You get $$\angle BAA_1+\angle A_1AC=\angle BAC=\angle A_1B_1C_1=\angle A_1AC_1=\angle A_1AB+\angle BAC_1=\angle A_1AB+\angle BCC_1$$so $\angle A_1AC=\angle BCC_1$ so $G$ is the $A$ HM point. Hence, $G$ lies on $(BHC).$ SImilarly $G$ lies on $(AHB), (CHA)$ and so $G \equiv H$ and done.

Well probably the same solution as above: The conditions imply that $C_1B = A_1C = B_1A$, so $\angle GAC = \angle GCB$, and $G$ lies on $A-$ median so $G$ is the $A-$ HM point which lies on $\odot(BHC)$ where $H$ is orthocenter. Similarly $G$ lies on $\odot(AHB)$ and $\odot(AHC)$ which finishes the problem.

Tearfulstage wrote: No it is not necessary that g is a hm point For that we also need gbc=gab $G$ lies on the $A$ median and $\angle BCG=\angle CAA_1=\angle BCX_A$ if $X_A$ is the $A$ hm point. Since $X_A$ also lies on the $A$ median, we must have $G \equiv X_A.$ Edit: Wait you just deleted your post

I had understood it

This was downright the easiest problem on the test Yes @below

Delta0001 wrote: This was downright the easiest problem on the test Then I guess your paper started with P2

I proved $A,E,F,G$ concyclic like #2, but then proceeded using INMO 2018 P2.

By angle chasing, $\angle GAC = \angle GBA = \angle GCB$, so $G$ is a Brocard point of $\triangle ABC$ with barycentric coordinates $(c^{-2} : a^{-2} : b^{-2})$. Yet by definition $G = (1 : 1 : 1)$, so $a = b = c$.

Straight up angle chasing, thanks to the paper setter for putting this on the test- just angle chase to get a bunch of arcs of equal length, then angle chase a bit more to get all angles equal- if I’m not wrong, the centroid’s only purpose was to ensure concurrency...

Could this be proved by showing that G is the circumcenter of the triangle and hence it will be equilateral

Equal chord subtend equal angle + congruency

Isosceles trapeziums

NO Doubt... THIS IS THE EASIEST PROBLEM IN TODAY'S RMO. Consider all the intersections $AB\cap B_1C_1=A'$ and $AC\cap B_1C_1=B'$, $AB\cap A_1C_1=P'$ and $BC\cap A_1C_1=Q'$, $B_1A_1\cap BC=E'$ and $B_1A_1\cap AC=E'$. So, $\angle B_1A'B=\angle AD'A_1=\angle CQ'C=\angle P'D'E'$. This is enough to show, $\angle BAC=\angle BCA=\angle ABC\implies ABC$ and $A_1B_1C_1$ are equilateral. The Geos were Suuuuper Easy, I don't know why? @below, I didn't give today's RMO, that's why I'm feeling very sad when I'm solving the questions from today's paper, I feel like crying. , and definitely I'm not a pro at all, you are seroiusly making me laugh.

amar_04 wrote: The Geos were super Easy, I don't know why? Cuz you're Pro at Geo. I could just prove both triangles congruent. That's it

In the second question I proved AB1BC1CA1 a regular hexagon so that each angles becomes 120 degree but i couldnot write furthur as the time was up how many marks can i get in it

ubermensch wrote: Straight up angle chasing, thanks to the paper setter for putting this on the test- just angle chase to get a bunch of arcs of equal length, then angle chase a bit more to get all angles equal- if I’m not wrong, the centroid’s only purpose was to ensure concurrency... Hmm, my solution used centroid a lot! @2below VERY similar solution

