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Was puzzling with the following (home made) puzzle:

Given the square $ABCD$ with $A = (1,1)$, $B = (1,-1)$, $C = (-1,-1)$ and $D = (-1,1)$

And given point $E = (0,2)$

What is the smallest (by area) quadrilateral $EFGH$ that contains the square $ABCD$?

sorry I only created this puzzle, I don't know the answer myself , but maybe you would like a puzzle

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  • $\begingroup$ You specifically require that E is one of the four vertices rather than allowing it to be a point on one of the four edges ? $\endgroup$ – Tom Collinge May 10 '14 at 14:59
  • $\begingroup$ originally it had to be a vertices, I think i'll keep it that way $\endgroup$ – Willemien May 10 '14 at 19:25
  • $\begingroup$ I can get an area of 7. Dunno whether that's minimal, though. $\endgroup$ – user2357112 supports Monica May 23 '14 at 12:43
  • $\begingroup$ I think you are right , funny that it is an asymmetrical solution $\endgroup$ – Willemien May 23 '14 at 18:04
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Here's a sketch proof that the answer is 7. I worked through the details, although I may have slipped up.

First note that the four edges of the quadrilateral must run though the four vertices of the square. Hence we know the direction of the lines $EF$ and $EH$. In fact, we are only left to choose the position of $G$.

Assume $G$ is at $(x,y)$. We know $x\in[-1,1]$ and $y\leq -1$. Using basic angle formulae and trig, find the area of the quadrilateral in terms of $x$ and $y$.

Fix $y<-1$, and do partial differentiation with respect to $x$. Since everything is symmetrical, there must be a turning point at $x=0$. In fact this is the only turning point for $x\in[-1,1]$, and it is a maximum. Thus the minimum occurs at $x\in\{\pm 1\}$.

Now we can fix $x=1$, and just vary $y$. The area of the quadrilateral is then smallest when $y$ is closest to $-1$.

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