# Find parameters of a circular arc inscribed into isosceles triangle

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?

• $AB$ and $AC$ are tangents or $AB$ and $BC$ are tangents? Assuming latter, draw perp to $AB$ and $BC$ at points $A$ and $C$. Where they intersect is the center of the circle. Radius can be obtained knowing length $AB$ and $\angle B$ Jan 2 at 19:18
• The tangents are $AB$ and $BC$. Excelent, it is the simple, obvious and correct solution! Could you write it as an answer? Jan 2 at 19:24

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:

$$\sin \frac {\alpha}2=\frac12\sqrt{4-\frac{b^2}{c^2}}$$