Points and Distances!

We have that the segment $AE$ must be equal to 19. Since the second longest segment is 17, it follows that one of the segments is of length 2 and is made of two adjacent points. This means it's either $AB$ or $DE$ is 2. WLOG, $AB$ is 2.

The third longest length is 15. This results in one of the segments being length 4. Since \ $AB$ is already length 2 only one segment has length 2, it means that this segment must be made of adjacent points. This could be either $BC$ , $CD$ or $DE$ . It cannot be $BC$ because otherwise, the length of $AC$ is 6, which is not on the list. We have that either $CD$ or $DE$ of length 4. Since this is made of adjacent points, it follows that the segment of length 15 is segment $AD$ , which implies segment $DE$ is 4.

The third shortest segment is 5 long. Since no pair of numbers before it add to 5 in the list, we have that this segment must be made from 2 adjacent points. This can be either $BC$ or $CD$ . If the segment is $BC$ , then segment $AC$ is 7, which is on the list, so this segment is valid. If the segment is $CD$ , then segment $CE$ is 9, which may or may not be possible. We will leave this as such for the time being.

Note that we already know the lengths of 3 out of the 4 segments made from adjacent points (2, 4, 5) and we know the total length is 19. This implies that the last segment made from adjacent points has length 8. Either segments $BC$ or $CD$ have a length of 8. If $BC$ is 8, we have $CE$ is 9 and $AC$ is 10. But this is a contradiction! We can only have one value between 8 and 13, and yet both segments $CE$ and $AC$ have length between these two numbers. Therefore, $CD$ is 8 and $BC$ is 5. The value of $K$ is $CE$ which is 12.

Therefore, the answer is 12.

$\text{There are at least two possibilities with the same internal structure. We shall see one of this.}\\ \text{Since maximum distance is 19, } \color{#3D99F6}{AE=19.} \\ \text{To have 17, we take }\color{#3D99F6}{ AD=17, \ DE=2}., \text{. OR.. BE=17, AB=2 also possible.}\\ \text{Since AD=17,... DE=2,.. next try if AC=15. Then CD=2, but 2 is used by DE.} \\ \text{So we try BD=15 then AB=2, but 2 is used by DE.} \\ \text{So we try } \color{#3D99F6}{ BE=15, \ then\ AB=4}.\\ \text{So BD=BC+CD=13. From the list only possible solution is BC=5, OR BC= 8.} \\ \text{Try BC=5, then AC=9, and CE=10. But 9 and 10 both can not be out of list .} \\ Try \ \color{#3D99F6}{ BC=8,\ then\ AC=12, \ only\ one\ not\ in\ the\ list,\ and\ CE=7 }\ a\ number\ in\ the\ list.\\ \color{#3D99F6}{ BD=15,\ \ ,CD=5.}\\ \text{So all ten in the list has been accounted for in blue.}\\ \text{So K= }\huge \color{#D61F06}{12} .$

$Options$

We have taken the numbers left to right, and distance from A as AB=4, AC=12, AD=17, AE=19. We could have just the reverse order. AB=2, AC=7, AD=15, AE=19.

The sum of all segment lengths is equal to $S = 4x_5+2x_4-2x_2-4x_1=4(x_5-x_1)+2(x_4-x_2)$ , where $x_i$ are the coordinates of point $i$ , with $x_{i+1} > x_i$ . We know that $x_5-x_1=19$ , and from adding all segments lengths that $S=90+K$ , therefore $K=2(x_4-x_2)-14$ . Notice that $x_4-x_2$ is a segment length itself, and therefore must appear on the list (it may also be the missing number $K$ ).

So we are looking for a number between 8 and 13 that satisfies the above conditions. $K$ cannot be 9, 11 or 13, because then $K+14$ is not even. If $K$ where 8 or 10, there should be a segment of length $(8+14)/2=11$ or $(10+14)/2=12$ respectively, none of which are in the list. The only case that the above conditions hold is when $K$ equals 12, for which $x_4-x_2=13$ .

We can therefore conclude that $K=12$ .

NOTE: I did not prove that such an arrangement of segment lengths is possible, I assumed it to be correct.

The ten distances will be: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE. The longest distance is 19, so this must be AE. There are 3 pairs of distances equal to AE - these are AB + BE, AC + CE and AD + DE. Each of these pairs must equal 19, so from the values available we have 2 + 17 and 4 + 15. Since the value list is in order from smallest to largest there is no 13 + 6 pair or 5 + 14 pair so the last pair will be K + 7 or K + 8, so K is either 11 or 12. The remaining distances are BC, CD and BD. The remaining values are 13, 5 and either 7 or 8 depending on value of K. Since BD = BC + CD and we have 13 and 5 to assign we can determine that BD = 13 = 8 + 5. This means that K = 19 - 7 so K is 12.