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
Circles
6 classical · 2 methods · 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 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.
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.
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.
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.
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.
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.
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.
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.
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.