IMO 1960 Problem 6
The geometry becomes transparent after reducing the three-dimensional configuration to a two-dimensional meridian section through the axis of the cone.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 3m00s
Problem
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 $V_2$ be the volume of the cylinder.
a) Prove that $V_1 \neq V_2$;
b) Find the smallest number $k$ for which $V_1 = kV_2$; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.
Exploration
The geometry becomes transparent after reducing the three-dimensional configuration to a two-dimensional meridian section through the axis of the cone. The cone becomes an isosceles triangle, the inscribed sphere becomes the incircle of that triangle, and the circumscribed cylinder becomes a rectangle whose vertical sides are tangent to the incircle.
Let the cone have height $h$, base radius $R$, and insphere radius $r$. Since the sphere is tangent to the base plane and to the lateral surface, the center of the sphere lies on the axis at distance $r$ from the base plane. The circumscribed cylinder then has radius $r$ and height $2r$, because its lower base lies in the base plane of the cone and the sphere is tangent to both bases of the cylinder.
The first task is to express $V_1/V_2$ in terms of a single parameter. The standard formula for the inradius of a cone should emerge from the meridian triangle. If the slant height is $s=\sqrt{R^2+h^2}$, then the area of the meridian triangle is $Rh$, while its semiperimeter is $R+s$. Hence
$$r=\frac{Rh}{R+s}.$$
Substituting into
$$V_1=\frac13\pi R^2h,\qquad V_2=2\pi r^3$$
should yield a ratio depending only on $x=h/R$.
Compute:
$$\frac{V_1}{V_2} =\frac{R^2h}{6r^3} =\frac{(R+s)^3}{6Rh^2}.$$
Setting $x=h/R$ and $s=R\sqrt{1+x^2}$ gives
$$\frac{V_1}{V_2} =\frac{(1+\sqrt{1+x^2})^3}{6x^2}.$$
The problem becomes a one-variable minimization.
Try several values:
$$x=1 \implies \frac{V_1}{V_2}\approx0.971,$$
$$x=2 \implies \frac{V_1}{V_2}\approx0.873,$$
$$x=3 \implies \frac{V_1}{V_2}\approx0.879.$$
The minimum appears near $x=\sqrt2$:
$$x=\sqrt2 \implies \frac{V_1}{V_2} =\frac{(1+\sqrt3)^3}{12}.$$
A direct derivative is possible, but there may be a cleaner substitution. Let
$$t=\sqrt{1+x^2}.$$
Then $x^2=t^2-1$, so
$$\frac{V_1}{V_2} =\frac{(1+t)^3}{6(t^2-1)} =\frac{(1+t)^2}{6(t-1)}.$$
Expanding,
$$\frac{(1+t)^2}{t-1} =t+3+\frac{4}{t-1}.$$
Differentiating gives
$$1-\frac4{(t-1)^2}=0,$$
hence $t=3$, so $x^2=8$. This contradicts the earlier numerical guess. Recheck the algebra carefully.
Indeed,
$$\frac{(1+t)^3}{t^2-1} =\frac{(1+t)^2}{t-1},$$
which is correct. Testing $t=3$ gives
$$\frac{V_1}{V_2}=\frac{16}{12}=\frac43,$$
far larger than previous computations. The earlier numerical computations must therefore contain an error.
Recompute at $x=1$:
$$\frac{(1+\sqrt2)^3}{6}\approx2.35,$$
not $0.97$. So the minimum should indeed exceed $1$, which would establish part (a).
Now minimize correctly:
$$f(t)=\frac{(1+t)^2}{6(t-1)}.$$
Then
$$f'(t)=\frac{(t+1)(t-3)}{6(t-1)^2}.$$
Thus the minimum occurs at $t=3$, namely
$$\sqrt{1+x^2}=3,\qquad x^2=8.$$
Hence
$$\frac hR=2\sqrt2.$$
The final geometric construction asks for the angle subtended by a diameter of the base at the vertex. In the meridian section this is exactly the apex angle $2\alpha$ of the cone, where
$$\tan\alpha=\frac Rh.$$
Since $h/R=2\sqrt2$,
$$\tan\alpha=\frac1{2\sqrt2}.$$
Then
$$\sin\alpha=\frac13,$$
because in a right triangle with opposite side $1$, adjacent side $2\sqrt2$, the hypotenuse is $3$. Hence the required angle is
$$2\alpha=2\arcsin\frac13.$$
A cleaner exact relation comes from
$$\cos(2\alpha)=1-2\sin^2\alpha =1-\frac29 =\frac79.$$
The delicate point is the formula for $r$. Any mistake there propagates through the entire optimization.
Problem Understanding
The problem concerns a right circular cone containing its inscribed sphere. Around this sphere one constructs a cylinder tangent to the sphere, with the lower base of the cylinder lying in the base plane of the cone. The cone has volume $V_1$ and the cylinder has volume $V_2$.
Part (a) asks for a proof that the two volumes are never equal. Part (b) asks for the smallest possible constant $k$ such that
$$V_1=kV_2,$$
together with the geometric configuration for which equality occurs. Since this is an optimization problem with an extremal value to determine and an equality case to construct, the problem is of Type C.
