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Rational Angle Bisection Problem

Updated 7 January 2026
  • The Rational Angle Bisection Problem is the challenge of identifying pairs of lines with rational slopes whose angle bisectors are rational, employing Diophantine equations, Pell sequences, and Pythagorean triples for explicit parametrizations.
  • It links classical Euclidean geometry with number theory by reformulating the criterion into a negative Pell equation and utilizing symmetry transformations to organize the solution space.
  • Recent advancements extend the problem to higher dimensions while providing concrete examples and characterizations of rational triangle centers and bisector lengths in geometrical constructions.

The Rational Angle Bisection Problem concerns the identification and classification of all pairs of lines in the Euclidean plane whose slopes, together with the slopes of their angle bisectors, are all rational numbers. Its solution is fundamentally linked with the arithmetic of Diophantine equations, notably the negative Pell equation, and extends to deep connections with classical number theory, geometry (including centers of triangles), and higher-dimensional generalizations. The recent literature provides a comprehensive and explicit characterization of both integral and rational solutions, parametrized via Pell sequences and Pythagorean triples.

1. Problem Statement and Fundamental Equation

Given two lines through the origin in the plane with rational slopes a,bQa, b \in \mathbb{Q}, the Rational Angle Bisection Problem asks for all such pairs %%%%1%%%% for which at least one angle bisector also has rational slope. If cRc\in\mathbb{R} is a slope of an angle bisector of the angle between y=axy = a\,x and y=bxy = b\,x, it is classically known via distance-equality or tangent addition that

acb2+1=bca2+1.|a-c|\,\sqrt{b^2+1} = |b-c|\,\sqrt{a^2+1}.

Squaring and reorganizing yields the central Diophantine condition

$(a - c)^2\, (b^2 + 1) = (b - c)^2\, (a^2 + 1). \tag{%%%%5%%%%}$

The other bisector, perpendicular to the first, has slope c1-c - 1. Trivial solutions correspond to a=b|a| = |b| (the lines coincide or are perpendicular), which are excluded in classification of proper solutions (Hirotsu, 2022, 2305.01091).

2. Characterization of Rational and Integral Solutions

The complete set of solutions to ()(\star) is obtained by transforming the problem into the negative Pell equation. For nontrivial integer solutions, a,bZa, b \in \mathbb{Z}, one obtains

a2+1=dα2,b2+1=dβ2a^2 + 1 = d\,\alpha^2, \quad b^2 + 1 = d\,\beta^2

for a unique square-free d>1d > 1, with integers α,β>0\alpha, \beta > 0, reducing to solving

x2dy2=1,x^2 - d y^2 = -1,

for both (a,α)(a, \alpha) and (b,β)(b, \beta).

Parametrization via Pell Sequences

Let (fn,gn)(f_n, g_n) denote the nn-th positive solution to x2dy2=±1x^2 - d y^2 = \pm 1, ordered by increasing yy. For any odd m=2m01>0m = 2m_0 - 1 > 0 and n>0n > 0,

(a,b,c)=(fmn(d),fm(n+1)(d),gm(n+1)(d)/gm(d))(a, b, c) = (f_{m n}^{(d)},\, f_{m(n+1)}^{(d)},\, g_{m(n+1)}^{(d)}/g_{m}^{(d)})

produces an integral solution to ()(\star) (Hirotsu, 2022).

An exceptional family occurs for d=2d=2: (a,b,c)=(f2n1(2),f2n+1(2),g2n(2)).(a, b, c) = (f_{2n-1}^{(2)},\, -f_{2n+1}^{(2)},\, g_{2n}^{(2)}). The half-companion Pell numbers and Pell numbers satisfy

fn+1=fn+dgn,gn+1=fn+gn,fn2dgn2=(1)n,f_{n+1} = f_n + d g_n, \quad g_{n+1} = f_n + g_n, \quad f_n^2 - d\,g_n^2 = (-1)^n,

with f0=1,g0=0;f1=1,g1=1f_0 = 1, g_0 = 0; f_1 = 1, g_1 = 1.

Rational Solutions Via Pythagorean Triples

Purely rational solutions (not obtainable via integral Pell solutions) are parametrized by ordered pairs of primitive Pythagorean triples sharing a leg: (u,v,w),(u,v,w),(u, v, w), \qquad (u', v, w'), with u2v2=w2u^2 - v^2 = w^2, (u)2v2=(w)2(u')^2 - v^2 = (w')^2 and ww fixed. Setting a=u/w,b=u/wa = u/w,\, b = u'/w, one obtains

c=v+wuw±vw×w,c = \frac{v + w}{u'w \pm v w} \times w,

leading, after simplification, to explicit rational values for cc (Hirotsu, 2022).

