Paul Revere is currently at $\left(x_0, y_0\right)$ in the Cartesian plane, which is inside a triangle-shaped ship with vertices at $\left(-\dfrac{7}{25},\dfrac{24}{25}\right),\left(-\dfrac{4}{5},\dfrac{3}{5}\right)$, and $\left(\dfrac{4}{5},-\dfrac{3}{5}\right)$. Revere has a tea crate in his hands, and there is a second tea crate at $(0,0)$. He must walk to a point on the boundary of the ship to dump the tea, then walk back to pick up the tea crate at the origin. He notices he can take 3 distinct paths to walk the shortest possible distance. Find the ordered pair $(x_0, y_0)$. [i]Proposed by Derek Zhao[/i] [hide=Solution][i]Solution.[/i] $\left(-\dfrac{7}{25},\dfrac{6}{25}\right)$ Let $L$, $M$, and $N$ be the midpoints of $BC$, $AC$, and $AB$, respectively. Let points $D$, $E$, and $F$ be the reflections of $O = (0,0)$ over $BC$, $AC$, and $AB$, respectively. Notice since $MN \parallel BC$, $BC \parallel EF$. Therefore, $O$ is the orthocenter of $DEF$. Notice that $(KMN)$ is the nine-point circle of $ABC$ because it passes through the midpoints and also the nine-point circle of $DEF$ because it passes through the midpoints of the segments connecting a vertex to the orthocenter. Since $O$ is both the circumcenter of $ABC$ and the orthocenter of $DEF$ and the triangles are $180^\circ$ rotations of each other, Revere is at the orthocenter of $ABC$. The answer results from adding the vectors $OA +OB +OC$, which gives the orthocenter of a triangle.[/hide]

Let [i]Revolution[/i]$(x) = x^3 +Ux^2 +Sx + A$, where $U$, $S$, and $A$ are all integers and $U +S + A +1 = 1773$. Given that [i]Revolution[/i] has exactly two distinct nonzero integer roots $G$ and $B$, find the minimum value of $|GB|$. [i]Proposed by Jacob Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{392}$ Notice that $U + S + A + 1$ is just [i]Revolution[/i]$(1)$ so [i]Revolution[/i]$(1) = 1773$. Since $G$ and $B$ are integer roots we write [i]Revolution[/i]$(X) = (X-G)^2(X-B)$ without loss of generality. So Revolution$(1) = (1-G)^2(1-B) = 1773$. $1773$ can be factored as $32 \cdot 197$, so to minimize $|GB|$ we set $1-G = 3$ and $1-B = 197$. We get that $G = -2$ and $B = -196$ so $|GB| = \boxed{392}$. [/hide]

Sam dumps tea for $6$ hours at a constant rate of $60$ tea crates per hour. Eddie takes $4$ hours to dump the same amount of tea at a different constant rate. How many tea crates does Eddie dump per hour? [i]Proposed by Samuel Tsui[/i] [hide=Solution] [i]Solution.[/i] $\boxed{90}$ Sam dumps a total of $6 \cdot 60 = 360$ tea crates and since it takes Eddie $4$ hours to dump that many he dumps at a rate of $\dfrac{360}{4}= \boxed{90}$ tea crates per hour. [/hide]

The number $2020$ has three different prime factors. What is their sum?

Consider the addition $\begin{tabular}{cccc} & O & N & E \\ + & T & W & O \\ \hline F & O & U & R \\ \end{tabular}$ where different letters represent different nonzero digits. What is the smallest possible value of the four-digit number $FOUR$?

Kanye West's favorite positive integer this year is $c$, and last year it was $c-t=20011$ (a prime), for some positive integer $t$ relatively prime to $c$. His two most streamed albums got $a$ and $b$ streams this year and $a-t$ and $b-t$ streams last year with $a > b > c$. Suppose $a \le 1.6 \times 10^9$ and his favorite integer in each year divides the number of streams for both albums in the corresponding year. Find the largest possible value of $c$.

