# Geometrical question just for fun

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

• 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 ? – Tom Collinge May 10 '14 at 14:59
• originally it had to be a vertices, I think i'll keep it that way – Willemien May 10 '14 at 19:25
• I can get an area of 7. Dunno whether that's minimal, though. – user2357112 supports Monica May 23 '14 at 12:43
• I think you are right , funny that it is an asymmetrical solution – Willemien May 23 '14 at 18:04

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 $-1$.