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int64 1
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6,601
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_4
| 1
|
The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]
|
Call the squares' side lengths from smallest to largest $a_1,\ldots,a_9$ , and let $l,w$ represent the dimensions of the rectangle.
The picture shows that \begin{align*} a_1+a_2 &= a_3\\ a_1 + a_3 &= a_4\\ a_3 + a_4 &= a_5\\ a_4 + a_5 &= a_6\\ a_2 + a_3 + a_5 &= a_7\\ a_2 + a_7 &= a_8\\ a_1 + a_4 + a_6 &= a_9\\ a_6 + a_9 &= a_7 + a_8.\end{align*}
Expressing all terms 3 to 9 in terms of $a_1$ and $a_2$ and substituting their expanded forms into the previous equation will give the expression $5a_1 = 2a_2$
We can guess that $a_1 = 2$ . (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives $a_9 = 36$ $a_6=25$ $a_8 = 33$ , which gives us $l=61,w=69$ . These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\boxed{260}$
| 260
|
6,602
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_4
| 2
|
The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]
|
We can just list the equations: \begin{align*} s_3 &= s_1 + s_2 \\ s_4 &= s_3 + s_1 \\ s_5 &= s_4 + s_3 \\ s_6 &= s_5 + s_4 \\ s_7 &= s_5 + s_3 + s_2 \\ s_8 &= s_7 + s_2 \\ s_9 &= s_8 + s_2 - s_1 \\ s_9 + s_8 &= s_7 + s_6 + s_5 \end{align*} We can then write each $s_i$ in terms of $s_1$ and $s_2$ as follows \begin{align*} s_4 &= 2s_1 + s_2 \\ s_5 &= 3s_1 +2s_2 \\ s_6 &= 5s_1 + 3s_2 \\ s_7 &= 4s_1 + 4s_2 \\ s_8 &= 4s_1 + 5s_2 \\ s_9 &= 3s_1 + 6s_2 \\ \end{align*} Since $s_9 + s_8 = s_7 + s_6 + s_5 \implies (3s_1 + 6s_2) + (4s_1 + 5s_2) = (4s_1 + 4s_2) + (5s_1 + 3s_2) + (3s_1 + 2s_2),$ \[2s_2 = 5s_1 \implies \frac{2}{5}s_2 = s_1.\] Since the side lengths of the rectangle are relatively prime, we can see that $s_1 = 2$ and $s_2 = 5.$ Therefore, $2(2s_9 + s_6 + s_8) = 30s_1 + 40s_2 = \boxed{260}.$ ~peelybonehead
| 260
|
6,603
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_4
| 3
|
The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]
|
We set the side length of the smallest square to 1, and set the side length of square $a_4$ in the previous question to a. We do some "side length chasing" and get $4a - 4 = 2a + 5$ . Solving, we get $a = 4.5$ and the side lengths are $61$ and $69$ . Thus, the perimeter of the rectangle is $2(61 + 69) = \boxed{260}.$
| 260
|
6,604
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6
| 1
|
For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$
|
\begin{eqnarray*} \frac{x+y}{2} &=& \sqrt{xy} + 2\\ x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\ \sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}
Because $y > x$ , we only consider $+2$
For simplicity, we can count how many valid pairs of $(\sqrt{x},\sqrt{y})$ that satisfy our equation.
The maximum that $\sqrt{y}$ can be is $\sqrt{10^6} - 1 = 999$ because $\sqrt{y}$ must be an integer (this is because $\sqrt{y} - \sqrt{x} = 2$ , an integer). Then $\sqrt{x} = 997$ , and we continue this downward until $\sqrt{y} = 3$ , in which case $\sqrt{x} = 1$ . The number of pairs of $(\sqrt{x},\sqrt{y})$ , and so $(x,y)$ is then $\boxed{997}$
| 997
|
6,605
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6
| 2
|
For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$
|
Let $a^2$ $x$ and $b^2$ $y$ , where $a$ and $b$ are positive.
Then \[\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2\] \[a^2 + b^2 = 2ab + 4\] \[(a-b)^2 = 4\] \[(a-b) = \pm 2\]
This makes counting a lot easier since now we just have to find all pairs $(a,b)$ that differ by 2.
Because $\sqrt{10^6} = 10^3$ , then we can use all positive integers less than 1000 for $a$ and $b$
We know that because $x < y$ , we get $a < b$
We can count even and odd pairs separately to make things easier*:
Odd: \[(1,3) , (3,5) , (5,7) . . . (997,999)\]
Even: \[(2,4) , (4,6) , (6,8) . . . (996,998)\]
This makes 499 odd pairs and 498 even pairs, for a total of $\boxed{997}$ pairs.
| 997
|
6,606
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6
| 3
|
For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$
|
Since the arithmetic mean is 2 more than the geometric mean, $\frac{x+y}{2} = 2 + \sqrt{xy}$ . We can multiply by 2 to get $x + y = 4 + 2\sqrt{xy}$ . Subtracting 4 and squaring gives \[((x+y)-4)^2 = 4xy\] \[((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy\] \[x^2 - 2xy + y^2 + 16 - 8x - 8y = 0\]
Notice that $((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y$ , so the problem asks for solutions of \[(x-y-4)^2 = 16y\] Since the left hand side is a perfect square, and 16 is a perfect square, $y$ must also be a perfect square. Since $0 < y < (1000)^2$ $y$ must be from $1^2$ to $999^2$ , giving at most 999 options for $y$
However if $y = 1^2$ , you get $(x-5)^2 = 16$ , which has solutions $x = 9$ and $x = 1$ . Both of those solutions are not less than $y$ , so $y$ cannot be equal to 1. If $y = 2^2 = 4$ , you get $(x - 8)^2 = 64$ , which has 2 solutions, $x = 16$ , and $x = 0$ . 16 is not less than 4, and $x$ cannot be 0, so $y$ cannot be 4. However, for all other $y$ , you get exactly 1 solution for $x$ , and that gives a total of $999 - 2 = \boxed{997}$ pairs.
| 997
|
6,607
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6
| 4
|
For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$
|
Rearranging our conditions to
\[x^2-2xy+y^2+16-8x-8y=0 \implies\] \[(y-x)^2=8(x+y-2).\]
Thus, $4|y-x.$
Now, let $y = 4k+x.$ Plugging this back into our expression, we get
\[(k-1)^2=x-1.\]
There, a unique value of $x, y$ is formed for every value of $k$ . However, we must have
\[y<10^6 \implies (k+1)^2< 10^6-1\]
and
\[x=(k-1)^2+1>0.\]
Therefore, there are only $\boxed{997}$ pairs of $(x,y).$
| 997
|
6,608
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_7
| 2
|
Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques
|
Let $r = \frac{m}{n} = z + \frac {1}{y}$
\begin{align*} (5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ &=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\ &=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\ &=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\ &=2 + 5 + 29 + r\\ &=36 + r \end{align*}
Thus $145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}$ . So $m + n = 1 + 4 = \boxed{5}$
| 5
|
6,609
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_10
| 1
|
sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$
|
Let the sum of all of the terms in the sequence be $\mathbb{S}$ . Then for each integer $k$ $x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k$ . Summing this up for all $k$ from $1, 2, \ldots, 100$
\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &= 1 + 2 + \cdots + 100\\ 100\mathbb{S} - 2\mathbb{S} &= \frac{100 \cdot 101}{2} = 5050\\ \mathbb{S}&=\frac{2525}{49}\end{align*}
Now, substituting for $x_{50}$ , we get $2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}$ , and the answer is $75+98=\boxed{173}$
| 173
|
6,610
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_10
| 2
|
sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$
|
Consider $x_k$ and $x_{k+1}$ . Let $S$ be the sum of the rest 98 terms. Then $x_k+k=S+x_{k+1}$ and $x_{k+1}+(k+1)=S+x_k.$ Eliminating $S$ we have $x_{k+1}-x_k=-\dfrac{1}{2}.$ So the sequence is arithmetic with common difference $-\dfrac{1}{2}.$
In terms of $x_{50},$ the sequence is $x_{50}+\dfrac{49}{2}, x_{50}+\dfrac{48}{2},\cdots,x_{50}+\dfrac{1}{2}, x_{50}, x_{50}-\dfrac{1}{2}, \cdots, x_{50}-\dfrac{49}{2}, x_{50}-\dfrac{50}{2}.$ Therefore, $x_{50}+50=99x_{50}-\dfrac{50}{2}$
Solving, we get $x_{50}=\dfrac{75}{98}.$ The answer is $75+98=\boxed{173}.$
| 173
|
6,611
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_11
| 1
|
Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$
|
Since all divisors of $1000 = 2^35^3$ can be written in the form of $2^{m}5^{n}$ , it follows that $\frac{a}{b}$ can also be expressed in the form of $2^{x}5^{y}$ , where $-3 \le x,y \le 3$ . Thus every number in the form of $a/b$ will be expressed one time in the product
\[(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^2 + 2^3)(5^{-3} + 5^{-2} +5^{-1} + 5^{0} + 5^{1} + 5^2 + 5^3)\]
Using the formula for a geometric series , this reduces to $S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}$ , and $\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}$
| 248
|
6,612
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_11
| 2
|
Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$
|
Essentially, the problem asks us to compute \[\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b}\] which is pretty easy: \[\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b} = \sum_{a=-3}^3 2^a \sum_{b=-3}^3 \frac{1}{5^b} = \sum_{a=-3}^3 2^a 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) \sum_{a=-3}^3 2^a = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg)2^{-3} \bigg( \frac{1-2^7}{1-2} \bigg) = 2480 + \frac{437}{1000}\] so our answer is $\left\lfloor \frac{2480 + \frac{437}{1000}}{10} \right\rfloor = \boxed{248}$
| 248
|
6,613
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_11
| 4
|
Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$
|
We can organize the fractions and reduce them in quantities to reach our answer. First, separate the fractions with coprime parts into those that are combinations of powers of 2 and 5, and those that are a combination of a 1 and another divisor.
To begin with the first list, list powers of 2 and 5 from 0 to 3. In this specific case I find it easier to augment every denominator to 1000 and then divide by 1000. To do this, write the corresponding divisor under each power. e.g. 2 - 500, 4 - 250, 5 - 200, etc. Call this the "partner" of any divisor.
Now we now the amount to multiply the numerator if given number is in the denominator. Now we simply combine and reduce these groups. If the powers of 2 are on the denominator, then every power of five will be multiplied by the partner of the power of 2. Essentially, all we have to do is a large scale distributive property application. There is nothing complicated about this except to be careful.
Add all powers of 2: 15
Add their partners: 1875
Add all powers of 5: 156
Add their partners: 1248
Then, follow this formula: (sum of powers * sum opposite power's partners)+(sum of powers * sum opposite power's partners)
Or, $156*1875+15*1248$ $=311220$
Now, divide by 1000 to compensate for the denominator. $311.22$
Finally, we have to calculate the other list of fractions with 1 and another divisor. e.g. 1 - 250, 1 - 20 etc. (these all count)
This time we need to list all divisors of 1000, including 1. Remove all powers of 2 or 5, because we already included those in the other list. Now, notice there are two cases. 1: 1 is in the denominator, making the fraction an integer. 2: 1 is in the numerator.
Adding all the integers in the first case gives us 2169. The second case can actually be discarded, but still can be found. Realize that if we include the powers of 2 and 5, then the second case is (sum of all divisors)/1000. Remove all partners of powers of 2 and 5, and we'll get exactly 217/1000, or 0.22.
Finally, add all the numbers together: $311.22 + 2169 + 0.22 = 2480.44$
And divide by 10: $248.044$
After an odyssey of bashing: $\boxed{248}$
| 248
|
6,614
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_12
| 1
|
Given a function $f$ for which \[f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)\] holds for all real $x,$ what is the largest number of different values that can appear in the list $f(0),f(1),f(2),\ldots,f(999)?$
|
\begin{align*}f(2158 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\ f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{align*}
Since $\mathrm{gcd}(1056, 1760) = 352$ we can conclude that (by the Euclidean algorithm
\[f(x) = f(352 + x)\]
So we need only to consider one period $f(0), f(1), ... f(351)$ , which can have at most $352$ distinct values which determine the value of $f(x)$ at all other integers.
But we also know that $f(x) = f(46 - x) = f(398 - x)$ , so the values $x = 24, 25, ... 46$ and $x = 200, 201, ... 351$ are repeated. This gives a total of
\[352 - (46 - 24 + 1) - (351 - 200 + 1) = \boxed{177}\]
| 177
|
6,615
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_13
| 1
|
In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is $m/n$ square miles, where $m$ and $n$ are relatively prime positive integers. Find $m + n$
|
Let the intersection of the highways be at the origin $O$ , and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.
