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Difference between revisions of "User:Rowechen"
Line 3: Line 3:
  
 
Here's the AIME compilation I will be doing:
 
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 <math> \frac mn, </math> where <math> m </math> and <math> n </math> are relatively prime integers, find <math> m+n. </math>
 
  
−
[[2005 AIME II Problems/Problem 2|Solution]]
+
== Problem 1 ==
−
== Problem 4 ==
+
Maya lists all the positive divisors of <math>2010^2</math>. She then randomly selects two distinct divisors from this list. Let <math>p</math> be the probability that exactly one of the selected divisors is a perfect square. The probability <math>p</math> can be expressed 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>.
−
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>.
 
  
−
[[2012 AIME II Problems/Problem 4|Solution]]
+
[[2010 AIME I Problems/Problem 1|Solution]]
−
==Problem 4==
+
==Problem 5==
−
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>.
 
  
−
[[2013 AIME II Problems/Problem 4|Solution]]
+
Real numbers <math>r</math> and <math>s</math> are roots of <math>p(x)=x^3+ax+b</math>, and <math>r+4</math> and <math>s-3</math> are roots of <math>q(x)=x^3+ax+b+240</math>. Find the sum of all possible values of <math>|b|</math>.
 +
 
 +
 
 +
[[2014 AIME II Problems/Problem 5|Solution]]
 +
==Problem 6==
 +
Let <math>N</math> be the number of complex numbers <math>z</math> with the properties that <math>|z|=1</math> and <math>z^{6!}-z^{5!}</math> is a real number. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 +
 
 +
[[2018 AIME I Problems/Problem 6 | Solution]]
 
== Problem 7 ==
 
== 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>.
+
Triangle <math>ABC</math> has <math>AB=21</math>, <math>AC=22</math> and <math>BC=20</math>. Points <math>D</math> and <math>E</math> are located on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, such that <math>\overline{DE}</math> is parallel to <math>\overline{BC}</math> and contains the center of the inscribed circle of triangle <math>ABC</math>. Then <math>DE=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 7|Solution]]
+
[[2001 AIME I Problems/Problem 7|Solution]]
−
== Problem 8 ==
+
== Problem 9 ==
−
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>.
+
Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is <math>\frac {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 8|Solution]]
+
[[2001 AIME II Problems/Problem 9|Solution]]
 
== Problem 9 ==
 
== 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>.
+
Let <math> ABC </math> be a triangle with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 rectangle. A segment is drawn to divide triangle <math> ABC </math> into a triangle <math> U_1 </math> and a trapezoid <math> V_1 </math> and another segment is drawn to divide rectangle <math> DEFG </math> into a triangle <math> U_2 </math> and a trapezoid <math> V_2 </math> such that <math> U_1 </math> is similar to <math> U_2 </math> and <math> V_1 </math> is similar to <math> V_2. </math> The minimum value of the area of <math> U_1 </math> can be written in the form <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 9|Solution]]
+
[[2004 AIME I Problems/Problem 9|Solution]]
−
== Problem 10 ==
+
== Problem 12 ==
−
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>?
+
Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is <math>8: 7</math>. Find the minimum possible value of their common perimeter.
  
−
[[2001 AIME II Problems/Problem 10|Solution]]
+
[[2010 AIME II Problems/Problem 12|Solution]]
−
== Problem 10 ==
+
== Problem 12 ==
−
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>
+
Six men and some number of women stand in a line in random order.  Let <math>p</math> be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man.  Find the least number of women in the line such that <math>p</math> does not exceed 1 percent.
  
−
[[2004 AIME I Problems/Problem 10|Solution]]
+
[[2011 AIME I Problems/Problem 12|Solution]]
 
== Problem 10 ==
 
== 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>
+
A circle with center <math>O</math> has radius 25. Chord <math>\overline{AB}</math> of length 30 and chord <math>\overline{CD}</math> of length 14 intersect at point <math>P</math>. The distance between the midpoints of the two chords is 12. The quantity <math>OP^2</math> can be represented as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find the remainder when <math>m + n</math> is divided by 1000.
  
