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