.
Let 
.
Let 
.
Let 
such that
Proposed by Ngo Van Trang, Vietnam
s.t. 
and 
be positiv integer numbers.
and 
are the feet of the angle bisector and altitude from 
,
and 
.
.
be a positive integer.Prove that there is a polynomial 
with integer coefficients so that 
,![$$a^{2n+1}+b^{2n+1}+c^{2n+1}=abc[P(a,b)+P(b,c)+P(a,c)]$$](http://latex.artofproblemsolving.com/5/2/0/520813bebb2f297928208f7b7b2eea76d24043ac.png)
.
.
Y by FaThEr-SqUiRrEl, tigerzhang, samrocksnature, icematrix2, srisainandan6, megarnie, centslordm
Carina has three pins, labeled 
, and 
, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance 
away. What is the least number of moves that Carina can make in order for triangle 
to have area 2021?
(A lattice point is a point
in the coordinate plane where 
and 
are both integers, not necessarily positive.)
, and 
, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance 
away. What is the least number of moves that Carina can make in order for triangle 
to have area 2021?(A lattice point is a point
in the coordinate plane where 
and 
are both integers, not necessarily positive.)This post has been edited 2 times. Last edited by bobthegod78, Apr 15, 2021, 6:40 PM
.
,
,
where one of 
is 
and one of 
is 
,
)
,
,
.
achieved by moving 
moving 
to 
and moving 
and 
respectively.
We may assume WLOG that 
Suppose for the sake of contradiction that 
and 
Then, Carina performed at least 
horizontal moves. If Carina had instead moved the pins to 
and 
(which wouldn't affect their relative positions), then she would have performed 
horizontal moves.
can be dealt with similarly, so it is always optimal for 
respectively. By Shoelace, ![$$[ABC]=\frac{1}{2}\left|\underline{(x_1y_2+x_3y_1)-(x_3y_2+x_1y_3)}\right|.$$](http://latex.artofproblemsolving.com/a/6/4/a64af125ee7b16b025c16adda118140b6424d149.png)
Immediately after Carina has performed all horizontal moves, the underlined expression is 
Moreover,
by 
by 
by 
Carina must perform at least 
vertical moves.
moves.
as desired.
can be in in the same direction in any axis, and then there were only 3 cases: the one where 
;
,
,
.
,
,
.
,
,
,![\[\lvert-1\rvert+|54|+|63|+|0|+|0|+\lvert-10\rvert=128.\]](http://latex.artofproblemsolving.com/f/2/e/f2e5d0938630f9567d0266af9d32e4f4857c8ba1.png)
We claim that 
with sides parallel to the axes such that it's the smallest rectangle that contains all of the points 
,
parallel to the y-axis from 
parallel to the x-axis from 
be our starting point, so that it's a distance 
from 
from 
,
.
where 
is the width of 
is the length (or height) of 
and let 
.![$[AP_1CB]=[ABP_2C]=[R]/2$](http://latex.artofproblemsolving.com/e/8/c/e8ca266eefca2b25b432d67d9d6e2a9526729880.png)
,![$[AP_1CB] = [ABC]+[AP_1C]$](http://latex.artofproblemsolving.com/0/6/f/06f9d43b7f7f190bcc269a327a8f7a89baa2599b.png)
and ![$[ABP_2C]=[ABC]+[BP_2C]$](http://latex.artofproblemsolving.com/f/e/6/fe615ac909b108345d401360a3a5ff444306a5ec.png)
,![$[ABC]\leq [R]/2$](http://latex.artofproblemsolving.com/4/6/1/46120bc97c15ae997c0b0ae383762cebf877372e.png)
,
,![$[ABC]\leq 64^2/2 = 2048$](http://latex.artofproblemsolving.com/b/0/2/b0272d242fd767e89e7fc9218d78c06d2ea39fab.png)
,
.
