Difference between revisions of "Mock AIME 4 2005-2006/Problems"

Latest revision as of 03:44, 26 February 2007

http://www.mathlinks.ro/Forum/latexrender/pictures/2ea0ce047615395691113a82d6c190b3.gif

Problem 1

1. A 5-digit number is leet if and only if the sum of the first 2 digits, the sum of the last 2 digits and the middle digit are equal. How many 5-digit leet numbers exist? Solution

Problem 2

2. Qin Shi Huang wants to count the number of warriors he has to invade China. He puts his warriors into lines with the most people such that they have even length. The people left over are the remainder. He makes 2 lines, with a remainder of 1, 3 lines with a remainder of 2, 4 lines with a remainder of 3, 5 lines with a remainder of 4, and 6 lines with a remainder of 5. Find the minimum number of warriors that he has. Solution

Problem 3

3. $T_1$ is a regular tetrahedron. Tetrahedron $T_2$ is formed by connecting the centers of the faces of $T_1$ . Generally, a new tetrahedron $T_{n+1}$ is formed by connecting the centers of the faces of $T_n$ . $V_n$ is the volume of tetrahedron $T_n$ . $\frac{V_{2006}}{V_1}=\frac{m}{n}$ where $m$ and $n$ are coprime positive integers, find the remainder when $m+n$ is divided by $1000$ . Solution

Problem 4

4. Let $P(x)=\sum_{i=1}^{20}(-x)^{20-i}(x+i)^i$ . Let $K$ be the product of the roots. How many digits are does $\lfloor K \rfloor$ have where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ ? Solution

Problem 5

5. A parabola $P: y=x^2$ is rotated $135$ degrees clockwise about the origin to $P'$ . This image is translated upward $\frac{8+\sqrt{2}}{2}$ to $P''$ . Point $A: (0,0)$ , $B: (256,0)$ , and $C$ is in Quadrant I, on $P''$ . If the area of $\triangle ABC$ is at a maximum, it is $a\sqrt{b}+c$ where $a$ , $b$ and $c$ are integers and $b$ is square free, find $a+b+c$ . Solution

Problem 6

6. Define a sequence $a_0=2006$ and $a_{n+1}=(n+1)^{a_n}$ for all positive integers $n$ . Find the remainder when $a_{2007}$ is divided by $1000$ . Solution

Problem 7

7. $f(x)$ is a function that satisfies $3f(x)=2x+1-f(\frac{1}{1-x})$ for all defined $x$ . Suppose that the sum of the zeros of $f(x)=\frac{m}{n}$ where $m$ and $n$ are coprime positive integers, find $m^2+n^2$ . Solution

Problem 8

8. $R$ is a solution to $x+\frac{1}{x}=\frac{ \sin210^{\circ} }{\sin285^{\circ} }$ . Suppose that $\frac{1}{R^{2006}}+R^{2006}=A$ find $\lfloor A^{10} \rfloor$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ . Solution

Problem 9

9. Zeus, Athena, and Posideon arrive at Mount Olympus at a random time between 12:00 pm and 12:00 am, and stay for 3 hours. All three hours does not need to fall within 12 pm to 12 am. If any of the 2 gods see each other during 12 pm to 12 am, it will be a good day. The probability of it being a good day is $\frac{m}{n}$ where $m$ and $n$ are coprime positive integers, find $m+n$ . Solution

Problem 10

10. Define $S= \tan^2{1^{\circ}}+\tan^2{3^{\circ}}+\tan^2{5^{\circ}}+...+\tan^2{87^{\circ}}+\tan^2{89^{\circ}}$ . Find the remainder when $S$ is divided by $1000$ . Solution

Problem 11

11. $\triangle ABC$ is isosceles with $\angle C= 90^{\circ}$ . A point $P$ lies inside the triangle such that $AP=33$ , $CP=28\sqrt{2}$ , and $BP=65$ . Let $A$ be the area of $\triangle ABC$ . Find the remainder when $2A$ is divided by $1000$ . Solution

