,0)--(r*unit,r*unit,0)--(0,r*unit,0)--P); draw(P--(r*unit,r*unit,0)); draw((r*unit,0,0)--(0,0,0)--(0,r*unit,0)); draw(P--(0,0,unit)--(unit,0,unit)--(unit,0,0)--(r*unit,0,0)--P,dashed+blue+linewidth(0.8)); label("$x$",(0,0,unit+unit/(r-1)/2),WSW); label("$1$",(unit/2,0,unit),N); label("$1$",(unit,0,unit/2),W); label("$1$",(unit/2,0,0),N); label("$6$",(unit*(r+1)/2,0,0),N); label("$7$",(unit*r,unit*r/2,0),SW); [/asy] (Figure not to scale) The area of the square shadow base is $48 + 1 = 49$ , and so the sides of the shadow are $7$ . Using the similar triangles in blue, $\frac {x}{1} = \frac {1}{6}$ , and $\left\lfloor 1000x \right\rfloor = \boxed{166}$ | null | 166 |
2159f67e06e548eaf9a96fb637f4bee8 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_6 | In a five-team tournament, each team plays one game with every other team. Each team has a $50\%$ chance of winning any game it plays. (There are no ties.) Let $\dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integers. Find $m+n$ | We can use complementary counting : finding the probability that at least one team wins all games or at least one team loses all games.
No more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games.
Now we use PIE
The probability that one team wins all games is $5\cdot \left(\frac{1}{2}\right)^4=\frac{5}{16}$
Similarity, the probability that one team loses all games is $\frac{5}{16}$
The probability that one team wins all games and another team loses all games is $\left(5\cdot \left(\frac{1}{2}\right)^4\right)\left(4\cdot \left(\frac{1}{2}\right)^3\right)=\frac{5}{32}$
$\frac{5}{16}+\frac{5}{16}-\frac{5}{32}=\frac{15}{32}$
Since this is the opposite of the probability we want, we subtract that from 1 to get $\frac{17}{32}$
$17+32=\boxed{049}$ | null | 049 |
2159f67e06e548eaf9a96fb637f4bee8 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_6 | In a five-team tournament, each team plays one game with every other team. Each team has a $50\%$ chance of winning any game it plays. (There are no ties.) Let $\dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integers. Find $m+n$ | There are $\dbinom{5}{2} = 10$ games in total, and every game can either end in a win or a loss. Therefore, there are $2^{10} = 1024$ possible outcomes.
Now, computing this probability directly seems a little hard, so let's compute the complement -- the probability that there is an undefeated team, a winless team, or both.
The number of ways that we can have an undefeated team is $5 \cdot 2^6,$ as there are five ways to choose the team itself and $2^6$ outcomes for the games that are not concerning the undefeated team. Reversing all the wings to losses yields a winless team, so this situation is symmetrical to the first one. Therefore, there are $5 \cdot 2^6 + 5 \cdot 2^6$ ways for an undefeated or winless team.
In the process, however, we have accidentally overcounted the scenario of both an undefeated and winless team. There are $5 \cdot 4 \cdot 2^3$ ways for this scenario to happen, because there are five ways to choose the undefeated team, four ways for the winless, and three games that don't concern either of the teams. Therefore, there are $5 \cdot 2^6 + 5 \cdot 2^6 - 5 \cdot 4 \cdot 2^3 = 480$ ways to have an undefeated and/or winless team, so there are $1024 - 480 = 544$ to not have any.
Our probability is thus $\dfrac{544}{1024},$ which simplifies to $\dfrac{17}{32},$ for our answer of $17 + 32 = \boxed{049}.$ | null | 049 |
c362de78cc425b19dfff2b2a84003b6f | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_7 | Two squares of a $7\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? | There are ${49 \choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards.
Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\frac{49-1}{2}=24$ such pairs. There are then ${49 \choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each of the ${49 \choose 2}-24$ pairs, there are three other pairs that yield an equivalent board.
Thus, the number of inequivalent boards is $\frac{{49 \choose 2} - 24}{4} + \frac{24}{2} = \boxed{300}$ . For a $(2n+1) \times (2n+1)$ board, this argument generalizes to $n(n+1)(2n^2+2n+1)$ inequivalent configurations. | null | 300 |
c362de78cc425b19dfff2b2a84003b6f | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_7 | Two squares of a $7\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible? | There are 4 cases:
1. The center square is occupied, in which there are $12$ cases.
2. The center square isn't occupied and the two squares that are opposite to each other with respect to the center square, in which there are $12$ cases.
3. The center square isn't occupied and the two squares can rotate to each other with a $90^{\circ}$ rotation with each other and with respect to the center square, in which case there are $12$ cases.
4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are $\dbinom{12}{2} \cdot \frac{16}{4} = 264$ cases.
Add up all the values for each case to get $\boxed{300}$ as your answer. | null | 300 |
4612c7bb354928eba37dec831f69431e | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_8 | The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers $(x,y)$ with $x<y$ is the harmonic mean of $x$ and $y$ equal to $6^{20}$ | The harmonic mean of $x$ and $y$ is equal to $\frac{1}{\frac{\frac{1}{x}+\frac{1}{y}}2} = \frac{2xy}{x+y}$ , so we have $xy=(x+y)(3^{20}\cdot2^{19})$ , and by SFFT $(x-3^{20}\cdot2^{19})(y-3^{20}\cdot2^{19})=3^{40}\cdot2^{38}$ . Now, $3^{40}\cdot2^{38}$ has $41\cdot39=1599$ factors, one of which is the square root ( $3^{20}2^{19}$ ). Since $x<y$ , the answer is half of the remaining number of factors, which is $\frac{1599-1}{2}= \boxed{799}$ | null | 799 |
1e147589dc3ac1ba026e35b6438fa174 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_9 | A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$ . He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens? | On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$ , leaving only lockers $2 \pmod{8}$ and $6 \pmod{8}$ . Then he goes ahead and opens all lockers $2 \pmod {8}$ , leaving lockers either $6 \pmod {16}$ or $14 \pmod {16}$ . He then goes ahead and opens all lockers $14 \pmod {16}$ , leaving the lockers either $6 \pmod {32}$ or $22 \pmod {32}$ . He then goes ahead and opens all lockers $6 \pmod {32}$ , leaving $22 \pmod {64}$ or $54 \pmod {64}$ . He then opens $54 \pmod {64}$ , leaving $22 \pmod {128}$ or $86 \pmod {128}$ . He then opens $22 \pmod {128}$ and leaves $86 \pmod {256}$ and $214 \pmod {256}$ . He then opens all $214 \pmod {256}$ , so we have $86 \pmod {512}$ and $342 \pmod {512}$ , leaving lockers $86, 342, 598$ , and $854$ , and he is at where he started again. He then opens $86$ and $598$ , and then goes back and opens locker number $854$ , leaving locker number $\boxed{342}$ untouched. He opens that locker. | null | 342 |
1e147589dc3ac1ba026e35b6438fa174 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_9 | A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$ . He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens? | We can also solve this with recursion. Let $L_n$ be the last locker he opens given that he started with $2^n$ lockers. Let there be $2^n$ lockers. After he first reaches the end of the hallway, there are $2^{n-1}$ lockers remaining. There is a correspondence between these unopened lockers and if he began with $2^{n-1}$ lockers. The locker $y$ (if he started with $2^{n-1}$ lockers) corresponds to the locker $2^n+2-2y$ (if he started with $2^n$ lockers). It follows that $L_{n} = 2^{n} +2 -2L_{n-1}$ as they are corresponding lockers. We can compute $L_1=2$ and use the recursion to find $L_{10}=\boxed{342}$ | null | 342 |
1e147589dc3ac1ba026e35b6438fa174 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_9 | A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$ . He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens? | List all the numbers from $1$ through $1024$ , then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get $\boxed{342}$ | null | 342 |
f9758a48edd1749d1c4363811396e9c8 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_10 | Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ | $\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{186^{\circ}}+\sin{96^{\circ}}}{\sin{186^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{(141^{\circ}+45^{\circ})}+\sin{(141^{\circ}-45^{\circ})}}{\sin{(141^{\circ}+45^{\circ})}-\sin{(141^{\circ}-45^{\circ})}} =$ $\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}$
The period of the tangent function is $180^\circ$ , and the tangent function is one-to-one over each period of its domain.
