Cute! but i think this problem would be hard for imo p2 and easy for imo p3

Rumour has it that this problem was too good to be wasted on the IMO.

Since $O_1A_1\parallel O_2A_2$ and $O$ is the center of negative homothety mapping $\gamma_2$ to $\gamma_1$, if we let $X'$ be the second intersection of $A_2O$ with $\gamma_1$, we get that $X'$ is the antipode of $A_1$ in $\gamma_1$, thus implying that $A_1XYA_2$ is cyclic (and has diameter $A_1A_2$). Let $\{R\}=A_1X\cap A_2Y$. $O$ is the orthocenter of $\triangle RA_1A_2$. Let us attach a Cartesian coordinate system with origin in the foot of the perpendicular from $R$ to $A_1A_2$, $R$ on the $y$ axis and $A_1A_2$ the $x$ axis. Let $R=(0,a)$, $A_1=(b,0)$ and $A_2=(c,0)$. We get $O=(0,-\frac{bc}{a})$. $Y$ lies at the intersection of the lines $RA_1$ and $OA_2$. Thus, the coordinates of $Y$ are the solutions of $$\begin{cases}\frac{x}{b}+\frac{y}{a}=1\\ \frac{x}{c}-\frac{y}{\frac{bc}{a}}=1\end{cases}$$and so $Y=\left(\frac{b(a^2+bc)}{a^2+b^2},\frac{ab(b-c)}{a^2+b^2}\right)$. A similar expression is obtained for $X$ by swapping $b$ and $c$. Let $r_1$ be the radius of $\gamma_1$. Since $O_1=(b,r_1)$ and $O_1Y=r_1$ we have $$r_1^2=(x_Y-b)^2+(y_Y-r_1^2)\iff r_1=\frac{(x_Y-b)^2+y_Y^2}{2y_Y}=\frac{b(b-c)}{2a}.$$Of course, a similar expression is for $r_2$ as well. Let $\{K\}=O_1B_1\cap RO$ and $\{L\}=O_2C_2\cap RO$. By Thales's Theorem, the conclusion is equivalent to $\frac{AK}{r_1}=\frac{AL}{r_2}$. If $\alpha_1$ is the slope of $OB_1$, line $O_1B_1$ has slope $-\frac{1}{\alpha_1}=\frac{y_K-r_1}{0-b}$. Using the analogous relation for $y_L$, we need to show that $\frac{b}{r_1\alpha_1}-\frac{a}{r_1}=\frac{c}{r_2\alpha_2}-\frac{a}{r_2}$ or equivalently, $$\frac{1}{\alpha_1}+\frac{1}{\alpha_2}=\frac{a}{b}+\frac{a}{c}.$$ Suppose lines $OB_1$ and $OC_2$ have equations $y=\alpha_{1,2} x -\frac{bc}{a}$. The distance from $O_1$ to this line is $r_1$, so using the distance formula we have $$r_1=\frac{|\alpha_1 b - r_1 -\frac{bc}{a}|}{\sqrt{\alpha_1^2+1}}.$$Substituting the expression for $r_1$ and squaring gives $$((b-c)^2-4a^2)\alpha_1^2 +4a(b+c)\alpha_1 -4bc =0$$(which is the same for $\alpha_2$ as well). Thus, $$\frac{1}{\alpha_{1,2}}=\frac{a(b+c)\pm\sqrt{a^2(b+c)^2+bc((b-c)^2-4a^2)}}{2bc},$$from which the conclusion follows as described in the previous paragraph.

