12 circles

The shortcut for this question is to start with an equilateral triangle and calculate $r_1 = r_3 = r_5 = \frac{\sqrt{3 - 1}}{4}r$ (for $r = AO$ ) and $R_1 = R_3 = R_5 = \frac{3\sqrt{3 - 1}}{4}r$ . Then that would mean $\frac{\frac{1}{r_1} + \frac{1}{r_3} + \frac{1}{r_5}}{\frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_5}} = \boxed{3}$ .

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For a more general solution for other triangles, let $A'$ be the intersection of $AO$ and $BC$ , $B'$ be the intersection of $BO$ and $AC$ , and $C'$ be the intersection of $CO$ and $AB$ . Let $r = AO = BO = CO$ .

Since an inscribed angle is half of a central angle that subtends the same arc, $\angle BOC = 2A$ , so $\angle BOC' = \angle COB' = 180° - 2A$ (from the straight line), $\angle CBO = \angle BCO = 90° - A$ (from the isosceles triangle $\triangle BOC$ ), and $BC = 2r \sin A$ (from a circle chord). Similarly, $\angle AOC' = \angle COA' = 180° - 2B$ and $\angle AOB' = \angle BOA' = 180° - 2C$ , $\angle ACO = \angle CAO = 90° - B$ and $\angle ABO = \angle BAO = 90° - C$ , and $AB = 2r \sin C$ and $AC = 2r \sin B$ .

Using the law of sines on $\triangle BOA'$ gives $OA' = \frac{r \cos A}{\cos (B - C)}$ and $BA' = \frac{r \sin 2C}{\cos(B - C)}$ . Similarly, $OB' = \frac{r \cos B}{\cos (A - C)}$ and $OC' = \frac{r \cos C}{\cos (A - B)}$ , and $CA' = \frac{r \sin 2B}{\cos(B - C)}$ , $CB' = \frac{r \sin 2A}{\cos(A - C)}$ , $AB' = \frac{r \sin 2C}{\cos(A - C)}$ , $AC' = \frac{r \sin 2B}{\cos(A - B)}$ , and $BC' = \frac{r \sin 2A}{\cos(A - B)}$ .

Since the radius $r$ of an incircle is the quotient of the area of its triangle and the semiperimeter of its triangle, $r_1 = \frac{\frac{1}{2} \cdot r \cdot \frac{r\sin 2B}{\cos(A - B)} \cdot \sin(90° - C)}{\frac{1}{2}(r + \frac{r\sin 2B}{\cos(A - B)} + \frac{r\cos C}{\cos(A - B)})} = \frac{r \cos A \cos B \cos C}{\cos A \sin A + \cos A \cos B}$ .

Similarly:

$r_3 = \frac{r \cos A \cos B \cos C}{\cos C \sin C + \cos C \cos A}$

$r_5 = \frac{r \cos A \cos B \cos C}{\cos B \sin B + \cos B \cos C}$

$r_2 = \frac{r \cos A \cos B \cos C}{\cos A \sin A + \cos A \cos C}$

$r_4 = \frac{r \cos A \cos B \cos C}{\cos C \sin C + \cos C \cos B}$

$r_6 = \frac{r \cos A \cos B \cos C}{\cos B \sin B + \cos B \cos A}$

Therefore,

$\frac{1}{r_1} + \frac{1}{r_3} + \frac{1}{r_5} = \frac{1}{r_2} + \frac{1}{r_4} + \frac{1}{r_6} = \frac{\cos A \sin A + \cos B \sin B + \cos C \sin C + \cos A \cos B + \cos B \cos C + \cos A \cos C}{r \cos A \cos B \cos C}$

Since the radius $R$ of an excircle is the quotient of the area of its triangle and the difference of the semiperimeter of its triangle and the side it is tangent to, $R_1 = \frac{\frac{1}{2} \cdot 2r \sin B \cdot (r + \frac{r \cos C}{\cos (A - B)}) \cdot \sin(90° - B)}{\frac{1}{2}(2r \sin B + r + \frac{r \cos C}{\cos (A - B)} + \frac{r \sin 2B}{\cos (A - B)}) - \frac{r \sin 2B}{\cos (A - B)}} = \frac{2r\sin A \sin B \sin C \cos A \cos B \cos C}{2 \sin A \sin B \sin C \cos A \cos B + \sin A \sin B \cos A \cos B - \sin B \cos A \cos B \cos C - \sin B \cos A \cos^2 B}$ .

Similarly:

$R_3 = \frac{2r\sin A \sin B \sin C \cos A \cos B \cos C}{2 \sin A \sin B \sin C \cos A \cos C + \sin A \sin C \cos A \cos C - \sin A \cos A \cos B \cos C - \sin A \cos C \cos^2 A}$

$R_5 = \frac{2r\sin A \sin B \sin C \cos A \cos B \cos C}{2 \sin A \sin B \sin C \cos B \cos C + \sin B \sin C \cos B \cos C - \sin C \cos A \cos B \cos C - \sin C \cos B \cos^2 C}$

$R_2 = \frac{2r\sin A \sin B \sin C \cos A \cos B \cos C}{2 \sin A \sin B \sin C \cos A \cos C + \sin A \sin C \cos A \cos C - \sin C \cos A \cos B \cos C - \sin C \cos A \cos^2 C}$

$R_4 = \frac{2r\sin A \sin B \sin C \cos A \cos B \cos C}{2 \sin A \sin B \sin C \cos B \cos C + \sin B \sin C \cos B \cos C - \sin B \cos A \cos B \cos C - \sin B \cos C \cos^2 B}$

$R_6 = \frac{2r\sin A \sin B \sin C \cos A \cos B \cos C}{2 \sin A \sin B \sin C \cos A \cos B + \sin A \sin B \cos A \cos B - \sin A \cos A \cos B \cos C - \sin A \cos B \cos^2 A}$

Careful examination shows that:

$\frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_5} = \frac{1}{R_2} + \frac{1}{R_4} + \frac{1}{R_6}$

only for acute isosceles (and equilateral) triangles, but not for scalene triangles.

Finally, using the above equations, $\frac{\frac{1}{r_1} + \frac{1}{r_3} + \frac{1}{r_5}}{\frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_5}} = 3$ for equilateral triangles, but $\frac{\frac{1}{r_1} + \frac{1}{r_3} + \frac{1}{r_5}}{\frac{1}{R_1} + \frac{1}{R_3} + \frac{1}{R_5}} > 3$ for all other acute triangles. In either case, the value of the integer part is $\boxed{3}$ .