Proposed by Anant Mudgal

here's a sketch of my solution from the test

math_comb01 wrote: Proposed by Anant Mudgal

I hope people attending the Mumbai INMOTC paid attention in my Transformations in Geometry class. Here's a more detailed solution: Let $\tau_{123} = \tau_1 \circ \tau_2 \circ \tau_3 = \mathrm{Rot}(A_1, \angle A_1) \circ \mathrm{Rot}(A_2, \angle A_2) \circ \mathrm{Rot}(A_3, \angle A_3)$. It is well known (see here for my old blog post regarding this) that the composition of $\mathrm{Rot}(O_1, \theta) \circ \mathrm{Rot}(O_2, \phi)$ is equal to $\mathrm{Rot}(X, \angle A_1 + \angle A_2)$ where $X$ is the unique point so that $\angle XO_1O_2 = \theta/2$ and $\angle XO_2O_1 = \phi/2$ (where angles are taken with sign). This immediately implies that $$\mathrm{Rot}(A_1, \angle A_1) \circ \mathrm{Rot}(A_2, \angle A_2) = \mathrm{Rot}(I, \angle A_1 + \angle A_2)$$where $I$ is the incentre of $\triangle A_1A_2A_3$ and applying this again we get $$\mathrm{Rot}(I, \angle A_1 + \angle A_2) \circ \mathrm{Rot}(A_3, \angle A_3) = \mathrm{Rot}(B_2, 180^{\circ}),$$where $B_2$ is the foot of perpendicular from $I$ onto $A_1A_3$. Thus, if we denote rotation by $180^{\circ}$ about $T$ i.e. Halfturns about $T$ as $H_T$, then $\tau_{123} = H_{B_2}, \tau_{231} = H_{B_3}, \tau_{312} = H_{B_1}$ where $B_1$ and $B_3$ are the foot of perpendicular from $I$ onto $A_2A_3$ and $A_1A_2$. Now we see that the homothety at $P$ with ratio $2$ takes $\triangle B_1B_2B_3$ to $\triangle P_1P_2P_3$ since $P_i = H_{B_i}(P)$. This immediately implies that circumradius of $\triangle P_1P_2P_3$ is twice the circumradius of $\triangle B_1B_2B_3$. On the other hand, the circumcircle of $\triangle B_1B_2B_3$ is the incircle of $\triangle A_1A_2A_3$, so the desired follows from Euler's Inequality $2r \le R$ applied to $\triangle A_1A_2A_3$.

Well, Interloop was right after all This problem destroyed me

My favourite from the test (its nice to see a problem involving geometric transformations): here goes the sketch; Let $\{X, \theta \}(P)$ denote rotation of point $P$ with respect to $X$ with an angle of $\theta$ (counter-clockwise). Its well-known that $\{ Y, \theta_1 \}(P) \cdot \{ Z, \theta_2 \} = \{ X, \theta_1+\theta_2 \}(P)$ where $X$ is a point such that $\angle XYZ = \theta_1/2, \angle XZY = \theta_2/2$. Let $I$ denote the incenter of $\triangle A_1 A_2 A_3$ and let its incircle touch $A_2 A_3, A_3 A_1, A_1 A_2$ at $T_1, T_2, T_3$ respectively. Let $\alpha_i$ denote the angle of vertex $A_i$ in $\triangle A_1 A_2 A_3$. Notice that: $$\{ A_2, \alpha_2 \}(P) \cdot \{A_1, \alpha_1 \} = \{ I, \alpha_1+ \alpha_2 \}(P)$$$$\{ I, \alpha_1+ \alpha_2 \}(P) \cdot \{ A_3, \alpha_3 \} = \{ T_1, 180^\circ \}(P)$$Thus, $\triangle P_1 P_2 P_3 \sim \triangle T_1 T_2 T_3$. Notice that, the circumradius of $\triangle P_1P_2P_3$ is $2r$, which is at most $R$ by Euler's Geometric Inequality : $R \ge 2r$.

