Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6);  draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight);  triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2));  draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4"));  draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy]

Solution: Since the diagonal is perpendicular to the base of the cylinder, all three edges and faces can be treated symmetrically.
The cross-section of the cube with the top face of the cylinder is an equilateral triangle inscribed in a circle with radius $4$ . This means the medians of the triangle have length $\frac{3}{2} \cdot 4=6$ , because the circumcenter is also the centroid, and the centroid divides the medians into lengths of ratio $2:1$ . Using $30-60-90$ triangles, the side length of the triangle is $4\sqrt{3}$ , and its area is $\frac{(4\sqrt{3})^2\sqrt{3}}{4}=12\sqrt{3}$
Next, consider the submerged triangular sections of the faces. Each is a $45-45-90$ triangle with leg length $x$ . The area of each is then $\frac{x^2}{2}$ . By De Gua's Theorem on the submerged pyramid (which we can apply because it has a right-angled corner), $3\left( \frac{x^2}{2} \right) ^2=(12\sqrt{3})^2$ . Solving yields $x=2\sqrt{6}$
The height of the pyramid is then $\sqrt{(2\sqrt{6})^2-4^2}=2\sqrt{2}$ , by the Pythagorean Theorem (using the slant height and circumradius). The volume is then $v=\frac{1}{3}\cdot 12\sqrt{3} \cdot 2\sqrt{2}=8\sqrt{6}$ , and the requested answer is $v^2=(8\sqrt{6})^2=\boxed{384}$ . ~bad_at_mathcounts

Answer:  384
Problem: Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$ . For example, 53421 and 14253 are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$ , but 45123 is not. Find the number of quasi-increasing permutations of the integers $1, 2, \ldots, 7$

Solution: The simple recurrence can be found.
When inserting an integer $n$ into a string with $n - 1$ integers, we notice that the integer $n$ has 3 spots where it can go: before $n - 1$ , before $n - 2$ , and at the very end.
EXAMPLE: 
Putting 4 into the string 123:
4 can go before the 2: 1423,
Before the 3: 1243,
And at the very end: 1234.
Only the addition of the next number, n, will change anything.
Thus the number of permutations with n elements is three times the number of permutations with $n-1$ elements.
Start with $n=3$ since all $6$ permutations work. And go up: $18, 54, 162, 486$
Thus for $n=7$ there are $2*3^5=\boxed{486}$ permutations.

Answer:  486
Problem: The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$ , respectively. Let $OB = r$ . Notice that $\triangle{OMB} \sim \triangle{QOB}$ because both are right triangles, and $\angle{OBQ} \cong \angle{OBM}$ . By $\frac{MB}{BO}=\frac{BO}{BQ}$ $MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}$ . However, since $O$ is the circumcenter of triangle $ABC$ $OM$ is a perpendicular bisector by the definition of a circumcenter. Hence, $\frac{r^2}{4.5} = 2 \implies r = 3$ . Since we know $BN=\frac{5}{2}$ and $\triangle BOP \sim \triangle BNO$ , we have $\frac{BP}{3} = \frac{3}{\frac{5}{2}}$ . Thus, $BP = \frac{18}{5}$ $m + n=\boxed{023}$

Answer:  023
Problem: The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Minor arc $BC = 2A$ so $\angle{BOC}=2A$ . Since $\triangle{BOC}$ is isosceles ( $BO$ and $OC$ are radii), $\angle{CBO}=(180-2A)/2=90-A$ $\angle{CBO}=90-A$ , so $\angle{BQO}=A$ . From this we get that $\triangle{BPQ}\sim \triangle{BCA}$ . So $\dfrac{BP}{BC}=\dfrac{BQ}{BA}$ , plugging in the given values we get $\dfrac{BP}{4}=\dfrac{4.5}{5}$ , so $BP=\dfrac{18}{5}$ , and $m+n=\boxed{023}$

