Berkeley Math Circle: Monthly Contest 7

Solution:

Let $a = BC$, $b = CA$, $c = AB$. It is well-known (and follows from, say, Stewart's Theorem) that $AM = \frac{bc}{a + c}$ and $AN = \frac{bc}{a + b}$.

Now, the distances from $O$ to $BC$, $AC$, $AB$ are $R \cos \alpha$, $R \cos \beta$, $R \cos \gamma$, respectively, where $\alpha, \beta, \gamma$ are the angles of $\triangle ABC$, and $R$ is the circumradius of $ABC$.

So, $O$ is on line $MN$ if and only if
$$\begin{aligned} {[ANM]} & = [ANO] + [AOM] \\ \Longleftrightarrow [ABC] \cdot \frac{AN}{AB} \cdot \frac{AM}{AC} & = [ANO] + [AOM] \\ \Longleftrightarrow \left(\frac{1}{2} a R \cos \alpha + \frac{1}{2} b R \cos \beta + \frac{1}{2} c R \cos \gamma\right) \cdot \frac{b}{a + b} \cdot \frac{c}{a + c} \end{aligned}$$
(Here we use $[\mathcal{P}]$ for the area of polygon $\mathcal{P}$.) Next, recall that $AE = c \cos \alpha$, $AF = b \cos \alpha$. Thus $I$ is on $EF$ if and only if
$$\begin{aligned} {[AFE]} & = [AFI] + [AIE] \\ \Longleftrightarrow [ABC] \cdot \frac{AF}{AB} \cdot \frac{AE}{AC} & = \frac{1}{2} r \cdot c \cos \alpha + \frac{1}{2} r \cdot b \cos \alpha \\ \Longleftrightarrow \frac{1}{2} r(a + b + c) \cos^2 \alpha & = \frac{1}{2} r(b + c) \cos \alpha \\ \Longleftrightarrow (a + b + c) \cos \alpha & = b + c. \end{aligned}$$
Because $AC = AE + EC$, we know $b = c \cos \alpha + a \cos \gamma$. Similarly, $c = b \cos \alpha + a \cos \beta$. Thus $I$ is on $EF$ if and only if
$$\begin{aligned} (a + b + c) \cos \alpha & = (c \cos \alpha + a \cos \gamma) + b(\cos \alpha + a \cos \beta) \\ \Longleftrightarrow \cos \alpha & = \cos \beta + \cos \gamma. \end{aligned}$$
This implies the result.