Three circles of radius $r$ $r$ units are drawn inside an equilateral triangle of side $a$ $a$ units such that each circle touches the other two circles and two sides of the triangle. What is the relation between $r$ $r$ and $a$ $a$ ?

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Answer 1 · 2016-11-25T15:40:47

$\frac{r}{a} = \frac{1}{2 (\sqrt{3} + 1)}$ $r/a=1/(2(sqrt(3)+1)$

We know that

$a = 2 x + 2 r$ $a = 2x+2r$ with $\frac{r}{x} = tan (30^{\circ})$ $r/x=tan(30^@)$

$x$ $x$ is the distance between the left bottom vertice and the vertical projection foot of the left bottom circle center.

because if an equilateral triangle's angle has $60^{\circ}$ $60^@$ , the bisector has $30^{\circ}$ $30^@$ then

$a = 2 r (\frac{1}{tan (30^{\circ})} + 1)$ $a = 2r(1/tan(30^@)+1)$

so

$\frac{r}{a} = \frac{1}{2 (\sqrt{3} + 1)}$ $r/a=1/(2(sqrt(3)+1)$

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Three circles of radius rr units are drawn inside an equilateral triangle of side aa units such that each circle touches the other two circles and two sides of the triangle. What is the relation between rr and aa?