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 1 ==
 
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>.
 
  
[[2010 AIME I Problems/Problem 1|Solution]]
+
== Problem 3 ==
 +
Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.
 +
 
 +
[[2012 AIME I Problems/Problem 3|Solution]]
 +
==Problem 4==
 +
 
 +
In equiangular octagon <math>CAROLINE</math>, <math>CA = RO = LI = NE =</math> <math>\sqrt{2}</math> and <math>AR = OL = IN = EC = 1</math>. The self-intersecting octagon <math>CORNELIA</math> enclosed six non-overlapping triangular regions. Let <math>K</math> be the area enclosed by <math>CORNELIA</math>, that is, the total area of the six triangular regions. Then <math>K = \frac{a}{b}</math>, where <math>a</math> and <math>b</math> are relatively prime positive integers. Find <math>a + b</math>.
 +
 
 +
[[2018 AIME II Problems/Problem 4 | Solution]]
 
==Problem 5==
 
==Problem 5==
  
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>.
+
Suppose that <math>x</math>, <math>y</math>, and <math>z</math> are complex numbers such that <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then there are real numbers <math>a</math> and <math>b</math> such that <math>x + y + z = a + bi</math>. Find <math>a^2 + b^2</math>.
  
 +
[[2018 AIME II Problems/Problem 5 | Solution]]
 +
== Problem 7 ==
 +
Let <math>S_i</math> be the set of all integers <math>n</math> such that <math>100i\leq n < 100(i + 1)</math>.  For example, <math>S_4</math> is the set <math>{400,401,402,\ldots,499}</math>.  How many of the sets <math>S_0, S_1, S_2, \ldots, S_{999}</math> do not contain a perfect square?
  
[[2014 AIME II Problems/Problem 5|Solution]]
+
[[2008 AIME I Problems/Problem 7|Solution]]
==Problem 6==
+
== Problem 7 ==
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>.
+
Define an ordered triple <math>(A, B, C)</math> of sets to be <math>\textit{minimally intersecting}</math> if <math>|A \cap B| = |B \cap C| = |C \cap A| = 1</math> and <math>A \cap B \cap C = \emptyset</math>. For example, <math>(\{1,2\},\{2,3\},\{1,3,4\})</math> is a minimally intersecting triple. Let <math>N</math> be the number of minimally intersecting ordered triples of sets for which each set is a subset of <math>\{1,2,3,4,5,6,7\}</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  
[[2018 AIME I Problems/Problem 6 | Solution]]
+
'''Note''': <math>|S|</math> represents the number of elements in the set <math>S</math>.
 +
 
 +
[[2010 AIME I Problems/Problem 7|Solution]]
 
== Problem 7 ==
 
== Problem 7 ==
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>.
+
At each of the sixteen circles in the network below stands a student. A total of <math>3360</math> coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.
 +
 
 +
<center><asy>
 +
import cse5;
 +
unitsize(6mm);
 +
defaultpen(linewidth(.8pt));
 +
dotfactor = 8;
 +
pathpen=black;
 +
 
 +
pair A = (0,0);
 +
pair B = 2*dir(54), C = 2*dir(126), D = 2*dir(198), E = 2*dir(270), F = 2*dir(342);
 +
pair G = 3.6*dir(18), H = 3.6*dir(90), I = 3.6*dir(162), J = 3.6*dir(234), K = 3.6*dir(306);
 +
pair M = 6.4*dir(54), N = 6.4*dir(126), O = 6.4*dir(198), P = 6.4*dir(270), L = 6.4*dir(342);
 +
pair[] dotted = {A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P};
 +
 
 +
D(A--B--H--M);
 +
D(A--C--H--N);
 +
D(A--F--G--L);
 +
D(A--E--K--P);
 +
D(A--D--J--O);
 +
D(B--G--M);
 +
D(F--K--L);
 +
D(E--J--P);
 +
D(O--I--D);
 +
D(C--I--N);
 +
D(L--M--N--O--P--L);
 +
 
 +
dot(dotted);
  
[[2001 AIME I Problems/Problem 7|Solution]]
+
</asy></center>
== 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 <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
 