the centroid’s only purpose was to ensure concurrency i am with ubermensch

MathPassionForever wrote: Let $ABC$ be a triangle with circumcircle $\Omega$ and let $G$ be the centroid of triangle $ABC$. Extend $AG, BG$ and $CG$ to meet the circle $\Omega$ again in $A_1, B_1$ and $C_1$. Suppose $\angle BAC = \angle A_1 B_1 C_1, \angle ABC = \angle A_1 C_1 B_1$ and $ \angle ACB = B_1 A_1 C_1$. Prove that $ABC$ and $A_1 B_1 C_1$ are equilateral triangles. Use cyclicity and problem condition and chase some angles to get that $\angle{BCG} = \angle{CAG} = \angle{ABG}$. Now, let midpoints of $BC,CA$ and $AB$ be $D,E$ and $F$ respectively. Again, midpoint theorem gives that $DCEG, EAFG$ and $FCDG$ are cyclic. Now, PoP gives $AF. AB = AG.AD = AE.AC$, to get that $BFEC$ is cyclic. So, $BFEC$ is a cyclic trapezoid $\implies$ $BF = EC$ or $AB = AC$. Similar chases finish.

............. I like to write $A',B',C'$ instead of $A_1,B_1,C_1$. let us assume a symbol that, if AB and CD intersect in point P. we sholud write , $AB\cap CD=P$. now go back to problem, $AB\cap B'C'=A_1'$. and $AC\cap B'C'=B_1$, $AB\cap A'C'=C_1$ and, $BC\cap A'C'=D_1$, $B'A'\cap BC=E_1$, and $B'A'\cap AC=F_1$. So, $\angle B'A_1B=\angle AC_1A'=\angle CD_1C'=\angle AF_1A'$. this can emply , $\angle BAC=\angle BCA=\angle ABC$. which imply $ABC$ and $A_1B_1C_1$ are equilateral.

Functional_equation wrote:

Observe that the trilinear co-ordinate you've given is of second Brocard point whereas I truly believe that first Brocard point is in the problem. But not difficult to fix $\left (\frac{c}{b}, \frac{a}{c}, \frac{b}{a}\right ) = (bc, ac, ab) \implies a = b = c$ and you're still done.

green_leaf wrote: I am bad at geo, so i hope this is correct. Clearly, $ABC$ is congruent to $A_1B_1C_1$. Let $\angle CB_1C_1 = x$. Then $\angle C_1BC = \angle BAA_1 = \angle BC_1A_1 = \angle ACB_1 = x$. Now $\angle BCA_1 = BC_1A_1 = x$ and $\angle C_1A_1C = \angle C_1BC = x$. So, we can see that $BC = A_1C_1$ but $BC = B_1C_1$ so $B_1C_1 = A_1C_1$. So we get all sides of $A_1B_1C_1$ are equal. Hence $ABC$ and $A_1B_1C_1$ are equilateral. I think this solution is wrong (or at least incomplete). Can you please tell how did you conclude that $BC = B_1C_1$. As a side note, you could have obtained that $BC = A_1C_1$ more easily by observing the fact that $\triangle ABC$ is congruent to $\triangle B_1C_1A_1$ (since these two triangles are similar and they also have the same circumradius)

I have a solution which only uses angle chasing and no advanced stuff. So here goes my solution to the above problem: Let $D, E, F$ be the mid-points of sides $BC, CA, AB$ respectively. By simple angle chasing, it is easy to obtain that $\angle GAE = \angle GBF = \angle GCD$. Now, by mid-point theorem, $BC||FE$ and hence we obtain: $\angle GAE = \angle GCD = \angle FCB = \angle CFE = \angle GFE$ Thus, quadrilateral $AEGF$ is cyclic and hence we obtain $\angle BGC = \angle EGF = 180^\circ - \angle A$. Thus, if $H$ is the orthocentre of $\triangle ABC$ then $G$ lies on $\odot(BHC)$ (since $\angle BHC = 180^\circ - \angle A$). Analogously, $G$ lies on $\odot(AHB)$ and $\odot(AHC)$. Now since two distinct circles can intersect at atmost two points, So $\odot(AHB)$ and $\odot(AHC)$ intersect at points $H, A$ only. Now since $G$ lies on both of $\odot(AHB)$ and $\odot(AHC)$ , thus $G \equiv H$ or $G \equiv A$. Now it is clear that $G$ and $A$ are distinct points (since $G$ is the centroid of $\triangle ABC$). Hence, $G \equiv H$. So the centroid and orthocentre of $\triangle ABC$ coincide which is enough to imply that $\triangle ABC$ is equilateral. Now since $\triangle CAB$ is similar to $\triangle A_1B_1C_1$ (and $\triangle CAB$ is equilateral), thus $\triangle A_1B_1C_1$ is also equilateral. Hence proved.