−
[[2004 AIME II Problems/Problem 10|Solution]]
+
[[2011 AIME II Problems/Problem 10|Solution]]
 
== Problem 11 ==
 
== Problem 11 ==
−
Define a <i>T-grid</i> to be a <math>3\times3</math> matrix which satisfies the following two properties:
+
A frog begins at <math>P_0 = (0,0)</math> and makes a sequence of jumps according to the following rule: from <math>P_n = (x_n, y_n),</math> the frog jumps to <math>P_{n+1},</math> which may be any of the points <math>(x_n + 7, y_n + 2),</math> <math>(x_n + 2, y_n + 7),</math> <math>(x_n - 5, y_n - 10),</math> or <math>(x_n - 10, y_n - 5).</math> There are <math>M</math> points <math>(x, y)</math> with <math>|x| + |y| \le 100</math> that can be reached by a sequence of such jumps. Find the remainder when <math>M</math> is divided by <math>1000.</math>
  
−
<OL>
+
[[2012 AIME I Problems/Problem 11|Solution]]
−
<LI>Exactly five of the entries are <math>1</math>'s, and the remaining four entries are <math>0</math>'s.</LI>
+
== Problem 14 ==
−
<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>
+
Let <math>A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8</math> be a regular octagon.  Let <math>M_1</math>, <math>M_3</math>, <math>M_5</math>, and <math>M_7</math> be the midpoints of sides <math>\overline{A_1 A_2}</math>, <math>\overline{A_3 A_4}</math>, <math>\overline{A_5 A_6}</math>, and <math>\overline{A_7 A_8}</math>, respectively.  For <math>i = 1, 3, 5, 7</math>, ray <math>R_i</math> is constructed from <math>M_i</math> towards the interior of the octagon such that <math>R_1 \perp R_3</math>, <math>R_3 \perp R_5</math>, <math>R_5 \perp R_7</math>, and <math>R_7 \perp R_1</math>.  Pairs of rays <math>R_1</math> and <math>R_3</math>, <math>R_3</math> and <math>R_5</math>, <math>R_5</math> and <math>R_7</math>, and <math>R_7</math> and <math>R_1</math> meet at <math>B_1</math>, <math>B_3</math>, <math>B_5</math>, <math>B_7</math> respectively.  If <math>B_1 B_3 = A_1 A_2</math>, then <math>\cos 2 \angle A_3 M_3 B_1</math> can be written in the form <math>m - \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m + n</math>.
  
−
Find the number of distinct <i>T-grids</i>.
+
[[2011 AIME I Problems/Problem 14|Solution]]
 +
== Problem 15 ==
 +
Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}</math> , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are positive integers. Find <math>a + b + c + d + e</math>.
  
 +
[[2011 AIME II Problems/Problem 15|Solution]]
 +
== Problem 13 ==
 +
Equilateral <math>\triangle ABC</math> has side length <math>\sqrt{111}</math>. There are four distinct triangles <math>AD_1E_1</math>, <math>AD_1E_2</math>, <math>AD_2E_3</math>, and <math>AD_2E_4</math>, each congruent to <math>\triangle ABC</math>,
 +
with <math>BD_1 = BD_2 = \sqrt{11}</math>. Find <math>\sum_{k=1}^4(CE_k)^2</math>.
  
−
[[2010 AIME II Problems/Problem 11|Solution]]
+
[[2012 AIME II Problems/Problem 13|Solution]]
−
== Problem 15 ==
+
== Problem 14 ==
−
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?
+
In a group of nine people each person shakes hands with exactly two of the other people from the group. Let <math>N</math> be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  
−
[[2004 AIME II Problems/Problem 15|Solution]]
+
[[2012 AIME II Problems/Problem 14|Solution]]
 
== Problem 14 ==
 
== Problem 14 ==
−
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.
+
For <math>\pi \le \theta < 2\pi</math>, let
  
−
[[2005 AIME I Problems/Problem 14|Solution]]
+
<cmath> P=\dfrac12\cos\theta-\dfrac14\sin2\theta-\dfrac18\cos3\theta+\dfrac1{16}\sin4\theta+\dfrac1{32}\cos5\theta-\dfrac1{64}\sin6\theta-\dfrac1{128}\cos7\theta+\ldots
−
== Problem 15 ==
+
</cmath>
−
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>
+
 
 +
and
  
−
[[2005 AIME II Problems/Problem 15|Solution]]
+
<cmath> Q=1-\dfrac12\sin\theta-\dfrac14\cos2\theta+\dfrac1{8}\sin3\theta+\dfrac1{16}\cos4\theta-\dfrac1{32}\sin5\theta-\dfrac1{64}\cos6\theta+\dfrac1{128}\sin7\theta
−
== Problem 15 ==
+
+\ldots </cmath>
−
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>.
 
