Question

The distance between the circumcenter and the orthocenter of triangle $ABC$ is

• A. $R\sqrt{1-8\cos A\cos B\cos C}$
• B. $2R\sqrt{1-4\cos A\cos B\cos C}$
• C. $R\sqrt{1-8\sin A\sin B\sin C}$
• D. $2R\sqrt{1-4\sin A\sin B\sin C}$

Answer 1

The correct option is A $R\sqrt{1-8\cos A\cos B\cos C}$

Let $O$ and $H$ be the circumcenter and the orthocenter respectively.

If $OF$ is the perpendicular to $AB$, we have

$\angle OAF={90}^{\circ }-\angle AOF={90}^{\circ }-C$

Also $\angle HAL={90}^{\circ }-C$

Hence $\angle OAH=A-\angle OAF-\angle HAL$ $=A-2\left({90}^{\circ -C}\right)$

$=A+2C-(A+B+C)=C-B$

Also $OA=R$

and $HA=2R\cos A$

Now in $\Delta AOH$

$O{H}^2=O{A}^2+H{A}^2-2OAHA\cos \left(\angle OAH\right)$

$=R^2+4R^2-{\cos }^{2}A-4R^2/cosA\cos (C-B)$

$=R^2+4R^2\cos A[/cosA-\cos (C-B\left)\right]$

$=R^2-4R^2\cos A\left[/cosA\right(B+C)+\cos (C-B\left)\right]$

$=R^2-8R^2\cos A\cos B\cos C$

Hence,$OH=R\sqrt{1-8\cos A\cos B\cos C}$