Problem Analysis
An almost equilateral triangle has two equal sides of length a and a third side that differs from them by exactly one unit, so the base is either a-1 or a+1. The challenge is to require that both the side lengths and the area be integers. The natural tool is Heron's formula. For an isosceles triangle with equal sides a and base b, dropping a perpendicular from the apex to the midpoint of the base splits the figure into two congruent right triangles, and the area becomes A = (b/2) * sqrt(a^2 - (b/2)^2). For A to be a positive integer, the quantity a^2 - (b/2)^2 under the square root must itself be a perfect square. That is the integrality condition that drives everything.
Substituting b = a+1 (and noting a must be odd so that b/2 stays rational), then clearing denominators, the perfect-square requirement collapses into a Pell-like equation - a Diophantine equation of the form x^2 - D*y^2 = N for small integer constants D and N, here built around D = 3. The case b = a-1 yields a sibling equation. The decisive fact about Pell equations is that their solutions, though infinite in number, are not scattered at random: they obey a linear recurrence relation. Each valid triangle can therefore be generated from the previous one by a fixed arithmetic update, instead of testing every integer side length up to the bound. Because the valid side lengths grow geometrically (roughly by a factor of 4 per step), fewer than twenty almost-equilateral triangles exist below a perimeter of one billion. The algorithm runs in O(log N) time in the perimeter bound N - effectively instantaneous.
Solution Approach
Recurrence-Based Pell Equation Enumeration (Python and C)
The Python and C solutions implement the same algorithm and differ only in language syntax. The state is five variables: s (the running sum of perimeters), s1 and s2 (two consecutive terms of the recurrence for the equal side length, both initialised to 1), m (a sign alternator initialised to +1), and p (the current candidate perimeter, initialised to 0).
The loop runs while p does not exceed the bound. Each iteration first folds the current perimeter into the accumulator (s + p), then advances the recurrence. The next side length is computed as 4*s2 - s1 + 2*m. The factor of 4 is the multiplier coming from the fundamental solution of the underlying Pell equation x^2 - 3*y^2 = 1; the +2*m term is a small correction whose sign alternates between the two families of triangles (base a+1 versus a-1). The sign alternator then flips with m = -m, and the new perimeter follows directly as p = 3*s2 - m. When the base is a+1 we have m = -1 so p = 3a+1; when it is a-1 we have m = +1 so p = 3a-1. A single expression 3*s2 - m captures both cases.
The two families are interleaved into one monotonically increasing sequence. The loop therefore terminates the instant a generated perimeter exceeds the bound, with no further searching. Because each new term is roughly four times the previous, the loop executes only a logarithmic number of times.
The Python version updates all four state variables in one parallel-assignment statement, exploiting the fact that the entire right-hand side is evaluated before any name is rebound. The C version computes the two dependent values (new_s2 and new_m) into temporaries before writing back, which is the idiomatic way to express the same simultaneous update in a language without parallel assignment. The implementations are logically identical, and both complete in unmeasurable time (recorded as 0.0000 s).