Proving the Minimum Total Length for a Convex Quadrilateral of Area 1

In this article, we delve into a proof that for any convex quadrilateral of area 1, the sum of the lengths of all sides and diagonals is not less than 4√8. This problem is a fascinating application of geometry and optimization. The assertion is true, and we will explore the steps to prove it.

Introduction

The proof revolves around the idea of transforming a given convex quadrilateral into simpler shapes while ensuring that the area remains constant at 1. We start by utilizing geometric transformations that decrease the total length while preserving the area, ultimately leading us to a square, which has the minimum total length among all quadrilaterals of area 1.

Geometric Transformations

Consider an arbitrary convex quadrilateral ABCD of area 1, with vertices cyclically in that order. Let (l_1) be the line through B parallel to the line AC, and (l_2) be the line through D parallel to the line AC. We can replace B with a point B' on (l_1) and D with a point D' on (l_2). This replacement ensures that the areas of the triangles remain unchanged, thus preserving the area of the quadrilateral.

To minimize the perimeter, we choose B' and D' to be the points where the perpendicular bisector of segment AC intersects (l_1) and (l_2) respectively. This transformation makes the combined lengths of AB and BC shorter than the original, unless B and B' coincide. Similarly, the combined lengths of AD and DC are also shortened, unless D and D' coincide. This ensures that the perimeter of the quadrilateral is decreased unless no change is necessary.

Therefore, we denote the new quadrilateral as ABCD, where triangles ABC and ADC are isosceles and the quadrilateral ABCD becomes a kite.

Further Simplification

Next, we replace quadrilateral ABCD with ABCD where A (resp. C) is on the line parallel to the line BD through A (resp. C) and also on the perpendicular bisector of the segment BD. This process ensures that the diagonals AC and BD are perpendicular bisectors of each other, transforming the quadrilateral into a rhombus.

The diagonals of the rhombus cut it into four right triangles. If we let r and s denote the lengths of the legs of any one of these triangles, then the area of the rhombus is 2rs and the total length L of sides plus diagonals is L 2r 2s 2√(r^2 s^2).

Optimization Through Lagrange Multipliers

To find the minimum of L, we use the constraint that rs 1/2 and r, s > 0. By applying the method of Lagrange multipliers, we set up the system of equations:

2 4r/√(r^2s^2) λs

2 4s/√(r^2s^2) λr

rs 1/2

Multiplying the first equation by r and the second by s, and subtracting, we obtain:

2r - s * 4(rs/√(r^2s^2)) 0

Simplifying, we find that the only solution is where r s, and thus r s 1/√2. This gives L 2(1/√2)(1/√2) 2√((1/√2)^2 (1/√2)^2) 2(1/2) 2√(1/2) √2 √2 4√2.

Therefore, the minimum total length L 4√2, which occurs only for a square with area 1.