After going $x$ miles, $t=\frac{d}{r}=\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\frac{1}{10}-\frac{x}{50}$ hours, or $d=rt=14t=1.4-\frac{7x}{25}$ miles. It can end up anywhere off the highway in a circle with this radius centered at $(x,0)$ . All these circles are homothetic with respect to a center at $(5,0)$
Now consider the circle at $(0,0)$ . Draw a line tangent to it at $A$ and passing through $B (5,0)$ . By the Pythagorean Theorem $AB^2+AO^2=OB^2 \Longrightarrow AB=\sqrt{OB^2-AO^2}=\sqrt{5^2-1.4^2}=\frac{24}{5}$ . Then $\tan(\angle ABO)=\frac{OA}{AB}=\frac{7}{24}$ , so the slope of line $AB$ is $\frac{-7}{24}$ . Since it passes through $(5,0)$ its equation is $y=\frac{-7}{24}(x-5)$
This line and the x and y axis bound the region the truck can go if it moves in the positive x direction. Similarly, the line $y=5-\frac{24}{7}x$ bounds the region the truck can go if it moves in positive y direction. The intersection of these two lines is $\left(\frac{35}{31},\frac{35}{31}\right)$ . The bounded region in Quadrant I is made up of a square and two triangles. $A=x^2+x(5-x)=5x$ . By symmetry, the regions in the other quadrants are the same, so the area of the whole region is $20x=\frac{700}{31}$ so the answer is $700+31=\boxed{731}$
| 731
|
6,616
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14
| 1
|
In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$
|
Let $\angle QPB=x^\circ$ . Because $\angle AQP$ is exterior to isosceles triangle $PQB$ its measure is $2x$ and $\angle PAQ$ has the same measure. Because $\angle BPC$ is exterior to $\triangle BPA$ its measure is $3x$ . Let $\angle PBC = y^\circ$ . It follows that $\angle ACB = x+y$ and that $4x+2y=180^\circ$ . Two of the angles of triangle $APQ$ have measure $2x$ , and thus the measure of $\angle APQ$ is $2y$ . It follows that $AQ=2\cdot AP\cdot \sin y$ . Because $AB=AC$ and $AP=QB$ , it also follows that $AQ=PC$ . Now apply the Law of Sines to triangle $PBC$ to find \[\frac{\sin 3x}{BC}=\frac{\sin y}{PC}=\frac{\sin y}{2\cdot AP\cdot \sin y}= \frac{1}{2\cdot BC}\] because $AP=BC$ . Hence $\sin 3x = \tfrac 12$ . Since $4x<180$ , this implies that $3x=30$ , i.e. $x=10$ . Thus $y=70$ and \[r=\frac{10+70}{2\cdot 70}=\frac 47,\] which implies that $1000r = 571 + \tfrac 37$ . So the answer is $\boxed{571}$
| 571
|
6,617
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14
| 2
|
In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$
|
Let point $R$ be in $\triangle ABC$ such that $QB = BR = RP$ . Then $PQBR$ is a rhombus , so $AB \parallel PR$ and $APRB$ is an isosceles trapezoid . Since $\overline{PB}$ bisects $\angle QBR$ , it follows by symmetry in trapezoid $APRB$ that $\overline{RA}$ bisects $\angle BAC$ . Thus $R$ lies on the perpendicular bisector of $\overline{BC}$ , and $BC = BR = RC$ . Hence $\triangle BCR$ is an equilateral triangle
Now $\angle ABR = \angle BAC = \angle ACR$ , and the sum of the angles in $\triangle ABC$ is $\angle ABR + 60^{\circ} + \angle BAC + \angle ACR + 60^{\circ} = 3\angle BAC + 120^{\circ} = 180^{\circ} \Longrightarrow \angle BAC = 20^{\circ}$ . Then $\angle APQ = 140^{\circ}$ and $\angle ACB = 80^{\circ}$ , so the answer is $\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = \boxed{571}$
| 571
|
6,618
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14
| 3
|
In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$
|
Again, construct $R$ as above.
Let $\angle BAC = \angle QBR = \angle QPR = 2x$ and $\angle ABC = \angle ACB = y$ , which means $x + y = 90$ $\triangle QBC$ is isosceles with $QB = BC$ , so $\angle BCQ = 90 - \frac {y}{2}$ .
Let $S$ be the intersection of $QC$ and $BP$ . Since $\angle BCQ = \angle BQC = \angle BRS$ $BCRS$ is cyclic , which means $\angle RBS = \angle RCS = x$ .
Since $APRB$ is an isosceles trapezoid, $BP = AR$ , but since $AR$ bisects $\angle BAC$ $\angle ABR = \angle ACR = 2x$
Therefore we have that $\angle ACB = \angle ACR + \angle RCS + \angle QCB = 2x + x + 90 - \frac {y}{2} = y$ .
We solve the simultaneous equations $x + y = 90$ and $2x + x + 90 - \frac {y}{2} = y$ to get $x = 10$ and $y = 80$ $\angle APQ = 180 - 4x = 140$ $\angle ACB = 80$ , so $r = \frac {80}{140} = \frac {4}{7}$ $\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}$
| 571
|
6,619
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_14
| 4
|
In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent . Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP = PQ = QB = BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find $\lfloor 1000r \rfloor$
|
Let $\angle BAC= 2\theta$ and $AP=PQ=QB=BC=x$ $\triangle APQ$ is isosceles, so $AQ=2x\cos 2\theta =2x(1-2\sin^2\theta)$ and $AB= AQ+x=x\left(3-4\sin^2\theta\right)$ $\triangle{ABC}$ is isosceles too, so $x=BC=2AB\sin\theta$ . Using the expression for $AB$ , we get \[1=2\left(3\sin\theta-4\sin^3\theta\right)=2\sin3\theta\] by the triple angle formula! Thus $\theta=10^\circ$ and $\angle A = 2\theta=20^\circ$ .
It follows now that $\angle APQ=140^\circ$ $\angle ACB=80^\circ$ , giving $r=\tfrac{4}{7}$ , which implies that $1000r = 571 + \tfrac 37$ . So the answer is $\boxed{571}$
| 571
|
6,620
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_15
| 2
|
A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$
|
To simplify matters, we want a power of $2$ . Hence, we will add $48$ 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position $1024$ in a $2048$ card stack, where the fake cards towards the front.
Let the fake cards have positions $1, 3, 5, \cdots, 95$ . Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the $2000$ card case, where all of them are below $1999$ . From this, we know that the cards from positions $1$ to $96$ alternate in fake-real-fake-real, where we have the correct order of cards once the first $96$ have moved and we can start putting real cards on the table. Hence, $1999$ is in position $1024 - 96 = 928$ , so $\boxed{927}$ cards are above it. - Spacesam
| 927
|
6,621
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_15
| 4
|
A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$
|
Let us treat each run through the deck as a separate "round". For example, in round one, you would go through all of the $2000$ cards initially in the deck once, in round two, you would go through all $1000$ cards initially in the deck once, so on and so forth. For each round, let us record what the initial and final actions are ( $r$ for moving the card to the right, $b$ for moving the card to the bottom), the number of cards moved to the right, the number of cards left, and what the position of the cards moved to the right were in the original $2000$ card deck (as $n = a + ck$ where $n$ is the position, $c$ is the spacing of the cards moved, $a$ is an integer such that the correct first card is moved, and $k$ is an integer greater than or equal to $1$ which represents which card out of all the cards moved to the right you are finding the position of). Then,
Round 1: $r$ to $b$ $1000$ to right, $1000$ left in deck, $n = -1 + 2k$
Round 2: $r$ to $b$ $500$ to right, $500$ left in deck, $n = -2 + 4k$
Round 3: $r$ to $b$ $250$ to right, $250$ left in deck, $n = -4 + 8k$
Round 4: $r$ to $b$ $125$ to right, $125$ left in deck, $n = -8 + 16k$
Let us treat the remaining deck of $125$ cards as a totally independent deck, note that the positions of card in this deck are $n = 16k$ . Also note that the first action of a new round is never the same action as the last action of the previous round because actions alternate. Also note that for every new round, the spacing between cards moved doubles, and that the cards remaining in the beginning of a new round have position $n = a + c/2 + ck$ for the values $a, c$ of the previous round. Also note that if there are an odd number of cards in an initial deck, the first and last actions are the same. Then,
Round 5: $r$ to $r$ $63$ to right, $62$ left in deck, $n = -1 + 2k$
Round 6: $b$ to $r$ $31$ to right, $31$ left in deck, $n = 4k$ , because $n = 2k$ positioned cards are left at the beginning of this round and the first card is sent to the bottom, only every second card is sent to the right, and because spacing doubles every round,
Round 7: $b$ to $b$ $15$ to right, $16$ left in deck, $n = -2 + 8k$ , because $n = 2 + 4k$ positioned cards are left at the beginning and only every second card is sent to the right, and because spacing doubles every round,
Round 8: $r$ to $b$ $8$ to right, $8$ left in deck, $n = -14 + 16k$ , by similar reasoning, since the cards $n = 2 + 8k$ are left and spacing doubles every round, from here on things get real easy,
Round 9: $r$ to $b$ $4$ to right, $4$ left in deck, $n = -22 + 32k$
Round 10: $r$ to $b$ $2$ to right, $2$ left in deck, $n = -38 + 64k$
Round 11: $r$ to $b$ $1$ to right, $1$ left in deck, $n = 58$ , since $58 = -38 + 64/2 + 64$
This card must be labeled 1999 since it is second to last. Then, since it is the $58th$ in the deck of $125$ , it must be the $58 * 8 = 928th$ card in the original deck, and because we've been numbering from the top of the deck to the bottom (also because AIME answers are 000-999), there were $928 - 1 = \boxed{927}$ cards above it in the deck - Romulus and minimaxweii
| 927
|
6,622
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_3
| 1
|
A deck of forty cards consists of four $1$ 's, four $2$ 's,..., and four $10$ 's. A matching pair (two cards with the same number) is removed from the deck. Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
|
There are ${38 \choose 2} = 703$ ways we can draw two cards from the reduced deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in $9{4 \choose 2} = 54$ ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in $1$ way. Thus, the answer is $\frac{54+1}{703} = \frac{55}{703}$ , and $m+n = \boxed{758}$
| 758
|
6,623
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_4
| 1
|
What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?
|
We use the fact that the number of divisors of a number $n = p_1^{e_1}p_2^{e_2} \cdots p_k^{e_k}$ is $(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$ . If a number has $18 = 2 \cdot 3 \cdot 3$ factors, then it can have at most $3$ distinct primes in its factorization.
