Sir,
A trapezoid is a quadrilateral with one pair of parallel sides.
I don't find a solution on your question. But I found 3 solutions for a quadrilateral. Suppose we have the quadrilateral ABCD.
1) Let AB=13, BC=5, CD+11, and DA=8 (the numbers are in the order as in
your question). Then I found the diagonals AC=15 and BD=12. And I think
this is the solution you ask.
2) Let AB=13, BC=5, CD=8, and DA=11; the diagonals are AC=16 and BD=9.
3) Let AB=13, BC=5, CD=11, and DA=8; the diagonals are AC=17 and BD=9.
This third quadrilateral is almost a trapezoid.
Best greetings.
Gustaaf Lahousse, Belgium.
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Hallo, Ed.
Concerning the Reid's problem on quadrilaterals with integer sides and
diagonal, I suspect you did not say "all" the truth. In particular, I refer
to the question:
<< Does the quadrilateral need to be convex? >>
If the answer is not, there is a smaller solution showed in the enclosed
file "Reid.gif". On the other hand, the quadruplet of side-lengths
cyclically given by [39, 25, 52, 60] can be splitted into two couples of
triangles in (essentially) two ways:
(39,25) and (52,60), with a common side of length 56:
the other diagonal can be 53 (convex) or 33;
With my best regards. Claudio Baiocchi
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For trapezoid ABCD, AB=13, BC=5, CD=11, AD=8, AC=17, BD=9
Alan Lemm
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The six numbers are 5,11,8,13, and 9,17.
The sides of length 5 and 8 are parallel, giving the altitude of the
trapezoid as (3/2)(35)^(1/2). The two diagonals then have integer
lengths 9 and 17, if my pencil and mental work late at night didn't goof
up.
At first I thought the sides 11 and 13 were parallel, but that didn't
work.
It shouldn't be too hard to write a program to loop thru the 4
parameters and check for the two perfect squares, thus generating a
complete list of all small solutions. Not sure if I have the time to do
that right now, tho. Nice problem!
Gary Mulkey
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Let AB = 13, BC=5, CD=11, DA=8, AC=a and BD=b
Area ABC = Sq root ( ((13+5+a)/2) ((13+5+a)/2-13) ((13+5+a)/2-5)
((13+5+a/2)-a) )
Area ABC = Sq root ( (9+a/2) (a/2-4) (a/2+4) (9-a/2) )
where a/2-4 > 0 and 9-a/2 > 0 i.e. 18 > a > 8
Area ACD = Sq root ( ((11+8+a)/2) ((11+8+a)/2-11) ((11+8+a)/2-8)
((11+8+a/2)-a) )
Area ACD = Sq root ( ((19+a)/2) ((a-3)/2) ((a+3)/2) ((19-a)/2) )
where a-3 > 0 and 19-a > 0 i.e. 19 > a > 3
Area ABD = Sq root ( ((13+8+b)/2) ((13+8+b)/2-13) ((13+8+b)/2-8)
((13+8+b)/2-b) )
Area ABD = Sq root ( ((21+b)/2) ((b-5)/2) ((b+5)/2) ((21-b)/2) )
where b-5 > 0 and 21-b > 0 i.e. 21 > b > 5
Area BCD = Sq root ( ((11+5+b)/2) ((11+5+b)/2-11) ((11+5+b)/2-5)
((11+5+b)/2-b) )
Area BCD = Sq root ( (8+b/2) (b/2-3) (b/2+3) (8-b/2) )
where b/2-3 > 0 and 8-b/2 > 0 i.e. 16 > b > 6
Area ABC + Area ACD = Area ABD + Area BCD where 18 > a > 8 and 16 > b > 6
By substituting a = 9, 10, ..., 17 and b = 7, 8, ..., 15, I found that when
a=17 and b = 9, area of ABC + area of ACD is equal to area of ABD + area of
BCD. Therefore, the diagonals are 17 and 9.
cedric Lo
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The diagonals are 9 and 17 long.
I used some trigonometry and spreadsheet to calculate it.
Stan Ponca
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Ed -
Great puzzle! I really enjoyed it. I had trouble with it until I cut drinking straws to the necessary lengths... this physical modeling then enabled me to successfully do the math.
Diagonals are of lengths 17 and 9.
Dave DeBrota
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The diagonals have lengths 9,17.
I have no nice proof of this.
L.T. Pebody
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I come up with diagonals of 9 and 17 for sides 8 and 5 parallel; from base
counterclockwise 8, 13, 5, 11.
Trickier than I expected at first blush. Thanks for another amusing puzzle.
If the restriction on different numbers is removed, here are some of the
possibilities:
3,2,4,2 diagonals = 4
3,4,3,4 (degenerate case) diagonals are of course 5
3,8,12,8 diagonals are 10
3,9,8,11 diagonals are 9 and 13 (smallest with four distinct side lengths)
3,10,7,10 diagonals are 11
4,5,6,5 diagonals are 7
4,6,7,6 diagonals are 8
4,7,8,7 diagonals are 9
4,8,9,8 diagonals are 10
4,9,10,9 diagonals are 11
4,10,11,10 diagonals are 12
4,11,12,11 diagonals are 13
4,12,13,12 diagonals are 14 (quite a surprising run of 8 'consecutive')
8,11,12,13 diagonals are 11 and 19
Be Well!
Dane Brooke
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the diagonals have lengths 17 and 9. You start by seeing that the 8 and 5
must be the parallel sides because 8,5,2 can't be the sides of a triangle.
George Adams
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