Answer:  023
Problem: The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $r=BO$ . Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$ . From there, \[OM=\sqrt{r^2-4}\] Thus, \[OQ=\frac{\sqrt{4r^2+9}}{2}\] Using $\triangle{BOQ}$ , we get $r=3$ . Now let's find $NP$ . After some calculations with $\triangle{BON}$ $\triangle{OPN}$ ${NP=11/10}$ . Therefore, \[BP=\frac{5}{2}+\frac{11}{10}=18/5\] $18+5=\boxed{023}$

Answer:  023
Problem: The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $\angle{BQO}=\alpha$ . Extend $OB$ to touch the circumcircle at a point $K$ . Then, note that $\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha$ . But since $BK$ is a diameter, $\angle{KAB}=90^\circ$ , implying $\angle{CAB}=\alpha$ . It follows that $APCQ$ is a cyclic quadrilateral.
Let $BP=x$ . By Power of a Point, \[5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.\] The answer is $18+5=\boxed{023}$

Answer:  023
Problem: The circumcircle of acute $\triangle ABC$ has center $O$ . The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ at $P$ and $Q$ , respectively. Also $AB=5$ $BC=4$ $BQ=4.5$ , and $BP=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: $\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}$
Denote the circumradius of $ABC$ to be $R$ , the circumcircle of $ABC$ to be $O$ , and the shortest distance from $Q$ to circle $O$ to be $x$
Using Power of a Point on $Q$ relative to circle $O$ , we get that $x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}$ . Using Pythagorean Theorem on triangle $QOB$ to get $(x + r)^2 + r^2 = \frac{81}{4}$ . Subtracting the first equation from the second, we get that $2r^2 = 18$ and therefore $r = 3$ . Now, set $\cos{ABC} = y$ . Using law of cosines on $ABC$ to find $AC$ in terms of $y$ and plugging that into the extended law of sines, we get $\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6$ . Squaring both sides and cross multiplying, we get $36x^2 - 40x + 5 = 0$ . Now, we get $x = \frac{10 \pm \sqrt{55}}{18}$ using quadratic formula. If you drew a decent diagram, $B$ is acute and therefore $x = \frac{10 + \sqrt{55}}{18}$ (You can also try plugging in both in the end and seeing which gives a rational solution). Note that $BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.$ Using the cosine addition formula and then plugging in what we know about $QBO$ , we get that $BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}$ . Now, the hard part is to find what $\sin{B}$ is. We therefore want $\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}$ . For the numerator, by inspection $(a + b\sqrt{55})^2$ will not work for integers $a$ and $b$ . The other case is if there is $(a\sqrt{5} + b\sqrt{11})^2$ . By inspection, $5\sqrt{5} - 2\sqrt{11}$ works. Therefore, plugging all this in yields the answer, $\frac{18}{5} \rightarrow \boxed{23}$ . Solution by hyxue

Answer:  23
Problem: There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.

Solution: Let $a_{n}$ be the number of ways to form $n$ -letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical.
Note that, at the end of each $n$ -letter string, there are $3$ possibilities for the last letter chain: it must be either $1$ $2$ , or $3$ letters long. Removing this last chain will make a new string that is $n-1$ $n-2$ , or $n-3$ letters long, respectively.
Therefore we can deduce that $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$
We can see that \[a_{1}=2\] \[a_{2}=2^{2}=4\] \[a_{3}=2^{3}=8\] so using our recursive relation we find \[a_{4}=14\] \[a_{5}=26\] \[a_{6}=48\] \[a_{7}=88\] \[a_{8}=162\] \[a_{9}=298\] \[a_{10}=\boxed{548}\]

Answer:  548
Problem: There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.

