Circumradius of a triangle

Question

Lengths of the sides of a triangle, $a$, $b$ and $c$ are given. We are required to find the circumradius of the triangle formed by the vertices $A$, $B$ and $G$, where $G$ is the centroid of the triangle $ABC$. The following is a picture of the problem: (w/o the given values for I don't want this to come across as a homework question)

I approached this problem by finding out the lengths of the sides $AG$ and $BG$ of the triangle $GAB$ by using the formula to find the length of the median drawn from any vertex of a triangle. After obtaining these two values(I already know the length $AB$), I used the formula for circumradius given as ABC/4∆ (Where ∆ is the area of the triangle) but I have not obtained the correct answer. I have checked and rechecked for calculation mistakes but nope! None seem to be there.

Is my method correct? Any help would be tremendously appreciated. Thanks so much in advance :) Regards.

Answer 1

Recall that $G$ divides the median starting from $A$ and ending at $BC$ in the ratio $2:1$. Furthermore, there is a standard formula for the length of a median provided by Apollonius. Combining these, $$AG = \frac{2}{3}\cdot\frac{1}{2}\sqrt{AC^2-BC^2+AB^2}.$$ Similarly, $$BG=\frac{2}{3}\cdot\frac{1}{2}\sqrt{AB^2+BC^2-AC^2}.$$

Now, recall that the area $|ABG|$ of the triangle $ABG$ can be specified in two ways. By Heron's formula, $$ |ABG| = \sqrt{s(s-AB)(s-AG)(s-BG)}, \text{ where } s= \frac{AB+AG+BG}{2}, $$ or $|ABG| = \frac{AB\cdot BG\cdot AG}{4R}$, where $R$ denotes the circumradius of the triangle $ABG$. By equating these two formulas, we find that $$\bbox[border:2px solid blue]{R = \frac{AB\cdot BG\cdot AG}{4\sqrt{s(s-AB)(s-BG)(s-AG)}}, \text{ where } s=\frac{AG+BG+AB}{2}.}$$

Answer 2

By Euler's formula $R=\frac{abc}{4\Delta}$, the circumradius of $ABG$ is given by

$$ R_{ABG} = \frac{c\left(\frac{2}{3}m_a\right)\left(\frac{2}{3}m_b\right)}{4[ABG]} = \color{red}{\frac{c\, m_a\, m_b}{3[ABC]}} = \frac{4}{3}R\left(\frac{m_a}{a}\right)\left(\frac{m_b}{b}\right)$$ where $$ m_a = \frac{1}{2}\sqrt{2b^2+2c^2-a^2} $$ follows from Stewart's theorem.