Integer Sided Triangles with Integral Area/perimeter Ratio

Consider the triangle with sides $6$, $8$, and $10$. It can be seen that the perimeter and the area are both equal to $24$.
So the area/perimeter ratio is equal to $1$.
Consider also the triangle with sides $13$, $14$ and $15$. The perimeter equals $42$ while the area is equal to $84$.
So for this triangle the area/perimeter ratio is equal to $2$.

Find the sum of the perimeters of all integer sided triangles for which the area/perimeter ratios are equal to positive integers not exceeding $1000$.

For a given triangle with side lengths $a, b, c$, the area and the semi-perimeter are given by:

– Semi-Perimeter = $s = \frac{a + b + c}{2}$
– Area = $A = \sqrt{s(s – a)(s – b)(s – c)}$

where $s$ in the area equation is the semi-perimeter.

The requested value is the sum of the perimeters of all triangles satisfying:

– $s = kA$ for integer $k = 2, 3, \ldots, 1000$,
– $a, b, c$ being integer sides of each triangle.

This equation simplifies to:

– $\frac{a + b + c}{2} = k\sqrt{\frac{a + b + c}{2}\left(\frac{a + b + c}{2} – a\right)\left(\frac{a + b + c}{2} – b\right)\left(\frac{a + b + c}{2} – c\right)}$

Squares on both sides:

– $\frac{(a + b + c)^2}{4} = k^2\left(\frac{a + b + c}{2}\right)^2\left(1 – \frac{2a}{a + b + c}\right)\left(1 – \frac{2b}{a + b + c}\right)\left(1 – \frac{2c}{a + b + c}\right)$

Simplified, becomes:

– $16k^2abc(a + b + c) = (a^2b^2 + b^2c^2 + c^2a^2) – (a^3b + b^3c + c^3a)$

This is a Heronian formula related to Heronian triangles, where all sides and area are integers.
For all Heronian triangles, above equation is satisfied for some $k$.

So, the problem becomes “find the sum of the perimeters of all Heronian triangles such that area/perimeter ratio is an integer from 2 to 1000 inclusive”.

A Heronian triangle has area and sides that are rational numbers which can be written as $n^2a, n^2b, n^2c$ and $n^3A$ for some Pythagorean triple $(a, b, c)$ and $n$. In this case, the triangle will have an area-to-perimeter ratio of $k/n$ for some integer $k$.

Let’s take a=3, b=4, c=5 since (3,4,5) is a primitive Pythagorean triple. Then if we take $n=1$ we get the triangle (3,4,5) with area 6 and perimeter 12, so $k=6/1=6$.
Now if we take $n=2$ we get the triangle (6,8,10) with area 24 and perimeter 24, so $k=24/2=12$.

For each primitive Pythagorean triple $(a, b, c)$ there will be a value of $k$ such that taking $n=k$ gives a triangle with the required property, and this gives $n \leq 1000$ (since $k$ is the same as our previously defined area/perimeter ratio which must be $\leq 1000$) . And conversely, as $k$ counts upwards through the integers, every number in [2, 1000] appears exactly once.

Therefore, we find all primitive Pythagorean triples, multiply the perimeters by each $k$ from 2 to 1000, and add these up.

Sadly, I can’t give a numerical answer because this is a complicated iterative process. It requires you to find all Pythagorean triples (up to reordering and multiplying by a constant), compute the perimeter, multiply this by $k$, and sum this for all $k$ from 2 to 1000. This involves a significant amount of computation!

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