Problem 2062

Solution:

In this solution we will use a method called Inversion in the Plane.
We invert with respect to point $O$ with an arbitrary radius $r$. We will label the images of objects (points, circles, lines, segments) under this inversion by putting a bar over them. By properties of inversion, the three given circles through $O$ will invert to three lines not through $O$. For instance, circle $k_{1}$ will invert to a line $\overline{k_{1}}$ through points $\bar{B}$, $\bar{C}$, and $\overline{A'}$, and similarly for circles $k_{2}$ and $k_{3}$. On the other hand, the original line $AOA'$ will invert to itself (as it passes through the center of inversion $O$); now, points $A$ and $A'$ will move to, perhaps different, points $\bar{A}$ and $\overline{A'}$ on this same line, while point $O$ will stay where it is (we shall not apply inversion here to the center of inversion $O$). Analogous situations occur for lines $BOB'$ and $COC'$.

Diagram before inversion

Diagram after inversion, with perpendiculars to $\overline{BC}$

We are ready to describe the new inverted picture. Point $O$ is inside triangle $\overline{ABC}$. Three points $\overline{A'}$, $\overline{B'}$, and $\overline{C'}$ are chosen on the triangle's sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively, so that the three segments $\overline{AA'}$, $\overline{BB'}$, and $\overline{CC'}$ all intersect in point $O$ (such segments are called cevians in $\triangle \overline{ABC}$). Two distance formulas relating new to old distances under inversion tell us:
$$|AO| = \frac{r^2}{|O\bar{A}|}, \text{ and } |AA'| = \frac{r^2 \cdot |\overline{AA'}|}{|O\bar{A}| \cdot |O\overline{A'}|}$$
Dividing these two expressions and canceling $r^2$ and $|O\bar{A}|$ re-expresses one of the desired ratios completely in terms of the new inverted picture:
$$\frac{|AO|}{|AA'|} = \frac{|O\overline{A'}|}{|\overline{AA'}|}$$
Now drop perpendiculars $OH_1$ and $\bar{A}H_2$ to side $\overline{BC}$ as shown in the inverted picture, and call their lengths $h_1$ and $h_2$. This creates two similar triangles: $\triangle O\overline{A'}H_1 \sim \triangle \overline{AA'}H_2$: they share an angle and have another right angle each. Hence, the ratios of corresponding sides are equal:
$$\frac{|O\overline{A'}|}{|\overline{AA'}|} = \frac{|OH_1|}{|\bar{A}H_2|} = \frac{h_1}{h_2} = \frac{h_1 \cdot |\overline{BC}|/2}{h_2 \cdot |\overline{BC}|/2} = \frac{S_{\triangle O\overline{BC}}}{S_{\triangle \overline{ABC}}}$$
where $h_1$ and $h_2$ are the lengths of the drawn altitudes $OH_1$ and $\bar{A}H_2$ in $\triangle O\overline{BC}$ and $\triangle \overline{ABC}$, respectively. Along the way, we multiplied by $|\overline{BC}|/2$ to recreate the standard formulas for the areas of $O\overline{BC}$ and $\triangle \overline{ABC}$, and denoted correspondingly those areas by $S_{\triangle}$ in the last ratio.

Of course, we can repeat the above discussion for the other two ratios $\frac{|BO|}{|BB'|}$ and $\frac{|CO|}{|CC'|}$, and end up rewriting the desired sum in a completely different way:
$$\frac{|AO|}{|AA'|} + \frac{|BO|}{|BB'|} + \frac{|CO|}{|CC'|} = \frac{S_{\triangle O\overline{BC}}}{S_{\triangle \overline{ABC}}} + \frac{S_{\triangle O\overline{CA}}}{S_{\triangle \overline{ABC}}} + \frac{S_{\triangle O\overline{AB}}}{S_{\triangle \overline{ABC}}} = \frac{S_{\triangle O\overline{BC}} + S_{\triangle O\overline{CA}} + S_{\triangle O\overline{AB}}}{S_{\triangle \overline{ABC}}} = \frac{S_{\triangle \overline{ABC}}}{S_{\triangle \overline{ABC}}} = 1$$
Here we used the fact that $O$ is inside $\triangle \overline{ABC}$ so that the three triangles with vertex $O$, $\triangle O\overline{BC}$, $\triangle O\overline{CA}$, and $\triangle O\overline{AB}$, make up the whole big $\triangle \overline{ABC}$, and hence their areas add up to the area of this big triangle.