Detailed mathematical approach

Problem Summary

For an integer-sided triangle \(\Delta\), let \(s(\Delta)\) be the side length of the smallest square that can contain \(\Delta\) after the triangle is rotated and translated in the plane. Project Euler 998 asks for the sum of the perimeters of all distinct integer-sided triangles whose minimum enclosing square has side length at most \(10^6\). The examples \(T(40)=346\), \(T(400)=76402\), and \(T(2000)=3237036\) are used as checkpoints.

Mathematical Approach and Geometric Model

Fix a square of side length \(L\), and rotate the whole configuration so the square is axis-aligned. A triangle that is tight for this square has vertices on the boundary. If one of its sides crosses the square with horizontal or vertical span \(L\), that side is the hypotenuse of a right triangle whose legs are \(L\) and some offset \(p\). Therefore its length is integral exactly when

\[ h^2=L^2+p^2. \]

This is the first major reduction: instead of searching arbitrary triangles, we enumerate Pythagorean segments attached to each possible square side \(L\). For every \(L\), the program stores pairs \((p,h)\), where \(0\le p\le L\) and \(h=\sqrt{L^2+p^2}\) is an integer. The restriction \(p\le L\) loses nothing because the complementary orientation is symmetric.

Formal Derivation and Normal Form

The enumeration rests on a normal-form statement. Let \(Q\) be a minimum square for a triangle. After rotating the plane, \(Q=[0,L]\times[0,L]\). A non-degenerate minimum cannot have all vertices strictly inside a smaller homothetic square, so at least two opposite supporting lines or two adjacent supporting lines are active. For a triangle, the supporting vertices can change only when a side direction becomes parallel to a square side. Thus every critical placement can be represented by boundary chords whose coordinate spans are determined by the side length \(L\) and by one boundary offset.

Lemma 1. If a side of a critical placement connects two boundary lines separated by \(L\), then its integral length is equivalent to a Pythagorean condition \(h^2=L^2+p^2\). This follows immediately from orthogonal projection: the side vector has components \(L\) and \(p\), up to sign. Conversely, every such Pythagorean segment can be placed on the square boundary and is therefore a legitimate building block for a candidate triangle.

Lemma 2. Pairing two boundary segments of the same square side \(L\) leaves only two combinatorial closures. Either the remaining side lies on a square boundary and has length \(p+q\), or it joins the two opposite residual endpoints and has squared length \((L-p)^2+(L-q)^2\). These are exactly the two families tested in the program. No third closure exists because the three vertices of the triangle occupy three chosen boundary endpoints, and after the first two side chords are fixed the last side is determined by which pair of residual endpoints is joined.

The inequality \(pq\ge L(L-p-q)\) is the algebraic form of the first closure's inside-square condition. It is obtained by writing the two boundary chords in coordinates, intersecting the two supporting rays from the boundary endpoints, and requiring the intersection coordinate to remain inside the square. This converts the geometric feasibility statement into one integer comparison, avoiding any floating-point decision in the first family.

Pythagorean Enumeration

The offsets are generated with Euclid's formula. Primitive triples are

\[ a=m^2-n^2,\qquad b=2mn,\qquad c=m^2+n^2, \]

with \(\gcd(m,n)=1\) and \(m-n\) odd. Each primitive triple is then scaled by \(k\). Either leg may play the role of the square side \(L\), and the other leg becomes the offset. Hence the code inserts both \((L, p, h)=(ka,kb,kc)\) and \((kb,ka,kc)\) whenever the offset is not larger than the side. Since the relevant leg is at most \(L\le 10^6\), the Euclid parameters only need to be tested up to \(\sqrt{2L}+2\). Duplicates are removed after sorting each list for a fixed \(L\).

First Family of Triangles

Choose two Pythagorean boundary segments for the same square side \(L\): \((p,h_p)\) and \((q,h_q)\). The first family is obtained when the remaining side of the triangle lies along one side of the square. Its length is

\[ b=p+q. \]

For the endpoints to lie on the boundary segment, we require \(b\le L\). The third vertex must also be on the correct side of the square rather than outside the \(L\times L\) box. In the boundary coordinates used by the implementation, this feasibility condition simplifies to

\[ pq\ge L(L-b). \]

When both inequalities hold, \((b,h_p,h_q)\) is a valid integer-sided triangle whose enclosing square has side length \(L\). Its perimeter is added after sorting the three side lengths into a canonical key.

