IMO 1962 · Problem 7

The tetrahedron \(SABC\) has the following property: there exist five spheres, each tangent to the edges \(SA, SB, SC, BCCA, AB\), or to their extensions.

(a) Prove that the tetrahedron \(SABC\) is regular.

(b) Prove conversely that for every regular tetrahedron five such spheres exist.

Official text of the problem. Official problem paper (PDF)

Solution

The printed \(BCCA\) denotes the two edges \(BC,CA\); thus all six edge lines are involved.

(a) We first classify the possible spheres and show that there are at most five.

Consider a sphere with center \(O\) and radius \(r>0\), tangent to all six edge lines. Every vertex \(V\) lies strictly outside the sphere. Indeed, it lies on a tangent line, so it cannot lie inside. If it lay on the sphere, the three incident edge lines would all lie in its tangent plane at \(V\), contradicting the noncoplanarity of the tetrahedron.

Consequently, the tangent segments from \(V\) have the same positive length \[ t_V=\sqrt{|OV|^2-r^2}. \] Call an edge contact internal if its tangency point lies between the endpoints, and external otherwise. No contact occurs at a vertex. For every edge \(UV\), \[ UV= \begin{cases} t_U+t_V,&\text{if its contact is internal},\\ |t_U-t_V|,&\text{if its contact is external}. \end{cases} \tag{1} \]

Choose a vertex \(V_0\) for which \(t_{V_0}\) is largest, and denote the other vertices by \(V_1,V_2,V_3\). Write \(t_i=t_{V_i}\). Then \[ V_0V_i=t_0+\varepsilon_i t_i,\qquad \varepsilon_i\in\{1,-1\}, \] where the sign is positive for an internal contact and negative for an external contact.

If \(\varepsilon_i\) and \(\varepsilon_j\) differed, then \[ |V_0V_i-V_0V_j|=t_i+t_j\ge V_iV_j, \] contrary to the strict triangle inequality in \(V_0V_iV_j\). Thus all three signs are equal. Moreover, \[ V_iV_j>|V_0V_i-V_0V_j|=|t_i-t_j|, \] so (1) forces every contact on the opposite face \(V_1V_2V_3\) to be internal.

Therefore every sphere has one of exactly five possible contact patterns:

  • all six contacts are internal;
  • for one vertex \(V\), the three incident contacts are external and the three contacts on the opposite face are internal.

We next prove that each pattern admits at most one sphere. Choose the exceptional vertex \(V\) for an external pattern, or any vertex for the all-internal pattern. Let the opposite face be \(UWZ\). Its internal contacts determine \[ t_U=\frac{UW+UZ-WZ}{2}, \] and similarly determine \(t_W,t_Z\). The remaining tangent length is \[ t_V= \begin{cases} VU-t_U,&\text{for the all-internal pattern},\\ VU+t_U,&\text{for the pattern external at }V. \end{cases} \] The second formula follows because the exceptional vertex is the vertex of largest tangent length in the classification above. Thus the pattern determines all four tangent lengths.

Suppose two spheres, with centers \(O,O'\) and radii \(r,r'\), have these same tangent lengths. At every vertex \(V\), \[ |O-V|^2-r^2=t_V^2=|O'-V|^2-r'^2. \] Subtracting the identities at \(V_0\) and \(V_i\) gives \[ (O-O')\cdot(V_i-V_0)=0\qquad(i=1,2,3). \] The three edge vectors are linearly independent, so \(O=O'\), and then \(r=r'\). This proves uniqueness.

Since five distinct spheres exist, every one of the five patterns must occur. Let \[ x_S,x_A,x_B,x_C>0 \] be the tangent lengths for the all-internal sphere. Thus \[ UV=x_U+x_V \tag{2} \] for every edge.

Now consider the sphere whose external contacts are on \(SA,SB,SC\). Its contacts on \(ABC\) are internal, so its tangent lengths at \(A,B,C\) are respectively \(x_A,x_B,x_C\), since those lengths are uniquely determined by the three sides of \(ABC\). If its tangent length at \(S\) is \(y\), then \[ SA=y-x_A=x_S+x_A, \] and similarly for \(SB,SC\). Hence \[ y=x_S+2x_A=x_S+2x_B=x_S+2x_C, \] giving \(x_A=x_B=x_C\).

Applying the same argument to the sphere whose external contacts are on the three edges through \(A\) gives \(x_S=x_B=x_C\). Therefore all four \(x\)'s are equal. Equation (2) now shows that all six edges are equal, so \(SABC\) is regular.

(b) Similarities preserve tangency, so it suffices to consider the regular tetrahedron \[ S=(1,1,1),\quad A=(1,-1,-1),\quad B=(-1,1,-1),\quad C=(-1,-1,1). \] Every edge has length \(2\sqrt2\).

The sphere centered at the origin with radius \(1\) is tangent to every edge: the six edge midpoints are \[ (\pm1,0,0),\quad(0,\pm1,0),\quad(0,0,\pm1), \] and each midpoint’s radius is perpendicular to its edge.

For a second sphere, take center \[ Q=(-2,-2,-2)=-2S \] and radius \(3\). The line \(SA\) consists of points \((1,t,t)\), whose squared distances from \(Q\) are \[ 9+2(t+2)^2. \] Its distance from \(Q\) is therefore \(3\). The line \(AB\) consists of points \((t,-t,-1)\), whose squared distances from \(Q\) are \[ (t+2)^2+(2-t)^2+1=2t^2+9. \] Its distance from \(Q\) is also \(3\).

Permuting the coordinates fixes \(S\) and \(Q\), and permutes \(A,B,C\). Thus the sphere is tangent to all three lines through \(S\) and all three lines of the opposite face.

Finally, changing the signs of any two coordinates preserves the tetrahedron and produces three further spheres, each of radius \(3\), with centers \[ -2A,\qquad -2B,\qquad -2C. \] These four centers and the origin are distinct. Hence we have constructed five distinct spheres with the required tangencies.

The solution was checked against John Scholes' IMO solution archive (Kalva).