Let's complex bash this one. Identify $\gamma_1$ with the unit circle, and let $a,b,c$ be the coordinates of points $A_1,B_1,C_1$, so that $$|a|=|b|=|c|=1$$$$o_1 = 0$$$$o = \frac{2bc}{b+c}$$Now we find the coordinate of $O_2$. Since $O_2$ lies on the line through $O_1$ and $O$, we have $$\overline{o_2} = \frac{o_2}{bc}$$Since $O_2$ is also the same distance from the tangent at $A_1$ as it is from the tangent at $B_1$, we have $$\frac12\left|o_2-2a+a^2\overline{o_2}\right| = \frac12\left|o_2-2b+b^2\overline{o_2}\right|$$$$|a^2o_2+bco_2-2abc|^2 = |bo_2+co_2-2bc|^2$$$$(a^2o_2+bco_2-2abc)\left(\frac{o_2}{a^2bc}+\frac{o_2}{b^2c^2}-\frac2{abc}\right) = (bo_2+co_2-2bc)\left(\frac{o_2}{b^2c}+\frac{o_2}{bc^2}-\frac2{bc}\right)$$$$(a^2o_2+bco_2-2abc)^2 = a^2(bo_2+co_2-2bc)^2$$$$a^2o_2+bco_2-2abc = \pm(abo_2+aco_2-2abc)$$If the $\pm$ is a $+$, we just have $o_2 = 0 = o_1$, which is to be expected since $O_1$ is one of the two points that is equidistant from these three tangent lines. $O_2$ is the other, so we have $$a^2o_2+bco_2-2abc = -abo_2-aco_2+2abc$$and thus $$o_2 = \frac{4abc}{(a+b)(a+c)}$$Then $$a_2 = \frac{2a+o_2-a^2\overline{o_2}}2 = \frac{a(ab+ac+3bc-a^2)}{(a+b)(a+c)}$$$$b_2 = \frac{2b+o_2-b^2\overline{o_2}}2 = \frac{b(a^2+3ac-ab+bc)}{(a+b)(a+c)}$$$$c_2 = \frac{2c+o_2-c^2\overline{o_2}}2 = \frac{c(a^2+3ab-ac+bc)}{(a+b)(a+c)}$$Now we consider $X$. We have $|x|=1$ (since $X$ lies on $\gamma_1$) and also $A_2, O, X$ are collinear. Now $$a_2-o = \frac{a^2b^2+a^2c^2+ab^2c+abc^2-a^3b-a^3c-2b^2c^2}{(a+b)(a+c)(b+c)} = -\frac{(a-b)(a-c)(ab+ac+2bc)}{(a+b)(a+c)(b+c)}$$So we have $$\frac{x-o}{a_2-o} \in \mathbb{R}$$$$\frac{(a+b)(a+c)(bx+cx-2bc)}{(a-b)(a-c)(ab+ac+2bc)} = \frac{a(a+b)(a+c)(b+c-2x)}{x(a-b)(a-c)(2a+b+c)}$$$$x(bx+cx-2bc)(2a+b+c) = a(ab+ac+2bc)(b+c-2x)$$$$x^2(b+c)(2a+b+c) - 2bcx(2a+b+c) + 2ax(ab+ac+2bc) - a(b+c)(ab+ac+2bc) = 0$$$$x^2(b+c)(2a+b+c) + 2x(b+c)(a^2-bc) - a(b+c)(ab+ac+2bc) = 0$$$$x^2(2a+b+c) + 2x(a^2-bc) - a(ab+ac+2bc) = 0$$$$(x+a)\left[x(2a+b+c) - (ab+ac+2bc)\right] = 0$$Now $x\neq -a$, because $A_1,B_1,C_1$ all lie on the same semicircle of $\gamma_1$ and the point with coordinate $-a$ lies opposite that semicircle, and thus satisfies $OX > OB_1$. Thus $$x = \frac{ab+ac+2bc}{2a+b+c}$$Let the perpendicular to $OX$ at $X$ intersect $\gamma_1$ again at $X'$, so that $$x' = -\frac{x-o}{x\overline{o}-1} = -\frac{\frac{ab+ac+2bc}{2a+b+c}-\frac{2bc}{b+c}}{\frac{2(ab+ac+2bc)}{(b+c)(2a+b+c)}-1} = -\frac{a(b-c)^2}{-(b-c)^2} = a$$Then $$p = \frac{ax(b-b) - b(-b)(a+x)}{ax-b(-b)} = \frac{b^2(a+x)}{b^2+ax} = \frac{2b^2(a+b)(a+c)}{(a+b)^2(b+c)} = \frac{2b^2(a+c)}{(a+b)(b+c)}$$Now define the affine transform $$z^* = \frac{z(a+b)(a+c) - 4abc}{(a-b)(a-c)}$$which sends $\gamma_2$ onto the unit circle: \begin{align*} o_2^* &= 0 \\ a_2^* &= -a \\ b_2^* &= b \\ c_2^* &= c \end{align*}Then by symmetry, we have $$q^* = \frac{2(c^*)^2(a^*+b^*)}{(a^*+c^*)(b^*+c^*)} = \frac{2c^2(a-b)}{(a-c)(b+c)}$$and so $$q = \frac{q^*(a-b)(a-c) + 4abc}{(a+b)(a+c)} = \frac{2c^2(a-b)^2 + 4abc(b+c)}{(a+b)(a+c)(b+c)} = \frac{2c(a^2c+2ab^2+b^2c)}{(a+b)(a+c)(b+c)}$$Then we compute $$p-q = \frac{2b^2(a+c)^2 - 2c(a^2c+2ab^2+b^2c)}{(a+b)(a+c)(b+c)} = \frac{2(a^2b^2-a^2c^2)}{(a+b)(a+c)(b+c)} = \frac{2a^2(b-c)}{(a+b)(a+c)}$$$$a-a_2=\frac{2a(a^2-bc)}{(a+b)(a+c)}$$and thus $$\frac{p-q}{a-a_2} = \frac{a(b-c)}{a^2-bc}$$which is real. $\blacksquare$

