
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
The sides of an equilateral triangle with sides of length  $n$  have been divided into equal parts, each of length  $1$ , and lines have been drawn through the points of division parallel to the sides of the triangle, thus dividing the large triangle into many small triangles. Nils has a pile of rhombic tiles, each of side  $1$  and angles  $60^\circ$  and  $120^\circ$ , and wants to tile most of the triangle using these, so that each tile covers two small triangles with no overlap.  In the picture, three tiles are placed somewhat arbitrarily as an illustration. How many tiles can Nils fit inside the triangle?
[asy]
/* original code by fedja: https://artofproblemsolving.com/community/c68h207503p1220868
modified by Klaus-Anton: https://artofproblemsolving.com/community/c2083h3267391_draw_me_a_grid_of_regular_triangles
*/
size(5cm);
int n=6;
pair A=(1,0), B=dir(60);
path P=A--B--(0,0)--cycle;
path Pp=A--shift(A)*B--B--cycle;

/*
label(" $A$ ",A,S);
label(" $B$ ",B,dir(120));
label(" $(0,0)$ ",(0,0),dir(210));

fill(shift(2*A-1+2*B-1)*P,yellow+white);
fill(shift(2*A-1+2*B-0)*P,yellow+white);
fill(shift(2*A-1+2*B+1)*P,yellow+white);
fill(shift(2*A-1+2*B+2)*P,yellow+white);

fill(shift(1*A-1+1*B)*P,blue+white);
fill(shift(2*A-1+1*B)*P,blue+white);
fill(shift(3*A-1+1*B)*P,blue+white);
fill(shift(4*A-1+1*B)*P,blue+white);
fill(shift(5*A-1+1*B)*P,blue+white);

fill(shift(0*A+0*B)*P,green+white);
fill(shift(0*A+1+0*B)*P,green+white);
fill(shift(0*A+2+0*B)*P,green+white);
fill(shift(0*A+3+0*B)*P,green+white);
fill(shift(0*A+4+0*B)*P,green+white);
fill(shift(0*A+5+0*B)*P,green+white);

fill(shift(2*A-1+3*B-1)*P,magenta+white);
fill(shift(3*A-1+3*B-1)*P,magenta+white);
fill(shift(4*A-1+3*B-1)*P,magenta+white);

fill(shift(5*A+5*B-5)*P,heavyred+white);

fill(shift(4*A+4*B-4)*P,palered+white);
fill(shift(4*A+4*B-3)*P,palered+white);


fill(shift(0*A+0*B)*Pp,gray);
fill(shift(0*A+1+0*B)*Pp,gray);
fill(shift(0*A+2+0*B)*Pp,gray);
fill(shift(0*A+3+0*B)*Pp,gray);
fill(shift(0*A+4+0*B)*Pp,gray);

fill(shift(1*A+1*B-1)*Pp,lightgray);
fill(shift(1*A+1*B-0)*Pp,lightgray);
fill(shift(1*A+1*B+1)*Pp,lightgray);
fill(shift(1*A+1*B+2)*Pp,lightgray);

fill(shift(2*A+2*B-2)*Pp,red);
fill(shift(2*A+2*B-1)*Pp,red);
fill(shift(2*A+2*B-0)*Pp,red);

fill(shift(3*A+3*B-2)*Pp,blue);
fill(shift(3*A+3*B-3)*Pp,blue);

fill(shift(4*A+4*B-4)*Pp,cyan);

fill(shift(0*A+1+0*B)*Pp,gray);
fill(shift(0*A+2+0*B)*Pp,gray);
fill(shift(0*A+3+0*B)*Pp,gray);
fill(shift(0*A+4+0*B)*Pp,gray);
*/

fill(Pp, rgb(244, 215, 158));
fill(shift(dir(60))*P, rgb(244, 215, 158));

fill(shift(1.5,(-sqrt(3)/2))*shift(2*dir(60))*Pp, rgb(244, 215, 158));
fill(shift(1.5,(-sqrt(3)/2))*shift(2*dir(60))*P, rgb(244, 215, 158));

fill(shift(-.5,(-sqrt(3)/2))*shift(4*dir(60))*Pp, rgb(244, 215, 158));
fill(shift(.5,(-sqrt(3)/2))*shift(4*dir(60))*P, rgb(244, 215, 158));

for(int i=0;i<n;++i){
for(int j;j<n-i;++j)
{draw(shift(i*A+j*B)*P);}}

shipout(bbox(2mm,Fill(white)));
[/asy]

Present the answer in LaTex format: \boxed{Your answer}
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