Let $I$ be the incenter of a triangle $ABC$. The lines passing through $A$ and parallel to $BI, CI$ meet the perpendicular bisector to $AI$ at points $S, T$ respectively. Let $Y$ be the common point of $BT$ and $CS$, and $A^*$ be a point such that $BICA^*$ is a parallelogram. Prove that the midpoint of segment $YA^*$ lies on the excircle of the triangle touching the side $BC$.

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]

The diagonals of a cyclic quadrilateral $ABCD$ meet at point $P$. The bisector of angle $ABD$ meets $AC$ at point $E$, and the bisector of angle $ACD$ meets $BD$ at point $F$. Prove that the lines $AF$ and $DE$ meet on the median of triangle $APD$.

The incircle of a right-angled triangle $ABC$ touches the hypothenuse $AB$ at point $T$. The squares $ATMP$ and $BTNQ$ lie outside the triangle. Prove that the areas of triangles $ABC$ and $TPQ$ are equal.

Let $ABC$ be a triangle with $\angle A=60^\circ$; $AD$, $BE$, and $CF$ be its bisectors; $P, Q$ be the projections of $A$ to $EF$ and $BC$ respectively; and $R$ be the second common point of the circle $DEF$ with $AD$. Prove that $P, Q, R$ are collinear.

Let points $P$ and $Q$ be isogonally conjugated with respect to a triangle $ABC$. The line $PQ$ meets the circumcircle of $ABC$ at point $X$. The reflection of $BC$ about $PQ$ meets $AX$ at point $E$. Prove that $A, P, Q, E$ are concyclic.

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]

Points $A, B, C, D$ on the plane do not form a rectangle. Let the sidelengths of triangle $T$ equal $AB+CD$, $AC+BD$, $AD+BC$. Prove that the triangle $T$ is acute-angled.

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]

Let $ABC$ be a triangle, and let $a$, $b$, and $c$ be the lengths of the sides opposite vertices $A$, $B$, and $C$, respectively. Let $R$ be its circumradius and $r$ its inradius. Suppose that $b + c = 2a$ and $R = 3r$. The excircle relative to vertex $A$ intersects the circumcircle of $ABC$ at points $P$ and $Q$. Let $U$ be the midpoint of side $BC$, and let $I$ be the incenter of $ABC$. Prove that $U$ is the centroid of triangle $QIP$.

The incircle of a triangle $ABC$ centered at $I$ touches the sides $BC, CA$, and $AB$ at points $A_1, B_1, $ and $C_1$ respectively. The excircle centered at $J$ touches the side $AC$ at point $B_2$ and touches the extensions of $AB, BC$ at points $C_2, A_2$ respectively. Let the lines $IB_2$ and $JB_1$ meet at point $X$, the lines $IC_2$ and $JC_1$ meet at point $Y$, the lines $IA_2$ and $JA_1$ meet at point $Z$. Prove that if one of points $X, Y, Z$ lies on the incircle then two remaining points also lie on it.

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]

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 $ABC$ be an isosceles triangle $(AC = BC)$, $O$ be its circumcenter, $H$ be the orthocenter, and $P$ be a point inside the triangle such that $\angle APH = \angle BPO = \pi /2$. Prove that $\angle PAC = \angle PBA = \angle PCB$.

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]

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]

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]

Let $BE$ and $CF$ be the bisectors of a triangle $ABC$. Prove that $2EF \leq BF + CE$.

Young Guy likes to make friends with numbers, so he calls a number “friendly” if the sum of its digits is equal to the product of its digits. How many $3 \text{digit friendly numbers}$ are there?

What is the area of a square in square feet, if each of its diagonals is $4$ feet long?

For which greatest $n$ there exists a convex polyhedron with $n$ faces having the following property: for each face there exists a point outside the polyhedron such that the remaining $n - 1$ faces are seen from this point?

Let $R$ be the rectangle on the cartesian plane with vertices $(0,0)$, $(5,0)$, $(5,7)$, and $(0,7)$. Find the number of squares with sides parallel to the axes and vertices that are lattice points that lie within the region bounded by $R$. [i]Proposed by Boyan Litchev[/i] [hide=Solution][i]Solution[/i]. $\boxed{85}$ We have $(6-n)(8-n)$ distinct squares with side length $n$, so the total number of squares is $5 \cdot 7+4 \cdot 6+3 \cdot 5+2 \cdot 4+1\cdot 3 = \boxed{85}$.[/hide]

A point $P$ lies on one of medians of triangle $ABC$ in such a way that $\angle PAB =\angle PBC =\angle PCA$. Prove that there exists a point $Q$ on another median such that $\angle QBA=\angle QCB =\angle QAC$.

Let $R$ be the rectangle on the cartesian plane with vertices $(0,0)$, $(5,0)$, $(5,7)$, and $(0,7)$. Find the number of squares with sides parallel to the axes and vertices that are lattice points that lie within the region bounded by $R$. [i]Proposed by Boyan Litchev[/i] [hide=Solution][i]Solution[/i]. $\boxed{85}$ We have $(6-n)(8-n)$ distinct squares with side length $n$, so the total number of squares is $5 \cdot 7+4 \cdot 6+3 \cdot 5+2 \cdot 4+1\cdot 3 = \boxed{85}$.[/hide]