Thus, $19x \equiv 141 \pmod{180}$
Since $19^2 \equiv 361 \equiv 1 \pmod{180}$ , multiplying both sides by $19$ yields $x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}$
Therefore, the smallest positive solution is $x = \boxed{159}$ | null | 159 |
f9758a48edd1749d1c4363811396e9c8 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_10 | Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ | $\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \dfrac{1 + \tan{96^{\circ}}}{1-\tan{96^{\circ}}}$ which is the same as $\dfrac{\tan{45^{\circ}} + \tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}} = \tan{141{^\circ}}$
So $19x = 141 +180n$ , for some integer $n$ .
Multiplying by $19$ gives $x \equiv 141 \cdot 19 \equiv 2679 \equiv 159 \pmod{180}$ .
The smallest positive solution of this is $x = \boxed{159}$ | null | 159 |
f9758a48edd1749d1c4363811396e9c8 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_10 | Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ | It seems reasonable to assume that $\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \tan{\theta}$ for some angle $\theta$ . This means \[\dfrac{\alpha (\cos{96^{\circ}}+\sin{96^{\circ}})}{\alpha (\cos{96^{\circ}}-\sin{96^{\circ}})} = \frac{\sin{\theta}}{\cos{\theta}}\] for some constant $\alpha$ . We can set $\alpha (\cos{96^{\circ}}+\sin{96^{\circ}}) = \sin{\theta}$ .Note that if we have $\alpha$ equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since $\sin{45^{\circ}} = \cos{45^{\circ}} = \tfrac{\sqrt{2}}{2}$ , if $\alpha = \tfrac{\sqrt{2}}{2}$ we have \[\alpha (\cos{96^{\circ}} + \sin{96^{\circ}}) = \cos{96^{\circ}} \frac{\sqrt{2}}{2} + \sin{96^{\circ}} \frac{\sqrt{2}}{2} = \cos{96^{\circ}} \sin{45^{\circ}} + \sin{96^{\circ}} \cos{45^{\circ}} = \sin({45^{\circ} + 96^{\circ}}) = \sin{141^{\circ}}\] from the sine sum formula. For the denominator, from the cosine sum formula, we have \[\alpha (\cos{96^{\circ}} - \sin{96^{\circ}}) = \cos{96^{\circ}} \frac{\sqrt{2}}{2} + \sin{96^{\circ}} \frac{\sqrt{2}}{2} = \cos{96^{\circ}} \cos{45^{\circ}} + \sin{96^{\circ}} \sin{45^{\circ}} = \cos({96^{\circ} + 45^{\circ}}) = \cos{141^{\circ}}.\] This means $\theta = 141^{\circ},$ so $19x = 141 + 180k$ for some positive integer $k$ (since the period of tangent is $180^{\circ}$ ), or $19 x \equiv 141 \pmod{180}$ . Note that the inverse of $19$ modulo $180$ is itself as $19^2 \equiv 361 \equiv 1 \pmod {180}$ , so multiplying this congruence by $19$ on both sides gives $x \equiv 2679 \equiv 159 \pmod{180}.$ For the smallest possible $x$ , we take $x = \boxed{159}.$ | null | 159 |
c04a63a5d9cc0cec5d36155d361d14ca | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_11 | Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ | \begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \frac{z^5-1}{z-1}\\ 0 &=& \frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \frac{(z^2-z+1)(z^5-1)}{z-1} \end{eqnarray*}
Thus $z^5 = 1, z \neq 1 \Longrightarrow z = \mathrm{cis}\ 72, 144, 216, 288$
or $z^2 - z + 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{-3}}{2} = \mathrm{cis}\ 60, 300$
(see cis ).
Discarding the roots with negative imaginary parts (leaving us with $\mathrm{cis} \theta,\ 0 < \theta < 180$ ), we are left with $\mathrm{cis}\ 60, 72, 144$ ; their product is $P = \mathrm{cis} (60 + 72 + 144) = \mathrm{cis} \boxed{276}$ | null | 276 |
c04a63a5d9cc0cec5d36155d361d14ca | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_11 | Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ | Divide through by $z^3$ . We get the equation $z^3 + \frac {1}{z^3} + z + \frac {1}{z} + 1 = 0$ . Let $x = z + \frac {1}{z}$ . Then $z^3 + \frac {1}{z^3} = x^3 - 3x$ . Our equation is then $x^3 - 3x + x + 1 = x^3 - 2x + 1 = (x - 1)(x^2 + x - 1) = 0$ , with solutions $x = 1, \frac { - 1\pm\sqrt {5}}{2}$ . For $x = 1$ , we get $z = \text{cis}60,\text{cis}300$ . For $x = \frac { - 1 + \sqrt {5}}{2}$ , we get $z = \text{cis}{72},\text{cis}{292}$ (using exponential form of $\cos$ ). For $x = \frac { - 1 - \sqrt {5}}{2}$ , we get $z = \text{cis}144,\text{cis}216$ . The ones with positive imaginary parts are ones where $0\le\theta\le180$ , so we have $60 + 72 + 144 = \boxed{276}$ | null | 276 |
c04a63a5d9cc0cec5d36155d361d14ca | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_11 | Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ | We recognize that $z^6+z^5+z^4+z^3+z^2+z+1=\frac{z^7-1}{z-1}$ and strongly wish for the equation to turn into so. Thankfully, we can do this by simultaneously adding and subtracting $z^5$ and $z$ as shown below. $z^6+z^5+z^4+z^3+z^2+z+1-(z^5+z)=0\implies \frac{z^7-1}{z-1}-(z^5+z)=0$
Now, knowing that $z=1$ is not a root, we multiply by $z-1$ to obtain $z^7-1-(z-1)(z^5+z)=0\implies z^7-1-(z^6+z^2-z^5-z)=0\implies z^7-z^6+z^5-z^2+z-1=0$ Now, we see the $z^2+z-1$ and it reminds us of the sum of two cubes. Cleverly factoring, we obtain that.. $z^5(z^2-z)+z^5-(z^2-z+1)=0\implies z^5(z^2-z+1)-(z^2-z+1)=0\implies (z^5-1)(z^2-z+1)=0$
Now, it is clear that we have two cases to consider.
Case $1$ $z^5-1=0$ We obtain that $z^5=1$ or $z^5=e^{2\pi{n}{i}}$ Obviously, the answers to this case are $e^{ia}, a\in{\frac{2\pi}{5}, \frac{4\pi}{5}}$
Case $2$ $z^2-z+1=0$ Completing the square and then algebra allows us to find that $z=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}$ which has $\arg$ $\frac{\pi}{3}$
Hence, the answer is $\frac{18\pi}{15}+\frac{5\pi}{15}=\frac{23\pi}{15}\cdot\frac{180}{\pi}=\boxed{276}$ | null | 276 |
c04a63a5d9cc0cec5d36155d361d14ca | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_11 | Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ | Add 1 to both sides of the equation to get $x^6+x^4+x^3+x^2+1+1=1$ . We can rearrange to find that $(x^6+x^3+1)+(x^4+x^2+1)=1$ . Then, using sum of a geometric series, $\frac{x^9-1}{x^3-1}+\frac{x^6-1}{x^2-1}=1$
Combining the two terms of the LHS, we get that $\frac{x^{11}-x^9-x^2+1+x^9-x^6-x^3+1}{x^5-x^3-x^2+1}=1$ , so $x^{11}-x^6-x^3-x^2+2=x^5-x^3-x^2+1$ , and simplifying, we see that $x^{11}-x^6-x^5+1=0$ , so by SFFT, $(x^6-1)(x^5-1)=0$ . Then, the roots of our polynomial are the fifth and sixth roots of unity. However, looking back at our expression when it had fractions, we realize that if $x$ is a second or third root of unity, it would cause a denominator to be zero, so the roots of our polynomial are the fifth and sixth roots of unity that are not the second or third roots of unity. The only roots in this category are $\mathrm{cis} (60, 72, 144)$ , so our desired sum is $\boxed{276}$ , and we are done. | null | 276 |
794e56617d175f5780c6409562b10060 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_12 | For each permutation $a_1,a_2,a_3,\cdots,a_{10}$ of the integers $1,2,3,\cdots,10$ , form the sum
\[|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|.\]
The average value of all such sums can be written in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Because of symmetry, we may find all the possible values for $|a_n - a_{n - 1}|$ and multiply by the number of times this value appears. Each occurs $5 \cdot 8!$ , because if you fix $a_n$ and $a_{n + 1}$ there are still $8!$ spots for the others and you can do this $5$ times because there are $5$ places $a_n$ and $a_{n + 1}$ can be.