Solution: The solution is a simple recursion:
We have three cases for the ending of a string: three in a row, two in a row, and a single:
...AAA $(1)$ ...BAA $(2)$ ...BBA or ...ABA $(3)$
(Here, WLOG each string ends with A. This won't be the case when we actually solve for values in recursion.)
For case $(1)$ , we could only add a B to the end, making it a case $(3)$ . 
For case $(2)$ , we could add an A or a B to the end, making it a case $(1)$ if you add an A, or a case $(3)$ if you add a B.
For case $(3)$ , we could add an A or a B to the end, making it a case $(2)$ or a case $(3)$
Let us create three series to represent the number of permutations for each case: $\{a\}$ $\{b\}$ , and $\{c\}$ representing case $(1)$ $(2)$ , and $(3)$ respectively.
The series have the following relationship:
$a_n=b_{n-1}$ $b_n=c_{n-1}$ $c_n=c_{n-1}+a_{n-1}+b_{n-1}$
For $n=3$ $a_3$ and $b_3$ both equal $2$ $c_3=4$ . With some simple math, we have: $a_{10}=88$ $b_{10}=162$ , and $c_{10}=298$ .
Summing the three up we have our solution: $88+162+298=\boxed{548}$

Answer:  548
Problem: There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.

Solution: This is a recursion problem. Let $a_n$ be the number of valid strings of $n$ letters, where the first letter is $A$ .  Similarly, let $b_n$ be the number of valid strings of $n$ letters, where the first letter is $B$
Note that $a_n=b_{n-1}+b_{n-2}+b_{n-3}$ for all $n\ge4$
Similarly, we have $b_n=a_{n-1}+a_{n-2}+a_{n-3}$ for all $n\ge4$
Here is why:  every valid strings of $n$ letters $(n\ge4)$ where the first letter is $A$ must begin with one of the following:
$AAAB$ - and the number of valid ways is $b_{n-3}$
$AAB$ - and the number of valid ways is $b_{n-2}$
$AB$ - and there are $b_{n-1}$ ways.
We know that $a_1=1$ $a_2=2$ , and $a_3=4$ .  Similarly, we have $b_1=1$ $b_2=2$ , and $b_3=4$ .  We can quickly check our recursion to see if our recursive formula works.  By the formula, $a_4=b_3+b_2+b_1=7$ , and listing out all $a_4$ , we can quickly verify our formula.
Therefore, we have the following:
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\hline a & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\\b & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\end{tabular}$
The total number of valid $10$ letter strings is equal to $a_{10}+b_{10}=274+274=\boxed{548}$

Answer:  548
Problem: There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.

Solution: Playing around with strings gives this approach: We have a certain number of As, then Bs, and so on. Therefore, what if we denoted each solution with numbers like $(3,3,3,1)$ to denote AAABBBAAAB or vice versa (starting with Bs)? Every string can be represented like this!
We can have from 4 to 10 numbers in our parentheses. For each case, we will start with the largest number possible, usually a bunch of 3s, then go down systematically. Realize also that if we are left with just 2s and 1s, there is only one number of 2s and 1s that adds up to the leftover amount. Our final answer is the sum of all of these parenthetical sets [each set multiplied by its permutations, as order matters] multiplied by two [starting with either A or B, and alternating as we go along].
$4 \rightarrow (3,3,3,1) = 4, (3,3,2,2) = 6$
$5 \rightarrow (3, 3, 2, 1, 1) = 30, (3, 2, 2, 2, 1) = 20, (2,2,2,2,2)=1$
$6 \rightarrow (3,3,1,1,1,1) = 15, (3,2,2,1,1,1) = 60, (2,2,2,2,1,1) = 15$
$7 \rightarrow (3,2,1,1,1,1,1) = 42, (2,2,2,1,1,1,1) = 35$
$8 \rightarrow (3,1...1) = 8, (2,2,1...1) = 28$
$9 \rightarrow (2,1...1) = 9$
$10 \rightarrow (1,1....1) =1$
Adding them all up gives you 274; multiplying by 2 gives $\boxed{548}$

Answer:  548
Problem: There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.

