Given any non-isosceles triangle $ABC$, let denote with $AB$ its longest sides, and draw the two circles with centers in $A$ and $B$ and passing by $C$. They determine two additional points $E$ and $D$ on the side $AB$.
If we draw the circles with centers in $A$ and $B$ and passing by $D$ and $E$, respectively, we obtain two other points $F$ and $G$ on the sides $AC$ and $CB$, respectively.
The whole triangle results subdivided in three kinds of segments of lenght $\alpha$ (red), $\beta$ (blue) and $\gamma$ (green).
(In this post A conjecture related to a circle intrinsically bound to any triangle is shown that the points $DFCGE$ always determine a circle).
Given the lengths of the three sides of $ABC$, ($\overline{AB}=\alpha+\beta+\gamma$, $\overline{AC}=\alpha+\gamma$, and $\overline{CB}=\beta+\gamma$), the triangle $ABC$ is uniquely determined.
What is the general relation between $\alpha,\beta$ and $\gamma$?
So far, I observed that $\gamma=\sqrt{2\alpha\beta}$ for each right triangle, but I have troubles to find the function $\gamma=\gamma(\alpha,\beta)$ for a general triangle.
Thanks for your suggestions!
Best Answer
There exists no general relation independent of $\,a,b,c\,$, unless there are additional known conditions on $\,a,b,c\,$. The equations resolve to:
$$ \begin{align} \begin{cases} \alpha = c - a \\ \beta = c - b \\ \gamma = a+b-c \end{cases} \end{align} $$
If you know nothing else about $\,a,b,c\,$ then about all you can say is that $\,\alpha,\beta,\gamma \ge 0\,$.
But if you know that for example $\,c^2=a^2+b^2\,$ (i.e. $\,\triangle ABC\,$ is a right triangle) then you can indeed derive $\,\gamma^2=2 \alpha\beta\,$, which is the relation you found.