IMO 1960 · Problem 6

Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let \(V_1\) be the volume of the cone and \(V_2\) the volume of the cylinder.

(a) Prove that \(V_1 \ne V_2\).

(b) Find the smallest number \(k\) for which \(V_1 = kV_2\), for this case, construct the angle subtended by a diameter of the base of the cone at the vertex of the cone.

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

Answer

\[ \boxed{k_{\min}=\frac43}. \]

The corresponding vertex angle is \[ \boxed{2\arcsin\frac13=\arccos\frac79}. \] A straightedge-and-compass construction is given below.

Solution

Let the cone have base radius \(R\), height \(h\), and slant height \[ \ell=\sqrt{R^2+h^2}, \] and let the inscribed sphere have radius \(r\). All these lengths are positive, since a degenerate cone cannot contain a sphere of positive radius.

An axial section of the cone is an isosceles triangle with base \(2R\), height \(h\), equal sides \(\ell\), and incircle radius \(r\). Calculating its area by joining the incenter to its vertices gives \[ Rh=r(R+\ell). \] Consequently, \[ \frac hr=1+\frac{\ell}{R}>2, \] because \(\ell>R\). Rearranging the area identity and then squaring yields \[ r\ell=R(h-r), \qquad r^2(R^2+h^2)=R^2(h-r)^2. \] After cancellation, \[ R^2=\frac{r^2h}{h-2r}. \]

The cylinder circumscribed about the sphere has radius \(r\) and height \(2r\). Therefore \[ V_1=\frac{\pi R^2h}{3}, \qquad V_2=2\pi r^3, \] and hence \[ \frac{V_1}{V_2} =\frac{h^2}{6r(h-2r)} =\frac43+\frac{(h-4r)^2}{6r(h-2r)} \ge \frac43. \] The denominator is positive because \(h>2r\). Thus \(V_1>V_2\), proving (a).

Equality in this bound holds exactly when \(h=4r\), in which case \(R=\sqrt2\,r\). Conversely, a cone with height \(4r\) and base radius \(\sqrt2\,r\) has slant height \(3\sqrt2\,r\), so its inradius is \[ \frac{Rh}{R+\ell} =\frac{(\sqrt2\,r)(4r)}{\sqrt2\,r+3\sqrt2\,r} =r. \] Thus equality is attained, proving that the smallest value in (b) is \(k=4/3\).

To determine the angle, consider an axial section in the equality case. Let \(S\) be its vertex, \(I\) its incenter, and \(T\) a point where its incircle touches an equal side. The center \(I\) lies a distance \(r\) above the base, so \[ SI=h-r=3r. \] Since \(IT\perp ST\), the half-angle \(\theta=\angle IST\) satisfies \[ \sin\theta=\frac{IT}{SI}=\frac13. \] The angle subtended by a base diameter is therefore \[ 2\theta=2\arcsin\frac13=\arccos\frac79. \]

To construct this angle with straightedge and compass, choose any length \(u>0\):

  1. Construct a segment \(SI=3u\).
  2. Draw the circle centered at \(I\) with radius \(u\), and the circle with diameter \(SI\). Let their two intersection points be \(T_1,T_2\).
  3. Draw the rays \(ST_1\) and \(ST_2\).

By Thales’ theorem, \(IT_j\perp ST_j\), so these rays are tangents to the first circle. They are symmetric about \(SI\), and each makes an angle with \(SI\) whose sine is \(u/(3u)=1/3\). Hence \(\angle T_1ST_2\) is precisely the required angle.

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