Let $ABC$ be a triangle, $O$ its circumcenter and $R=1$ its circumradius. Let $G_1,G_2,G_3$ be the centroids of the triangles $OBC, OAC$ and $OAB.$ Prove that the triangle $ABC$ is equilateral if and only if $$AG_1+BG_2+CG_3=4$$
Problem
Source: Romania NMO - 2018
Tags: geometry, circumcircle, complex numbers
Wizard_32
12.04.2018 19:51
Consider the scene in the complex plane (vectors also work). Set $\odot(ABC)$ to be the unit circle. Hence, $O$ is the origin. Set $H$ to be the orthocenter of $\triangle ABC$. Hence, $|a|=|b|=|c|=1$ and $h=a+b+c$.. Now, clearly $g_1=\frac{b+c}{3}, g_2=\frac{c+a}{3}, g_3=\frac{a+b}{3}$ . Hence, $3|AG_1|=|a-g_1|=3\left| a-\frac{b+c}{3} \right|=| 3a-b-c|$ Similarly we get 2 other results. Now, note that $ABC$ is equilateral if and only if $a+b+c=0$, which gives $$3(|AG_1|+|BG_2|+|CG_2|)=|3a-b-c|+|3b-c-a|+|3c-a-b|=4|a|+4|b|+4|c|=12$$as desired. [I am still trying to prove the other implication]
Wizard_32
12.04.2018 20:05
Configuration Note: $AG_1, BG_2, CG_3$ are concurrent.
Let $D,E,F$ be the midpoints of $AO,BO,CO$ respectively.
Then clearly $G_1=CE \cap BF$, $G_2=AF \cap CD$ and $G_1=BD \cap AE$.
Hence, by $BG_3:G_3D=2:1=BG_1:G_1F$ gives $G_1G_3 \parallel DF$. Also, $DF$ is the $O-midline$ of $\triangle OAC$ and hence $DF \parallel AC$. Thus, $G_1G_3 \parallel AC$.
Similarly, we get that the sides of $\triangle G_1G_2G_3$ are parallel to that of $\triangle ABC$, and hence these are homothetic triangles. Thus we must have that $AG_1, BG_2, CG_3$ are concurrent, as desired.
[asy][asy]
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[/asy][/asy]
trumpeter
13.04.2018 01:28
Wizard_32 wrote: Configuration Note: $AG_1, BG_2, CG_3$ are concurrent. Another proof of this fact:
$O$ has barycentric coordinates $O=\left(a^2S_A:b^2S_B:c^2S_C\right)$ which has component sum $2S^2$, where $S=2\left[ABC\right]$. Midpoint $BC$ has coordinates $M=\left(0:S^2:S^2\right)$ with the same component sum, so $G_1$ is $\frac{O+2M}{3}=\left(a^2S_A:b^2S_B+2S^2:c^2S_C+2S^2\right)$. Clearly then $AG_1,BG_2,CG_3$ concur at $\left(a^2S_A+2S^2,b^2S_B+2S^2,c^2S_C+2S^2\right)$.
ThE-dArK-lOrD
13.04.2018 13:28
Continue from the solution in #2. We need to prove that if $a,b,c$ are complex numbers that $|a|=|b|=|c|=1$ and $$\sum_{cyc}{|3a-b-c|}=12,$$then $a+b+c=0$. Let $a+b+c=t=p+qi$ where $p,q\in \mathbb{R}$. We've $$144=\left( \sum_{cyc}{|4a-t|}\right)^2\leq 3\sum_{cyc}{|4a-t|^2}=3\sum_{cyc}{(16-4\overline{a}t-4a\overline{t}+|t|^2)}=144-12(\overline{a+b+c})t-12(a+b+c)\overline{t}+9|t|^2.$$Hence, $12(p-qi)(p+qi)+12(p+qi)(p-qi)\leq 9|t|^2\implies 24|t|^2\leq 9|t|^2\implies |t|=0$. This means $a+b+c=0$, done.
e_plus_pi
14.04.2018 10:20
Remark: A much similar problem appeared in 1990, Yugoslavia national olympiad: Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$. Let $P$ be the reflection of $H$ over $O$ and let $G_1,G_2,G_3 $ denote the centroids of $\triangle BCP , \triangle CAP , \triangle ABP $. Prove that $ AG_1 = BG_2 = CG_3 = \dfrac {4R}{3} $ where $ R = OA.$