# Fall 2018, problem 70

Prove that for any convex quadrilateral of area 1, the sum of the lengths of all sides and diagonals is not less than $4+\sqrt{8}$.

TME102475
3 years ago

Let $ABCD$ be a convex quadrilateral of unit area. By the Isoperimetric Theorem, among all quadrilaterals of a given area $\Rightarrow$ the square has the smallest perimeter. This implies $AB = BC = CD = DA = 1$, and the perimeter:

$AB + BC + CD + DA \geq 4$ (i).

Also, the area of $ABCD$ equals $\frac{1}{2} \cdot AC \cdot BD \cdot \sin(\theta)$, where $\theta$ is the included angle between diagonals $AC$ and $BD$. This is exactly half the area of the parallelogram $ACC'A'$, which is formed on the original diagonals $AC$, $BD$. Thus, the area of $ACC'A'$ equals $2$, which is a square of side length $\sqrt{2}$ and a minimum perimeter of $4\sqrt{2}$. This in turn yields:

$2(AC+BD) \geq 4\sqrt{2} \Rightarrow AC + BD \geq 2\sqrt{2}$ (ii).

Combining inequalities (i) & (ii) produces:

$(AB + BC + CD + DA) + (AC + BD) \geq 4 + 2\sqrt{2}$

with equality occuring when $ABCD$ is a square.

$\mathbb{Q}. \mathbb{E}. \mathbb{D}.$

Claudio
3 years ago

The claim will obviously follow from the following points 1 and 2:

1. For any quadrilateral of area 1 there exists a rhombus of area 1 for which the sum is less.

2. For any rhombus of area 1 the sum is greater than the one of unit square.

Property 1 has a simple geometrical proof.

Let ABCD be a quadrilateral of area 1; choose a diagonal (e.g. AC) and let $r_1,r_2$ be the lines parallel to AC passing throug B, D respectively; we call a,b the sides on the half-plane of B and c,d the other two sides.

Moving B along $r_1$ and D along $r_2$ the area remains the same; the length of BD is minimal when BD is orthogonal to AC and both the sums a+b, c+d are minimal when BD cuts AC on its middle point; thus with a=b, c=d.

Do the same with respect to the diagonal BD we get a=d, b=c.

Unfortunateli, concerning property 2, I do not see a simple geometrical proof; however the algebric proof is a (tedious but) trivial inequality.

Poojanpujara
3 years ago
• S=a+b+c+d. Â Â T=a+b+c+d+m+n
• let a,b,c,d be sides of a quadrilateral and m and n are lengths of diagonals.
• The angle between diagonals be â‚¬ and between sides of lengths a and d be Î± and between b and c be Î².(a,b,c,d are consecutive).
• 1=m nsin(â‚¬)/2...m n=2cosec(â‚¬)...mn>=2.
• 1=[a dsin(Î±)+b csin(Î²)]/2
• since a d>=ad sin(Î±) and bc>=b csin(Î²)...
• a d+bc >=2
• let a d=x^2 and bc=y^2. We have to find min. of a+b+c+d+m+n since n>=2/m (adding m both the sides) m+n>=m+2/m...m+n >=2âˆš2(AM -GM enequality)and write a+b+c+d as (a+d)+(b+c). Again a+d>=2(âˆša âˆšd) Â Â Â Â Â b+c>=2(âˆšbâˆšc) ...adding these S>=2(x+y). We have to minimize x+y given x^2+y^2>=2 Off course it will be then when line x+y=c(say) will tangent to circle x^2+y^2=2(note that x and y are positive ) it gives c=2 . hence S>=4 and m+n >=âˆš8 Which will give T>=4+âˆš8
wgw
3 years ago

Coolidge's formula for the area of a convex planar quadrilateral is:

$K^2 = \frac{(a^2 + b^2 + c^2 + d^2)^2 + 8abcd -2(a^4 + b^4 + c^4 +d^4) -4(ac + bd +pq)(ac +bd -pq)}{16}$

where $K$ is the area and $a, b, c, d, p, q$ are the lengths of the 4 sides and the two diagonals, respectively.

The problem is to find the minimum of the function:

$a + b + c + d + p + q$

subject to the constraint that the above area equals 1. (Obviously, geometric lengths are non-negative, so that, additionally, all these variables are greater than or equal to zero.) So, just use the Method of Langrange Multipliers for constrained optimization problems from Advanced Calculus. Define:

$h(a,b,c,d,p,q) = a + b + c + d + p + q - m( (a^2 + b^2 + c^2 + d^2)^2 + 8abcd -2(a^4 + b^4 + c^4 +d^4) -4(ac + bd +pq)(ac +bd -pq)-16$)

This function is to be minimized over the the non-negative values of the six variables. (Help me out -- what generalizes "octant" in this context?) Anyway, just take the six partial derivatives and set them to zero.

Clearly, these equations are symmetric in $a, b, c, d$ and also in $p, q$. Common sense suggests taking the $p, q$ derivatives first, and, eliminating $m$ (as usual), quickly gives $p = q$. After taking two of the $a, b, c, d$ partial derivatives (really, just one with the symmetry) and eliminating $m$ gives an algebraic equation quite close to having, say, $(a - b)$ as a factor, where the only problem is another term with $(c - d)$ as a factor.
This should suggest $a = b$ and $c = d$, and if you carry on solving in this way, you can show that, and eventually $a = b = c = d$. Consequently, the minimizing quadrilateral is a square and the result follows immediately.