The key geometric objects are the cone, its insphere, and the cylinder tangent to that sphere. The main difficulty is that the dimensions are not independent. The sphere simultaneously touches the base and the lateral surface of the cone, so its radius is constrained by the cone dimensions. A naive approach using arbitrary parameters produces too many variables. The essential reduction is to pass to a meridian section and use the inradius formula for a triangle.
The expected answer is that the ratio $V_1/V_2$ has a positive minimum strictly larger than $1$. Intuitively, the cone always wastes more space near the vertex than the cylinder does, so the cone volume should always exceed the cylinder volume. The extremal configuration should occur for a special proportion between height and base radius.
Proof Architecture
The proof proceeds through four claims.
Lemma 1 states that if the cone has base radius $R$, height $h$, and slant height $s=\sqrt{R^2+h^2}$, then the insphere radius is
$$r=\frac{Rh}{R+s}.$$
This follows by passing to the meridian section and applying the formula for the inradius of a triangle.
Lemma 2 states that
$$\frac{V_1}{V_2} =\frac{(1+\sqrt{1+x^2})^3}{6x^2}, \qquad x=\frac hR.$$
This is obtained by substituting the expression for $r$ into the volume formulas.
Lemma 3 states that for
$$t=\sqrt{1+x^2}>1,$$
the ratio becomes
$$\frac{V_1}{V_2} =\frac{(1+t)^2}{6(t-1)},$$
and this quantity is minimized uniquely at $t=3$. Differentiation gives the minimum.
Lemma 4 states that equality in the minimum case corresponds to
$$\frac hR=2\sqrt2,$$
and the angle subtended by a diameter of the base at the cone vertex equals
$$2\arcsin\frac13.$$
The most delicate step is Lemma 1. A small algebraic error in the inradius computation changes the optimization completely.
Solution
Let the cone have base radius $R$, height $h$, and slant height
$$s=\sqrt{R^2+h^2}.$$
Let $r$ denote the radius of the inscribed sphere.
The cylinder circumscribed about the sphere has radius $r$. Since one base of the cylinder lies in the base plane of the cone and the sphere is tangent to both bases of the cylinder, the height of the cylinder equals $2r$.
Hence
$$V_1=\frac13\pi R^2h, \qquad V_2=\pi r^2(2r)=2\pi r^3.$$
Lemma 1
The radius $r$ of the inscribed sphere satisfies
$$r=\frac{Rh}{R+s}.$$
Proof.
Consider the meridian section through the axis of the cone. This section is an isosceles triangle whose base equals $2R$, whose equal sides equal $s$, and whose altitude equals $h$.
The inscribed sphere becomes the incircle of this triangle. Hence $r$ is the inradius of the triangle.
The area of the triangle is
$$\Delta=\frac12\cdot 2R\cdot h=Rh.$$
Its semiperimeter is
$$p=\frac{2R+2s}{2}=R+s.$$
For any triangle,
$$\Delta=pr,$$
because the triangle decomposes into three triangles having common altitude $r$ to the three sides.
Therefore
$$r=\frac{\Delta}{p} =\frac{Rh}{R+s}.$$
This proves the lemma. ∎
This establishes the precise dependence of the sphere radius on the cone dimensions; replacing this step by an unproved geometric intuition would leave the optimization unsupported.
Lemma 2
If
$$x=\frac hR,$$
then
$$\frac{V_1}{V_2} = \frac{(1+\sqrt{1+x^2})^3}{6x^2}.$$
Proof.
From Lemma 1,
$$r=\frac{Rh}{R+s}.$$
Hence
$$r^3=\frac{R^3h^3}{(R+s)^3}.$$
Using the volume formulas,
$$\frac{V_1}{V_2} = \frac{\frac13\pi R^2h}{2\pi r^3} = \frac{R^2h}{6r^3}.$$
Substituting the expression for $r^3$ gives
$$\frac{V_1}{V_2} = \frac{R^2h}{6}\cdot \frac{(R+s)^3}{R^3h^3} = \frac{(R+s)^3}{6Rh^2}.$$
Now write
$$h=xR, \qquad s=\sqrt{R^2+h^2}=R\sqrt{1+x^2}.$$
Then
$$R+s = R\left(1+\sqrt{1+x^2}\right).$$
Substituting,
$$\frac{V_1}{V_2} = \frac{ R^3(1+\sqrt{1+x^2})^3 }{ 6R(x^2R^2) } = \frac{(1+\sqrt{1+x^2})^3}{6x^2}.$$
This proves the lemma. ∎
This reduces the three-dimensional geometry to a one-variable function; skipping this normalization obscures the extremal structure.
Lemma 3
The quantity
$$\frac{V_1}{V_2}$$
is always greater than $1$, and its minimum value equals
$$\frac43.$$
Proof.
