Find a triangle, quadrilateral and pentagon with integer side lengths whose areas form a set of three consecutive positive integers. Make the areas as small as possible subject to these constraints. Report you answer by giving the areas of each figure with the side lengths of those figure?

We use a rectangle for our quadrilateral with base $= 7$ and height $= 1$ to get an area of $7$.

We can use a triangle with $b = 1$, $h =12$ and legs $= 12$ to get an area of $6$.

We use a pentagon with side length $= 1$ and height $= 2$ to get an area of $5$.

This would leave us with our consecutive integers of $5, 6, 7$.

Does the height for the pentagon matter in this case?

Can we make these any smaller? The problem seems too simple to solve am I missing something concerning the geometric properties of these shapes?

  • $\begingroup$ The pentagon claim does not seem right $\endgroup$ – Moti Feb 17 '16 at 5:24
  • $\begingroup$ Sorry, side length = 1, and height = 2. I broke it into 5 triangles and calculated the are of one and multiplied it by 5. $\endgroup$ – John Feb 17 '16 at 5:28
  • $\begingroup$ In this case the height can not be 2. Did you try to draw it as a pentagon? The height is less than 1. $\endgroup$ – Moti Feb 17 '16 at 5:31
  • $\begingroup$ Yes, I see I neglected to calculate the actual height, Wouldn't the height never be an integer though therefore never making the answer an integer? $\endgroup$ – John Feb 17 '16 at 5:52
  • 1
    $\begingroup$ What is the source of this problem, please? $\endgroup$ – Gerry Myerson Feb 17 '16 at 6:12

I can do 4, 5, 6.

4 = a 2×2 square 6 = a 3-4-5 triangle

5 = the above triangle with a 1×1 square chipped out of the right angle vertex making a pentagon. The chip fits in the interior of the triangle.

enter image description here

  • 1
    $\begingroup$ Alternatively, a $3-4-5$ triangle,a $1\times 5$ rectangle, and a pentagon by deleting a $1\times 2$ rectangle from the corner of a $3-4-5$ triangle. $\endgroup$ – Ng Chung Tak Feb 17 '16 at 13:04
  • $\begingroup$ Since your triangle seems to be the Heronian triangle of smallest area (I don't have found a proof for this) one cannot do better. $\endgroup$ – Christian Blatter Feb 18 '16 at 13:23
  • $\begingroup$ The 3-4-5 triangle is in fact the smallest Heron triangle. See fir instance mathworld.wolfram.com/HeronianTriangle.html. As Chung Tak points out, the minimal solution is not unique. $\endgroup$ – Oscar Lanzi Feb 20 '16 at 2:40

My attempt is using the lowest primitive Pythagorean triples $(3,4,5)$ enter image description here


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.