Solution: We are going to build the string, 1 character at a time. And, we are going to only care about the streak of letters at the end of the string.
Let $a_n$ be the number of strings of length n that satisfy the problem statement and also has a "streak" of length 1 at the end. ABABBA has a streak of length 1.
Let $b_n$ be the number of strings of length n that satisfy the problem statement and also has of length 2 at the end. ABABBAA has a streak of length 2. There are 2 "A" s at the end of the string.
Let $c_n$ be the number of string of length n that that satisfy the problem statement and also has a "streak" of length 3 at the end. ABABBAAA has a streak of length 3. There are 3 "A" s at the end of the string.
Let's establish a recursive relationship. $a_{n+1} = a_n+b_n+c_n$ , since you can simply break the streak. $b_{n+1} = a_n$ , and $c_{n+1} = b_n$ Since you can just add to the streak.
We can log everything using a table.
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\hline a & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 & 298\\b & 0 & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 \\c & 0 & 0 & 2 & 2 &4 & 8 & 14 & 26 & 48 & 88\end{tabular}$
Adding $a_{10}$ $b_{10}$ $c_{10}$ gets the total number of numbers that doesn't have more than 3 concecutive letters.
That gets a total of $298+162+88 = \boxed{548}$

Answer:  548
Problem: There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.

Solution: Let $S_n$ be the number of n-letter strings satisfying the problem criteria. Then we can easily see $S_1 = 2, S_2 = 4, S_3 = 8, S_4 = 16-2 =14$ . For $n \ge 4$ consider the last three elements of the list. For example to get $S_5$ we could try to take $S_4$ and multiply by 2 since for each $S_4$ string we have two choices for the fifth character. However we have to be careful if the last three characters of $S_4$ are all the same, as in that case we would be overcounting, (i.e if we have a string $\dots aaa$ , we can only add $b$ to the end so that the problem criteria is satisfied. Additionally strings such as $\dots aaa$ or $\dots bbb$ should only be counted once as we only have one choice for the $n$ th character to add ( $b$ and $a$ respectively).
Thus to compute $S_n$ we start with $S_{n-1}$ then take out the strings ending in $\dots bbb$ or $\dots aaa$ . There are $S_{n-4}$ remaining valid strings (i.e if we pick any of the $S_{n-4}$ valid strings, examine the last character, if it is a $b$ then we append $aaa$ to it, and if if it an $a$ we append $bbb$ to it, hence the number of $S_{n-1}$ strings are in one-to-one correspondence with the number of strings in $S_{n-4}$ . Thus we have the recursion $S_n = 2(S_{n-1} - S_{n-4}) + S_{n-4}$ (where we are first taking away $S_{n-4}$ , doubling the result, then adding $S_{n-4}$ back in signifying that we only count it once, as described above).
The recursion simplifies to $S_n = 2S_{n-1} - S_{n-4}$ and we can now quickly compute the remaining values: \[S_5 = 2S_4 - S_1 = 26, S_6 = 2(26)-4 = 48, S_7 = 2(48) - 8 = 88, S_8 = 2(88) - 14 = 162, S_9 = 2(162) - 26 = 298, S_{10} = 2S_9 - 48 = \boxed{548}\]

Answer:  548
Problem: Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$

Solution: If $n = 1$ $a_n = \sin(1) > 0$ .  Then if $n$ satisfies $a_n < 0$ $n \ge 2$ , and \[a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{2\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{2\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].\] Since $2\sin 1$ is positive, it does not affect the sign of $a_n$ .  Let $b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)$ .  Now since $\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)$ and $\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)$ $b_n$ is negative if and only if $\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)$ , or when $n \in [2k\pi - 1, 2k\pi]$ .  Since $\pi$ is irrational, there is always only one integer in the range, so there are values of $n$ such that $a_n < 0$ at $2\pi, 4\pi, \cdots$ .  Then the hundredth such value will be when $k = 100$ and $n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}$

Answer:  628
Problem: Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$

Solution: Notice that $a_n$ is the imaginary part of $\sum_{k=1}^n e^{ik}$ , by Euler's formula. Using the geometric series formula, we find that this sum is equal to \[\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}\] We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have \[\frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}\] We only need to look at the imaginary part, which is \[\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}\] Since $\cos 1 < 1$ $2-2 \cos 1 > 0$ , so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have $\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(n + \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),$ by sum to product. This only holds when $n$ is between $2\pi k - 1$ and $2\pi k$ for integer $k$ [continuity proof here], and since this has exactly one integer solution for every such interval, the $100$ th such $n$ is $\lfloor 200\pi \rfloor = \boxed{628}$

