A Solution to Crux Mathematicorum Problem 5098
Crux Mathematicorum Problem 5098, followed by a unit-circle complex proof that reduces two geometric conditions to the same algebraic equation.
Topic
Algebra
5 classical · 1 method · 1 solution
Barycentric Coordinates
Points, lines, and circumcircle tangents
A method note on homogeneous barycentric coordinates, the circumcircle equation, and tangent lines obtained from their linear terms.
Napoleon and Van Aubel through Complex Numbers
Two regular side constructions encoded by fixed rotations
Complex multiplication encodes the external equilateral triangles of Napoleon’s theorem and the external squares of Van Aubel’s theorem, reducing both conclusions to rotation identities.
Radical Axes and Orthogonal Parabolas
Two short proofs from quadratic equations
Subtracting circle equations produces the radical axes; adding normalized equations of two parabolas with perpendicular axes produces a circle through their four intersections.
Tangents, Angle Bisectors, and a Common Point
A barycentric route to X(55)
A barycentric calculation locates the common point of three lines joining tangent intersections to angle-bisector traces, then identifies it as the isogonal conjugate of the Gergonne point.
The Centroid through Ceva and Vectors
Two proofs and the further reach of each method
Ceva’s theorem proves the concurrence of the medians and leads naturally to isotomic conjugation. Vectors locate the centroid and reveal why another triangle can share it.
Two Views of the Orthocenter
Cyclic quadrilaterals, vectors, and the Euler line
Two cyclic quadrilaterals explain why the third altitude is forced through the intersection of the first two, while a circumcenter-based vector formula reveals the Euler line.