The wedding dates are April 6th, May 12th, and August 1st.
Bonus question: ~ 6.86 miles.
Explanation:
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All villages are crossings. Here are the steps to
determine the number of routes to reach any point:
1) First set the starting point to one.
2) Select the unvisited crossing that is adjacent to a
visited crossing and is closer to the destination,
but which is the farthest such unvisited crossing from
the destination.
3) Set this crossing to the sum of the values for all
of the adjacent visited crossings. (If you do this
inthe correct order, all the adjacent visited
crossings are farther from the destination.)
4) If the last selected crossing is not the
destination then continue with step two.
Which points are closer or farther from the
destination can be determined using the right
triangles formed by the intersecting paths. The
shortest path from a point to a line is perpendicular
to the line. Moving toward the right angle formed by
the path and the line is moving closer to the point;
moving away from this right angle is moving farther
from the point.
Note that when moving from a village to the opposite
village that all of the crossings to the left of the
direct path between the villages are symmetrical with
the corresponding crossing to the right.
The routes:
This is a little hard to follow without a diagram, so
you might want to create one. The diagram is a regular
hexagon with lines connecting each pair of vertices.
There are three intersections created along each of
these connecting lines, with the three lines
connecting opposite vertices meeting in the middle.
Label one vertex A and moving clockwise, label the
remaining vertices B-F. Label the middle intersection
M. Label the intersection along the path from A to M
(path AM) as Ai; label the similar intersections for
the remaining vertices as Bi to Fi. Label the
intersection of paths AC and BF as ABi. Label the
similar intersections BCi, CDi, DEi, EFi and FAi.
Routes to an adjacent village (A to B):
To determine the routes from A to B first set
A to 1.
The farthest point from B that is closer than A and adjacent to A is Ai. Ai can only be reached from A, so
set Ai to 1.
The next farthest point is ABi. ABi can be reached
from A and Ai so set ABi to 1+1=2.
The next farthest point is Bi. Bi c an only be reached
from ABi, so set Bi to 2.
The next farthest point is the destination B. B can be
reached from A, ABi and Bi, so set B to 1+2+2=5.
There are five routes to an adjacent village, so that
son will make his last trip on April 5th and the wedding will be the next day on April 6th. Routes to the second village along the coast (A to C):
A = 1
AFi = A = 1
Fi = AFi = 1
Ai = A + AFi = 2
ABi = Ai + A = 3
M and B are equidistant from C, so in either order:
M = Ai + Fi = 3
B = A + ABi = 4
Bi and Di are equidistant from C, so in either order:
Bi = B + ABi + M = 10
Di = M = 3
BCi and CDi are equidistant from C, so in either
order:
BCi = B + Bi = 14
CDi = Di = 3
Ci = BCi + CDi + M = 20
C = B + BCi + CDi + Ci = 41
There are 41 routes from A to C, so that son will make his last trip on May 11th and the wedding will be
the next day on May 12th.
Routes to the opposite village (A to D):
A = 1
F, B = A = 1
FAi, ABi = A + B = 2
Ai = A + ABi + FAi = 5
Fi, Bi = ABi + B = 3
EFi, BCi = B + Bi = 4
M, C and E are equidistant from D, so in any order:
M = Ai + Fi + Bi = 11
E, C = B + BCi = 5
Ei, Ci = BCi + M + C = 20
DEi, CDi = C + Ci = 25
Di = M + CDi + DEi = 61
D = C + E + CDi + DEi + Di = 121
There are 121 routes from A to D, so that son will make his last trip on July 31st and the wedding will
be the next day on August 1st.
Bonus Question:
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The longest route is a route to the opposite village.
Given the above diagram, the longest routes the son will have to travel from A to D is:
A-B-ABi-C-Ci-CDi-Di-D
and the mirror route
A-F-FAi-E-Ei-DEi-Di-D
(ABi-C is the same as ABi-Bi-BCi-C)
If the circumference is 10 miles then the diameter is
10/Pi miles and the radius is 5/Pi miles. Since a
regular hexagon is formed from six equilateral triangles, the length of a side of the hexagon is
equal to its radius.
The path A-B is then 5/Pi miles. The path B-D bisects the path C-M, so both C-Ci and Di-D are half the
radius. A-B + C-Ci + Di-D is then equal to the diameter 10/Pi.
The paths B-ABi, ABi-BCi, and BCi-C are all sides of equilateral triangles with a height of one half of the
radius. The paths Ci-CDi and CDi-Di are each half of
the side of one of these same triangles. The side of
one of these equilateral triangles is:
So the total distance traveled is:
10/Pi + 4*(5/(sqrt(3)*Pi)) = 10/Pi + 20/(sqrt(3)*Pi)
~6.86 miles |
