[/quote]
,
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
be an integer, let 
be a set of 
elements, and let 
,
such that ![\[ A_i \cap A_j \not= \emptyset, A_i \cap A_k \not= \emptyset, A_j \cap A_k \not= \emptyset, \;\textrm{imply}\ \;A_i \cap A_j \cap A_k \not= \emptyset \ .\]](http://latex.artofproblemsolving.com/c/1/1/c113f5cdce00dfec6714c382a8e1bc74f188f025.png)
Show that 
.
such that 
or 
,![\[
1 \leq a_1 < a_2 < \cdots < a_{(p-1)/2} < p
\]](http://latex.artofproblemsolving.com/6/8/e/68e3d73edec43b2cc480832adf23bbbb5f4c7bee.png)
such that![\[
\prod_{\substack{1 \leq i < j \leq (p-1)/2}} (a_i + a_j)^2 \equiv 1 \pmod{p}.
\]](http://latex.artofproblemsolving.com/2/9/e/29edf6ab8cee2a526e7753a140c5a291e987856a.png)
denote the maximum number of 
-
board such that each 
such that![\[
f(N) \leq cN^2
\]](http://latex.artofproblemsolving.com/7/4/8/748109c3d63203f9a67a167434e10a909f38e83f.png)
for all positive integers 
.
be a graph colored using 
colors. We say that a vertex is forced if it has neighbors in all the other 
colors.
-
colors such that at least 
of the colors have a forced vertex of that color.
vertices of the same color may be adjacent.
,
,
,
for all 
,
be an isosceles triangle with 
.
be a point on 
.
be a point inside the triangle such that 
and![\[
CL \cdot BD = BL \cdot CD.
\]](http://latex.artofproblemsolving.com/1/6/7/16771838c95f86e92f79b8d049e46ab473e6287d.png)
Prove that the circumcenter of triangle 
lies on line 
.
triangle with 
.
,
,
.
,
lines intersect at point 
.
,
of functions by 
for 
.
is a polynomial with integer coefficients.
be a convex pentagon such that![\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]](http://latex.artofproblemsolving.com/d/3/d/d3d9a82f4318190298a8f008d417a364e03f1fca.png)
The diagonals 
and 
meet at 
.
bisects the side 
.
,
,
are fixed on a line in this order. Let 
be a circle passing through 
.
at 
.
and the line 
be positive real numbers such that 
.
.
q
answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?
,
,
where 
Y by OronSH, Therealway, CatinoBarbaraCombinatoric, ohiorizzler1434
Let 
be a positive integer, and consider an 
grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence.
Prove that the cells of the grid cannot be partitioned into less than right-down or right-up paths. For example, the following partition of the 
grid uses 
paths.
![[asy]
size(4cm);
draw((5,-1)--(0,-1)--(0,-2)--(5,-2)--(5,-3)--(0,-3)--(0,-4)--(5,-4),gray+linewidth(0.5)+miterjoin);
draw((1,-5)--(1,0)--(2,0)--(2,-5)--(3,-5)--(3,0)--(4,0)--(4,-5),gray+linewidth(0.5)+miterjoin);
draw((0,0)--(5,0)--(5,-5)--(0,-5)--cycle,black+linewidth(2.5)+miterjoin);
draw((0,-1)--(3,-1)--(3,-2)--(1,-2)--(1,-4)--(4,-4)--(4,-3)--(2,-3)--(2,-2),black+linewidth(2.5)+miterjoin);
draw((3,0)--(3,-1),black+linewidth(2.5)+miterjoin);
draw((1,-4)--(1,-5),black+linewidth(2.5)+miterjoin);
draw((4,-3)--(4,-1)--(5,-1),black+linewidth(2.5)+miterjoin);
[/asy]](//latex.artofproblemsolving.com/4/7/7/477145f5dd40b84a4daff1c2fdd46ae90f013fb7.png)