To find all possible values for $|a_n - a_{n - 1}|$ we have to compute \begin{eqnarray*} |1 - 10| + |1 - 9| + \ldots + |1 - 2|\\ + |2 - 10| + \ldots + |2 - 3| + |2 - 1|\\ + \ldots\\ + |10 - 9| \end{eqnarray*}
This is equivalent to
\[2\sum\limits_{k = 1}^{9}\sum\limits_{j = 1}^{k}j = 330\]
The total number of permutations is $10!$ , so the average value is $\frac {330 \cdot 8! \cdot 5}{10!} = \frac {55}{3}$ , and $m+n = \boxed{058}$ | null | 058 |
794e56617d175f5780c6409562b10060 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_12 | For each permutation $a_1,a_2,a_3,\cdots,a_{10}$ of the integers $1,2,3,\cdots,10$ , form the sum
\[|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|.\]
The average value of all such sums can be written in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Without loss of generality , let $a_1 > a_2;\, a_3 > a_4;\, \ldots ;\, a_9 > a_{10}$ . We may do this because all sums obtained from these paired sequences are also obtained in another $2^5-1$ ways by permuting the adjacent terms $\{a_1,a_2\},\{a_3,a_4\}, \cdots , \{a_9, a_{10}\}$ , and thus are canceled when the average is taken.
So now we only have to form the sum $S= (a_1 + a_3 + a_5 + a_7 + a_9) - (a_2 + a_4 + a_6 + a_8 + a_{10})$ . Due to the symmetry of this situation, we only need to compute the expected value of the result. $10$ must always be the greatest number in its pair; $9$ will be the greater number in its pair $\frac{8}{9}$ of the time and the lesser number $\frac 19$ of the time; $8$ will be the greater number in its pair $\frac 79$ of the time and the lesser $\frac 29$ of the time; and so forth. Each number either adds or subtracts from the sum depending upon whether it is one of the five greater or five lesser numbers in the pairs, respectively. Thus
\begin{align*} \overline{S} &= 10 + \left(\frac{8}{9} \cdot 9\right) - \left(\frac{1}{9} \cdot 9\right) + \left(\frac{7}{9} \cdot 8\right) - \left(\frac 29 \cdot 8\right) + \cdots + \left(\frac{1}{9} \cdot 2\right) - \left(\frac{8}{9} \cdot 2\right) - 1 \\ &= \frac{9 \cdot 10 + 7 \cdot 9 + 5 \cdot 8 + 3 \cdot 7 + 1 \cdot 6 - 1 \cdot 5 - 3 \cdot 4 - 5 \cdot 3 - 7 \cdot 2 - 9 \cdot 1}{9} \\ &= \frac{55}{3} \end{align*}
And the answer is $m+n = \boxed{058}$ | null | 058 |
794e56617d175f5780c6409562b10060 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_12 | For each permutation $a_1,a_2,a_3,\cdots,a_{10}$ of the integers $1,2,3,\cdots,10$ , form the sum
\[|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|.\]
The average value of all such sums can be written in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Similar to Solution 1, we can find the average value of $|a_2 - a_1|$ , and multiply this by 5 due to symmetry. And again due to symmetry, we can arbitrarily choose $a_2 > a_1$ . Thus there are $\binom{10}{2} = 45$ ways to pick the two values of $a_2$ and $a_1$ from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ such that $a_2 > a_1$ . First fix $a_2 = 10$ , and vary $a_1$ from $1$ to $9$ . Then fix $a_2 = 9$ , and vary $a_1$ from $1$ to $8$ . Continue, and you find that the sum of these $45$ ways to pick $|a_2 - a_1|$ is:
$\sum\limits_{k = 1}^{9}\sum\limits_{j = 1}^{k}j = 45+36+28+21+15+10+6+3+1 = 165$
Thus, each term contributes on average $\frac{165}{45}$ , and the sum will be five times this, or $\frac{165}{9} = \frac{55}{3}$
The final answer is $p+q = \boxed{058}$ | null | 058 |
794e56617d175f5780c6409562b10060 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_12 | For each permutation $a_1,a_2,a_3,\cdots,a_{10}$ of the integers $1,2,3,\cdots,10$ , form the sum
\[|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|.\]
The average value of all such sums can be written in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | We use expected value of one of the sums, say $|a_2 - a_1|$ , since all five are similar, so the average or expected value of the sum is just 5 times the expected value of one of them.
To pick two random expected numbers from 1 to 10, we use a well known expected value trick by tying the two ends 1 and 10 with an extra number 0.* This gives 11 spaces, and we distribute 3 gaps evenly to choose our two numbers. We get $\frac{11}{3}$ and $\frac{22}{3}$ giving an absolute difference of $\frac{11}{3}$ . Therefore, the average/expected value of the sum of the five would be $\frac{55}{3}$ . This gives $55+3=\boxed{058}$ | null | 058 |
7a5b42a10d1734414c39c2583d6409fb | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_13 | In triangle $ABC$ $AB=\sqrt{30}$ $AC=\sqrt{6}$ , and $BC=\sqrt{15}$ . There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$ , and $\angle ADB$ is a right angle. The ratio $\frac{[ADB]}{[ABC]}$ can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $E$ be the midpoint of $\overline{BC}$ . Since $BE = EC$ , then $\triangle ABE$ and $\triangle AEC$ share the same height and have equal bases, and thus have the same area. Similarly, $\triangle BDE$ and $BAE$ share the same height, and have bases in the ratio $DE : AE$ , so $\frac{[BDE]}{[BAE]} = \frac{DE}{AE}$ (see area ratios ). Now,
By Stewart's Theorem $AE = \frac{\sqrt{2(AB^2 + AC^2) - BC^2}}2 = \frac{\sqrt {57}}{2}$ , and by the Pythagorean Theorem on $\triangle ABD, \triangle EBD$
\begin{align*} BD^2 + \left(DE + \frac {\sqrt{57}}2\right)^2 &= 30 \\ BD^2 + DE^2 &= \frac{15}{4} \\ \end{align*}
Subtracting the two equations yields $DE\sqrt{57} + \frac{57}{4} = \frac{105}{4} \Longrightarrow DE = \frac{12}{\sqrt{57}}$ . Then $\frac mn = \frac{1}{2} + \frac{DE}{2AE} = \frac{1}{2} + \frac{\frac{12}{\sqrt{57}}}{2 \cdot \frac{\sqrt{57}}{2}} = \frac{27}{38}$ , and $m+n = \boxed{065}$ | null | 065 |
7a5b42a10d1734414c39c2583d6409fb | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_13 | In triangle $ABC$ $AB=\sqrt{30}$ $AC=\sqrt{6}$ , and $BC=\sqrt{15}$ . There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$ , and $\angle ADB$ is a right angle. The ratio $\frac{[ADB]}{[ABC]}$ can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Because the problem asks for a ratio, we can divide each side length by $\sqrt{3}$ to make things simpler. We now have a triangle with sides $\sqrt{10}$ $\sqrt{5}$ , and $\sqrt{2}$
We use the same graph as above.