  
[[2001 AIME II Problems/Problem 9|Solution]]
+
[[2012 AIME I Problems/Problem 7|Solution]]
== Problem 9 ==
+
== Problem 11 ==
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>
+
Ms. Math's kindergarten class has 16 registered students. The classroom has a very large number, ''N'', of play blocks which satisfies the conditions:
  
[[2004 AIME I Problems/Problem 9|Solution]]
+
(a) If 16, 15, or 14 students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
 +
 
 +
(b) There are three integers <math>0 < x < y < z < 14</math> such that when <math>x</math>, <math>y</math>, or <math>z</math> students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
 +
 
 +
Find the sum of the distinct prime divisors of the least possible value of ''N'' satisfying the above conditions.
 +
 
 +
[[2013 AIME I Problems/Problem 11|Solution]]
 
== Problem 12 ==
 
== 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 <math>8: 7</math>. Find the minimum possible value of their common perimeter.
+
Let <math>\bigtriangleup PQR</math> be a triangle with <math>\angle P = 75^o</math> and <math>\angle Q = 60^o</math>. A regular hexagon <math>ABCDEF</math> with side length 1 is drawn inside <math>\triangle PQR</math> so that side <math>\overline{AB}</math> lies on <math>\overline{PQ}</math>, side <math>\overline{CD}</math> lies on <math>\overline{QR}</math>, and one of the remaining vertices lies on <math>\overline{RP}</math>. There are positive integers <math>a, b, c, </math> and <math>d</math> such that the area of <math>\triangle PQR</math> can be expressed in the form <math>\frac{a+b\sqrt{c}}{d}</math>, where <math>a</math> and <math>d</math> are relatively prime, and c is not divisible by the square of any prime. Find <math>a+b+c+d</math>.
  
[[2010 AIME II Problems/Problem 12|Solution]]
+
[[2013 AIME I Problems/Problem 12|Solution]]
== Problem 12 ==
+
==Problem 11==
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.
+
Let <math>A = \{1, 2, 3, 4, 5, 6, 7\}</math>, and let <math>N</math> be the number of functions <math>f</math> from set <math>A</math> to set <math>A</math> such that <math>f(f(x))</math> is a constant function. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 +
 
 +
[[2013 AIME II Problems/Problem 11|Solution]]
 +
==Problem 11==
 +
In <math>\triangle RED</math>, <math>\measuredangle DRE=75^{\circ}</math> and <math>\measuredangle RED=45^{\circ}</math>. <math> RD=1</math>. Let <math>M</math> be the midpoint of segment <math>\overline{RD}</math>. Point <math>C</math> lies on side <math>\overline{ED}</math> such that <math>\overline{RC}\perp\overline{EM}</math>. Extend segment <math>\overline{DE}</math> through <math>E</math> to point <math>A</math> such that <math>CA=AR</math>. Then <math>AE=\frac{a-\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer. Find <math>a+b+c</math>.
 +
 
 +
[[2014 AIME II Problems/Problem 11|Solution]]
 +
==Problem 15==
 +
Let <math>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions (a) <math>0\le A<B<C\le99</math>, (b) there exist integers <math>a</math>, <math>b</math>, and <math>c</math>, and prime <math>p</math> where <math>0\le b<a<c<p</math>, (c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and (d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</math>.
  
[[2011 AIME I Problems/Problem 12|Solution]]
+
[[2013 AIME I Problems/Problem 15|Solution]]
== Problem 10 ==
 
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.
 
  
[[2011 AIME II Problems/Problem 10|Solution]]
+
{{AIME box|year=2013|n=I|before=[[2012 AIME II Problems]]|after=[[2013 AIME II Problems]]}}
== Problem 11 ==
 
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>
 
  
[[2012 AIME I Problems/Problem 11|Solution]]
+
{{MAA Notice}}
== Problem 14 ==
+
==Problem 14==
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>.
+
For positive integers <math>n</math> and <math>k</math>, let <math>f(n, k)</math> be the remainder when <math>n</math> is divided by <math>k</math>, and for <math>n > 1</math> let <math>F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)</math>. Find the remainder when <math>\sum\limits_{n=20}^{100} F(n)</math> is divided by <math>1000</math>.
  
