Team Solutions

1. Geoboard problem.

Parts (a) and (c): When you construct polygons so that they have 4 boundary pegs, the area
of your polygons should be:

• A = 2 when there is one interior peg

• A = 3 when there are two interior pegs

• A = 4 when there are three interior pegs

Note that each time we add one interior peg, the area increases by one as well { this implies
a linear relationship with a slope of 1. This indicates that, in general, the area of a polygon
with 4 boundary pegs will be i + 1, where i is the number of interior pegs.

Parts (b) and (d): When you construct polygons so that they have 2 interior pegs, the area
of your polygons should be:

• A = 3 when there are four boundary pegs

• A = 3:5 when there are five boundary pegs

• A = 4 when there are six boundary pegs

Note that each time we add one boundary peg, the area increases by one half { this implies a
linear relationship with a slope of 1/2. This indicates that, in general, the area of a polygon
with 4 boundary pegs will be
b/2 + 1, where b is the number of boundary pegs.

In general, the area of a geoboard figure is , where i is the number of interior pegs
and b is the number of boundary pegs.
(For more information on this type of problem, search for \Pick's Theorem" on the internet.)

2. In order to determine how many different teams must exist, it makes life easier to call the
teams something generic like T ₁; T₂; etc. (That is, just identify different teams by subscripts,
rather than inventing new names for each.)

So, since Ed has three favorite teams, let's call these teams T₁; T₂ and T₃. Now, Ed and Fran
must have exactly one favorite team in common; let's say that team is team T₁. Then, Fran's
other two teams must be teams not in Ed's list { say, T₄ and T₅.
Thus, so far we have:

• Ed's teams: T₁; T₂; T₃

• Fran's teams: T₁; T₄; T₅

Now we come to Gus. To prove that there must be more than five elbowball teams, we must
establish that Gus has a favorite team that has not already been included in Ed's or Fran's
list of favorites. But with a bit of thought, we see that this is obvious: Gus has one favorite
team in common with Ed , and one favorite in common with Fran, accounting for (at most)
two of his three favorite teams. Therefore, Gus must have at least one favorite team that is
in neither Ed's nor Fran's list. Thus, there are more than five elbowball teams in all.
(Note: This actually shows that the number of elbowball teams must be 6 or 7. There are 7
teams if Gus has team T₁ as a favorite, but there are only 6 teams if Gus has, for example,
teams T₂ and T₄ as favorites.)

3. Correct answer: 79 coconuts were gathered.
Let N be the number of coconuts gathered.

• The first man divides these into three equal groups, with one left over. He keeps one of
these groups and leaves the other two groups in the pile (and gives the remaining one
coconut to the monkey).

Let x₁ denote the size of each of the three groups of coconuts. Then, N = 3x₁ + 1. The
first man keeps x₁ coconuts, and the remaining pile contains 2x₁ coconuts.

• Later, the second man finds the pile containing 2x₁ coconuts. He divides these into three
equal groups, with one left over. He keeps one of these groups and leaves the other two
groups in the pile (and gives the remaining one coconut to the monkey).

Let x₂ denote the size of each of these three equal groups of coconuts. Then, 2x₁ = 3x₂+1.
(Take a moment to make sure you understand this step! It just writes the number of
coconuts in the current pile in two different ways - once in terms of x ₁, and then again
in terms of x₂.) The second man keeps x₂coconuts, and the remaining pile contains 2x₂
coconuts.

Now, let's solve for N in terms of x₂. We know and we have from the
preceding step that N = 3x₁ + 1. By substitution, then, we have

• Still later, the third man finds the pile containing 2x₂ coconuts. He divides these into
three equal groups, with one left over. He keeps one of these groups and leaves the other
two groups in the pile (and gives the remaining one coconut to the monkey).

Let x₃ denote the size of each of these three equal groups of coconuts. Then, 2x₂ = 3x₃+1.
(Noticing a pattern?) The second man keeps x3 coconuts, and the remaining pile contains
2x₃ coconuts.

Now, let's solve for N in terms of x₃. We know and we have from the
preceding step that By substitution, then, we have

• Finally, in the morning, the three men find the pile with the remaining 2x₃ coconuts.
They divide these coconuts into three equal groups, with one left over for the monkey.

Let x₄ denote the size of each of these equal groups of coconuts. Then, 2x₃ = 3x₄ + 1.
Each of the three men keep x₄ coconuts, and finally all coconuts are accounted for.

Now (one more time!), let's solve for N in terms of x₄. We know and we
have from the preceding step that By substitution, then, we have

This will al low us to work backwards - our goal is to find \small" whole number values of x ₄
for which N will be a whole number. For example, x₄ can't equal 1, since substituting x ₄ = 1
would give us N = 146/8, which is not a whole number.

By trial-and-error, we find that the smallest whole number value of x₄ for which N is a whole
number is x₄ = 7 - in this case, N = 632/8 = 79. This is less than 100, so it should be our
answer!