We use complex numbers for this problem. We assume that points $a_1,a_2,a_3$ represent the points $A_1,A_2,A_3$ in the complex plane. The centre of $\tau$ is the centre of the Argand plane, and the radius is taken to be $1$ unit by trivializing. We take the subscripts with variable $k$ instead of $i$, so that we don't confuse it with the square root of -1 ($i$). Let $p_1,p_2,p_3,p$ denote points $P_1,P_2,P_3,P$ on the complex plane. We use the following Lemma: For complex numbers $x,y,z$ representing points $X,Y,Z$, if $Z$ is the reflection of $X$ about $Y$ by an angle $\theta$, then we have: \[(z-y)=(x-y)e^{i\theta},\]where $i$ is the square root of -1 and $e$ is Euler's constant. Key Claim: The area of triangle $P_1P_2P_3$ in independent of choice of $P$. This means that\[\tau_{k+1}(p)=pe^{iA_{k+1}}-a_{k+1}(e^{iA_{k+1}}-1)\]\[\tau_{k}(\tau_{k+1}(p))=pe^{iA_{k+1}}e^{iA_k}-a_{k+1}(e^{iA_{k+1}}-1)e^{iA_k}-a_{k}(e^{iA_{k}}-1)\]\[\tau_{k+2}(\tau_{k+1}(\tau_k(p)))=pe^{iA_{k+1}}e^{iA_k}e^{iA_{k+2}}\]\[-a_{k+1}(e^{iA_{k+1}}-1)e^{iA_k}e^{iA_{k+2}}-a_{k}(e^{iA_{k}}-1)e^{iA_{k+2}}-a_{k+2}(e^{iA_{k+2}}-1)\]We know that $A_k+A_{k+1}+A_{k+2}=\pi$ and that $e^{i\pi}=-1$. So, we may write $p_k=-p+c_k$, where $c_k$ is completely independent of $p$. So, the lengths of triangle $P_1P_2P_3$ are independent of the choice of $P$. This means that area of $P_1P_2P_3$ is also independent of $P$, as claimed. Q.E.D. Notice that if $D,E,F$ are the in-touch points of triangle $A_1A_2A_3$, with $D$ on $A_2A_3$, $E$ on $A_3A_1$, $F$ on $A_1A_2$. Note that under $\tau_{2}$, $D$ goes to $E$ and under $\tau_1$, $E$ goes to $F$, and under $\tau_{3}$, $F$ goes to $D$. It means that if $d$ denotes the complex number for $D$, then $d=-d+c_1$. So, $c_1=2d$. Similarly, $c_2=2e$ and $c_3=2f$. It follows that $p_1+p=2d$, so $P_1$ is the reflection of $P$ about $D$. Similarly, $P_2$ is reflection of $E$ about $P$ and $P_3$ is reflection of $P$ about $F$. The triangle thus obtained is basically from homothety at $P$ with factor $2$. So, circumradius of $P_1P_2P_3$ is twice the circumradius of $DEF$, which itself is the inradius of $A_1A_2A_3$. From Euler's inequality, we have that $R\geq2r$, where $R$ is circumradius of $A_1A_2A_3$ and $r$ is it's inradius. Hence, circumradius of $P_1P_2P_3$ is at most the circumradius of $A_1A_2A_3$.

Consider a vector $PQ$. After making a rotation, u preserve its length, so infact that is true for the composition of all three rotations. But, after we rotate thrice anti-clockwise each time in the angles of $\triangle A_1A_2A_3$, we rotate by a total of $180^\circ$. So, the vector $PQ$ gets mapped to $P_iQ_i = -PQ$. So, we have that $P_1Q_1 = P_2Q_2 = P_3Q_3$ for any two points $P$ and $Q$. So, $P_1Q_1Q_2P_2$ is a parallelogram. Therefore, $P_1P_2 = Q_1Q_2$. Similarly, $P_2P_3 = Q_2Q_3$ and $P_3P_1 = Q_3Q_1$ and thus triangles $P_1P_2P_3$ and $Q_1Q_2Q_3$ are congruent. So it suffices to show this result for one point $P$ What I could not find in the exam however is that the most suitable point is the incentre. Anyways funny story, this is my first time giving INMO, so I had no idea whether I could quote spiral similarity, so I ended up proving it lol