Second Family of Triangles

The same two Pythagorean segments can also close through the opposite pair of endpoints. The remaining displacement has legs \(L-p\) and \(L-q\), so the third side is integral exactly when

\[ r^2=(L-p)^2+(L-q)^2. \]

This gives a candidate triangle \((h_p,h_q,r)\). Unlike the first family, the construction alone does not prove that \(L\) is the minimum enclosing square side. The same side triple might fit into a smaller square after a different rotation. Therefore every candidate from this family is passed to an independent minimum-square verifier, and it is accepted only when the verified value is at least \(L\), up to a small floating-point tolerance.

Minimum Enclosing Square Verification

For a triangle with side lengths \(a\le b\le c\), place the side \(c\) on the \(x\)-axis and compute the third vertex by the law of cosines:

\[ x=\frac{a^2+c^2-b^2}{2c},\qquad y=\sqrt{a^2-x^2}. \]

For a rotation angle \(\theta\), project the three vertices onto the rotated axes \(u\) and \(v\). The side length of the smallest axis-aligned square at that angle is

\[ w(\theta)=\max\{\max u-\min u,\ \max v-\min v\}. \]

The vertices that realize the maxima and minima change only when an edge becomes parallel to one of the square axes. Thus the interval \([0,\pi/2)\) is split at the edge directions. Inside each interval, both widths are fixed sinusoidal expressions. The optimum is found either at a boundary or at an interior angle where the two widths are equal. The routine minimum_square evaluates exactly these candidates, which is the standard rotating-calipers idea specialized to three points.

Deduplication and Summation

A triangle may be produced by several square sides, by both endpoint orderings, or by both geometric families. The problem asks for distinct triangles, so the program sorts every side triple and packs it into one integer key. The side lengths are below \(2^{21}\), so the key

\[ (a\ll 42)\,|\,(b\ll 21)\,|\,c \]

is collision-free for the target range. The perimeter is added only the first time a key appears.

Correctness Argument

Soundness. Every accepted triangle is integer-sided by construction. In the first family, the side lengths are \(p+q\), \(h_p\), and \(h_q\), with the Pythagorean equations ensuring integral hypotenuses and the feasibility inequality ensuring a placement inside the square. In the second family, the extra square test ensures that \(r\) is integral, and minimum_square verifies that the side triple does not belong to a smaller enclosing square. Therefore every inserted triangle satisfies \(s(\Delta)\le L\le n\).

Completeness. Take any integer-sided triangle with \(s(\Delta)\le n\), and place it in a minimum square of side \(L=s(\Delta)\). By the normal-form discussion, at least two of the tight side chords are Pythagorean boundary segments for the same \(L\). Their remaining endpoints close in one of the two possible combinatorial ways described in Lemma 2. Hence the triangle is generated by the corresponding iteration of the algorithm. If it appears through several placements, the canonical key keeps only one copy.

Numerical robustness. The enumeration, square tests, perimeter sums, and deduplication keys are exact integer operations. Floating point is used only in the secondary verifier for the second family. That verifier evaluates all critical angles of a three-point set and compares against \(L\) with a small tolerance; the published checkpoints provide an end-to-end guard against both missed candidates and false positives.

How the Code Works

pythagorean_offsets builds the list of all admissible Pythagorean boundary segments for every \(L\le n\). The main solve loop then considers all unordered pairs of segments attached to the same \(L\). It tests the first family with the algebraic inequality above, tests the second family with a square check and minimum_square, and inserts every accepted side triple into the global set. The three published checkpoints are asserted before the target value is printed.

Complexity Analysis

Let \(d_L\) be the number of Pythagorean offsets stored for side \(L\). The running time after triple generation is

\[ O\!\left(\sum_{L\le n} d_L^2\right), \]

because every unordered pair of offsets for the same \(L\) is tested once. The memory usage is \(O(\sum d_L+u)\), where \(u\) is the number of distinct accepted triangles. In practice the lists \(d_L\) are sparse, so \(n=10^6\) is feasible in compiled code.

References

• Problem page: Project Euler 998
• Wolfram MathWorld: Pythagorean Triple for Euclid's formula and primitive triples.
• Toussaint, Solving Geometric Problems with the Rotating Calipers for the support-line method used by the verifier.
• Wolfram MathWorld: Law of Cosines for reconstructing a triangle from its side lengths.