Proposed by Zixiang Zhou, Canada
be a positive integer, and consider an 
grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence.Prove that the cells of the
grid cannot be partitioned into less than right-down or right-up paths. For example, the following partition of the 
grid uses 
paths.![[asy]
size(4cm);
draw((5,-1)--(0,-1)--(0,-2)--(5,-2)--(5,-3)--(0,-3)--(0,-4)--(5,-4),gray+linewidth(0.5)+miterjoin);
draw((1,-5)--(1,0)--(2,0)--(2,-5)--(3,-5)--(3,0)--(4,0)--(4,-5),gray+linewidth(0.5)+miterjoin);
draw((0,0)--(5,0)--(5,-5)--(0,-5)--cycle,black+linewidth(2.5)+miterjoin);
draw((0,-1)--(3,-1)--(3,-2)--(1,-2)--(1,-4)--(4,-4)--(4,-3)--(2,-3)--(2,-2),black+linewidth(2.5)+miterjoin);
draw((3,0)--(3,-1),black+linewidth(2.5)+miterjoin);
draw((1,-4)--(1,-5),black+linewidth(2.5)+miterjoin);
draw((4,-3)--(4,-1)--(5,-1),black+linewidth(2.5)+miterjoin);
[/asy]](http://latex.artofproblemsolving.com/4/7/7/477145f5dd40b84a4daff1c2fdd46ae90f013fb7.png)
Proposed by Zixiang Zhou, Canada
This post has been edited 8 times. Last edited by Scilyse, Jan 23, 2025, 12:41 PM
Reason: add "proposed by"
Reason: add "proposed by"
![[asy]
usepackage("tikz");
label("
\begin{tikzpicture}
\fill[purple] (0,0)--(0,1)--(1,2)--(1,1)--cycle;
\fill[purple] (2,0)--(3,1)--(4,1)--(3,0)--cycle;
\fill[purple] (5,1)--(5,2)--(6,1)--(6,0)--cycle;
\fill[purple] (7,1)--(8,1)--(9,0)--(8,0)--cycle;
\end{tikzpicture}
");
[/asy]](http://latex.artofproblemsolving.com/a/c/6/ac6eca9988976b89caa7ef37c17df6657edeebd6.png)
rightdown or rightup paths, we can find a corresponding packing of good parallelograms leaving an area of ![[asy]
usepackage("tikz");
label("
\begin{tikzpicture}
\fill[yellow] (0,0) rectangle (1,1);
\fill[yellow] (0,1) rectangle (1,2);
\fill[yellow] (0,2) rectangle (1,3);
\fill[yellow] (0,3) rectangle (1,4);
\fill[red] (0,4) rectangle (1,5);
\fill[orange] (1,0) rectangle (2,1);
\fill[green] (1,1) rectangle (2,2);
\fill[green] (1,2) rectangle (2,3);
\fill[yellow] (1,3) rectangle (2,4);
\fill[red] (1,4) rectangle (2,5);
\fill[orange] (2,0) rectangle (3,1);
\fill[green] (2,1) rectangle (3,2);
\fill[blue] (2,2) rectangle (3,3);
\fill[yellow] (2,3) rectangle (3,4);
\fill[red] (2,4) rectangle (3,5);
\fill[orange] (3,0) rectangle (4,1);
\fill[green] (3,1) rectangle (4,2);
\fill[blue] (3,2) rectangle (4,3);
\fill[blue] (3,3) rectangle (4,4);
\fill[blue] (3,4) rectangle (4,5);
\fill[orange] (4,0) rectangle (5,1);
\fill[orange] (4,1) rectangle (5,2);
\fill[orange] (4,2) rectangle (5,3);
\fill[orange] (4,3) rectangle (5,4);
\fill[blue] (4,4) rectangle (5,5);
\draw (0,0) grid (5,5);
\end{tikzpicture}
");
label("
\begin{tikzpicture}
\filldraw[color=black,fill=yellow,thick] (1,0)--(0,1)--(0,2)--(1,1)--cycle;
\filldraw[color=black,fill=yellow,thick] (1,1)--(0,2)--(0,3)--(1,2)--cycle;
\filldraw[color=black,fill=yellow,thick] (1,2)--(0,3)--(0,4)--(1,3)--cycle;
\filldraw[color=black,fill=yellow,thick] (0,4)--(1,4)--(2,3)--(1,3)--cycle;
\filldraw[color=black,fill=yellow,thick] (1,4)--(2,4)--(3,3)--(2,3)--cycle;
