Counting Isosceles Acute Triangles in a Regular n‑gon – Analysis and Java Solution

The problem asks for the number of isosceles acute‑angled triangles that can be formed using vertices of a regular n‑gon, where two triangles are considered different if any vertex differs. The input range is 3 ≤ n ≤ 10^7.

Analysis – even n : By arranging the polygon on a circle, each vertex can form a certain number of acute‑angled isosceles triangles. The count reduces to the number of horizontal lines below the right angle, which is ⌊(n‑2)/4⌋ per vertex. Multiplying by n gives total = n * ((n-2)/4) .

Analysis – odd n : For odd n the same reasoning leads to ⌈(n‑1)/2⌉ lines that can serve as bases, yielding total = n * (((n-1)/2 + 1)/2) .

Duplicate triangles : When n is a multiple of 3, some counted triangles are equilateral and thus counted three times. To remove duplicates we subtract (n/3) * 2 from the total.

Final formula (before duplicate correction):

if (n % 2 == 0) {
    total = n * ((n - 2) / 4);
} else {
    total = n * (((n - 1) / 2 + 1) / 2);
}
if (n % 3 == 0) {
    total -= (n / 3) * 2;
}

Implementation pitfalls : Using int for intermediate calculations overflows for n up to 10^7, so long must be used and the cast applied before the multiplication. Also, integer division truncates, so explicit casting to long (or double when needed) is required.

Correct Java code:

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    long total;
    if (n % 2 == 0) {
        total = (long) n * ((n - 2) / 4);
    } else {
        total = (long) n * (((n - 1) / 2 + 1) / 2);
    }
    if (n % 3 == 0) {
        total -= (n / 3) * 2;
    }
    System.out.println(total);
}

The article concludes that such combinatorial geometry problems appear occasionally in coding interviews, and accumulating experience with them is valuable.