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

Solution: First note that \[\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2\] is a constant not depending on $X$ , so by $[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2$ it suffices to minimize $(AI_1)(AI_2)$ .  Let $a = BC$ $b = AC$ $c = AB$ , and $\alpha = \angle AXB$ .  Remark that \[\angle AI_1B = 180^\circ - (\angle I_1AB + \angle I_1BA) = 180^\circ - \tfrac12(180^\circ - \alpha) = 90^\circ + \tfrac\alpha 2.\] Applying the Law of Sines to $\triangle ABI_1$ gives \[\frac{AI_1}{AB} = \frac{\sin\angle ABI_1}{\sin\angle AI_1B}\qquad\Rightarrow\qquad AI_1 = \frac{c\sin\frac B2}{\cos\frac\alpha 2}.\] Analogously one can derive $AI_2 = \tfrac{b\sin\frac C2}{\sin\frac\alpha 2}$ , and so \[[AI_1I_2] = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{2\cos\frac\alpha 2\sin\frac\alpha 2} = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{\sin\alpha}\geq bc\sin\frac A2 \sin\frac B2\sin\frac C2,\] with equality when $\alpha = 90^\circ$ , that is, when $X$ is the foot of the perpendicular from $A$ to $\overline{BC}$ .  In this case the desired area is $bc\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2$ .  To make this feasible to compute, note that \[\sin\frac A2=\sqrt{\frac{1-\cos A}2}=\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.\] Applying similar logic to $\sin \tfrac B2$ and $\sin\tfrac C2$ and simplifying yields a final answer of \begin{align*}bc\sin\frac A2\sin\frac B2\sin\frac C2&=bc\cdot\dfrac{(a-b+c)(b-c+a)(c-a+b)}{8abc}\\&=\dfrac{(30-32+34)(32-34+30)(34-30+32)}{8\cdot 32}=\boxed{126}

Answer:  126
Problem: Let $\triangle ABC$ have side lengths $AB=30$ $BC=32$ , and $AC=34$ . Point $X$ lies in the interior of $\overline{BC}$ , and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$ , respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$

Solution: First, instead of using angles to find $[AI_1I_2]$ , let's try to find the area of other, simpler figures, and subtract that from $[ABC]$ . However, to do this, we need to be able to figure out the length of the inradii, and so, we need to find $AX$
To minimize $[AI_1I_2]$ , intuitively, we should try to minimize the length of $AX$ , since, after using the $rs=A$ formula for the area of a triangle, we'll be able to minimize the inradii lengths, and thus, eventually minimize the area of $[AI_1I_2]$ . (Proof needed here).
We need to minimize $AX$ . Let $AX=d$ $BX=s$ , and $CX=32-s$ . After an application of Stewart's Theorem, we will get that \[d=\sqrt{s^2-24s+900}\] To minimize this quadratic, $s=12$ whereby we conclude that $d=6\sqrt{21}$
From here, draw perpendiculars down from $I_1$ and $I_2$ to $AB$ and $AC$ respectively, and label the foot of these perpendiculars $D$ and $E$ respectively. After, draw the inradii from $I_1$ to $BX$ , and from $I_2$ to $CX$ , and draw in $I_1I_2$
Label the foot of the inradii to $BX$ and $CX$ $F$ and $G$ , respectively. From here, we see that to find $[AI_1I_2]$ , we need to find $[ABC]$ , and subtract off the sum of $[DBCEI_2I_1], [ADI_1],$ and $[AEI_2]$
$[DBCEI_2I_1]$ can be found by finding the area of two quadrilaterals $[DBFI_1]+[ECGI_2]$ as well as the area of a trapezoid $[FGI_2I_1]$ . If we let the inradius of $ABX$ be $r_1$ and if we let the inradius of $ACX$ be $r_2$ , we'll find, after an application of basic geometry and careful calculations on paper, that $[DBCEI_2I_1]=13r_1+19r_2$
The area of two triangles can be found in a similar fashion, however, we must use $XYZ$ substitution to solve for $AD$ as well as $AE$ . After doing this, we'll get a similar sum in terms of $r_1$ and $r_2$ for the area of those two triangles which is equal to \[\frac{(9+3\sqrt{21})(r_1)}{2} + \frac{(7+3\sqrt{21})(r_2)}{2}.\]
Now we're set. Summing up the area of the Hexagon and the two triangles and simplifying, we get that the formula for $[AI_1I_2]$ is just \[[ABC]-\left(\frac{(35+3\sqrt{21})(r_1)}{2}+\frac{(45+3r_2\sqrt{21})(r_2)}{2}\right).\]
Using Heron's formula, $[ABC]=96\sqrt{21}$ . Solving for $r_1$ and $r_2$ using Heron's in $ABX$ and $ACX$ , we get that $r_1=3\sqrt{21}-9$ and $r_2=3\sqrt{21}-7$ . From here, we just have to plug into our above equation and solve.
Doing so gets us that the minimum area of $AI_1I_2=\boxed{126}.$

