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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==
+
==Problem 3==
−
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.  
+
A triangle has vertices <math>A(0,0)</math>, <math>B(12,0)</math>, and <math>C(8,10)</math>. The probability that a randomly chosen point inside the triangle is closer to vertex <math>B</math> than to either vertex <math>A</math> or vertex <math>C</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
  
−
<asy>
+
[[2017 AIME II Problems/Problem 3 | Solution]]
−
size(200);
+
== Problem 4 ==
−
defaultpen(linewidth(0.7));
+
Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are <math>60</math>, <math>84</math>, and <math>140</math> years.  The three planets and the star are currently collinear.  What is the fewest number of years from now that they will all be collinear again?
−
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]]
 
  
 +
[[2007 AIME I Problems/Problem 4|Solution]]
 
==Problem 5==
 
==Problem 5==
 +
5. If
  
−
Compute, to the nearest integer, the area of the region enclosed by the graph of
+
<cmath>\frac{1}{0!10!}+\frac{1}{1!9!}+\frac{1}{2!8!}+\frac{1}{3!7!}+\frac{1}{4!6!}+\frac{1}{5!5!}</cmath>
  
−
<cmath>13x^2-20xy+52y^2-10x+52y=563</cmath>
+
is written as a common fraction reduced to lowest terms, the result is <math>\frac{m}{n}</math>. Compute the sum of the prime divisors of <math>m</math> plus the sum of the prime divisors of <math>n</math>.
  
−
==Problem 6==
+
==Problem 9==
−
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>.
+
Let <math>a_{10} = 10</math>, and for each integer <math>n >10</math> let <math>a_n = 100a_{n - 1} + n</math>. Find the least <math>n > 10</math> such that <math>a_n</math> is a multiple of <math>99</math>.
  
−
[[2020 AIME I Problems/Problem 6 | Solution]]
+
[[2017 AIME I Problems/Problem 9 | Solution]]
−
== Problem 9 ==
+
==Problem 8==
−
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>.
+
Two real numbers <math>a</math> and <math>b</math> are chosen independently and uniformly at random from the interval <math>(0, 75)</math>. Let <math>O</math> and <math>P</math> be two points on the plane with <math>OP = 200</math>. Let <math>Q</math> and <math>R</math> be on the same side of line <math>OP</math> such that the degree measures of <math>\angle POQ</math> and <math>\angle POR</math> are <math>a</math> and <math>b</math> respectively, and <math>\angle OQP</math> and <math>\angle ORP</math> are both right angles. The probability that <math>QR \leq 100</math> is equal to <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
  
−
[[2012 AIME II Problems/Problem 9|Solution]]
+
[[2017 AIME I Problems/Problem 8 | Solution]]
−
==Problem 9==
+
==Problem 7==
−
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>.
 
  
−
[[2014 AIME I Problems/Problem 9|Solution]]
+
Triangle <math>ABC</math> has side lengths <math>AB = 9</math>, <math>BC =</math> <math>5\sqrt{3}</math>, and <math>AC = 12</math>. Points <math>A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B</math> are on segment <math>\overline{AB}</math> with <math>P_{k}</math> between <math>P_{k-1}</math> and <math>P_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>, and points <math>A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C</math> are on segment <math>\overline{AC}</math> with <math>Q_{k}</math> between <math>Q_{k-1}</math> and <math>Q_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>. Furthermore, each segment <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2449</math>, is parallel to <math>\overline{BC}</math>. The segments cut the triangle into <math>2450</math> regions, consisting of <math>2449</math> trapezoids and <math>1</math> triangle. Each of the <math>2450</math> regions has the same area. Find the number of segments <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2450</math>, that have rational length.
−
==Problem 9==
 
−
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>.
 
  
−
[[2015 AIME I Problems/Problem 9|Solution]]
+
[[2018 AIME II Problems/Problem 7 | Solution]]
 +
==Problem 10==
  
−
-------------
+
Find the number of functions <math>f(x)</math> from <math>\{1, 2, 3, 4, 5\}</math> to <math>\{1, 2, 3, 4, 5\}</math> that satisfy <math>f(f(x)) = f(f(f(x)))</math> for all <math>x</math> in <math>\{1, 2, 3, 4, 5\}</math>.
−
==Problem 12==
 
−
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>.  
 
