# How many pentagons can you make with four corners of $120^\circ$ and five sides whose lengths are consecutive integers (but not necessarily in order)?

I’m trying to prepare for a national math competition, but I’m stuck at this and can’t figured anything out. The challenge is to prove your answer.

How many pentagons can you make with four corners of $$120^\circ$$ and five sides whose lengths are consecutive integers (but not necessarily in order)?

I’ve only come as far as to realize that the shape will be very similar to a hexagon, and that you will probably need to use the Pythagorean theorem to solve it.

• Does "$5$ sides with consecutive lengths" mean that the lengths are consecutive integers? – Blue Mar 27 at 19:09
• Blue it means that every side is 1 unit longer than another, like 3;4;5;6;7. This doesn’t have to be ordered this way. The lengths need to be 5 consecutive natural numbers, in any order. – Andy Mar 27 at 19:14

Let $$a$$, $$b$$, $$c$$ be consecutive sides not adjacent to the pentagon's $$60^\circ$$ angle, as shown:

By appending a side-$$b$$ equilateral triangle, the pentagon becomes a parallelogram, and we see that the pentagon's remaining sides must be $$a+b$$ and $$b+c$$. As in my previous answer, we take $$m$$ to be the smallest side (necessarily, the smallest of $$a$$, $$b$$, $$c$$) and define $$a^\prime := a-m$$, $$b^\prime := b-m$$, $$c^\prime := c-m$$. Then the five sides, reduced by $$m$$, are $$( a^\prime, \;b^\prime, \;c^\prime, \;a^\prime+b^\prime+m, \;b^\prime+c^\prime+m) \tag{1}$$ which must be a permutation of $$(0,1,2,3,4)$$. Note that $$0$$ must be one of the first three terms; and, since two values are larger than $$m$$ (which itself is non-zero), we must have that $$m=1$$ or $$m=2$$. Moreover, if $$m=2$$, then $$b^\prime=0$$, as there is no other way for the last two values to be less than $$5$$.

Thus, we have the following possibilities:

$$\begin{array}{cc:ccccc:l} m=1 & a^\prime = 0 & 0 & b^\prime & c^\prime & b^\prime+1 & b^\prime+c^\prime+1 & (0,2,1,3,4) \\ & b^\prime = 0 & a^\prime & 0 & c^\prime & a^\prime+1 & c^\prime+1 & (1,0,3,2,4)\;(3,0,1,4,2) \\ & c^\prime = 0 & a^\prime & b^\prime & 0 & a^\prime+b^\prime+1 & b^\prime+1 & (1,2,0,4,3)\\ \hline m=2 & b^\prime = 0 & a^\prime & 0 & c^\prime & a^\prime+2 & c^\prime+2 & (1,0,2,3,4)\;(2,0,1,4,3) \end{array}\tag{2}$$

In addition to the obvious $$a^\prime \leftrightarrow c^\prime$$ symmetry, I suspect there may be ways to simplify the consideration of cases. In any event, we arrive at the same six results as in my previous answer (with sides cyclically permuted):

\begin{align} (1,3,2,4,5) \quad (3,2,4,5,6) \quad (2,1,4,3,5) \\ (2,3,1,5,4) \quad (4,2,3,6,5) \quad (4,1,2,5,3) \end{align} \tag{\star}

As before, if reflections are ignored, then there are three solutions (one from each column of $$(\star)$$). $$\square$$

Let the side-lengths be $$a$$, $$b$$, $$c$$, $$d$$, $$e$$, as shown in the figure:

We see that we must have \begin{align} a+b &= d+e \tag{1} \\ a+e &= b+2c+d \tag{2} \end{align}

Let $$m$$ be the minimum side-length, and define $$a^\prime := a-m$$, etc. Then $$\left(a^\prime, b^\prime, c^\prime, d^\prime, e^\prime\right)$$ is some permutation of $$\left(0, 1, 2, 3, 4\right)$$, so that $$a'+b'+c'+d'+e'=10$$. We can re-write $$(1)$$ and $$(2)$$ as \begin{align} a^\prime + b^\prime &= d^\prime + e^\prime \tag{1'} \\ 2m &= 3a'+b'+d'+3e'-20\tag{2'} \end{align} Now, $$(1')$$ has limited solutions, arising from permuting the terms and sides of $$0+3=1+2$$, $$0+4=1+3$$, and $$1+4=2+3$$; very few of these give rise to feasible values of $$m$$. There aren't unreasonably-many cases to check, but a simple observation can save some work: If neither $$a'$$ nor $$e'$$ is $$4$$, then $$3a'+b'+d'+3e'$$ is at most $$3\cdot 3+3\cdot 2+4+1=20$$, and even this value is unattainable in light of $$(1')$$; but that sum must be at least $$20$$ for a valid $$m$$ by $$(2')$$, so we must have that either $$a'$$ or $$e'$$ is $$4$$. This leaves the following solutions:

\begin{align}(a',b',d',e')\;\text{or}\;(e',d',b',a') &= (4,0,1,3) \quad\to\quad m = \phantom{-}1 \quad\checkmark \\ &= (4,0,3,1) \quad\to\quad m = -1 \\ &= (4,1,2,3) \quad\to\quad m = \phantom{-}2 \quad\checkmark \\ &= (4,1,3,2) \quad\to\quad m = \phantom{-}1 \quad\checkmark \end{align} \tag{3}

Consequently, there are six pentagons with sides $$(a,b,c,d,e)$$.

\begin{align} (5,1,3,2,4) \quad (6,3,2,4,5) \quad (5,2,1,4,3) \\ (4,2,3,1,5) \quad (5,4,2,3,6) \quad (3,4,1,2,5) \end{align} \tag{\star}

If reflections are ignored, there are only three (one from each column of $$(\star)$$). $$\square$$