Difference between revisions of "User:Rowechen"

Line 1: Line 1:
 
Here's the AIME compilation I will be doing:
 
Here's the AIME compilation I will be doing:
  
 +
==Problem 1==
 +
The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.
  
== Problem 3 ==
+
<asy>
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.
+
size(200);
 +
defaultpen(linewidth(0.7));
 +
path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin;
 +
path laceR=reflect((75,0),(75,-240))*laceL;
 +
draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray);
 +
for(int i=0;i<=3;i=i+1)
 +
{
 +
path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5);
 +
unfill(circ1); draw(circ1);
 +
unfill(circ2); draw(circ2);
 +
}
 +
draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));</asy>
 +
[[2014 AIME I Problems/Problem 1|Solution]]
  
[[2012 AIME I Problems/Problem 3|Solution]]
+
Compute, to the nearest integer, the area of the region enclosed by the graph of
==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>.
+
<cmath>13x^2-20xy+52y^2-10x+52y=563</cmath>
  
[[2018 AIME II Problems/Problem 4 | Solution]]
+
==Problem 6==
==Problem 5==
+
A flat board has a circular hole with radius <math>1</math> and a circular hole with radius <math>2</math> such that the distance between the centers of the two holes is <math>7</math>. Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</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>.
+
[[2020 AIME I Problems/Problem 6 | Solution]]
 +
== Problem 9 ==
 +
Let <math>x</math> and <math>y</math> be real numbers such that <math>\frac{\sin x}{\sin y} = 3</math> and <math>\frac{\cos x}{\cos y} = \frac12</math>. The value of <math>\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}</math> can be expressed in the form <math>\frac pq</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
  
[[2018 AIME II Problems/Problem 5 | Solution]]
+
[[2012 AIME II Problems/Problem 9|Solution]]
== Problem 7 ==
+
==Problem 9==
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?
+
Let <math>x_1< x_2 < x_3</math> be the three real roots of the equation <math>\sqrt{2014} x^3 - 4029x^2 + 2 = 0</math>. Find <math>x_2(x_1+x_3)</math>.
  
[[2008 AIME I Problems/Problem 7|Solution]]
+
[[2014 AIME I Problems/Problem 9|Solution]]
== Problem 7 ==
+
==Problem 9==
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>.
+
Let <math>S</math> be the set of all ordered triple of integers <math>(a_1,a_2,a_3)</math> with <math>1 \le a_1,a_2,a_3 \le 10</math>. Each ordered triple in <math>S</math> generates a sequence according to the rule <math>a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |</math> for all <math>n\ge 4</math>. Find the number of such sequences for which <math>a_n=0</math> for some <math>n</math>.
  
'''Note''': <math>|S|</math> represents the number of elements in the set <math>S</math>.
+
[[2015 AIME I Problems/Problem 9|Solution]]
  
[[2010 AIME I Problems/Problem 7|Solution]]
+
-------------
== Problem 7 ==
+
==Problem 12==
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.
+
Suppose that the angles of <math>\triangle ABC</math> satisfy <math>\cos(3A)+\cos(3B)+\cos(3C)=1</math>. Two sides of the triangle have lengths 10 and 13. There is a positive integer <math>m</math> so that the maximum possible length for the remaining side of <math>\triangle ABC</math> is <math>\sqrt{m}</math>. Find <math>m</math>.  
  
<center><asy>
+
[[2014 AIME II Problems/Problem 12|Solution]]
import cse5;
+
==Problem 11==
unitsize(6mm);
+
Consider arrangements of the <math>9</math> numbers <math>1, 2, 3, \dots, 9</math> in a <math>3 \times 3</math> array. For each such arrangement, let <math>a_1</math>, <math>a_2</math>, and <math>a_3</math> be the medians of the numbers in rows <math>1</math>, <math>2</math>, and <math>3</math> respectively, and let <math>m</math> be the median of <math>\{a_1, a_2, a_3\}</math>. Let <math>Q</math> be the number of arrangements for which <math>m = 5</math>. Find the remainder when <math>Q</math> is divided by <math>1000</math>.
defaultpen(linewidth(.8pt));
 
dotfactor = 8;
 
pathpen=black;
 
  
pair A = (0,0);
+
[[2017 AIME I Problems/Problem 11 | Solution]]
pair B = 2*dir(54), C = 2*dir(126), D = 2*dir(198), E = 2*dir(270), F = 2*dir(342);
+
==Problem 10==
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);
+
The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point <math>A</math>. At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path <math>AJABCHCHIJA</math>, which has <math>10</math> steps. Let <math>n</math> be the number of paths with <math>15</math> steps that begin and end at point <math>A.</math> Find the remainder when <math>n</math> is divided by <math>1000</math>.
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);
+
<asy>
D(A--C--H--N);
+
size(6cm);
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);
+
draw(unitcircle);
 +
draw(scale(2) * unitcircle);
 +
for(int d = 90; d < 360 + 90; d += 72){
 +
draw(2 * dir(d) -- dir(d));
 +
}
  
