The sides of a trapezoid are 5, 8, 11, and 13, and its diagonals are also integer numbers; what are they?

The subject should've been No Solution (Proof by Hand) but I didn't want to give it away that there is no solution. The first step is to correctly determine the orientation of the sides.

I'm going to try my best to describe the symbols I use. I labeled the sides a,b,c,d. b and d are the parallel sides with d being at the bottom and the longer of the two. a is left side while b is the right side.

I dropped perpediculars from the endpoints for the side of length b to the side of length d. This basically splits the trapezoid into 2 right triangles and a rectangle between them.

I labelled the little segment on the left x1 and the one on the right x2. And the altitude is h.

LEFT RIGHT TRIANGLE
hypotenuse a, legs x1 and h

RIGHT TRIANGLE
hypotenuse c, legs x2 and h

 

EQUATIONS
x1^2+h^2=a^2 (Pythagorean Theorem)
x2^2+h^2=c^2 (Pythagorean Theorem)
x1+x2+b=d (trapezoid and how we split up the segments)

Subtracting the first two equations, we have
x1^2-x2^2=a^2-c^2.

Then substituting x2=d-b-x1 into that and solving for x1, we get
x1=(a^2+b^2-c^2+d^2-2bd)/(2d-2b).
x2=d-b-x1

Neglecting symmetry, there are 24 ways to pick (a,b,c,d) from 5, 8, 11, and 13. Of these 24 ways, there are only two options where both x1 and x2 are positive.

(a,b,c,d) = (8,5,11,13) and (11,5,8,13).

Now the next step is to see if the diagonals are of integer lengths or not. It turns out that they aren't.

Let the diagonal from the top left to the bottom right be of length D1. The diagonal forms a right triangle with the corresponding altitude. The diagonal is the hypotenuse. h is the altitude. (d-x1) is the base. So we have

D1^2=h^2+(d-x1)^2
        =a^2-x1^2+(d-x1)^2
        =a^2+d^2-2dx1
        =a^2+d^2-2d*(a^2+b^2-c^2+d^2-2bd)/(2d-2b).

Plugging in numbers and solving for D1 gives 14.887 and 9.663 for the two orientations.

HENCE, NO SOLUTION! Q.E.D.

Edited on January 17, 2005, 8:38 pm

Posted by np_rt on 2005-01-17 20:37:24