# Putting three different sided regular polygon together

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

Reference

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

## 1 Answer

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.

• 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. – Mick Aug 6 '16 at 19:49
• The example in the remark is $(4,6,12)$, different-sided. – André Nicolas Aug 6 '16 at 21:04
• Sorry for misunderstanding your remark. – Mick Aug 7 '16 at 4:11