$60$-degree Triangle Inscribed Circles

Let’s call an integer sided triangle with exactly one angle of $60$ degrees a $60$-degree triangle.
Let $r$ be the radius of the inscribed circle of such a $60$-degree triangle.
There are $1234$ $60$-degree triangles for which $r \le 100$.
Let $T(n)$ be the number of $60$-degree triangles for which $r \le n$, so
$T(100) = 1234$, $T(1000) = 22767$, and $T(10000) = 359912$.
Find $T(1053779)$.

This problem is dealing with Heronian triangles which are integer-sided triangles with integer area. However since the angles of such triangle are specific that it is 60 degrees, the triangle can be given in a simpler form in relation to equilateral triangles.

We start with an equilateral triangle of side length $n$. Notice that among the three ways to partition this into two triangles using a line from one vertex to the other side, there will be a way such that the partitioned triangles have all integer sides, namely a 60-degree triangle.

This triangle will have a longest side length of $n$ and a shorter side length of $\frac{n\sqrt(3)}{2} = \frac{n\sqrt(3)}{2} \cdot \frac{2\sqrt(3)}{2\sqrt(3)} = \frac{n\sqrt(3)^2}{4} = \frac{n*3}{4} = \frac{3n}{4}$, which is an integer when $n$ is a multiple of 4.

The radius of the in-circle is $\frac{A}{s}$ where $A$ is the area and $s$ is the semi-perimeter. For a 60 degree triangle with longest side $n$ and shorter side $\frac{3n}{4}$, the area is $\frac{1}{2} * base * height = \frac{1}{2} * n * \frac{\sqrt(3)}{2} * \frac{3n}{4} = \frac{3n^2\sqrt(3)}{16}$ and the semi-perimeter is $\frac{1}{2} * (n + n + \frac{3n}{4}) = \frac{5n}{4}$.

Dividing the area by the semi-perimeter gives the radius of the in-circle (r), which equals $\frac{3n\sqrt(3)}{20}$. This is an integer when $n$ is a multiple of 20.

Since $n$ must be a multiple of both 4 and 20 for the two conditions to be met, it has to be a multiple of the least common multiple of 4 and 20, which is 20. Therefore for any such length of n, there exists one and only one 60 degree triangle meeting the two conditions.

To find $T(r)$ where r is the radius of the in-circle, we can find the largest $n$ so that it satisfies $r (radius) \ge \frac{3n\sqrt(3)}{20}$ which results to $n \le \frac{20r}{3\sqrt(3)}$.

Due to the condition that n has to be multiple of 20, we get $n \le 20 * floor(\frac{20r}{60}) = 20 * floor(\frac{r}{3})$ where floor function rounds down to the nearest integer. The total number of such possible n’s (and hence, such 60 degree triangles) is $T(r) = floor(\frac{r}{3})$ because for every 20 units increment in n, a new Heronian 60-degree triangle forms.

Plug in $r = 1053779$ to get $T(1053779) = floor(\frac{1053779}{3}) = 351259$.

So, the final answer is 351259.

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