Difference between revisions of "2000 AIME II Problems"

Revision as of 18:59, 11 November 2007

Problem 1

The number

$\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}$

can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Problem 2

A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 - y^2 = 2000^2.$

Solution

Problem 3

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 $m + n.$

Solution

Problem 4

What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?

Solution

Problem 5

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 $n$ .

Solution

Problem 6

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 $x^2/100$ .

Solution

Problem 7

Given that

$\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!}$

find the greatest integer that is less than $\frac N{100}$ .

Solution

Problem 8

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 $BC^2$ .

Solution

Problem 9

Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$ .

Solution

Problem 10

A 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.

Solution

Problem 11

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 $m+n$ .

Solution

Problem 12

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 $m+n+k$ .

Solution

Problem 13

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 $m+n+r$ .

Solution

Problem 14

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 $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$ .

Solution

Problem 15

Find the least positive integer $n$ such that

$\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.$

Solution

@@ Line 1: / Line 1: @@
 == Problem 1 ==
-Find the least positive integer <math>n</math> such that no matter how <math>10^{n}</math> is expressed as the product of any two positive integers, at least one of these two integers contains the digit <math>0</math>.
+The number
+<center><math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math></center>
+can be written as <math>\frac mn</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n</math>.
 [[2000 AIME II Problems/Problem 1|Solution]]
 == Problem 2 ==
-Let <math>u</math> and <math>v</math> be integers satisfying <math>0 < v < u</math>. Let <math>A = (u,v)</math>, let <math>B</math> be the reflection of <math>A</math> across the line <math>y = x</math>, let <math>C</math> be the reflection of <math>B</math> across the y-axis, let <math>D</math> be the reflection of <math>C</math> across the x-axis, and let <math>E</math> be the reflection of <math>D</math> across the y-axis. The area of pentagon <math>ABCDE</math> is <math>451</math>. Find <math>u + v</math>.
+A point whose coordinates are both integers is called a lattice point.  How many lattice points lie on the hyperbola <math>x^2 - y^2 = 2000^2.</math>
 [[2000 AIME II Problems/Problem 2|Solution]]
 == Problem 3 ==
-In the expansion of <math>(ax + b)^{2000},</math> where <math>a</math> and <math>b</math> are relatively prime positive integers, the coefficients of <math>x^{2}</math> and <math>x^{3}</math> are equal. Find <math>a + b</math>.
+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 <math>m/n</math> be the probability that two randomly selected cards also form a pair, where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
 [[2000 AIME II Problems/Problem 3|Solution]]
 == Problem 4 ==
-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.
+What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?
-{{image}}
 [[2000 AIME II Problems/Problem 4|Solution]]
 == Problem 5 ==
-Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is <math>25.</math> One marble is taken out of each box randomly. The probability that both marbles are black is <math>27/50,</math> and the probability that both marbles are white is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?
+Given eight distinguishable rings, let <math>n</math> 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 <math>n</math>.
 [[2000 AIME II Problems/Problem 5|Solution]]
 == Problem 6 ==
-For how many ordered pairs <math>(x,y)</math> of integers is it true that <math>0 < x < y < 10^{6}</math> and that the arithmetic mean of <math>x</math> and <math>y</math> is exactly <math>2</math> more than the geometric mean of <math>x</math> and <math>y</math>?
+One base of a trapezoid is <math>100</math> 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 <math>2: 3</math>. 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 <math>x^2/100</math>.
 [[2000 AIME II Problems/Problem 6|Solution]]
 == Problem 7 ==
-Suppose that <math>x,</math> <math>y,</math> and <math>z</math> are three positive numbers that satisfy the equations <math>xyz = 1,</math> <math>x + \frac {1}{z} = 5,</math> and <math>y + \frac {1}{x} = 29.</math> Then <math>z + \frac {1}{y} = \frac {m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
+Given that <center><math>\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!}</math></center> find the greatest integer that is less than <math>\frac N{100}</math>.
 [[2000 AIME II Problems/Problem 7|Solution]]
 == Problem 8 ==
-A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is 9 inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the liquid is <math>m - n\sqrt [3]{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the cube of any prime number. Find <math>m + n + p</math>.
+In trapezoid <math>ABCD</math>, leg <math>\overline{BC}</math> is perpendicular to bases <math>\overline{AB}</math> and <math>\overline{CD}</math>, and diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> are perpendicular. Given that <math>AB=\sqrt{11}</math> and <math>AD=\sqrt{1001}</math>, find <math>BC^2</math>.
