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Find the quadrilateral of maximal perimeter of fixed area $15$ enclosed in a square of side $4$.

One (trivial) remark: we cannot take a concave quadrilateral because any concave quadrilateral is contained in a triangle, which must in turn be contained in the square by convexity, and a triangle contained in a square of side $4$ has maximum area $8<15$. Therefore the quadrilateral must be convex. Hence the perimeter cannot be larger than that of the square ($16$). Also, we might thus decompose the quadrilateral as the union of $4$ triangles by drawing the diagonals.

See the square as $[-2,2]^2\subset \mathbb{R}^2$.

My idea: the sought quadrilateral is a rhombus having two vertices $A=(-2,-2)$, $C=(2,2)$ (on one diagonal of the square) and the other two $D=(a,-a)$ and $B=(-a,a)$ (on the other diagonal of the square, equally distant from the center), with the parameter $a\in (0,2)$ chosen so that the area is actually $15$ (actually $a=15/8$, which yields the perimeter $\sqrt{1+31^2}/2\approx 15.51$).

I would say that because without the condition of being enclosed in a square, we can arbitrarily increase the perimeter of a quadrilateral with a given area by making the angles on two opposite vertices $A,C$ vanish (while the others tend to $\pi$), and their distance increase to infinity. Inside the square, we can maximize the distance between $A$ and $C$ by taking them to be opposite vertices of the square, i.e. $A=(-2,-2)$ and $C=(2,2)$.

With this assumption, the situation is simpler. But I still need another assumption: the two triangles $ABC$ and $ACD$ have same area. Then for the area to be fixed, $B$ and $D$ must lie on a specific line parallel to the diagonal. At this point, it is easy (one-variable calculus) to show that the perimeter of $ABC$ is maximized if $B$ lies on the other diagonal of the square, and the same goes for $D$. EDIT: This is wrong and lead me to wrong conclusions. The isosceles triangle minimizes, not maximizes, the perimeter! As a result, the point on the line which maximizes the perimeter is the furthest away point of intersection with the border of the square. In particular, all vertices of the quadrilateral must lie on the border of the square.

Since the areas must be the same, the distances of $B$ and $D$ from the center must also be the same.

Either way, I have no idea how to make my assumptions more formal. I'm not even sure that my solution is correct.

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  • $\begingroup$ Can you justify why you assume that the angles must be all equal to $\pi/2$? To me the square being even close to a solution is counterintuitive because in the whole plane it minimizes the perimeter amongst quadrilaterals of a given area. Why would it maximize it in this case? $\endgroup$ – Lorenzo Quarisa Dec 4 '17 at 17:17
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    $\begingroup$ That is usually true when there is a single constraint, but here the double constraint (on the area and the enclosure in the square) breaks such expected symmetry. Additionally: the maxima of convex or subharmonic functions lie on the boundary, so this lack of symmetry is not that strange in this peculiar situation. $\endgroup$ – Jack D'Aurizio Dec 4 '17 at 17:40
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    $\begingroup$ That is some useful insight, thank you. $\endgroup$ – Lorenzo Quarisa Dec 4 '17 at 17:44
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    $\begingroup$ Exactly. So if we have a convex quadrilateral $ABCD$ we may slide $A$ along a parallel to $BD$ till meeting the boundary of the square. The area is preserved, the perimeter increases. So we just have to study the quadrilaterals with all their vertices on the boundary, and this is not terribly difficult. $\endgroup$ – Jack D'Aurizio Dec 4 '17 at 18:32
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    $\begingroup$ Actually it seems harder to me now. To do the same procedure as before (move vertices in a way such that the area is fixed but the perimeter increases), I need to move two vertices at the same time. I would like to show that I can move one vertex of the quadrilateral to a vertex of the square, but the computations required to verify this seem too rough. Anyway I can see where we are going - if this actually works, and intuitively it does, since it's similar to what we did previously - I can move up to 3 vertices of the quadrilateral to vertices of the square, which leads to your solution. $\endgroup$ – Lorenzo Quarisa Dec 4 '17 at 19:18

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