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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?

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    $\begingroup$ $AB$ 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 $\angle B$ $\endgroup$
    – Math Lover
    Commented Jan 2, 2022 at 19:18
  • $\begingroup$ The tangents are $AB$ and $BC$. Excelent, it is the simple, obvious and correct solution! Could you write it as an answer? $\endgroup$
    – Ivan Bunin
    Commented Jan 2, 2022 at 19:24

1 Answer 1

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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:

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

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