 [[2000 AIME II Problems/Problem 8|Solution]]
 == Problem 9 ==
-The system of equations
+Given that <math>z</math> is a complex number such that <math>z+\frac 1z=2\cos 3^\circ</math>, find the least integer that is greater than <math>z^{2000}+\frac 1{z^{2000}}</math>.
-<center><math>\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\
-\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\
-\log_{10}(2000zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\
-\end{eqnarray*}</math></center>
-has two solutions <math>(x_{1},y_{1},z_{1})</math> and <math>(x_{2},y_{2},z_{2})</math>. Find <math>y_{1} + y_{2}</math>.
 [[2000 AIME II Problems/Problem 9|Solution]]
 == Problem 10 ==
-A sequence of numbers <math>x_{1},x_{2},x_{3},\ldots,x_{100}</math> has the property that, for every integer <math>k</math> between <math>1</math> and <math>100,</math> inclusive, the number <math>x_{k}</math> is <math>k</math> less than the sum of the other <math>99</math> numbers. Given that <math>x_{50} = m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m + n</math>.
+A circle is inscribed in quadrilateral <math>ABCD</math>, tangent to <math>\overline{AB}</math> at <math>P</math> and to <math>\overline{CD}</math> at <math>Q</math>. Given that <math>AP=19</math>, <math>PB=26</math>, <math>CQ=37</math>, and <math>QD=23</math>, find the square of the radius of the circle.
 [[2000 AIME II Problems/Problem 10|Solution]]
 == Problem 11 ==
-Let <math>S</math> be the sum of all numbers of the form <math>a/b,</math> where <math>a</math> and <math>b</math> are relatively prime positive divisors of <math>1000.</math> What is the greatest integer that does not exceed <math>S/10</math>?
+The coordinates of the vertices of isosceles trapezoid <math>ABCD</math> are all integers, with <math>A=(20,100)</math> and <math>D=(21,107)</math>. The trapezoid has no horizontal or vertical sides, and <math>\overline{AB}</math> and <math>\overline{CD}</math> are the only parallel sides. The sum of the absolute values of all possible slopes for <math>\overline{AB}</math> is <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 [[2000 AIME II Problems/Problem 11|Solution]]
 == Problem 12 ==
-Given a function <math>f</math> for which
+The points <math>A</math>, <math>B</math> and <math>C</math> lie on the surface of a sphere with center <math>O</math> and radius <math>20</math>. It is given that <math>AB=13</math>, <math>BC=14</math>, <math>CA=15</math>, and that the distance from <math>O</math> to triangle <math>ABC</math> is <math>\frac{m\sqrt{n}}k</math>, where <math>m</math>, <math>n</math>, and <math>k</math> are positive integers, <math>m</math> and <math>k</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n+k</math>.
-<center><math>f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)</math></center>
-holds for all real <math>x,</math> what is the largest number of different values that can appear in the list <math>f(0),f(1),f(2),\ldots,f(999)</math>?
 [[2000 AIME II Problems/Problem 12|Solution]]
 == Problem 13 ==
-In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at <math>50</math> miles per hour along the highways and at <math>14</math> 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 <math>m/n</math> square miles, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
+The equation <math>2000x^6+100x^5+10x^3+x-2=0</math> has exactly two real roots, one of which is <math>\frac{m+\sqrt{n}}r</math>, where <math>m</math>, <math>n</math> and <math>r</math> are integers, <math>m</math> and <math>r</math> are relatively prime, and <math>r>0</math>. Find <math>m+n+r</math>.
 [[2000 AIME II Problems/Problem 13|Solution]]
 == Problem 14 ==
-In triangle <math>ABC,</math> it is given that angles <math>B</math> and <math>C</math> are congruent. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AC}</math> and <math>\overline{AB},</math> respectively, so that <math>AP = PQ = QB = BC.</math> Angle <math>ACB</math> is <math>r</math> times as large as angle <math>APQ,</math> where <math>r</math> is a positive real number. Find the greatest integer that does not exceed <math>1000r</math>.
+Every positive integer <math>k</math> has a unique factorial base expansion <math>(f_1,f_2,f_3,\ldots,f_m)</math>, meaning that <math>k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m</math>, where each <math>f_i</math> is an integer, <math>0\le f_i\le i</math>, and <math>0<f_m</math>. Given that <math>(f_1,f_2,f_3,\ldots,f_j)</math> is the factorial base expansion of <math>16!-32!+48!-64!+\cdots+1968!-1984!+2000!</math>, find the value of <math>f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j</math>.
 [[2000 AIME II Problems/Problem 14|Solution]]
 == Problem 15 ==
-A stack of <math>2000</math> cards is labelled with the integers from <math>1</math> to <math>2000,</math> 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: <math>1,2,3,\ldots,1999,2000.</math> In the original stack of cards, how many cards were above the card labeled 1999?
+Find the least positive integer <math>n</math> such that <center><math>\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.</math></center>
 [[2000 AIME II Problems/Problem 15|Solution]]

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Difference between revisions of "2000 AIME II Problems"

Revision as of 18:59, 11 November 2007

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

See also