  
−
[[2010 AIME I Problems/Problem 15|Solution]]
+
so that <math>\frac{P}{Q} = \frac{2\sqrt2}{7}</math>. Then <math>\sin\theta = -\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
−
== Problem 15 ==
 
−
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>.
 
  
−
[[2010 AIME II Problems/Problem 15|Solution]]
+
[[2013 AIME I Problems/Problem 14|Solution]]

Revision as of 08:15, 29 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 1

Maya lists all the positive divisors of $2010^2$. She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square. The probability $p$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Problem 5

Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$, and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$. Find the sum of all possible values of $|b|$.


Solution

Problem 6

Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$.

Solution

Problem 7

Triangle $ABC$ has $AB=21$, $AC=22$ and $BC=20$. Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$, respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$. Then $DE=m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Problem 9

Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Problem 9

Let $ABC$ be a triangle with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 rectangle. A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is similar to $V_2.$ The minimum value of the area of $U_1$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

Problem 12

Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$. Find the minimum possible value of their common perimeter.

Solution

Problem 12

Six men and some number of women stand in a line in random order. Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent.

Solution

Problem 10

A circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$. The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000.

Solution

Problem 11

A frog begins at $P_0 = (0,0)$ and makes a sequence of jumps according to the following rule: from $P_n = (x_n, y_n),$ the frog jumps to $P_{n+1},$ which may be any of the points $(x_n + 7, y_n + 2),$ $(x_n + 2, y_n + 7),$ $(x_n - 5, y_n - 10),$ or $(x_n - 10, y_n - 5).$ There are $M$ points $(x, y)$ with $|x| + |y| \le 100$ that can be reached by a sequence of such jumps. Find the remainder when $M$ is divided by $1000.$

Solution

Problem 14

Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$, $M_3$, $M_5$, and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$, $\overline{A_3 A_4}$, $\overline{A_5 A_6}$, and $\overline{A_7 A_8}$, respectively. For $i = 1, 3, 5, 7$, ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$, $R_3 \perp R_5$, $R_5 \perp R_7$, and $R_7 \perp R_1$. Pairs of rays $R_1$ and $R_3$, $R_3$ and $R_5$, $R_5$ and $R_7$, and $R_7$ and $R_1$ meet at $B_1$, $B_3$, $B_5$, $B_7$ respectively. If $B_1 B_3 = A_1 A_2$, then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$, where $m$ and $n$ are positive integers. Find $m + n$.

Solution

Problem 15

Let $P(x) = x^2 - 3x - 9$. A real number $x$ is chosen at random from the interval $5 \le x \le 15$. The probability that $\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$, $b$, $c$, $d$, and $e$ are positive integers. Find $a + b + c + d + e$.

Solution

Problem 13

Equilateral $\triangle ABC$ has side length $\sqrt{111}$. There are four distinct triangles $AD_1E_1$, $AD_1E_2$, $AD_2E_3$, and $AD_2E_4$, each congruent to $\triangle ABC$, with $BD_1 = BD_2 = \sqrt{11}$. Find $\sum_{k=1}^4(CE_k)^2$.

Solution

Problem 14

In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by $1000$.

Solution

Problem 14

For $\pi \le \theta < 2\pi$, let

\[P=\dfrac12\cos\theta-\dfrac14\sin2\theta-\dfrac18\cos3\theta+\dfrac1{16}\sin4\theta+\dfrac1{32}\cos5\theta-\dfrac1{64}\sin6\theta-\dfrac1{128}\cos7\theta+\ldots\]

and

\[Q=1-\dfrac12\sin\theta-\dfrac14\cos2\theta+\dfrac1{8}\sin3\theta+\dfrac1{16}\cos4\theta-\dfrac1{32}\sin5\theta-\dfrac1{64}\cos6\theta+\dfrac1{128}\sin7\theta +\ldots\]

so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$. Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

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