Dividing the greatest power of $2$ from $n$ , we have an odd integer with six positive divisors, which indicates that it either is ( $6 = 2 \cdot 3$ ) a prime raised to the $5$ th power, or two primes, one of which is squared. The smallest example of the former is $3^5 = 243$ , while the smallest example of the latter is $3^2 \cdot 5 = 45$
Suppose we now divide all of the odd factors from $n$ ; then we require a power of $2$ with $\frac{18}{6} = 3$ factors, namely $2^{3-1} = 4$ . Thus, our answer is $2^2 \cdot 3^2 \cdot 5 = \boxed{180}$
| 180
|
6,624
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_4
| 2
|
What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?
|
Somewhat similar to the first solution, we see that the number $n$ has two even factors for every odd factor. Thus, if $x$ is an odd factor of $n$ , then $2x$ and $4x$ must be the two corresponding even factors. So, the prime factorization of $n$ is $2^2 3^a 5^b 7^c...$ for some set of integers $a, b, c, ...$
Since there are $18$ factors of $n$ , we can write:
$(2+1)(a+1)(b+1)(c+1)... = 18$
$(a+1)(b+1)(c+1)... = 6$
Since $6$ only has factors from the set $1, 2, 3, 6$ , either $a=5$ and all other variables are $0$ , or $a=3$ and $b=2$ , with again all other variables equal to $0$ . This gives the two numbers $2^2 \cdot 3^5$ and $2^2 \cdot 3^2 \cdot 5$ . The latter number is smaller, and is equal to $\boxed{180}$
| 180
|
6,625
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_5
| 1
|
Given eight distinguishable rings, let $n$ be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost three nonzero digits of $n$
|
There are $\binom{8}{5}$ ways to choose the rings, and there are $5!$ distinct arrangements to order the rings [we order them so that the first ring is the bottom-most on the first finger that actually has a ring, and so forth]. The number of ways to distribute the rings among the fingers is equivalent the number of ways we can drop five balls into 4 urns, or similarly dropping five balls into four compartments split by three dividers. The number of ways to arrange those dividers and balls is just $\binom {8}{3}$
Multiplying gives the answer: $\binom{8}{5}\binom{8}{3}5! = 376320$ , and the three leftmost digits are $\boxed{376}$
| 376
|
6,626
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_6
| 1
|
One base of a trapezoid is $100$ units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3$ . Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed $x^2/100$
|
Let the shorter base have length $b$ (so the longer has length $b+100$ ), and let the height be $h$ . The length of the midline of the trapezoid is the average of its bases, which is $\frac{b+b+100}{2} = b+50$ . The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height $h/2$ . Then,
\[\frac{\frac 12 (h/2) (b + b+50)}{\frac 12 (h/2) (b + 50 + b + 100)} = \frac{2}{3} \Longrightarrow \frac{b + 75}{b + 25} = \frac 32 \Longrightarrow b = 75\]
Construct the perpendiculars from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle and two triangles; it also splits the desired line segment into three partitions with lengths $x_1, 75, x_2$ . By similar triangles, we easily find that $\frac{x - 75}{100} = \frac{x_1+x_2}{100} = \frac{h_1}{h}$
The area of the region including the shorter base must be half of the area of the entire trapezoid, so
\[2 \cdot \frac 12 h_1 (75 + x) = \frac 12 h (75 + 175) \Longrightarrow x = 125 \cdot \frac{h}{h_1} - 75\]
Substituting our expression for $\frac h{h_1}$ from above, we find that
\[x = \frac{12500}{x-75} - 75 \Longrightarrow x^2 - 75x = 5625 + 12500 - 75x \Longrightarrow x^2 = 18125\]
The answer is $\left\lfloor\frac{x^2}{100}\right\rfloor = \boxed{181}$
| 181
|
6,627
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_7
| 1
|
Given that
find the greatest integer that is less than $\frac N{100}$
|
Multiplying both sides by $19!$ yields:
\[\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11!}+\frac {19!}{9!10!}=\frac {19!N}{1!18!}.\]
\[\binom{19}{2}+\binom{19}{3}+\binom{19}{4}+\binom{19}{5}+\binom{19}{6}+\binom{19}{7}+\binom{19}{8}+\binom{19}{9} = 19N.\]
Recall the Combinatorial Identity $2^{19} = \sum_{n=0}^{19} {19 \choose n}$ . Since ${19 \choose n} = {19 \choose 19-n}$ , it follows that $\sum_{n=0}^{9} {19 \choose n} = \frac{2^{19}}{2} = 2^{18}$
Thus, $19N = 2^{18}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124$
So, $N=\frac{262124}{19}=13796$ and $\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}$
| 137
|
6,628
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_7
| 2
|
Given that
find the greatest integer that is less than $\frac N{100}$
|
Let $f(x) = (1+x)^{19}.$ Applying the binomial theorem gives us $f(x) = \binom{19}{19} x^{19} + \binom{19}{18} x^{18} + \binom{19}{17} x^{17}+ \cdots + \binom{19}{0}.$ Since $\frac 1{2!17!}+\frac 1{3!16!}+\dots+\frac 1{8!11!}+\frac 1{9!10!} = \frac{\frac{f(1)}{2} - \binom{19}{19} - \binom{19}{18}}{19!},$ $N = \frac{2^{18}-20}{19}.$ After some fairly easy bashing, we get $\boxed{137}$ as the answer.
| 137
|
6,629
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_7
| 3
|
Given that
find the greatest integer that is less than $\frac N{100}$
|
Convert each denominator to $19!$ and get the numerators to be $9,51,204,612,1428,2652,3978,4862$ (refer to note). Adding these up we have $13796$ therefore $\boxed{137}$ is the desired answer.
| 137
|
6,630
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8
| 1
|
In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$
|
Let $x = BC$ be the height of the trapezoid, and let $y = CD$ . Since $AC \perp BD$ , it follows that $\triangle BAC \sim \triangle CBD$ , so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$
Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$ . Then $AE = x$ , and $ADE$ is a right triangle . By the Pythagorean Theorem
\[x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0\]
The positive solution to this quadratic equation is $x^2 = \boxed{110}$
| 110
|
6,631
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8
| 2
|
In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$
|
Let $BC=x$ . Dropping the altitude from $A$ and using the Pythagorean Theorem tells us that $CD=\sqrt{11}+\sqrt{1001-x^2}$ . Therefore, we know that vector $\vec{BD}=\langle \sqrt{11}+\sqrt{1001-x^2},-x\rangle$ and vector $\vec{AC}=\langle-\sqrt{11},-x\rangle$ . Now we know that these vectors are perpendicular, so their dot product is 0. \[\vec{BD}\cdot \vec{AC}=-11-\sqrt{11(1001-x^2)}+x^2=0\] \[(x^2-11)^2=11(1001-x^2)\] \[x^4-11x^2-11\cdot 990=0.\] As above, we can solve this quadratic to get the positve solution $BC^2=x^2=\boxed{110}$
| 110
|
6,632
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8
| 3
|
In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$
|
Let $BC=x$ and $CD=y+\sqrt{11}$ . From Pythagoras with $AD$ , we obtain $x^2+y^2=1001$ . Since $AC$ and $BD$ are perpendicular diagonals of a quadrilateral, then $AB^2+CD^2=BC^2+AD^2$ , so we have \[\left(y+\sqrt{11}\right)^2+11=x^2+1001.\] Substituting $x^2=1001-y^2$ and simplifying yields \[y^2+\sqrt{11}y-990=0,\] and the quadratic formula gives $y=9\sqrt{11}$ . Then from $x^2+y^2=1001$ , we plug in $y$ to find $x^2=\boxed{110}$
| 110
|
6,633
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_8
| 4
|
In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$
|
Let $E$ be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have \begin{align*} BC^2&=BE^2+CE^2 \\ &=(AB^2-AE^2)+(CD^2-DE^2) \\ &=CD^2+\sqrt{11}^2-(AE^2+DE^2) \\ &=CD^2+11-AD^2 \\ &=CD^2-990 \end{align*} Followed by dropping the perpendicular like in solution 1, we obtain system of equation \[BC^2=CD^2-990\] \[BC^2+CD^2-2\sqrt{11}CD=990\] Rearrange the first equation yields \[BC^2-CD^2=990\] Equating it with the second equation we have \[BC^2-CD^2=BC^2+CD^2-2\sqrt{11}CD\] Which gives $CD^2=\frac{BC^2}{11}$ .
Substituting into equation 1 obtains the quadratic in terms of $BC^2$ \[(BC^2)^2-11BC^2-11\cdot990=0\] Solving the quadratic to obtain $BC^2=\boxed{110}$
| 110
|
6,634
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_10
| 1
|
circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle.
|
Call the center of the circle $O$ . By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.
Thus, $\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180$ , or $(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180$
Take the $\tan$ of both sides and use the identity for $\tan(A+B)$ to get \[\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.\]
Use the identity for $\tan(A+B)$ again to get \[\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0.\]
Solving gives $r^2=\boxed{647}$
| 647
|
6,635
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_10
| 2
|
circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle.
|
Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on ( $a, b, c,$ and $d$ are the tangent lengths, not the side lengths). \[A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}\] $r^2=\frac{A^2}{(a+b+c+d)^2} = \boxed{647}$
| 647
|
6,636
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_10
| 3
|
circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle.
|
Using the formulas established in solution 2, one notices: \[r^2=\frac{A^2}{(a+b+c+d)^2}\] \[r^2=\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}\] \[r^2=\frac{abc+bcd+acd+abd}{a+b+c+d}\] \[r^2=\boxed{647}\]
| 647
|
6,637
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_11
| 1
|
The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$ . The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\overline{AB}$ is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$ . Suppose $B$ has integer coordinates; then $\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\overline{CD}$ , and let $D' = (a,b)$ be the reflection of $D$ across that perpendicular. Then $ABCD'$ is a parallelogram , and $\overrightarrow{AB} = \overrightarrow{D'C}$ . Thus, for $C$ to have integer coordinates, it suffices to let $D'$ have integer coordinates.
Let the slope of the perpendicular be $m$ . Then the midpoint of $\overline{DD'}$ lies on the line $y=mx$ , so $\frac{b+7}{2} = m \cdot \frac{a+1}{2}$ . Also, $AD = AD'$ implies that $a^2 + b^2 = 1^2 + 7^2 = 50$ . Combining these two equations yields
\[a^2 + \left(7 - (a+1)m\right)^2 = 50\]
Since $a$ is an integer, then $7-(a+1)m$ must be an integer. There are $12$ pairs of integers whose squares sum up to $50,$ namely $( \pm 1, \pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)$ . We exclude the cases $(\pm 1, \pm 7)$ because they lead to degenerate trapezoids (rectangle, line segment, vertical and horizontal sides). Thus we have
\[7 - 8m = \pm 1, \quad 7 + 6m = \pm 1, \quad 7 - 6m = \pm 5, 7 + 4m = \pm 5\]
These yield $m = 1, \frac 34, -1, -\frac 43, 2, \frac 13, -3, - \frac 12$ . Therefore, the corresponding slopes of $\overline{AB}$ are $-1, -\frac 43, 1, \frac 34, -\frac 12, -3, \frac 13$ , and $2$ . The sum of their absolute values is $\frac{119}{12}$ . The answer is $m+n= \boxed{131}$
| 131
|
6,638
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_11
| 2
|
The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$ . The trapezoid has no horizontal or vertical sides, and $\overline{AB}$ and $\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\overline{AB}$ is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
A very natural solution:
. Shift $A$ to the origin. Suppose point $B$ was $(x, kx)$ . Note $k$ is the slope we're looking for. Note that point $C$ must be of the form: $(x \pm 1, kx \pm 7)$ or $(x \pm 7, kx \pm 1)$ or $(x \pm 5, kx \pm 5)$ . Note that we want the slope of the line connecting $D$ and $C$ so also be $k$ , since $AB$ and $CD$ are parallel.
Instead of dealing with the 12 cases, we consider
point $C$ of the form $(x \pm Y, kx \pm Z)$ where
we plug in the necessary values for $Y$ and $Z$ after simplifying.
