Ken's POTW 970206

Ken's POTW

Quadrilaterals
A quadrilateral is a polygon with four sides. I am interested in finding a few representative quadrilaterals (ABCD) with all the following characteristics:

All four sides are integers (AB, BC, CD, DA).
The two diagonals (lines between opposite vertices; AC, BD) are integers.
The above six integers are all distinct.
The perimeter of the quadrilateral is a minimum.

Special Case 1:
- The quadrilateral contains any angles to meet the requirements.
Special Case 2:
- The quadrilateral contains two right angles, adjacent to each other (DAB, CDA).
Special Case 3:
- The quadrilateral contains two right angles, opposite one another (DAB, BCD).

Source: Original.

Solution for Case 1 was received 11/2/97 from Philippe Fondanaiche:

Let consider the quadrilateral ABCD with the 2 diagonals AC and BD.
The quadrilateral which meets  the 4 characteristics is the following:

AB = 5  BC = 13  CD = 8 DA = 11 AC = 17 BD = 9 

Perimeter of the quadrilateral = 37

It is easy to check that these values are adequate with the following
identities :
c1 = cos(ABD) = ( 25 + 81 -121 ) / 90 = -1 / 6
s1 = sin(ABD) = sqr(35) / 6
c2 = cos(CBD) = ( 169 +81 - 64 ) / 234 = 31 / 39
s2 = sin(CBD) = 3sqr(35) / 26
Hence cos(ABD) = c1*c2 - s1*s2 = - 19 / 26 
    and AC^2 = 25 + 169 +130x19/26 = 289 ,so AC = 17
 
We can add another constraint. For example, if E is at the intersection of AC
and BD ,the 8 segments AB,BC,CD,DA,AE,BE,CE,DE must be integers.
The corresponding solution is AB = 16, BC = 26, CD = 11, DA = 13, AE = 6, BE
= 18, CE = 12, DE = 9.

Wilfred Theunissen sent solutions to Case 2 on 10/22/97:

AD      AB      BC      CD      BD      AC      Perimeter
60      91      100     11      109     61      262
60      91      61      80      109     100     292
120     182     200     22      218     122     524
120     182     122     160     218     200     584
120     209     218     27      241     123     574
120     209     169     90      241     150     588
120     209     150     119     241     169     598
120     209     123     182     241     218     634
240     161     267     44      289     244     712
240     161     244     117     289     267     762

Radu Ionescu sent a solution to Case 3:
I have begun with the smallest Pythagorean triples: (3,4,5) & (5,12,13) The hypotenuse must be at least 5*13=65 =>BD=65, DC=39, BC=52, AD=25, AB=60 In compliance with Ptolemy's Theorem (If a Quadrilateral is inscribed in a circle (i.e., for a cyclic quadrilateral), the sum of the products of the two pairs of opposite sides equals the product of the diagonals)=>

AC=(AD*BC+AB*DC)/DB=56

(25,39,52,56,60,65)
[The area of this quadrilateral is 1764, and thus is a solution nine months ahead of its time for the MIT Technology Review's Novemeber/December issue's PuzzleCorner, problem 1. - KD 11/18/97]

[Ken's additions to Case 3]: Here are other solutions I found. The first all have 65 as the largest diagonal. The next have 125 as the largest diagonal.

16, 63, 60, 25 / 65, 39  [Area = 1254, added 11/18/97 - KD]
16, 63, 39, 52 / 65, 60 
25, 60, 52, 39 / 65, 56
25, 60, 39, 52 / 65, 63
25, 60, 56, 33 / 65, 52
39, 52, 56, 33 / 65, 60
35, 120, 117, 44 / 125, 75
35, 120, 100, 75 / 125, 117
44, 117, 75, 100 / 125, 120

The Macalester Problem 839 was similar to this but did not require unique numbers for each of the lengths. The solution: an isosceles trapezoid with sides 2,3,2,4 with diagonals of 4,4.

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