Let we have an isosceles triangle ABC where $AB=BC$.
How to find parameters of a circular arc that (1) pass through points A and C, (2) such as AB and AC are tangents to this arc.
The arc parameters we need are: $R$ – radius of a circle that produces arc, and $\alpha$ – arc angle.
A possible solution can be to prolong the triangle sides and to find the circle inscribed into this bigger triangle that touches it into points $A$ and $C$. But how to deside to which length we nedd to prolong these sides?
Best Answer
In figure H is the foot of altitude BH, $AO=R$ is radius of circle and $\angle AOC=\alpha$. In right angle triangle BAH and AHO we respectively have:
$BH^2=c^2-(\frac b2)^2$
$HO^2=R^2-(\frac b2)^2$
In right angle triangle ABO we have:
$BH\times HO=AH^2=(\frac b2)^2$
Multiplying the first two relations we get:
$BH^2\times HO^2=(c^2-\frac{b^2}4)(R^2-\frac{b^2}4)=AH^4=\frac {b^4}{16}$
which finally gives:
$c^2R^2=\frac{b^2}4(c^2+R^2)$
which gives:
$R=\frac{bc}{\sqrt{4c^2-b^2}}$
In right angle triangle AOH we have:
$\sin \frac{\alpha}2=\frac {AH=\frac b2}{AO=R}$
substituting values we finally get:
$\sin \frac {\alpha}2=\frac12\sqrt{4-\frac{b^2}{c^2}}$