Problem 12

12. There exists a line $L$ with points $D$ , $E$ , $F$ with $E$ in between $D$ and $F$ . Point $A$ , not on the line is such that $\overline{AF}=6$ , $\overline{AD}=\frac{36}{7}$ , $\overline{AE}=\frac{12}{\sqrt{7}}$ with $\angle AEF > 90$ . Construct $E'$ on ray $AE$ such that $(\overline{AE})(\overline{AE'})=36$ and $\overline{FE'}=3$ . Point $G$ is on ray $AD$ such that $\overline{AG}=7$ . If $2*(\overline{E'G})=a+\sqrt{b}$ where $a$ and $b$ are integers, then find $a+b$ . Solution

Problem 13

13. $\triangle VA_0A_1$ is isosceles with base $\overline{{A_1A_0}}$ . Construct $A_2$ on segment $\overline{{A_0V}}$ such that $\overline{A_0A_1}=\overline{A_1A_2}=b$ . Construct $A_3$ on $\overline{A_1V}$ such that $b=\overline{A_2A_3}$ . Contiue this pattern: construct $\overline{A_{2n}A_{2n+1}}=b$ with $A_{2n+1}$ on segment $\overline{VA_1}$ and $\overline{A_{2n+1}A_{2n+2}}=b$ with $A_{2n+2}$ on segment $\overline{VA_0}$ . The points $A_n$ do not coincide and $\angle VA_1A_0=90-\frac{1}{2006}$ . Suppose $A_k$ is the last point you can construct on the perimeter of the triangle. Find the remainder when $k$ is divided by $1000$ . Solution

Problem 14

14. $P$ is the probability that if you flip a fair coin, $20$ heads will occur before $19$ tails. If $P=\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, find the remainder when $m+n$ is divided by $1000$ . Solution

Problem 15

15. A regular 61-gon with verticies $A_1$ , $A_2$ , $A_3$ ,... $A_{61}$ is inscribed in a circle with a radius of $r$ . Suppose $(\overline{A_1A_2})(\overline{A_1A_3})(\overline{A_1A_4})...(\overline{A_1A_{61}})=r$ . If $r^{2006}=\frac{p}{q}$ where $p$ and $q$ are coprime positive integers, find the remainder when $p+q$ is divided by $1000$ . Solution

