TITLE: Unit square inside triangle.
QUESTION [16 upvotes]: 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 $S \ge 2$.

REPLY [3 votes]: If some edge, say $BC$ of a triangle $ABC$, does not touch the square, by moving $BC$ towards $A$, we may obtain a smaller similar triangle such that the side parallel to the original $BC$ touches the square.
Therefore, we may assume that all three sides of the triangle touch the square. Consequently, the triangle can have at most one vertex (say, possibly $A$) that lies on the open interiors of the NW, NE, SW or SE regions to the four corners (because, if there are two such vertices, the edge join them will not touch the square). So we get something like the figure below:
A       :       :
        :       :
     NW :       : NE
    ~~~~.-------.~~~~
        |       |      B
        |       |
        |       |
    ~~~~.-------.~~~~
     SW :       : SE
        :       :
        : C     :

Drop a perpendicular of $B$ to its adjacent side of the square. Let $P$ be where the extension of this line segment intersects the other side of the square (see the figure below) and let $D$ be where this line extension meets $AC$. Drop a perpendicular of $CQ$ to $PB$. Also, extend the side of the square opposite to $B$. Let this extended line intersects $AB$ at $E$:
A       
 \      E
  \     |
   \    .-------.
    D---P-Q-----|------B
     \  | |     |
      \ | |     |
       \.-------.
        \ |
         \|
          C

Now the area of $ABC$ is greater than the total area of the triangles $CQD, CQB$ and $BPE$. So, it suffices to prove the following proposition:

if $R$ is an inscribed rectangle of a right-angled triangle $T$ such the two geometric entities share a common vertex (which must be the right angle of $T$), then the area of $T$ is at least double the area of $R$.

But the truthfulness or falseness of the above proposition is invariant if we scale the whole figure along the direction of any side of $R$, because both the areas of $R$ and $T$ scale by the same proportion. Therefore we may further assume that $R$ is a square. Now, borrowing the words of Hagen von Eitzen in another answer here, it is obvious that if we flip the outside tips of $T$ inside, the two flipped tips of $T$ will always "envelope" the interior of the square seamlessly, with the longer tip poking out. Hence the proposition is true.