IMO 1960 Problem 7
The problem concerns a right circular cone inscribed in a sphere and a cylinder circumscribed about the same sphere.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 5m59s
Problem
A regular cone is inscribed in a sphere. Around the sphere a cylinder is circumscribed so that its base lies in the same plane as the base of the cone. Let $V_1$ be the volume of the cone and $V_2$ that of the cylinder.
(a) Prove that $V_1 = V_2$ is impossible.
(b) Find the smallest $k$ for which $V_1 = k V_2$, and in this case construct the half-angle at the vertex of the cone.
Exploration
The problem concerns a right circular cone inscribed in a sphere and a cylinder circumscribed about the same sphere. Since the cylinder is tangent to the sphere and has its base in the same plane as the cone’s base, both solids are controlled by the same geometric parameters. The natural parameter is the cone’s half-angle at the vertex, denoted by $\alpha$, or equivalently the height of the cone relative to the sphere’s radius.
Let the sphere have radius $R$, and let the cone have height $h$. If the base plane lies at distance $R-h$ from the center of the sphere, then the cone’s base radius is
$$r=\sqrt{R^2-(R-h)^2}=\sqrt{h(2R-h)}.$$
Hence
$$V_1=\frac13\pi r^2h =\frac13\pi h^2(2R-h).$$
The circumscribed cylinder about the sphere has radius $R$ and height $2R$, since every tangent cylinder around a sphere has these dimensions. Thus
$$V_2=2\pi R^3.$$
The ratio becomes
$$\frac{V_1}{V_2} = \frac{h^2(2R-h)}{6R^3}.$$
Introducing
$$x=\frac{h}{R}, \qquad 0<x<2,$$
gives
$$\frac{V_1}{V_2} = \frac{x^2(2-x)}{6}.$$
Part (a) asks whether this ratio can equal $1$. Since
$$x^2(2-x)\le ?$$
it seems likely that the maximum is strictly below $6$. Trying values:
$$x=1 \implies \frac16, \qquad x=\frac43 \implies \frac{16}{81}, \qquad x=2 \implies 0.$$
The value $\frac{16}{81}$ appears naturally from optimizing a cubic, suggesting the minimum admissible $k$ in
$$V_1=kV_2$$
is the reciprocal:
$$k_{\min}=\frac{V_1}{V_2}_{\max}=\frac{16}{81}.$$
Since the problem phrases it as “smallest $k$ for which $V_1=kV_2$”, one must interpret this carefully. Equality can hold only for ratios actually attained. Because $V_1<V_2$, the intended meaning is the smallest constant $k$ satisfying $V_1=kV_2$, namely the largest attainable ratio. The critical point of
$$x^2(2-x)$$
comes from
$$4x-3x^2=x(4-3x)=0,$$
yielding $x=\frac43$. Then
$$\cos\alpha = \frac{h-R}{R} = x-1 = \frac13.$$
Hence
$$\alpha=\arccos\frac13.$$
A geometric construction of this angle should come from constructing a right triangle with adjacent side $1$ and hypotenuse $3$.
The delicate point is interpreting the parameter correctly and avoiding an algebraic slip in maximizing the cubic.
Problem Understanding
We are given a sphere with an inscribed regular cone and a cylinder tangent to the sphere. The plane of the cylinder’s base coincides with the plane of the cone’s base. Let $V_1$ denote the cone’s volume and $V_2$ the cylinder’s volume.
The task first asks for a proof that the equality
$$V_1=V_2$$
cannot occur. Then it asks for the smallest constant $k$ such that
$$V_1=kV_2$$
for some admissible cone, and requests the construction of the cone’s half-angle at the vertex in the extremal case.
This is a Type C problem, because one must determine an extremal value of a ratio. The proof must establish two facts: first, the claimed value is attained by an explicit cone; second, no better value is possible.
The mathematical objects are a sphere, a right circular cone inscribed in it, and a tangent cylinder. The core difficulty lies in reducing the geometry to a single parameter and proving a sharp extremal bound. A naive geometric comparison of volumes gives no immediate information because the cone and cylinder depend on the same sphere in different ways.
