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Source: BMO 2024 SL G7

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#1 h Apr 27, 2025, 1:08 PM

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Let $f: \pi \to \mathbb R$ be a function from the Euclidean plane to the real numbers such that $f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$ for any acute triangle $\Delta ABC$ with circumcenter $O$ , centroid $G$ and orthocenter $H$ . Prove that $f$ is constant.

This post has been edited 1 time. Last edited by GreekIdiot, Apr 27, 2025, 1:08 PM

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#2 h Apr 30, 2025, 3:50 AM

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Consider $\triangle ABC$ with circumcenter $O$ , centroid $G$ and orthocenter $H$ .
$D,E,F,M_{OG},M_{OH},M_{HG}$ is midpoint of $BC,CA,AB,OG,OH,HG$
Define $O_A,G_A,H_A$ be circumcenter, centroid and orthocenter of $\triangle AEF$ .
and $H_B,O_B,G_B,H_C,O_C,G_C$ analogously
by $\triangle ABC, \triangle AFE , \triangle BFD , \triangle CDE , \triangle DEF$ and some calculation
known that $f(O)+f(G)=f(M_{OG})+f(M_{HG})$
Let $M'$ be reflection of $M_{HG}$ wrt point $M_{OG}$ notice that $f(M_{HG}=f(M')$
so f is constant. done $\square$

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#3 h Apr 30, 2025, 8:29 PM

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Use complex numbers:

For a point $A(a)$ we use $f(a)=f(A)$ .

WLOG suppose $f(0)=0$ .

Let $\omega=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ .

We have $f(a)+f(\omega a)+f(\omega^2 a)=3\cdot f(0)=0, \forall a\in \mathbb{C}$ .

We also have $f(x)+f(\omega a)+f(\omega^2 a)=0+f\left(\dfrac{x-a}{3}\right)+f(x-a), \forall a,x\in \mathbb{C}$ with $|a|=|x|>|a-x|$ (so that the triangle is acute).

So $f(x)-f(a)=f\left(\dfrac{x-a}{3}\right)+f(x-a)$ .

Because $f(x)-f(y)$ depends on just $x-y$ (for $x,y$ complex with $|x|=|y|$ ), we can consider function $g$ such that $g(x-y)=f(x)-f(y)$ .

So $g(x-a)=f\left(\dfrac{x-a}{3}\right)+f(x-a)$ .

Notice that $g$ is additive: $g(x-y)+g(y-z)=f(x)-f(y)+f(y)-f(z)=f(x)-f(z)=g(x-z)$ .

So $g(nx)=ng(x)$ , $\forall n\in \mathbb{Z}$ and $x\in\mathbb{C}$ .

So $g(x)+g(-x)=0\implies g(a-x)+g((-a)-(-x))=0\implies f(a)-f(x)+f(-a)-f(-x)=0\implies$
$\implies f(x)+f(-x)=f(a)+f(-a)$ , so $f(x)+f(-x)$ is a constant $c$ .

But $f(1)+f(\omega)+f(\omega^2)=0=-(f(-1)+f(-\omega)+f(-\omega^2))\implies 3c=0\implies c=0\implies f$ is odd.

$f(x)-f(y)=g(x-y)\implies f(x)+f(-y)=g(x-y)\implies f(x)+f(y)=g(x+y)$ , where $|x|=|y|$ .

Choose $x=\omega a$ and $y=\omega^2 a$ in the above to get $-f(a)=f(\omega a)+f(\omega^2 a)=g(-a)=-g(a)\implies f\equiv g$ .

But we know that $g(x)=f(x)+f\left(\frac{x}{3}\right)\implies f\left(\frac{x}{3}\right)=0, \forall x\in \mathbb{C}$ , so we're done.

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