Project Euler

Integer Sided Triangles for which Area/Perimeter Ratio is Integral

Find the sum of perimeters of all integer-sided triangles for which A/P = k is a positive integer with 1 <= k <= 1000, where A is the area and P is the perimeter.

Source sync Apr 19, 2026

Problem #0283

Level Level 18

Solved By 750

Languages C++, Python

Answer 28038042525570324

Length 307 words

geometrymodular_arithmeticanalytic_math

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 $1000$.

Problem 283: Integer Sided Triangles for which Area/Perimeter Ratio is Integral

Mathematical Foundation

Theorem (Reduction to Diophantine Equation). Let $a, b, c$ be the integer side lengths of a triangle with semi-perimeter $s = (a+b+c)/2$ . If $A/P = k$ for a positive integer $k$ , then defining $x = s-a$ , $y = s-b$ , $z = s-c$ with $0 < x \le y \le z$ , we have

xyz = 4k^2(x + y + z),

the sides are $a = y+z$ , $b = x+z$ , $c = x+y$ , the perimeter is $P = 2(x+y+z)$ , and $x, y, z$ are either all integers (even perimeter) or all half-integers (odd perimeter).

Proof. By Heron’s formula, $A = \sqrt{s(s-a)(s-b)(s-c)}$ . The condition $A = kP = 2ks$ gives $A^2 = 4k^2 s^2$ , so $s(s-a)(s-b)(s-c) = 4k^2 s^2$ , hence $(s-a)(s-b)(s-c) = 4k^2 s$ . Substituting $x = s-a$ , $y = s-b$ , $z = s-c$ yields $xyz = 4k^2(x+y+z)$ .

Since $a, b, c$ are positive integers, $2s = a+b+c$ is a positive integer. If $2s$ is even, then $s$ is an integer and $x, y, z$ are positive integers. If $2s$ is odd, then $s$ is a half-integer and $x, y, z$ are positive half-integers.

The triangle inequality is automatically satisfied: $a + b - c = 2z > 0$ , etc. The ordering $x \le y \le z$ corresponds to $a \ge b \ge c$ (up to reindexing). $\quad\square$

Lemma (Enumeration Bounds). In the equation $xyz = 4k^2(x+y+z)$ with $0 < x \le y \le z$ :

$x \le 2k\sqrt{3}$ ,
$y \le \frac{4k^2(x+y)}{xy - 4k^2}$ requires $xy > 4k^2$ ,
$z = \frac{4k^2(x+y)}{xy - 4k^2}$ .

Proof. From $x \le y \le z$ , we get $x^3 \le xyz = 4k^2(x+y+z) \le 4k^2 \cdot 3z$ , and also $z \le xyz/x^2 = 4k^2(x+y+z)/x^2$ . More directly, since $z \ge y \ge x$ , we have $xyz \ge x^3$ and $x+y+z \le 3z$ , giving $x^2 y \le 12k^2$ , so $x^3 \le 12k^2$ , hence $x \le (12k^2)^{1/3} \le 2k\sqrt{3}$ .

For $z$ , solving $xyz = 4k^2(x+y+z)$ gives $z(xy - 4k^2) = 4k^2(x+y)$ , requiring $xy > 4k^2$ (otherwise $z \le 0$ or undefined), and $z = \frac{4k^2(x+y)}{xy - 4k^2}$ . The condition $z \ge y$ then constrains $y$ . $\quad\square$

Lemma (Odd Perimeter Case). When $x, y, z$ are half-integers, setting $X = 2x$ , $Y = 2y$ , $Z = 2z$ (odd positive integers with $X \le Y \le Z$ ) yields

XYZ = 16k^2(X + Y + Z), \quad Z = \frac{16k^2(X+Y)}{XY - 16k^2}.

Proof. Direct substitution into $xyz = 4k^2(x+y+z)$ with $x = X/2$ , etc. $\quad\square$

Editorial

For triangle with integer sides a,b,c, let s=(a+b+c)/2, A=sqrt(s(s-a)(s-b)(s-c)). Need A/(2s) = k (positive integer, 1 <= k <= 1000). Substituting x=s-a, y=s-b, z=s-c (with 0<x<=y<=z): xyz = 4k^2(x+y+z), P = 2(x+y+z) Case 1 (even P): x,y,z positive integers. z = 4k^2(x+y) / (xy - 4k^2) Case 2 (odd P): x=X/2, y=Y/2, z=Z/2, X,Y,Z odd positive integers. XYZ = 16k^2(X+Y+Z), Z = 16k^2(X+Y)/(XY-16k^2) Sum all perimeters of valid triangles. We case 1: even perimeter (x, y, z positive integers). Finally, case 2: odd perimeter (X, Y, Z odd positive integers).