The expected answer is
$$k=\frac{16}{81},$$
attained when the cone’s height is $\frac43$ times the sphere radius. The half-angle $\alpha$ at the vertex satisfies
$$\cos\alpha=\frac13.$$
This value is plausible because optimizing the cone volume inside a sphere produces a nontrivial cubic extremum.
Proof Architecture
We will proceed through three lemmas.
Lemma 1. If the sphere has radius $R$ and the cone has height $h$, then the cone volume equals
$$V_1=\frac13\pi h^2(2R-h).$$
This follows by expressing the cone’s base radius through the Pythagorean theorem in the sphere.
Lemma 2. The circumscribed cylinder has volume
$$V_2=2\pi R^3.$$
This follows because a sphere tangent to a right circular cylinder determines the cylinder radius and height uniquely.
Lemma 3. For
$$0<x<2,$$
the function
$$f(x)=x^2(2-x)$$
satisfies
$$f(x)\le \frac{32}{27},$$
with equality exactly when
$$x=\frac43.$$
This gives the maximal possible ratio $V_1/V_2$.
The hardest direction is proving sharp optimality, because one must guarantee that no larger ratio can occur. The most error-prone step is maximizing the cubic expression without overlooking endpoint behavior or the admissibility of the equality case.
Solution
Let the sphere have radius $R$, and let the cone have height $h$. Since the cone is inscribed in the sphere, its vertex and every point of its base circle lie on the sphere.
Choose coordinates so that the center of the sphere is at the origin and the cone’s axis is vertical. Let the vertex of the cone be the highest point of the sphere, and let the base plane lie at height $R-h$. If $r$ denotes the base radius of the cone, then every point of the base circle lies on the sphere, so
$$r^2+(R-h)^2=R^2.$$
Hence
$$r^2 = R^2-(R-h)^2 = 2Rh-h^2 = h(2R-h).$$
Lemma 1
The cone volume is
$$V_1=\frac13\pi h^2(2R-h).$$
Proof
The volume of a right circular cone equals
$$V_1=\frac13\pi r^2h.$$
Substituting
$$r^2=h(2R-h)$$
gives
$$V_1 = \frac13\pi h\bigl(h(2R-h)\bigr) = \frac13\pi h^2(2R-h).$$
∎
This establishes the cone volume in terms of a single parameter $h$; replacing $r$ independently would fail because $r$ and $h$ are constrained by the sphere.
Lemma 2
The cylinder volume is
$$V_2=2\pi R^3.$$
Proof
A cylinder circumscribed about a sphere of radius $R$ is tangent to the sphere along its lateral surface, so its radius equals $R$. Since the sphere touches both bases, the cylinder height equals the sphere diameter $2R$.
Hence
$$V_2 = \pi R^2(2R) = 2\pi R^3.$$
∎
This establishes that the cylinder volume is independent of the cone; treating it as variable would obscure the optimization.
We now introduce
$$x=\frac{h}{R}.$$
Since the cone lies inside the sphere,
$$0<h<2R,$$
so
$$0<x<2.$$
From Lemmas 1 and 2,
$$\frac{V_1}{V_2} = \frac{\frac13\pi h^2(2R-h)}{2\pi R^3} = \frac{h^2(2R-h)}{6R^3}.$$
Substituting $h=xR$ yields
$$\frac{V_1}{V_2} = \frac{x^2(2-x)}{6}.$$
Lemma 3
For
$$0<x<2,$$
one has
$$x^2(2-x)\le \frac{32}{27},$$
with equality if and only if
$$x=\frac43.$$
Proof
Define
$$f(x)=x^2(2-x).$$
Differentiation gives
$$f'(x)=4x-3x^2=x(4-3x).$$
The critical points are
$$x=0, \qquad x=\frac43.$$
Only
$$x=\frac43$$
lies in the interval $(0,2)$.
The second derivative is
$$f''(x)=4-6x.$$
Thus
$$f''!\left(\frac43\right) = 4-8 = -4<0,$$
so
$$x=\frac43$$
gives a local maximum.