Answer:  126
Problem: Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$ . From any vertex of the heptagon except $E$ , the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$ , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that end at $E$

Solution: This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line $E \leftrightarrow P_4 \leftrightarrow P_5 \leftrightarrow S \leftrightarrow P_1 \leftrightarrow P_2 \leftrightarrow P_3 \leftrightarrow E$ . We can count the number of left/right (L/R) paths of length $\le 11$ that start at $S$ and end at either $P_4$ or $P_3$
We can visualize the paths using the common grid counting method by starting at the origin $(0,0)$ , so that a right (R) move corresponds to moving 1 in the positive $x$ direction, and a left (L) move corresponds to moving 1 in the positive $y$ direction. Because we don't want to move more than 2 units left or more than 3 units right, our path must not cross the lines $y = x+2$ or $y = x-3$ . Letting $p(x,y)$ be the number of such paths from $(0,0)$ to $(x,y)$ under these constraints, we have the following base cases:
$p(x,0) = \begin{cases} 1 & x \le 3 \\ 0 & x > 3 \end{cases} \qquad p(0,y) = \begin{cases} 1 & y \le 2 \\ 0 & y > 2 \end{cases}$
and recursive step $p(x,y) = p(x-1,y) + p(x,y-1)$ for $x,y \ge 1$
The filled in grid will look something like this, where the lower-left $1$ corresponds to the origin:
$\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline 0 & 0 & 0 & 0 & \textbf{89} & & & \\ \hline 0 & 0 & 0 & \textbf{28} & 89 & & & \\ \hline 0 & 0 & \textbf{9} & 28 & 61 & 108 & 155 & \textbf{155} \\ \hline 0 & \textbf{3} & 9 & 19 & 33 & 47 & \textbf{47} & 0 \\ \hline \textbf{1} & 3 & 6 & 10 & 14 & \textbf{14} & 0 & 0 \\ \hline 1 & 2 & 3 & 4 & \textbf{4} & 0 & 0 & 0 \\ \hline 1 & 1 & 1 & \textbf{1} & 0 & 0 & 0 & 0 \\ \hline \end{tabular}$
The bolded numbers on the top diagonal represent the number of paths from $S$ to $P_4$ in 2, 4, 6, 8, 10 moves, and the numbers on the bottom diagonal represent the number of paths from $S$ to $P_3$ in 3, 5, 7, 9, 11 moves. We don't care about the blank entries or entries above the line $x+y = 11$ . The total number of ways is $1+3+9+28+89+1+4+14+47+155 = \boxed{351}$

Answer:  351
Problem: Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$ . From any vertex of the heptagon except $E$ , the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$ , the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that end at $E$