\filldraw[color=black,fill=red,thick] (0,5)--(1,5)--(2,4)--(1,4)--cycle;
\filldraw[color=black,fill=red,thick] (1,5)--(2,5)--(3,4)--(2,4)--cycle;
\filldraw[color=black,fill=blue,thick] (2,3)--(3,3)--(4,2)--(3,2)--cycle;
\filldraw[color=black,fill=blue,thick] (3,3)--(3,4)--(4,3)--(4,2)--cycle;
\filldraw[color=black,fill=blue,thick] (3,4)--(3,5)--(4,4)--(4,3)--cycle;
\filldraw[color=black,fill=blue,thick] (3,5)--(4,5)--(5,4)--(4,4)--cycle;
\filldraw[color=black,fill=green,thick] (2,3)--(1,2)--(1,1)--(2,2)--cycle;
\filldraw[color=black,fill=green,thick] (1,1)--(2,2)--(3,2)--(2,1)--cycle;
\filldraw[color=black,fill=green,thick] (2,1)--(3,2)--(4,2)--(3,1)--cycle;
\filldraw[color=black,fill=orange,thick] (1,1)--(2,0)--(3,0)--(2,1)--cycle;
\filldraw[color=black,fill=orange,thick] (2,1)--(3,0)--(4,0)--(3,1)--cycle;
\filldraw[color=black,fill=orange,thick] (3,1)--(4,0)--(5,0)--(4,1)--cycle;
\filldraw[color=black,fill=orange,thick] (4,1)--(4,2)--(5,1)--(5,0)--cycle;
\filldraw[color=black,fill=orange,thick] (4,2)--(4,3)--(5,2)--(5,1)--cycle;
\filldraw[color=black,fill=orange,thick] (4,3)--(4,4)--(5,3)--(5,2)--cycle;
\draw (0,0) rectangle (5,5);
\end{tikzpicture}
",(5,0),E);
[/asy]](http://latex.artofproblemsolving.com/4/7/4/474d6b77ac2aa274612c4525594fedb25f7cb0b8.png)
or a 
in each cell such that it does not intersect any of the good parallelograms. Now, view these segments as mirrors, and consider a laser fired from each of the 
boundary edges, bouncing along these mirrors until it exits at some other edge. It is well known that this forms a bijection between the boundary edges and these laser beams altogether visit every triangular region at most once. Consider such a pair of boundary edges. There are several cases:
corresponding to this pair.
corresponding to this pair.
times and only for vertical edges. Then out of the remaining boundary edges, there are 
horizontal boundary edges and 
vertical boundary edges. It follows that there must be at least ![\[
1(t)+0.5(2N-2t) = N
\]](http://latex.artofproblemsolving.com/8/7/6/876fa1bd2fee54d255483b6792956b248e63401e.png)
as required.
having slope 1. For each right-up path, cover the path by parallelograms with sides of length 
.![[asy]
size(0, 3.5cm);
usepackage("amsmath");
int n=6;
draw(box((0,0), (n,n)), linewidth(1));
for(int i=1; i<n; ++i){
draw((i,0) -- (i,n), mediumgray);
draw((0,i) -- (n,i), mediumgray);
}
pen p = linewidth(1);
draw((1,5) -- (1,1) -- (6,1), p);
draw((2,1) -- (2,4) -- (4,4) -- (4,6), p);
draw((0,5) -- (4,5), p);
draw((2,2) -- (5,2) -- (5,5) -- (6,5), p);
draw((3,4) -- (3,3) -- (5,3), p);
picture pic;
pen f = mediumgray;
fill(pic,(0,4)--(0,5)--(1,5)--cycle,f);
fill(pic,(5,0)--(6,0)--(6,1)--cycle,f);
fill(pic,(1,1)--(2,1)--(1,2)--cycle,f);
fill(pic,(4,4)--(4,5)--(3,5)--cycle,f);
fill(pic,(0,5)--(0,6)--(1,6)--cycle,f);
fill(pic,(4,5)--(4,6)--(3,5)--cycle,f);
fill(pic,(2,4)--(3,4)--(2,3)--cycle,f);
fill(pic,(4,2)--(5,2)--(5,3)--cycle,f);
fill(pic,(3,3)--(4,3)--(3,4)--cycle,f);
