Have you ever complained about the price of tolls? Thank your lucky stars you’re not living in the kingdom of mad King Fengel of Rohan.
The motorway network of the kingdom of Rohan looks like a hexagonal star, with six straight roads. There are 12 toll plazas on these motorways, indicated by the circles on this map.
King Fengel of Rohan decided that the toll at each plaza should be a whole-number value between 1 Rohan dollar (RD$) and RD$12 (ie RD$1, RD$2, RD$3 … RD$12). No two plazas can have the same toll.
He ruled that the total toll to be paid along each of the six straight roads – the sum of what’s paid at the four plazas along that stretch – should be the same.
In a moment of madness, he also ruled that the sum of the tolls due at the six plazas located at the vertices (points) of the star should also equal the amount to be paid along each of the six straight roads.
Is it possible to satisfy these rules, or is King Fengel’s crown a little too tight for his head?
Scroll down to see the solution.
This week’s maths puzzle comes courtesy of Dr Bernd Kreussler, lecturer in mathematics at Mary Immaculate College in Limerick, who is actively involved in mathematics enrichment classes at the University of Limerick and the Irish Mathematical Olympiad.
Yes, it is possible.
To make our lives a little easier, let’s use the letters A, B, C, …, L as stand-ins for the amount of toll to be paid at each of the 12 plazas, as shown here:
From here, we can make a few observations based on Fengel’s rules.
If we allow S to stand for the amount of toll to be paid along each of the six straight roads (and the total amount to be paid at the vertices of the star), the rules lead to the following seven equations:
A + G + H + C = S
C + I + J + E = S
E + K + L + A = S
B + H + I + D = S
D + J + K + F = S
F + L + G + B = S
A + B + C + D + E + F = S
Because A, B, C, …, L are the numbers 1, 2, 3, …, 12 in some order, we know that
A + B + … + L = 1 + 2 + … + 12 = 78
When we add the first three equations above, we get
2(A + C + E) + (G + H+ I + J + K + L) = 3S
And, if we add the second three equations, we get
2(B + D + F) + (G + H + I + J + K + L) = 3S
This implies that A + C + E = B + D + F, and hence
2(A + C + E) + (G + H + I + J + K + L) = A + B + … + L = 78 = 3S
Therefore, we know S = 26 and, as they are half of S, A + C + E = B + D + F = 13.
Armed with this information, it is not so hard to find solutions. One solution can be seen here, although there are many more.
Updated at 10.40 on 8 August to include a very important part of the puzzle that had been omitted.
Main image via Shutterstock