Solution: Let $E_n$ denotes the number of sequences with length $n$ that ends at $E$ . Define similarly for the other vertices. We seek for a recursive formula for $E_n$ \begin{align*} E_n&=P_{3_{n-1}}+P_{4_{n-1}} \\ &=P_{2_{n-2}}+P_{5_{n-2}} \\ &=P_{1_{n-3}}+P_{3_{n-3}}+S_{n-3}+P_{4_{n-3}} \\ &=(P_{3_{n-3}}+P_{4_{n-3}})+S_{n-3}+P_{1_{n-3}} \\ &=E_{n-2}+S_{n-3}+P_{1_{n-3}} \\ &=E_{n-2}+P_{1_{n-4}}+P_{5_{n-4}}+S_{n-4}+P_{2_{n-4}} \\ &=E_{n-2}+(S_{n-4}+P_{1_{n-4}})+P_{5_{n-4}}+P_{2_{n-4}} \\ &=E_{n-2}+(E_{n-1}-E_{n-3})+S_{n-5}+P_{4_{n-5}}+P_{1_{n-5}}+P_{3_{n-5}} \\ &=E_{n-2}+(E_{n-1}-E_{n-3})+(S_{n-5}+P_{1_{n-5}})+(P_{4_{n-5}}+P_{3_{n-5}}) \\ &=E_{n-2}+(E_{n-1}-E_{n-3})+(E_{n-2}-E_{n-4})+E_{n-4} \\ &=E_{n-1}+2E_{n-2}-E_{n-3} \\ \end{align*} Computing a few terms we have $E_0=0$ $E_1=0$ $E_2=0$ $E_3=1$ , and $E_4=1$
Using the formula yields $E_5=3$ $E_6=4$ $E_7=9$ $E_8=14$ $E_9=28$ $E_{10}=47$ $E_{11}=89$ , and $E_{12}=155$
Finally adding yields $\sum_{k=0}^{12}E_k=\boxed{351}$

Answer:  351
Problem: David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$ , which can each be inscribed in a circle with radius $1$ . Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$ , and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\tfrac{2}{3}$ $\sin\varphi_B=\tfrac{3}{5}$ , and $\sin\varphi_C=\tfrac{6}{7}$ . All three quadrilaterals have the same area $K$ , which can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Suppose our four sides lengths cut out arc lengths of $2a$ $2b$ $2c$ , and $2d$ , where $a+b+c+d=180^\circ$ . Then, we only have to consider which arc is opposite $2a$ . These are our three cases, so \[\varphi_A=a+c\] \[\varphi_B=a+b\] \[\varphi_C=a+d\] Our first case involves quadrilateral $ABCD$ with $\overarc{AB}=2a$ $\overarc{BC}=2b$ $\overarc{CD}=2c$ , and $\overarc{DA}=2d$
Then, by Law of Sines, $AC=2\sin\left(\frac{\overarc{ABC}}{2}\right)=2\sin(a+b)$ and $BD=2\sin\left(\frac{\overarc{BCD}}{2}\right)=2\sin(a+d)$ . Therefore,
\[K=\frac{1}{2}\cdot AC\cdot BD\cdot \sin(\varphi_A)=2\sin\varphi_A\sin\varphi_B\sin\varphi_C=\frac{24}{35},\] so our answer is $24+35=\boxed{059}$

