f(M)=MA^n+MB^n+MC^n

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Source: Moldova TST 2021

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#1 h Sep 20, 2021, 4:43 AM

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Let $ABC$ be an equilateral triangle. Find all positive integers $n$ , for which the function $f$ , defined on all points $M$ from the circle $S$ circumscribed to triangle $ABC$ , defined by the formula $f:S \rightarrow R, f(M)=MA^n+MB^n+MC^n$ , is a constant function.

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#3 h Sep 20, 2021, 7:13 PM • 1 Y

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Nice and easy.
Let $a$ be the sidelength of the triangle and $R$ is the radius of $S$ .
Claim 01. $n= 2,4$ are only possibilities.
Proof. Taking $M=A$ , thus $f(A)=2a^n$ . Taking $M=X$ as the midpoint of arc $BC$ , we have $f(X)=2R^n+(2R)^n=R^n(2+2^n)$ . By law of cosines, $a^2=2R^2-2R^2\cos{120^\circ}=3R^2$ . Therefore,
$4a^{2n}3^n=a^{2n}(2+2^n)^2\implies 3^n=(2^{n-1}+1)^2.$ Hence, $n=2k$ , $3^k=2^{2k-1}+1.$ By the Catalan's conjecture, either $k=1,2$ and they both work.

Firstly, note that by Ptolemy's theorem, $MA=MB+MC$ , where $M$ is the point on the arc $BC$ .

For $n=2$ , we have
$\begin{align*} c_2=MA^2+MB^2+MC^2&=2(MB^2+MC^2+MB\cdot MC)\\&=8R^2(\sin^2{\alpha}+\sin^2{(60^\circ-\alpha)}+\sin{\alpha}\cdot \sin{(60^\circ-\alpha)})\\&=8R^2(\sin^2{\alpha}+(\frac{\sqrt{3}}{2}\cos{\alpha}-\frac{1}{2}\sin{\alpha})^2+\sin{\alpha}(\frac{\sqrt{3}}{2}\cos{\alpha}-\frac{1}{2}\sin{\alpha}))\\&=8R^2(\sin^2{\alpha}+\frac{3}{4}\cos^2{\alpha}+\frac{1}{4}\sin^2{\alpha}-\frac{1}{2}\sin^2{\alpha})=\\&=6R^2, \end{align*}$ where we used the fact that $\sin{(60^\circ-\alpha)}=\frac{\sqrt{3}}{2}\cos{\alpha}-\frac{1}{2}\sin{\alpha}$ .

Now it might be unexpected but $n=4$ also works. We deploy barycentric coordinates in order to prove that.
Claim 02. $S_A^2+S_B^2+S_C^2=S^2$ for any point $M$ on $S$ , where $S_A=[MBC]$ , $S_B=[MAC]$ , $S_C=[MAB]$ and $S=[ABC]$ .
Proof.
Let $A=(1,0,0)$ , $B=(0,1,0)$ and $C=(0,0,1)$ as usual. Let $M=(x,y,z)$ .
As $M$ lies on the circumcircle of $\triangle ABC$ , the following must hold $a^2yz+b^2xz+c^2xy=0$ and as $a=b=c$ , we have $xy+yz+xz=0$ . Also note that $x+y+z=1$ . Therefore $x^2+y^2+z^2=1\implies S_A^2+S_B^2+S_C^2=S^2$ for any $M$ on $S$ .

Now we use the claim in order to finish the problem, note that
$\begin{align*} c_4=MA^4+MB^4+MC^4&=(MA^2+MB^2+MC^2)^2-2(MA^2\cdot MB^2+MB^2\cdot MC^2+MC^2\cdot MA^2)\\&=c_2^2-\frac{32}{3}\left(MA^2\cdot MB^2\cdot \frac{\sin^2{60^\circ}}{2^2}+MB^2\cdot MC^2\cdot \frac{\sin^2{60^\circ}}{2^2}+MC^2\cdot MA^2\cdot \frac{\sin^2{60^\circ}}{2^2}\right)\\&=c_2^2-\frac{32}{3}(S_A^2+S_B^2+S_C^2)\\&=c_2^2-\frac{32}{3}S^2, \end{align*}$ which shows that $c_4$ is also a constant, we are done.

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#4 h Sep 20, 2021, 7:29 PM

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2017 Mediterranean Mathematics Olympiad p1 and Mathematical Reflections

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