,![$[R]/2$](http://latex.artofproblemsolving.com/4/2/2/422504d07256ed2508c5e114b5ee8b46f12ef591.png)
would be 
,
,
.
by AM-GM. \qed
of 
.
.
and the width 
,
moves, implying that 
.
.
,
.
,
,
with bounding rectangle of dimensions 
.
This time around I had to look at hints from ARCH :skull:
so the area of the triangle is at most equal to 
This must be greater than or equal to 
so 
is 
the area of the surrounding rectangle after all n moves is at most 
.
,
and 
.
by say, AM-GM. The minimum positive integer 
,
,
.
as the smallest rectangle with sides parallel to the axes containing all three vertices. I claim that a triangle's area is at most twice the area of its bounding box. We have two cases; the case where one of the sides of the triangle coincides with a side of the bounding box is trivial to see. The other case is not much harder:![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(5cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -5.320243582333011, xmax = 13.300111666175955, ymin = -2.57973075125075, ymax = 6.877837547794034; /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0);
draw((0,0)--(0,3)--(4,3)--(4,0)--cycle, linewidth(1) + zzttqq);
draw((4,0)--(0.9561426524876215,3.02725616889723)--(0,1.2831331838785818)--cycle, linewidth(1) + zzttqq);
/* draw figures */
draw((0,0)--(0,3), linewidth(1) + zzttqq);
draw((0,3)--(4,3), linewidth(1) + zzttqq);
draw((4,3)--(4,0), linewidth(1) + zzttqq);
draw((4,0)--(0,0), linewidth(1) + zzttqq);
draw((4,0)--(0.9561426524876215,3.02725616889723), linewidth(1) + zzttqq);
draw((0.9561426524876215,3.02725616889723)--(0,1.2831331838785818), linewidth(1) + zzttqq);
draw((0,1.2831331838785818)--(4,0), linewidth(1) + zzttqq);
/* dots and labels */
dot((0,0),linewidth(4pt) + dotstyle);
label("$F$", (0.047233491280721025,0.09786650743985113), NE * labelscalefactor);
dot((0,3),dotstyle);
label("$E$", (0.047233491280721025,3.1193753406411715), NE * labelscalefactor);
dot((4,3),dotstyle);
label("$D$", (4.0513468230841,3.1193753406411715), NE * labelscalefactor);
dot((4,0),dotstyle);
label("$A$", (4.0513468230841,0.12243161990490252), NE * labelscalefactor);
dot((0.9561426524876215,3.02725616889723),dotstyle);
label("$B$", (1.0052728774177258,3.1562230093387487), NE * labelscalefactor);
dot((0,1.2831331838785818),dotstyle);
label("$C$", (-0.2,1.15), W * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/a/7/3/a7339a0a38ff00eb529d89fd2850cbdfebc2860c.png)
,
,
,
.![$[ABC]=ab-\left(\frac{xy+a(b-x)+b(a-y)}{2}\right)=\frac{ax+by-xy}{2}=\frac{ab}2-\frac{(a-y)(b-x)}{2}<\frac{ab}2$](http://latex.artofproblemsolving.com/2/1/6/21633d33516a87fa030bf34c4c6f131b347db20a.png)
,
(less moves will produce a smaller bounding box).
,
,
,
,
are the smallest and largest 
and 
are defined similarly with respect to ![$[ABC] \leq \frac{1}{2} R$](http://latex.artofproblemsolving.com/3/1/d/31dcd5ff2cdf9ff91028759db07d722d84a337ce.png)
,
,![$[ABC] \leq \frac{P^2}{32}$](http://latex.artofproblemsolving.com/3/0/e/30ec01c0f1cdf54bfac5d59a224a3819a0f0b0a3.png)
.![$[ABC] = 2021$](http://latex.artofproblemsolving.com/9/0/2/902d21d1fccf064cb4d61d49ae501181a86b8788.png)
,
,
,
.
Notice that moving a pin by 1 will increase the width/height by at most 1. By AM-GM, that is the max.
Hence, 
A construction is 