Answer:  059
Problem: David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$ , which can each be inscribed in a circle with radius $1$ . Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$ , and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\tfrac{2}{3}$ $\sin\varphi_B=\tfrac{3}{5}$ , and $\sin\varphi_C=\tfrac{6}{7}$ . All three quadrilaterals have the same area $K$ , which can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Suppose the four side lengths of the quadrilateral cut out arc lengths of $2a$ $2b$ $2c$ , and $2d$ $a+b+c+d=180^\circ$ . 
Therefore, without losing generality,
\[\varphi_A=a+b\] \[\varphi_B=b+c\] \[\varphi_C=a+c\]
$(1)+(3)-(2)$ $(1)+(2)-(3)$ , and $(2)+(3)-(1)$ yields
\[2a=\varphi_A+\varphi_C-\varphi_B\] \[2b=\varphi_A+\varphi_B-\varphi_C\] \[2c=\varphi_B+\varphi_C-\varphi_A\]
Because $2d=360^\circ-2a-2b-2c,$ Therefore,
\[2d=360^\circ-\varphi_A-\varphi_B-\varphi_C\]
Using the sum-to-product identities , our area of the quadrilateral $K$ then would be
\begin{align*}  K&=\frac{1}{2}(\sin(2a)+\sin(2b)+\sin(2c)+\sin(2d))\\ &=\frac{1}{2}(\sin(\varphi_A+\varphi_B-\varphi_C)+\sin(\varphi_B+\varphi_C-\varphi_A)+\sin(\varphi_C+\varphi_A-\varphi_B)-\sin(\varphi_A+\varphi_B+\varphi_C))\\ &=\frac{1}{2}(2\sin\varphi_B\cos(\varphi_A-\varphi_C)-2\sin\varphi_B\cos(\varphi_A+\varphi_C))\\ &=\frac{1}{2}\cdot2\cdot2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\ &=2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\ &=\frac{24}{35}\\ \end{align*}
Therefore, our answer is $24+35=\boxed{059}$

Answer:  059
Problem: David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$ , which can each be inscribed in a circle with radius $1$ . Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$ , and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\tfrac{2}{3}$ $\sin\varphi_B=\tfrac{3}{5}$ , and $\sin\varphi_C=\tfrac{6}{7}$ . All three quadrilaterals have the same area $K$ , which can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let the four stick lengths be $a$ $b$ $c$ , and $d$ . WLOG, let’s say that quadrilateral $A$ has sides $a$ and $d$ opposite each other, quadrilateral $B$ has sides $b$ and $d$ opposite each other, and quadrilateral $C$ has sides $c$ and $d$ opposite each other. The area of a convex quadrilateral can be written as $\frac{1}{2} d_1 d_2 \sin{\theta}$ , where $d_1$ and $d_2$ are the lengths of the diagonals of the quadrilateral and $\theta$ is the angle formed by the intersection of $d_1$ and $d_2$ . By Ptolemy's theorem $d_1 d_2 = ad+bc$ for quadrilateral $A$ , so, defining $K_A$ as the area of $A$ \[K_A = \frac{1}{2} (ad+bc)\sin{\varphi_A}\] Similarly, for quadrilaterals $B$ and $C$ \[K_B = \frac{1}{2} (bd+ac)\sin{\varphi_B}\] and \[K_C = \frac{1}{2} (cd+ab)\sin{\varphi_C}\] Multiplying the three equations and rearranging, we see that \[K_A K_B K_C = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\sin{\varphi_A}\sin{\varphi_B}\sin{\varphi_B}\] \[K^3 = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\left(\frac{2}{3}\right) \left(\frac{3}{5}\right) \left(\frac{6}{7}\right)\] \[\frac{70}{3}K^3 = (ab+cd)(ac+bd)(ad+bc)\] The circumradius $R$ of a cyclic quadrilateral with side lengths $a$ $b$ $c$ , and $d$ and area $K$ can be computed as $R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}$ .
Inserting what we know, \[1 = \frac{\sqrt{\frac{70}{3}K^3}}{4K}\quad \Rightarrow \quad 16K^2 = \frac{70}{3}K^3\quad \Rightarrow \quad \frac{24}{35} = K\] So our answer is $24 + 35 = \boxed{059}$

