Problem 1
Vasquez 1
Andrew Vasquez
Geoboard Triangles
When we did the geoboard problem in class we were given a geoboard and some rubber bands.
The geoboard we were given was a square block of wood with 25 nails placed in it that were evenly
spaced to make a 5 by 5 square. So it looked like this from the top:
The very first question we were asked was, how many squares can we make with the rubber bands by
wrapping them around the nails. Our group quickly counted up every 1x1, 2x2, 3x3 and 4x4 square,
where a length of 1 is the distance from one nail to a nail directly adjacent to it. We found that there are
16 1x1 squares, 9 2x2 squares, 4 3x3 squares, and 1 4x4 square. That made a total of 30 squares which
we thought was the answer to the question. Then someone in class pointed out that a square can be
made with the sides going at an angle across the board. An example would look like this:
In this square, all of the sides have a length of √2. This length is easily and naturally found by using the
Pythagorean Theorem where the side of the square is the hypotenuse to the triangle with legs that are
both 1. So the question then became what other lengths could we make for a side of a square. We
specialized for a little while and then came about a much more systematic approach. We had already
found the length from a right triangle with legs of length 1 and 1. So then we tried a right triangle with
legs of length 1 and 2. This gave us squares with sides of length √5. Then we tried a right triangle using
1 and 3 as the legs. This gave us squares with sides of length √10. The next triangle we would try is one
with legs of length 1 and 4. When constructing this with rubber bands on the geoboard it became
immediately obvious that it couldn’t fit on the board that we were given. For a more mathematical
reason, if we carried out the calculations we would get a square with a length of √17. This would have
an area of 17 which is more than the area of the board we are working with. So we found all of the
squares using 1 as a leg for the right triangle. Next we tried a triangle with legs 2 and2, then 2 and 3, and
so on. The legs of length 2 and 2 gave us a square with sides of length equal to √8. All other triangles
with longer lengths could not fit on the geoboard that was given to us. So for
the √2x√2, √5x√5, √8x√8, and the √10x√10 squares we counted 9, 8, 1, and 2 possibilities
respectively. So our total was 50 squares.
Problem 1
Vasquez 2
The next question that was proposed was as follows: Suppose that you have a 100 x 100
geoboard. You want to make an equilateral triangle on this geoboard. What are the possible sizes of an
equilateral triangle? If no equilateral triangle is possible, prove there are no solutions. The answer is that
it is impossible to make an equilateral triangle on this geoboard. That is the answer plain and simple.
The reason for this requires a proof. Assume you have an equilateral triangle on the geoboard. Then you
can draw a rectangle around it so that its vertices are each on one of the sides of the rectangle. So it
would look as follows:
R
c
Q
b
P
a
A
The length of a, b, and c have to be an integer because it denotes how many pegs to the right and how
many pegs up one point goes to the next. So the area of the rectangle around the triangle must be equal
to a(b+c). Since a, b, and c are integers this must also be an integer. The area can also be found by
finding the area of the four parts the triangle breaks up the square into. The area of the right triangle
with sides a and b must be
𝑎𝑏
.
2
This area has to be a rational number. The same logic applies to all the
right triangles outside of the equilateral triangle. They must all have rational areas because the lengths
of the sides are integers. The length of PQ must be equal to √𝑎2 + 𝑏 2 . If you take this to be the base of
1
the triangle then the height must be equal to 2 √3√𝑎2 +𝑏 2. This is because all angles of an equilateral
ℎ𝑒𝑖𝑔ℎ𝑡
triangle have to be 60 degrees and tan(60) = (.5) 𝑏𝑎𝑠𝑒. Thus the height must equal √3(
𝑏𝑎𝑠𝑒
).
2
So the area
1
of this triangle must be 4 √3√𝑎2 +𝑏 2 . Since the area of the entire square is rational and the 3 other
regions of the square are rational, then the area of this triangle must be rational. In order for this area to
be rational a2 + b2 must equal a number that has a prime factorization that contains a 3 with an odd
power and every other prime has an even power. However, the numbers that can be written as the sum
of two squares cannot satisfy this property due to one of Fermat’s theorems which states: A number N is
expressible as a sum of 2 squares if and only if in the prime factorization of N, every prime of the form
(4k+3) occurs an even number of times. Therefore the area of this triangle must be irrational. This is a
contradiction which implies that you cannot make an equilateral triangle on any size geoboard.