@@ Line 1: / Line 1: @@
+http://www.mathlinks.ro/Forum/latexrender/pictures/2ea0ce047615395691113a82d6c190b3.gif
 == Problem 1 ==
-Suppose <math>n</math> is a positive integer. Let <math>f(n)</math> be the sum of the distinct positive prime divisors of <math>n</math> less than <math>50</math> (e.g. <math>f(12) = 2+3 = 5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</math>.
+. A 5-digit number is leet if and only if the sum of the first 2 digits, the sum of the last 2 digits and the middle digit are equal. How many 5-digit leet numbers exist?
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 1|Solution]]
 == Problem 2 ==
-A circle <math>\omega_1</math> of radius <math>6\sqrt{2}</math> is internally tangent to a larger circle <math>\omega_2</math> of radius <math>12\sqrt{2}</math> such that the center of <math>\omega_2</math> lies on <math>\omega_1</math>. A diameter <math>AB</math> of <math>\omega_2</math> is drawn tangent to <math>\omega_1</math>. A second line <math>l</math> is drawn from <math>B</math> tangent to <math>\omega_1</math>. Let the line tangent to <math>\omega_2</math> at <math>A</math> intersect <math>l</math> at <math>C</math>. Find the area of <math>\triangle ABC</math>.
+. Qin Shi Huang wants to count the number of warriors he has to invade China. He puts his warriors into lines with the most people such that they have even length. The people left over are the remainder. He makes 2 lines, with a remainder of 1, 3 lines with a remainder of 2, 4 lines with a remainder of 3, 5 lines with a remainder of 4, and 6 lines with a remainder of 5. Find the minimum number of warriors that he has.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 2|Solution]]
 == Problem 3 ==
-A <math>\emph hailstone</math> number <math>n = d_1d_2 \cdots d_k</math>, where <math>d_i</math> denotes the <math>i</math>th digit in the base-<math>10</math> representation of <math>n</math> for <math>i = 1,2, \ldots,k</math>, is a positive integer with distinct nonzero digits such that <math>d_m < d_{m-1}</math> if <math>m</math> is even and <math>d_m > d_{m-1}</math> if <math>m</math> is odd for <math>m = 1,2,\ldots,k</math> (and <math>d_0 = 0</math>). Let <math>a</math> be the number of four-digit hailstone numbers and <math>b</math> be the number of three-digit hailstone numbers. Find <math>a+b</math>.
+. <math>T_1</math> is a regular tetrahedron. Tetrahedron <math>T_2</math> is formed by connecting the centers of the faces of <math>T_1</math>. Generally, a new tetrahedron <math>T_{n+1}</math> is formed by connecting the centers of the faces of <math>T_n</math>. <math>V_n</math> is the volume of tetrahedron <math>T_n</math>. <math>\frac{V_{2006}}{V_1}=\frac{m}{n}</math> where <math>m</math> and <math>n</math> are coprime positive integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 3|Solution]]
 == Problem 4 ==
-Let <math>m</math> and <math>n</math> be integers such that <math>1 < m \le 10</math> and <math>m < n \le 100</math>. Given that <math>x = \log_m{n}</math> and <math>y = \log_n{m}</math>, find the number of ordered pairs <math>(m,n)</math> such that <math> \displaystyle \lfloor x \rfloor = \lceil y \rceil</math>. (<math>\lfloor a \rfloor</math> is the greatest integer less than or equal to <math>a</math> and <math>\lceil a \rceil</math> is the least integer greater than or equal to <math>a</math>).
+. Let <math>P(x)=\sum_{i=1}^{20}(-x)^{20-i}(x+i)^i</math>. Let <math>K</math> be the product of the roots. How many digits are does <math>\lfloor K \rfloor</math> have where <math>\lfloor x \rfloor</math> denotes the greatest integer less than or equal to <math>x</math>?
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 4|Solution]]
 == Problem 5 ==
-Find the largest prime divisor of <math>25^2+72^2</math>.
