,
,
,
,
or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
,
isn't even rigorously defined in this expression. What exactly do we mean by 
.
,
does not exist in the first place, although you will see why in a calculus course. So the point is that 
?
indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that 
,![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function 
given by the mapping 
is undefined for 
.
is undefined because dividing by ![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In ![\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by ![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by ![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and 
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with ![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by 
(i dont remember exactly if it was 8032038 or some other number, so correct me on this)
(i posted this at 12:29) lol
?
feet and area 
![$\sqrt[2]{2}$](http://latex.artofproblemsolving.com/1/8/6/1869ac018fb72a26a12ab14499cbf1b1c13b4a10.png)
,![$\sqrt[3]{3}$](http://latex.artofproblemsolving.com/1/7/a/17a6bb5dc11696b8650b6d3a11e770f861f4c326.png)
,![$\sqrt[5]{5}$](http://latex.artofproblemsolving.com/a/b/9/ab90f2d42dc24065e2a64988fe76f22eb9359799.png)
.
such that![\[
\sqrt{x - 2011} + \sqrt{2011 - x} + 10
\]](http://latex.artofproblemsolving.com/4/8/b/48b82d13b733947420a0d763c6af74636770835a.png)
is an integer.
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 
is 
,
)
,
,
.
achieved by moving 
to 
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 
.
moves, the bounding box has area at most 
.
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)
,
,
,
.