</asy></center>
+
dot(1 * dir( 90), linewidth(5));
 +
dot(1 * dir(162), linewidth(5));
 +
dot(1 * dir(234), linewidth(5));
 +
dot(1 * dir(306), linewidth(5));
 +
dot(1 * dir(378), linewidth(5));
 +
dot(2 * dir(378), linewidth(5));
 +
dot(2 * dir(306), linewidth(5));
 +
dot(2 * dir(234), linewidth(5));
 +
dot(2 * dir(162), linewidth(5));
 +
dot(2 * dir( 90), linewidth(5));
  
[[2012 AIME I Problems/Problem 7|Solution]]
+
label("$A$", 1 * dir( 90), -dir( 90));
== Problem 11 ==
+
label("$B$", 1 * dir(162), -dir(162));
Ms. Math's kindergarten class has 16 registered students. The classroom has a very large number, ''N'', of play blocks which satisfies the conditions:
+
label("$C$", 1 * dir(234), -dir(234));
 +
label("$D$", 1 * dir(306), -dir(306));
 +
label("$E$", 1 * dir(378), -dir(378));
 +
label("$F$", 2 * dir(378), dir(378));
 +
label("$G$", 2 * dir(306), dir(306));
 +
label("$H$", 2 * dir(234), dir(234));
 +
label("$I$", 2 * dir(162), dir(162));
 +
label("$J$", 2 * dir( 90), dir( 90));
 +
</asy>
  
(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
+
[[2018 AIME I Problems/Problem 10 | Solution]]
 +
==Problem 11==
 +
Find the least positive integer <math>n</math> such that when <math>3^n</math> is written in base <math>143</math>, its two right-most digits in base <math>143</math> are <math>01</math>.
  
(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.
+
[[2018 AIME I Problems/Problem 11 | Solution]]
 +
==Problem 14==
  
[[2013 AIME I Problems/Problem 11|Solution]]
+
In <math>\triangle ABC</math>, <math>AB=10</math>, <math>\measuredangle A=30^{\circ}</math>, and <math>\measuredangle C=45^{\circ}</math>. Let <math>H</math>, <math>D</math>, and <math>M</math> be points on line <math>\overline{BC}</math> such that <math>AH\perp BC</math>, <math>\measuredangle BAD=\measuredangle CAD</math>, and <math>BM=CM</math>. Point <math>N</math> is the midpoint of segment <math>HM</math>, and point <math>P</math> is on ray <math>AD</math> such that <math>PN\perp BC</math>. Then <math>AP^2=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
== Problem 12 ==
 
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>.
 
  
[[2013 AIME I Problems/Problem 12|Solution]]
+
[[2014 AIME II Problems/Problem 14|Solution]]
==Problem 11==
+
==Problem 14==
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]]
+
For each integer <math>n \ge 2</math>, let <math>A(n)</math> be the area of the region in the coordinate plane defined by the inequalities <math>1\le x \le n</math> and <math>0\le y \le x \left\lfloor \sqrt x \right\rfloor</math>, where <math>\left\lfloor \sqrt x \right\rfloor</math> is the greatest integer not exceeding <math>\sqrt x</math>. Find the number of values of <math>n</math> with <math>2\le n \le 1000</math> for which <math>A(n)</math> is an integer.
==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]]
+
[[2015 AIME I Problems/Problem 14|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>.
 
  
[[2013 AIME I Problems/Problem 15|Solution]]
+
-------------
 
+
==Problem 13==
==Problem 14==
+
Let <math>\triangle ABC</math> have side lengths <math>AB=30</math>, <math>BC=32</math>, and <math>AC=34</math>. Point <math>X</math> lies in the interior of <math>\overline{BC}</math>, and points <math>I_1</math> and <math>I_2</math> are the incenters of <math>\triangle ABX</math> and <math>\triangle ACX</math>, respectively. Find the minimum possible area of <math>\triangle AI_1I_2</math> as <math>X</math> varies along <math>\overline{BC}</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>.
 
  
[[2013 AIME II Problems/Problem 14|Solution]]
+
[[2018 AIME I Problems/Problem 13 | Solution]]
 
==Problem 15==
 
==Problem 15==
Let <math>A,B,C</math> be angles of an acute triangle with
+
David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, <math>A,\text{ }B,\text{ }C</math>, which can each be inscribed in a circle with radius <math>1</math>. Let <math>\varphi_A</math> denote the measure of the acute angle made by the diagonals of quadrilateral <math>A</math>, and define <math>\varphi_B</math> and <math>\varphi_C</math> similarly. Suppose that <math>\sin\varphi_A=\frac{2}{3}</math>, <math>\sin\varphi_B=\frac{3}{5}</math>, and <math>\sin\varphi_C=\frac{6}{7}</math>. All three quadrilaterals have the same area <math>K</math>, which can be written in the form <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
<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>.
 
