Problem 138: Special Isosceles Triangles

Mathematical Foundation

Theorem 1. The Pythagorean relation $L^2 = h^2 + (b/2)^2$ combined with $|b - h| = 1$ and $b$ even (say $b = 2m$) leads to the Pell equation $X^2 - 5L^2 = -1$, where $X = 5m \pm 2$.

Proof. The height bisects the base, forming right triangles with legs $h$ and $m = b/2$, and hypotenuse $L$:

For $b/2 = m$ to be an integer, $b$ must be even. (If $b$ were odd, $L^2 = h^2 + (b/2)^2$ would require $4L^2 = 4h^2 + b^2$; checking modular conditions shows $b$ must be even for integer solutions.) Write $b = 2m$, so $h = 2m \pm 1$.

Multiply by 5:

Setting $X = 5m \pm 2$:

$\square$

Theorem 2. The fundamental solution of $X^2 - 5Y^2 = -1$ is $(X, Y) = (2, 1)$. All positive solutions are given by $(X_n + Y_n\sqrt{5}) = (2 + \sqrt{5})^{2n-1}$ for $n = 1, 2, 3, \ldots$.

Proof. The continued fraction expansion of $\sqrt{5} = [2; \overline{4}]$ has period 1, which is odd, so $x^2 - 5y^2 = -1$ is solvable. The convergent $p_0/q_0 = 2/1$ gives $4 - 5 = -1$, confirming $(2, 1)$ as the fundamental solution. All solutions of $X^2 - 5Y^2 = -1$ are obtained by taking odd powers of the fundamental element $2 + \sqrt{5}$. Equivalently, using the fundamental solution $(9, 4)$ of $X^2 - 5Y^2 = 1$ (obtained as $(2 + \sqrt{5})^2$), successive solutions of the $-1$ equation satisfy:

$\square$

Lemma 1. The $Y$-components (i.e., the $L$-values) satisfy the linear recurrence $L_{n+1} = 18L_n - L_{n-1}$ with characteristic polynomial $\lambda^2 - 18\lambda + 1 = 0$.

Proof. The recurrence matrix for $(X, Y)$ has eigenvalues $9 \pm 4\sqrt{5}$. Since $Y_n$ is a linear combination of $(9 + 4\sqrt{5})^n$ and $(9 - 4\sqrt{5})^n$, it satisfies the recurrence with characteristic equation $\lambda^2 - 18\lambda + 1 = 0$ (trace = 18, determinant = 1 of the matrix $\begin{pmatrix}9 & 20 \\ 4 & 9\end{pmatrix}$). $\square$

Lemma 2. From each Pell solution $(X_n, L_n)$, the value $m = (X_n \mp 2)/5$ must be a positive integer. This requires $X_n \equiv \pm 2 \pmod{5}$.

Proof. Since $X = 5m \pm 2$, we need $m = (X \mp 2)/5 \in \mathbb{Z}_{>0}$. Checking: $X_1 = 2 \equiv 2 \pmod{5}$. By the recurrence, $X_{n+1} = 9X_n + 20L_n \equiv 4X_n \pmod{5}$. Starting from $X_1 \equiv 2$: $X_2 \equiv 3$, $X_3 \equiv 2$, $X_4 \equiv 3$, … So $X_n \equiv 2$ or $3 \pmod{5}$, i.e., $X_n \equiv \pm 2 \pmod{5}$ always. Every Pell solution yields a valid triangle. $\square$

Editorial

The L values satisfy the recurrence L_{n+1} = 18*L_n - L_{n-1} with L_1 = 17, L_2 = 305. Sum the first 12 values. We advance to next solution. We then check m = (X - 2)/5 or (X + 2)/5 is positive integer. Finally, by Lemma 2, one always works.

Pseudocode

count = 12
Fundamental solution of X^2 - 5L^2 = -1
Advance to next solution
Check m = (X - 2)/5 or (X + 2)/5 is positive integer
By Lemma 2, one always works
else

Complexity Analysis

• Time: $O(k)$ arithmetic operations for $k = 12$ terms. With big-integer arithmetic, bit complexity is $O(k^2)$ due to growing number sizes.
• Space: $O(k)$ bits to store the current values.

Answer

#include <bits/stdc++.h>
using namespace std;

int main() {
    // L values satisfy L_{n+1} = 18*L_n - L_{n-1}
    // L_1 = 17, L_2 = 305
    // Sum the first 12 L values.

    long long L_prev = 17;
    long long L_curr = 305;
    long long total = L_prev + L_curr;

    for (int i = 3; i <= 12; i++) {
        long long L_next = 18 * L_curr - L_prev;
        total += L_next;
        L_prev = L_curr;
        L_curr = L_next;
    }

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

"""
Problem 138: Special Isosceles Triangles

The L values satisfy the recurrence L_{n+1} = 18*L_n - L_{n-1}
with L_1 = 17, L_2 = 305.
Sum the first 12 values.
"""

def solve():
    L_prev, L_curr = 17, 305
    total = L_prev + L_curr
    for _ in range(10):  # 10 more to get 12 total
        L_prev, L_curr = L_curr, 18 * L_curr - L_prev
        total += L_curr
    return total

if __name__ == "__main__":
    answer = solve()
assert answer == 1118049290473932
print(answer)