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2000 AIME II Problems

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2000 AIME II (Answer Key)
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Instructions

  1. This is a 15-question, 3-hour examination. All answers are integers ranging from 000 to 999, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
  2. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators are not permitted.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Contents

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.

Solution

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...
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{...

Solution

See also

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