Since the slopes of $AB$ and $CD$ must both be $k$ $\frac{7 - kx \pm Z}{1 - x \pm Y} = k \implies k = \frac{7 \pm Z}{1 \pm Y}$ . Plugging in the possible values of $\pm 7, \pm 1, \pm 5$ in their respective pairs and ruling out degenerate cases, we find the sum is $\frac{119}{12} \implies m + n = \boxed{131}$ - whatRthose
| 131
|
6,639
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_12
| 1
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The points $A$ $B$ and $C$ lie on the surface of a sphere with center $O$ and radius $20$ . It is given that $AB=13$ $BC=14$ $CA=15$ , and that the distance from $O$ to $\triangle ABC$ is $\frac{m\sqrt{n}}k$ , where $m$ $n$ , and $k$ are positive integers, $m$ and $k$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+k$
|
Let $D$ be the foot of the perpendicular from $O$ to the plane of $ABC$ . By the Pythagorean Theorem on triangles $\triangle OAD$ $\triangle OBD$ and $\triangle OCD$ we get:
\[DA^2=DB^2=DC^2=20^2-OD^2\]
It follows that $DA=DB=DC$ , so $D$ is the circumcenter of $\triangle ABC$
By Heron's Formula the area of $\triangle ABC$ is (alternatively, a $13-14-15$ triangle may be split into $9-12-15$ and $5-12-13$ right triangles ):
\[K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-15)(21-14)(21-13)} = 84\]
From $R = \frac{abc}{4K}$ , we know that the circumradius of $\triangle ABC$ is:
\[R = \frac{abc}{4K} = \frac{(13)(14)(15)}{4(84)} = \frac{65}{8}\]
Thus by the Pythagorean Theorem again,
\[OD = \sqrt{20^2-R^2} = \sqrt{20^2-\frac{65^2}{8^2}} = \frac{15\sqrt{95}}{8}.\]
So the final answer is $15+95+8=\boxed{118}$
| 118
|
6,640
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_12
| 2
|
The points $A$ $B$ and $C$ lie on the surface of a sphere with center $O$ and radius $20$ . It is given that $AB=13$ $BC=14$ $CA=15$ , and that the distance from $O$ to $\triangle ABC$ is $\frac{m\sqrt{n}}k$ , where $m$ $n$ , and $k$ are positive integers, $m$ and $k$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+k$
|
We know the radii to $A$ $B$ , and $C$ form a triangular pyramid $OABC$ . We know the lengths of the edges $OA = OB = OC = 20$ . First we can break up $ABC$ into its two component right triangles $5-12-13$ and $9-12-15$ . Let the $y$ axis be perpendicular to the base and $x$ axis run along $BC$ , and $z$ occupy the other dimension, with the origin as $C$ . We look at vectors $OA$ and $OC$ . Since $OAC$ is isoceles we know the vertex is equidistant from $A$ and $C$ , hence it is $7$ units along the $x$ axis. Hence for vector $OC$ , in form $<x,y,z>$ it is $<7, h, l>$ where $h$ is the height (answer) and $l$ is the component of the vertex along the $z$ axis. Now on vector $OA$ , since $A$ is $9$ along $x$ , and it is $12$ along $z$ axis, it is $<-2, h, 12- l>$ . We know both vector magnitudes are $20$ . Solving for $h$ yields $\frac{15\sqrt{95} }{8}$ , so Answer = $\boxed{118}$
| 118
|
6,641
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_13
| 1
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The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$ , where $m$ $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$ . Find $m+n+r$
|
We may factor the equation as:
\begin{align*} 2000x^6+100x^5+10x^3+x-2&=0\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\ \end{align*}
Now $100x^4+10x^2+1\ge 1>0$ for real $x$ . Thus the real roots must be the roots of the equation $20x^2+x-2=0$ . By the quadratic formula the roots of this are:
\[x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}.\]
Thus $r=\frac{-1+\sqrt{161}}{40}$ , and so the final answer is $-1+161+40 = \boxed{200}$
| 200
|
6,642
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_13
| 2
|
The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$ , where $m$ $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$ . Find $m+n+r$
|
It would be really nice if the coefficients were symmetrical. What if we make the substitution, $x = -\frac{i}{\sqrt{10}}y$ . The the polynomial becomes
$-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2$
It's symmetric! Dividing by $y^3$ and rearranging, we get
$-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})$
Now, if we let $z = y + \frac{1}{y}$ , we can get the equations
$z = y + \frac{1}{y}$
$z^2 - 2 = y^2 + \frac{1}{y^2}$
and
$z^3 - 3z = y^3 + \frac{1}{y^3}$
(These come from squaring $z$ and subtracting $2$ , then multiplying that result by $z$ and subtracting $z$ )
Plugging this into our polynomial, expanding, and rearranging, we get
$-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})$
Now, we see that the two $i$ terms must cancel in order to get this polynomial equal to $0$ , so what squared equals 3? Plugging in $z = \sqrt{3}$ into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying $z = -\sqrt{3}$ , we see that it also works! Great, we use long division on the polynomial by
$(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)$ and we get
$2z -(\frac{i}{\sqrt{10}}) = 0$
We know that the other two solutions for z wouldn't result in real solutions for $x$ since we have to solve a quadratic with a negative discriminant, then multiply by $-(\frac{i}{\sqrt{10}})$ . We get that $z = (\frac{i}{-2\sqrt{10}})$ . Solving for $y$ (using $y + \frac{1}{y} = z$ ) we get that $y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}$ , and multiplying this by $-(\frac{i}{\sqrt{10}})$ (because $x = -(\frac{i}{\sqrt{10}})y$ ) we get that $x = \frac{-1 \pm \sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = \boxed{200}$
| 200
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6,643
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https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_13
| 3
|
The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$ , where $m$ $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$ . Find $m+n+r$
|
Observe that the given equation may be rearranged as $2000x^6-2+(100x^5+10x^3+x)=0$ .
The expression in parentheses is a geometric series with common factor $10x^2$ . Using the geometric sum formula, we rewrite as $2000x^6-2+\frac{1000x^7-x}{10x^2-1}=0, 10x^2-1\neq0$ .
Factoring a bit, we get $2(1000x^6-1)+(1000x^6-1)\frac{x}{10x^2-1}=0, 10x^2-1\neq0 \implies$ $(1000x^6-1)(2+\frac{x}{10x^2-1})=0, 10x^2-1\neq0$ .
Note that setting $1000x^6-1=0$ gives $10x^2-1=0$ , which is clearly extraneous.
Hence, we set $2+\frac{x}{10x^2-1}=0$ and use the quadratic formula to get the desired root $x=\frac{-1+\sqrt{161}}{40} \implies -1+161+40=\boxed{200}$
| 200
|
6,644
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_14
| 1
|
Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$ , meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$ , where each $f_i$ is an integer, $0\le f_i\le i$ , and $0<f_m$ . Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$ , find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$
|
Note that $1+\sum_{k=1}^{n-1} {k\cdot k!} = 1+\sum_{k=1}^{n-1} {((k+1)\cdot k!- k!)} = 1+\sum_{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \cdots + (n! - (n-1)!)) = n!$
Thus for all $m\in\mathbb{N}$
$(32m+16)!-(32m)! = \left(1+\sum_{k=1}^{32m+15} {k\cdot k!}\right)-\left(1+\sum_{k=1}^{32m-1} {k\cdot k!}\right) = \sum_{k=32m}^{32m+15}k\cdot k!.$
So now,
\begin{align*} 16!-32!+48!-64!+\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\cdots+(2000!-1984!)\\ &=16! +\sum_{m=1}^{62}(32m+16)!-(32m)!\\ &=16! +\sum_{m=1}^{62}\sum_{k=32m}^{32m+15}k\cdot k! \end{align*}
Therefore we have $f_{16} = 1$ $f_k=k$ if $32m\le k \le 32m+15$ for some $m=1,2,\ldots,62$ , and $f_k = 0$ for all other $k$
Therefore we have:
\begin{align*} f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j &= (-1)^{17}\cdot 1 + \sum_{m=1}^{62}\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\ &= -1 + \sum_{m=1}^{62}\left[\sum_{j=16m}^{16m+7}(-1)^{2j+1}2j+\sum_{j=16m}^{16m+7}(-1)^{2j+2}(2j+1)\right]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[(-1)^{2j+1}2j+(-1)^{2j+2}(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[-2j+(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}1\\ &= -1 + \sum_{m=1}^{62}8\\ &= -1 + 8\cdot 62\\ &= \boxed{495}
| 495
|
6,645
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_14
| 2
|
Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$ , meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$ , where each $f_i$ is an integer, $0\le f_i\le i$ , and $0<f_m$ . Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$ , find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$
|
This is equivalent to Solution 1. I put up this solution merely for learners to see the intuition.
Let us consider a base $n$ number system. It’s a well known fact that when we take the difference of two integral powers of $n$ , (such as $10000_{10} - 100_{10}$ ) the result will be an integer in base $n$ composed only of the digits $n - 1$ and $0$ (in this example, $9900$ ). More specifically, the difference $(n^k)_n - (n^j)_n$ $j<k$ , is an integer $k$ digits long (note that $(n^k)_n$ has $k + 1$ digits). This integer is made up of $(k-j)$ $(n - 1)$ ’s followed by $j$ $0$ ’s.
It should make sense that this fact carries over to the factorial base, albeit with a modification. Whereas in the general base $n$ , the largest digit value is $n - 1$ , in the factorial base, the largest digit value is the argument of the factorial in that place. (for example, $321_!$ is a valid factorial base number, as is $3210_!$ . However, $31_!$ is not, as $3$ is greater than the argument of the second place factorial, $2$ $31_!$ should be represented as $101_!$ , and is $7_{10}$ .) Therefore, for example, $1000000_! - 10000_!$ is not $990000_!$ , but rather is $650000_!$ . Thus, we may add or subtract factorials quite easily by converting each factorial to its factorial base expression, with a $1$ in the argument of the factorial’s place and $0$ ’s everywhere else, and then using a standard carry/borrow system accounting for the place value.
With general intuition about the factorial base system out of the way, we may tackle the problem. We use the associative property of addition to regroup the terms as follows: $(2000! - 1984!) + (1968! - 1952!) + \cdots + (48! - 32!) + 16!$ we now apply our intuition from paragraph 2. $2000!_{10}$ is equivalent to $1$ followed by $1999$ $0$ ’s in the factorial base, and $1984!$ is $1$ followed by $1983$ $0$ ’s, and so on. Therefore, $2000! - 1984! = (1999)(1998)(1997)\cdots(1984)$ followed by $1983$ $0$ ’s in the factorial base. $1968! - 1952! = (1967)(1966)\cdots(1952)$ followed by $1951$ $0$ ’s, and so on for the rest of the terms, except $16!$ , which will merely have a $1$ in the $16!$ place followed by $0$ ’s. To add these numbers, no carrying will be necessary, because there is only one non-zero value for each place value in the sum. Therefore, the factorial base place value $f_k$ is $k$ for all $32m \leq k \leq 32m+15$ if $1\leq m \in\mathbb{Z} \leq 62$ $f_{16} = 1$ , and $f_k = 0$ for all other $k$
Therefore, to answer, we notice that $1999 - 1998 = 1997 - 1996 = 1$ , and this will continue. Therefore, $f_{1999} - f_{1998} + \cdots - f_{1984} = 8$ . We have 62 sets that sum like this, and each contains $8$ pairs of elements that sum to $1$ , so our answer is almost $8 \cdot 62$ . However, we must subtract the $1$ in the $f_{16}$ place, and our answer is $8 \cdot 62 - 1 = \boxed{495}$
| 495
|
6,646
|
https://artofproblemsolving.com/wiki/index.php/2000_AIME_II_Problems/Problem_14
| 3
|
Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$ , meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$ , where each $f_i$ is an integer, $0\le f_i\le i$ , and $0<f_m$ . Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$ , find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$
|
Let $S = 16!-32!+\cdots-1984!+2000!$ . Note that since $|S - 2000!| << 2000!$ (or $|S - 2000!| = 1984! + \cdots$ is significantly smaller than $2000!$ ), it follows that $1999! < S < 2000!$ . Hence $f_{2000} = 0$ . Then $2000! = 2000 \cdot 1999! = 1999 \cdot 1999! + 1999!$ , and as $S - 2000! << 1999!$ , it follows that $1999 \cdot 1999! < S < 2000 \cdot 1999!$ . Hence $f_{1999} = 1999$ , and we now need to find the factorial base expansion of
\[S_2 = S - 1999 \cdot 1999! = 1999! - 1984! + 1962! - 1946! + \cdots + 16!\]
Since $|S_2 - 1999!| << 1999!$ , we can repeat the above argument recursively to yield $f_{1998} = 1998$ , and so forth down to $f_{1985} = 1985$ . Now $S_{16} = 1985! - 1984! + 1962! + \cdots = 1984 \cdot 1984! + 1962! + \cdots$ , so $f_{1984} = 1984$
The remaining sum is now just $1962! - 1946! + \cdots + 16!$ . We can repeatedly apply the argument from the previous two paragraphs to find that $f_{16} = 1$ , and $f_k=k$ if $32m\le k \le 32m+15$ for some $m=1,2,\ldots,62$ , and $f_k = 0$ for all other $k$
Now for each $m$ , we have $-f_{32m} + f_{32m+1} - f_{32m+2} + \cdots + f_{32m + 31}$ $= -32m + (32m + 1) - (32m + 2) + \cdots - (32m - 30) + (32 m + 31)$ $= 1 + 1 + \cdots + 1 + 1$ $= 8$ . Thus, our answer is $-f_{16} + 8 \cdot 62 = \boxed{495}$
| 495
|
6,647
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_2
| 1
|
Consider the parallelogram with vertices $(10,45)$ $(10,114)$ $(28,153)$ , and $(28,84)$ . A line through the origin cuts this figure into two congruent polygons . The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n$
|
Let the first point on the line $x=10$ be $(10,45+a)$ where a is the height above $(10,45)$ . Let the second point on the line $x=28$ be $(28, 153-a)$ . For two given points, the line will pass the origin if the coordinates are proportional (such that $\frac{y_1}{x_1} = \frac{y_2}{x_2}$ ). Then, we can write that $\frac{45 + a}{10} = \frac{153 - a}{28}$ . Solving for $a$ yields that $1530 - 10a = 1260 + 28a$ , so $a=\frac{270}{38}=\frac{135}{19}$ . The slope of the line (since it passes through the origin) is $\frac{45 + \frac{135}{19}}{10} = \frac{99}{19}$ , and the solution is $m + n = \boxed{118}$
| 118
|
6,648
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_2
| 2
|
Consider the parallelogram with vertices $(10,45)$ $(10,114)$ $(28,153)$ , and $(28,84)$ . A line through the origin cuts this figure into two congruent polygons . The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n$
|
You can clearly see that a line that cuts a parallelogram into two congruent pieces must go through the center of the parallelogram. Taking the midpoint of $(10,45)$ , and $(28,153)$ gives $(19,99)$ , which is the center of the parallelogram. Thus the slope of the line must be $\frac{99}{19}$ , and the solution is $\boxed{118}$
| 118
|
6,649
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_2
| 3
|
Consider the parallelogram with vertices $(10,45)$ $(10,114)$ $(28,153)$ , and $(28,84)$ . A line through the origin cuts this figure into two congruent polygons . The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n$
|
Note that the area of the parallelogram is equivalent to $69 \cdot 18 = 1242,$ so the area of each of the two trapezoids with congruent area is $621.$ Therefore, since the height is $18,$ the sum of the bases of each trapezoid must be $69.$
The points where the line in question intersects the long side of the parallelogram can be denoted as $(10, \frac{10m}{n})$ and $(28, \frac{28m}{n}),$ respectively. We see that $\frac{10m}{n} - 45 + \frac{28m}{n} - 84 = 69,$ so $\frac{38m}{n} = 198 \implies \frac{m}{n} = \frac{99}{19} \implies \boxed{118}.$
| 118
|
6,650
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_2
| 4
|
Consider the parallelogram with vertices $(10,45)$ $(10,114)$ $(28,153)$ , and $(28,84)$ . A line through the origin cuts this figure into two congruent polygons . The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n$
|
$(\Sigma x_i /4, \Sigma y_i /4)$ is the centroid, which generates $(19,99)$ , so the answer is $\boxed{118}$ . This is the fastest way because you do not need to find the opposite vertices by drawing.
