Taiwan IMO Selection Camp

As shown in the figure, let the incircle touch the side $BC$ at the point $D$, let $GH$ meet $BI, CI$ at the points $C'$, $B'$ respectively, and let $D'$ be a point on $GH$ such that $ID' \perp GH$.

1.

a. Since $\angle C'ID' = \frac{1}{2}(\angle A + \angle B)$, $\angle IC'B' = \frac{1}{2}\angle C$. Similarly, $\angle IB'C' = \frac{1}{2}\angle B$. From this we know $\triangle IBC \sim \triangle IB'C'$.

b. Since $G, H$ are respectively the reflection points of $E, F$ with respect to $I$, $G, H$ lie on the incircle, and $EF \parallel GH$. Therefore $\angle IGD' = \angle IEF = \frac{1}{2}\angle A$ (since $A, E, I, F$ are concyclic). Hence
$$\frac{IB'}{IB} = \frac{IC'}{IC} = \frac{ID'}{ID} = \frac{ID'}{IG} = \sin \angle D'IG = \sin \frac{1}{2}\angle A.$$

2. Applying Menelaus' theorem to $\triangle IBC, B'C'$ gives
$$\frac{BQ}{QC} \cdot \frac{CB'}{B'I} \cdot \frac{IC'}{C'B} = -1 \quad \text{or} \quad \frac{BM + MQ}{BM - MQ} = \frac{BQ}{QC} = \frac{IB'}{CB'} \cdot \frac{C'B}{IC'}$$
Therefore
$$\begin{align*} \frac{BM + MQ}{IB' \cdot C'B} &= \frac{BM - MQ}{CB' \cdot IC'} \\ &= \frac{2BM}{IB' \cdot C'B + CB' \cdot IC'} \\ &= \frac{2MQ}{IB' \cdot C'B - CB' \cdot IC'} \end{align*}$$

3. From 1. we get
$$\begin{aligned} IB' \cdot C'B + CB' \cdot IC' &= IB \sin \frac{\angle A}{2} (IB - IC') + (IB' - IC)IC \sin \frac{\angle A}{2} \\ &= \sin \frac{\angle A}{2} [IB(IB - IC \sin \frac{\angle A}{2}) + (IB \sin \frac{\angle A}{2} - IC)IC] \\ &= \sin \frac{\angle A}{2} (IB^2 - IC^2), \end{aligned}$$
and
$$\begin{aligned} IB' \cdot C'B - CB' \cdot IC' &= \sin \frac{\angle A}{2} [IB(IB - IC \sin \frac{\angle A}{2}) - (IB \sin \frac{\angle A}{2} - IC)IC] \\ &= \sin \frac{\angle A}{2} (IB^2 + IC^2 - 2 IB \cdot IC \sin \frac{\angle A}{2}). \end{aligned}$$
Since $IB^2 = ID^2 + BD^2, IC^2 = ID^2 + CD^2,$
$$IB^2 - IC^2 = BD^2 - CD^2 = (BM + MD)^2 - (BM - MD)^2 = 4BM \cdot MD.$$
Since $BC^2 = IB^2 + IC^2 - 2 IB \cdot IC \cos(90^\circ + \frac{\angle A}{2}) = IB^2 + IC^2 + 2 IB \cdot IC \sin \frac{\angle A}{2},$
$$IB' \cdot C'B - CB' \cdot IC' = \sin \frac{\angle A}{2} (2IB^2 + 2IC^2 - BC^2) = 4IM^2 \sin \frac{\angle A}{2}.$$

4. From 2. and 3. we get $\frac{BM}{BM \cdot MD} = \frac{MQ}{IM^2}$, that is, $\frac{IM}{MD} = \frac{MQ}{IM}$. From this we know $\triangle IMD \sim \triangle QMI$. Therefore $\angle QIM = \angle IQM = 90^\circ$, which completes the proof.