fill(pic,(6,6)--(5,6)--(6,5)--cycle,f);
fill(pic,(2,1)--(3,1)--(2,2)--cycle,f);
fill(pic,(6,4)--(6,5)--(5,5)--cycle,f);
draw(pic,box((0,0), (n,n)), linewidth(1));
pen p = linewidth(1);
draw(pic,(1,5) -- (1,1) -- (6,1), p);
draw(pic,(2,1) -- (2,4) -- (4,4) -- (4,6), p);
draw(pic,(0,5) -- (4,5), p);
draw(pic,(2,2) -- (5,2) -- (5,5) -- (6,5), p);
draw(pic,(3,4) -- (3,3) -- (5,3), p);
void up(real x, real y) {draw(pic,(x,y) -- (x+1,y+1), p);}
void down(real x, real y) {draw(pic,(x+1,y) -- (x,y+1), p);}
up(1,0); up(2,0); up(3,0); up(4,0); up(5,0);
up(0,0); up(0,1); up(0,2); up(0,3); up(0,4);
up(0,5); up(1,5); up(2,5); up(3,5);
down(1,1); down(1,2); down(1,3); down(1,4); down(2,4); down(3,4);
down(2,1); down(3,1); down(4,1); down(5,1); down(5,2); down(5,3); down(5,4);
up(2,3); up(2,2); up(3,2); up(4,2);
down(3,3); down(4,3); down(4,4); down(4,5); down(5,5);
label("$\implies$", (7,3));
add(shift(8,0)*pic);
[/asy]](http://latex.artofproblemsolving.com/a/7/9/a7964a1d2a09367f9f9a7f76e09b2f5c8a41d81e.png)
![[asy]
size(3.5cm,0);
int n=6;
draw(box((0,0), (n,n)), linewidth(1));
pen p = linewidth(1);
void up(real x, real y) {draw((x,y) -- (x+1,y+1), p);}
void down(real x, real y) {draw((x+1,y) -- (x,y+1), p);}
up(1,0); up(2,0); up(3,0); up(4,0); up(5,0);
up(0,0); up(0,1); up(0,2); up(0,3); up(0,4);
up(0,5); up(1,5); up(2,5); up(3,5);
down(1,1); down(1,2); down(1,3); down(1,4); down(2,4); down(3,4);
down(2,1); down(3,1); down(4,1); down(5,1); down(5,2); down(5,3); down(5,4);
up(2,3); up(2,2); up(3,2); up(4,2);
down(3,3); down(4,3); down(4,4); down(4,5); down(5,5);
[/asy]](http://latex.artofproblemsolving.com/2/e/f/2ef509965e557b62c0f2c9dd1d7c21dc84f85b28.png)
is uncovered.
.
square. It turns out the other parts that are split gets glued back.
share an endpoint and the endpoint is blue or red (each one ends next to the spot from which a path enters the red cell). There are at most 
blue cells, so there is an "arrow path" from the first column to the last. But each cell of it has to be visited by a different path, contradiction.
must hold.
and 
)
or some other set of distinct real numbers, so we'll enumerate with real numbers. We start with the observation that path A lies over path B if in at least one column which they both intersect A lies over B (it follows from the property of the paths). Now we finally start enumerating. We start with the left-most column. We enumerate the paths crossing it with some real numbers in the right order. Now we go to the next column. The new paths that cross this column we enumerate with real numbers accordingly. and so on. Because of the property already mentioned this enumeration works. Now convert the real numbers into 
.
step by step. We will prove the following. Right after the 
-
-
cells which are not yet occupied by paths for 
(
-
,
it is obvious. Assume it is proven for 
be the unoccupied cells in the 
be the cell right above 
.
and 
then 
would all be contained in the path, the impossibility of which is easily seen. Now we are done.
