# Brocard Point and a Relation of Circumradii

### Proof 1

Let $\omega\,$ denote the Brocard angle. Then

$\displaystyle\frac{c}{2\sin(\pi-\omega-(B-\omega))}=\frac{2R\sin C}{2\sin B}.$

Two more relations are obtained in the same manner. The product of the three solves the problem:

$\displaystyle \prod_{cycl}R_a=\prod_{cycl}\frac{R\sin C}{\sin B}=R^3.$

### Proof 2

WLOG, assume $A=(-2u, 0),\,$ $B=(2v, 0),\,$ $C=(0,2).\,$ It's well known that $R^2=(1+u^2)(1+v^2).\,$ Let's find $R_a\,$ first. The midpoint of $AB\,$ is $(v-u,0);\,$ the perpendicular at $B\,$ to $BC\,$ has the equation $vx-y=2v^2,\,$ so that the center of $(\Omega BC)\,$ is at $(v-u, v(u+v)),\,$ implying $R_a^2=(u+v)^2(1+v^2).\,$ Similarly, $\displaystyle R_b^2=\frac{(1+u^2)(1+v^2)^2}{(u+v)^2}\,$ and $R_c=(1+u^2)^2.\,$ Clearly, $R_aR_bR_c=R^3.$

### Proof 3

Clearly, $\angle B\Omega C=180^{\circ}-C,\,$ $\angle C\Omega A=180^{\circ}-A,\,$ $\angle A\Omega B=180^{\circ}-B.\,$ Using the Law of Sines in triangles in $\Omega BC,\,$ $\Omega CA,\,$ $\Omega AB,\,$ we get

$\displaystyle\frac{a}{\sin (180^{\circ}-C)}=2R_a,\\ \displaystyle\frac{b}{\sin (180^{\circ}-A)}=2R_b,\\ \displaystyle\frac{c}{\sin (180^{\circ}-B)}=2R_c,\,$

so that

$\displaystyle\frac{a}{\sin C}=2R_a,\,$ $\displaystyle\frac{b}{\sin A}=2R_b,\,$ $\displaystyle\frac{c}{\sin B}=2R_c.$

In $\Delta ABC,\,$ $\displaystyle R=\frac{a}{2\sin A}=\frac{b}{2\sin B}=\frac{c}{2\sin C}.\,$

Combining the finds,

$\displaystyle R_aR_bR_c=\frac{a}{2\sin A}\frac{b}{2\sin B}\frac{c}{2\sin C}=R^3,$

as desired.

### Acknowledgment

The problem has been kindly posted at the CutTheKnotMath facebook page Mehmet Sahin (Turkey). Proof 1 is by Dan Sitaru; Proof 2 is by Leo Giugiuc; Proof 3 is by Mehmet Sahin.