At the endpoints,
$$f(0)=0, \qquad f(2)=0.$$
Therefore the interior critical point gives the global maximum on $[0,2]$:
$$f!\left(\frac43\right) = \left(\frac43\right)^2 \left(2-\frac43\right) = \frac{16}{9}\cdot\frac23 = \frac{32}{27}.$$
∎
This establishes the sharp upper bound; checking only the derivative without endpoint analysis would leave open the possibility of a larger boundary value.
For part (a), suppose
$$V_1=V_2.$$
Then
$$\frac{V_1}{V_2}=1.$$
However, Lemma 3 gives
$$\frac{V_1}{V_2} = \frac{x^2(2-x)}{6} \le \frac{32}{162} = \frac{16}{81} <1.$$
Hence
$$V_1=V_2$$
is impossible.
For part (b), the largest possible ratio is
$$\frac{16}{81}.$$
Hence the smallest constant $k$ for which
$$V_1=kV_2$$
can occur is
$$k=\frac{16}{81},$$
and equality holds exactly when
$$x=\frac43, \qquad h=\frac43R.$$
Let $\alpha$ denote the half-angle at the cone’s vertex. In the axial cross section of the cone,
$$\tan\alpha=\frac{r}{h}.$$
Using
$$r^2=h(2R-h)$$
and
$$h=\frac43R,$$
we obtain
$$r^2 = \frac43R\cdot\frac23R = \frac89R^2,$$
so
$$r=\frac{2\sqrt2}{3}R.$$
The distance from the sphere center to the vertex equals $R$, and the distance from the center to the base plane equals
$$h-R=\frac13R.$$
In the axial right triangle,
$$\cos\alpha = \frac{h-R}{R} = \frac13.$$
Hence
$$\alpha=\arccos\frac13.$$
To construct this angle, construct a right triangle with hypotenuse $3$ and adjacent side $1$; the angle opposite the remaining side satisfies
$$\cos\alpha=\frac13.$$
Thus the extremal value is
$$\boxed{k=\frac{16}{81}},$$
with equality when the cone half-angle satisfies
$$\boxed{\cos\alpha=\frac13}.$$
Verification of Key Steps
The first delicate step is the relation
$$r^2=h(2R-h).$$
Re-deriving it directly, the base plane lies at signed distance $R-h$ from the center. Since the base circle lies on the sphere,
$$r^2+(R-h)^2=R^2.$$
Expanding gives
$$r^2 = R^2-\bigl(R^2-2Rh+h^2\bigr) = 2Rh-h^2.$$
A careless argument may incorrectly use the distance $h$ instead of $R-h$, producing an impossible formula.
The second delicate step is maximizing
$$x^2(2-x).$$
An independent derivation uses algebra:
$$x^2(2-x) = \frac{32}{27} - \left(x-\frac43\right)^2 \left(x+\frac23\right).$$
Since
$$x+\frac23>0,$$
the correction term is nonnegative, so
$$x^2(2-x)\le \frac{32}{27},$$
with equality only at
$$x=\frac43.$$
Ignoring endpoint values after differentiation could silently miss a larger boundary value.
The third delicate step is computing the half-angle. Since the center lies on the cone axis and the vertex is on the sphere, the adjacent side in the axial triangle equals
$$h-R=\frac13R,$$
while the slant segment from vertex to center equals $R$. Using $h$ instead of $R$ in the denominator would incorrectly give
$$\cos\alpha=\frac14.$$
Alternative Approaches
A purely geometric approach avoids calculus. After writing
$$\frac{V_1}{V_2} = \frac{x^2(2-x)}{6},$$
one may rewrite
$$x^2(2-x) = \frac{32}{27} - \left(x-\frac43\right)^2 \left(x+\frac23\right).$$
Since the second term is nonnegative on $(0,2)$, the maximum follows immediately. This proof is shorter and algebraic.
Another method parametrizes the cone by the half-angle $\alpha$. Since the sphere center lies on the axis,
$$h=R(1+\cos\alpha), \qquad r=R\sin\alpha.$$
Then
$$V_1 = \frac13\pi R^3 (1+\cos\alpha)^2(1-\cos\alpha^2),$$
which simplifies to a cubic in $\cos\alpha$. Optimizing that expression again yields
$$\cos\alpha=\frac13.$$
The height parametrization is preferable because it reduces the geometry to one variable immediately and keeps the algebra shorter.