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Difference between revisions of "User:Rowechen"
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Here's the AIME compilation I will be doing:
 
Here's the AIME compilation I will be doing:
−
 
 
== Problem 2 ==
 
== Problem 2 ==
−
Find the number of [[ordered pair]]s <math>(x,y)</math> of positive integers that satisfy <math>x \le 2y \le 60</math> and <math>y \le 2x \le 60</math>.
+
A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is <math> \frac mn, </math> where <math> m </math> and <math> n </math> are relatively prime integers, find <math> m+n. </math>
  
−
[[1998 AIME Problems/Problem 2|Solution]]
+
[[2005 AIME II Problems/Problem 2|Solution]]
 +
== Problem 4 ==
 +
Ana, Bob, and Cao bike at constant rates of <math>8.6</math> meters per second, <math>6.2</math> meters per second, and <math>5</math> meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point <math>D</math> on the south edge of the field. Cao arrives at point <math>D</math> at the same time that Ana and Bob arrive at <math>D</math> for the first time. The ratio of the field's length to the field's width to the distance from point <math>D</math> to the southeast corner of the field can be represented as <math>p : q : r</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are positive integers with <math>p</math> and <math>q</math> relatively prime. Find <math>p+q+r</math>.
  
−
== Problem 5 ==
+
[[2012 AIME II Problems/Problem 4|Solution]]
−
Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 3 by 4 by 5 units. Given that the volume of this set is <math> \frac{m + n \pi}{p}, </math> where <math> m, n, </math> and <math> p </math> are positive integers, and <math> n </math> and <math> p </math> are relatively prime, find <math> m + n + p. </math>
+
==Problem 4==
 +
In the Cartesian plane let <math>A = (1,0)</math> and <math>B = \left( 2, 2\sqrt{3} \right)</math>. Equilateral triangle <math>ABC</math> is constructed so that <math>C</math> lies in the first quadrant.  Let <math>P=(x,y)</math> be the center of <math>\triangle ABC</math>.  Then <math>x \cdot y</math> can be written as <math>\tfrac{p\sqrt{q}}{r}</math>, where <math>p</math> and <math>r</math> are relatively prime positive integers and <math>q</math> is an integer that is not divisible by the square of any prime.  Find <math>p+q+r</math>.
  
−
[[2003 AIME I Problems/Problem 5|Solution]]
+
[[2013 AIME II Problems/Problem 4|Solution]]
−
== Problem 6 ==
+
== Problem 7 ==
−
The cards in a stack of <math> 2n </math> cards are numbered consecutively from 1 through <math> 2n </math> from top to bottom. The top <math> n </math> cards are removed, kept in order, and form pile <math> A. </math> The remaining cards form pile <math> B. </math> The cards are then restacked by taking cards alternately from the tops of pile <math> B </math> and <math> A, </math> respectively. In this process, card number <math> (n+1) </math> becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles <math> A </math> and <math> B </math> are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. Find the number of cards in the magical stack in which card number 131 retains its original position.
+
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>.
  
−
[[2005 AIME II Problems/Problem 6|Solution]]
+
[[2000 AIME II Problems/Problem 7|Solution]]
 
== Problem 8 ==
 
== Problem 8 ==
−
How many different <math>4\times 4</math> arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?
+
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>.
  
−
[[1997 AIME Problems/Problem 8|Solution]]
+
[[2000 AIME II Problems/Problem 8|Solution]]
 
== Problem 9 ==
 
== Problem 9 ==
−
Given a nonnegative real number <math>x</math>, let <math>\langle x\rangle</math> denote the fractional part of <math>x</math>; that is, <math>\langle x\rangle=x-\lfloor x\rfloor</math>, where <math>\lfloor x\rfloor</math> denotes the greatest integer less than or equal to <math>x</math>. Suppose that <math>a</math> is positive, <math>\langle a^{-1}\rangle=\langle a^2\rangle</math>, and <math>2<a^2<3</math>. Find the value of <math>a^{12}-144a^{-1}</math>.
+
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>.
  
−
[[1997 AIME Problems/Problem 9|Solution]]
+
[[2000 AIME II Problems/Problem 9|Solution]]
 
== Problem 10 ==
 
== Problem 10 ==
−
Let <math>S</math> be the set of points whose coordinates <math>x,</math> <math>y,</math> and <math>z</math> are integers that satisfy <math>0\le x\le2,</math> <math>0\le y\le3,</math> and <math>0\le z\le4.</math>  Two distinct points are randomly chosen from <math>S.</math>  The probability that the midpoint of the segment they determine also belongs to <math>S</math> is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
+
How many positive integer multiples of 1001 can be expressed in the form <math>10^{j} - 10^{i}</math>, where <math>i</math> and <math>j</math> are integers and <math>0\leq i < j \leq 99</math>?
  
