Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$

Solution: Note that $AB=2\sqrt3$ $DF=11$ , and $EF=13$ . If we take a homothety of the parallelogram with respect to $A$ , such that $F$ maps to $G$ , we see that $\frac{[ABG]}{[ACG]}=\frac{11}{13}$ . Since $\angle AGB=\angle AGC=60^{\circ}$ , from the sine area formula we have $\frac{BG}{CG}=\frac{11}{13}$ . Let $BG=11k$ and $CG=13k$
By Law of Cosines on $\triangle BGC$ , we have \[12=k^2(11^2+11\cdot13+13^2)=433k^2\implies k^2=\frac{12}{433}\] Thus, $[CBG]=\frac12 (11k)(13k)\sin 120^{\circ} = \frac{\sqrt3}{4}\cdot 143\cdot \frac{12}{433}=\frac{429\sqrt3}{433}\implies\boxed{865}$

Answer:  865
Problem: How many integers $N$ less than $1000$ can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$

Solution: Let the first odd integer be $2n+1$ $n\geq 0$ . Then the final odd integer is $2n+1 + 2(j-1) = 2(n+j) - 1$ . The odd integers form an arithmetic sequence with sum $N = j\left(\frac{(2n+1) + (2(n+j)-1)}{2}\right) = j(2n+j)$ . Thus, $j$ is a factor of $N$
Since $n\geq 0$ , it follows that $2n+j \geq j$ and $j\leq \sqrt{N}$
Since there are exactly $5$ values of $j$ that satisfy the equation, there must be either $9$ or $10$ factors of $N$ . This means $N=p_1^2p_2^2$ or $N=p_1p_2^4$ . Unfortunately, we cannot simply observe prime factorizations of $N$ because the factor $(2n+j)$ does not cover all integers for any given value of $j$
Instead we do some casework:
\[(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)\]
\[N = 4k(n+k) \Longrightarrow \frac{N}{4} = k(n+k)\]
\[\frac{N}{4} = (2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13), (3^4\cdot2)\]
The total number of integers $N$ is $5 + 10 = \boxed{15}$

Answer:  15
Problem: How many integers $N$ less than $1000$ can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$

Solution: Let the largest odd number below the sequence be the $q$ th positive odd number, and the largest odd number in the sequence be the $p$ th positive odd number. Therefore, the sum is $p^2-q^2=(p+q)(p-q)$ by sum of consecutive odd numbers. Note that $p+q$ and $p-q$ have the same parity, and $q$ can equal $0$ . We then perform casework based on the parity of $p-q$
If $p-q$ is odd, then $p^2-q^2$ must always be odd.  Therefore, to have 5 pairs of odd factors, we must have either $9$ (in which case the number is a perfect square) or $10$ factors. Considering the upper bound, the only way this can happen is $p_1^4\cdot{p_2}$ or $p_1^2\cdot{p_2^2}$ . N must then be one of \[(3^2\cdot5^2),\ (3^2\cdot7^2),\ (3^4\cdot5),\ (3^4\cdot7),\ (3^4\cdot 11)\] So, there are $5$ solutions when $(p+q)(p-q)$ is odd.
If $p-q$ is even, then $(p+q)(p-q)$ must have at least two factors of $2$ , so we can rewrite the expression as $4(k)(k-q)$ where $k=\frac{p+q}{2}$ . We can disregard the $4$ by dividing by $4$ and restricting our upper bound to $250$ . Since $k$ and $q$ don't have to be the same parity, we can include all cases less than 250 that have 9 or 10 factors. We then have \[(2^2\cdot3^2),(2^2\cdot5^2),(2^2\cdot7^2), (3^2\cdot5^2), (2^4\cdot3), (2^4\cdot5), (2^4\cdot7), (2^4\cdot11), (2^4\cdot13), (3^4\cdot2)\] as the possibilities.
Therefore, there are $10+5=\boxed{015}$ possibilities for $p^2-q^2=N$ ~sigma

Answer:  015
Problem: Let $S_n$ be the sum of the reciprocals of the non-zero digits of the integers from $1$ to $10^n$ inclusive. Find the smallest positive integer $n$ for which $S_n$ is an integer.

