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Difference between revisions of "2006 AIME I Problems"
m (Replacing another PNG, this time with the asy that already exists in /Problem7, but with a constraint on the width)
 
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== Problem 5 ==
 
== Problem 5 ==
−
The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </math> where <math> a, b, </math> and <math> c </math> are positive integers. Find <math> a\cdot b\cdot c.  </math>
+
The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </math> where <math> a, b, </math> and <math> c </math> are positive integers. Find <math> abc.  </math>
  
 
[[2006 AIME I Problems/Problem 5|Solution]]
 
[[2006 AIME I Problems/Problem 5|Solution]]
Line 32: Line 32:
  
 
== Problem 7 ==
 
== Problem 7 ==
−
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region <math> \mathcal{C} </math> to the area of shaded region <math> \mathcal{B} </math> is 11/5. Find the ratio of shaded region <math> \mathcal{D} </math> to the area of shaded region <math> \mathcal{A}.  </math>
+
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region <math>C</math> to the area of shaded region <math>B</math> is <math>\frac{11}{5}</math>. Find the ratio of shaded region <math>D</math> to the area of shaded region <math>A</math>.
  
−
[[Image:2006AimeA7.PNG]]
+
<asy>
 +
size(6cm);
 +
defaultpen(linewidth(0.7)+fontsize(10));
 +
for(int i=0; i<4; i=i+1) {
 +
fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray);
 +
}
 +
pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10);
 +
draw(B--A--C);
 +
fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white);
 +
clip(B--A--C--cycle);
 +
for(int i=0; i<9; i=i+1) {
 +
draw((i,1)--(i,6));
 +
}
 +
label("$\mathcal{A}$", A+0.2*dir(-17), S);
 +
label("$\mathcal{B}$", A+2.3*dir(-17), S);
 +
label("$\mathcal{C}$", A+4.4*dir(-17), S);
 +
label("$\mathcal{D}$", A+6.5*dir(-17), S);
 +
</asy>
  
 
[[2006 AIME I Problems/Problem 7|Solution]]
 
[[2006 AIME I Problems/Problem 7|Solution]]
  
 
== Problem 8 ==
 
== Problem 8 ==
−
Hexagon <math> ABCDEF </math> is divided into five rhombuses, <math> \mathcal{P, Q, R, S,} </math> and <math> \mathcal{T,} </math> as shown. Rhombuses <math> \mathcal{P, Q, R,} </math> and <math> \mathcal{S} </math> are congruent, and each has area <math> \sqrt{2006}. </math> Let <math> K </math> be the area of rhombus <math> \mathcal{T}. </math> Given that <math> K </math> is a positive integer, find the number of possible values for <math> K. </math>  
+
Hexagon <math> ABCDEF </math> is divided into five rhombuses, <math>P, Q, R, S,</math> and <math>T</math>, as shown. Rhombuses <math>P, Q, R,</math> and <math>S</math> are congruent, and each has area <math> \sqrt{2006}. </math> Let <math> K </math> be the area of rhombus <math>T</math>. Given that <math> K </math> is a positive integer, find the number of possible values for <math> K. </math>  
  
−
[[Image:2006AimeA8.PNG]]
+
<asy>
 +
// TheMathGuyd
 +
size(8cm);
 +
pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6);
 +
draw(A--B--C--D--EE--F--cycle);
 +
draw(F--G--(2.1,0));
 +
draw(C--H--(2.1,0));
 +
draw(G--(2.1,-3.2));
 +
draw(H--(2.1,-3.2));
 +
label("$\mathcal{T}$",(2.1,-1.6));
 +
label("$\mathcal{P}$",(0,-1),NE);
 +
label("$\mathcal{Q}$",(4.2,-1),NW);
 +
label("$\mathcal{R}$",(0,-2.2),SE);
 +
label("$\mathcal{S}$",(4.2,-2.2),SW);
 +
</asy>
  
 
[[2006 AIME I Problems/Problem 8|Solution]]
 
[[2006 AIME I Problems/Problem 8|Solution]]

Latest revision as of 02:31, 4 January 2023

2006 AIME I (Answer Key)
Printable version | AoPS Contest Collections • PDF

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 and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Problem 1

In quadrilateral $ABCD , \angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD}, AB=18, BC=21,$ and $CD=14.$ Find the perimeter of $ABCD.$

Solution

Problem 2

Let set $\mathcal{A}$ be a 90-element subset of $\{1,2,3,\ldots,100\},$ and let $S$ be the sum of the elements of $\mathcal{A}.$ Find the number of possible values of $S.$

Solution

Problem 3

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $1/29$ of the original integer.

Solution

Problem 4

Let $N$ be the number of consecutive 0's at the right end of the decimal representation of the product $1!2!3!4!\cdots99!100!.$ Find the remainder when $N$ is divided by 1000.

Solution

Problem 5

The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers. Find $abc.$

Solution

Problem 6

Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits. Find the sum of the elements of $\mathcal{S}.$

Solution

Problem 7

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is $\frac{11}{5}$. Find the ratio of shaded region $D$ to the area of shaded region $A$.

[asy] size(6cm); defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S); [/asy]

Solution

Problem 8

Hexagon $ABCDEF$ is divided into five rhombuses, $P, Q, R, S,$ and $T$, as shown. Rhombuses $P, Q, R,$ and $S$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $T$. Given that $K$ is a positive integer, find the number of possible values for $K.$

[asy] // TheMathGuyd size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); label("$\mathcal{T}$",(2.1,-1.6)); label("$\mathcal{P}$",(0,-1),NE); label("$\mathcal{Q}$",(4.2,-1),NW); label("$\mathcal{R}$",(0,-2.2),SE); label("$\mathcal{S}$",(4.2,-2.2),SW); [/asy]

Solution

Problem 9

The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$

Solution

Problem 10

Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$'s equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$

[asy] unitsize(0.50cm); draw((0,-1)--(0,6)); draw((-1,0)--(6,0)); draw(shift(1,1)*unitcircle); draw(shift(1,3)*unitcircle); draw(shift(1,5)*unitcircle); draw(shift(3,1)*unitcircle); draw(shift(3,3)*unitcircle); draw(shift(3,5)*unitcircle); draw(shift(5,1)*unitcircle); draw(shift(5,3)*unitcircle); [/asy]

Solution

Problem 11

A collection of 8 cubes consists of one cube with edge-length $k$ for each integer $k, 1 \le k \le 8.$ A tower is to be built using all 8 cubes according to the rules:

  • Any cube may be the bottom cube in the tower.
  • The cube immediately on top of a cube with edge-length $k$ must have edge-length at most $k+2.$

Let $T$ be the number of different towers than can be constructed. What is the remainder when $T$ is divided by 1000?

Solution

Problem 12

Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x,$ where $x$ is measured in degrees and $100< x< 200.$

Solution

Problem 13

For each even positive integer $x,$ let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square.

Solution

Problem 14

A tripod has three legs each of length 5 feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 4 feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$)

Solution

Problem 15

Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$

Solution

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
2005 AIME II Problems 		Followed by
2006 AIME II Problems
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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