| 118
|
6,651
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_3
| 1
|
Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square
|
If $n^2-19n+99=x^2$ for some positive integer $x$ , then rearranging we get $n^2-19n+99-x^2=0$ . Now from the quadratic formula,
Because $n$ is an integer, this means $4x^2-35=q^2$ for some nonnegative integer $q$ . Rearranging gives $(2x+q)(2x-q)=35$ . Thus $(2x+q, 2x-q)=(35, 1)$ or $(7,5)$ , giving $x=3$ or $9$ . This gives $n=1, 9, 10,$ or $18$ , and the sum is $1+9+10+18=\boxed{38}$
| 38
|
6,652
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_3
| 2
|
Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square
|
When $n \geq 12$ , we have \[(n-10)^2 < n^2 -19n + 99 < (n-8)^2.\]
So if $n \geq 12$ and $n^2 -19n + 99$ is a perfect square, then \[n^2 -19n + 99 = (n-9)^2\]
or $n = 18$
For $1 \leq n < 12$ , it is easy to check that $n^2 -19n + 99$ is a perfect square when $n = 1, 9$ and $10$ ( using the identity $n^2 -19n + 99 = (n-10)^2 + n - 1.)$
We conclude that the answer is $1 + 9 + 10 + 18 = \boxed{38}.$
| 38
|
6,653
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_4
| 1
|
The two squares shown share the same center $O_{}$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n.$
[asy] //code taken from thread for problem real alpha = 25; pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin; pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z; draw(W--X--Y--Z--cycle^^w--x--y--z--cycle); pair A=intersectionpoint(Y--Z, y--z), C=intersectionpoint(Y--X, y--x), E=intersectionpoint(W--X, w--x), G=intersectionpoint(W--Z, w--z), B=intersectionpoint(Y--Z, y--x), D=intersectionpoint(Y--X, w--x), F=intersectionpoint(W--X, w--z), H=intersectionpoint(W--Z, y--z); dot(O); label("$O$", O, SE); label("$A$", A, dir(O--A)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$D$", D, dir(O--D)); label("$E$", E, dir(O--E)); label("$F$", F, dir(O--F)); label("$G$", G, dir(O--G)); label("$H$", H, dir(O--H));[/asy]
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Triangles $AOB$ $BOC$ $COD$ , etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length $1$ in the circumcircle of the squares pass through $B$ , etc.), and each area is $\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}$ . Since the area of a triangle is $bh/2$ , the area of all $8$ of them is $\frac{86}{99}$ and the answer is $\boxed{185}$
| 185
|
6,654
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_4
| 2
|
The two squares shown share the same center $O_{}$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n.$
[asy] //code taken from thread for problem real alpha = 25; pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin; pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z; draw(W--X--Y--Z--cycle^^w--x--y--z--cycle); pair A=intersectionpoint(Y--Z, y--z), C=intersectionpoint(Y--X, y--x), E=intersectionpoint(W--X, w--x), G=intersectionpoint(W--Z, w--z), B=intersectionpoint(Y--Z, y--x), D=intersectionpoint(Y--X, w--x), F=intersectionpoint(W--X, w--z), H=intersectionpoint(W--Z, y--z); dot(O); label("$O$", O, SE); label("$A$", A, dir(O--A)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$D$", D, dir(O--D)); label("$E$", E, dir(O--E)); label("$F$", F, dir(O--F)); label("$G$", G, dir(O--G)); label("$H$", H, dir(O--H));[/asy]
|
Define the two possible distances from one of the labeled points and the corners of the square upon which the point lies as $x$ and $y$ . The area of the octagon (by subtraction of areas) is $1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy$
By the Pythagorean theorem \[x^2 + y^2 = \left(\frac{43}{99}\right)^2\]
Also, \begin{align*}x + y + \frac{43}{99} &= 1\\ x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{align*}
Substituting, \begin{align*}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \\ 2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{align*}
Thus, the area of the octagon is $1 - \frac{13}{99} = \frac{86}{99}$ , so $m + n = \boxed{185}$
| 185
|
6,655
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_5
| 1
|
For any positive integer $x_{}$ , let $S(x)$ be the sum of the digits of $x_{}$ , and let $T(x)$ be $|S(x+2)-S(x)|.$ For example, $T(199)=|S(201)-S(199)|=|3-19|=16.$ How many values of $T(x)$ do not exceed 1999?
|
For most values of $x$ $T(x)$ will equal $2$ . For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take $T(a999)$ as an example, \[|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|\] And in general, the values of $T(x)$ will then be in the form of $|2 - 9n| = 9n - 2$ . From $7$ to $1999$ , there are $\left\lceil \frac{1999 - 7}{9}\right\rceil = 222$ solutions; including $2$ and there are a total of $\boxed{223}$ solutions.
| 223
|
6,656
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_6
| 1
|
A transformation of the first quadrant of the coordinate plane maps each point $(x,y)$ to the point $(\sqrt{x},\sqrt{y}).$ The vertices of quadrilateral $ABCD$ are $A=(900,300), B=(1800,600), C=(600,1800),$ and $D=(300,900).$ Let $k_{}$ be the area of the region enclosed by the image of quadrilateral $ABCD.$ Find the greatest integer that does not exceed $k_{}.$
|
\begin{eqnarray*}A' = & (\sqrt {900}, \sqrt {300})\\ B' = & (\sqrt {1800}, \sqrt {600})\\ C' = & (\sqrt {600}, \sqrt {1800})\\ D' = & (\sqrt {300}, \sqrt {900}) \end{eqnarray*}
First we see that lines passing through $AB$ and $CD$ have equations $y = \frac {1}{3}x$ and $y = 3x$ , respectively. Looking at the points above, we see the equations for $A'B'$ and $C'D'$ are $y^2 = \frac {1}{3}x^2$ and $y^2 = 3x^2$ , or, after manipulation $y = \frac {x}{\sqrt {3}}$ and $y = \sqrt {3}x$ , respectively, which are still linear functions. Basically the square of the image points gives back the original points and we could plug them back into the original equation to get the equation of the image lines.
Now take a look at $BC$ and $AD$ , which have the equations $y = - x + 2400$ and $y = - x + 1200$ . The image equations hence are $x^2 + y^2 = 2400$ and $x^2 + y^2 = 1200$ , respectively, which are the equations for circles
1999 AIME-6.png
To find the area between the circles (actually, parts of the circles), we need to figure out the angle of the arc . This could be done by $\arctan \sqrt {3} - \arctan \frac {1}{\sqrt {3}} = 60^\circ - 30^\circ = 30^\circ$ . So the requested areas are the area of the enclosed part of the smaller circle subtracted from the area enclosed by the part of the larger circle = $\frac {30^\circ}{360^\circ}(R^2\pi - r^2\pi) = \frac {1}{12}(2400\pi - 1200\pi) = 100\pi$ . Hence the answer is $\boxed{314}$
| 314
|
6,657
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_7
| 1
|
There is a set of 1000 switches, each of which has four positions, called $A, B, C$ , and $D$ . When the position of any switch changes, it is only from $A$ to $B$ , from $B$ to $C$ , from $C$ to $D$ , or from $D$ to $A$ . Initially each switch is in position $A$ . The switches are labeled with the 1000 different integers $(2^{x})(3^{y})(5^{z})$ , where $x, y$ , and $z$ take on the values $0, 1, \ldots, 9$ . At step i of a 1000-step process, the $i$ -th switch is advanced one step, and so are all the other switches whose labels divide the label on the $i$ -th switch. After step 1000 has been completed, how many switches will be in position $A$
|
For each $i$ th switch (designated by $x_{i},y_{i},z_{i}$ ), it advances itself only one time at the $i$ th step; thereafter, only a switch with larger $x_{j},y_{j},z_{j}$ values will advance the $i$ th switch by one step provided $d_{i}= 2^{x_{i}}3^{y_{i}}5^{z_{i}}$ divides $d_{j}= 2^{x_{j}}3^{y_{j}}5^{z_{j}}$ . Let $N = 2^{9}3^{9}5^{9}$ be the max switch label. To find the divisor multiples in the range of $d_{i}$ to $N$ , we consider the exponents of the number $\frac{N}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}$ . In general, the divisor-count of $\frac{N}{d}$ must be a multiple of 4 to ensure that a switch is in position A:
We consider the cases where the 3 factors above do not contribute multiples of 4.
The number of switches in position A is $1000-125-225 = \boxed{650}$
| 650
|
6,658
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9
| 1
|
A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n.$
|
Suppose we pick an arbitrary point on the complex plane , say $(1,1)$ . According to the definition of $f(z)$ \[f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i,\] this image must be equidistant to $(1,1)$ and $(0,0)$ . Thus the image must lie on the line with slope $-1$ and which passes through $\left(\frac 12, \frac12\right)$ , so its graph is $x + y = 1$ . Substituting $x = (a-b)$ and $y = (a+b)$ , we get $2a = 1 \Rightarrow a = \frac 12$
By the Pythagorean Theorem , we have $\left(\frac{1}{2}\right)^2 + b^2 = 8^2 \Longrightarrow b^2 = \frac{255}{4}$ , and the answer is $\boxed{259}$
| 259
|
6,659
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9
| 2
|
A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n.$
|
Plugging in $z=1$ yields $f(1) = a+bi$ . This implies that $a+bi$ must fall on the line $Re(z)=a=\frac{1}{2}$ , given the equidistant rule. By $|a+bi|=8$ , we get $a^2 + b^2 = 64$ , and plugging in $a=\frac{1}{2}$ yields $b^2=\frac{255}{4}$ . The answer is thus $\boxed{259}$
| 259
|
6,660
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9
| 3
|
A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n.$
|
We are given that $(a + bi)z$ is equidistant from the origin and $z.$ This translates to \begin{eqnarray*} |(a + bi)z - z| & = & |(a + bi)z| \\ |z(a - 1 + bi)| & = & |z(a + bi)| \\ |z|\cdot|(a - 1) + bi| & = & |z|\cdot|a + bi| \\ |(a - 1) + bi| & = & |a + bi| \\ (a - 1)^2 + b^2 & = & a^2 + b^2 \\ & \Rightarrow & a = \frac 12 \end{eqnarray*} Since $|a + bi| = 8,$ $a^2 + b^2 = 64.$ Because $a = \frac 12,$ thus $b^2 = \frac {255}4.$ So the answer is $\boxed{259}$
| 259
|
6,661
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9
| 4
|
A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n.$
|
Similarly to in Solution 3, we see that $|(a + bi)z - z| = |(a + bi)z|$ . Letting the point $z = c + di$ , we have $\sqrt{(ab+bc-d)^2+(ac-bd-c)^2} = \sqrt{(ac-bd)^2+(ad+bc)^2}$ . Expanding both sides of this equation (after squaring, of course) and canceling terms, we get $(d^2+c^2)(-2a+1) = 0$ . Of course, $(d^2+c^2)$ can't be zero because this property of the function holds for all complex $z$ . Therefore, $a = \frac{1}{2}$ and we proceed as above to get $\boxed{259}$
| 259
|
6,662
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9
| 5
|
A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n.$
|
This is a solution that minimizes the use of complex numbers, turning this into an introductory algebra analytic geometry problem.
