2000年AIME II卷真题及答案 - American Mathematics Competitions

2000年AIME II 真题：

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 $x^2 - y^2 = 2000^2$ ?

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 $m + n.$

Problem 4

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

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 $n$ .

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 $x^2/100$ .

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 $\frac N{100}$ .

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 $BC^2$ .