Draw perpendicular from $C$ to $AE$ . Denote this point as $F$ . We know that $DE = EF = x$ and $BD = CF = z$ and also let $AE = y$
Using Pythagorean theorem, we get three equations,
Adding the first and second, we obtain $x^2 + y^2 + z^2 = 6$ , and then subtracting the third from this we find that $y = \frac{\sqrt{19}}{2}$ . (Note, we could have used Stewart's Theorem to achieve this result).
Subtracting the first and second, we see that $xy = 2$ , and then we find that $x = \frac{4}{\sqrt{19}}$
Using base ratios, we then quickly find that the desired ratio is $\frac{27}{38}$ so our answer is $\boxed{065}$ | null | 065 |
f35eaf92b4c75eaf6532bfc188ced5c1 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_14 | $150\times 324\times 375$ rectangular solid is made by gluing together $1\times 1\times 1$ cubes. An internal diagonal of this solid passes through the interiors of how many of the $1\times 1\times 1$ cubes | Place one corner of the solid at $(0,0,0)$ and let $(a,b,c)$ be the general coordinates of the diagonally opposite corner of the rectangle, where $a, b, c \in \mathbb{Z_{+}}$
We consider the vector equation of the diagonal segment represented by: $(x, y, z) =m (a, b, c)$ , where $m \in \mathbb{R}, 0 < m \le 1$
Consider a point on the diagonal with coordinates $(ma, mb, mc)$ . We have 3 key observations as this point moves from $(0,0,0)$ towards $(a,b,c)$
The number of cubes the diagonal passes is equal to the number of points on the diagonal that has one or more positive integers as coordinates.
If we slice the solid up by the $x$ -planes defined by $x=1,2,3,4, \ldots, a$ , the diagonal will cut these $x$ -planes exactly $a$ times (plane of $x=0$ is not considered since $m \ne 0$ ). Similar arguments for slices along $y$ -planes and $z$ -planes give diagonal cuts of $b$ , and $c$ times respectively. The total cuts by the diagonal is therefore $a+b+c$ , if we can ensure that no more than $1$ positive integer is present in the x, y, or z coordinate in all points $(ma,mb,mc)$ on the diagonal. Note that point $(a,b,c)$ is already one such exception.
But for each diagonal point $(ma,mb,mc)$ with 2 (or more) positive integers occurring at the same time, $a+b+c$ counts the number of cube passes as $2$ instead of $1$ for each such point. There are $\gcd(a,b)+\gcd(b,c)+\gcd(c,a)$ points in such over-counting. We therefore subtract one time such over-counting from $a+b+c$
And for each diagonal point $(ma,mb,mc)$ with exactly $3$ integers occurring at the same time, $a+b+c$ counts the number of cube passes as $3$ instead of $1$ ; ie $a+b+c$ over-counts each of such points by $2$ . But since $\gcd(a,b)+\gcd(b,c)+\gcd(c,a)$ already subtracted three times for the case of $3$ integers occurring at the same time (since there are $3$ of these gcd terms that represent all combinations of 3 edges of a cube meeting at a vertex), we have the final count for each such point as $1 = 3-3+k \Rightarrow k=1$ , where $k$ is our correction term. That is, we need to add $k=1$ time $\gcd(a,b,c)$ back to account for the case of 3 simultaneous integers.
Therefore, the total diagonal cube passes is: $D = a+b+c-\left[ \gcd(a,b)+\gcd(b,c)+\gcd(c,a) \right]+\gcd(a,b,c)$
For $(a,b,c) = (150, 324, 375)$ , we have: $\gcd(150,324)=6$ $\gcd(324,375)=3$ $\gcd(150,375)=75$ $\gcd(150,324,375)=3$
Therefore $D = 150+324+375-(6+3+75)+3 = \boxed{768}$ | null | 768 |
f35eaf92b4c75eaf6532bfc188ced5c1 | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_14 | $150\times 324\times 375$ rectangular solid is made by gluing together $1\times 1\times 1$ cubes. An internal diagonal of this solid passes through the interiors of how many of the $1\times 1\times 1$ cubes | Consider a point travelling across the internal diagonal, and let the internal diagonal have a length of $d$ . The point enters a new unit cube in the $x,y,z$ dimensions at multiples of $\frac{d}{150}, \frac{d}{324}, \frac{d}{375}$ respectively. We proceed by using PIE.
The point enters a new cube in the $x$ dimension $150$ times, in the $y$ dimension $324$ times and in the $z$ dimension, $375$ times.
The point enters a new cube in the $x$ and $y$ dimensions whenever a multiple of $\frac{d}{150}$ equals a multiple of $\frac{d}{324}$ . This occurs $\gcd(150, 324)$ times. Similarly, a point enters a new cube in the $y,z$ dimensions $\gcd(324, 375)$ times and a point enters a new cube in the $z,x$ dimensions $\gcd(375, 150)$ times.
The point enters a new cube in the $x,y$ and $z$ dimensions whenever some multiples of $\frac{d}{150}, \frac{d}{324}, \frac{d}{375}$ are equal. This occurs $\gcd(150, 324, 375)$ times.
The total number of unit cubes entered is then $150+324+375-[\gcd(150, 324)+\gcd(324, 375)+\gcd(375, 150)] + \gcd(150, 324, 375) = \boxed{768}$ | null | 768 |
717e8186c3b331342bafd12d3bfd787d | https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_15 | In parallelogram $ABCD$ , let $O$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}$ . Angles $CAB$ and $DBC$ are each twice as large as angle $DBA$ , and angle $ACB$ is $r$ times as large as angle $AOB$ . Find $\lfloor 1000r \rfloor$ | Let $\theta = \angle DBA$ . Then $\angle CAB = \angle DBC = 2\theta$ $\angle AOB = 180 - 3\theta$ , and $\angle ACB = 180 - 5\theta$ . Since $ABCD$ is a parallelogram, it follows that $OA = OC$ . By the Law of Sines on $\triangle ABO,\, \triangle BCO$
Dividing the two equalities yields
\[\frac{\sin 2\theta}{\sin \theta} = \frac{\sin (180 - 5\theta)}{\sin 2\theta} \Longrightarrow \sin^2 2\theta = \sin 5\theta \sin \theta.\]
Pythagorean and product-to-sum identities yield
\[1 - \cos^2 2 \theta = \frac{\cos 4\theta - \cos 6 \theta}{2},\]
and the double and triple angle ( $\cos 3x = 4\cos^3 x - 3\cos x$ ) formulas further simplify this to
\[4\cos^3 2\theta - 4\cos^2 2\theta - 3\cos 2\theta + 3 = (4\cos^2 2\theta - 3)(\cos 2\theta - 1) = 0\]
The only value of $\theta$ that fits in this context comes from $4 \cos^2 2\theta - 3 = 0 \Longrightarrow \cos 2\theta = \frac{\sqrt{3}}{2} \Longrightarrow \theta = 15^{\circ}$ . The answer is $\lfloor 1000r \rfloor = \left\lfloor 1000 \cdot \frac{180 - 5\theta}{180 - 3\theta} \right\rfloor = \left \lfloor \frac{7000}{9} \right \rfloor = \boxed{777}$ | null | 777 |
3eb8c296ab0ccf7cd81d6f728110a505 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_1 | Square $S_{1}$ is $1\times 1.$ For $i\ge 1,$ the lengths of the sides of square $S_{i+1}$ are half the lengths of the sides of square $S_{i},$ two adjacent sides of square $S_{i}$ are perpendicular bisectors of two adjacent sides of square $S_{i+1},$ and the other two sides of square $S_{i+1},$ are the perpendicular bisectors of two adjacent sides of square $S_{i+2}.$ The total area enclosed by at least one of $S_{1}, S_{2}, S_{3}, S_{4}, S_{5}$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m-n.$
AIME 1995 Problem 1.png | The sum of the areas of the squares if they were not interconnected is a geometric sequence
Then subtract the areas of the intersections, which is $\left(\frac{1}{4}\right)^2 + \ldots + \left(\frac{1}{32}\right)^2$
The majority of the terms cancel, leaving $1 + \frac{1}{4} - \frac{1}{1024}$ , which simplifies down to $\frac{1024 + \left(256 - 1\right)}{1024}$ . Thus, $m-n = \boxed{255}$ | null | 255 |
a6a9b5558d54236b147bbc828ba0fb89 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_2 | Find the last three digits of the product of the positive roots of $\sqrt{1995}x^{\log_{1995}x}=x^2$ | Instead of taking $\log_{1995}$ , we take $\log_x$ of both sides and simplify:
$\log_x(\sqrt{1995}x^{\log_{1995}x})=\log_x(x^{2})$
$\log_x\sqrt{1995}+\log_x x^{\log_{1995}x}=2$
$\dfrac{1}{2} \log_x 1995 + \log_{1995} x = 2$
We know that $\log_x 1995$ and $\log_{1995} x$ are reciprocals, so let $a=\log_{1995} x$ . Then we have $\dfrac{1}{2}\left(\dfrac{1}{a}\right) + a = 2$ . Multiplying by $2a$ and simplifying gives us $2a^2-4a+1=0$ , as shown above.