−
[[2001 AIME I Problems/Problem 10|Solution]]
+
[[2001 AIME II Problems/Problem 10|Solution]]
 +
== Problem 10 ==
 +
A circle of radius 1 is randomly placed in a 15-by-36 rectangle <math> ABCD </math> so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal <math> AC </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m + n. </math>
  
 +
[[2004 AIME I Problems/Problem 10|Solution]]
 +
== Problem 10 ==
 +
Let <math> S </math> be the set of integers between 1 and <math> 2^{40} </math> whose binary expansions have exactly two 1's. If a number is chosen at random from <math> S, </math> the probability that it is divisible by 9 is <math> p/q, </math> where <math> p </math> and <math> q </math> are relatively prime positive integers. Find <math> p+q. </math>
 +
 +
[[2004 AIME II Problems/Problem 10|Solution]]
 
== Problem 11 ==
 
== Problem 11 ==
−
In a rectangular array of points, with 5 rows and <math>N</math> columns, the points are numbered consecutively from left to right beginning with the top row.  Thus the top row is numbered 1 through <math>N,</math> the second row is numbered <math>N + 1</math> through <math>2N,</math> and so forth.  Five points, <math>P_1, P_2, P_3, P_4,</math> and <math>P_5,</math> are selected so that each <math>P_i</math> is in row <math>i.</math>  Let <math>x_i</math> be the number associated with <math>P_i.</math>  Now renumber the array consecutively from top to bottom, beginning with the first column.  Let <math>y_i</math> be the number associated with <math>P_i</math> after the renumbering.  It is found that <math>x_1 = y_2,</math> <math>x_2 = y_1,</math> <math>x_3 = y_4,</math> <math>x_4 = y_5,</math> and <math>x_5 = y_3.</math>  Find the smallest possible value of <math>N.</math>
+
Define a <i>T-grid</i> to be a <math>3\times3</math> matrix which satisfies the following two properties:
−
 
 
−
[[2001 AIME I Problems/Problem 11|Solution]]
 
−
 
 
−
== Problem 12 ==
 
−
A sphere is inscribed in the tetrahedron whose vertices are <math>A = (6,0,0), B = (0,4,0), C = (0,0,2),</math> and <math>D = (0,0,0).</math>  The radius of the sphere is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
 
  
−
[[2001 AIME I Problems/Problem 12|Solution]]
+
<OL>
−
== Problem 9 ==
+
<LI>Exactly five of the entries are <math>1</math>'s, and the remaining four entries are <math>0</math>'s.</LI>
−
The system of equations
+
<LI>Among the eight rows, columns, and long diagonals (the long diagonals are <math>\{a_{13},a_{22},a_{31}\}</math> and <math>\{a_{11},a_{22},a_{33}\})</math>, no more than one of the eight has all three entries equal.</LI></OL>
−
<cmath>\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\
 
−
\log_{10}(2yz) - (\log_{10}y)(\log_{10}z) & = & 1 \\
 
−
\log_{10}(zx) - (\log_{10}z)(\log_{10}x) & = & 0 \\
 
−
\end{eqnarray*}</cmath>
 
  
−
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>.
+
Find the number of distinct <i>T-grids</i>.
  
−
[[2000 AIME I Problems/Problem 9|Solution]]
 
  
−
== Problem 12 ==
+
[[2010 AIME II Problems/Problem 11|Solution]]
−
Given a function <math>f</math> for which
+
== Problem 15 ==
−
<center><math>f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)</math></center>
+
A long thin strip of paper is 1024 units in length, 1 unit in width, and is divided into 1024 unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a 512 by 1 strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a 256 by 1 strip of quadruple thickness. This process is repeated 8 more times. After the last fold, the strip has become a stack of 1024 unit squares. How many of these squares lie below the square that was originally the 942nd square counting from the left?
−
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 I Problems/Problem 12|Solution]]
+
[[2004 AIME II Problems/Problem 15|Solution]]
 
== Problem 14 ==
 
== Problem 14 ==
−
There are <math>2n</math> complex numbers that satisfy both <math>z^{28} - z^{8} - 1 = 0</math> and <math>|z| = 1</math>. These numbers have the form <math>z_{m} = \cos\theta_{m} + i\sin\theta_{m}</math>, where <math>0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360</math> and angles are measured in degrees. Find the value of <math>\theta_{2} + \theta_{4} + \ldots + \theta_{2n}</math>.
+
Consider the points <math> A(0,12), B(10,9), C(8,0), </math> and <math> D(-4,7). </math> There is a unique square <math> S </math> such that each of the four points is on a different side of <math> S. </math> Let <math> K </math> be the area of <math> S. </math> Find the remainder when <math> 10K </math> is divided by 1000.
  
