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

The figure is shown below:

Sketch of the problem

The alternatives given are:

$\begin{array}{ll} 1.&54\\ 2.&60\\ 3.&50\\ 4.&64\\ \end{array}$

I tried several ways to fill up this pentagon but none did really came up with a clear solution. Does it exist a method, strategy or something that can be used in an effective manner to solve this riddle?. I'd appreciate someone could help me with this. I'm stuck at the very beginning hence I can't offer more. Since I believe this would require the use of maths, please try to include some visual aid and the most detailed way to solve this problem as I mentioned I don't know what can be done to solve this.

  • 1
    $\begingroup$ 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? $\endgroup$ – SamRoy Oct 13 '19 at 13:42
  • $\begingroup$ @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. $\endgroup$ – Daniel Mathias 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 $16+20+8+6=\color{blue}{50}$

  • $\begingroup$ Nice approach. +1 $\endgroup$ – SamRoy Oct 15 '19 at 4:00

The answer is:


Because the filled-up pentagon looks like:

enter image description here

The sum in each edge is 38. This is the maximum I got after several trial & errors.

  • $\begingroup$ Indeed, this is correct. See my answer for details on how to arrive at this result more directly. $\endgroup$ – Daniel Mathias Oct 13 '19 at 19:06

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