Answer:  059
Problem: David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$ , which can each be inscribed in a circle with radius $1$ . Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$ , and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\tfrac{2}{3}$ $\sin\varphi_B=\tfrac{3}{5}$ , and $\sin\varphi_C=\tfrac{6}{7}$ . All three quadrilaterals have the same area $K$ , which can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let the sides of the quadrilaterals be $a,b,c,$ and $d$ in some order such that $A$ has $a$ opposite of $c$ $B$ has $a$ opposite of $b$ , and $C$ has $a$ opposite of $d$ . Then, let the diagonals of $A$ be $e$ and $f$ . Similarly to solution $2$ , we get that $\tfrac{2}{3}(ac+bd)=\tfrac{3}{5}(ab+cd)=\tfrac{6}{7}(ad+bc)=2K$ , but this is also equal to $2\cdot\tfrac{eab+ecd}{4(1)}=2\cdot\tfrac{fad+fbc}{4(1)}$ using the area formula for a triangle using the circumradius and the sides, so $\tfrac{e(ab+cd)}{2}=\tfrac{3}{5}(ab+cd)$ and $\tfrac{f(ad+bc)}{2}=\tfrac{6}{7}(ad+bc)$ . Solving for $e$ and $f$ , we get that $e=\tfrac{6}{5}$ and $f=\tfrac{12}{7}$ , but $K=\tfrac{1}{2}\cdot\tfrac{2}{3}\cdot{}ef$ , similarly to solution $2$ , so $K=\tfrac{24}{35}$ and the answer is $24+35=\boxed{059}$

Answer:  059
Problem: Points $A$ $B$ , and $C$ lie in that order along a straight path where the distance from $A$ to $C$ is $1800$ meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at $A$ and running toward $C$ , Paul starting at $B$ and running toward $C$ , and Eve starting at $C$ and running toward $A$ . When Paul meets Eve, he turns around and runs toward $A$ . Paul and Ina both arrive at $B$ at the same time. Find the number of meters from $A$ to $B$

Solution: We know that in the same amount of time given, Ina will run twice the distance of Eve, and Paul would run quadruple the distance of Eve. Let's consider the time it takes for Paul to meet Eve: Paul would've run 4 times the distance of Eve, which we can denote as $d$ . Thus, the distance between $B$ and $C$ is $4d+d=5d$ . In that given time, Ina would've run twice the distance $d$ , or $2d$ units.
Now, when Paul starts running back towards $A$ , the same amount of time would pass since he will meet Ina at his starting point. Thus, we know that he travels another $4d$ units and Ina travels another $2d$ units.
Therefore, drawing out the diagram, we find that $2d+2d+4d+d=9d=1800 \implies d=200$ , and distance between $A$ and $B$ is the distance Ina traveled, or $4d=4(200)=\boxed{800}$

Answer:  800
Problem: Points $A$ $B$ , and $C$ lie in that order along a straight path where the distance from $A$ to $C$ is $1800$ meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at $A$ and running toward $C$ , Paul starting at $B$ and running toward $C$ , and Eve starting at $C$ and running toward $A$ . When Paul meets Eve, he turns around and runs toward $A$ . Paul and Ina both arrive at $B$ at the same time. Find the number of meters from $A$ to $B$

Solution: Let $x$ be the distance from $A$ to $B$ . Then the distance from $B$ to $C$ is $1800-x$ . Since Eve is the slowest, we can call her speed $v$ , so that Ina's speed is $2v$ and Paul's speed is $4v$
For Paul and Eve to meet, they must cover a total distance of $1800-x$ which takes them a time of $\frac{1800-x}{4v+v}$ . Paul must run the same distance back to $B$ , so his total time is $\frac{2(1800-x)}{5v}$
For Ina to reach $B$ , she must run a distance of $x$ at a speed of $2v$ , taking her a time of $\frac{x}{2v}$
Since Paul and Ina reach $B$ at the same time, we know that $\frac{2(1800-x)}{5v} = \frac{x}{2v}$ (notice that $v$ cancels out on both sides). Solving the equation gives $x = \boxed{800}$

Answer:  800
Problem: Let $a_{0} = 2$ $a_{1} = 5$ , and $a_{2} = 8$ , and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$ $a_{n-1}$ $+$ $a_{n-2}$ $+$ $a_{n-3}$ ) is divided by $11$ . Find $a_{2018} \cdot a_{2020} \cdot a_{2022}$