Parametric Description Table

Solution Type Parametrization Constraints
Integral (Pell, d>1d>1) fn,gnf_n, g_n sequence as above a2dα2=1a^2-d\alpha^2=-1
Rational (Pythagoras) Shared leg vv with triples (u,v,w),(u,v,w)(u, v, w), (u', v, w') u2v2=w2u^2-v^2=w^2, u2v2=(w)2u'^2-v^2=(w')^2

3. Group-Theoretic and Symmetric Structure

The solution set to ()(\star) exhibits involutive symmetries:

  • (a,b,c)(a,b,c1)(a, b, c) \mapsto (a, b, -c-1) swaps the two bisectors
  • (a,b,c)(a,b,c)(a, b, c) \mapsto (-a, -b, -c) reverses orientation
  • (a,b,c)(b,a,c)(a, b, c) \mapsto (b, a, c) swaps labels

These symmetries, together with sign conventions, imply that families constructed from the Pell and Pythagorean parametrizations already describe all possible solutions modulo trivialities. No nontrivial integer solution admits zero slope for aa or bb (Hirotsu, 2022).

4. Higher-Dimensional Generalization

The problem admits extension to knk^n for kk a subfield of R\mathbb{R}. For direction vectors a,bkn\mathbf{a}, \mathbf{b} \in k^n, existence of a rational-direction bisector ckn\mathbf{c} \in k^n is equivalent to

a2b2(modk×2).\|\mathbf{a}\|^2 \equiv \|\mathbf{b}\|^2 \pmod{k^{\times 2}}.

For integral or algebraic-integer vectors, this is further characterized by existence of nontrivial integral solutions to

x12++xn2=dxn+12x_1^2 + \cdots + x_n^2 = d x_{n+1}^2

with prescribed congruence conditions, where dd is a square-free positive integer (Hirotsu, 31 Dec 2025).

5. Worked Examples and Illustrations

A canonical integral example is for d=2d = 2, whose Pell solutions are (f1,g1)=(1,1)(f_1, g_1) = (1,1), (f2,g2)=(3,2)(f_2, g_2) = (3,2), (f3,g3)=(7,5)(f_3, g_3) = (7,5), etc.

  • With (a,b,c)=(f1,f3,g3/g1)=(1,7,5)(a, b, c) = (f_1, f_3, g_3/g_1) = (1, 7, 5), both a,b,ca, b, c are integral and satisfy ()(\star).
  • Rational (non-integral) solution: by employing Pythagorean triples with common leg v=12v=12 (e.g., (5,12,13)(5,12,13) and (35,12,37)(35,12,37)), one finds (a,b,c)=(5/13,35/13,25/16)(a, b, c) = (5/13,\, 35/13,\, 25/16) (Hirotsu, 2022).

6. Connections to Triangle Centers and Geometric Constructions

An important geometric implication is the link to the rationality of triangle centers. For a triangle with side lengths a,b,ca, b, c defined over kk, the incenter is kk-rational if and only if all a2,b2,c2a^2, b^2, c^2 represent the same class in k×/k×2k^\times / k^{\times 2}. In the plane over Q\mathbb{Q}, this provides a direct criterion for when triangles with rational vertices and incenters exist. The same square-class condition ensures the rationality of excenters, the Gergonne and Nagel points, and their isotomic and isogonal conjugates. Every such triangle is homothetic to a Heronian triangle (i.e., with rational side lengths and area), up to the appropriate scaling (Hirotsu, 31 Dec 2025).

7. Specialized Case: Triangles with Bisector Lengths

In the special instance where a triangle ABCABC has side lengths a,b,ca, b, c with angle B=2AB = 2A, integrality of all elements and the bisector length constrains the triple (a,b,c)(a,b,c) to solutions of b2=a(a+c)b^2 = a(a + c), together with explicit parameterizations: a=wvu2,b=wuv2,c=wv(v2u2),d=wu(v2u2)a = w v u^2,\, b = w u v^2,\, c = w v (v^2 - u^2),\, d = w u (v^2 - u^2) for (u,v,w)(u, v, w) positive integers with gcd(u,v)=1\gcd(u,v) = 1 and strict inequalities u2<v2<2u2u^2 < v^2 < 2u^2 or 2u2<v2<4u22u^2 < v^2 < 4u^2 (Zelator, 2012).


The Rational Angle Bisection Problem is thus fundamentally resolved by explicit parametrizations in terms of negative Pell equations for integral solutions and by Pythagorean configurations for rational ones, with algebraic and geometric structure arising from the interplay between number theory and classical Euclidean geometry. The problem has been extended to arbitrary dimensions with analogous arithmetic classifications, yielding a complete description of possible rationality patterns for bisectors and geometric centers (Hirotsu, 2022, 2305.01091, Hirotsu, 31 Dec 2025, Zelator, 2012).

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