India — Team Selection Test

Solution A
Observe that $\angle PIQ = 90 - \frac{\angle A}{2}$, $\angle IQP = 90 - \frac{\angle C}{2}$ and $\angle IPQ = 90 - \frac{\angle B}{2}$. Thus, if $\triangle DEF$ is the orthic triangle of $\triangle IPQ$, then it is similar to $\triangle ABC$.
Let $\ell$ be the $I$ midline in $\triangle IPQ$ and $O$ be the circumcenter. Now, let $X = \ell \cap AO$ and $K'$ be the reflection of $I$ across $X$. Then, we just want that $K', K$ coincide or equivalently that $I, M, X$ are collinear.
Now, with respect to triangle $IPQ$, $AI$ is tangent to the circumcircle of $IPQ$ as $\angle AIQ = 90 - \frac{\angle B}{2}$ as required.
Let $H$ be the orthocenter of $\triangle IPQ$ and $N$ be the midpoint of $PQ$. Then, we have that $\angle AIM = \angle NHD$ so we just want $\angle NHD + \angle AIX = 180^\circ$. But $\angle AIX = 90^\circ + \angle OIX$ and $\angle NHD = 90^\circ - \angle HND$. Thus, we just want that $\angle HND = \angle OIX$
Taking homothety with dilation factor $+2$ from $I$, we have $X$ going to $K'$ which is now the intersection of $PQ$ and the line through the antipode of $I$ in $IPQ$ and the point $R$ on $(IPQ)$ such that $AR$ is tangent to $(IPQ)$.
Now let $HN \cap (IPQ) = S_1, S_2$ where $S_2$ is the antipode of $I$ and let $H'$ be the reflection of $H$ in $PQ$.
Now, we just want $\angle HND = \angle HS_2H' = \angle S_1IH$. Thus, we want $\angle S_1IH = \angle OIK'$ or that $IK'$ and $IS_1$ are isogonal in $\angle QIP$.
Now, performing $\sqrt{bc}$ and reflection in $\triangle IPQ$, we get that $S_1$ and $K'$ interchange. Thus, $IK'$ and $IS_1$ are isogonal in $\angle QIP$ as required. Thus, we are done. $\square$

Let $O$ be the circumcenter of $\triangle PIQ$, $S$ be the point on circumcircle of $\triangle PIQ$ so that $IS \perp AJ$, $R$ be the antipode of $I$ in the circumcircle of $\triangle PIQ$. I will use the following:
Lemma 1.1 Let $T$ be a point on the line through $I$ perpendicular to $AI$. Then $(IB, IC; IT, IM) = -1$.
Proof. Let $I_B, I_C$ be the $B, C$-excenters. Then we can project onto $I_B I_C$ and reduce to showing that $M$ is the midpoint of $I_B I_C$ which is known. $\square$
Now, we begin investigating the problem at hand:
Lemma 1.2 $IO \perp AI$.
Proof. This is direct: $\angle AIP = 90^\circ - B/2$ and $\angle PIO = 90^\circ - \angle PQO = 90^\circ - (90^\circ - B/2) = B/2$. $\square$
Proposition 1.3 $S, R, K$ are collinear.
Proof. Let the circumcircle of $\triangle PIQ$ be denoted by $\omega$. From Lemma 1.2, $AI$ is tangent to $\omega$. Therefore, $IS$ is the $I$-symmedian of $\triangle PIQ$. Hence, $(P, Q; S, I) = -1$. Projecting from $R$, we get $(RP, RQ; RS, RI) = -1$, hence if $RI \cap PQ = X$ and $RS \cap PQ = K'$, then $(P, Q; X, K') = -1$. On the other hand, we know $(IB, IC; IR, IM) = -1$ from Lemma 1.1, so projecting onto $PQ$ we get $(P, Q; X, K) = -1$. Thus $K = K'$ which finishes the proof. $\square$
Proposition 1.4 $AO$ bisects $IK$.
Proof. $RS \perp SI$ since $RI$ is the diameter, hence $RK \parallel AO$. But $O$ is the midpoint of $RI$, hence $AO$ bisects $IK$ by midpoint theorem. $\square$

Solution C
Let $J$ be the $A$-excentre of $\triangle ABC$ and $L$ the point on $BC$ with line $LJ$ is tangent to the $A$-excircle. Let $N$ be the midpoint of arc $BC$ not containing $A$ of the circumcircle of $ABC$. Let $S$ be the point on $BC$ with $\angle ANS = 90^\circ$. Let $T$ be the point on $BC$ with $TN \perp MI$.
Claim. $S$ is the midpoint of $LT$.

Proof. Note that by the Incenter-Excenter Lemma, $N$ is the midpoint of $IJ$ so $JL \perp AI$ and by standard mixtilinear facts, $TI \perp AI$. Thus, $L, S, T$ are obtained by drawing perpendicular lines to line $AI$ from the points $J, N, I$ respectively and intersecting with line $BC$. Since $N$ is the midpoint of $JI$, then $S$ is the midpoint of $LT$ as desired. $\square$
Now observe that $\triangle IPQ \sim \triangle JBC$ so the spiral similarity mapping $IPQ$ to $JBC$ consists of a 90° rotation as $PQ \perp BC$. Since $AI$ is tangent to $(PIQ)$ and $A$ lies on $PQ$, the spiral similarity sends $A$ to $L$ and the line passing through $A$ and the circumcentre of $PIQ$ to the line $LN$. So it suffices to show that the line through $A$ perpendicular to $LN$ bisects $IK$.
Thus we want cross ratio of lines $AP, AI$, line through $A$ parallel to $IK$ and line through $A$ perpendicular to $LN$ form a harmonic bundle. Taking directions perpendicular to this pencil from $N$, we get the four lines $N\infty_{BC}$ (as $BC \perp AP$), $NS$ (as $NS \perp AI$), $NL$ and the line $NT$ as $NT \perp KI$. Projecting this pencil at $N$ to $BC$ we get that this cross ratio equals since $(\infty_{BC}S; LT) = -1$ as $S$ is the midpoint of $LT$. This concludes the proof. $\square$

Solution D
Let $D, E, F$ be the foot of perpendiculars from $I$ to $BC, CA, AB$ respectively. Let $N$ be the midpoint of $BC$. Suppose the foot of perpendicular from $N$ onto $EF$ is $L$. We use that $I, L, M$ are collinear. This can be proven in many ways: one sketch of the argument goes as follows: If $X$ (resp. $Y$) is the foot of perpendicular of $B$ (resp. $C$) onto $CI$ (resp. $BI$), then $X, Y$ lie on $EF$, and $N$ is the midpoint of $XY$. On the other hand, if $I_B, I_C$ are the $B$-, $C$-excenters, then $M$ is the midpoint of $I_B I_C$. The result then follows on noting that $XY \parallel I_B I_C$, and considering a homothety at $I$.
Now toss the diagram on the complex plane. Let the incircle of $\triangle ABC$ be the unit circle, so $I$ is the origin, and let $D = 1$, $E = e$, $F = f$ without loss of generality.
Now, it is well known that the intersection of tangents to the unit circle at points $x, y$ is $\frac{2xy}{x+y}$.
Therefore, $A = a = \frac{2ef}{e+f}$, $B = b = \frac{2f}{f+1}$, $C = c = \frac{2e}{e+1}$. Thus, the midpoint of $BC$ is $N = \frac{e}{e+1} + \frac{f}{f+1}$.
Hence, the foot of perpendicular from $N$ to $EF$ is
$$L = l = \frac{1}{2} \left( e + f + \frac{e}{e+1} + \frac{f}{f+1} - ef \left( \frac{1}{e+1} + \frac{1}{f+1} \right) \right) = \frac{(e+f)(e+f+2)}{2(e+1)(f+1)}.$$
We know $D = 1$, therefore the line perpendicular to $BC$ from $A$ is the set of points $z$ with $\operatorname{Im}(z) = \operatorname{Im}(A)$, i.e. $z - \bar{z} = \frac{2(ef - 1)}{e + f}$.
Now, $P$ lies on $BI$, therefore $p = \lambda b$ for some $\lambda \in \mathbb{R}$. Note that $\bar{b} = \frac{1}{f} \cdot b$, therefore, $\bar{p} = \frac{1}{f} \cdot p$. Thus,
$$p - \bar{p} = \frac{(f-1)}{f} \cdot p \implies p = \frac{2f(ef-1)}{(e+f)(f-1)} \cdot \text{Similarly, } q = \frac{2e(ef-1)}{(e+f)(e-1)}.$$
With the same idea, one checks that $\bar{l} = \frac{(e+f)(2ef+e+f)}{2ef(e+1)(f+1)} = l \cdot \frac{(2ef+e+f)}{ef(e+f+2)}$, therefore,
$$k - \bar{k} = \frac{(ef - 1)(e + f)}{ef(e + f + 2)} \cdot k \implies k = \frac{2ef(e + f + 2)}{(e + f)^2}.$$
We know that the circumcenter of a circle passing through origin, $p, q$ is given by $\frac{pq(\bar{p} - \bar{q})}{\bar{p}q - p\bar{q}}$. Note that $p = \frac{2(ef - 1)}{(e+f)} \cdot p_1$, and $q = \frac{2(ef - 1)}{(e+f)} \cdot q_1$ where $p_1, q_1$ are $\frac{f}{f-1}, \frac{e}{e-1}$ respectively. Thus the circumcenter of $\triangle IPQ$ is $\frac{2(ef - 1)}{(e+f)} \cdot t$ where $t$ is the circumcenter of $\triangle IP_1Q_1$. This simplifies calculation, and we get
$$\text{The circumcenter of } \triangle IPQ = \frac{2(ef-1)}{(e+f)} \cdot \frac{\frac{ef}{(e-1)(f-1)} \left( \frac{1}{1-f} - \frac{1}{1-e} \right)}{\frac{e}{(1-f)(e-1)} - \frac{f}{(1-e)(f-1)}} = \frac{2ef(ef-1)}{(e+f)(e-1)(f-1)}.$$
Thus, we need to show that $\frac{2ef}{(e+f)}$, $\frac{2ef(ef-1)}{(e+f)(e-1)(f-1)}$, $\frac{ef(e+f+2)}{(e+f)^2}$ are collinear. Multiplying by $\frac{(e+f)}{2ef}$, we reduce to proving 1, $\frac{ef-1}{(e-1)(f-1)}$, $\frac{e+f+2}{2(e+f)}$ are collinear for which we consider:
$$\frac{\frac{ef-1}{(e-1)(f-1)} - 1}{\frac{e+f+2}{2(e+f)} - 1} = \frac{\frac{e+f-2}{(e-1)(f-1)}}{\frac{2-e-f}{2(e+f)}} = -\frac{2(e+f)}{(e-1)(f-1)}$$
The above quantity may be easily checked to equal the conjugate, so we are done.