Bulgarian National Mathematical Olympiad

Consider two arbitrary points $X$ and $Y$ and let $x$ and $y$ be the corresponding numbers. Choose points $I$ and $J$ on the line $XY$ such that $XY=YI=IJ$. Let $X^{\prime }$ on the line $XY$ be such that $XI=J{X}^{\prime }$. Consider the plane $\lambda$ perpendicular to $IJ$ and passing through the midpoint of $IJ$. Let $ABCX$ and $ABC{X}^{\prime }$ be two equal regular triangular pyramids with base $ABC$ in the plane $\lambda$ having incenters $I$ and $J$.

Since $n_A{n}_B{n}_C{n}_X=n_I$ and $n_A{n}_B{n}_C{n}_{X^{\prime }}=n_J$ we have that $$n_X=\frac{n_{X^{\prime }}\cdot n_I}{n_J}.$$ Move the plane $\lambda$ towards point $I$ and consider the spheres $S_I$ and $S_J$ with centers $I$ and $J$ respectively that are tangent to $\lambda$. Let $A_1{B}_1{C}_1{X}^{\prime }$ be a regular triangular pyramid with base $A_1{B}_1{C}_1$ in $\lambda$ and insphere $S_J$. The regular triangular pyramid with base $A_1{B}_1{C}_1$ and insphere $S_I$ has vertex $X_1$. It follows from the above that $$n_{X_1}=\frac{n_{X^{\prime }}\cdot n_I}{n_J}=n_X.$$ When $\lambda$ moves towards $I$ the radius of the sphere $S_I$ tends to zero and $△A_1{B}_1{C}_1$ tends to a triangle which is the base of a regular triangular pyramid with vertex $X^{\prime }$ and inscribed sphere with center $J$ and radius $IJ$.

We conclude that $X_1$ tends to $I$. Continuity arguments show that all inner points on the segment $XI$ (with point $X$) are assigned with the same number.

Thus, $x=y$ and all numbers are equal. It follows from $x^4=x$ and $x\ne 0$ that $x=1$.