+. A parabola <math>P: y=x^2</math> is rotated <math>135</math> degrees clockwise about the origin to <math>P'</math>. This image is translated upward <math>\frac{8+\sqrt{2}}{2}</math> to <math>P''</math>. Point <math>A: (0,0)</math>, <math>B: (256,0)</math>, and <math>C</math> is in Quadrant I, on <math>P''</math>. If the area of <math>\triangle ABC</math> is at a maximum, it is <math>a\sqrt{b}+c</math> where <math>a</math>, <math>b</math> and <math>c</math> are integers and <math>b</math> is square free, find <math>a+b+c</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 5|Solution]]
 == Problem 6 ==
-<math>P_1</math>, <math>P_2</math>, and <math>P_3</math> are polynomials defined by:
+. Define a sequence <math>a_0=2006</math> and <math>a_{n+1}=(n+1)^{a_n}</math> for all positive integers <math>n</math>. Find the remainder when <math>a_{2007}</math> is divided by <math>1000</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-: <math>P_1(x) = 1+x+x^3+x^4+\cdots+x^{96}+x^{97}+x^{99}+x^{100}</math>
-: <math>P_2(x) = 1-x+x^2-\cdots-x^{99}+x^{100}</math>
-: <math>P_3(x) = 1+x+x^2+\cdots+x^{66}+x^{67}</math>
-Find the number of distinct complex roots of <math>P_1 \cdot P_2 \cdot P_3</math>.
-[[Mock AIME 5 2005-2006/Problem 6|Solution]]
 == Problem 7 ==
-A coin of radius <math>1</math> is flipped onto an <math>500 \times 500</math> square grid divided into <math>2500</math> equal squares. Circles are inscribed in <math>n</math> of these <math>2500</math> squares. Let <math>P_n</math> be the probability that, given that the coin lands completely within one of the smaller squares, it also lands completely within one of the circles. Let <math>P</math> be the probability that, when flipped onto the grid, the coin lands completely within one of the smaller squares. Let <math>n_0</math> smallest value of <math>n</math> such that <math>P_n > P</math>. Find the value of <math>\displaystyle \left\lfloor \frac{n_0}{3} \right\rfloor</math>.
+. <math>f(x)</math> is a function that satisfies <math>3f(x)=2x+1-f(\frac{1}{1-x})</math> for all defined <math>x</math>. Suppose that the sum of the zeros of <math>f(x)=\frac{m}{n}</math> where <math>m</math> and <math>n</math> are coprime positive integers, find <math>m^2+n^2</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 7|Solution]]
 == Problem 8 ==
-Let <math>P</math> be a polyhedron with <math>37</math> faces, all of which are equilateral triangles, squares, or regular pentagons with equal side length. Given there is at least one of each type of face and there are twice as many pentagons as triangles, what is the sum of all the possible number of vertices <math>P</math> can have?
+. <math>R</math> is a solution to <math>x+\frac{1}{x}=\frac{ \sin210^{\circ} }{\sin285^{\circ} }</math>. Suppose that <math>\frac{1}{R^{2006}}+R^{2006}=A</math> find <math>\lfloor A^{10} \rfloor</math> where <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 8|Solution]]
 == Problem 9 ==
-<math>13</math> nondistinguishable residents are moving into <math>7</math> distinct houses in Conformistville, with at least one resident per house. In how many ways can the residents be assigned to these houses such that there is at least one house with <math>4</math> residents?
+. Zeus, Athena, and Posideon arrive at Mount Olympus at a random time between 12:00 pm and 12:00 am, and stay for 3 hours. All three hours does not need to fall within 12 pm to 12 am. If any of the 2 gods see each other during 12 pm to 12 am, it will be a good day. The probability of it being a good day is <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are coprime positive integers, find <math>m+n</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 9|Solution]]
 == Problem 10 ==
-Find the smallest positive integer <math>n</math> such that <math>\displaystyle {2n \choose n}</math> is divisible by all the primes between <math>10</math> and <math>30</math>.
+. Define <math>S= \tan^2{1^{\circ}}+\tan^2{3^{\circ}}+\tan^2{5^{\circ}}+...+\tan^2{87^{\circ}}+\tan^2{89^{\circ}}</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 10|Solution]]
 == Problem 11 ==