[[2011 AIME I Problems/Problem 14|Solution]]
+
[[2013 AIME II Problems/Problem 14|Solution]]
== Problem 15 ==
+
==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>.
+
Let <math>A,B,C</math> be angles of an acute triangle with
 +
<cmath> \begin{align*}
 +
\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\
 +
\cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9}
 +
\end{align*} </cmath>
 +
There are positive integers <math>p</math>, <math>q</math>, <math>r</math>, and <math>s</math> for which <cmath> \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s}, </cmath> where <math>p+q</math> and <math>s</math> are relatively prime and <math>r</math> is not divisible by the square of any prime. Find <math>p+q+r+s</math>.
  
[[2011 AIME II Problems/Problem 15|Solution]]
+
[[2013 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>.
 
  
[[2012 AIME II Problems/Problem 13|Solution]]
+
{{AIME box|year=2013|n=II|before=[[2013 AIME I Problems]]|after=[[2014 AIME I Problems]]}}
== Problem 14 ==
 
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>.
 
  
[[2012 AIME II Problems/Problem 14|Solution]]
+
{{MAA Notice}}
== Problem 14 ==
+
==Problem 13==
For <math>\pi \le \theta < 2\pi</math>, let
+
On square <math>ABCD</math>, points <math>E,F,G</math>, and <math>H</math> lie on sides <math>\overline{AB},\overline{BC},\overline{CD},</math> and <math>\overline{DA},</math> respectively, so that <math>\overline{EG} \perp \overline{FH}</math> and <math>EG=FH = 34</math>. Segments <math>\overline{EG}</math> and <math>\overline{FH}</math> intersect at a point <math>P</math>, and the areas of the quadrilaterals <math>AEPH, BFPE, CGPF,</math> and <math>DHPG</math> are in the ratio <math>269:275:405:411.</math> Find the area of square <math>ABCD</math>.
  
<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
+
<asy>
</cmath>
+
pair A = (0,sqrt(850));
 +
pair B = (0,0);
 +
pair C = (sqrt(850),0);
 +
pair D = (sqrt(850),sqrt(850));
 +
draw(A--B--C--D--cycle);
 +
dotfactor = 3;
 +
dot("$A$",A,dir(135));
 +
dot("$B$",B,dir(215));
 +
dot("$C$",C,dir(305));
 +
dot("$D$",D,dir(45));
 +
pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850));
 +
pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);
 +
dot("$H$",H,dir(90));
 +
dot("$F$",F,dir(270));
 +
draw(H--F);
 +
pair E = (0,(sqrt(850)-6)/2);
 +
pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2);
 +
dot("$E$",E,dir(180));
 +
dot("$G$",G,dir(0));
 +
draw(E--G);
 +
pair P = extension(H,F,E,G);
 +
dot("$P$",P,dir(60));
 +
label("$w$", intersectionpoint( A--P, E--H ));
 +
label("$x$", intersectionpoint( B--P, E--F ));
 +
label("$y$", intersectionpoint( C--P, G--F ));
 +
label("$z$", intersectionpoint( D--P, G--H ));</asy>
  
and
+
[[2014 AIME I Problems/Problem 13|Solution]]
 +
==Problem 15==
 +
In <math>\triangle ABC, AB = 3, BC = 4,</math> and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B, \overline{BC}</math> at <math>B</math> and <math>D,</math> and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4},</math> length <math>DE=\frac{a\sqrt{b}}{c},</math> where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer not divisible by the square of any prime. Find <math>a+b+c</math>.
  
<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
+
[[2014 AIME I Problems/Problem 15|Solution]]
+\ldots </cmath>
 
  
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>.
+
{{AIME box|year=2014|n=I|before=[[2013 AIME II Problems]]|after=[[2014 AIME II Problems]]}}
  
[[2013 AIME I Problems/Problem 14|Solution]]
+
{{MAA Notice}}

Revision as of 16:38, 29 May 2020

Here's the AIME compilation I will be doing:


Problem 3

Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.

Solution

Problem 4

In equiangular octagon $CAROLINE$, $CA = RO = LI = NE =$ $\sqrt{2}$ and $AR = OL = IN = EC = 1$. The self-intersecting octagon $CORNELIA$ enclosed six non-overlapping triangular regions. Let $K$ be the area enclosed by $CORNELIA$, that is, the total area of the six triangular regions. Then $K = \frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find $a + b$.

