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]

A positive $n$ is called [i]sigma rizz[/i] if the sum of its digits is equal to two times the number of digits it has. Find the number of sigma rizz numbers less than $1000.$

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]

In Brawl Stars, Rico can shoot his opponent directly or make his bullet bounce off the wall at the same angle. His opponent is $15$ feet in front of him and there are infinitely long walls $1$ feet to the left and right of Rico. If Rico's bullet travels $d$ feet before hitting the opponent, find the sum of all possible integer values of $d$.

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 $XS$.

On day $1$ of the new year, John Adams and Samuel Adams each drink one gallon of tea. For each positive integer $n$, on the $n$th day of the year, John drinks $n$ gallons of tea and Samuel drinks $n^2$ gallons of tea. After how many days does the combined tea intake of John and Samuel that year first exceed $900$ gallons? [i]Proposed by Aidan Duncan[/i] [hide=Solution] [i]Solution. [/i] $\boxed{13}$ The total amount that John and Samuel have drank by day $n$ is $$\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}=\dfrac{n(n+1)(n+2)}{3}.$$ Now, note that ourdesired number of days should be a bit below $\sqrt[3]{2700}$. Testing a few values gives $\boxed{13}$ as our answer. [/hide]

Determine the least integer $n$ such that for any set of $n$ lines in the 2D plane, there exists either a subset of $1001$ lines that are all parallel, or a subset of $1001$ lines that are pairwise nonparallel. [i]Proposed by Samuel Wang[/i] [hide=Solution][i]Solution.[/i] $\boxed{1000001}$ Since being parallel is a transitive property, we note that in order for this to not exist, there must exist at most $1001$ groups of lines, all pairwise intersecting, with each group containing at most $1001$ lines. Thus, $n = 1000^2 + 1 = \boxed{1000001}$.[/hide]

In Survev.io, Calvin observes that he has exactly twice as much blue ammo as red ammo. After firing one blue bullet and $9$ red bullets, he remarks that the amount of blue ammo he has is divisible by $5$ and the amount of red ammo he has is divisible by $7$. Find the least amount of red ammo he could have started with.

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 $c$.

Evin and Jerry are playing a game with a pile of marbles. On each players' turn, they can remove $2$, $3$, $7$, or $8$ marbles. If they can’t make a move, because there's $0$ or $1$ marble left, they lose the game. Given that Evin goes first and both players play optimally, for how many values of $n$ from $1$ to $1434$ does Evin lose the game? [i]Proposed by Evin Liang[/i] [hide=Solution][i]Solution.[/i] $\boxed{573}$ Observe that no matter how many marbles a one of them removes, the next player can always remove marbles such that the total number of marbles removed is $10$. Thus, when the number of marbles is a multiple of $10$, the first player loses the game. We analyse this game based on the number of marbles modulo $10$: If the number of marbles is $0$ modulo $10$, the first player loses the game If the number of marbles is $2$, $3$, $7$, or $8$ modulo $10$, the first player wins the game by moving to $0$ modulo 10 If the number of marbles is $5$ modulo $10$, the first player loses the game because every move leads to $2$, $3$, $7$, or $8$ modulo $10$ In summary, the first player loses if it is $0$ mod 5, and wins if it is $2$ or $3$ mod $5$. Now we solve the remaining cases by induction. The first player loses when it is $1$ modulo $5$ and wins when it is $4$ modulo $5$. The base case is when there is $1$ marble, where the first player loses because there is no move. When it is $4$ modulo $5$, then the first player can always remove $3$ marbles and win by the inductive hypothesis. When it is $1$ modulo $5$, every move results in $3$ or $4$ modulo $5$, which allows the other player to win by the inductive hypothesis. Thus, Evin loses the game if n is $0$ or $1$ modulo $5$. There are $\boxed{573}$ such values of $n$ from $1$ to $1434$.[/hide]

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]

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]

Travis Scott says "FEIN'' every $0.8$ seconds. Find the tens digit of the number of times he says "FEIN'' in $1$ minute.

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

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]

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]

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 $FOUR$?

On day $1$ of the new year, John Adams and Samuel Adams each drink one gallon of tea. For each positive integer $n$, on the $n$th day of the year, John drinks $n$ gallons of tea and Samuel drinks $n^2$ gallons of tea. After how many days does the combined tea intake of John and Samuel that year first exceed $900$ gallons? [i]Proposed by Aidan Duncan[/i] [hide=Solution] [i]Solution. [/i] $\boxed{13}$ The total amount that John and Samuel have drank by day $n$ is $$\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}=\dfrac{n(n+1)(n+2)}{3}.$$ Now, note that ourdesired number of days should be a bit below $\sqrt[3]{2700}$. Testing a few values gives $\boxed{13}$ as our answer. [/hide]

In Pokemon, there are $10$ indistinguishable Poke Beans in a pile. Pikachu eats a prime number of Poke Beans. Charmander eats an even number of Poke Beans. Snorlax eats an odd number of Poke Beans. Find the number of ways for the three Pokemon to eat all $10$ Poke Beans.

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]

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]

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]

Eminem is trying to find the real Slim Shady in a row of $2025$ indistinguishable Slim Shady clones, one of which is the real Slim Shady. Eminem randomly guesses, and if he guesses wrong, a new clone joins the row and all the clones randomly rearrange themselves. He keeps guessing as more identical clones are added, trying to find the real Slim Shady. Find the probability that he will eventually find him within $15$ guesses.

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]

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 $MEW$.