Solution: When given a sequence problem, one good thing to do is to check if the sequence repeats itself or if there is a pattern.
After computing more values of the sequence, it can be observed that the sequence repeats itself every 10 terms starting at $a_{0}$
$a_{0} = 2$ $a_{1} = 5$ $a_{2} = 8$ $a_{3} = 5$ $a_{4} = 6$ $a_{5} = 10$ $a_{6} = 7$ $a_{7} = 4$ $a_{8} = 7$ $a_{9} = 6$ $a_{10} = 2$ $a_{11} = 5$ $a_{12} = 8$ $a_{13} = 5$
We can simplify the expression we need to solve to $a_{8}\cdot a_{10} \cdot a_{2}$
Our answer is $7 \cdot 2 \cdot 8$ $= \boxed{112}$

Answer:  112
Problem: Let $a_{0} = 2$ $a_{1} = 5$ , and $a_{2} = 8$ , and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$ $a_{n-1}$ $+$ $a_{n-2}$ $+$ $a_{n-3}$ ) is divided by $11$ . Find $a_{2018} \cdot a_{2020} \cdot a_{2022}$

Solution: Notice that the characteristic polynomial of this is $x^3-4x^2-4x-4\equiv 0\pmod{11}$
Then since $x\equiv1$ is a root, using Vieta's formula, the other two roots $r,s$ satisfy $r+s\equiv3$ and $rs\equiv4$
Let $r=7+d$ and $s=7-d$
We have $49-d^2\equiv4$ so $d\equiv1$ . We found that the three roots of the characteristic polynomial are $1,6,8$
Now we want to express $a_n$ in an explicit form as $a(1^n)+b(6^n)+c(8^n)\pmod{11}$
Plugging in $n=0,1,2$ we get
$(*)$ $a+b+c\equiv2,$
$(**)$ $a+6b+8c\equiv5,$
$(***)$ $a+3b+9c\equiv8$
$\frac{(***)-(*)}{2}$ $\implies b+4c\equiv3$ and $(***)-(**)$ $\implies -3b+c\equiv3$
so $a\equiv6,$ $b\equiv1,$ and $c\equiv6$
Hence, $a_n\equiv 6+(6^n)+6(8^n)\equiv(2)^{-n\pmod{10}}+(2)^{3n-1\pmod{10}}-5\pmod{11}$
Therefore
$a_{2018}\equiv4+8-5=7$
$a_{2020}\equiv1+6-5=2$
$a_{2022}\equiv3+10-5=8$
And the answer is $7\times2\times8=\boxed{112}$

Answer:  112
Problem: Find the sum of all positive integers $b < 1000$ such that the base- $b$ integer $36_{b}$ is a perfect square and the base- $b$ integer $27_{b}$ is a perfect cube.

Solution: The first step is to convert $36_{b}$ and $27_{b}$ into base-10 numbers. Then, we can write \[36_{b} = 3b + 6\] and \[27_{b} = 2b + 7\] . It should also be noted that $8 \leq b < 1000$
Because there are less perfect cubes than perfect squares for the restriction we are given on $b$ , it is best to list out all the perfect cubes. Since the maximum $b$ can be is 1000 and $2$ $1000 + 7 = 2007$ , we can list all the perfect cubes less than 2007.
Now, $2b + 7$ must be one of \[3^3, 4^3, ... , 12^3\] . However, $2b + 7$ will always be odd, so we can eliminate the cubes of the even numbers and change our list of potential cubes to \[3^3, 5^3, 7^3, 9^3\text{, and }11^3\]
Because $3b + 6$ is a perfect square and is clearly divisible by 3, it must be divisible by 9, so $b + 2$ is divisible by 3. Thus the cube, which is \[2b + 7 = 2(b + 2) + 3\] , must also be divisible by 3. Therefore, the only cubes that $2b + 7$ could potentially be now are $3^3$ and $9^3$
We need to test both of these cubes to make sure $3b + 6$ is a perfect square.
$\textbf{Case 1:}$ If we set \[3^3 = 2b + 7\] so \[b = 10\] . If we plug this value of b into $3b + 6$ , the expression equals $36$ , which is indeed a perfect square.
$\textbf{Case 2:}$ If we set \[9^3 = 2b + 7\] so \[b = 361\] . If we plug this value of b into $3b + 6$ , the expression equals $1089$ , which is $33^2$
We have proven that both $b = 10$ and $b = 361$ are the only solutions, so \[10 + 361 = \boxed{371}\]