Consider any complex number $z=c+di$ . Let $z$ denote point $P$ on the complex plane. Then $P=(c,d)$ on the complex plane. The equation for the line $OP$ is $y=\frac{d}{c}x$
Let the image of point $P$ be $Q$ , after the point undergoes the function. Since each image is equidistant from the preimage and the origin, $Q$ must be on the perpendicular bisector of $OP$ .Given $z=c+di$ $f(z)=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ . Then $Q=(ac-bd,ad+bc)$ . The midpoint of $OP$ is $(0.5c, 0.5d)$ . Since the slopes of two respectively nonvertical and nonhorizontal lines have a product of $-1$ , using the point-slope form, the equation of the perpendicular line to $OP$ is $y-0.5d=-\frac{c}{d}(x-0.5c)$ . Rearranging, we have $y=-\frac{cx}{d}+\frac{c^2}{2d}+\frac{d}{2}$
Since we know that $Q=(ac-bd,ad+bc)$ , thus we plug in $Q$ into the line: $ad+bc=-\frac{ac^2-bcd}{d}+\frac{c^2}{2d}+\frac{d}{2}$
Let's start canceling. $2ad^2+2bcd=-2ac^2+2bcd+c^2+d^2$ . Subtracting, $c^2+d^2-2ac^2=2ad^2$ . Thus $c^2+d^2=2ac^2+2ad^2$ . Since this is an identity for any $(c,d)$ , thus $2a=1$ $a=\frac{1}{2}$ . Since $|a+bi|=8$ , thus $a^2+b^2=64$ (or simply think of $a+bi$ as the point $(a,b)$ , and $|a+bi|$ being the distance of $(a,b)$ to the origin). Thus plug in $a=\frac{1}{2}, b^2=\frac{255}{4}$ . Since $255$ and $4$ are relatively prime, the final result is $255+4=\boxed{259}$
| 259
|
6,663
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_10
| 1
|
Ten points in the plane are given, with no three collinear . Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n.$
|
Note that 4 points can NEVER form 2 triangles. Therefore, we just need to multiply the probability that the first three segments picked form a triangle by 4. We can pick any segment for the first choice, then only segments that share an endpoint with the first one, then the one segment that completes the triangle. Note that the fourth segment doesn't matter in this case.
Note that there are $(9 - 1) \times 2 = 16$ segments that share an endpoint with the first segment. The answer is then $4 \times \frac{16}{44} \times \frac{1}{43} = \frac{16}{11} \times \frac{1}{43} = \frac{16}{473} \implies m + n = \boxed{489}$ -whatRthose
| 489
|
6,664
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_10
| 2
|
Ten points in the plane are given, with no three collinear . Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers . Find $m+n.$
|
Instead of working with the four segments, let's focus on their endpoints. When we select these segments, we are working with $4, 5, 6, 7,$ or $8$ endpoints in total.
If we have $6, 7,$ or $8$ endpoints, it is easy to see that we cannot form a triangle by drawing four segments between them, because at least one point will be "left out".
However, if we have $5$ endpoints, we can use three segments form a triangle with three of the points and connect the remaining two points with the last segment. There are ${10\choose5}$ ways to select these $5$ points from the original $10,$ and ${5\choose3}$ ways to decide which three points are in the triangle.
Finally, if we have $4$ endpoints, we can also form a triangle with three of the points, then use the remaining segment to connect the last point to either of the previous three. We have ${10\choose4}$ ways to select the $4$ points and ${4\choose3}$ ways to choose three points for the triangle. Finally, we must connect the last point to one vertex of the triangle; we can do this in $3$ ways.
As in Solution 1, there are ${45\choose4}$ total ways to select four segments. So, our desired probability is
\[\dfrac{{10\choose5}{5\choose3}+{10\choose4}{4\choose3}\cdot 3}{{45\choose4}}=\dfrac{16}{473} \implies m + n = \boxed{489}.\]
| 489
|
6,665
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_11
| 1
|
Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$
|
Let $s = \sum_{k=1}^{35}\sin 5k = \sin 5 + \sin 10 + \ldots + \sin 175$ . We could try to manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to telescope the sum. Using the identity $\sin a \sin b = \frac 12(\cos (a-b) - \cos (a+b))$ , we can rewrite $s$ as
\begin{align*} s \cdot \sin 5 = \sum_{k=1}^{35} \sin 5k \sin 5 &= \sum_{k=1}^{35} \frac{1}{2}(\cos (5k - 5)- \cos (5k + 5))\\ &= 0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 \ldots + \cos 165 - \cos 175+ \cos 170 - \cos 180) \end{align*}
This telescopes to \[s = \frac{\cos 0 + \cos 5 - \cos 175 - \cos 180}{2 \sin 5} = \frac{1 + \cos 5}{\sin 5}.\] Manipulating this to use the identity $\tan x = \frac{1 - \cos 2x}{\sin 2x}$ , we get \[s = \frac{1 - \cos 175}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},\] and our answer is $\boxed{177}$
| 177
|
6,666
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_11
| 2
|
Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$
|
We note that $\sin x = \mbox{Im } e^{ix}\text{*}$ . We thus have that \begin{align*} \sum_{k = 1}^{35} \sin 5k &= \sum_{k = 1}^{35} \mbox{Im } e^{5ki}\\ &= \mbox{Im } \sum_{k = 1}^{35} e^{5ki}\\ &= \mbox{Im } \frac{e^{5i}(1 - e^{180i})}{1 - e^{5i}}\\ &= \mbox{Im } \frac{2\cos5 + 2i \sin 5}{(1 - \cos 5) - i \sin 5}\\ &= \mbox{Im } \frac{(2 \cos 5 + 2i \sin 5)[(1 - \cos 5) + i \sin 5]}{(1 - \cos 5)^2 + \sin^2 5}\\ &= \frac{2 \sin 5}{2 - 2 \cos 5}\\ &= \frac{\sin 5}{1 - \cos 5}\\ &= \frac{\sin 175}{1 + \cos 175} \\ &= \tan \frac{175}{2}.\\ \end{align*} The desired answer is thus $175 + 2 = \boxed{177}$
| 177
|
6,667
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_11
| 3
|
Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m_{}$ and $n_{}$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$
|
Let $x=e^{\frac{i\pi}{36}}$ . By Euler's Formula, $\sin{5k^\circ}=\frac{x^k-\frac{1}{x^{k}}}{2i}$
The sum we want is thus $\frac{x-\frac{1}{x}}{2i}+\frac{x^2-\frac{1}{x^{2}}}{2i}+\cdots+\frac{x^{35}-\frac{1}{x^{35}}}{2i}$
We factor the $\frac{1}{2i}$ and split into two geometric series to get $\frac{1}{2i}\left(\frac{-\frac{1}{x^{35}}(x^{35}-1)}{x-1}+\frac{x(x^{35}-1)}{x-1}\right)$
However, we note that $x^{36}=-1$ , so $-\frac{1}{x^{35}}=x$ , so our two geometric series are actually the same. We combine the terms and simplify to get $\frac{1}{i}\left(\frac{x^{36}-x}{x-1}\right)$
Apply Euler's identity and simplify again to get $\frac{1}{i}\left(\frac{-x-1}{x-1}\right)$
Now, we need to figure out how to express this as the tangent of something. We note that $\tan(5k^\circ)=\frac{\sin(5k^\circ)}{\cos(5k^\circ)}=\frac{\frac{x^k-\frac{1}{x^k}}{2i}}{\frac{x^k+\frac{1}{x^k}}{2}}=\frac{1}{i}\frac{x^{2k}-1}{x^{2k}+1}$
So, we set the two equal to each other to solve for $k$ . Cross multiplying gets $(-x-1)(x^{2k}+1)=(x-1)(x^{2k}-1)$ . Expanding yields $-x^{2k+1}-x-x^{2k}-1=x^{2k+1}-x-x^{2k}+1$ . Simplifying yields $x^{2k+1}=-1$ . Since $2k+1=36$ is the smallest solution, we have $k=\frac{35}{2}$ , and the argument of tangent is $5k=\frac{175}{2}$ . The requested sum is $175+2=\boxed{177}$
| 177
|
6,668
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_12
| 1
|
The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is $21$ . Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.
|
Let $Q$ be the tangency point on $\overline{AC}$ , and $R$ on $\overline{BC}$ . By the Two Tangent Theorem $AP = AQ = 23$ $BP = BR = 27$ , and $CQ = CR = x$ . Using $rs = A$ , where $s = \frac{27 \cdot 2 + 23 \cdot 2 + x \cdot 2}{2} = 50 + x$ , we get $(21)(50 + x) = A$ . By Heron's formula $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(50+x)(x)(23)(27)}$ . Equating and squaring both sides,
\begin{eqnarray*} [21(50+x)]^2 &=& (50+x)(x)(621)\\ 441(50+x) &=& 621x\\ 180x = 441 \cdot 50 &\Longrightarrow & x = \frac{245}{2} \end{eqnarray*}
We want the perimeter, which is $2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}$
| 345
|
6,669
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_12
| 2
|
The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is $21$ . Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.
|
Let the incenter be denoted $I$ . It is commonly known that the incenter is the intersection of the angle bisectors of a triangle. So let $\angle ABI = \angle CBI = \alpha, \angle BAI = \angle CAI = \beta,$ and $\angle BCI = \angle ACI = \gamma.$
We have that \begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\ \tan \beta & = & \frac {21}{23} \\ \tan \gamma & = & \frac {21}x. \end{eqnarray*} So naturally we look at $\tan \gamma.$ But since $\gamma = \frac \pi2 - (\beta + \alpha)$ we have \begin{eqnarray*} \tan \gamma & = & \tan\left(\frac \pi2 - (\beta + \alpha)\right) \\ & = & \frac 1{\tan(\alpha + \beta)} \\ \Rightarrow \frac {21}x & = & \frac {1 - \frac {21\cdot 21}{23\cdot 27}}{\frac {21}{27} + \frac {21}{23}} \end{eqnarray*} Doing the algebra, we get $x = \frac {245}2.$
The perimeter is therefore $2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.$
| 345
|
6,670
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_13
| 1
|
Forty teams play a tournament in which every team plays every other team exactly once. No ties occur, and each team has a $50 \%$ chance of winning any game it plays. The probability that no two teams win the same number of games is $\frac mn,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $\log_2 n.$
|
There are ${40 \choose 2} = 780$ total pairings of teams, and thus $2^{780}$ possible outcomes. In order for no two teams to win the same number of games, they must each win a different number of games. Since the minimum and maximum possible number of games won are 0 and 39 respectively, and there are 40 teams in total, each team corresponds uniquely with some $k$ , with $0 \leq k \leq 39$ , where $k$ represents the number of games the team won. With this in mind, we see that there are a total of $40!$ outcomes in which no two teams win the same number of games. Further, note that these are all the valid combinations, as the team with 1 win must beat the team with 0 wins, the team with 2 wins must beat the teams with 1 and 0 wins, and so on; thus, this uniquely defines a combination.
The desired probability is thus $\frac{40!}{2^{780}}$ . We wish to simplify this into the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime. The only necessary step is to factor out all the powers of 2 from $40!$ ; the remaining number is clearly relatively prime to all powers of 2.
The number of powers of 2 in $40!$ is $\left \lfloor \frac{40}{2} \right \rfloor + \left \lfloor \frac{40}{4} \right \rfloor + \left \lfloor \frac{40}{8} \right \rfloor + \left \lfloor \frac{40}{16} \right \rfloor + \left \lfloor \frac{40}{32} \right \rfloor = 20 + 10 + 5 + 2 + 1 = 38.$
$780-38 = \boxed{742}$
| 742
|
6,671
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14
| 1
|
Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
|
Drop perpendiculars from $P$ to the three sides of $\triangle ABC$ and let them meet $\overline{AB}, \overline{BC},$ and $\overline{CA}$ at $D, E,$ and $F$ respectively.
