Let $f$ be a function from the set of all points on the plane to the nonzero real numbers. Suppose that for any triangle $ABC$ with incenter $I$, we have that $f(I)=f(A)f(B)f(C)$. What are the possibilities for $f$?
Clearly, $f=1$ and $f=-1$ work. Are there other ones?
Let $D$ be the incenter of the triangle $ABI$
Let's draw the picture:
Thus $f(D)=f(I)f(A)f(B)=f(A)^2f(B)^2f(C)$
Now, consider pictures symmetric pictures with respect to the line $IC$.
We get points $A',B',D'$ for which we have
1) $f(A')f(B')=f(A)f(B)=f(I)/f(C)$
2) $f(D')=f(A')^2f(B')^2f(C)=f(A)^2f(B)^2f(C)=f(D)$
Thus, we have so proved that $f(D)=f(D')$
Now, given two points $X,Y$ in the plane, let $L$ be their axis of simmetry. That is to say, $L$ is the line of points equidistant from $X,Y$.
Up to scaling and rotating we can suppose that $X=D$, $Y=D'$ and $L$ is the line $CI$.
In other words, given any two $X,Y$ in the plane, we can construct a triangle $A,B,C$ so that $X,Y$ play the role of $D,D'$ above. Thus $f(X)=f(Y)$ for any $X,Y$.
This proves that $f$ is constant. Given that $f$ is constant it must bu a square root of $1$ hence $\pm1$