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$f:\mathbb{R}^3\to \mathbb{R}^{\ast}$ is such that for any non-degenerate tetrahedron $ABCD$ with $O$ the center of the inscribed sphere, we have : $$f(O)=f(A)f(B)f(C)f(D) $$ Prove that $f(X)=1$ for all points $X$. Here $\mathbb{R}^{\ast}=\mathbb{R}\setminus \{0\}$.

How would someone go on proving this? I have seen the problem

If a function $f:\mathbb{R}^2\to \mathbb{R}$ satisfies $$\sum_{A\in \mathcal{P}}f(A)=0$$ where $\mathcal{P}$ is a regular $n$-gon then $f(X)=0$ for all $X$.

But I have no idea about it. I have been told it is from a olympiad but I don't know which. Could someone help me? I tried to mimic the solution to the second problem but that didn't really help. Thanks.

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Let $ABCD$ be some tetrahedron with incenter $O$, and let $E$ be the incenter of the tetrahedron $OBCD$.

Draw a small sphere around $O$ and select $B'C'D'$ such that this sphere is inscribed in $AB'C'D'$. This is always possible, by selecting three tangent planes through $A$ that enclose the sphere, and intersecting them with a tangent plane behind the sphere. Let $E'$ be the incenter of $OB'C'D'$. It is clear that things can be chosen such that $E\ne E'$, e.g. by making the small sphere small enough that $E$ is outside $OB'C'D'$.

Now, by assumption we have $$ f(O)=f(A)f(B)f(C)f(D) \qquad f(O)f(B)f(C)f(D)=f(E) $$ and by multiplying these equations and canceling common factors (since by assumption $f(X)\ne 0$) we get $$ f(O)^2 = f(A)f(E) $$ Similarly, $$ f(O)^2 = f(A)f(E') $$ Therefore, canceling again, $f(E)=f(E')$.

However, by appropriate similarity transforms, we can put the entire configuration such that $E$ and $E'$ are anywhere we please, and this calculation still shows that $f(E)=f(E')$.

So $f$ must be a constant function.

The equation $f(O)=f(A)f(B)f(C)f(D)$ implies that the common value of $f(X)$ must be a cube root of $1$, and since the values of $f$ are specified to be real, $f(X)=1$ is the only solution.

If the values are allowed to be complex, we get two additional solutions with $f(X)=e^{\pm 2\pi/3 i}$.


The problem specified that $f(X)\ne 0$ everywhere, but relaxing this gives us only one additional solution, the constant function $f(X)=0$. If there's any point $A$ with $f(A)=0$, then for an arbitrary $O\ne A$ we can find $B'C'D'$ as above such that $O$ is the incenter of $AB'C'D'$. Then $f(O)=f(A)\times\cdots=0$ too.

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This is a beautiful solution, but since I am not too used to 3D geometry, could you tell me about the similarity transforms you were talking about, which could vary $A,A^{\prime}$ over the whole space? I am not sure I follow. Anyways about your complex solutions, well if we consider $\mathbb{C}^{\ast}$ your proof works right? You haven't used the fact that $f$ takes real values except $f(O)^6=1$. So checking for $6$-th roots of unity will give us a solution. What I think is, how about allowing $f$ to take the value zero? Anyways a clarification on the similarity part would be helpful. Thanks. –  shadow10 Jul 6 at 5:02
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@shadow10: First scale everything such that the distance between $A$ and $A'$ becomes the desired one. Then translate the space such that $A$ gets the position you want. Finally rotate about $A$ to put $A'$ where you want it. –  Henning Makholm Jul 6 at 8:10
    
Oh, thanks a lot. It is clear now. –  shadow10 Jul 6 at 8:11
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@shadow10: I found a simpler proof that handles the complex case too. I have removed some now-obsolete comments. –  Henning Makholm Jul 6 at 10:25
    
Quite nice, if only I could again give it a +1.. thanks again. –  shadow10 Jul 6 at 10:36
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