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
Transformations
4 classical · 4 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 5021
Crux Mathematicorum Problem 5021, followed by a direct-similarity generalization proved with Pappus's theorem and a radical axis.
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.
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.