
4. Let $A B C D E$ be a convex pentagon in which $\angle A=\angle B=\angle C=\angle D=120^{\circ}$ and whose side lengths are 5 consecutive integers in some order. Find all possible values of $A B+B C+C D$.


Solution 1: Let $A B=a, B C=b$, and $C D=c$. By symmetry, we may assume that $c<a$. We show that $D E=a+b$ and $E A=b+c$.

![](https://cdn.mathpix.com/cropped/2024_06_05_bf078063ce163dc741bfg-3.jpg?height=393&width=333&top_left_y=1533&top_left_x=885)

Draw a line parallel to $B C$ through $D$. Extend $E A$ to meet this line at $F$. Draw a line parallel to $C D$ through $B$ and let it intersect $D F$ in $G$. Let $A B$ intersect $D F$ in $H$. We have $\angle F D E=60^{\circ}$ and $\angle E=60^{\circ}$. Hence $E F D$ is an equilateral triangle. Similarly $A F H$ and $B G H$ are also equilateral triangles. Hence $H G=G B=c$. Moreover, $D G=b$. Therefore $H D=b+c$. But $H D=A E$ since $F H=F A$ and $F D=F E$. Also $A H=a-B H=$ $a-B G=a-c$. Hence $E D=E F=E A+A F=b+c+A H=(b+c)+(a-c)=b+a$.

We have five possibilities:

(1) $b<c<a<b+c<a+b$;

(2) $c<b<a<b+c<a+b$;

(3) $c<a<b<b+c<a+b$;

(4) $b<c<b+c<a<a+b$;

(5) $c<b<b+c<a<a+b$.

In (1), we see that $c<a<b+c$ are three consecutive integers provided $b=2$. Hence we get $c=3$ and $a=4$. In this case $b+c=5$ and $a+b=6$ so that we have five consecutive integrs $2,3,4,5,6$ as side lengths. In (2), $b<a<b+c$ form three consecutive integrs only when $c=2$. Hence $b=3, a=4$. But then $b+c=5$ and $a+b=7$. Thus the side lengths are $2,3,4,6,7$ which are not consecutive integers. In case (3), $b<b+c$ are two consecutive integrs so that $c=1$. Hence $a=2$ and $b=3$. We get $b+c=4$ and $a+b=5$ so that the consecutive integers $1,2,3,4,5$ form the side lengths. In case (4), we have $c<b+c$ as two consecutive integers and hence $b=1$. Therefore $c=2, b+c=3, a=4$ and $a+b=5$ which is admissible. Finally, in case (5) we have $b<b+c$ as two consecutive integers, so that $c=1$. Thus $b=2, b+c=3, a=4$ and $a+b=6$. We do not get consecutive integers.

Therefore the only possibilities are $(a, b, c)=(4,2,3),(2,3,1)$ and $(4,1,2)$. This shows that $a+b+c=9,6$ or 7 . Thus there are three possible sums $A B+B C+C A$, namely, 6,7 or 9 .