  
−
[[2014 AIME II Problems/Problem 12|Solution]]
+
[[2018 AIME II Problems/Problem 10 | Solution]]
 
==Problem 11==
 
==Problem 11==
−
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>.
 
  
−
[[2017 AIME I Problems/Problem 11 | Solution]]
+
Find the number of permutations of <math>1, 2, 3, 4, 5, 6</math> such that for each <math>k</math> with <math>1</math> <math>\leq</math> <math>k</math> <math>\leq</math> <math>5</math>, at least one of the first <math>k</math> terms of the permutation is greater than <math>k</math>.
−
==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 <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>.
 
  
−
<asy>
+
[[2018 AIME II Problems/Problem 11 | Solution]]
−
size(6cm);
+
==Problem 14==
  
−
draw(unitcircle);
+
The incircle <math>\omega</math> of triangle <math>ABC</math> is tangent to <math>\overline{BC}</math> at <math>X</math>. Let <math>Y \neq X</math> be the other intersection of <math>\overline{AX}</math> with <math>\omega</math>. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>\overline{PQ}</math> is tangent to <math>\omega</math> at <math>Y</math>. Assume that <math>AP = 3</math>, <math>PB = 4</math>, <math>AC = 8</math>, and <math>AQ = \dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
−
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));
+
[[2018 AIME II Problems/Problem 14 | Solution]]
−
dot(1 * dir(162), linewidth(5));
+
== Problem 10 ==
−
dot(1 * dir(234), linewidth(5));
+
Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac{p\sqrt{r}}{q}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p+q+r</math>.
−
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));
+
[[2009 AIME II Problems/Problem 10|Solution]]
−
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>
 
  
−
[[2018 AIME I Problems/Problem 10 | Solution]]
 
 
==Problem 11==
 
==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>.
+
<math>10</math> lines and <math>10</math> circles divide the plane into at most <math>n</math> disjoint regions. Compute <math>n</math>.
  
 +
==Problem 15==
  
−
[[2018 AIME I Problems/Problem 11 | Solution]]
+
Find the number of functions <math>f</math> from <math>\{0, 1, 2, 3, 4, 5, 6\}</math> to the integers such that <math>f(0) = 0</math>, <math>f(6) = 12</math>, and
−
==Problem 14==
 
  
−
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>.
+
<cmath>|x - y|  \leq  |f(x) - f(y)|  \leq  3|x - y|</cmath>
  
−
[[2014 AIME II Problems/Problem 14|Solution]]
+
for all <math>x</math> and <math>y</math> in <math>\{0, 1, 2, 3, 4, 5, 6\}</math>.
−
==Problem 14==
 
  
−
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.
+
[[2018 AIME II Problems/Problem 15 | Solution]]
  
−
[[2015 AIME I Problems/Problem 14|Solution]]
+
== Problem 14 ==
 +
The sequence <math>(a_n)</math> satisfies <math>a_0=0</math> and <math>a_{n + 1} = \frac{8}{5}a_n + \frac{6}{5}\sqrt{4^n - a_n^2}</math> for <math>n \geq 0</math>. Find the greatest integer less than or equal to <math>a_{10}</math>.
  
−
-------------
+
[[2009 AIME II Problems/Problem 14|Solution]]
−
==Problem 13==
+
== Problem 15 ==
−
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>.
+
Let <math>\overline{MN}</math> be a diameter of a circle with diameter <math>1</math>. Let <math>A</math> and <math>B</math> be points on one of the semicircular arcs determined by <math>\overline{MN}</math> such that <math>A</math> is the midpoint of the semicircle and <math>MB=\dfrac 35</math>. Point <math>C</math> lies on the other semicircular arc. Let <math>d</math> be the length of the line segment whose endpoints are the intersections of diameter <math>\overline{MN}</math> with the chords <math>\overline{AC}</math> and <math>\overline{BC}</math>. The largest possible value of <math>d</math> can be written in the form <math>r-s\sqrt t</math>, where <math>r</math>, <math>s</math>, and <math>t</math> are positive integers and <math>t</math> is not divisible by the square of any prime. Find <math>r+s+t</math>.
  
−
[[2018 AIME I Problems/Problem 13 | Solution]]
+
[[2009 AIME II Problems/Problem 15|Solution]]
−
==Problem 15==
+
==Problem 14==
−
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>.
 