Because $a=\log_{1995} x$ $x=1995^a$ . By the quadratic formula, the two roots of our equation are $a=\frac{2\pm\sqrt2}{2}$ . This means our two roots in terms of $x$ are $1995^\frac{2+\sqrt2}{2}$ and $1995^\frac{2-\sqrt2}{2}.$ Multiplying these gives $1995^2$
$1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}$ , so our answer is $\boxed{025}$ | null | 025 |
a6a9b5558d54236b147bbc828ba0fb89 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_2 | Find the last three digits of the product of the positive roots of $\sqrt{1995}x^{\log_{1995}x}=x^2$ | Let $y=\log_{1995}x$ . Rewriting the equation in terms of $y$ , we have \[\sqrt{1995}\left(1995^y\right)^y=1995^{2y}\] \[1995^{y^2+\frac{1}{2}}=1995^{2y}\] \[y^2+\frac{1}{2}=2y\] \[2y^2-4y+1=0\] \[y=\frac{4\pm\sqrt{16-\left(4\right)\left(2\right)\left(1\right)}}{4}=\frac{4\pm\sqrt{8}}{4}=\frac{2\pm\sqrt{8}}{2}={1\pm\sqrt{2}}\] Thus, the product of the positive roots is $\left(1995^{\frac{2+\sqrt{8}}{2}}\right)\left(1995^{\frac{2-\sqrt{8}}{2}}\right)=1995^2=\left(2000-5\right)^2$ , so the last three digits are $\boxed{025}$ | null | 025 |
ce6414f81d26187f889b896d147d18a4 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3 | Starting at $(0,0),$ an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let $p$ be the probability that the object reaches $(2,2)$ in six or fewer steps. Given that $p$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | It takes an even number of steps for the object to reach $(2,2)$ , so the number of steps the object may have taken is either $4$ or $6$
If the object took $4$ steps, then it must have gone two steps and two steps , in some permutation. There are $\frac{4!}{2!2!} = 6$ ways for these four steps of occuring, and the probability is $\frac{6}{4^{4}}$
If the object took $6$ steps, then it must have gone two steps and two steps , and an additional pair of moves that would cancel out, either N/S or W/E . The sequences N,N,N,E,E,S can be permuted in $\frac{6!}{3!2!1!} = 60$ ways. However, if the first four steps of the sequence are N,N,E,E in some permutation, it would have already reached the point $(2,2)$ in four moves. There are $\frac{4!}{2!2!}$ ways to order those four steps and $2!$ ways to determine the order of the remaining two steps, for a total of $12$ sequences that we have to exclude. This gives $60-12=48$ sequences of steps. There are the same number of sequences for the steps N,N,E,E,E,W , so the probability here is $\frac{2 \times 48}{4^6}$
The total probability is $\frac{6}{4^4} + \frac{96}{4^6} = \frac{3}{64}$ , and $m+n= \boxed{067}$ | null | 067 |
ce6414f81d26187f889b896d147d18a4 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3 | Starting at $(0,0),$ an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let $p$ be the probability that the object reaches $(2,2)$ in six or fewer steps. Given that $p$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | Let's let the object wander for 6 steps so we get a constant denominator of $4^{6}$
In the first case, we count how many ways the object can end at (2,2), at the end of 6 steps. We will also count it even if we go to (2,2), and go back to (2,2). So, there are 2 different paths for the object to end at (2,2):
1.To go a permutation of R,R,R,U,U,L or
2.To go a permutation of R,R,U,U,U,D.
There are 60 ways to permute for each case, giving a total of 120 ways for the object to succeed and end at (2,2). In these 120 ways the object could reach (2,2) first and then come back to (2,2). This will be a factor in our second case.
In the second case, the object can get to (2,2) first in 4 moves, then move away from (2,2) with the remaing 2 moves. So, there are 6 ways to get to (2,2) in 4 moves, then there are 16 ways the object can "move around", but 4 of the ways will return the object back to (2,2). Those 4 ways were already counted in the first case, so we should only count 12 of the 16 ways to prevent over-counting. Thus, there are $12*6 =$ 72 ways in the second case.
So, in all, there are 120+72 ways for the object to achieve it's goal of moving to (2,2). Put that over our denominator, we get $\frac{192}{4^{6}} = \frac{3}{4^{3}} = \frac{3}{64}$ , in which adding the numerator and denominator get us an answer of $\boxed{067}$ | null | 067 |
9c26dc3ec751baff97e80fed85edaa8c | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_4 | Circles of radius $3$ and $6$ are externally tangent to each other and are internally tangent to a circle of radius $9$ . The circle of radius $9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord. | We label the points as following: the centers of the circles of radii $3,6,9$ are $O_3,O_6,O_9$ respectively, and the endpoints of the chord are $P,Q$ . Let $A_3,A_6,A_9$ be the feet of the perpendiculars from $O_3,O_6,O_9$ to $\overline{PQ}$ (so $A_3,A_6$ are the points of tangency ). Then we note that $\overline{O_3A_3} \parallel \overline{O_6A_6} \parallel \overline{O_9A_9}$ , and $O_6O_9 : O_9O_3 = 3:6 = 1:2$ . Thus, $O_9A_9 = \frac{2 \cdot O_6A_6 + 1 \cdot O_3A_3}{3} = 5$ (consider similar triangles). Applying the Pythagorean Theorem to $\triangle O_9A_9P$ , we find that \[PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}\] | null | 224 |
4f2511ab2a4fcafdae9bfc51511528f2 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_5 | For certain real values of $a, b, c,$ and $d_{},$ the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots. The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$ Find $b.$ | Since the coefficients of the polynomial are real, it follows that the non-real roots must come in complex conjugate pairs. Let the first two roots be $m,n$ . Since $m+n$ is not real, $m,n$ are not conjugates, so the other pair of roots must be the conjugates of $m,n$ . Let $m'$ be the conjugate of $m$ , and $n'$ be the conjugate of $n$ . Then, \[m\cdot n = 13 + i,m' + n' = 3 + 4i\Longrightarrow m'\cdot n' = 13 - i,m + n = 3 - 4i.\] By Vieta's formulas , we have that $b = mm' + nn' + mn' + nm' + mn + m'n' = (m + n)(m' + n') + mn + m'n' = \boxed{051}$ | null | 051 |
4f2511ab2a4fcafdae9bfc51511528f2 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_5 | For certain real values of $a, b, c,$ and $d_{},$ the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots. The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$ Find $b.$ | Let's assume that the 2 roots multiplied together are p+qi, and r+si, and the two roots added together are the conjugates of the previous roots. Using Vieta, we get \[b = (p+qi)(r+si) + (p+qi)(r-si) + (p-qi)(r+si) + (p-qi)(r-si) + (p+qi)(p-qi) + (r+si)(r-si) =\]
\[(p+qi+p-qi)(r+si+r-si) + (p+qi)(p-qi) + (r+si)(r-si) =\]
\[(2p)(2r) + p^2 + q^2 + r^2 + s^2 = 4pr + p^2 + q^2 + r^2 + s^2 = (p+r)^2 + 2pr + q^2 + s^2\]
We are now stuck. We can't simplify further. But, we look back to the problem and see that the two roots that are multiplied together give a product of $13+i$ , and the two roots that are added give $3+4i$ . This gets three equations necessary for solving the problem. \[p+r = 3\] \[pr-qs = 13\] \[-q-s = 4\] So, alright. Let's use the first equation to get that $(p+r)^2 = 9$ , and substitute that in. Now, the equation becomes:
\[b = 9 + 2pr + q^2 + s^2\]
We wish that we can turn the 2pr into 2qs. Fortunately, we can do that. By using the second equation, we can manipulate it to be $pr = 13+qs$ , and substitute that in.