Solution: Let $K = \sum_{i=1}^{9}{\frac{1}{i}}$ .  Examining the terms in $S_1$ , we see that $S_1 = K + 1$ since each digit $n$ appears once and 1 appears an extra time.  Now consider writing out $S_2$ .  Each term of $K$ will appear 10 times in the units place and 10 times in the tens place (plus one extra 1 will appear), so $S_2 = 20K + 1$
In general, we will have that
because each digit will appear $10^{n - 1}$ times in each place in the numbers $1, 2, \ldots, 10^{n} - 1$ , and there are $n$ total places.
The denominator of $K$ is $D = 2^3\cdot 3^2\cdot 5\cdot 7$ .  For $S_n$ to be an integer, $n10^{n-1}$ must be divisible by $D$ .  Since $10^{n-1}$ only contains the factors $2$ and $5$ (but will contain enough of them when $n \geq 3$ ), we must choose $n$ to be divisible by $3^2\cdot 7$ .  Since we're looking for the smallest such $n$ , the answer is $\boxed{063}$

Answer:  063
Problem: Given that $x, y,$ and $z$ are real numbers that satisfy: \begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*} and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$

Solution: Let $\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$ , and suppose this triangle is acute (so all altitudes are in the interior of the triangle).
Let the altitude to the side of length $x$ be of length $h_x$ , and similarly for $y$ and $z$ . Then we have by two applications of the Pythagorean Theorem we that \[x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}\] As a function of $h_x$ , the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once.  Thus we must have that $h_x^2 = \frac1{16}$ and so $h_x = \frac{1}4$ and similarly $h_y = \frac15$ and $h_z = \frac16$
The area of the triangle must be the same no matter how we measure it; therefore $x\cdot h_x = y\cdot h_y = z \cdot h_z$ gives us $\frac x4 = \frac y5 = \frac z6 = 2A$ and $x = 8A, y = 10A$ and $z = 12A$
The semiperimeter of the triangle is $s = \frac{8A + 10A + 12A}{2} = 15A$ so by Heron's formula we have \[A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}\]
Thus, $A = \frac{1}{15\sqrt{7}}$ and $x + y + z = 30A = \frac2{\sqrt{7}}$ and the answer is $2 + 7 = \boxed{009}$

Answer:  009
Problem: Given that $x, y,$ and $z$ are real numbers that satisfy: \begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*} and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$

Solution: Note that none of $x,y,z$ can be zero.
Each of the equations is in the form \[a=\sqrt{b^2-d^2}+\sqrt{c^2-d^2}\]
Isolate a radical and square the equation to get \[b^2-d^2=a^2-2a\sqrt{c^2-d^2}+c^2-d^2\]
Now cancel, and again isolate the radical, and square the equation to get \[a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2\]
Rearranging gives \[a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2\]
Now note that everything is cyclic but the last term (i.e. $-4a^2d^2$ ), which implies \[-4x^2\cdot\frac1{16}=-4y^2\cdot\frac1{25}=-4z^2\cdot\frac1{36}\]
Or
\[x: y: z=4: 5: 6 \implies x=\frac{4y}5 \textrm{ and } z=\frac{6y}5\]
Plug these values into the middle equation to get \[\frac{256y^4+625y^4+1296y^4}{625}=\frac{800y^4}{625}+\frac{1800y^4}{625}+\frac{1152y^4}{625}-\frac{100y^2}{625}\]
Simplifying gives \[1575y^4=100y^2 \textrm{ but } y \neq 0 \implies y^2=\frac4{63} \textrm{ or } y=\frac2{3\sqrt7}\]
Substituting the value of $y$ for $x$ and $z$ gives \[x+y+z = \frac{4y+5y+6y}5 = 3y = 3 \cdot \frac{2}{3\sqrt7} = \frac{2}{\sqrt7}\]
And thus the answer is $\boxed{009}$

Answer:  009
Problem: Six congruent circles form a ring with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius 30. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $\lfloor K \rfloor$ (the floor function ).
2005 AIME I Problem 1.png

