MHB Does the Triangle's Centroid Lie Inside the Circle with Diameter OA?

[anemone]

Here is this week's POTW:

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Let $ABC$ be a triangle with centroid $G$ and circumcenter $O$. Prove that if $BC$ is its largest side, then $G$ lies in the interior of the circle with diameter $OA$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week's problem.(Sadface)

You can find the suggested solution as follows:

Spoiler

View attachment 6469
Let $BC=a,\,CA=b,\,AB=c$, $M$ be the midpoint of $BC$ and $D$ be the second intersection point between $AM$ and the circumcircle of triangle $ABC$.

By the Power of A Point theorem, we have $AM\cdot MD=BM \cdot MC$. Thus we get

$\begin{align*}AD&=AM+MD\\&=m_a+\dfrac{\dfrac{a}{2}\cdot \dfrac{a}{2}}{m_a}\\&=\dfrac{4m_a^2+a^2}{4m_a}\\&=\dfrac{b^2+c^2}{2m_a}\end{align*}$

Now, taking into consideration that the circle of diameter $OA$ is the locus of midpoints of chords of $O$ that pass through $A$, we have $G$ lies in the interior of the circle of diameter $OA$ if and only if $AG<\dfrac{AD}{2}$, or equivalently

$\dfrac{2}{3}m_a<\dfrac{1}{2}\left(\dfrac{b^2+c^2}{2m_a}\right)$

$8m_a^2<3(b^2+c^2)$

$b^2+c^2<2a^2$

Since $BC$ is the greatest side of triangle $ABC$, the last inequality holds so we are done.