# An Inequality of the Areas of Triangles Formed by Circumcenter And Orthocenter

### Solution

Let $AD,BE,CF\,$ be the altitudes of the triangle. $2[\Delta HBC]=BC\cdot HD,\,$ $2[\Delta HAB]=AB\cdot HF,\,$ $2[\Delta HAC]=AC\cdot HE.\,$ On the other hand,

\displaystyle \begin{align} 2[\Delta OAB]&=\frac{OA\cdot OB\cdot\sin\angle AOB}{2}=\frac{R^2\sin 2C}{2},\\ 2[\Delta OBC]&=\frac{R^2\sin 2A}{2},\\ 2[\Delta OAC]&=\frac{R^2\sin 2B}{2}. \end{align}

Thus we have

\displaystyle \begin{align} 2[\Delta HAB]+2[\Delta HAC]&=2[\Delta ABC]-2[\Delta HBC]\\ &=AD\cdot BC-BC\cdot HD=BC\cdot AH\\ &=2R\sin A\cdot 2R\cos A=2R^2\sin 2A\\ &=4[\Delta OBC]. \end{align}

To sum up,

\displaystyle \begin{align} [\Delta HAB]+[\Delta HAC]&=2[\Delta OBC]\;\text{and, similarly,}\\ [\Delta HAC]+[\Delta HBC]&=2[\Delta OAB],\\ [\Delta HBC]+[\Delta HAB]&=2[\Delta OAC]. \end{align}

It follows that

\displaystyle \begin{align} [\Delta HBC]&=[\Delta OAB]+[\Delta OAC]-[\Delta OBC],\\ [\Delta HAC]&=[\Delta OBC]+[\Delta OAB]-[\Delta OAC],\\ [\Delta HAB]&=[\Delta OAC]+[\Delta OBC]-[\Delta OAB].\\ \end{align}

The function $f(x)=\sqrt[n]{x}\,$ is concave; so, by definition,

\displaystyle \begin{align} f(x)+f(y)+f(z)&=\frac{f(x)+f(y)}{2}+\frac{f(y)+f(z)}{2}+\frac{f(z)+f(x)}{2}\\ &\le f\left(\frac{x+y}{2}\right)+f\left(\frac{y+z}{2}\right)+f\left(\frac{z+x}{2}\right). \end{align}

Taking $x=[\Delta OAB]+[\Delta OAC]-[\Delta OBC],\,$ $y=[\Delta OBC]+[\Delta OAB]-[\Delta OAC],\,$ $z=[\Delta OAC]+[\Delta OBC]-[\Delta OAB]\,$ we obtain the required inequality.

### Acknowledgment

This problem by Dan Mariancescu and Leo Goigiuc has been kindly posted at the CutTheKnotMath facebook page by Leo with a link to the original post at the Peru Geometrico facebook group. The solution is by Marian Dinca.