Solution: Define the radii of the six congruent circles as $r$ . If we draw all of the radii to the points of external tangency, we get a regular hexagon . If we connect the vertices of the hexagon to the center of the circle $C$ , we form several equilateral triangles . The length of each side of the triangle is $2r$ . Notice that the radius of circle $C$ is equal to the length of the side of the triangle plus $r$ . Thus, the radius of $C$ has a length of $3r = 30$ , and so $r = 10$ $K = 30^2\pi - 6(10^2\pi) = 300\pi$ , so $\lfloor 300\pi \rfloor = \boxed{942}$

Answer:  942
Problem: For each positive integer $k$ , let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$ .  For example, $S_3$ is the sequence $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term $2005$

Solution: Suppose that the $n$ th term of the sequence $S_k$ is $2005$ . Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2\cdot 3\cdot 167$ .  The ordered pairs $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$ $(2,1002)$ $(3,668)$ $(4,501)$ $(6,334)$ $(12,167)$ $(167,12)$ $(334,6)$ $(501,4)$ $(668,3)$ $(1002,2)$ and $(2004,1)$ , and each of these gives a possible value of $k$ . Thus the requested number of values is $12$ , and the answer is $\boxed{012}$

Answer:  012
Problem: For each positive integer $k$ , let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$ .  For example, $S_3$ is the sequence $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term $2005$

Solution: Any term in the sequence $S_k$ can be written as 1+kx.  If this is to equal 2005, then the remainder when 2005 is divided by k is 1.
Now all we have to do is find the numbers of factors of 2004.  There are $(2 + 1)(1 + 1)(1 + 1) = \boxed{012}$ divisors of $2^2\cdot 3^1\cdot 167^1$

Answer:  012
Problem: How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50?

Solution: Suppose $n$ is such an integer . Because $n$ has $3$ proper divisors, it must have $4$ divisors,, so $n$ must be in the form $n=p\cdot q$ or $n=p^3$ for distinct prime numbers $p$ and $q$
In the first case, the three proper divisors of $n$ are $1$ $p$ and $q$ .  Thus, we need to pick two prime numbers less than $50$ . There are fifteen of these ( $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$ and $47$ ) so there are ${15 \choose 2} =105$ ways to choose a pair of primes from the list and thus $105$ numbers of the first type.
In the second case, the three proper divisors of $n$ are 1, $p$ and $p^2$ .  Thus we need to pick a prime number whose square is less than $50$ .  There are four of these ( $2, 3, 5,$ and $7$ ) and so four numbers of the second type.
Thus there are $105+4=\boxed{109}$ integers that meet the given conditions.

Answer:  109
Problem: The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.

Solution: If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$ .  If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$ .  Thus, if the number of columns is $n$ , the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$ .  In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$ , so this number works and no larger number can.  Thus, the answer is $\boxed{294}$

Answer:  294
Problem: The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.

Solution: Let there be $m$ members and $n$ members for the square and $c$ for the number of columns of the other formation. We have $n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}\right) \implies (2n-2c-7)(2n+2n+7) = -69.$
To maximize this we let $2n+2c+7 = 68$ and $2n-2c-7 = 1.$ Solving we find $n = 17$ so the desired number of members is $17^2 + 5 = \boxed{294}.$

Answer:  294
Problem: The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.

Solution: Think of the process of moving people from the last column to new rows. Since there are less columns than rows, for each column removed, there are people discarded to the "extra" pile to be placed at the end. To maximize the number of "extra" people to fill in the last few rows. We remove 3 columns and add 4 rows. For the first new row, one more person will be discarded. For the second, 1 extra person are added since there is one more row now, and there are 2 less columns. Thus, there are 3 extra people discarded. Similarly, 5 extra people are discarded for the third column. Now there are $5+1+3+5=14$ people in the extra pile to put as the last row, so there are $14(14+7)=\boxed{294}$ people.

Answer:  294
Problem: The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.