Answer:  628
Problem: Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$

Solution: Similar to solution 2, we set a complex number $z=\cos 1+i\sin 1$ . We start from $z$ instead of $1$ because $k$ starts from $1$ : be careful.
The sum of $z+z^2+z^3+z^4+z^5\dots=\frac{z-z^{n+1}}{1-z}=\frac{z}{z-1}\left(z^n-1\right)$
We are trying to make $n$ so that the imaginary part of this expression is negative.
The argument of $z$ is $1$ . The argument of $z-1$ , however, is a little more tricky. $z-1$ is on a circle centered on $(-1,0)$ with radius $1$ . The change in angle due to $z$ is $1$ with respect to the center, but the angle that $z-1$ makes with the $y$ -axis is $half$ the change, due to Circle Theorems (this intercepted arc is the argument of $z$ ), because the $y$ - axis is tangent to the circle at the origin. So $\text{arg}(z-1)=\frac{\pi+1}{2}$ . Dividing $z$ by $z-1$ subtracts the latter argument from the former, so the angle of the quotient with the $x$ -axis is $\frac{1-\pi}{2}$
We want the argument of the whole expression $-\pi<\theta<0$ . This translates into $\frac{-\pi-1}{2}<\text{arg}\left(z^n-1\right)<\frac{\pi-1}{2}$ $z^n-1$ also consists of points on the circle centered at $(-1,0)$ , so we deal with this argument similarly: the argument of $z^n$ is twice the angle $z^n-1$ makes with the $y$ -axis. Since $z^n-1$ is always negative, $\frac{-3\pi}{2}<\text{arg}\left(z^n-1\right)<\frac{-\pi}{2}$ , and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a $\frac{\pi}{2}-\frac{\pi-1}{2}=\frac{-1}{2}$ angle with the $y$ -axis both ways.
So the argument of $z^n$ must be in the bound $-1<\theta<0$ by doubling, namely the last $z^n$ negative before another rotation. Since there is always one $z^n$ in this category per rotation because $\pi$ is irrational, $n_{100}\equiv z^{628}$ and the answer is $\boxed{628}$

Answer:  628
Problem: 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: The expression we want to find is $2(x^3+y^3) + x^3y^3$
Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x^3+y^3)=945$ , respectively. Dividing the latter by the former equation yields $\frac{x^2-xy+y^2}{xy} = \frac{945}{810}$ . Adding 3 to both sides and simplifying yields $\frac{(x+y)^2}{xy} = \frac{25}{6}$ . Solving for $x+y$ and substituting this expression into the first equation yields $\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810$ . Solving for $xy$ , we find that $xy = 3\sqrt[3]{2}$ , so $x^3y^3 = 54$ . Substituting this into the second equation and solving for $x^3+y^3$ yields $x^3+y^3=\frac{35}{2}$ . So, the expression to evaluate is equal to $2 \times \frac{35}{2} + 54 = \boxed{089}$

Answer:  089
Problem: 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: Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x+y)(x^2-xy+y^2)=945$ , respectively. By the first equation, $x+y=\frac{810}{x^4y^4}$ . Plugging this in to the second equation and simplifying yields $(\frac{x}{y}-1+\frac{y}{x})=\frac{7}{6}$ . Now substitute $\frac{x}{y}=a$ . Solving the quadratic in $a$ , we get $a=\frac{x}{y}=\frac{2}{3}$ or $\frac{3}{2}$ As both of the original equations were symmetric in $x$ and $y$ , WLOG, let $\frac{x}{y}=\frac{2}{3}$ , so $x=\frac{2}{3}y$ . Now plugging this in to either one of the equations, we get the solutions $y=\frac{3(2^{\frac{2}{3}})}{2}$ $x=2^{\frac{2}{3}}$ . Now plugging into what we want, we get $8+54+27=\boxed{089}$

Answer:  089