\[b = 9 + 2(13+qs) + q^2 + s^2 = 9 + 26 + 2qs + q^2 + s^2 = 35 + (q+s)^2\]
We can square both sides of the third equation, and get $(q+s)^2 = 16$ We substitute that in and we get
\[b = 35+16 = \boxed{051}\] | null | 051 |
39568281b79de8b7115621f4d80e2360 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_6 | Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$ | We know that $n^2 = 2^{62}3^{38}$ must have $(62+1)\times (38+1)$ factors by its prime factorization . If we group all of these factors (excluding $n$ ) into pairs that multiply to $n^2$ , then one factor per pair is less than $n$ , and so there are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$ . There are $32\times20-1 = 639$ factors of $n$ , which clearly are less than $n$ , but are still factors of $n$ . Therefore, using complementary counting, there are $1228-639=\boxed{589}$ factors of $n^2$ that do not divide $n$ | null | 589 |
39568281b79de8b7115621f4d80e2360 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_6 | Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$ | Let $n=p_1^{k_1}p_2^{k_2}$ for some prime $p_1,p_2$ . Then $n^2$ has $\frac{(2k_1+1)(2k_2+1)-1}{2}$ factors less than $n$
This simplifies to $\frac{4k_1k_2+2k_1+2k_2}{2}=2k_1k_2+k_1+k_2$
The number of factors of $n$ less than $n$ is equal to $(k_1+1)(k_2+1)-1=k_1k_2+k_1+k_2$
Thus, our general formula for $n=p_1^{k_1}p_2^{k_2}$ is
Number of factors that satisfy the above $=(2k_1k_2+k_1+k_2)-(k_1k_2+k_1+k_2)=k_1k_2$
Incorporating this into our problem gives $19\times31=\boxed{589}$ | null | 589 |
39568281b79de8b7115621f4d80e2360 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_6 | Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$ | Consider divisors of $n^2: a,b$ such that $ab=n^2$ .
WLOG, let $b\ge{a}$ and $b=\frac{n}{a}$
Then, it is easy to see that $a$ will always be less than $b$ as we go down the divisor list of $n^2$ until we hit $n$
Therefore, the median divisor of $n^2$ is $n$
Then, there are $(63)(39)=2457$ divisors of $n^2$ . Exactly $\frac{2457-1}{2}=1228$ of these divisors are $<n$
There are $(32)(20)-1=639$ divisors of $n$ that are $<n$
Therefore, the answer is $1228-639=\boxed{589}$ | null | 589 |
309ab70ab52a2b4424874abcc6cabbae | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_7 | Given that $(1+\sin t)(1+\cos t)=5/4$ and
where $k, m,$ and $n_{}$ are positive integers with $m_{}$ and $n_{}$ relatively prime , find $k+m+n.$ | From the givens, $2\sin t \cos t + 2 \sin t + 2 \cos t = \frac{1}{2}$ , and adding $\sin^2 t + \cos^2t = 1$ to both sides gives $(\sin t + \cos t)^2 + 2(\sin t + \cos t) = \frac{3}{2}$ . Completing the square on the left in the variable $(\sin t + \cos t)$ gives $\sin t + \cos t = -1 \pm \sqrt{\frac{5}{2}}$ . Since $|\sin t + \cos t| \leq \sqrt 2 < 1 + \sqrt{\frac{5}{2}}$ , we have $\sin t + \cos t = \sqrt{\frac{5}{2}} - 1$ . Subtracting twice this from our original equation gives $(\sin t - 1)(\cos t - 1) = \sin t \cos t - \sin t - \cos t + 1 = \frac{13}{4} - \sqrt{10}$ , so the answer is $13 + 4 + 10 = \boxed{027}$ | null | 027 |
309ab70ab52a2b4424874abcc6cabbae | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_7 | Given that $(1+\sin t)(1+\cos t)=5/4$ and
where $k, m,$ and $n_{}$ are positive integers with $m_{}$ and $n_{}$ relatively prime , find $k+m+n.$ | Let $(1 - \sin t)(1 - \cos t) = x$ . Multiplying $x$ with the given equation, $\frac{5x}{4} = (1 - \sin^2 t)(1 - \cos^2 t) = \sin^2 t \cos ^2 t$ , and $\frac{\sqrt{5x}}{2} = \sin t \cos t$ . Simplifying and rearranging the given equation, $\sin t + \cos t = \frac{5}{4} - (\sin^2 t + \cos^2 t) - \sin t \cos t = \frac{1}{4} - \frac{\sqrt{5x}}{2}$ . Notice that $(1 + \sin t)(1 + \cos t) - 2(\sin t + \cos t) = x$ , and substituting, $x = \frac{5}{4} - 2( \frac{1}{4} - \frac{\sqrt{5x}}{2}) = \frac{3}{4} + \sqrt{5x}$ . Rearranging and squaring, $5x = x^2 - \frac{3}{2} x + \frac{9}{16}$ , so $x^2 - \frac{13}{2} x + \frac{9}{16} = 0$ , and $x = \frac{13}{4} \pm \sqrt{10}$ , but clearly, $0 \leq x < 4$ . Therefore, $x = \frac{13}{4} - \sqrt{10}$ , and the answer is $13 + 4 + 10 = \boxed{027}$ | null | 027 |
309ab70ab52a2b4424874abcc6cabbae | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_7 | Given that $(1+\sin t)(1+\cos t)=5/4$ and
where $k, m,$ and $n_{}$ are positive integers with $m_{}$ and $n_{}$ relatively prime , find $k+m+n.$ | We want $1+\sin t \cos t-\sin t-\cos t$ . However, note that we only need to find $\sin t+\cos t$
Let $y = \sin t+\cos t \rightarrow y^2 = \sin^2 t + \cos^2 t + 2\sin t \cos t = 1 + 2\sin t \cos t$
From this we have $\sin t \cos t = \frac{y^2-1}{2}$ and $\sin t + \cos t = y$
Substituting, we have $2y^2+4y-3=0 \rightarrow y = \frac {-2 \pm \sqrt{10}}{2}$
$\frac{5}{4} - 2(\frac{-2+\sqrt{10}}{2}) = \frac{13}{4}-\sqrt{10} \rightarrow 13+10+4=\boxed{027}$ | null | 027 |
a020cf57483e68a565424a6498b9b802 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_8 | For how many ordered pairs of positive integers $(x,y),$ with $y<x\le 100,$ are both $\frac xy$ and $\frac{x+1}{y+1}$ integers? | Since $y|x$ $y+1|x+1$ , then $\text{gcd}\,(y,x)=y$ (the bars indicate divisibility ) and $\text{gcd}\,(y+1,x+1)=y+1$ . By the Euclidean algorithm , these can be rewritten respectively as $\text{gcd}\,(y,x-y)=y$ and $\text{gcd}\, (y+1,x-y)=y+1$ , which implies that both $y,y+1 | x-y$ . Also, as $\text{gcd}\,(y,y+1) = 1$ , it follows that $y(y+1)|x-y$
Thus, for a given value of $y$ , we need the number of multiples of $y(y+1)$ from $0$ to $100-y$ (as $x \le 100$ ). It follows that there are $\left\lfloor\frac{100-y}{y(y+1)} \right\rfloor$ satisfactory positive integers for all integers $y \le 100$ . The answer is
\[\sum_{y=1}^{99} \left\lfloor\frac{100-y}{y(y+1)} \right\rfloor = 49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.\] | null | 085 |
a020cf57483e68a565424a6498b9b802 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_8 | For how many ordered pairs of positive integers $(x,y),$ with $y<x\le 100,$ are both $\frac xy$ and $\frac{x+1}{y+1}$ integers? | We know that $x \equiv 0 \mod y$ and $x+1 \equiv 0 \mod y+1$
Write $x$ as $ky$ for some integer $k$ . Then, $ky+1 \equiv 0\mod y+1$ . We can add $k$ to each side in order to factor out a $y+1$ . So, $ky+k+1 \equiv k \mod y+1$ or $k(y+1)+1 \equiv k \mod y+1$ . We know that $k(y+1) \equiv 0 \mod y+1$ . We finally achieve the congruence $1-k \equiv 0 \mod y+1$
We can now write $k$ as $(y+1)a+1$ . Plugging this back in, if we have a value for $y$ , then $x = ky = ((y+1)a+1)y = y(y+1)a+y$ . We only have to check values of $y$ when $y(y+1)<100$ . This yields the equations $x = 2a+1, 6a+2, 12a+3, 20a+4, 30a+5, 42a+6, 56a+7, 72a+8, 90a+9$
Finding all possible values of $a$ such that $y<x<100$ , we get $49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.$ | null | 085 |
a020cf57483e68a565424a6498b9b802 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_8 | For how many ordered pairs of positive integers $(x,y),$ with $y<x\le 100,$ are both $\frac xy$ and $\frac{x+1}{y+1}$ integers? | We use casework:
For $y=1$ , we have $3,5,\cdots ,99$ , or $49$ cases.