-Let <math>A</math> be a subset of consecutive elements of <math>S = \{n, n+1, \ldots, n+999\}</math> where <math>n</math> is a positive integer. Define <math>\displaystyle \mu(A) = \sum_{k \in A} \tau(k)</math>, where <math>\tau(k) = 1</math> if <math>k</math> has an odd number of divisors and <math>\tau(k) = 0</math> if <math>k</math> has an even number of divisors. For how many <math>n \le 1000</math> does there exist an <math>A</math> such that <math>|A| = 620</math> and <math>\mu(A) = 11</math>? (<math>|X|</math> denotes the cardinality of the set <math>X</math>, or the number of elements in <math>X</math>)
+. <math>\triangle ABC</math> is isosceles with <math>\angle C= 90^{\circ}</math>. A point <math>P</math> lies inside the triangle such that <math>AP=33</math>, <math>CP=28\sqrt{2}</math>, and <math>BP=65</math>. Let <math>A</math> be the area of <math>\triangle ABC</math>. Find the remainder when <math>2A</math> is divided by <math>1000</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 11|Solution]]
 == Problem 12 ==
-Let <math>ABC</math> be a triangle with <math>AB = 13</math>, <math>BC = 14</math>, and <math>AC = 15</math>. Let <math>D</math> be the foot of the altitude from <math>A</math> to <math>BC</math> and <math>E</math> be the point on <math>BC</math> between <math>D</math> and <math>C</math> such that <math>BD = CE</math>. Extend <math>AE</math> to meet the circumcircle of <math>ABC</math> at <math>F</math>. If the area of triangle <math>FAC</math> is <math>\displaystyle \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m+n</math>.
+. There exists a line <math>L</math> with points <math>D</math>,<math>E</math>,<math>F</math> with <math>E</math> in between <math>D</math> and <math>F</math>. Point <math>A</math>, not on the line is such that <math>\overline{AF}=6</math>, <math>\overline{AD}=\frac{36}{7}</math>, <math>\overline{AE}=\frac{12}{\sqrt{7}}</math> with <math>\angle AEF > 90</math>. Construct <math>E'</math> on ray <math>AE</math> such that <math>(\overline{AE})(\overline{AE'})=36</math> and <math>\overline{FE'}=3</math>. Point <math>G</math> is on ray <math>AD</math> such that <math>\overline{AG}=7</math>. If <math>2*(\overline{E'G})=a+\sqrt{b}</math> where <math>a</math> and <math>b</math> are integers, then find <math>a+b</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 12|Solution]]
 == Problem 13 ==
-Let <math>S</math> be the set of positive integers with only odd digits satisfying the following condition: any <math>x \in S</math> with <math>n</math> digits must be divisible by <math>5^n</math>. Let <math>A</math> be the sum of the <math>20</math> smallest elements of <math>S</math>. Find the remainder upon dividing <math>A</math> by <math>1000</math>.
+. <math>\triangle VA_0A_1</math> is isosceles with base <math>\overline{{A_1A_0}}</math>. Construct <math>A_2</math> on segment <math>\overline{{A_0V}}</math> such that <math>\overline{A_0A_1}=\overline{A_1A_2}=b</math>. Construct <math>A_3</math> on <math>\overline{A_1V}</math> such that <math>b=\overline{A_2A_3}</math>. Contiue this pattern: construct <math>\overline{A_{2n}A_{2n+1}}=b</math> with <math>A_{2n+1}</math> on segment <math>\overline{VA_1}</math> and <math>\overline{A_{2n+1}A_{2n+2}}=b</math> with <math>A_{2n+2}</math> on segment <math>\overline{VA_0}</math>. The points <math>A_n</math> do not coincide and <math>\angle VA_1A_0=90-\frac{1}{2006}</math>. Suppose <math>A_k</math> is the last point you can construct on the perimeter of the triangle. Find the remainder when <math>k</math> is divided by <math>1000</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 13|Solution]]
 == Problem 14 ==
-Let <math>ABC</math> be a triangle such that <math>AB = 68</math>, <math>BC = 100</math>, and <math>\displaystyle CA = 112</math>. Let <math>H</math> be the orthocenter of <math>\triangle ABC</math> (intersection of the altitudes). Let <math>D</math> be the midpoint of <math>BC</math>, <math>E</math> be the midpoint of <math>CA</math>, and <math>F</math> be the midpoint of <math>AB</math>. Points <math>X</math>, <math>Y</math>, and <math>Z</math> are constructed on <math>HD</math>, <math>HE</math>, and <math>HF</math>, respectively, such that <math>D</math> is the midpoint of <math>XH</math>, <math>E</math> is the midpoint of <math>YH</math>, and <math>F</math> is the midpoint of <math>ZH</math>. Find <math>AX+BY+CZ</math>.