Problem 1
Vasquez 3
Another way to find the answer goes very similar to the last argument except this one doesn’t
use the area but rather the lengths of the sides. Say you start trying to make an equilateral triangle on
the geoboard. The best place to start is by constructing one of the sides first. Referring back to the
diagram, let PQ be the line that is constructed on the geoboard. In order to make an equilateral triangle
we have to find a point R on the geoboard such that the length of PQ is the same as PR and the angle
QPR is 60 degrees. So we start by making the length the same. Since the length of PQ must be equal to
√𝑎2 + 𝑏 2 then that must also be the length of PR. So we can make R by going up a length of a and to the
right a length of b from the point P. Our second condition is that the angle QPR is 60 degrees. This would
mean that the angle APQ would have to be 15 degrees since going to the right and going up from P form
𝑏
𝑎
a right angle. This would mean that tan(15) = . By using the half angle formula for tangent, we can
compute that tan(15) =
1−cos(30)
=
sin(30)
2 - √3. This is an irrational number but
𝑏
𝑎
must be a rational number
since b and a are integers. Thus we’ve reached a contradiction. Angle QPR cannot be equal to 60
degrees and thus triangle PQR cannot be an equilateral triangle.
So an interesting generalization that we can take away from this is that you can’t make an
equilateral triangle on a coordinate plain where all of the vertices are integer points. That is the direct
result of what we just proved. An extension of this problem can look more into the numbers that can be
written as the sum of two squares. In the first proof I stated a theorem about the sums of two integer
squares. We were also starting to see how the sums of two squares were coming into place when
finding the different lengths of a side of a square on the geoboard. This is a very interesting concept and
if you explore more of Fermat’s theorem you can go on to show that a prime can be written as the sum
of two squares if it congruent to 1 mod 4. These findings do require knowledge of abstract algebra to
prove and to really understand why they are true but they are still very interesting none the less.
Another way to think about this concept is that if a number can be written as the sum of two squares
then it can be factored in the complex field Zi. In other words if a number N can be written as the sum of
two squares then N = (a + bi)(a-bi) for some integers a and b and i = √−1. This shows that some of the
primes of the integers can be expressed as a product of two complex numbers. This tells you about
primes in the complex field and the different forms they might take on. I’m sure there are many more
things that you can go into about finding primes in the complex field but this is about all I know.
When my group was trying to solve the question about the equilateral triangle, we came across
a lot of difficulties. We started off trying to make an equilateral triangle very much like the procedure in
my method 2; however, it wasn’t that formalized. We started at the bottom of the geoboard and we
made PQ horizontal. We soon came across the realization that if the length of PQ is odd then the point
in the middle where R had to lie would have to be between two pegs. Then we tried it when the length
of PQ is even. Clearly there was a peg that R can lie on in the middle but did any of them make an
equilateral triangle was the question we needed to figure out. After doing some calculations to find
what the height of an equilateral triangle must be, we found that the height would have to be an
irrational number. So from there we concluded that it must be impossible and we stopped working on
the question. Then when the professor came around he told us that this only proves the case for a
family of triangles where the base is an integer. By this time the group had already stopped working and
Problem 1
Vasquez 4
it was very hard to concentrate on starting over again. We were also running out of time for the class
period and we did not explore anymore options. Then at the end of class we were given a brief
explanation of what I now used as method 1. After I saw the answer I realized that I could have gotten
to the answer right from the start. In my very first reaction to the new question, I thought to myself,
“How can I inscribe an equilateral triangle inside of a square?” Since the answer to this question was not
immediately obvious to me I decided to leave the question alone and look for another way to solve the
question. If only I had pursued this thought I might have been able to understand method 1. The only
difference in my first thought and the first method was that I was assuming I had a square and I wanted
to put an equilateral triangle in it. In method 1 you have to assume that you have an equilateral triangle
and then put a rectangle around it. If I would have just changed my thought in this way it would have
led me down a path to finding a proof by contradiction. This is exactly like what we talked about in class
the very same day and what we had for the readings. Sometimes the first thought is the right one but if
we are stuck on how to proceed with it then we often try to find another way. If I wrote down my idea
when I first thought about it, then maybe after I got stuck going down a different path I would have
thought to look back at it. This problem was very difficult because of the way it had to be proven. We
kept thinking we had the answer when really we only proved it for a special case. Thinking you have it
right and then realizing you’re not done yet can be very frustrating for any problem solver. I think we
never came to the final answer because we weren’t thinking abstract enough and kept creating special
cases without really thinking about what assumptions we were making. Getting a glimpse at the solution
at the end of class was very relieving and satisfying for all of us though. It especially encouraged me to
trust my first instinct a little better.
Problem 1
Vasquez 5
Geoboard triangles.
Suppose that you have a 100 x 100 geoboard. You want to make an equilateral triangle on this geoboard.
What are the possible sizes of an equilateral triangle? If no equilateral triangle is possible, prove there are
no solutions.