Is it possible to put three different sided regular polygon side by side together as shown?

enter image description here

I assume that is possible. Then, $$\dfrac {(p – 2)180}{p} + \dfrac {(q – 2)180}{q} + \dfrac {(r – 2)180}{r} = 360$$

$$(p -2)qr + (q – 2)rp + (r – 2)pq = 2pqr$$

$$pqr = 2(pq + qr + rp)$$

From that, how can I get a set of integral solution?


  • $\begingroup$ What about $3,7,42$? $\endgroup$ – André Nicolas Aug 6 '16 at 18:35
  • $\begingroup$ You may try first with $p=q=r$. Do you want to find all solutions? $\endgroup$ – Bernard Aug 6 '16 at 18:36
  • $\begingroup$ You can also further modify it to: $$\frac{1}{2}=\frac1p + \frac1q +\frac1r$$ $\endgroup$ – Fimpellizieri Aug 6 '16 at 18:40
  • $\begingroup$ @AndréNicolas I got 3, 10, 15 too, but by trial.. $\endgroup$ – Mick Aug 6 '16 at 19:31
  • $\begingroup$ @Bernard Yes and preferably is a deduced result rather than trials. $\endgroup$ – Mick Aug 6 '16 at 19:33

Rewrite as $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{2}$.

Without loss of generality we may assume that $p\lt q\lt r$.

Let $p=3$. Then $q=7$ and $r=42$ works. We can get to this with a little patience. For if $p=3$, we need $\frac{1}{q}+\frac{1}{r}=\frac{1}{6}$. The only conceivable candidates for $q$ are $7$ to $11$. Try them all. After getting a "hit" with $q=7$, when you try $q=8$ you will find that $r=24$ works.

We cannot have $p\ge 6$, so we only need to explore the possibilities $p=4$ and $p=5$. In each case there is only a short list of candidates for $q$. For example, with $p=4$ all $q$ greater than $7$ are automatically ruled out for reasons of size.

Remark: Probably one could get the example square, hexagon, dodecagon without the above machinery.

  • $\begingroup$ I think you solution can isolate all the possible integral solution sets. Indeed, the examples in the remark can save us a lot of trouble but they don't fit the requirement of different sided polygons. $\endgroup$ – Mick Aug 6 '16 at 19:49
  • $\begingroup$ The example in the remark is $(4,6,12)$, different-sided. $\endgroup$ – André Nicolas Aug 6 '16 at 21:04
  • $\begingroup$ Sorry for misunderstanding your remark. $\endgroup$ – Mick Aug 7 '16 at 4:11

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.