A thing about most of the solutions above which can be easily missed- the composition of two rotations is a rotation **ONLY IF** the two angles don’t sum to a multiple of $2 \pi$ (possibly $0$). Worth max a few points but still. Because if that’s the case, the transformation is a TRANSLATION. And that derails the solution.

everythingpi3141592 wrote: Consider a vector $PQ$. After making a rotation, u preserve its length, so infact that is true for the composition of all three rotations. But, after we rotate thrice anti-clockwise each time in the angles of $\triangle A_1A_2A_3$, we rotate by a total of $180^\circ$. So, the vector $PQ$ gets mapped to $P_iQ_i = -PQ$. So, we have that $P_1Q_1 = P_2Q_2 = P_3Q_3$ for any two points $P$ and $Q$. So, $P_1Q_1Q_2P_2$ is a parallelogram. Therefore, $P_1P_2 = Q_1Q_2$. Similarly, $P_2P_3 = Q_2Q_3$ and $P_3P_1 = Q_3Q_1$ and thus triangles $P_1P_2P_3$ and $Q_1Q_2Q_3$ are congruent. So it suffices to show this result for one point $P$ What I could not find in the exam however is that the most suitable point is the incentre. Anyways funny story, this is my first time giving INMO, so I had no idea whether I could quote spiral similarity, so I ended up proving it lol Another suitable point is $P \equiv A_1$ and it is easy to show the problem in this case lol.

Master_of_Aops wrote: A thing about most of the solutions above which can be easily missed- the composition of two rotations is a rotation **ONLY IF** the two angles don’t sum to a multiple of $2 \pi$ (possibly $0$). Worth max a few points but still. Because if that’s the case, the transformation is a TRANSLATION. And that derails the solution. How is that a problem? The total rotation angle at each turn is $\leq 180^{\circ}$ counter clockwise. How will it ever reach (even close to) $360^{\circ}$?

(Edited) nvm the composition of a translation and a rotation is a rotation too

math_comb01 wrote: everythingpi3141592 wrote: Consider a vector $PQ$. After making a rotation, u preserve its length, so infact that is true for the composition of all three rotations. But, after we rotate thrice anti-clockwise each time in the angles of $\triangle A_1A_2A_3$, we rotate by a total of $180^\circ$. So, the vector $PQ$ gets mapped to $P_iQ_i = -PQ$. So, we have that $P_1Q_1 = P_2Q_2 = P_3Q_3$ for any two points $P$ and $Q$. So, $P_1Q_1Q_2P_2$ is a parallelogram. Therefore, $P_1P_2 = Q_1Q_2$. Similarly, $P_2P_3 = Q_2Q_3$ and $P_3P_1 = Q_3Q_1$ and thus triangles $P_1P_2P_3$ and $Q_1Q_2Q_3$ are congruent. So it suffices to show this result for one point $P$ What I could not find in the exam however is that the most suitable point is the incentre. Anyways funny story, this is my first time giving INMO, so I had no idea whether I could quote spiral similarity, so I ended up proving it lol Another suitable point is $P \equiv A_1$ and it is easy to show the problem in this case lol. Yeah, I actually tried this point. But iirc it gets mapped to nice points for 2 of the transformations (and nice in the sense lying on the sides of the triangle) but for the 3rd it goes somewhere random. Of course looking at the official solution shows that it was reflection about the touchpoints, but it was hard to think of that in contest

A translation is a rotation centered at a point at infinity.

Observe that from the Theory of Composition of Rotations, $\tau_{i+2}(\tau_{i}(\tau_{i+1}))$ denotes a rotation of angle $\angle A_2 + \angle A_1 + \angle A_3 = \pi$, i. e. this function gives the reflection of P about some point. From the special case of the incentre, we see that the points in question are the intouch points. Now observe that we obtain $P_1P_2P_3$ by taking a homothety of the inradius with centre $P$ and scale factor $2$, as such the circumradius of $\triangle P_1P_2P_3$ is twice the inradius of $\triangle ABC$, and we are done from Euler's inequality. $\square$

I have discussed this question with a nice complex coordinates approach in my INMO 2024 video on my channel "little fermat". Here is the Link