Unit square inside triangle. Some time ago I saw this beautiful problem and think it is worth to post it:
Let $S$ be the area of triangle that covers a square of side $1$. Prove that $S \ge 2$.

Intuitively when vertices of square
$S:=v_1v_2v_3v_4$ of side length $1$ are in sides of triangle
$T=\Delta\ ABC$, then ${\rm area}\ T$ is smallest
Notation : Edge $xy$ is $[xy]$ And length of edge $xy$ is $|xy|$
Here we have two cases :
(1) $ v_1\in [AB],\ v_2\in [AC]$ and edge $[v_3v_4]$ is in edge
$[BC]$
(2) $v_1\in [AB],\ v_2\in [AC],\ v_3\in [BC]  $ That is $v_4$ is in
interior of $T$
Consider case (1) : Consider a reflection point $A'$ of $A$ wrt edge
$[v_1v_2]$
(1.1) Assume that $A'$ has distance larger than $1$ from $[v_1v_2]$
Define
$$ [A'v_1]\cap [BC]:=\{ v_4'\},\ [A'v_2]\cap [BC]:=\{ v_3'\}
$$
so that $$ 2{\rm area}\ S +2 {\rm area}\ \Delta\ A'v_3'v_4' ={\rm
area}\ T $$
(1.2) Assume that $A'$ has distance smaller than $1$ from $[v_1v_2]$
If line $v_1 + t(A'-v_1)$ meets at $v_4'$ with $BC$, then note that
$$ {\rm area}\ \Delta A'v_1v_2={\rm area}\ \Delta Av_1v_2 $$
$$
{\rm area}\ \Delta v_1Bv_4 ={\rm area}\ \Delta v_1v_4 v_4' $$
$$  {\rm area}\ (S- \Delta A'v_1v_2-\Delta v_1v_4v_4' )<
{\rm area}\ \Delta v_2Cv_3 $$
Consider case (2) : WLOG assume that $[Av_2]<[v_2C]$
(2.1) Now consider $A'$ has distance larger than $1$ from $[v_1v_2]$
Note that $\Delta v_2 A' v_2'$ contains $v_3$ where $$ v_2'\in
[v_2C],\ |v_2v_2'|=|Av_2| $$
Here $[v_2v_3]$ is angle bisector of $\angle A'v_2v_2'$ so that
$\overrightarrow{v_2'v_3}$ enters into square So $Cv_3$ enters into
square This contradicts to definition of $v_3$
(2.2) Assume that we have $A'$ in interior of the square $S$
Define $$ v_1'\in [v_1B],\ |v_1v_1'|=|Av_1| $$
$$ v_2'\in v_2C,\ |v_2v_2'|=|Av_2| $$
$$v_3'\in [Bv_3],\ |v_3'v_3 |=|v_3v_2'| $$
Note that $Av_1'v_4v_3'v_3v_2'$ has area $2$
