Find a convex quadrilateral such that the quadrilateral has the smallest area and all four of its sides **(a, b, c, d)**, two diagonals **(e, f)** and area **K** are :

(1) distinct integers

(2) distinct prime integers

In a convex quadrilateral, the two diagonals cut the interior into four triangles, each with integer area.

What do you notice about the product of these areas?

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## About benvitalis

math grad - Interest: Number theory

This might be the one for part a

{a, b, c, d} = {52,25,39,60}

{e, f} = {63, 56}

area = 1764 sqr units

Here is a link to a diagram

Paul

The product of areas is 360^4

P.

(2) The area is always divisible by 6. so it is not possible to find a quadrilateral with a prime area

(1) The area of triangle (10,17,21) is 84

The area of triangle (21,28,35) is 294

So the area of quadrilateral (10,17,28,35) is

84 + 294 = 378

but one of the diagonals isn’t integer, its 36.5 in that arrangement.:)

P.

This is what I get:

(10,17,21) –> 84

(21,28,35) –> 294

(10,35,39) –> 168

(17,28,39) –> 210

sides (10,17,28,35)

diagonals (21,39)

Yes you are right, I drew the opposite orientation of one of the triangles.