Bamal, Halvan, and Zuca are playing [i]The Game[/i]. To start, they‘re placed at random distinct vertices on regular hexagon $ABCDEF$. Two or more players collide when they‘re on the same vertex. When this happens, all the colliding players lose and the game ends. Every second, Bamal and Halvan teleport to a random vertex adjacent to their current position (each with probability $\dfrac{1}{2}$), and Zuca teleports to a random vertex adjacent to his current position, or to the vertex directly opposite him (each with probability $\dfrac{1}{3}$). What is the probability that when [i]The Game[/i] ends Zuca hasn‘t lost? [i]Proposed by Edwin Zhao[/i] [hide=Solution][i]Solution.[/i] $\boxed{\dfrac{29}{90}}$ Color the vertices alternating black and white. By a parity argument if someone is on a different color than the other two they will always win. Zuca will be on opposite parity from the others with probability $\dfrac{3}{10}$. They will all be on the same parity with probability $\dfrac{1}{10}$. At this point there are $2 \cdot 2 \cdot 3$ possible moves. $3$ of these will lead to the same arrangement, so we disregard those. The other $9$ moves are all equally likely to end the game. Examining these, we see that Zuca will win in exactly $2$ cases (when Bamal and Halvan collide and Zuca goes to a neighboring vertex). Combining all of this, the answer is $$\dfrac{3}{10}+\dfrac{2}{9} \cdot \dfrac{1}{10}=\boxed{\dfrac{29}{90}}$$ [/hide]

For a real number $x$, the minimum value of the expression $$\frac{2x^2 + x - 3}{x^2 - 2x + 3}$$ can be written in the form $\frac{a-\sqrt{b}}{c}$, where $a, b$, and $c$ are positive integers such that $a$ and $c$ are relatively prime. Find $a + b + c$

Sam dumps tea for $6$ hours at a constant rate of $60$ tea crates per hour. Eddie takes $4$ hours to dump the same amount of tea at a different constant rate. How many tea crates does Eddie dump per hour? [i]Proposed by Samuel Tsui[/i] [hide=Solution] [i]Solution.[/i] $\boxed{90}$ Sam dumps a total of $6 \cdot 60 = 360$ tea crates and since it takes Eddie $4$ hours to dump that many he dumps at a rate of $\dfrac{360}{4}= \boxed{90}$ tea crates per hour. [/hide]

In triangle $ABC$, $AB = 13$, $BC = 14$, and $CA = 15$. Let $M$ be the midpoint of side $AB$, $G$ be the centroid of $\triangle ABC$, and $E$ be the foot of the altitude from $A$ to $BC$. Compute the area of quadrilateral $GAME$. [i]Proposed by Evin Liang[/i] [hide=Solution][i]Solution[/i]. $\boxed{23}$ Use coordinates with $A = (0,12)$, $B = (5,0)$, and $C = (-9,0)$. Then $M = \left(\dfrac{5}{2},6\right)$ and $E = (0,0)$. By shoelace, the area of $GAME$ is $\boxed{23}$.[/hide]

Evaluate $\dbinom{6}{0}+\dbinom{6}{1}+\dbinom{6}{4}+\dbinom{6}{3}+\dbinom{6}{4}+\dbinom{6}{5}+\dbinom{6}{6}$ [i]Proposed by Jonathan Liu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{64}$ We have that $\dbinom{6}{4}=\dbinom{6}{2}$, so $\displaystyle\sum_{n=0}^{6} \dbinom{6}{n}=2^6=\boxed{64}.$ [/hide]

Suppose $h$, $i$, $o$ are real numbers that satisfy the products $hi = 12$, $ooh = 18$, and $hohoho = 27$. Find the value of the product $ohio$.

How many distinct triangles are there with prime side lengths and perimeter $100$? [i]Proposed by Muztaba Syed[/i] [hide=Solution][i]Solution.[/i] $\boxed{0}$ As the perimeter is even, $1$ of the sides must be $2$. Thus, the other $2$ sides are congruent by Triangle Inequality. Thus, for the perimeter to be $100$, both of the other sides must be $49$, but as $49$ is obviously composite, the answer is thus $\boxed{0}$.[/hide]

The number $2020$ has three different prime factors. What is their sum?