Pseudocode

Case 1: even perimeter (x, y, z positive integers)
Case 2: odd perimeter (X, Y, Z odd positive integers)

Complexity Analysis

Time: $O(K^2 \log K)$ where $K = 1000$ . For each $k$ , the inner loops run $O(k \log k)$ iterations on average (bounded by the divisor structure).
Space: $O(1)$ .

Answer

\boxed{28038042525570324}

C++ project_euler/problem_283/solution.cpp

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef __int128 lll;

int main() {
    ll total = 0;

    for (ll k = 1; k <= 1000; k++) {
        ll k2 = k * k;

        // Case 1: even perimeter, x,y,z positive integers
        // xyz = 4k^2(x+y+z), x <= y <= z
        // z = 4k^2(x+y) / (xy - 4k^2)
        ll c = 4 * k2;
        ll x_max = (ll)sqrt(3.0 * c) + 2;
        for (ll x = 1; x <= x_max; x++) {
            ll y_start = max(x, c / x + 1);
            // y upper bound from z >= y
            // y <= (c + sqrt(c^2 + c*x^2)) / x
            double disc = (double)c * ((double)c + (double)x * x);
            ll y_end = (ll)((c + sqrt(disc)) / x) + 2;

            for (ll y = y_start; y <= y_end; y++) {
                ll d = x * y - c;
                if (d <= 0) continue;
                ll n = c * (x + y);
                if (n % d != 0) continue;
                ll z = n / d;
                if (z < y) break;
                total += 2 * (x + y + z);
            }
        }

        // Case 2: odd perimeter, X,Y,Z odd positive integers
        // XYZ = 16k^2(X+Y+Z), X <= Y <= Z
        // Z = 16k^2(X+Y) / (XY - 16k^2)
        ll c2 = 16 * k2;
        ll X_max = (ll)sqrt(3.0 * c2) + 2;
        for (ll X = 1; X <= X_max; X += 2) {
            ll Y_start = max(X, c2 / X + 1);
            if (Y_start % 2 == 0) Y_start++;

            double disc2 = (double)c2 * ((double)c2 + (double)X * X);
            ll Y_end = (ll)((c2 + sqrt(disc2)) / X) + 2;

            for (ll Y = Y_start; Y <= Y_end; Y += 2) {
                ll d = X * Y - c2;
                if (d <= 0) continue;
                ll n = c2 * (X + Y);
                if (n % d != 0) continue;
                ll Z = n / d;
                if (Z < Y) break;
                if (Z % 2 == 0) continue;
                total += X + Y + Z;  // P = X+Y+Z
            }
        }
    }

    cout << total << endl;
    return 0;
}

Python project_euler/problem_283/solution.py

"""
Project Euler Problem 283: Integer Sided Triangles for which Area/Perimeter is Integral

For triangle with integer sides a,b,c, let s=(a+b+c)/2, A=sqrt(s(s-a)(s-b)(s-c)).
Need A/(2s) = k (positive integer, 1 <= k <= 1000).

Substituting x=s-a, y=s-b, z=s-c (with 0<x<=y<=z):
  xyz = 4k^2(x+y+z), P = 2(x+y+z)

Case 1 (even P): x,y,z positive integers.
  z = 4k^2(x+y) / (xy - 4k^2)

Case 2 (odd P): x=X/2, y=Y/2, z=Z/2, X,Y,Z odd positive integers.
  XYZ = 16k^2(X+Y+Z), Z = 16k^2(X+Y)/(XY-16k^2)

Sum all perimeters of valid triangles.
"""

from math import isqrt, sqrt

def solve():
    total = 0

    for k in range(1, 1001):
        k2 = k * k

        # Case 1: even perimeter
        c = 4 * k2
        x_max = int(sqrt(3.0 * c)) + 2
        for x in range(1, x_max + 1):
            y_start = max(x, c // x + 1)
            disc = c * (c + x * x)
            y_end = int((c + sqrt(disc)) / x) + 2

            for y in range(y_start, y_end + 1):
                d = x * y - c
                if d <= 0:
                    continue
                n = c * (x + y)
                if n % d != 0:
                    continue
                z = n // d
                if z < y:
                    break
                total += 2 * (x + y + z)

        # Case 2: odd perimeter
        c2 = 16 * k2
        X_max = int(sqrt(3.0 * c2)) + 2
        for X in range(1, X_max + 1, 2):
            Y_start = max(X, c2 // X + 1)
            if Y_start % 2 == 0:
                Y_start += 1
            disc2 = c2 * (c2 + X * X)
            Y_end = int((c2 + sqrt(disc2)) / X) + 2

            for Y in range(Y_start, Y_end + 1, 2):
                d = X * Y - c2
                if d <= 0:
                    continue
                n = c2 * (X + Y)
                if n % d != 0:
                    continue
                Z = n // d
                if Z < Y:
                    break
                if Z % 2 == 0:
                    continue
                total += X + Y + Z

    print(total)

solve()