−
 
 
−
[[2018 AIME I Problems/Problem 15 | 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 <math>\dfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+
Let <math>x</math> and <math>y</math> be real numbers satisfying <math>x^4y^5+y^4x^5=810</math> and <math>x^3y^6+y^3x^6=945</math>. Evaluate <math>2x^3+(xy)^3+2y^3</math>.
  
−
[[2018 AIME II Problems/Problem 13 | Solution]]
+
[[2015 AIME II Problems/Problem 14 | Solution]]

Revision as of 15:02, 31 May 2020

Here's the AIME compilation I will be doing:

Problem 3

A triangle has vertices $A(0,0)$, $B(12,0)$, and $C(8,10)$. The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

Problem 4

Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are $60$, $84$, and $140$ years. The three planets and the star are currently collinear. What is the fewest number of years from now that they will all be collinear again?

Solution

Problem 5

5. If

\[\frac{1}{0!10!}+\frac{1}{1!9!}+\frac{1}{2!8!}+\frac{1}{3!7!}+\frac{1}{4!6!}+\frac{1}{5!5!}\]

is written as a common fraction reduced to lowest terms, the result is $\frac{m}{n}$. Compute the sum of the prime divisors of $m$ plus the sum of the prime divisors of $n$.

Problem 9

Let $a_{10} = 10$, and for each integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least $n > 10$ such that $a_n$ is a multiple of $99$.

Solution

Problem 8

Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$. Let $O$ and $P$ be two points on the plane with $OP = 200$. Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Problem 7

Triangle $ABC$ has side lengths $AB = 9$, $BC =$ $5\sqrt{3}$, and $AC = 12$. Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$, and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_{k}$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, ..., 2449$. Furthermore, each segment $\overline{P_{k}Q_{k}}$, $k = 1, 2, ..., 2449$, is parallel to $\overline{BC}$. The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_{k}Q_{k}}$, $k = 1, 2, ..., 2450$, that have rational length.

Solution

Problem 10

Find the number of functions $f(x)$ from $\{1, 2, 3, 4, 5\}$ to $\{1, 2, 3, 4, 5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1, 2, 3, 4, 5\}$.

Solution

Problem 11

Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$, at least one of the first $k$ terms of the permutation is greater than $k$.

Solution

Problem 14

The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$. Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$. Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$, respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$. Assume that $AP = 3$, $PB = 4$, $AC = 8$, and $AQ = \dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Problem 10

Four lighthouses are located at points $A$, $B$, $C$, and $D$. The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$, the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$, and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$. To an observer at $A$, the angle determined by the lights at $B$ and $D$ and the angle determined by the lights at $C$ and $D$ are equal. To an observer at $C$, the angle determined by the lights at $A$ and $B$ and the angle determined by the lights at $D$ and $B$ are equal. The number of kilometers from $A$ to $D$ is given by $\frac{p\sqrt{r}}{q}$, where $p$, $q$, and $r$ are relatively prime positive integers, and $r$ is not divisible by the square of any prime. Find $p+q+r$.

Solution

Problem 11

$10$ lines and $10$ circles divide the plane into at most $n$ disjoint regions. Compute $n$.

Problem 15

Find the number of functions $f$ from $\{0, 1, 2, 3, 4, 5, 6\}$ to the integers such that $f(0) = 0$, $f(6) = 12$, and

\[|x - y| \leq |f(x) - f(y)| \leq 3|x - y|\]

for all $x$ and $y$ in $\{0, 1, 2, 3, 4, 5, 6\}$.

Solution

Problem 14

The sequence $(a_n)$ satisfies $a_0=0$ and $a_{n + 1} = \frac{8}{5}a_n + \frac{6}{5}\sqrt{4^n - a_n^2}$ for $n \geq 0$. Find the greatest integer less than or equal to $a_{10}$.

Solution

Problem 15

Let $\overline{MN}$ be a diameter of a circle with diameter $1$. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\dfrac 35$. Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with the chords $\overline{AC}$ and $\overline{BC}$. The largest possible value of $d$ can be written in the form $r-s\sqrt t$, where $r$, $s$, and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$.

Solution

Problem 14

Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$. Evaluate $2x^3+(xy)^3+2y^3$.

Solution

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