Let $NAS$ be a triangle such that $NA=NS=5$ and $AS=6$. Let $D$ be the foot of the altitude from $N$ to $AS$ and $E$ the foot of the altitude from $A$ to $NS$. Point $X$ lies on line $DE$ outside the triangle such that $XA=\tfrac{18}{5}$. Find $XS$.

The equation of line $\ell_1$ is $24x-7y = 319$ and the equation of line $\ell_2$ is $12x-5y = 125$. Let $a$ be the number of positive integer values $n$ less than $2023$ such that for both $\ell_1$ and $\ell_2$ there exists a lattice point on that line that is a distance of $n$ from the point $(20,23)$. Determine $a$. [i]Proposed by Christopher Cheng[/i] [hide=Solution][i]Solution. [/i] $\boxed{6}$ Note that $(20,23)$ is the intersection of the lines $\ell_1$ and $\ell_2$. Thus, we only care about lattice points on the the two lines that are an integer distance away from $(20,23)$. Notice that $7$ and $24$ are part of the Pythagorean triple $(7,24,25)$ and $5$ and $12$ are part of the Pythagorean triple $(5,12,13)$. Thus, points on $\ell_1$ only satisfy the conditions when $n$ is divisible by $25$ and points on $\ell_2$ only satisfy the conditions when $n$ is divisible by $13$. Therefore, $a$ is just the number of positive integers less than $2023$ that are divisible by both $25$ and $13$. The LCM of $25$ and $13$ is $325$, so the answer is $\boxed{6}$.[/hide]

A rectangular tea bag $PART$ has a logo in its interior at the point $Y$ . The distances from $Y$ to $PT$ and $PA$ are $12$ and $9$ respectively, and triangles $\triangle PYT$ and $\triangle AYR$ have areas $84$ and $42$ respectively. Find the perimeter of pentagon $PARTY$. [i]Proposed by Muztaba Syed[/i] [hide=Solution] [i]Solution[/i]. $\boxed{78}$ Using the area and the height in $\triangle PYT$, we see that $PT = 14$, and thus $AR = 14$, meaning the height from $Y$ to $AR$ is $6$. This means $PA = TR = 18$. By the Pythagorean Theorem $PY=\sqrt{12^2+9^2} = 15$ and $YT =\sqrt{12^2 +5^2} = 13$. Combining all of these gives us an answer of $18+14+18+13+15 = \boxed{78}$. [/hide]

The Den has two deals on chicken wings. The first deal is $4$ chicken wings for $3$ dollars, and the second deal is $11$ chicken wings for $ 8$ dollars. If Jeremy has $18$ dollars, what is the largest number of chicken wings he can buy?

Evaluate $\dbinom{6}{0}+\dbinom{6}{1}+\dbinom{6}{4}+\dbinom{6}{3}+\dbinom{6}{4}+\dbinom{6}{5}+\dbinom{6}{6}$ [i]Proposed by Jonathan Liu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{64}$ We have that $\dbinom{6}{4}=\dbinom{6}{2}$, so $\displaystyle\sum_{n=0}^{6} \dbinom{6}{n}=2^6=\boxed{64}.$ [/hide]