Solution: Note: Only do this if you have a LOT of time (and you've memorized all your perfect squares up to 1000).
We can see that the number of members in the band must be of the form $n(n + 7)$ for some positive integer $n$ . When $n = 28$ , this product is $980$ , and since AIME answers are nonnegative integers less than $1000$ , we don't have to check any higher $n$ . Also, we know that this product must be 5 more than a perfect square, so we can make a table as shown and bash: \[\begin{tabular}{|c|c|c|} n & n(n+7) & 5 more than a perfect square?\\ \hline 1 & 8 & no\\ \hline 2 & 18 & no\\ \hline 3 & 30 & yes\\ \hline 4 & 44 & no\\ \hline 5 & 60 & no\\ \hline 6 & 78 & no\\ \hline 7 & 98 & no\\ \hline 8 & 120 & no\\ \hline 9 & 144 & no\\ \hline 10 & 170 & no\\ \hline 11 & 198 & no\\ \hline 12 & 228 & no\\ \hline 13 & 260 & no\\ \hline 14 & 294 & yes\\ \hline 15 & 330 & no\\ \hline 16 & 368 & no\\ \hline 17 & 408 & no\\ \hline 18 & 450 & no\\ \hline 19 & 494 & no\\ \hline 20 & 540 & no\\ \hline 21 & 588 & no\\ \hline 22 & 638 & no\\ \hline 23 & 690 & no\\ \hline 24 & 744 & no\\ \hline 25 & 800 & no\\ \hline 26 & 858 & no\\ \hline 27 & 918 & no\\ \hline 28 & 980 & no\\ \hline \end{tabular}\] Thus, we can see that our largest valid $n(n+7)$ is $\boxed{294}$

Answer:  294
Problem: Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangements of the 8 coins.

Solution: There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.
There are ${8\choose4} = 70$ ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.
Create a string of letters H and T to denote the orientation of the top of the coin. To avoid making two faces touch, we cannot have the arrangement HT. Thus, all possible configurations must be a string of tails followed by a string of heads, since after the first H no more tails can appear. The first H can occur in a maximum of eight times different positions, and then there is also the possibility that it doesn’t occur at all, for $9$ total configurations. Thus, the answer is $70 \cdot 9 = \boxed{630}$

Answer:  630
Problem: Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangements of the 8 coins.

Solution: First, we can break this problem up into two parts: the amount of ways to order the coins based on color, and which side is facing up for each coin (assuming they are indistinguishable). In the end, we multiply these values together.
First, there are obviously $\binom{8}{4}$ ways to order the coins based on color.
Next, we set up a recurrence. Let $a_n$ be the number of ways $n$ indistinguishable coins to be stacked such that no heads are facing each other.
Consider the top coin. If it is facing up, then there are $a_{n-1}$ ways for the rest of the coins below it to be arranged. If it is facing down however, there will only be one way to arrange the coins. We can the recurrence: \[a_n=a_{n-1}+1.\] Note $a_1=2$ , so $a_8 = 9$
Finally, to get the result, we do $70\times 9$ to get $\boxed{630}$

Answer:  630
Problem: Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$

Solution: The left-hand side of that equation is nearly equal to $(x - 1)^4$ .  Thus, we add 1 to each side in order to complete the fourth power and get $(x - 1)^4 = 2006$
Let $r = \sqrt[4]{2006}$ be the positive real fourth root of 2006.  Then the roots of the above equation are $x = 1 + i^n r$ for $n = 0, 1, 2, 3$ .  The two non-real members of this set are $1 + ir$ and $1 - ir$ .  Their product is $P = 1 + r^2 = 1 + \sqrt{2006}$ $44^2 = 1936 < 2006 < 2025 = 45^2$ so $\lfloor P \rfloor = 1 + 44 = \boxed{045}$

Answer:  045
Problem: Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$

Solution: Starting like before, $(x-1)^4= 2006$ This time we apply differences of squares. $(x-1)^4-2006=0$ so $((x-1)^2+\sqrt{2006})((x-1)^2 -\sqrt{2006})=0$ If you think of each part of the product as a quadratic, then $((x-1)^2+\sqrt{2006})$ is bound to hold the two  non-real roots since the other definitely crosses the x-axis twice since it is just $x^2$ translated down and right.
Therefore $P$ is the product of the roots of $((x-1)^2+\sqrt{2006})$ or $P=1+\sqrt{2006}$ so
$\lfloor P \rfloor = 1 + 44 = \boxed{045}$

Answer:  045
Problem: Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$

Solution: If we don't see the fourth power, we can always factor the LHS to try to create a quadratic substitution. Checking, we find that $x=0$ and $x=2$ are both roots. Synthetic division gives $(x^2-2x)(x^2-2x+2)=2005$ . We now have our quadratic substitution of $y=x^2-2x+1=(x-1)^2$ , giving us $(y-1)(y+1)=2005$ . From here we proceed as in Solution 1 to get $\boxed{045}$

