A Solution to BMO1 2025, Problem 4
A synthetic solution to BMO1 2025, Problem 4 using tangent–chord angles, cyclic symmetry, and power of a point.
Topic
Triangle geometry
10 classical · 2 methods · 1 original problem · 6 solutions
A Solution to Crux Mathematicorum Problem 4941
Crux Mathematicorum Problem 4941, followed by an incircle-inversion proof using the medial triangle of the contact triangle.
A Solution to Crux Mathematicorum Problem 5031
Crux Mathematicorum Problem 5031, followed by an incircle-inversion proof using contact-chord midpoints and a cyclic quadrilateral.
A Solution to Crux Mathematicorum Problem 5088
Crux Mathematicorum Problem 5088, followed by an angle-chasing proof through tangency and cyclic quadrilaterals.
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.
A Solution to Crux Mathematicorum Problem 5109
Crux Mathematicorum Problem 5109, followed by a synthetic fixed-point argument and a projective generalization.
Crux Mathematicorum Problem 5145
An isogonal configuration leading to concurrency
A problem proposed by Chikara Tsugawa and published as Crux Mathematicorum Problem 5145: an isogonal configuration whose associated lines are concurrent or parallel.
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.
From Ceva to Kiepert and Jacobi
Side ratios, angle ratios, and external concurrence
Trigonometric Ceva, derived by the sine rule, proves Kiepert’s, Jacobi’s, and Kariya’s concurrence theorems, with the Gergonne point as Kariya’s boundary case.
Directed Angles
Angle arithmetic modulo a half-turn
A consistent convention for angle addition, cyclic quadrilaterals, and collinearity, illustrated by the perpendicular feet in Simson’s theorem.
The Euler Line from the Medial Triangle
A homothety centered at the centroid
The circumcenter of a triangle is the orthocenter of its medial triangle, and a centroid-centered homothety carries the circumcenter to the original orthocenter.
Four Orthocenters of a Cyclic Quadrilateral
A hidden half-turn
For a cyclic quadrilateral, the four orthocenters obtained by omitting one vertex at a time are the images of the original vertices under a single half-turn.
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.
The Nine-Point Circle from the Circumcircle
Three formulas revealing one homothety
In the unit-circle model, three formulas for the side midpoints, altitude feet, and vertex–orthocenter midpoints reveal a single homothety carrying the circumcircle to the nine-point circle.
Simson’s Theorem
Perpendicular feet and Carnot’s equal-angle extension
Two cyclic quadrilaterals align the perpendicular feet of a point on the circumcircle. The same angle argument gives Carnot’s extension to oblique projections.
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.
The Incenter through Trigonometric Ceva
Angle-bisector concurrence and isogonal conjugation
Trigonometric Ceva proves that the three internal angle bisectors concur and shows why isogonal reflection preserves concurrence.
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.