For $y=2$ , we have $8,14,\cdots ,98$ , or $16$ cases.
For $y=3$ , we have $15,27,\cdots ,99$ , or $8$ cases.
For $y=4$ , we have $24,44\cdots ,84$ , or $4$ cases.
For $y=5$ , we have $35,65,95$ , or $3$ cases.
For $y=6$ , we have $48,90$ , or $2$ cases.
For $y=7$ , we have $63$ , or $1$ case.
For $y=8$ , we have $80$ , or $1$ case.
For $y=9$ , we have $99$ , or $1$ case.
Adding, we get our final result of $\boxed{085}.$ | null | 085 |
6d191f5729923bba77a40556b417b088 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(1,1.63),W); dot((1,3),ds); label("$A$",(1,3),N); dot((0,0),ds); label("$B$",(0,0),SW); dot((2,0),ds); label("$C$",(2,0),SE); dot((1,0),ds); label("$M$",(1,0),S); dot((1,0.7),ds); label("$D$",(1,0.7),NE); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy] | Let $x=\angle CAM$ , so $3x=\angle CDM$ . Then, $\frac{\tan 3x}{\tan x}=\frac{CM/1}{CM/11}=11$ . Expanding $\tan 3x$ using the angle sum identity gives \[\tan 3x=\tan(2x+x)=\frac{3\tan x-\tan^3x}{1-3\tan^2x}.\] Thus, $\frac{3-\tan^2x}{1-3\tan^2x}=11$ . Solving, we get $\tan x= \frac 12$ . Hence, $CM=\frac{11}2$ and $AC= \frac{11\sqrt{5}}2$ by the Pythagorean Theorem . The total perimeter is $2(AC + CM) = \sqrt{605}+11$ . The answer is thus $a+b=\boxed{616}$ | null | 616 |
6d191f5729923bba77a40556b417b088 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(1,1.63),W); dot((1,3),ds); label("$A$",(1,3),N); dot((0,0),ds); label("$B$",(0,0),SW); dot((2,0),ds); label("$C$",(2,0),SE); dot((1,0),ds); label("$M$",(1,0),S); dot((1,0.7),ds); label("$D$",(1,0.7),NE); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy] | In a similar fashion, we encode the angles as complex numbers, so if $BM=x$ , then $\angle BAD=\text{Arg}(11+xi)$ and $\angle BDM=\text{Arg}(1+xi)$ . So we need only find $x$ such that $\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)$ . This will happen when $\frac{363x-x^3}{1331-33x^2}=x$ , which simplifies to $121x-4x^3=0$ . Therefore, $x=\frac{11}{2}$ . By the Pythagorean Theorem, $AB=\frac{11\sqrt{5}}{2}$ , so the perimeter is $11+11\sqrt{5}=11+\sqrt{605}$ , giving us our answer, $\boxed{616}$ | null | 616 |
6d191f5729923bba77a40556b417b088 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(1,1.63),W); dot((1,3),ds); label("$A$",(1,3),N); dot((0,0),ds); label("$B$",(0,0),SW); dot((2,0),ds); label("$C$",(2,0),SE); dot((1,0),ds); label("$M$",(1,0),S); dot((1,0.7),ds); label("$D$",(1,0.7),NE); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy] | Let $\angle BAD=\alpha$ , so $\angle BDM=3\alpha$ $\angle BDA=180-3\alpha$ , and thus $\angle ABD=2\alpha.$ We can then draw the angle bisector of $\angle ABD$ , and let it intersect $\overline{AM}$ at $E.$ Since $\angle BAE=\angle ABE$ $AE=BE.$ Let $AE=x$ . Then we see by the Pythagorean Theorem, $BM=\sqrt{BE^2-ME^2}=\sqrt{x^2-(11-x)^2}=\sqrt{22x-121}$ $BD=\sqrt{BM^2+1}=\sqrt{22x-120}$ $BA=\sqrt{BM^2+121}=\sqrt{22x}$ , and $DE=10-x.$ By the angle bisector theorem, $BA/BD=EA/ED.$ Substituting in what we know for the lengths of those segments, we see that \[\frac{\sqrt{22x}}{\sqrt{22x-120}}=\frac{x}{10-x}.\] multiplying by both denominators and squaring both sides yields \[22x(10-x)^2=x^2(22x-120)\] which simplifies to $x=\frac{55}{8}.$ Substituting this in for x in the equations for $BA$ and $BM$ yields $BA=\frac{\sqrt{605}}{2}$ and $BM=\frac{11}{2}.$ Thus the perimeter is $11+\sqrt{605}$ , and the answer is $\boxed{616}$ | null | 616 |
6d191f5729923bba77a40556b417b088 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(1,1.63),W); dot((1,3),ds); label("$A$",(1,3),N); dot((0,0),ds); label("$B$",(0,0),SW); dot((2,0),ds); label("$C$",(2,0),SE); dot((1,0),ds); label("$M$",(1,0),S); dot((1,0.7),ds); label("$D$",(1,0.7),NE); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy] | The triangle is symmetrical so we can split it in half ( $\triangle ABM$ and $\triangle ACM$ ).