Answer:  045
Problem: Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$

Solution: Realizing that if we add 1 to both sides we get $x^4-4x^3+6x^2-4x+1=2006$ which can be factored as $(x-1)^4=2006$ . Then we can substitute $(x-1)$ with $y$ which leaves us with $y^4=2006$ . Now subtracting 2006 from both sides we get some difference of squares $y^4-2006=0 \rightarrow (y-\sqrt[4]{2006})(y+\sqrt[4]{2006})(y^2+\sqrt{2006})=0$ . The question asks for the product of the complex roots so we only care about the last factor which is equal to zero. From there we can solve $y^2+\sqrt{2006}=0$ , we can substitute $(x-1)$ for $y$ giving us $(x-1)^2+\sqrt{2006}=0$ , expanding this we get $x^2-2x+1+\sqrt{2006}=0$ . We know that the product of a quadratics roots is $\frac{c}{a}$ which leaves us with $\frac{1+\sqrt{2006}}{1}=1+\sqrt{2006}\approx\boxed{045}$

Answer:  045
Problem: Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$

Solution: As in solution 1, we find that $(x-1)^4 = 2006$ . Now $x-1=\pm \sqrt[4]{2006}$ so $x_1 = 1+\sqrt[4]{2006}$ and $x_2 = 1-\sqrt[4]{2006}$ are the real roots of the equation. Multiplying, we get $x_1 x_2 = 1 - \sqrt{2006}$ . Now transforming the original function and using Vieta's formula, $x^4-4x^3+6x^2-4x-2005=0$ so $x_1 x_2 x_3 x_4 = \frac{-2005}{1} = -2005$ . We find that the product of the nonreal roots is $x_3 x_4 = \frac{-2005}{1-\sqrt{2006}} \approx 45.8$ and we get $\boxed{045}$

Answer:  045
Problem: Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$

Solution: As all the other solutions, we find that $(x-1)^4 = 2006$ . Thus $x=\sqrt[4]{2006}+1$ . Thus $x= \sqrt[4]{2006}(\cos(\frac{2\pi(k)}{4}+i\sin(\frac{2\pi(k)}{4}))+1$ when $k=0,1,2,3$ . The complex values of $x$ are the ones where $i\sin(\frac{2\pi(k)}{4})$ does not equal 0. These complex roots are $1+\sqrt[4]{2006}(i)$ and $1-\sqrt[4]{2006}(i)$ . The product of these two nonreal roots is ( $1+\sqrt[4]{2006}(i)$ )( $1-\sqrt[4]{2006}(i)$ ) which is equal to $1+\sqrt {2006}$ . The floor of that value is $\boxed{045}$

Answer:  045
Problem: In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers , find $p+q.$

Solution: [asy]draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label("$A$",(0,0),SW); label("$B$",(20.87,0),SE); label("$E$",(15.87,8.66),NE); label("$D$",(5,8.66),NW); label("$P$",(5,0),S); label("$Q$",(15.87,0),S); label("$C$",(16.87,7),E); label("$12$",(10.935,7.794),S); label("$10$",(2.5,4.5),W); label("$10$",(18.37,4.5),E); [/asy]
Draw line segment $DE$ such that line $DE$ is concurrent with line $BC$ . Then, $ABED$ is an isosceles trapezoid so $AD=BE=10$ , and $BC=8$ and $EC=2$ . We are given that $DC=12$ . Since $\angle CED = 120^{\circ}$ , using Law of Cosines on $\bigtriangleup CED$ gives \[12^2=DE^2+4-2(2)(DE)(\cos 120^{\circ})\] which gives \[144-4=DE^2+2DE\] . Adding $1$ to both sides gives $141=(DE+1)^2$ , so $DE=\sqrt{141}-1$ $\bigtriangleup DAP$ and $\bigtriangleup EBQ$ are both $30-60-90$ , so $AP=5$ and $BQ=5$ $PQ=DE$ , and therefore $AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141} \rightarrow (p,q)=(9,141) \rightarrow \boxed{150}$

Answer:  150