Paul Revere is currently at $\left(x_0, y_0\right)$ in the Cartesian plane, which is inside a triangle-shaped ship with vertices at $\left(-\dfrac{7}{25},\dfrac{24}{25}\right),\left(-\dfrac{4}{5},\dfrac{3}{5}\right)$, and $\left(\dfrac{4}{5},-\dfrac{3}{5}\right)$. Revere has a tea crate in his hands, and there is a second tea crate at $(0,0)$. He must walk to a point on the boundary of the ship to dump the tea, then walk back to pick up the tea crate at the origin. He notices he can take 3 distinct paths to walk the shortest possible distance. Find the ordered pair $(x_0, y_0)$. [i]Proposed by Derek Zhao[/i] [hide=Solution][i]Solution.[/i] $\left(-\dfrac{7}{25},\dfrac{6}{25}\right)$ Let $L$, $M$, and $N$ be the midpoints of $BC$, $AC$, and $AB$, respectively. Let points $D$, $E$, and $F$ be the reflections of $O = (0,0)$ over $BC$, $AC$, and $AB$, respectively. Notice since $MN \parallel BC$, $BC \parallel EF$. Therefore, $O$ is the orthocenter of $DEF$. Notice that $(KMN)$ is the nine-point circle of $ABC$ because it passes through the midpoints and also the nine-point circle of $DEF$ because it passes through the midpoints of the segments connecting a vertex to the orthocenter. Since $O$ is both the circumcenter of $ABC$ and the orthocenter of $DEF$ and the triangles are $180^\circ$ rotations of each other, Revere is at the orthocenter of $ABC$. The answer results from adding the vectors $OA +OB +OC$, which gives the orthocenter of a triangle.[/hide]

Consider the addition $\begin{tabular}{cccc} & O & N & E \\ + & T & W & O \\ \hline F & O & U & R \\ \end{tabular}$ where different letters represent different nonzero digits. What is the smallest possible value of the four-digit number $FOUR$?

The equation of line $\ell_1$ is $24x-7y = 319$ and the equation of line $\ell_2$ is $12x-5y = 125$. Let $a$ be the number of positive integer values $n$ less than $2023$ such that for both $\ell_1$ and $\ell_2$ there exists a lattice point on that line that is a distance of $n$ from the point $(20,23)$. Determine $a$. [i]Proposed by Christopher Cheng[/i] [hide=Solution][i]Solution. [/i] $\boxed{6}$ Note that $(20,23)$ is the intersection of the lines $\ell_1$ and $\ell_2$. Thus, we only care about lattice points on the the two lines that are an integer distance away from $(20,23)$. Notice that $7$ and $24$ are part of the Pythagorean triple $(7,24,25)$ and $5$ and $12$ are part of the Pythagorean triple $(5,12,13)$. Thus, points on $\ell_1$ only satisfy the conditions when $n$ is divisible by $25$ and points on $\ell_2$ only satisfy the conditions when $n$ is divisible by $13$. Therefore, $a$ is just the number of positive integers less than $2023$ that are divisible by both $25$ and $13$. The LCM of $25$ and $13$ is $325$, so the answer is $\boxed{6}$.[/hide]

Kendrick Lamar and Drake are cutting their circular beef to share with their fans. The cuts must pass all the way from one side of the beef to the other, and no other modifications may be performed on the beef (e.g. folding, eating, stacking, etc.). Find the minimum number of cuts they will need to split their beef into $2024$ pieces.

Let $MEW$ and $MOG$ be isosceles right triangles such that $E$, $M$, $O$ are collinear in that order and $G$, $M$, $W$ are collinear in that order. Suppose $ME=MW=\sqrt{6-4\sqrt{2}}$ and $MO=MG=\sqrt{6+2\sqrt{2}}$. Find the least possible area of a circle which contains both triangles $MOG$ and $MEW$.

In triangle $ABC$, $AB = 13$, $BC = 14$, and $CA = 15$. Let $M$ be the midpoint of side $AB$, $G$ be the centroid of $\triangle ABC$, and $E$ be the foot of the altitude from $A$ to $BC$. Compute the area of quadrilateral $GAME$. [i]Proposed by Evin Liang[/i] [hide=Solution][i]Solution[/i]. $\boxed{23}$ Use coordinates with $A = (0,12)$, $B = (5,0)$, and $C = (-9,0)$. Then $M = \left(\dfrac{5}{2},6\right)$ and $E = (0,0)$. By shoelace, the area of $GAME$ is $\boxed{23}$.[/hide]