Let $A_2$, $D_2$ be the antipode of $A_1$, $D_1$ on their corresponding circles. Note that $O$ is the insimilicenter of $\gamma_1$ and $\gamma_2$ (the center of a homotethy $\theta$ with a negative factor mapping $\gamma_1 \mapsto \gamma_2$). Let $r_1$ and $r_2$ be the radius of the two circles. Since \[ \frac{D_1O_1}{O_1O} = \frac{r_1}{O_1O} = \frac{r_2}{O_2O} = \frac{A_2O_2}{O_2O}, \]we have $D_1$, $O$, $A_2$ are collinear, where $X$ is also on this line. Similarly, $A_1$, $O$, $Y$, $D_2$ are collinear. Let $C \equiv A_1X \cap A_2Y$. We have $A_1X \perp XD_1 \equiv XO$, and similarly $A_2Y \perp A_1Y$, so that $O$ is the orthocenter of $\triangle CA_1A_2$, i.e. $CO \perp A_1A_2$. This means, $A_1D_1 \parallel CO \parallel A_2D_2$. We need to prove $CP/PA_1 = CQ/QA_2$, which will solve the problem. Let $OA_2$ meet $\gamma_2$ for the second time at $M$. Note that the map $\theta$ maps the following pair of points: \[X \mapsto M, A \mapsto D_2, C \mapsto N = D_2M \cap OC. \]Therefore, $CP / PA_1 = NR / RD_2$. So, we need to prove: \[ \frac{NR}{RD_2} = \frac{CQ}{QA_2} \iff RQ \parallel A_2D_2. \]To prove the latter, we use the following lemma: Lemma Let $AB$ be a diameter of a circle $\Gamma$ with center $O$, $C$ be a point outside the circle, the tangents from $C$ to $\Gamma$ meet $\Gamma$ at $C_1$ and $C_2$. $CB$ and $CA$ meet $\Gamma$ for the second time at $D$ and $E$, respectively. Let $X = AD \cap OC_1$ and $Y = BE \cap OC_2$. Then, $XY \parallel AB$. Proof. Let $Z = BE \cap AD$. By Brokard, $Z$ is on the polar of $C$ w.r.t. $\Gamma$, which is $C_1C_2$, so $Z \in C_1C_2$. Let $T = BC_1 \cap AC_2$. By Pappus on $AOB$ and $C_1ZC_2$, we have $X$, $Y$, and $T$ are collinear. Next, $C$, $Z$, $T$ all lie on a line perpendicular to $AB$ (because they are located on the polar of the point $C_1C_2 \cap AB$ w.r.t. $\Gamma$). Let $CZ \cap AB = F$. We have: \[ BD \times BC = BF \times BA = BT \times BC_1, \]Which implies that $T$, $C_1$, $D$, $C$ are concyclic. But $XC_1 \perp C_1C$ and $XD \perp DC$, which means all five points $X$, $T$, $C_1$, $D$, $C$ lie on a circle with diameter $CX$. This means $CT \perp TX$, and similarly $CT \perp TY$, implying $XY \perp CT$, so $XY \parallel AB$, proving the lemma. $\blacksquare$ Finally, just apply the lemma by substituting $\Gamma \coloneq \gamma_2$, $C \coloneq O$, $AB \coloneq A_2D_2$. $\square$

Let $A_1'$ and $A_2'$ be the antipodal points of $A_1$ and $A_2$. First observe that $O$ is the center of the negative homothety swapping circles $\gamma_1$ and $\gamma_2$. Since this homothety rotates by $180^{\circ}$ we must have that $A_1$ and $A_2$ get mapped to $A_2'$ and $A_1'$. So $A_1$, $O$, $Y$, and $A_2'$ are collinear and $A_2$, $O$, $X$, and $A_1'$ are collinear. Define $T_1$ to be the intersection of the tangents at $A_1$ and $B_1$ on $\gamma_1$ and $T_2$ to be the intersection of the tangents at $A_2$ and $B_2$ on $\gamma_2$. We claim that $P$ is the orthocenter of triangle $OO_1T_1$. Notice that $\angle OB_1P = \angle OXP = 90^{\circ}$ so quadrilateral $OB_1PX$ is cyclic. As $O_1P \perp OT_1$ it suffices to show that $OP \perp O_1T_1$. $$180^{\circ} - \angle OPO_1 = \angle OPB_1 = \angle OXB_1 = \angle B_1A_1A_1' = 90^{\circ} - \angle T_1O_1P$$So $T_1P \perp O_1O_2$ and similarly $T_2Q \perp O_1O_2$. We claim that triangle $T_1PB_1$ and $T_2QC_2$ are congruent. Notice that $\angle T_1B_1P = \angle T_2C_2Q = 90^{\circ}$. Also think of $\gamma_1$ and $\gamma_2$ as excircles of $OT_1T_2$. Then it becomes apparent that $A_1T_1 = A_2T_2$ or that $T_1B_1 = T_2C_2$. To finish the claim simply note $$\angle B_1T_1P = \angle B_1O_1O = \angle B_2O_2O = \angle C_2O_2O = \angle C_2T_2Q$$ Now since $T_1P = T_2Q$ and $T_1P \parallel T_2Q$ we must have that $PQ \parallel T_1T_2$, as desired.

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