of the corners of the board (and thus consists of two directions) an amogus. We will induct down on ![[asy]
int W=5; int H=3; int T=7; for (int i=0; i<=W; ++i) { draw( (i,0)--(i,H), rgb("333333")+dashed );
draw( (i+T,0)--(i+T,H), rgb("333333")+dashed );
} for (int i=0; i<=H; ++i) { draw( (0,i)--(W,i), rgb("333333")+dashed ); draw( (0+T,i)--(W+T,i), rgb("333333")+dashed ); }
label("$\mapsto$", (6, 1.5));
pen p1 = blue+white+white; pen p2 = green+white+white; filldraw((0,0)--(0,1)--(2,1)--(2,0)--cycle, p1); filldraw((2,0)--(2,3)--(3,3)--(3,1)--(5,1)--(5,0)--cycle, p2); filldraw((3,1)--(5,1)--(5,2)--(4,2)--(4,3)--(3,3)--cycle, p2);
filldraw((T+0,0)--(T+0,1)--(T+3,1)--(T+3,0)--cycle, p1); filldraw((T+2,1)--(T+2,3)--(T+3,3)--(T+3,2)--(T+4,2)--(T+4,1)--(T+5,1)--(T+5,0)--(T+3,0)--(T+3,1)--cycle, p2); filldraw((T+3,2)--(T+3,3)--(T+5,3)--(T+5,1)--(T+4,1)--(T+4,2)--cycle, p2);
label("$A$", (1.5, 0.5)); label("$B$", (2.5, 0.5)); label("$C$", (3.5, 1.5));
label("$A$", (T+2.5, 0.5)); label("$B$", (T+3.5, 1.5)); label("$C$", (T+4.5, 2.5));
[/asy]](http://latex.artofproblemsolving.com/a/0/9/a09f215557e428b72b230e74dc171427b5611216.png)
and 
where ![[asy]
int W=8; int H=6; for (int i=0; i<=W; ++i) { draw( (i,0)--(i,H), rgb("333333")+dashed );
} for (int i=0; i<=H; ++i) { draw( (0,i)--(W,i), rgb("333333")+dashed ); }
pen p1 = orange+white+white; pen p2 = blue+white+white; pen p3 = green+white+white; pen p4 = red+white+white;
filldraw((0,0)--(2,0)--(2,3)--(1,3)--(1,1)--(0,1)--cycle, p1); filldraw((2,0)--(2,5)--(3,5)--(3,1)--(8,1)--(8,0)--cycle, p1);
label("$T_0$", (1.5,2.5)); label("$L_0$", (2.5,4.5)); [/asy]](http://latex.artofproblemsolving.com/5/d/8/5d8952b88f6cc83a5c061d3e772b76b7c21c09d0.png)
paths cutters. Call the edge between ![[asy]
int W=8; int H=6; for (int i=0; i<=W; ++i) { draw( (i,0)--(i,H), rgb("333333")+dashed );
} for (int i=0; i<=H; ++i) { draw( (0,i)--(W,i), rgb("333333")+dashed ); }
pen p1 = orange+white+white; pen p2 = blue+white+white; pen p3 = green+white+white; pen p4 = red+white+white;
filldraw((0,0)--(2,0)--(2,3)--(1,3)--(1,1)--(0,1)--cycle, p1); filldraw((2,0)--(2,5)--(3,5)--(3,1)--(8,1)--(8,0)--cycle, p1);
filldraw((1,5)--(1,6)--(8,6)--(8,4)--(7,4)--(7,5)--cycle, p3); draw((2,5.05)--(7,5.05), red); draw((1,5.05)--(2,5.05), purple);
label("$T_0$", (1.5,2.5)); label("$L_0$", (2.5,4.5)); label("$B$", (1.5, 5.5)); [/asy]](http://latex.artofproblemsolving.com/2/2/7/2271f935fd3281633e7b5704244981a888d09c96.png)
from left to right. This serves as a bounded monovariant.
before hitting a noncutter.
