On with the story. My initial guess was that there would be an infinite
number of these triangles and it
seemed to me that the obvious way to attack this problem was to use
Heron's formula for the area of a triangle.
the lengths of the three sides of the triangle and
. I let the sides of the triangle
represented by the three consecutive integers
. I then found, after simplification, that
Upon examination of this result it was clear that had to be
an even integer in order for to be an integer.
For if were odd, would be odd and
even if were a perfect square, would not
divide . Also, by observation, it is clear that must be equal to a number of the form . I did
a quick search and found that the integers
all worked. Since my assumption was that there were going to be an
infinite number of these triangles, I looked for a recurrence relation
for which were the first three terms that
generated and infinite sequence of values for satisfying
(1). I would then test this sequence of to see if the values always
determined a Super-Heronian triangle.
Looking at the three terms, , my guess was that
these values satisfied the sequence given by the recurrence relation
I used this sequence to generate more terms, all of which seemed to
work (see the table on the next page).
The sequence generated by the recurrence relation (2) can also be
Here are the first terms terms which do, in fact, generate
Super-Heronian triangles. The first row generates a line which
represents a degenerate triangle.
As one can see, the sides get large very fast. It is left to show that
every term of the sequence yields a solution;
that is, an area that is an integer. Let
Let us examine another interesting feature of these triangles. Suppose
we use the area formula . We then have the area of the triangle given
One might observe that the s can be generated on
their own from the recurrence relation
One can investigate these triangles and find many interesting properties. I will leave further investigation to the reader.
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