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]

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]

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?

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]

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.

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]

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]

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]

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 $I$ 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 $BC$.

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 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.

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.

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 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 $ABC$ 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 $\angle PAC = \angle PBA = \angle PCB$.

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]

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?

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 $QIP$.

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?

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.

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]

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.