Solution

Problem 5

Suppose that $x$, $y$, and $z$ are complex numbers such that $xy = -80 - 320i$, $yz = 60$, and $zx = -96 + 24i$, where $i$ $=$ $\sqrt{-1}$. Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$. Find $a^2 + b^2$.

Solution

Problem 7

Let $S_i$ be the set of all integers $n$ such that $100i\leq n < 100(i + 1)$. For example, $S_4$ is the set ${400,401,402,\ldots,499}$. How many of the sets $S_0, S_1, S_2, \ldots, S_{999}$ do not contain a perfect square?

Solution

Problem 7

Define an ordered triple $(A, B, C)$ of sets to be $\textit{minimally intersecting}$ if $|A \cap B| = |B \cap C| = |C \cap A| = 1$ and $A \cap B \cap C = \emptyset$. For example, $(\{1,2\},\{2,3\},\{1,3,4\})$ is a minimally intersecting triple. Let $N$ be the number of minimally intersecting ordered triples of sets for which each set is a subset of $\{1,2,3,4,5,6,7\}$. Find the remainder when $N$ is divided by $1000$.

Note: $|S|$ represents the number of elements in the set $S$.

Solution

Problem 7

At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.

[asy] import cse5; unitsize(6mm); defaultpen(linewidth(.8pt)); dotfactor = 8; pathpen=black;  pair A = (0,0); pair B = 2*dir(54), C = 2*dir(126), D = 2*dir(198), E = 2*dir(270), F = 2*dir(342); pair G = 3.6*dir(18), H = 3.6*dir(90), I = 3.6*dir(162), J = 3.6*dir(234), K = 3.6*dir(306); pair M = 6.4*dir(54), N = 6.4*dir(126), O = 6.4*dir(198), P = 6.4*dir(270), L = 6.4*dir(342); pair[] dotted = {A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P};  D(A--B--H--M); D(A--C--H--N); D(A--F--G--L); D(A--E--K--P); D(A--D--J--O); D(B--G--M); D(F--K--L); D(E--J--P); D(O--I--D); D(C--I--N); D(L--M--N--O--P--L);  dot(dotted);  [/asy]

Solution

Problem 11

Ms. Math's kindergarten class has 16 registered students. The classroom has a very large number, N, of play blocks which satisfies the conditions:

(a) If 16, 15, or 14 students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and

(b) There are three integers $0 < x < y < z < 14$ such that when $x$, $y$, or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.

Find the sum of the distinct prime divisors of the least possible value of N satisfying the above conditions.

Solution

Problem 12

Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^o$ and $\angle Q = 60^o$. A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$, side $\overline{CD}$ lies on $\overline{QR}$, and one of the remaining vertices lies on $\overline{RP}$. There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$, where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$.

Solution

Problem 11

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$, and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$.

Solution

Problem 11

In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\frac{a-\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$.

Solution

Problem 15

Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions (a) $0\le A<B<C\le99$, (b) there exist integers $a$, $b$, and $c$, and prime $p$ where $0\le b<a<c<p$, (c) $p$ divides $A-a$, $B-b$, and $C-c$, and (d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$.

Solution

2013 AIME I (ProblemsAnswer KeyResources)
Preceded by
2012 AIME II Problems
Followed by
2013 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

Problem 14

For positive integers $n$ and $k$, let $f(n, k)$ be the remainder when $n$ is divided by $k$, and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$. Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$.

Solution

Problem 15

Let $A,B,C$ be angles of an acute triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$, $q$, $r$, and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$.

Solution

2013 AIME II (ProblemsAnswer KeyResources)
Preceded by
2013 AIME I Problems
Followed by
2014 AIME I 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

Problem 13

On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$.

[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy]

Solution

Problem 15

In $\triangle ABC, AB = 3, BC = 4,$ and $CA = 5$. Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B, \overline{BC}$ at $B$ and $D,$ and $\overline{AC}$ at $F$ and $G$. Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4},$ length $DE=\frac{a\sqrt{b}}{c},$ where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$.

Solution

2014 AIME I (ProblemsAnswer KeyResources)
Preceded by
2013 AIME II Problems
Followed by
2014 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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