containing it is also right-down.![[asy]
int W=8; int H=6; for (int i=0; i<=W; ++i) { draw( (i,0)--(i,H), rgb("333333")+dashed ); } for (int i=0; i<=H; ++i) { draw( (0,i)--(W,i), rgb("333333")+dashed ); } pen p1 = orange+white+white; pen p2 = blue+white+white; pen p3 = green+white+white; pen p4 = red+white+white; filldraw((0,0)--(2,0)--(2,3)--(1,3)--(1,1)--(0,1)--cycle, p1); filldraw((2,0)--(2,5)--(3,5)--(3,1)--(8,1)--(8,0)--cycle, p1); filldraw((3,1)--(3,5)--(4,5)--(4,2)--(6,2)--(6,1)--cycle, p2); filldraw((4,2)--(4,5)--(5,5)--(5,3)--(6,3)--(6,2)--cycle, p2);
filldraw((1,5)--(1,6)--(8,6)--(8,4)--(7,4)--(7,5)--cycle, p3); draw((2,5.05)--(7,5.05), red); draw((1,5.05)--(2,5.05), purple); filldraw((5,4)--(5,5)--(7,5)--(7,3)--(6,3)--(6,4)--cycle, p4);
label("$T_0$", (1.5,2.5)); label("$L_0$", (2.5,4.5)); label("$L_1$", (3.5,4.5)); label("$L_2$", (4.5,4.5)); label("$S$", (5.5,4.5));
label("$B$", (1.5, 5.5)); [/asy]](http://latex.artofproblemsolving.com/d/d/1/dd136e7df2c280381aa7f2b7fc28ce32c8f829d8.png)
![[asy]
int W=8; int H=6; for (int i=0; i<=W; ++i) { draw( (i,0)--(i,H), rgb("333333")+dashed ); } for (int i=0; i<=H; ++i) { draw( (0,i)--(W,i), rgb("333333")+dashed ); } pen p1 = orange+white+white; pen p2 = blue+white+white; pen p3 = green+white+white; pen p4 = red+white+white; filldraw((0,0)--(2,0)--(2,3)--(1,3)--(1,1)--(0,1)--cycle, p1); filldraw((2,0)--(2,4)--(3,4)--(3,1)--(8,1)--(8,0)--cycle, p1); filldraw((3,1)--(3,4)--(4,4)--(4,2)--(6,2)--(6,1)--cycle, p2); filldraw((4,2)--(4,4)--(5,4)--(5,3)--(6,3)--(6,2)--cycle, p2);
filldraw((1,5)--(1,6)--(8,6)--(8,4)--(7,4)--(7,5)--cycle, p3); draw((2,5.05)--(7,5.05), red); draw((1,5.05)--(2,5.05), purple);
filldraw((2,4)--(2,5)--(7,5)--(7,3)--(6,3)--(6,4)--cycle, p4);
label("$T_0$", (1.5,2.5)); label("$L_0$", (2.5,3.5)); label("$L_1$", (3.5,3.5)); label("$L_2$", (4.5,3.5)); label("$S$", (2.5,4.5));
label("$B$", (1.5, 5.5)); [/asy]](http://latex.artofproblemsolving.com/f/b/0/fb0e73f0af4ffb1ac84a4c90e90f02880f78d4c9.png)
![[asy]
int W=8; int H=6; for (int i=0; i<=W; ++i) { draw( (i,0)--(i,H), rgb("333333")+dashed ); } for (int i=0; i<=H; ++i) { draw( (0,i)--(W,i), rgb("333333")+dashed ); } pen p1 = orange+white+white; pen p2 = blue+white+white; pen p3 = green+white+white; pen p4 = red+white+white; filldraw((0,0)--(2,0)--(2,3)--(1,3)--(1,1)--(0,1)--cycle, p1); filldraw((2,0)--(2,5)--(3,5)--(3,1)--(8,1)--(8,0)--cycle, p1); filldraw((3,1)--(3,5)--(4,5)--(4,2)--(6,2)--(6,1)--cycle, p2); filldraw((4,2)--(4,5)--(5,5)--(5,3)--(6,3)--(6,2)--cycle, p2);
filldraw((1,5)--(1,6)--(8,6)--(8,4)--(7,4)--(7,5)--cycle, p3); draw((2,5.05)--(7,5.05), red); draw((1,5.05)--(2,5.05), purple); filldraw((5,4)--(5,5)--(7,5)--(7,4)--(6,4)--(6,3)--(5,3)--cycle, p4);
label("$T_0$", (1.5,2.5)); label("$L_0$", (2.5,4.5)); label("$L_1$", (3.5,4.5)); label("$L_2$", (4.5,4.5)); label("$S$", (5.5,3.5));
label("$B$", (1.5, 5.5)); [/asy]](http://latex.artofproblemsolving.com/e/6/6/e660e93461a73969a7bfcf39b1a0f38afd22bae9.png)
,
,![[asy]