Answer:  371
Problem: Find the sum of all positive integers $b < 1000$ such that the base- $b$ integer $36_{b}$ is a perfect square and the base- $b$ integer $27_{b}$ is a perfect cube.

Solution: The conditions are: \[3b+6 = n^2\] \[2b+7 = m^3\] We can see $n$ is multiple is 3, so let $n=3k$ , then $b= 3k^2-2$ . Substitute $b$ into second condition and we get $m^3=3(2k^2+1)$ . Now we know $m$ is both a multiple of 3 and odd. Also, $m$ must be smaller than 13 for $b$ to be smaller than 1000. So the only two possible values for $m$ are 3 and 9. Test and they both work. The final answer is $10 + 361 =$ $\boxed{371}$ .
-Mathdummy

Answer:  371
Problem: Find the sum of all positive integers $b < 1000$ such that the base- $b$ integer $36_{b}$ is a perfect square and the base- $b$ integer $27_{b}$ is a perfect cube.

Solution: As shown above, let \[3b+6 = n^2\] \[2b+7 = m^3\] such that \[6b+12=2n^2\] \[6b+21=3m^3\]
Subtracting the equations we have \[3m^3-2n^2=9 \implies 3m^3-2n^2-9=0.\]
We know that $m$ and $n$ both have to be integers, because then the base wouldn't be an integer. Furthermore, any integer solution $m$ must divide $9$ by the Rational Root Theorem.
We can instantly know $m \neq -9,-3,-1,1$ since those will have negative solutions.
When $m=3$ we have $n=6$ , so then $b=10$
When $m=9$ we have $n=33$ , so then $b=361$
Therefore, the sum of all possible values of $b$ is \[10+361=\boxed{371}.\]

Answer:  371
Problem: In equiangular octagon $CAROLINE$ $CA = RO = LI = NE =$ $\sqrt{2}$ and $AR = OL = IN = EC = 1$ . The self-intersecting octagon $CORNELIA$ encloses six non-overlapping triangular regions. Let $K$ be the area enclosed by $CORNELIA$ , that is, the total area of the six triangular regions. Then $K =$ $\dfrac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a + b$

Solution: We can draw $CORNELIA$ and introduce some points.
2018 AIME II Problem 4.png
The diagram is essentially a 3x3 grid where each of the 9 squares making up the grid have a side length of 1.
In order to find the area of $CORNELIA$ , we need to find 4 times the area of $\bigtriangleup$ $ACY$ and 2 times the area of $\bigtriangleup$ $YZW$
Using similar triangles $\bigtriangleup$ $ARW$ and $\bigtriangleup$ $YZW$ (We look at their heights), $YZ$ $=$ $\frac{1}{3}$ . Therefore, the area of $\bigtriangleup$ $YZW$ is $\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}$ $=$ $\frac{1}{12}$
Since $YZ$ $=$ $\frac{1}{3}$ and $XY = ZQ$ $XY$ $=$ $\frac{1}{3}$ and $CY$ $=$ $\frac{4}{3}$
Therefore, the area of $\bigtriangleup$ $ACY$ is $\frac{4}{3}\cdot$ $1$ $\cdot$ $\frac{1}{2}$ $=$ $\frac{2}{3}$
Our final answer is $\frac{1}{12}$ $\cdot$ $2$ $+$ $\frac{2}{3}$ $\cdot$ $4$ $=$ $\frac{17}{6}$
$17 + 6 =$ $\boxed{023}$

Answer:  023