IMO TST Day 4

Let $N$ be the midpoint of arc $BAC$. Note that $Z, L, N$ are collinear; further, $AM \parallel LZ$ because they are reflections in the circumcenter $O$ of $\triangle ABC$. Therefore $\angle IML = \angle MLZ$. So it is sufficient to prove that $\angle IVM = \angle NLM$; in fact we will prove that $\triangle IVM \sim \triangle NLM$. Note that $\angle DIM = \angle MAZ = \angle MNZ$ because $AZ$ and perpendicular from $A$ to $BC$ are isogonal. Therefore it is sufficient to prove that
$$\frac{IV}{IM} = \frac{NL}{NM}$$
Let $r, R$ denote the inradius and circumradius of $\triangle ABC$ respectively. Note that $IV = \frac{r}{2}$, $NM = 2R$, and $NL = \frac{AI}{2}$. Also note that $AI \cdot IM$ is the power of $I$ w.r.t. $\odot(ABC)$, which is $R^2 - OI^2 = 2rR$. Therefore
$$\frac{IV}{IM} = \frac{r}{2IM} = \frac{r}{2 \cdot \frac{2rR}{AI}} = \frac{AI}{4R} = \frac{NL}{NM}$$
as required. □

Let the incircle touch $AB$ and $AC$ at $F$ and $E$ respectively. Let $X$ be the reflection of $F$ across $B$, and let $Y$ be the reflection of $E$ across $C$. Let $I_a$ be the $A$-excentre of $ABC$. Let $X_1$ and $Y_1$ be the midpoints of $AX$ and $AY$, and let $N$ be the midpoint of arc $BAC$. Let $A'$ be the reflection of $A$ across $N$.
Claim 1 $I_a$ is the circumcenter of $XDY$
Proof. Since $BD = BX$, the perpendicular bisector of $DX$ is the angle bisector of $XBD$ or the line $BI_a$. Similarly, $CI_a$ is the perpendicular bisector of $DY$. Thus the two perpendicular bisectors of $DX$ and $DY$ meet at $I_a$. □
Claim 2 $A'X = A'Y$
Proof. Observe that $BX_1 = AF/2 = AE/2 = CY_1$, so $\triangle NX_1B \cong NY_1C$ and $NX_1 = NY_1$. Dilating by 2 from $A$ gives us the desired result. As a corollary, observe that $I_aA'$ is perpendicular to $XY$. □
Claim 3 $\angle IDI_a = \pi - \angle (EF, XY)$
Proof. Let $DR$ and $DS$ be the altitudes from $D$ to $EF$ and $XY$. Observe that $\angle IDI_a = \angle CDI_a + \angle EDC + \angle IDE = \angle XDS + \angle FDX + \angle FDR = \angle RDS$ which gives the desired result. □
Claim 4 $L$ is the midpoint of $IA'$.
Proof. Observe that $Z, L, N$ are collinear, and that $NZ$ is the perpendicular bisector of $AA'$. This implies $LA = LI = LA'$, and that $L$ is the circumcentre of $AIZ$. Since $\angle IAZ = \frac{\pi}{2}$, $L$ is the midpoint of $IZ$. □
The key idea is that $\angle DVM = \pi - \angle IVM = \pi - \angle IDI_a = \angle (EF, XY)$. On the other hand, we also have $II_a \perp EF$ and $I_aA' \perp XY$, so $\angle IML = \angle II_aZ = \angle (EF, XY)$. This proves the problem. □

WLOG assume $AB < AC$. Let $U$ be the midpoint of $AI$. Then $\square ULZM$ is clearly a rectangle, so
$$\angle IML = \angle UML = \angle UZL = 90^\circ - \angle UZM$$
Let $I_a$ be the $A$-excenter. Homothety with ratio 2 at $I$ gives
$$\angle DVM = 180^\circ - \angle I_aDI = 90^\circ - \angle I_aDC$$
Therefore it is sufficient to prove that $\angle UZM = \angle I_aDC$. We use complex numbers for this.
Let $\odot(ABC)$ be the unit circle, and let coordinates of $A, B, C$ be $a^2, b^2, c^2$ respectively. Then it is well known that coordinates of $M, I, I_a$ are $-bc, -ab-ac-bc, ab+ac-bc$ respectively. Coordinate of $Z$ is $-a^2$ by reflection in the origin. Further, coordinate of $U$ is $\frac{a^2-ab-ac-bc}{2}$ by midpoint formula. By projection formula, we get the coordinate of $D$ as $\frac{(bc-a^2)(b+c)+a(b^2+c^2)}{2a}$. Also, $\angle I_aDC$ is the same as directed angle from $I_aD$ to $BC$, so we have to prove
$$\frac{M-Z}{U-Z} \bigg/ \frac{C-B}{I_a-D} \in \mathbb{R}$$
Putting the coordinates in, this is equivalent to proving $\frac{bc-a^2}{a(b-c)} \in \mathbb{R}$. If we let $y$ be that quantity, we have
$$\bar{y} = \frac{\frac{1}{bc} - \frac{1}{a^2}}{\frac{1}{ab} - \frac{1}{ac}} = \frac{bc - a^2}{a(b-c)} = y$$
Hence $y \in \mathbb{R}$, as needed. □

WLOG assume $AB < AC$. Let $N$ be midpoint of arc $BAC$, and let $MD$ intersect $\odot(ABC)$ again at $D'$.
Claim 1 $D', I, Z$ are collinear.
Proof. Consider an inversion with center $M$ and radius $MB$. $B, C, I$ are fixed since $MB = MC = MI$. The inversion interchanges line $BC$ and $\odot(ABC)$, so $D' \leftrightarrow D$. Also note that $MN \parallel ID$ and $NZ \parallel IM$, so
$$\angle MD'I = \angle DIM = \angle MM'Z = \angle MD'Z$$
Hence $D', I, Z$ are collinear. □
Let $U$ be the midpoint of $AI$.
Claim 2 MV and ZU intersect on $\odot(ABC)$.
Proof. We use phantom points and cross-ratios. Let $MV, ZU$ intersect $\odot(ABC)$ at $V', V''$ respectively. Since $V$ is midpoint of $ID$, we have $(I, D; V, P_\infty) = -1$ where $P_\infty$ is the point at infinity along $ID$. Since $MN \parallel ID$, projecting from $M$ onto $\odot(ABC)$ we get $(A, D'; V', N) = -1$. Similarly, since $U$ is midpoint of $AI$, we have $(A, I; U, P'_\infty) = -1$ where $P'_\infty$ is the point at infinity along $AI$. Since $ZN \parallel AI$, projecting from $M$ onto $\odot(ABC)$ we get $(A, D'; V'', N) = -1$.
$$\implies (A, D'; V', N) = (A, D'; V'', N)$$
Therefore $V' = V''$, as required. □
Note that $\square LZMU$ is a rectangle, and $ID \parallel MN$, so
$$\angle DVM = \angle VMN = \angle V'MN = \angle V'ZN = \angle L'ZL = \angle L'ML = \angle IML$$
as required. □