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


Inequality with areas of triangles formed by orthocenter and incenter, source


An Inequality of the Areas of Triangles Formed by Circumcenter And Orthocenter, problem


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.


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.


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