int W=8; int H=6; for (int i=0; i<=W; ++i) { draw( (i,0)--(i,H), rgb("333333")+dashed ); } for (int i=0; i<=H; ++i) { draw( (0,i)--(W,i), rgb("333333")+dashed ); } pen p1 = orange+white+white; pen p2 = blue+white+white; pen p3 = green+white+white; pen p4 = red+white+white; filldraw((0,0)--(2,0)--(2,3)--(1,3)--(1,1)--(0,1)--cycle, p1); filldraw((3,0)--(3,4)--(4,4)--(4,1)--(8,1)--(8,0)--cycle, p1); filldraw((4,1)--(4,4)--(5,4)--(5,2)--(6,2)--(6,1)--cycle, p2); filldraw((5,2)--(5,4)--(6,4)--(6,2)--cycle, p2);
filldraw((1,5)--(1,6)--(8,6)--(8,4)--(7,4)--(7,5)--cycle, p3); draw((2,5.05)--(7,5.05), red); draw((1,5.05)--(2,5.05), purple); filldraw((2,0)--(2,5)--(7,5)--(7,4)--(3,4)--(3,0)--cycle, p4);
label("$T_0$", (1.5,2.5)); label("$L_0$", (3.5,3.5)); label("$L_1$", (4.5,3.5)); label("$L_2$", (5.5,3.5)); label("$S$", (2.5,0.5));
label("$B$", (1.5, 5.5)); [/asy]](http://latex.artofproblemsolving.com/9/e/c/9ec929b9cc1140aa69425b680db6bc281b611cb7.png)
,
,
,
lie below it, we can just extend them up to get more cutters. If this still doesn't cover all of 
or 
.
triangles pasted together, cover the right-down and right-up paths with nice parallelograms leaving an isosceles right triangle at borders of the path, it suffices to show that in a tiling of nice parallelograms, at least an area of 
is left where the square has dimensions 
,
boundary edges, fire a laser from each of these boundary edges, it is clear that a laser can't visit a triangle more than once, now we break into cases,
to our sum.
times, there 
horizontal boundaries and 
vertical, it follows there are atleast 
grid, there are 
distinct edge paths, where 
is the number of grid paths.![[asy]
size(4cm);
draw((5,-1)--(0,-1)--(0,-2)--(5,-2)--(5,-3)--(0,-3)--(0,-4)--(5,-4),gray+linewidth(0.5)+miterjoin);
draw((1,-5)--(1,0)--(2,0)--(2,-5)--(3,-5)--(3,0)--(4,0)--(4,-5),gray+linewidth(0.5)+miterjoin);
draw((0,0)--(5,0)--(5,-5)--(0,-5)--cycle,black+linewidth(2.5)+miterjoin);
draw((0,-1)--(3,-1)--(3,-2)--(1,-2)--(1,-4)--(4,-4)--(4,-3)--(2,-3)--(2,-2),black+linewidth(2.5)+miterjoin);
draw((3,0)--(3,-1),black+linewidth(2.5)+miterjoin);
draw((1,-4)--(1,-5),black+linewidth(2.5)+miterjoin);
draw((4,-3)--(4,-1)--(5,-1),black+linewidth(2.5)+miterjoin);
filldraw((0, -1)--(3, -1)--(3, -2)--(1, -2)--(1, -5)--(0, -5)--cycle,white,white+linewidth(2.5)+miterjoin);
draw((0, -1)--(3, -1)--(3, -2)--(1, -2)--(1, -5),black+linewidth(2.5)+miterjoin);
draw((1, -5)--(0, -5)--(0, -1),grey+linewidth(1)+miterjoin+dashed);
filldraw(circle((3, 0), 0.15));
filldraw(circle((0, -1), 0.15));
filldraw(circle((3, -1), 0.15));
filldraw(circle((5, -1), 0.15));
filldraw(circle((2, -2), 0.15));
filldraw(circle((4, -3), 0.15));
filldraw(circle((1, -4), 0.15));
filldraw(circle((1, -5), 0.15));
[/asy]](http://latex.artofproblemsolving.com/d/3/c/d3c3755a57cfb6c23d6c0fdde29a00b8b06c59ea.png)
,
are part of the same path (for a degree 4 
the arms are part of the same path, and the tip and stem are part of two additional paths). Note that this way the only points edge paths turn are at degree 2 vertices.
