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]

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]

For which $n > 0$ it is possible to mark several different points and several different circles on the plane in such a way that: — exactly $n$ marked circles pass through each marked point; — exactly $n$ marked points lie on each marked circle; — the center of each marked circle is marked?

In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled. [i]Proposed by Jerry Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{21}$ Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows: Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane. [asy] import geometry; size(8cm); pair A = (0,sqrt(3)); pair B = (-1,0); pair C = (1,0); pair M = (0,0); for (int i = -1; i <= 2; ++i) { draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3))); draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3))); draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3))); } draw(A--B--C--A, red); dot(M); label("$A$",A+(0,0.25),N); label("$B$",B-(0.25,0),SW); label("$C$",C+(0.25,0),SE); label("$M$",M,S); [/asy] It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases: $1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that $$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case. By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution $$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$). We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution). From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are \begin{align*} (x,y)&=(0,1),(4,7),(56,97) \\ \longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97) \end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex). For $k=3$, using the multiplicative principle we get two new smaller solutions \begin{align*} (x,y)&=(0,3),(12,21) \\ \longrightarrow (n,m,l)&=(0,1,3),(12,1,21) \end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$. For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$. Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$. $2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that $$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions. Our final answer is thus $1+7+13=\boxed{21}$. [/hide]

In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled. [i]Proposed by Jerry Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{21}$ Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows: Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane. [asy] import geometry; size(8cm); pair A = (0,sqrt(3)); pair B = (-1,0); pair C = (1,0); pair M = (0,0); for (int i = -1; i <= 2; ++i) { draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3))); draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3))); draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3))); } draw(A--B--C--A, red); dot(M); label("$A$",A+(0,0.25),N); label("$B$",B-(0.25,0),SW); label("$C$",C+(0.25,0),SE); label("$M$",M,S); [/asy] It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases: $1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that $$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case. By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution $$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$). We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution). From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are \begin{align*} (x,y)&=(0,1),(4,7),(56,97) \\ \longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97) \end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex). For $k=3$, using the multiplicative principle we get two new smaller solutions \begin{align*} (x,y)&=(0,3),(12,21) \\ \longrightarrow (n,m,l)&=(0,1,3),(12,1,21) \end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$. For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$. Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$. $2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that $$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions. Our final answer is thus $1+7+13=\boxed{21}$. [/hide]

Let $H$ be the orthocenter of an acute-angled triangle $ABC$; $A_1, B_1, C_1$ be the touching points of the incircle with $BC, CA, AB$ respectively; $E_A, E_B, E_C$ be the midpoints of $AH, BH, CH$ respectively. The circle centered at $E_A$ and passing through $A$ meets for the second time the bisector of angle $A$ at $A_2$; points $B_2, C_2$ are defined similarly. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ are similar.

Let $(P, P')$ and $(Q, Q')$ be two pairs of points isogonally conjugated with respect to a triangle $ABC$, and $R$ be the common point of lines $PQ$ and $P'Q'$. Prove that the pedal circles of points $P$, $Q$, and $R$ are coaxial.

The common tangents to the circumcircle and an excircle of triangle $ABC$ meet $BC, CA,AB$ at points $A_1, B_1, C_1$ and $A_2, B_2, C_2$ respectively. The triangle $\Delta_1$ is formed by the lines $AA_1, BB_1$, and $CC_1$, the triangle $\Delta_2$ is formed by the lines $AA_2, BB_2,$ and $CC_2$. Prove that the circumradii of these triangles are equal.

Let $P$ and $Q$ be arbitrary points on the side $BC$ of triangle ABC such that $BP = CQ$. The common points of segments $AP$ and $AQ$ with the incircle form a quadrilateral $XYZT$. Find the locus of common points of diagonals of such quadrilaterals.

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]