Assign weights $m_{A1} = OO_1 = r - r_1, m_O = O_1A_1 = r_1$ to $A_1, O$, respectively, so that these 2 weights balance at $O_1$, which is then assigned weight $m_{O1} = m_{A1} + m_O = r - r_1 + r_1 = r$. Assign weight $m_{O2}$ to $O_2$, such that the weights $m_{O1}, m_{O2}$ balance at $A$: $\frac{m_{O2}}{m_{O1}} = \frac{O_1A}{O_2A} = \frac{r_1}{r_2}$ $m_{O2} = m_{O1}\ \frac{r_1}{r_2} = \frac{r r_1}{r_2}$ Assign weight $m_{A2}$ to $A_2$, so that the weights $m_O, m_{A2}$ balance at $O_2$: $\frac{m_{A2}}{m_O} = \frac{OO_2}{O_2A_2} = \frac{r - r_2}{r_2}$ $m_{A2} + m_O = m_{O2} = \frac{r r_1}{r_2}$ Solving for $m_{A2}$, $m_{A2} = m_O\ \frac{r - r_2}{r_2} = \left(\frac{r r_1}{r_2} - m_{A2}\right) \frac{r - r_2}{r_2}$ $m_{A2} \left(1 + \frac{r - r_2}{r_2}\right) = \frac{r r_1}{r_2} \cdot \frac{r - r_2}{r_2}$ $m_{A2} = \frac{r_1(r - r_2)}{r_2}$ Let $OA$ meet $A_1A_2$ at $P$. The weights $m_{A1}, m_{A2}$ now balance at $P$, i.e., $\frac{A_1P}{A_2P} = \frac{m_{A2}}{m_{A1}} = \frac{r_1(r - r_2)}{r_2(r - r_1)}$ Since we also have $\frac{OO_1}{A_1O_1} = \frac{r - r_1}{r_1},\ \frac{OO_2}{A_2O_2} = \frac{r - r_2}{r_2}$, it follows that $\frac{OO_1}{O_1A_1} \cdot \frac{A_1P}{PA_2} \cdot \frac{A_2O_2}{O_2O} = \frac{r - r_1}{r_1} \cdot \frac{r_1(r - r_2)}{r_2(r - r_1)} \cdot \frac{r_2}{r - r_2} = 1$ By Ceva's theorem, the cevians $A_2O_1,\ OA \equiv OP,\ A_1O_2$ in the triangle $\triangle OA_1A_2$ are concurrent.

in this problem, we just apply the ceva theorem with triangle $OO_1O_2$, with A in $O_1O_2$, $A_1$ in $OO_1$, $A_2$ in $OO_2$

shobber wrote: Prove that the three lines $OA$, $O_1 A_2$, and $O_2 A_1$ are concurrent. This result is also true if the three circles are externally tangent. That's what neverstop means and this is true because $\frac{s-a}{s-b} \cdot \frac{s-b}{s-c} \cdot \frac{s-c}{s-a} =1$

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1992Problem2 Vo Duc Dien

oh nice problem!! lemma 1:$O1,O2,A$ are on same line. proof:we know that $\angle O2AO =90$ $ \angle O1AO =90$ so $O1,A,O2$ are on same line lemma 2:$O,O2,A2$are on same line and $O,O1,A1$ are on same line. proof:we know that $O2A2$ is perpendicular to line that tangent to circle$C$ and we know that $OA2$ is perpendicular to line that tangent to circle$C$ so $A2,O2,O$ are on same line ! proof of problem: suppose $A2A1O2 = X$ and $O2A1O=Y$ and $OA2O1 = P$ AND $O1A2A1=Q$ we must proof: $\frac{\sin X}{\sin Y} . \frac{\sin AOO1}{\sin AOO2} . \frac{\sin P}{\sin Q} =1$ we know that : $\frac{\sin AOO1}{\sin AOO2} = \frac {r_1}{r_2} . \frac{OO2}{OO1}$ and we know that: $\frac{\sin X}{\sin Y} . \frac{AA2}{A1O} = \frac {r_2}{OO2}$ and also it is right for $\frac{\sin P}{\sin Q}$and we know that $OA2=OA1$ so the problem is proof! excuse me because im so lazy to right all of my solution but im sure that you know that my solution

Dear MLs It is simple. The concurrency point is this case is, obviously, the Lemoine point (symmedian point) of triangle AA1A2. r1 and r2 are two red herrings, while r isn't. This post is dedicated to both Mr. Referendum - G. Papandreou and Mr. M.I.T. - Darij, not necessarily in this order. Friendly, M.T.

@armpist: very fine observation, although not quite 'obvious'; I think it can bring new good questions about this configuration. Best regards, sunken rock

@sunken rock maybe the very first candidate should be the Sangaku of Gumma, even though there are plenty of solutions by now. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=18266 It may lead to a new approach. M.T.

Apply Ceva's theorem on $\triangle OO_1O_2$. The following statement is equivalent to the desired lines being concurrent: \[\frac{OA_1}{A_1O_1} \cdot \frac{O_1A}{AO_2} \cdot \frac{O_2A_2}{A_2O} = 1.\] Then, notice that we can write everything in terms of $r$, $r_1$ and $r_2$. Since \[\frac{r}{r_1} \cdot \frac{r_1}{r_2} \cdot \frac{r_2}{r} = 1,\] we are done. $\square$