Reworded UKMT Intermediate Challenge

Straight away, we see that these distances are along the road, so any possible curvatures and twists in the road are included, and we do not worry about the direct distances if the road is not straight. The distances are between pairs of towns.

This means that $20$ must be the distance between the first town and the last town (Alphaville and Echoton). The distance between any towns must not exceed $20$ . Now, we need to find out which combinations of the other numbers work and which don't.

$2$ and $18$ is an inevitable combination. The sole variation is the order of these numbers. Either the $2$ or the $18$ could be the distance between A and B with the other then being the distance between D and E. This also means that the $5, 6$ and, logically, $7$ must be together $(5+6+7=18)$ . This leaves us with the following possible number sequences $2,5,6,7; 2,5,7,6; 2,6,7,5; 2,7,6,5; 2,7,5,6;$ and $2,6,5,7,$ as well as $7,6,5,2; 6,7,5,2; 5,7,6,2; 5,6,7,2; 6,5,7,2;$ and $7,5,6,2$ .

5 and 15 is another inevitable combination. 5 cannot be in the middle (inside the 15) because to add up to 20, 15 would then need to be summed up with the 2 numbers on the side (which would be either 2 and 3 or 4 and 1), however, 1, 3 and 4 are not allowed. The 5 must also be on the opposite side from the two. For example, 5 cannot be the distance between A and C if the distance between A and B is 2. Otherwise, the distance between B and C would be 3, which is impossible in accordance with the problem’s terms.

We have deduced that 2 and 5 must be on the sides: either one could be the distance between A and B and the other then between D and E . We must now find the possible positions of 6. It can be the distance between either B and C or C and D. The 7 would fall back into the remaining position.

This, in turn, means that the 6, 7 and 2 must be together $(7+6+2=15)$ . Now, of the sequences mentioned above, we have only got 2,6,7,5; 2,7,6,5, as well as 5,7,6,2 and 5,6,7,2 left. Now we can calculate the distances between non-adjacent towns (for example, between A and C or C and E). They are: 8, 12 and 13 (for 5,7,6,2, and 2,6,7,5) or 9, 11 and 13 (for 5,6,7,2 and 2,7,6,5).

Therefore, there are two different possible sequences:

a) $2, 5, 6, 7, 8, 12, 13, 15, 18, 20$ .

b) $2, 5, 6, 7, 9, 11, 13, 15, 18, 20$ .

So, all we have to do is add $7+8+12+13+7+9+11+13$ .

Since the distances are measured along the road, you don't have to worry about any curves or connections between the cities through other roads (thank god.) So, the first thing that becomes clear is that if we know 20 is the longest distance along this road between any two of these five cities, then it must be the distance between Alphaville and Echoton as they are the two endpoints in the scenario.

We also know that Alphaville passes through Bravoton, Charlyville, and Deltatown in that order on its way to Echoton. So in other words, the full road from A to E is made up of four smaller paths (A to B, B to C, C to D, and D to E) which I will call "single paths."

Since the measurements we're given correspond to the distances along this road between the cities, and since the road from A to E can't be split into any smaller relevant pieces than those "single paths," the smallest of the 10 distances (2) must belong to one of those four paths. For the next two smallest distances (5 and 6,) it is seemingly possible for them to correspond either to two of the remaining "single paths" or to some composition of "single paths" (for example A to B + B to C = A to C) So to test out the possibilities, you can consider the following:

The distance of 5 must either be a "single path" or a composition of "single paths." However, to add up to 5, we would need something like distances of 2 and 3 or distances of 1 and 4. However, none of the possible options of adding up to 5 are allowed based on our list of the smallest distances. So the distance of 5 must refer to a "single path." Similarly, the distance of 6 must either be "single path" or a composition of "single paths," but, just like 5, we are unable to add up to 6 based on our list of the smallest distances. So, the distance of 6 must also refer to a "single path."

All of that means that the total distance of 20 must be composed of single paths of distances 2, 5, 6, and x. In other words, 2 + 5 + 6 + x = 20, and, therefore, x = 7. So, one of the missing distances must always be 7. The rest just comes down to a matter of how to arrange the four single paths.

Fortunately, we're given a clue about the possible arrangements because we know that 15 and 18 are the second longest distances between any of the two cities. Since we have determined the lengths of all four single paths, these distances must correspond to some combination of single paths: either a combination of two single paths or a combination of three. If 15 referred to a composition of just two single paths, that would mean the remaining two have to add up to just 5. However, this is impossible given the lengths of the four single paths. The same is true if 18 referred to a composition of just two single paths.

So 15 and 18 must refer instead to compositions of three single paths. In other words, either 15 or 18 must be the distance between A and D and the other must be the distance between B and E. By looking at the distances of the single paths, you can deduce that there is only one possible way to find numbers which sum to 18 (5, 6, and 7) and only one way to find numbers which sum to 15 (2, 6, and 7.) As both compositions share only the numbers 6 and 7, 6 and 7 must refer to the distances of the shared area of these two compositions: the paths from B to C and C to D. The rest of the problem just comes down to testing out possible combinations:

1.) AB = 2, BC = 6, CD = 7, and DE = 5, which gives the additional distances: 8, 12, and 13 (AC, CE, and BD respectively)

2.) AB = 2, BC = 7, CD = 6, and DE = 5, which gives the additional distances: 9, 11, and 13 (AC, CE, and BD respectively)

You could test out two additional arrangements by swapping distances 2 and 5, but they would give the same results.

So, the possible missing measurements are either: 7, 8, 12, 13; OR: 7, 9, 11, 13

7+8+12+13+7+9+11+13 = 80.

This is my first time writing a solution! I hope it isn't confusing or messy!