View source

{{AIME Problems|year=2000|n=II}}

== Problem 1 ==
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 ==
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 ==
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 ==
What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?

[[2000 AIME II Problems/Problem 4|Solution]]

== Problem 5 ==
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 ==
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 ==
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 ==
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 ==
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>.

[[2000 AIME II Problems/Problem 9|Solution]]

== Problem 10 ==
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 ==
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 ==
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>.

[[2000 AIME II Problems/Problem 12|Solution]]

== Problem 13 ==
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 ==
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 ==
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]]

== See also ==
* [[American Invitational Mathematics Examination]]
* [[AIME Problems and Solutions]]
* [[Mathematics competition resources]]