−
[[2001 AIME II Problems/Problem 14|Solution]]
+
[[2005 AIME I Problems/Problem 14|Solution]]
−
== Problem 13 ==
+
== Problem 15 ==
−
In triangle <math>ABC</math> the medians <math>\overline{AD}</math> and <math>\overline{CE}</math> have lengths 18 and 27, respectively, and <math>AB = 24</math>. Extend <math>\overline{CE}</math> to intersect the circumcircle of <math>ABC</math> at <math>F</math>.  The area of triangle <math>AFB</math> is <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime.  Find <math>m + n</math>.
+
Let <math> w_1 </math> and <math> w_2 </math> denote the circles <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest positive value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally tangent to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math>  
  
−
[[2002 AIME I Problems/Problem 13|Solution]]
+
[[2005 AIME II Problems/Problem 15|Solution]]
−
== Problem 14 ==
+
== Problem 15 ==
−
The perimeter of triangle <math>APM</math> is <math>152</math>, and the angle <math>PAM</math> is a right angle. A circle of radius <math>19</math> with center <math>O</math> on <math>\overline{AP}</math> is drawn so that it is tangent to <math>\overline{AM}</math> and <math>\overline{PM}</math>. Given that <math>OP=m/n</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m+n</math>.
+
In <math>\triangle{ABC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such that the incircles of <math>\triangle{ABM}</math> and <math>\triangle{BCM}</math> have equal radii. Let <math>p</math> and <math>q</math> be positive relatively prime integers such that <math>\frac {AM}{CM} = \frac {p}{q}</math>. Find <math>p + q</math>.
  
−
[[2002 AIME II Problems/Problem 14|Solution]]
+
[[2010 AIME I Problems/Problem 15|Solution]]
−
== Problem 13 ==
+
== Problem 15 ==
−
Let <math> N </math> be the number of positive integers that are less than or equal to 2003 and whose base-2 representation has more 1's than 0's. Find the remainder when <math> N </math> is divided by 1000.
+
In triangle <math>ABC</math>, <math>AC=13</math>, <math>BC=14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math>AC</math> with <math>AM=MC</math> and <math>\angle ABD = \angle DBC</math>. Points <math>N</math> and <math>E</math> lie on <math>AB</math> with <math>AN=NB</math> and <math>\angle ACE = \angle ECB</math>. Let <math>P</math> be the point, other than <math>A</math>, of intersection of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ratio <math>\frac{BQ}{CQ}</math> can be written in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m-n</math>.
−
 
 
−
[[2003 AIME I Problems/Problem 13|Solution]]
 
−
== Problem 14 ==
 
−
The decimal representation of <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers and <math> m < n, </math> contains the digits 2, 5, and 1 consecutively, and in that order. Find the smallest value of <math> n </math> for which this is possible.
 
  
−
[[2003 AIME I Problems/Problem 14|Solution]]
+
[[2010 AIME II Problems/Problem 15|Solution]]

Revision as of 09:28, 28 May 2020

Hey how did you get to this page? If you aren't me then I have to say hello. If you are me then I must be pretty conceited to waste my time looking at my own page. If you aren't me, seriously, how did you get to this page? This is pretty cool. Well, nice meeting you! I'm going to stop wasting my time typing this up and do some math. Gtg. Bye.

Here's the AIME compilation I will be doing:

Problem 2

A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$

Solution

Problem 4

Ana, Bob, and Cao bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point $D$ on the south edge of the field. Cao arrives at point $D$ at the same time that Ana and Bob arrive at $D$ for the first time. The ratio of the field's length to the field's width to the distance from point $D$ to the southeast corner of the field can be represented as $p : q : r$, where $p$, $q$, and $r$ are positive integers with $p$ and $q$ relatively prime. Find $p+q+r$.

Solution

Problem 4

In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$. Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$. Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$, where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$.

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

How many positive integer multiples of 1001 can be expressed in the form $10^{j} - 10^{i}$, where $i$ and $j$ are integers and $0\leq i < j \leq 99$?

Solution

Problem 10

A circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Solution

Problem 10

Let $S$ be the set of integers between 1 and $2^{40}$ whose binary expansions have exactly two 1's. If a number is chosen at random from $S,$ the probability that it is divisible by 9 is $p/q,$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution

Problem 11

Define a T-grid to be a $3\times3$ matrix which satisfies the following two properties:

  1. Exactly five of the entries are $1$'s, and the remaining four entries are $0$'s.
  2. Among the eight rows, columns, and long diagonals (the long diagonals are $\{a_{13},a_{22},a_{31}\}$ and $\{a_{11},a_{22},a_{33}\})$, no more than one of the eight has all three entries equal.

Find the number of distinct T-grids.


Solution

Problem 15

A long thin strip of paper is 1024 units in length, 1 unit in width, and is divided into 1024 unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a 512 by 1 strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a 256 by 1 strip of quadruple thickness. This process is repeated 8 more times. After the last fold, the strip has become a stack of 1024 unit squares. How many of these squares lie below the square that was originally the 942nd square counting from the left?

Solution

Problem 14

Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by 1000.

Solution

Problem 15

Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$

Solution

Problem 15

In $\triangle{ABC}$ with $AB = 12$, $BC = 13$, and $AC = 15$, let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii. Let $p$ and $q$ be positive relatively prime integers such that $\frac {AM}{CM} = \frac {p}{q}$. Find $p + q$.

Solution

Problem 15

In triangle $ABC$, $AC=13$, $BC=14$, and $AB=15$. Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$. Points $N$ and $E$ lie on $AB$ with $AN=NB$ and $\angle ACE = \angle ECB$. Let $P$ be the point, other than $A$, of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$. Ray $AP$ meets $BC$ at $Q$. The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m-n$.

Solution

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