-![[asy]
size(7cm);
draw((3,-2)--(3,-3)--(1,-3),black+linewidth(2.5)+miterjoin);
draw((0,0)--(0,-2)--(4,-2)--(4,-1),lightred+linewidth(2.5)+miterjoin);
draw((4.5,-1.5)--(5.5,-1.5),black,arrow=Arrow);
draw((6,0)--(6,-2)--(9,-2),black+linewidth(2.5)+miterjoin);
draw((10,-1)--(10,-2)--(9,-2)--(9,-3)--(7,-3),lightred+linewidth(2.5)+miterjoin);
[/asy]](http://latex.artofproblemsolving.com/5/7/0/5706f4f7d7a56b51bf20a0b286292ea29b47cea4.png)
![[asy]
size(4cm);
draw((5,-1)--(0,-1)--(0,-2)--(5,-2)--(5,-3)--(0,-3)--(0,-4)--(5,-4),gray+linewidth(0.5)+miterjoin);
draw((1,-5)--(1,0)--(2,0)--(2,-5)--(3,-5)--(3,0)--(4,0)--(4,-5),gray+linewidth(0.5)+miterjoin);
draw((0,0)--(5,0)--(5,-5)--(0,-5)--cycle,black+linewidth(2.3)+miterjoin);
draw((0,-1)--(3,-1)--(3,-2)--(1,-2)--(1,-4)--(4,-4)--(4,-3)--(2,-3)--(2,-2),black+linewidth(2.3)+miterjoin);
draw((3,0)--(3,-1),black+linewidth(2.3)+miterjoin);
draw((1,-4)--(1,-5),black+linewidth(2.3)+miterjoin);
draw((4,-3)--(4,-1)--(5,-1),black+linewidth(2.3)+miterjoin);
draw((0,-1)--(3,-1),lightred+linewidth(2.5)+miterjoin);
draw((3,0)--(3,-2)--(1,-2)--(1,-5),lightgreen+linewidth(2.5)+miterjoin);
draw((5,-1)--(4,-1)--(4,-4)--(1,-4),lightblue+linewidth(2.5)+miterjoin);
draw((2,-2)--(2,-3)--(4,-3),lightcyan+linewidth(2.5)+miterjoin);
[/asy]](http://latex.artofproblemsolving.com/b/1/2/b12275657b953651628279163eb468e3e00ed67a.png)
grid! To see this, we simultaneously take every instance of ![[asy]
size(9cm);
for(int i=1;i<=4;++i){
draw((i,0)--(i,-5),gray(0.9)+linewidth(0.5)+miterjoin);
draw((0,-i)--(5,-i),gray(0.9)+linewidth(0.5)+miterjoin);
}
draw((0,0)--(5,0)--(5,-5)--(0,-5)--cycle,gray(0.8)+linewidth(2.5)+miterjoin);
draw((1,-1)--(2,-1),black+linewidth(3)+miterjoin);
draw((3,-1)--(3,-2)--(1,-2)--(1,-4),black+linewidth(3)+miterjoin);
draw((2,-3)--(3,-3),black+linewidth(3)+miterjoin);
draw((2,-4)--(4,-4)--(4,-1),black+linewidth(3)+miterjoin);
draw((5.5,-2.5)--(6.5,-2.5),black,arrow=Arrow);
for(int i=1;i<=3;++i){
draw((7+i,-0.5)--(7+i,-4.5),gray+linewidth(0.5)+miterjoin);
draw((7,-0.5-i)--(11,-0.5-i),gray+linewidth(0.5)+miterjoin);
}
draw((7,-0.5)--(11,-0.5)--(11,-4.5)--(7,-4.5)--cycle,black+linewidth(2.5)+miterjoin);
draw((7,-1.5)--(9,-1.5)--(9,-0.5),black+linewidth(2.5)+miterjoin);
draw((10,-0.5)--(10,-2.5)--(8,-2.5)--(8,-4.5),black+linewidth(2.5)+miterjoin);
draw((8,-3.5)--(10,-3.5)--(10,-2.5),black+linewidth(2.5)+miterjoin);
draw((7,-1.5)--(9,-1.5)--(9,-0.5),black+linewidth(2.5)+miterjoin);
[/asy]](http://latex.artofproblemsolving.com/4/5/2/4523e76a29255b5a1a9b9a9e2d4e6d456247ff0b.png)
)