+. <math>P</math> is the probability that if you flip a fair coin, <math>20</math> heads will occur before <math>19</math> tails. If <math>P=\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 <math>1000</math>.
 [[Mock AIME 5 2005-2006/Problem 14|Solution]]
 == Problem 15 ==
-<math>2006</math> colored beads are placed on a necklace (circular ring) such that each bead is adjacent to two others. The beads are labeled <math>a_0</math>, <math>a_1</math>, <math>\ldots</math>, <math>a_{2005}</math> around the circle in order. Two beads <math>a_i</math> and <math>a_j</math>, where <math>i</math> and <math>j</math> are non-negative integers, satisfy <math>a_i = a_j</math> if and only if the color of <math>a_i</math> is the same as the color of <math>a_j</math>. Given that there exists no non-negative integer <math>m < 2006</math> and positive integer <math>n < 1003</math> such that <math>a_m = a_{m-n} = a_{m+n}</math>, where all subscripts are taken <math>\pmod{2006}</math>, find the minimum number of different colors of beads on the necklace.
+. A regular 61-gon with verticies <math>A_1</math>, <math>A_2</math>, <math>A_3</math>,...<math>A_{61}</math> is inscribed in a circle with a radius of <math>r</math>. Suppose <math>(\overline{A_1A_2})(\overline{A_1A_3})(\overline{A_1A_4})...(\overline{A_1A_{61}})=r</math>. If <math>r^{2006}=\frac{p}{q}</math> where <math>p</math> and <math>q</math> are coprime positive integers, find the remainder when <math>p+q</math> is divided by <math>1000</math>.
+[[Mock AIME 5 2005-2006/Problem 14|Solution]]
-[[Mock AIME 5 2005-2006/Problem 15|Solution]]
 == See also ==
-* [[Mock AIME 5 2005-2006]]
+* [[Mock AIME 4 2005-2006]]
 * [[Mock AIME]]
-* A [http://wangsblog.com/jeffrey/MockAIME5.pdf .pdf] version of the problems
+* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70988 Here is the page on AoPS]
-http://www.mathlinks.ro/Forum/latexrender/pictures/2ea0ce047615395691113a82d6c190b3.gif
-. A 5-digit number is leet if and only if the sum of the first 2 digits, the sum of the last 2 digits and the middle digit are equal. How many 5-digit leet numbers exist?
-. Qin Shi Huang wants to count the number of warriors he has to invade China. He puts his warriors into lines with the most people such that they have even length. The people left over are the remainder. He makes 2 lines, with a remainder of 1, 3 lines with a remainder of 2, 4 lines with a remainder of 3, 5 lines with a remainder of 4, and 6 lines with a remainder of 5. Find the minimum number of warriors that he has.
-. <math>T_1</math> is a regular tetrahedron. Tetrahedron <math>T_2</math> is formed by connecting the centers of the faces of <math>T_1</math>. Generally, a new tetrahedron <math>T_{n+1}</math> is formed by connecting the centers of the faces of <math>T_n</math>. <math>V_n</math> is the volume of tetrahedron <math>T_n</math>. <math>\frac{V_{2006}}{V_1}=\frac{m}{n}</math> where <math>m</math> and <math>n</math> are coprime positive integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>.
-. Let <math>P(x)=\sum_{i=1}^{20}(-x)^{20-i}(x+i)^i</math>. Let <math>K</math> be the product of the roots. How many digits are does <math>\lfloor K \rfloor</math> have where <math>\lfloor x \rfloor</math> denotes the greatest integer less than or equal to <math>x</math>?
-. A parabola <math>P: y=x^2</math> is rotated <math>135</math> degrees clockwise about the origin to <math>P'</math>. This image is translated upward <math>\frac{8+\sqrt{2}}{2}</math> to <math>P''</math>. Point <math>A: (0,0)</math>, <math>B: (256,0)</math>, and <math>C</math> is in Quadrant I, on <math>P''</math>. If the area of <math>\triangle ABC</math> is at a maximum, it is <math>a\sqrt{b}+c</math> where <math>a</math>, <math>b</math> and <math>c</math> are integers and <math>b</math> is square free, find <math>a+b+c</math>.
