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.

Thanks in advance!

  • $\begingroup$ Does "$5$ sides with consecutive lengths" mean that the lengths are consecutive integers? $\endgroup$ – Blue Mar 27 '19 at 19:09
  • 1
    $\begingroup$ 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. $\endgroup$ – Andy Mar 27 '19 at 19:14

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

enter image description here

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:

enter image description here

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 $(\star)$). $\square$


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.