10
$\begingroup$

Which is the concave quadrilateral with integer sides and integer diagonals with the smallest possible perimeter? This puzzle is my own creation.

$\endgroup$
0
6
$\begingroup$

I shamelessly exploited the fact that there is no tag and wrote a little proggy.

The best non-intersecting non-degenerate it finds has

sides 6,8,19,17 and diagonals 4 and 22. The perimeter is 50.

Picture

Left: solution as described, right: same set of vertices but reassigned so the longer diagonal is inside, here the perimeter sums to 51
![enter image description here

$\endgroup$
8
  • $\begingroup$ Are you sure? I found a smaller perimeter with my program. $\endgroup$
    – Vepir
    Jan 28 at 19:48
  • $\begingroup$ @Vepir no, I'm never sure with my programs ;-) $\endgroup$ Jan 28 at 19:50
  • 1
    $\begingroup$ Never mind, my bad. It was a "near example" due to mishandling float precision. After patching it, I can confirm that what you found is smallest. $\endgroup$
    – Vepir
    Jan 28 at 20:14
  • $\begingroup$ Nice! I found the same solution by computer search so I am quite confident that it is indeed the solution. The short diagonal is inside the quadrilateral, the long outside. What if we request to have the long diagonal inside? My best solution has perimeter 51 in this case. $\endgroup$ Jan 28 at 20:14
  • 1
    $\begingroup$ @PaulPanzerYou got it almost. Just connect the 4 corners in a different way... $\endgroup$ Jan 28 at 21:22
5
$\begingroup$

(edited) Here is a non-convex quadrilateral that meets your requirements:

AB = 5
BC = 3
Diagonals = AC = BD = 4 (BC² + AC² = AB²)

Perimeter = 3 + 5 + 3 + 5 = 16

Crossed quadrilateral

$\endgroup$
2
  • $\begingroup$ Selfintersecting quadrilaterals were not what I had in mind. Else the task would be quite trivial. The smallest convex quadrilateral with the integer properties is an isosceles trapezoid with parallel sides 4 and 3, legs 2 and 2 and diagonals 4 and 4. If we connect the corners in a different way we get an selfintersecting quadrilateral with perimeter 3+4+4+4=15. I changed non-convex to concave. $\endgroup$ Jan 28 at 19:15
  • 1
    $\begingroup$ Thanks for your comment. Out of curiosity, why not just ask for a concave quadrilateral? Since concave was not mentioned, I tried to think out of the box. $\endgroup$
    – xhienne
    Jan 28 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.