# How to fill up the numbers in a set of empty discs drawing a pentagon?

The following pentagon is composed of 10 discs. Each disc must be filled with one number. The first 10 even positive numbers are allowed. The sum of the numbers to be placed in each side of the pentagon must be the same and the maximum possible. Find the sum of the numbers adjacent to the discs filled with the numbers 2 and 18.

The figure is shown below: The choices given are:

1. 54
2. 60
3. 50
4. 64

I tried several ways to fill up this pentagon but did not find a clear solution. Does a strategy to solve this kind of puzzle exit? I'm stuck at the very beginning. I believe this requires mathematics. Please try to include some visual aid and the most detailed way to solve this problem.

• Need some clarification: In the question, you said to fill the pentagon with first ten positive even numbers, but at the end of first paragraph you ask the pentagon to be filled by only 2 and 18? Oct 13 '19 at 13:42
• @SamRoy Fill the circles with each of the even numbers $2$ to $20$ as described. Then take the sum of the numbers that are adjacent to the numbers $2$ and $18$ in the filled diagram. Oct 13 '19 at 18:31

The puzzle asks for a maximal sum for the numbers on the edges, so...

The sum of all five edges will include each number at a vertex twice, once for each edge that it is part of. We need to put the largest numbers at the vertices in order to maximize the sum. This will give us a maximum total sum of $$2(20+18+16+14+12)+10+8+6+4+2=190$$. Dividing the among the five edges gives us a maximum equal edge sum of $$\frac{190}{5}=38$$. In order to obtain equal sums on all edges, we need to balance the high and low numbers. Starting with $$20$$ at a vertex, we add $$2$$ and $$4$$ to either side of the $$20$$. Given the target sum of $$38$$, we complete the first two edges as $$20+2+\color{blue}{16}=38$$ and $$20+4+\color{blue}{14}=38$$. Now consider the other two vertices. We need to put $$12$$ and $$18$$ in those discs. If we put $$18$$ on the edge with $$16$$, we would need $$16+4+18=38$$, but we have already placed the $$4$$. So we place $$18$$ with $$14$$ and $$12$$ with $$16$$ which gives us $$16+10+12=38$$ and $$14+6+18=38$$, with the final number, $$8$$, we have $$18+8+12=38$$ and the diagram is complete. The numbers adjacent to $$2$$ and $$18$$, with their sum, are $$16+20+8+6=\color{blue}{50}$$

• Nice approach. +1 Oct 15 '19 at 4:00