-. Define a sequence <math>a_0=2006</math> and <math>a_{n+1}=(n+1)^{a_n}</math> for all positive integers <math>n</math>. Find the remainder when <math>a_{2007}</math> is divided by <math>1000</math>.
-. <math>f(x)</math> is a function that satisfies <math>3f(x)=2x+1-f(\frac{1}{1-x})</math> for all defined <math>x</math>. Suppose that the sum of the zeros of <math>f(x)=\frac{m}{n}</math> where <math>m</math> and <math>n</math> are coprime positive integers, find <math>m^2+n^2</math>.
-. <math>R</math> is a solution to <math>x+\frac{1}{x}=\frac{ \sin210^{\circ} }{\sin285^{\circ} }</math>. Suppose that <math>\frac{1}{R^{2006}}+R^{2006}=A</math> find <math>\lfloor A^{10} \rfloor</math> where <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.
-. Zeus, Athena, and Posideon arrive at Mount Olympus at a random time between 12:00 pm and 12:00 am, and stay for 3 hours. All three hours does not need to fall within 12 pm to 12 am. If any of the 2 gods see each other during 12 pm to 12 am, it will be a good day. The probability of it being a good day is <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are coprime positive integers, find <math>m+n</math>.
-. Define <math>S= \tan^2{1^{\circ}}+\tan^2{3^{\circ}}+\tan^2{5^{\circ}}+...+\tan^2{87^{\circ}}+\tan^2{89^{\circ}}</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.
-. <math>\triangle ABC</math> is isosceles with <math>\angle C= 90^{\circ}</math>. A point <math>P</math> lies inside the triangle such that <math>AP=33</math>, <math>CP=28\sqrt{2}</math>, and <math>BP=65</math>. Let <math>A</math> be the area of <math>\triangle ABC</math>. Find the remainder when <math>2A</math> is divided by <math>1000</math>.
-. There exists a line <math>L</math> with points <math>D</math>,<math>E</math>,<math>F</math> with <math>E</math> in between <math>D</math> and <math>F</math>. Point <math>A</math>, not on the line is such that <math>\overline{AF}=6</math>, <math>\overline{AD}=\frac{36}{7}</math>, <math>\overline{AE}=\frac{12}{\sqrt{7}}</math> with <math>\angle AEF > 90</math>. Construct <math>E'</math> on ray <math>AE</math> such that <math>(\overline{AE})(\overline{AE'})=36</math> and <math>\overline{FE'}=3</math>. Point <math>G</math> is on ray <math>AD</math> such that <math>\overline{AG}=7</math>. If <math>2*(\overline{E'G})=a+\sqrt{b}</math> where <math>a</math> and <math>b</math> are integers, then find <math>a+b</math>.
-. <math>\triangle VA_0A_1</math> is isosceles with base <math>\overline{{A_1A_0}}</math>. Construct <math>A_2</math> on segment <math>\overline{{A_0V}}</math> such that <math>\overline{A_0A_1}=\overline{A_1A_2}=b</math>. Construct <math>A_3</math> on <math>\overline{A_1V}</math> such that <math>b=\overline{A_2A_3}</math>. Contiue this pattern: construct <math>\overline{A_{2n}A_{2n+1}}=b</math> with <math>A_{2n+1}</math> on segment <math>\overline{VA_1}</math> and <math>\overline{A_{2n+1}A_{2n+2}}=b</math> with <math>A_{2n+2}</math> on segment <math>\overline{VA_0}</math>. The points <math>A_n</math> do not coincide and <math>\angle VA_1A_0=90-\frac{1}{2006}</math>. Suppose <math>A_k</math> is the last point you can construct on the perimeter of the triangle. Find the remainder when <math>k</math> is divided by <math>1000</math>.
-. <math>P</math> is the probability that if you flip a fair coin, <math>20</math> heads will occur before <math>19</math> tails. If <math>P=\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 <math>1000</math>.
-. A regular 61-gon with verticies <math>A_1</math>, <math>A_2</math>, <math>A_3</math>,...<math>A_{61}</math> is inscribed in a circle with a radius of <math>r</math>. Suppose <math>(\overline{A_1A_2})(\overline{A_1A_3})(\overline{A_1A_4})...(\overline{A_1A_{61}})=r</math>. If <math>r^{2006}=\frac{p}{q}</math> where <math>p</math> and <math>q</math> are coprime positive integers, find the remainder when <math>p+q</math> is divided by <math>1000</math>.
-[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70988 Here is the page on AoPS]

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