Let $BE = x, CF = y,$ and $AD = z$ . We have that \begin{align*}DP&=z\tan\theta\\ EP&=x\tan\theta\\ FP&=y\tan\theta\end{align*} We can then use the tool of calculating area in two ways \begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\\ &=\frac{1}{2}(13)(z\tan\theta)+\frac{1}{2}(14)(x\tan\theta)+\frac{1}{2}(15)(y\tan\theta)\\ &=\frac{1}{2}\tan\theta(13z+14x+15y)\end{align*} On the other hand, \begin{align*}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{21\cdot6\cdot7\cdot8}\\ &=84\end{align*} We still need $13z+14x+15y$ though. We have all these right triangles and we haven't even touched Pythagoras . So we give it a shot: \begin{align}x^2+x^2\tan^2\theta&=z^2\tan^2\theta+(13-z)^2\\ z^2+z^2\tan^2\theta&=y^2\tan^2\theta+(15-y)^2\\ y^2+y^2\tan^2\theta&=x^2\tan^2\theta+(14-x)^2\end{align} Adding $(1) + (2) + (3)$ gives \begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\\ \Rightarrow13z+14x+15y&=295\end{align*} Recall that we found that $[ABC]=\frac{1}{2}\tan\theta(13z+14x+15y)=84$ . Plugging in $13z+14x+15y=295$ , we get $\tan\theta=\frac{168}{295}$ , giving us $\boxed{463}$ for an answer.
| 463
|
6,672
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14
| 2
|
Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
|
Let $AB=c$ $BC=a$ $AC=b$ $PA=x$ $PB=y$ , and $PC=z$
So by the Law of Cosines , we have: \begin{align*}x^2 &= z^2 + b^2 - 2bz\cos{\theta}\\ y^2 &= x^2 + c^2 - 2cx\cos{\theta}\\ z^2 &= y^2 + a^2 - 2ay\cos{\theta}\end{align*} Adding these equations and rearranging, we have: \[a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}\qquad(1)\] Now $[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84$ , by Heron's formula
Now the area of a triangle, $[A] = \frac {mn\sin{\beta}}{2}$ , where $m$ and $n$ are sides on either side of an angle, $\beta$ . So, \begin{align*}[CAP] &= \frac {bz\sin{\theta}}{2}\\ [ABP] &= \frac {cx\sin{\theta}}{2}\\ [BCP] &= \frac {ay\sin{\theta}}{2}\end{align*} Adding these equations yields: \begin{align*}[ABC]= 84 &= \frac {(bz + cx + ay)\sin{\theta}}{2}\\ \Rightarrow 168&= (bz + cx + ay)\sin{\theta}\qquad (2)\end{align*} Dividing $(2)$ by $(1)$ , we have: \begin{align*}\frac {168}{a^2 + b^2 + c^2} &= \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}\\ \Rightarrow \tan{\theta} = \frac {336}{a^2 + b^2 + c^2} &= \frac {336}{14^2 + 15^2 + 13^2} = \frac {336}{590} = \frac {168}{295}\end{align*} Thus, $m + n = 168 + 295 = \boxed{463}$
| 463
|
6,673
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14
| 3
|
Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
|
Let $\angle{PAB} = \angle{PBC} = \angle{PCA} = x.$ Then, using Law of Cosines on the three triangles containing vertex $P,$ we have \begin{align*} b^2 &= a^2 + 169 - 26a \cos x \\ c^2 &= b^2 + 196 - 28b \cos x \\ a^2 &= c^2 + 225 - 30c \cos x. \end{align*} Add the three equations up and rearrange to obtain \[(13a + 14b + 15c) \cos x = 295.\] Also, using $[ABC] = \frac{1}{2}ab \sin \angle C$ we have \[[ABC] = [APB] + [BPC] + [CPA] = \dfrac{\sin x}{2}(13a + 14b + 15c) = 84 \iff (13a + 14b + 15c) \sin x = 168.\] Divide the two equations to obtain $\tan x = \frac{168}{295} \iff \boxed{463}.~\square$
| 463
|
6,674
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_14
| 4
|
Point $P_{}$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15,$ and the tangent of angle $PAB$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$
|
Firstly, denote angles $ABC$ $BCA$ , and $CAB$ as $B$ $A$ , and $C$ respectively. Let $\angle{PAB}=x$ .
Notice that by angle chasing that $\angle{BPC}=180-C$ and $\angle{BPA}=180-B$ .
Using the nice properties of the 13-14-15 triangle, we have $\sin B = \frac{12}{13}$ and $\sin C = \frac{4}{5}$ $\cos C$ is easily computed, so we have $\cos C=\frac{3}{5}$
Using Law of Sines, \begin{align*} \frac{BP}{\sin x} &= \frac{13}{\sin (180 - B)} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin (180 - C)} \end{align*} hence, \begin{align*} \frac{BP}{\sin x} &= \frac{13}{\sin B} \\ \frac{BP}{\sin (C - x)} &= \frac{14}{\sin C} \end{align*} Now, computation carries the rest. \begin{align*} \frac{13 \sin x}{\sin B} &= \frac{14 \sin (C-x)}{\sin C} \\ \frac{169 \sin x}{12} &= \frac{210 \sin (C-x)}{12} \\ 169 \sin x &= 210 (\sin C \cos x - \cos C \sin x) \\ 169 \sin x &= 210 (\frac{4}{5} \cos x - \frac{3}{5} \sin x) \\ 169 \sin x &= 168 \cos x - 126 \sin x \\ 295 \sin x &= 168 \cos x \\ \tan x &= \frac{168}{295} \end{align*} Extracting yields $168 + 295 = \boxed{463}$
| 463
|
6,675
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15
| 1
|
Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?
|
As shown in the image above, let $D$ $E$ , and $F$ be the midpoints of $\overline{BC}$ $\overline{CA}$ , and $\overline{AB}$ , respectively. Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $\triangle ABC$ . The crux of this problem is the following lemma.
Lemma: The point $O$ is the orthocenter of $\triangle ABC$
Proof. Observe that \[OF^2 - OE^2 = PF^2 - PE^2 = AF^2 - AE^2;\] the first equality follows by the Pythagorean Theorem, while the second follows from $AF = FP$ and $AE = EP$ . Thus, by the Perpendicularity Lemma, $AO$ is perpendicular to $FE$ and hence $BC$ . Analogously, $O$ lies on the $B$ -altitude and $C$ -altitude of $\triangle ABC$ , and so $O$ is, indeed, the orthocenter of $\triangle ABC$
To find the coordinates of $O$ , we need to find the intersection point of altitudes $BE$ and $AD$ . The equation of $BE$ is simply $x=16$ $AD$ is perpendicular to line $BC$ , so the slope of $AD$ is equal to the negative reciprocal of the slope of $BC$ $BC$ has slope $\frac{24-0}{16-34}=-\frac{4}{3}$ , therefore $y=\frac{3}{4} x$ . These two lines intersect at $(16,12)$ , so that's the base of the height of the tetrahedron.
Let $S$ be the foot of altitude $BS$ in $\triangle BPQ$ . From the Pythagorean Theorem $h=\sqrt{BS^2-SO^2}$ . However, since $S$ and $O$ are, by coincidence, the same point, $SO=0$ and $h=12$
The area of the base is $102$ , so the volume is $\frac{102*12}{3}=\boxed{408}$
| 408
|
6,676
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15
| 2
|
Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?
|
Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates $(17, 0, 0)$ $(8, 12, 0)$ , and $(25, 12, 0)$ . We can compute the area of this triangle as $102$ . Suppose $(x, y, z)$ are the coordinates of the vertex of the resulting pyramid. Call this point $V$ . Clearly, the height of the pyramid is $z$ . The desired volume is thus $\frac{102z}{3} = 34z$
We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, $VR = RA$ $VP = PB$ , and $VQ = QC$ . We then use distance formula to find the distances from $V$ to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding $z = 12$ . The desired volume is thus $34 \times 12 = \boxed{408}$
| 408
|
6,677
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15
| 3
|
Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?
|
The formed tetrahedron has pairwise parallel planar and oppositely equal length ( $4\sqrt{13},15,17$ ) edges and can be inscribed in a parallelepiped (rectangular box) with the six tetrahedral edges as non-intersecting diagonals of the box faces. Let the edge lengths of the parallelepiped be $p,q,r$ and solve (by Pythagoras)
$p^2+q^2=4^2\cdot{13}$
$q^2+r^2=15^2$
$r^2+p^2=17^2$
to find that $(p^2,q^2,r^2)=(153,136,72)=(3^2\cdot{17},2^3\cdot{17},2^3\cdot{3^2}).$
Use the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is $\tfrac{1}{3}$ and then the volume is
$\tfrac{1}{3}pqr=\tfrac{1}{3}\sqrt{2^6\cdot{3^4}\cdot{17^2}}=\boxed{408}$
| 408
|
6,678
|
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_15
| 4
|
Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?
|
Let $A = (0,0), B = (16, 24), C = (34,0).$ Then define $D,E,F$ as the midpoints of $BC, AC, AB$ . By Pythagorean theorem, $EF = \frac{1}{2} BC = 15, DE = \frac{1}{2}AB = 4 \sqrt{13}, DF = \frac{1}{2} AC = 17.$ Then let $P$ be the point in space which is the vertex of the tetrahedron with base $DEF$
Note that $\triangle DEP \cong \triangle EDF$ . Create point $F'$ on the plane of $DEF$ such that $\triangle DEP \cong \triangle DEF'$ (i.e by reflecting $F$ over the perpendicular bisector of $DE$ ). Project $F, P$ onto $DE$ as $X, Y$ . Note by the definition of $F'$ then $\angle PYF'$ is the dihedral angle between planes $DEP, DEF$
Now see that by Heron's, \[[DEP] = [DEF] = \sqrt{(16 + 2 \sqrt{13})(16 - 2 \sqrt{13})(1 + 2 \sqrt{13})(-1 + 2 \sqrt{13})} = 102.\] So $PY$ , the hypotenuse $DEP$ has length $\frac{102 \cdot 2}{4 \sqrt{13}} = \frac{51}{\sqrt{13}}$ . Similarly $F'Y = \frac{51}{\sqrt{13}}.$ Further from Pythagoras $DY = \sqrt{DP^2 - PY^2} = \frac{18}{\sqrt{13}}.$ Symmetrically $EX = \frac{18}{\sqrt{13}}.$ Therefore $XY = DE - DY - EX = \frac{16}{\sqrt{13}}.$
By Law of Cosines on $\triangle PYF'$ \begin{align*} PF'^2 &= PY^2 + F'Y^2 - 2 \cdot PY \cdot F'Y \cos{\angle PYF'} \\ PF^2 - XY^2 &= 2 (\frac{51}{\sqrt{13}})^2 \cos{\angle PYF'} \\ (4\sqrt{13})^2 - (\frac{16}{\sqrt{13}})^2 &= 2 (\frac{51}{\sqrt{13}})^2 \cos{\angle PYF'} \\ \cos{\angle PYF'} &= \frac{9}{17} \\ \sin{\angle PYF'} &= \frac{4 \sqrt{13}}{17}. \end{align*}.
Therefore the altitude of the tetrahedron from vertex $P$ to base $DEF$ is $PY \sin{\angle PYF'} = \frac{51}{\sqrt{13}} \frac{4 \sqrt{13}}{17} = 12.$ So the area is $\frac{1}{3}bh = \frac{1}{3} 12 \cdot 102 = \boxed{408}.$
| 408
|
6,679
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_1
| 1
|
For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6$ $8^8$ , and $k$
|
It is evident that $k$ has only 2s and 3s in its prime factorization, or $k = 2^a3^b$
The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. $[6^6,8^8] = 2^{24}3^6$ . Therefore $12^{12} = 2^{24}\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\max(24,a)}3^{\max(6,b)}$ , and $b = 12$ . Since $0 \le a \le 24$ , there are $\boxed{25}$ values of $k$
| 25
|
6,680
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_3
| 1
|
The graph of $y^2 + 2xy + 40|x|= 400$ partitions the plane into several regions. What is the area of the bounded region?
|
The equation given can be rewritten as:
We can split the equation into a piecewise equation by breaking up the absolute value
Factoring the first one: (alternatively, it is also possible to complete the square
Hence, either $y = -20$ , or $2x = 20 - y \Longrightarrow y = -2x + 20$
Similarily, for the second one, we get $y = 20$ or $y = -2x - 20$ . If we graph these four equations, we see that we get a parallelogram with base 20 and height 40. Hence the answer is $\boxed{800}$
| 800
|
6,681
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_6
| 1
|
Let $ABCD$ be a parallelogram . Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$ Given that $PQ = 735$ and $QR = 112,$ find $RC.$
|
We have $\triangle BRQ\sim \triangle DRC$ so $\frac{112}{RC} = \frac{BR}{DR}$ . We also have $\triangle BRC \sim \triangle DRP$ so $\frac{ RC}{847} = \frac {BR}{DR}$ . Equating the two results gives $\frac{112}{RC} = \frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=\boxed{308}$
| 308
|
6,682
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7
| 2
|
Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$
|
Define $x_i = 2y_i - 1$ . Then $2\left(\sum_{i = 1}^4 y_i\right) - 4 = 98$ , so $\sum_{i = 1}^4 y_i = 51$
So we want to find four natural numbers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50; thus the answer is $n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600$ , and $\frac n{100} = \boxed{196}$
| 196
|
6,683
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7
| 3
|
Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$
|
Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into $4$ boxes so that each box has an odd number of stones. We then proceed by placing one stone in each box to begin with, ensuring that we have a positive number in every box. Now we have $94$ stones left. Because we want an odd number in each box, we pair the stones, creating $47$ sets of $2$ . Every time we add a pair to one of the boxes, the number of stones in the box remains odd, because (an odd #) + (an even #) = (an odd #).