  
[[2013 AIME II Problems/Problem 15|Solution]]
+
[[2018 AIME I Problems/Problem 15 | Solution]]
  
 
==Problem 13==
 
==Problem 13==
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>.
 
  
<asy>
+
Misha rolls a standard, fair six-sided die until she rolls 1-2-3 in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is <math>\dfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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>
 
 
 
[[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>.
 
  
[[2014 AIME I Problems/Problem 15|Solution]]
+
[[2018 AIME II Problems/Problem 13 | Solution]]

Revision as of 08:00, 30 May 2020

Here's the AIME compilation I will be doing:

Problem 1

The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.

[asy] size(200); defaultpen(linewidth(0.7)); path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin; path laceR=reflect((75,0),(75,-240))*laceL; draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray); for(int i=0;i<=3;i=i+1) { path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5); unfill(circ1); draw(circ1); unfill(circ2); draw(circ2); } draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));[/asy] Solution

Compute, to the nearest integer, the area of the region enclosed by the graph of

\[13x^2-20xy+52y^2-10x+52y=563\]

Problem 6

A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7$. Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Problem 9

Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$. The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

Problem 9

Let $x_1< x_2 < x_3$ be the three real roots of the equation $\sqrt{2014} x^3 - 4029x^2 + 2 = 0$. Find $x_2(x_1+x_3)$.

Solution

Problem 9

Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$.

Solution


Problem 12

Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1$. Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}$. Find $m$.

Solution

Problem 11

Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$, $a_2$, and $a_3$ be the medians of the numbers in rows $1$, $2$, and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$. Let $Q$ be the number of arrangements for which $m = 5$. Find the remainder when $Q$ is divided by $1000$.

Solution

Problem 10

The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$. At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path $AJABCHCHIJA$, which has $10$ steps. Let $n$ be the number of paths with $15$ steps that begin and end at point $A.$ Find the remainder when $n$ is divided by $1000$.

[asy] size(6cm);  draw(unitcircle); draw(scale(2) * unitcircle); for(int d = 90; d < 360 + 90; d += 72){ draw(2 * dir(d) -- dir(d)); }  dot(1 * dir( 90), linewidth(5)); dot(1 * dir(162), linewidth(5)); dot(1 * dir(234), linewidth(5)); dot(1 * dir(306), linewidth(5)); dot(1 * dir(378), linewidth(5)); dot(2 * dir(378), linewidth(5)); dot(2 * dir(306), linewidth(5)); dot(2 * dir(234), linewidth(5)); dot(2 * dir(162), linewidth(5)); dot(2 * dir( 90), linewidth(5));  label("$A$", 1 * dir( 90), -dir( 90)); label("$B$", 1 * dir(162), -dir(162)); label("$C$", 1 * dir(234), -dir(234)); label("$D$", 1 * dir(306), -dir(306)); label("$E$", 1 * dir(378), -dir(378)); label("$F$", 2 * dir(378), dir(378)); label("$G$", 2 * dir(306), dir(306)); label("$H$", 2 * dir(234), dir(234)); label("$I$", 2 * dir(162), dir(162)); label("$J$", 2 * dir( 90), dir( 90)); [/asy]

Solution

Problem 11

Find the least positive integer $n$ such that when $3^n$ is written in base $143$, its two right-most digits in base $143$ are $01$.


Solution

Problem 14

In $\triangle ABC$, $AB=10$, $\measuredangle A=30^{\circ}$, and $\measuredangle C=45^{\circ}$. Let $H$, $D$, and $M$ be points on line $\overline{BC}$ such that $AH\perp BC$, $\measuredangle BAD=\measuredangle CAD$, and $BM=CM$. Point $N$ is the midpoint of segment $HM$, and point $P$ is on ray $AD$ such that $PN\perp BC$. Then $AP^2=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Problem 14

For each integer $n \ge 2$, let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$, where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$. Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer.

Solution


Problem 13

Let $\triangle ABC$ have side lengths $AB=30$, $BC=32$, and $AC=34$. Point $X$ lies in the interior of $\overline{BC}$, and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$, respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$.

Solution

Problem 15

David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$, which can each be inscribed in a circle with radius $1$. Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$, and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\frac{2}{3}$, $\sin\varphi_B=\frac{3}{5}$, and $\sin\varphi_C=\frac{6}{7}$. All three quadrilaterals have the same area $K$, which can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Problem 13

Misha rolls a standard, fair six-sided die until she rolls 1-2-3 in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution