IMO TST Day 2

Assume $AB \neq AC$ for all solutions as in the case that $AB = AC$, $X, Y$ are symmetric about $AD$ and the result is immediate.
Let $M$ be the midpoint of $BC$. Let $\omega = (BC)$ and $\Omega = (ABC)$.
We prove some preliminary results here which will be freely used across different proofs.
Let $T$ be the intersection of the $B$ and $C$ tangents to $\Omega$.

Claim 0.1 (BKOCT) is cyclic.
Proof. Let $T$ be a point such that $T$ is the intersection of the tangents from $B$ and $C$ to $(ABC)$. Now, observe that from the given angle conditions, that $\angle BKC = 180^\circ - A = \angle BOC$ where $O$ is the center of $\Omega$. Also, $\angle BTC = A$, thus, $(BKOCT)$ is cyclic. $\square$

Claim 0.2 The point $K$ lies on the $A$ symmedian in $\triangle ABC$.
Proof. From Claim 0.1 $\angle TKC = A$. Also, from the given angle conditions, we have that $\angle AKC = 180^\circ - A$. Thus, $A, K, T$ are collinear. This gives the desired result. $\square$

Claim 0.3 $\angle AKO = 90^\circ$.
Proof. As $\angle OKT = 90^\circ$, we have $\angle AKO = 90^\circ$. $\square$

Claim 0.4 Reflection of $A$ in $K$ lies on $\Omega$.
Proof. Follows immediately from Claim 0.3. $\square$

Let $K'$ be the reflection of $A$ in $K$, we have proved that $K' \in \Omega$.

Claim 0.5 The reflection of $K'$ in $BC$ lies on the $A$-median of $\triangle ABC$.
Proof. Let $K''$ be the point on the $A$ median such that $MA \cdot MK'' = MB^2$.
Thus, we get that
$$\angle K'CB = \angle K'AB = \angle MAC = \angle MCK'' = \angle BCK''$$
Similarly, $\angle K'BC = \angle CBK''$. Thus, $K''$ is the reflection of $K'$ in $BC$ as required. $\square$

We call the point $K''$, the $A$-HM point in $\triangle ABC$.

Claim 0.6 If $S$ is the intersection of the $A$ tangent with $BC$, then $\angle SKA = 90^\circ$.
Proof. Since $AK$ is the symmedian, the tangents from $S$ to $\Omega$ are at $A, K'$. Thus, $S, K, O$ are collinear as this is the perpendicular bisector of $AK'$. Thus, we get the desired result. $\square$

Let $D'$ be the point such that $(BC; DD') = -1$, or equivalently if $E, F$ are feet of perpendiculars from $B, C$ onto $AB$ and $AC$ respectively then $D' = EF \cap BC$.

Claim 0.7 $D'K'' \perp AM$
Proof. Observe that
$$\angle MDA = 90^\circ, \ MD \cdot MD' = MB^2 = MK'' \cdot MA \implies \angle MK''D' = 90^\circ$$
$\square$

Solution Path A
We wish to show that $D$ lies on the $A$-Appolonian circle in $\triangle XAY$. Thus, it suffices to show that there exists point $P$ on line $XY$ with $PA^2 = PX \cdot PY = PD^2$ since such a point $P$ would be the center of the $A$-Appolonian circle in $\triangle XAY$, and if $PA^2 = PD^2$, then $D$ would be on the Appolonian circle as well.

Solution A1
We will show that the perpendicular bisector of $AD$, the line $XY$ and the radical axis of point circle $\odot(A)$ and $\odot(BC)$ concur. The point of concurrence is the required point $P$. The latter radical axis is simply the midline of $A$ with its polar in $\odot(BC)$.
Claim 1 $A, K''$ are inverses in $\omega$.
Proof. Same as Claim 0.5 in Solution Preliminary
$\square$
By homothety at $A$ of ratio $+2$, the conclusion is same as showing that line $BC$ and the polar of $A$ in $\odot(BC)$, and the line through $K'$ perpendicular to $A'K'$ concur. However, the polar is simply the line through $K''$ perpendicular to $AK''$; the reflection of the other line in $BC$, and the concurrence is obvious.
$\square$

Solution A2
We claim that $P$ is the midpoint of $AD'$.
Claim 1 It is sufficient to prove that $\angle PKD = 90^\circ$.
Proof. Observe that $MB \cdot MC = MD \cdot MD'$ and thus $(ADD')$ and $(BC)$ are orthogonal. Thus, if $P, X$ and $Y$ are collinear, then $\text{pow}_\omega(P) = PX \cdot PY$ but since $(ADD')$ and $(BC)$ are orthogonal, we have that $\text{pow}_\omega(P) = PA^2$. Thus, we get that $PA^2 = PX \cdot PY = PD^2 = PD'^2$, and we will be done.
Thus, we just want $P, X, Y$ collinear that is $\angle PKD = 90^\circ$.
$\square$
$\square$
Taking homothety $+2$ from $A$, we get that if $A'$ is the reflection of $A$ in $BC$ and if $K'$ is reflection of $A$ in $K$, then we need that $\angle D'K'A' = 90^\circ$.
Reflecting in $BC$, this is the same as proving $\angle AK''D' = 90^\circ$ which is the same as Claim 0.7 in the preliminaries. Thus, we are done.
$\square$

Solution A3
The following proof is due to ideas from Adhitya MV in the TST. The proof has been revised and rewritten by the Problem Selection Committee.
We claim that $P$ is the midpoint of $AD'$.

Claim 1 It is sufficient to prove that $\angle PKD = 90^\circ$.
Proof. Same as Claim 1 in Solution A2
Now, let $R = KP \cap AD$. Now, it suffices to prove that $\angle RKD = 90^\circ$. Now, $\angle SKA = 90^\circ$ by Claim 0.6 from preliminaries. Thus, we just need $\angle RKA = \angle SKD$. But $(SDKA)$ is cyclic. Thus, $\angle SKD = \angle SAD = \angle SAR$.
Thus, we just need to show that circles $\Omega$ and $(SAK)$ are tangent. Thus, we just need that reflection of $A$ in $R$ is on $\Omega$. Taking homothety $+2$ from $A$, we need $AD \cap K'D' \in \Omega$.
Let $R' = AD \cap \Omega$.
Now, projecting $(BC; DD')$ onto $\Omega$ from $R'$, we get that $(B, C; A, D'R' \cap \Omega) = -1$ but $(B, C; A, K') = -1 \implies D'R' \cap \Omega = K'$. Thus, $R'$ is the reflection of $A$ in $R$ as required.

We claim that $P$ is the midpoint of $AD'$.
Claim 1 It is sufficient to prove that $\angle PKD = 90^\circ$.
Proof. Same as Claim 1 in Solution A2
Now, reflecting figure in $BC$ and taking homothety $\frac{1}{2}$ from $A$, we have $\triangle ABC \mapsto \triangle DLN$, $K'' \mapsto K$ and $D' \mapsto P$. Thus, $K$ is the $D$-HM point in $\triangle DLN$ and $P$ is the point such that $(K, L; P, AD \cap LN) = -1$. But then by Claim 0.7 from preliminaries, we get that $\angle PKD = 90^\circ$ and we are done.

The following proof is due to ideas from Atul Shatavart Nadig in the TST. The proof has been revised and rewritten by the Problem Selection Committee.
We claim that $P$ is point of intersection of $XY$ and perpendicular bisector of $AD$.
Let $Q$ be the foot of perpendicular from $P$ onto $BC$, and let $N$ be the midpoint of $AM$. Since $K''$ lies on $AM$ by Claim 0.5, taking reflection in $BC$, we get $A', K'$, $M$ collinear, and hence taking homothety with center $A$ and ratio $\frac{1}{2}$, we get that $D, K, N$ are collinear.
We already know $PA = PD$. Note that $PX \cdot PY = \text{pow}_\omega(P) = PM^2 - MB^2$. So it suffices to show
$$PD^2 = PM^2 - MB^2 \\ \iff MB^2 = PM^2 - PD^2$$
By the Pythagorean Theorem, since $PQ \perp DM$,
$$\begin{aligned} PM^2 - PD^2 &= QM^2 - QD^2 \\ &= (QM - QD)(QM + QD) \\ &= DM \cdot (2QD + DM) \end{aligned}$$
Let $x = QD$. Thus we need to show
$$\begin{aligned} MB^2 &= DM \cdot (2x + DM) \\ \iff x &= \frac{MB^2 - DM^2}{2DM} \\ \iff x + \frac{DM}{2} &= \frac{MB^2}{2DM} \end{aligned}$$
Let $R$ be the midpoint of $AD$. By midpoint theorem, $RN = \frac{DM}{2}$, so $x + \frac{DM}{2} = PR + RN = PN$. So it suffices to prove that
$$PN = \frac{MB^2}{2DM}$$
Let $\angle PND = \angle NDM = \angle NMD = \theta$. Then
$$PN \cos \theta = NK = \frac{MK'}{2} \implies PN = \frac{MK'}{2 \cos \theta}$$

But since $K'$ is the A'-HM point in $\triangle ABC$, $MK' \cdot MA' = MB^2$. So,
$$PN = \frac{MB^2}{2MA' \cos \theta} = \frac{MB^2}{2MA \cos \theta}$$
But in $\triangle AMD$, $\angle D = 90^\circ$ and $\angle AMD = \theta$, so $\cos \theta = \frac{DM}{AM}$. Therefore
$$PN = \frac{MB^2}{2MA \cdot \frac{DM}{AM}} = \frac{MB^2}{2DM}$$
as required, so we are done. $\square$

## Solution B
For the following two solutions, we begin by inverting from $A$ with radius $\sqrt{|AB||AC|}$ and reflecting across the $A$ angle bisector. Let this transformation be $\Phi$ and we write $Z \mapsto Z^*$ for any point $Z$ under this transformation.
Observe the following maps:
$$\bullet B \mapsto C$$
$$\bullet C \mapsto B$$
$\bullet K \mapsto K^*$ such that $K^*BAC$ is a parallelogram.
$$\bullet X \mapsto X^*$$
$$\bullet Y \mapsto Y^*$$
$\bullet D \mapsto D^*$ such $D^*$ is the antipode of $A$ in $\Omega$.
$\bullet \omega \mapsto \omega^*$ where $\omega^*$ is centered at $T$ and passes through $B, C$.
Now,
$$AX \cdot DY = AY \cdot DX \iff \frac{AB \cdot AC}{AX^*} \cdot \frac{AB \cdot AC \cdot D^*Y^*}{AD^*AY^*} = \frac{AB \cdot AC}{AY^*} \cdot \frac{AB \cdot AC \cdot D^*X^*}{AD^*AX^*} \iff X^*D^* = Y^*D^*$$
Let $(AX^*Y^*K^*) = \omega_1$. Let $O_1$ be the center of $\omega_1$. Now, we have that $XY$ is the common chord of $\omega_1$ and $\omega^*$. Thus, we just want that $D^* \in O_1T$.
Now, we know that $XY \perp DK$, thus $\omega^*$ and $(AD^*K^*)$ are orthogonal. Thus, $O_1$ is the intersection of the $D^*$ and $K^*$ tangents to $(AD^*K^*)$. Thus, we want that $D^*T$ is the symmedian in $\triangle AD^*K^*$. This is equivalent to $\angle AD^*M = \angle K^*D^*T$.

### Solution B1
We relabel the points, call $D^*$, $A'$ and call $K^*K'$.
Now, we just want $\angle AA'T = \angle K'A'M$. Let $H$ be the orthocenter of $\triangle ABC$. Then $\angle K'A'M = \angle AHM$ by reflection over $M$. Taking homothety half from $A'$, $\angle AHM = \angle OMA'$.
Thus, we just want $\angle OMA' = \angle OA'T$ but this simply follows from the fact that $OM \cdot OT = OA'^2$. $\square$

The following proof is due to ideas from Vedant Saini in the TST. The proof has been revised and rewritten by the Problem Selection Committee.
We now provide a complex bash solution to the required angle condition.
Let $A, B, C$ be on the unit circle. Then $m = \frac{b+c}{2}$, $k^* = b+c-a$, $d^* = -a$. $t = \frac{2bc}{b+c}$.

We now want that $\frac{a-d^*}{m-d^*} \cdot \frac{k^*-d^*}{t-d^*} \in \mathbb{R}$.
$$\begin{aligned} \frac{a-d^*}{m-d^*} \cdot \frac{k^*-d^*}{t-d^*} &= \frac{2a}{\frac{b+c+2a}{2}} \cdot \frac{b+c}{\frac{2bc+ab+ac}{b+c}} = \frac{4a(b+c)^2}{(b+c+2a)(2bc+ab+ac)} = \frac{\frac{4a(b+c)^2}{a^2b^2c^2}}{\frac{(2bc+ab+ac)(2a+b+c)}{a^2b^2c^2}} \\ \\ \frac{a-d^*}{m-d^*} \cdot \frac{k^*-d^*}{t-d^*} &= \frac{\frac{4}{a}\left(\frac{1}{b} + \frac{1}{c}\right)^2}{\left(\frac{2}{a} + \frac{2}{b} + \frac{2}{c}\right)\left(\frac{2}{bc} + \frac{1}{ab} + \frac{1}{ac}\right)} = \overline{\left(\frac{a-d^*}{m-d^*} \cdot \frac{k^*-d^*}{t-d^*}\right)} \end{aligned}$$
This gives the desired result and we are done. $\square$

## Solution C
The following proof is due to ideas from Bhavya Tiwari in the TST. The proof has been revised and rewritten by the Problem Selection Committee.
This is a coordinate bash solution. Take a coordinate system with the midpoint $M$ of $BC$ as the origin, $B = (-1, 0)$, $C = (1, 0)$ and $A = (d, h)$. Then we get $D = (d, 0)$.
Claim 1 The point $K$ has coordinates
$$\left( \frac{d}{2} \left( \frac{c^2+1}{c^2} \right) , \frac{h}{2} \left( \frac{c^2-1}{c^2} \right) \right)$$
where $c^2 = d^2 + h^2$.
Proof. We use complex numbers for this. Let $k, a, b, c$ denote the complex coordinates of $K, A, B, C$ respectively. Note that since $\triangle KBA \sim \triangle KAC$ by AA test, $K$ is the center of spiral similarity sending $BA$ to $AC$. Hence,
$$\begin{aligned} \frac{k-a}{k-b} &= \frac{k-c}{k-a} \\ \implies k^2 - 2ak + a^2 &= k^2 - (b+c)k + bc \\ \implies k &= \frac{bc-a^2}{b+c-2a} \end{aligned}$$
Putting $b=1$, $c=-1$ and $a=d+hi$, we get
$$\begin{aligned} k &= \frac{a^2 + 1}{2a} \\ &= \frac{a|a|^2 + \bar{a}}{2|a|^2} \\ &= \frac{d(|a|^2 + 1) + h(|a|^2 - 1)i}{2|a|^2} \end{aligned}$$
which gives us the required coordinates since $c^2 = |a|^2$ $\square$

Slope of line $DK$ is:
$$\frac{\frac{h}{2} \left( \frac{c^2-1}{c^2} \right)}{\frac{d}{2} \left( \frac{c^2+1}{c^2} \right) - d} = -\frac{h}{d}$$
Therefore slope of $XY$ is $\frac{d}{h}$ because $XY \perp DK$. Therefore, since $K \in XY$, the equation of line $XY$ is:
$$\begin{aligned} y - \frac{h}{2} \left( \frac{c^2 - 1}{c^2} \right) &= \frac{d}{h} \left( x - \frac{d}{2} \left( \frac{c^2 + 1}{c^2} \right) \right) \\ \implies y &= \frac{d}{h}x + \frac{c^2(h^2 - d^2) - h^2 - d^2}{2hc^2} \\ \implies y &= \frac{d}{h}x + \frac{h^2 - d^2 - 1}{2h} \end{aligned}$$

The problem is equivalent to proving $\frac{AX}{DX} = \frac{AY}{DY}$. We will in fact prove that $\frac{AX}{DX} = \sqrt{2} = \frac{AY}{DY}$. We will only prove the first equality, as the other one follows from symmetry. Let $X = (x_1, x_2)$. Note that $X$ satisfies equation of line $XY$. Thus:
$$\begin{aligned} \frac{AX}{DX} &= \sqrt{2} \\ \Leftrightarrow \frac{AX^2}{DX^2} &= 2 \\ \Leftrightarrow \frac{(x_1 - d)^2 + (x_2 - h)^2}{(x_1 - d)^2 + h^2} &= 2 \\ \Leftrightarrow x_1^2 - 2dx_1 + d^2 + x_2^2 - 2x_2h + h^2 &= 0 \end{aligned}$$
Putting $x_2$ obtained from equation of $XY$ into the above, we get that it is equivalent to:
$$\begin{aligned} x_1^2 - 2dx_1 + d^2 + \left(\frac{d}{h}x_1 + \frac{h^2 - d^2 - 1}{2h}\right)^2 &+ 2h\left(\frac{d}{h}x_1 + \frac{h^2 - d^2 - 1}{2h}\right) - h^2 = 0 \\ \Leftrightarrow \left(1 + \frac{d^2}{h^2}x_1\right) + \frac{d(h^2 - d^2 - 1)}{h^2}x_1 &+ \left(\frac{h^2 - d^2 - 1}{2h}\right)^2 - 1 = 0 \\ \Leftrightarrow x_1^2 + \left(\frac{d^2}{h^2}x_1 + 2\frac{d}{h}\left(\frac{h^2 - d^2 - 1}{2h}\right)x_1 + \left(\frac{h^2 - d^2 - 1}{2h}\right)^2\right) &= 1 \\ \Leftrightarrow x_1^2 + \left(\frac{d}{h}x_1 + \frac{h^2 - d^2 - 1}{2h}\right)^2 &= 1 \\ \Leftrightarrow x_1^2 + x_2^2 &= 1 \end{aligned}$$
because $(x_1, x_2)$ satisfies equation of $XY$. But the final equation is precisely the equation of the circle with diameter $BC$, which is satisfied by the points $XY$. Hence, all the implications holds, and we are done. $\square$

## Supplementary
We can in fact prove that $\frac{AX}{DX} = \frac{AY}{DY} = \sqrt{2}$. Let $U, V$ be points on $BC$ such that $DU = DA = DV$. Observe that the circle centered at $A'$ (reflection of $A$ in $BC$) passing through $U, V$ is Appolonius circle for ratio $\sqrt{2}$ wrt $A, D$. Let this circle be $\omega_1$.
Thus, we just want $X, Y$ to be on this circle. But $X, Y$ also lies on $(BC)$. Observe that $A'M \parallel KD$ as proven in Claim 0.5 of the preliminaries.
Thus, the line joining the centers of $\omega_1$ and $(BC)$ is perpendicular to $XY$. Thus, we if can find a point $P$ on $XY$ such that $\text{pow}_{\omega_1}(P) = \text{pow}_{(BC)}(P)$, we will be done, since the radical axis of $(BC)$ and $\omega_1$ is parallel to $XY$.
Observe that if we consider the point circle at $A$, $\omega_1$ and $(BC)$, their radical center is the intersection of the $A$-midline of $\triangle ABC$ and the midline of the polar of $A$ in $(BC)$. Thus, we just want the perpendicular bisector of $AD, XY$ and the midline of the polar of $A$ in $(BC)$ to concur.
This is exactly what we show in Solution A1! Thus, we are done. $\square$