Our problem can now be restated: how many ways are there to partition a line of $47$ stones? We can easily solve this by using $3$ sticks to separate the stones into $4$ groups, and this is the same as arranging a line of $3$ sticks and $47$ stones. \[\frac{50!}{47! \cdot 3!} = 19600\] \[\frac{50 * 49 * 48}{3 * 2} = 19600\] Our answer is therefore $\frac{19600}{100} = \boxed{196}$
| 196
|
6,684
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7
| 4
|
Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$
|
Let $x = a + b$ and $y = c + d$ . Then $x + y = 98$ , where $x, y$ are positive even integers ranging from $2-98$
We quickly see that the total number of acceptable ordered pairs $(a, b, c, d)$ is:
\begin{align*} &\mathrel{\phantom{=}} 1 \cdot 48 + 2 \cdot 47 + 3 \cdot 46 + ... + 48 \cdot 1\\ &= (24.5 - 23.5)(24.5) + (24.5 - 22.5)(24.5 + 22.5) + ... + (24.5 + 23.5)(24.5 - 23.5)\\ &= 48(24.5)^2 - 2(0.5^2 + 1.5^2 + ... + 23.5^2)\\ &= 28812 - \frac{1^2 + 3^2 + ... + 47^2}{2}\\ &= 28812 - \frac{1^2 + 2^2 + ... + 47^2 - 4(1^2 + 2^2 + ... + 23^2)}{2}\\ &= 28812 - \frac{\frac{47(47 + 1)(2(47) + 1)}{6} - \frac{4(23)(23 + 1)(2(23) + 1)}{6}}{2}\\ &= 19600 \end{align*}
Therefore, $\frac{n}{100} = \frac{19600}{100} = \boxed{196}$
| 196
|
6,685
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_7
| 5
|
Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$
|
We write the generating functions for each of the terms, and obtain $(x+x^3+x^5\cdots)^4$ as the generating function for the sum of the $4$ numbers. We seek the $x^{98}$ coefficient, or the $x^{94}$ coefficient in $(1+x^2+x^4...)^4.$ Now we simplify this as $\left(\frac{1}{1-x^2}\right)^4=\binom{3}{3} +\binom{4}{3}x^2+\binom{5}{3}x^4 \cdots$ and in general we want that the coefficient of $x^{2k}$ is $\binom{k+3}{3}.$ We see the $x^{94}$ coefficient so we let $k=47$ and so the coefficient is $\binom{50}{3}=19600$ in which $\frac{n}{100}=\boxed{196}.$
| 196
|
6,686
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_8
| 1
|
Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first negative term encounted. What positive integer $x$ produces a sequence of maximum length?
|
We can start to write out some of the inequalities now:
And in general,
It is apparent that the bounds are slowly closing in on $x$ , so we can just calculate $x$ for some large value of $n$ (randomly, 10, 11):
$x < \frac{F_{7}}{F_{8}} \cdot 1000 = \frac{13}{21} \cdot 1000 = 618.\overline{18}$
$x > \frac{F_{8}}{F_{9}} \cdot 1000 = \frac{21}{34} \cdot 1000 \approx 617.977$
Thus the sequence is maximized when $x = \boxed{618}.$
| 618
|
6,687
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_8
| 2
|
Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first negative term encounted. What positive integer $x$ produces a sequence of maximum length?
|
It is well known that $\lim_{n\rightarrow\infty} \frac{F_{n-1}}{F_n} = \phi - 1 =\frac{1 + \sqrt{5}}{2} - 1 \approx .61803$ , so $1000 \cdot \frac{F_{n-1}}{F_n}$ approaches $x = \boxed{618}.$
| 618
|
6,688
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_10
| 1
|
Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon . A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a +b\sqrt {c},$ where $a, b,$ and $c$ are positive integers , and $c$ is not divisible by the square of any prime . Find $a + b + c$
|
The key is to realize the significance that the figures are spheres, not circles . The 2D analogue of the diagram onto the flat surface will not contain 8 circles tangent to a ninth one; instead the circles will overlap since the middle sphere has a larger radius and will sort of “bulge” out.
1998 AIME-10a.png
Let us examine the relation between one of the outside 8 spheres and the center one (with radius $r$ ):
1998 AIME-10b.png
If we draw the segment containing the centers and the radii perpendicular to the flat surface, we get a trapezoid ; if we draw the segment parallel to the surface that connects the center of the smaller sphere to the radii of the larger, we get a right triangle. Call that segment $x$ . Then by the Pythagorean Theorem
\[x^2 + (r-100)^2 = (r+100)^2 \Longrightarrow x = 20\sqrt{r}\]
1998 AIME-10c.png
$x$ is the distance from one of the vertices of the octagon to the center, so the diagonal of the octagon is of length $2x =40\sqrt{r}$ . We can draw another right triangle as shown above. One leg has a length of $200$ . The other can be found by partitioning the leg into three sections and using $45-45-90 \triangle$ s to see that the leg is $100\sqrt{2} + 200 + 100\sqrt{2} = 200(\sqrt{2} + 1)$ . Pythagorean Theorem:
\begin{eqnarray*} (40\sqrt{r})^2 &=& 200^2 + [200(\sqrt{2}+1)]^2\\ 1600r &=& 200^2[(1 + \sqrt{2})^2 + 1] \\ r &=& 100 + 50\sqrt{2} \end{eqnarray*}
Thus $a + b + c = 100 + 50 + 2 = \boxed{152}$
| 152
|
6,689
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_10
| 2
|
Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon . A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a +b\sqrt {c},$ where $a, b,$ and $c$ are positive integers , and $c$ is not divisible by the square of any prime . Find $a + b + c$
|
Isolate a triangle, with base length $200$ (a side of the octagon). This triangle is obviously isoceles. Denote the other side length as $x$ . Since the interior angle is $45$ degrees (due to the shape being an octagon), then we can apply Law of Cosines to this triangle. We get: \begin{eqnarray*} 200^2 &=& 2x^2 - 2x^2*cos(45^\circ) \\ &=& 2x^2 - 2x^2*\frac{\sqrt{2}}{2} \\ &=& (2-\sqrt{2})x^2 \end{eqnarray*}
And thus \[x = \frac{200}{\sqrt{2-\sqrt{2}}}\]
From the above, $x = 20\sqrt{r}$ , so we get
\begin{eqnarray*} r &=& (\frac{200}{20(\sqrt{2-\sqrt{2}})})^2 \\ &=& (\frac{10}{\sqrt{2-\sqrt{2}})})^2 \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}} \\ &=& \frac{200 + 100\sqrt{2}}{2} \\ &=& 100 + 50\sqrt{2} \end{eqnarray*}
And hence the answer is $100 + 50 + 2 \Rightarrow \boxed{152}$
| 152
|
6,690
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_14
| 1
|
An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$
|
\[2mnp = (m+2)(n+2)(p+2)\]
Let’s solve for $p$
\[(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)\] \[[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)\] \[p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}\]
Clearly, we want to minimize the denominator, so we test $(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9$ . The possible pairs of factors of $9$ are $(1,9)(3,3)$ . These give $m = 3, n = 11$ and $m = 5, n = 5$ respectively. Substituting into the numerator, we see that the first pair gives $130$ , while the second pair gives $98$ . We now check that $130$ is optimal, setting $a=m-2$ $b=n-2$ in order to simplify calculations. Since \[0 \le (a-1)(b-1) \implies a+b \le ab+1\] We have \[p = \frac{2(a+4)(b+4)}{ab-8} = \frac{2ab+8(a+b)+32}{ab-8} \le \frac{2ab+8(ab+1)+32}{ab-8} = 10 + \frac{120}{ab-8} \le 130\] Where we see $(m,n)=(3,11)$ gives us our maximum value of $\boxed{130}$
| 130
|
6,691
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_14
| 2
|
An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$
|
Similarly as above, we solve for $p,$ but we express the denominator differently:
\[p=\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(m+n+2)}{(m+2)(n+2)}.\] Hence, it suffices to maximize $\dfrac{m+n+2}{(m+2)(n+2)},$ under the conditions that $p$ is a positive integer.
Then since $\dfrac{m+n+2}{(m+2)(n+2)}>\dfrac{1}{2}$ for $m=1,2,$ we fix $m=3.$ \[\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+5)}{5(n+2)}=\dfrac{n-10}{10(n+2)},\] where we simply let $n=11$ to achieve $p=\boxed{130}.$
| 130
|
6,692
|
https://artofproblemsolving.com/wiki/index.php/1998_AIME_Problems/Problem_15
| 1
|
Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which $(i,j)$ and $(j,i)$ do not both appear for any $i$ and $j$ . Let $D_{40}$ be the set of all dominos whose coordinates are no larger than 40. Find the length of the longest proper sequence of dominos that can be formed using the dominos of $D_{40}.$
|
We can draw a comparison between the domino a set of 40 points (labeled 1 through 40) in which every point is connected with every other point. The connections represent the dominoes.
You need to have all even number of segments coming from each point except 0 or 2 which have an odd number of segments coming from the point. (Reasoning for this: Everytime you go to a vertex, you have to leave the vertex, so every vertex reached is equivalent to adding 2 more segments. So the degree of each vertex must be even, with the exception of endpoints) Since there are 39 segments coming from each point it is impossible to touch every segment.
But you can get up to 38 on each segment because you go in to the point then out on a different segment. Counting going out from the starting and ending at the ending point we have:
$\frac{38\cdot 38 + 2\cdot 39}2 = \boxed{761}$
| 761
|
6,693
|
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_1
| 1
|
How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?
|
Notice that all odd numbers can be obtained by using $(a+1)^2-a^2=2a+1,$ where $a$ is a nonnegative integer. All multiples of $4$ can be obtained by using $(b+1)^2-(b-1)^2 = 4b$ , where $b$ is a positive integer. Numbers congruent to $2 \pmod 4$ cannot be obtained because squares are $0, 1 \pmod 4.$ Thus, the answer is $500+250 = \boxed{750}.$
| 750
|
6,694
|
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_2
| 1
|
The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles , of which $s$ are squares . The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
|
To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\choose 2} = 36$ . Similarly, there are ${9\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles.
For $s$ , there are $8^2$ unit squares $7^2$ of the $2\times2$ squares, and so on until $1^2$ of the $8\times 8$ squares. Using the sum of squares formula, that gives us $s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204$
Thus $\frac sr = \dfrac{204}{1296}=\dfrac{17}{108}$ , and $m+n=\boxed{125}$
| 125
|
6,695
|
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_3
| 1
|
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 $112 + 14 = \boxed{126}$
| 126
|
6,696
|
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_3
| 2
|
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 $\boxed{126}$
| 126
|
6,697
|
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_3
| 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 $112 + 14 = \boxed{126}$
| 126
|
6,698
|
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_4
| 1
|
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 $m+n = \boxed{17}$
| 17
|
6,699
|
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_5
| 1
|
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 $3095 - 2679 + 1 = \boxed{417}$
| 417
|
6,700
|
https://artofproblemsolving.com/wiki/index.php/1997_AIME_Problems/Problem_7
| 1
|
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 $-\frac{b}{2a} = \frac{110}{2 \cdot \frac{5}{18}} = \boxed{198}$
| 198
|
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