Let $\angle BAM = y$ and $\angle BDM = 3y$ . By the Law of Sines on triangle $BAD$ $\frac{10}{\sin 2y} = \frac{BD}{\sin y}$ . Using $\sin 2y = 2\sin y\cos y$ we can get $BD = \frac{5}{\cos y}$ . We can use this information to relate $BD$ to $DM$ by using the Law of Sines on triangle $BMD$
\[\frac{\frac{5}{\cos y}}{\sin BMD} = \frac{1}{\sin 90^\circ - 3y}\]
$\sin BMD = 1$ (as $\angle BMD$ is a right angle), so $\frac{1}{\sin 90^\circ - 3y} = \frac{5}{\cos y}$ . Using the identity $\sin 90^\circ - x = \cos x$ , we can turn the equation into::
\[\frac{1}{\cos 3y} = \frac{5}{\cos y}\]
\[5\cos 3y = \cos y\]
\[5(4\cos ^3 y - 3\cos y) = \cos y\]
\[20\cos ^3 y = 16 \cos y\]
\[5\cos ^3 y = 4\cos y\]
\[5\cos ^2 y = 4\]
\[\cos ^2 y = \frac{4}{5}\]
Now that we've found $\cos y$ , we can look at the side lengths of $BM$ and $AB$ (since they are symmetrical, the perimeter of $\triangle ABC$ is $2(BM + AB)$
We note that $BM = 11\tan y$ and $AB = 11\sec y$
\[\sin ^2 y = 1 - \cos ^2 y\]
\[\sin ^2 y = \frac{1}{5}\]
\[\tan ^2 y = \frac{1}{4}\]
\[\tan y = \frac{1}{2}\]
(Note it is positive since $BM > 0$ ).
\[\sec ^2 y = \frac{5}{4}\]
\[\sec y = \frac{\sqrt{5}}{2}\]
\[BM + AB = 11\frac{\sqrt{5}+1}{2}\]
\[2(BM + AB) = 11(\sqrt{5} + 1)\]
\[2(BM + AB) = 11\sqrt{5} + 11\]
\[2(BM + AB) = \sqrt{605} + 11\]
The answer is $\boxed{616}$ | null | 616 |
6d191f5729923bba77a40556b417b088 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_9 | Triangle $ABC$ is isosceles , with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(1,1.63),W); dot((1,3),ds); label("$A$",(1,3),N); dot((0,0),ds); label("$B$",(0,0),SW); dot((2,0),ds); label("$C$",(2,0),SE); dot((1,0),ds); label("$M$",(1,0),S); dot((1,0.7),ds); label("$D$",(1,0.7),NE); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy] | Suppose $\angle BAM=\angle CAM =x$ , since $\angle BDC=3\angle BAC$ , we have $\angle BDM=\angle MDC = 3x$ . Therefore, $\angle DBC=\angle DCB = 90^\circ -3x$ and $\angle ABD=\angle DCA=2x$ . As a result, $\triangle KAC$ is isosceles, $KC=KA$
Let $H$ be a point on the extension of $CD$ through $D$ such that $\overline{HB}\perp\overline{BC}$ and denote the intersection of $\overline{HC}$ and $\overline{AB}$ as $K$ . Then, $BH=2DM=2, \overline{HB}\parallel\overline{DM}$ , and $HD=DC$ by the Midpoint Theorem. So, $\angle HBA=x$ and $\angle CDM=\angle CHB=\angle HDA= 3x$
Consequently, $\triangle HBK\sim \triangle DAK$ \[\frac{BK}{KA}=\frac{HK}{KD}=\frac{1}{5}\] Assume $BK=a$ and $HK=b$ , then $KA=5a$ and $KD = 5b$ . Since $KC=KA, KC=5a$ , and since $HD=DC$ $KC=11b$ . Therefore, $a=\frac{11}{5}b$
In $\triangle BDM$ , by the Pythagorean Theorem $BM=\sqrt{36b^2-1}$ . Similarly in $\triangle BAM$ $BM=\sqrt{36a^2-121}$ . So \[\sqrt{36a^2-121}=\sqrt{36b^2-1}\] Since $a=\frac{11}{5}b$ , we have $b=\frac{5\sqrt{5}}{12}$ and $a=\frac{11\sqrt{5}}{12}$ . Consequently, $BM=\frac{11}{2}$ and $AB=\frac{11\sqrt{5}}{2}$ . Thus, the perimeter of $\triangle ABC$ is $11+\sqrt{605}$ , and the answer is $\boxed{616}$ | null | 616 |
ad03b17fa75cee0b5e320f174909059c | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_10 | What is the largest positive integer that is not the sum of a positive integral multiple of $42$ and a positive composite integer? | Let our answer be $n$ . Write $n = 42a + b$ , where $a, b$ are positive integers and $0 \leq b < 42$ . Then note that $b, b + 42, ... , b + 42(a-1)$ are all primes.
If $b$ is $0\mod{5}$ , then $b = 5$ because $5$ is the only prime divisible by $5$ . We get $n = 215$ as our largest possibility in this case.
If $b$ is $1\mod{5}$ , then $b + 2 \times 42$ is divisible by $5$ and thus $a \leq 2$ . Thus, $n \leq 3 \times 42 = 126 < 215$
If $b$ is $2\mod{5}$ , then $b + 4 \times 42$ is divisible by $5$ and thus $a \leq 4$ . Thus, $n \leq 5 \times 42 = 210 < 215$
If $b$ is $3\mod{5}$ , then $b + 1 \times 42$ is divisible by $5$ and thus $a = 1$ . Thus, $n \leq 2 \times 42 = 84 < 215$
If $b$ is $4\mod{5}$ , then $b + 3 \times 42$ is divisible by $5$ and thus $a \leq 3$ . Thus, $n \leq 4 \times 42 = 168 < 215$
Our answer is $\boxed{215}$ | null | 215 |
aae24b862004e67889b97a440c22bf3c | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_11 | A right rectangular prism $P_{}$ (i.e., a rectangular parallelpiped) has sides of integral length $a, b, c,$ with $a\le b\le c.$ A plane parallel to one of the faces of $P_{}$ cuts $P_{}$ into two prisms, one of which is similar to $P_{},$ and both of which have nonzero volume. Given that $b=1995,$ for how many ordered triples $(a, b, c)$ does such a plane exist? | Let $P'$ be the prism similar to $P$ , and let the sides of $P'$ be of length $x,y,z$ , such that $x \le y \le z$ . Then
\[\frac{x}{a} = \frac{y}{b} = \frac zc < 1.\]
Note that if the ratio of similarity was equal to $1$ , we would have a prism with zero volume. As one face of $P'$ is a face of $P$ , it follows that $P$ and $P'$ share at least two side lengths in common. Since $x < a, y < b, z < c$ , it follows that the only possibility is $y=a,z=b=1995$ . Then,
\[\frac{x}{a} = \frac{a}{1995} = \frac{1995}{c} \Longrightarrow ac = 1995^2 = 3^25^27^219^2.\]
The number of factors of $3^25^27^219^2$ is $(2+1)(2+1)(2+1)(2+1) = 81$ . Only in $\left\lfloor \frac {81}2 \right\rfloor = 40$ of these cases is $a < c$ (for $a=c$ , we end with a prism of zero volume). We can easily verify that these will yield nondegenerate prisms, so the answer is $\boxed{040}$ | null | 040 |
fc958d16f44e146b2b2253153b47ede8 | https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_12 | Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m_{}$ and $n_{}$ are integers, find $m+n.$ | The angle $\theta$ is the angle formed by two perpendiculars drawn to $BO$ , one on the plane determined by $OAB$ and the other by $OBC$ . Let the perpendiculars from $A$ and $C$ to $\overline{OB}$ meet $\overline{OB}$ at $P.$ Without loss of generality , let $AP = 1.$ It follows that $\triangle OPA$ is a $45-45-90$ right triangle , so $OP = AP = 1,$ $OB = OA = \sqrt {2},$ and $AB = \sqrt {4 - 2\sqrt {2}}.$ Therefore, $AC = \sqrt {8 - 4\sqrt {2}}.$
From the Law of Cosines $AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\cos \theta,$ so
\[8 - 4\sqrt {2} = 1 + 1 - 2\cos \theta \Longrightarrow \cos \theta = - 3 + 2\sqrt {2} = - 3 + \sqrt{8}.\]
Thus $m + n = \boxed{005}$ | null | 005 |