Krdytkyu

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.

The formula for length of $AG = frac{2}{3}cdotfrac{1}{2}sqrt{AC^2-BC^2+AB^2}$. Similarly, the length of $BG=frac{2}{3}cdotfrac{1}{2}sqrt{AB^2+BC^2-AC^2}$. Now, we can find the circumradius of the triangle is $$frac{ABcdot BGcdot AG}{4sqrt{s(s-AB)(s-BG)(s-AG)}}$$, where $s=frac{AG+BG+AB}{2}$.

If you are given $AB,BC,AC$, surely you cannot get the wrong answer unless there is a calculation error.

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

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