# The Brocard Centres

## The Brocard Centres of a Triangle

There are six circles that pass through two vertices of a triangle and are tangent to one of its sides. The tangency must occur at one of the vertices passed through. These six circles have two triple points of intersection, known as the Brocard Points of the triangle. They are always inside the triangle and coincide only when the triangle is regular, in which case the six circles reduce to three.

M is the intersection of the circles centred at U, V, W (where the perpendiculars from A, B, C are drawn clockwise) and M* of those centred at U*, V*, W* (anticlockwise). These centres lie in pairs on the external parts of the side-bisectors. The join MM* is the Brocard Line of the triangle. The six circles form three crescents with cusps at the vertices of the triangle. In the isosceles case the crescent at the base reduces to one arc, and in the regular case the six reduce to three.

There is a unique angle w such that lines drawn through A, B, C at the angle w to AB, BC, CA concur. It follows that the similar lines through A, B, C at angle w to AC, BA, CB also concur. The points of concurrence are the Brocard Points, and the Brocard angle w is given by cotw = cota + cotb + cotg.

We also have AM.BM.CM = AM*.BM*.CM*.

MM* is perpendicular to OO* where O is the circumcentre and O* the Lemoine centre. M and M* lie on the circle on OO* as diameter (the Brocard circle) which also passes through the points of intersection BM.CM*, CM.AM*, AM.BM*, which we denote as A*, B*, C*.

What is the angle subtended by MM* at O (and B*, C*) is it w?

The midpoint of MM* is the incentre.

## The Triplicate Ratio Centre???

If through the Lemoine centre O* lines are drawn parallel to the sides of the triangle, then the six points in which they meet the sides lie on a circle (the triplicate ratio circle) whose centre is the midpoint of OO* (O being the circumcentre). This coincides with the centre of the Brocard circle. This circle cuts off from the sides of the triangle segments whose lengths are in the ratios a³:b³:c³. If through the Lemoine centre O* lines are drawn anti-parallel to the sides of the triangle the six points in which they meet the sides also lie on a circle (the cosine circle) whose centre is O*. This circle cuts off from the sides segments whose lengths are in the ratios cosa: cosb: cosg.

???? The points in the diagram do not appear to fit onto circles. Has this been miscopied from some source? Does the theorem refer to some other point (not the Lemoine)?

## Quasi-Brocard Centres?

There are six circles that touch two sides of the triangle 'internally' and pass through one vertex. Are there any results analogous to the Brocard case? 