Why G lies on Circle through BHC??Can anyone please tell

Let $O$ be the circumcenter of $\triangle ABC, \triangle A_1B_1C_1$ then observe $\angle C_1OA_1=\angle BOC \implies C_1A_1=BC$ and as $\triangle ABC \sim \triangle A_1B_1C_1$ By AAA, with 1 equal sides so $\triangle ABC \cong \triangle A_1B_1C_1$ and $\angle OC_1B=2\angle C_1B_1A_1 -\angle BOA_1, \angle A_1OC=2\angle BAC -\angle BOA_1\implies \angle C_1OB=\angle A_1OC$ similarly we can prove $\angle A_1OC=\angle AOB_1=\angle C_1OB= 2x^\circ$ (say). Hence $\angle C_1A_1B=\angle A_1BC=x^\circ$ and also $AB_1=C_1B=A_1C$ So $C_1CA_1B$ is isosceles trapezoid. And as $CC_1$ is a median , $CC_1||BA_1\implies$ in $\triangle ABA_1$ , $G$ is midpoint of $AA_1$ Let $M_1$ be the mid point of $BC$ So $\triangle BA_1M_1\cong GCM_1 \implies \triangle BGC \cong \triangle BA_1C \implies BG=AC=C_1B$ Similarly $G$ is midpoint of $BB_1$ So $\triangle AGC \cong AB_1C \implies AB_1=GC=BG=BC_1=A_1C \implies $ all median of $\triangle ABC$ are equal. So $\triangle ABC \cong \triangle A_1B_1C_1$ is equilateral. $\blacksquare$. Its similar to the solution which I wrote at exam.. But I don't know why but examiner gave only $12$ marks for it :-(

Observe it's a six point star where the inside portion is a hexagon. So (6-2) ×180° so thus both the triangles are equilateral

green_leaf wrote:

splendid solution!

teomihai wrote: green_leaf wrote:

Note that triangle B1A1C1 is a rotation of ABC around O, so ∠AOB1 = ∠BOC1 = ∠COA1. Let this constant be 2θ Then ∠ABG = ∠ABB1 = ∠AGB1/2 = θ and similarly ∠BCG = ∠CAG = θ. To prove that ABC is equilateral, enough to show ∠ABC = ∠BCA = ∠CBA. Let M be the midpoint of AC. Since ∠MAG = ∠ABG, the line AC is tangent to (ABG). Powers of points, MG*MB = MA^2 = MC^2. This means that the straight line AC is also tangent to (CBG), so ∠MCG = ∠CBG. This proves ∠ACG = ∠CBG, and similarly ∠CBG = ∠BAG and ∠BAG = ∠ACG. Finally, ∠ABC = ∠BCA = ∠CBA

Instructive and easy. Claim 1: $\Delta ABC \cong \Delta B_{1}C_{1}A_{1}$ Proof: As equal chords subtend equal angles at the center thus by $SSS$ congruency criterion we get the desired claim. By angle chasing we get $EA$ is tangent to $(BAG)$. Thus by Power of Point we get, ${EA}^{2}=EG.EB$. Now using $AE=EC$ we get, ${EC}^{2}=EG.EB$. Thus $EC$ is tangent to $(BGC)$. Thus we get, $\angle{B_{1}BC}=\angle{GBC}=\angle{ECG}$. Also by angle chasing, $\angle{BCC_{1}}=\angle{ABB_{1}}$. Hence adding the two relations we get, $\angle{ABC}=\angle{ACB}$. Similarly we can also get $\angle{ACB}=\angle{BAC}$. Thus $\Delta ABC$ is equilateral.$\blacksquare$.

Angle Chase and use of pop finishes this problem