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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 $AB$.

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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.

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The whole triangle results subdivided in three kinds of segments of lenght $\alpha$ (red), $\beta$ (blue) and $\gamma$ (green).

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(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!

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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.

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  • $\begingroup$ You mean that such function can be found given an angle, e.g. given $A\hat{C}B$? $\endgroup$ – user559615 Aug 1 '18 at 21:43
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    $\begingroup$ @andrea.prunotto Yes, just write the law of cosines for the known angle and substitute $\,a,b,c\,$ in terms of $\,\alpha,\beta,\gamma\,$, then you get a relation between $\,\alpha,\beta,\gamma\,$ and $\,\cos C\,$. $\endgroup$ – dxiv Aug 1 '18 at 21:45
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    $\begingroup$ @andrea.prunotto shouldn't this been enough Enough for what? It is enough to calculate $\,\alpha,\beta,\gamma\,$ in terms of $\,a,b,c\,$, see the formulas above. It is not enough to calculate $\,\gamma\,$ in terms of $\,\alpha,\beta\,$, though. Or, think at it the other way around: if it were possible to find $\,\gamma=\gamma(\alpha,\beta)\,$ you could then determine all three of $\,a,b,c\,$ from $\,\beta,\gamma\,$ alone, but of course you can't (unless you use some additional condition). $\endgroup$ – dxiv Aug 1 '18 at 21:51
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    $\begingroup$ @andrea.prunotto The law of cosines gives something like $\,(1 - 2 \cos C)\gamma^2$ $- 2(\alpha+\beta) \cos C \,\gamma$ $-2\alpha\beta(1+ \cos C)=0\,$ (which of course reduces to $\gamma^2-2\alpha\beta=0$ when $\,\cos C = 0\,$). While not too pretty, it's still just a quadratic in $\,\gamma\,$. $\endgroup$ – dxiv Aug 1 '18 at 22:12
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    $\begingroup$ You're right. I will think about it. But thanks for your explanations: They are really helpful! $\endgroup$ – user559615 Aug 1 '18 at 22:14

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