Define
$$t=\sqrt{1+x^2}.$$
Since $x>0$, one has $t>1$ and
$$x^2=t^2-1.$$
By Lemma 2,
$$\frac{V_1}{V_2} = \frac{(1+t)^3}{6(t^2-1)} = \frac{(1+t)^2}{6(t-1)}.$$
Define
$$f(t)=\frac{(1+t)^2}{6(t-1)}, \qquad t>1.$$
Differentiate:
$$f'(t) = \frac{2(t+1)\cdot 6(t-1)-6(t+1)^2}{36(t-1)^2}.$$
Factor the numerator:
$$f'(t) = \frac{(t+1)\bigl(2(t-1)-(t+1)\bigr)}{6(t-1)^2} = \frac{(t+1)(t-3)}{6(t-1)^2}.$$
Since $(t-1)^2>0$ and $t+1>0$, the sign of $f'(t)$ is the sign of $t-3$.
Thus
$$f'(t)<0 \quad \text{for } 1<t<3,$$
and
$$f'(t)>0 \quad \text{for } t>3.$$
Hence $f$ attains its unique minimum at
$$t=3.$$
Evaluating,
$$f(3) = \frac{(1+3)^2}{6(3-1)} = \frac{16}{12} = \frac43.$$
Therefore
$$\frac{V_1}{V_2}\ge \frac43.$$
In particular,
$$V_1\ne V_2.$$
This proves the lemma. ∎
This establishes both the impossibility of equality and the sharp lower bound; a derivative computation without sign analysis would not justify minimality.
Lemma 4
Equality
$$\frac{V_1}{V_2}=\frac43$$
holds precisely when the angle subtended by a diameter of the cone base at the vertex equals
$$2\arcsin\frac13.$$
Proof.
Equality occurs exactly when
$$t=3.$$
Since
$$t=\sqrt{1+x^2},$$
this gives
$$1+x^2=9,$$
hence
$$x^2=8.$$
Because
$$x=\frac hR>0,$$
one obtains
$$\frac hR=2\sqrt2.$$
Let $2\alpha$ denote the angle subtended by a diameter of the base at the cone vertex. In the meridian section, $\alpha$ is the angle between the axis and a generator of the cone.
From the right triangle in the meridian section,
$$\tan\alpha=\frac Rh.$$
Using
$$\frac hR=2\sqrt2,$$
one gets
$$\tan\alpha=\frac1{2\sqrt2}.$$
Take a right triangle with opposite side $1$ and adjacent side $2\sqrt2$. Its hypotenuse equals
$$\sqrt{1+8}=3.$$
Therefore
$$\sin\alpha=\frac13.$$
Hence
$$\alpha=\arcsin\frac13,$$
and the required angle equals
$$2\alpha=2\arcsin\frac13.$$
This proves the lemma. ∎
This identifies the extremal cone uniquely; omitting the trigonometric reconstruction would leave the geometric construction incomplete.
Combining the lemmas yields
$$\frac{V_1}{V_2}\ge\frac43,$$
with equality exactly when
$$\frac hR=2\sqrt2.$$
Thus the smallest possible value of $k$ is
$$\boxed{\frac43},$$
and equality occurs when the angle subtended by a diameter of the base at the vertex is
$$\boxed{2\arcsin\frac13}.$$
Equality is achieved for the cone satisfying
$$\frac hR=2\sqrt2.$$
Verification of Key Steps
The first delicate step is the formula
$$r=\frac{Rh}{R+s}.$$
Derive it independently. In the meridian section, the triangle has side lengths
$$s,s,2R.$$
Its area is
$$Rh.$$
The incircle touches all three sides, so decomposing the triangle into three triangles with common altitude $r$ gives
$$Rh = \frac12rs+\frac12rs+\frac12r(2R) = r(s+R).$$
Hence
$$r=\frac{Rh}{R+s}.$$
A careless computation often replaces the semiperimeter by the perimeter, producing the incorrect factor
$$r=\frac{Rh}{2(R+s)}.$$
That error changes the final minimum by a factor of $8$.
The second delicate step is the minimization. Starting from
$$f(t)=\frac{(1+t)^2}{6(t-1)},$$
one must differentiate carefully:
$$f'(t) = \frac{2(t+1)(t-1)-(t+1)^2}{6(t-1)^2}.$$
Expanding the numerator,
$$2(t^2-1)-(t^2+2t+1) = t^2-2t-3 = (t+1)(t-3).$$
A sign error here incorrectly predicts a minimum at $t=1+\sqrt5$ or another spurious value.
The third delicate step is the reconstruction of the angle. Since
$$\tan\alpha=\frac Rh=\frac1{2\sqrt2},$$
one must compute
$$\sin\alpha = \frac{1}{\sqrt{1+8}} = \frac13.$$
Using instead
$$\sin\alpha=\frac Rh$$
would confuse tangent with sine and lead to an impossible value exceeding $1$ for other parameter choices.
Alternative Approaches
A different method avoids calculus. From
$$\frac{V_1}{V_2} = \frac{(1+t)^2}{6(t-1)}, \qquad t>1,$$
one may write
$$\frac{V_1}{V_2}-\frac43 = \frac{(1+t)^2-8(t-1)}{6(t-1)}.$$
Expanding the numerator gives
$$t^2-6t+9 = (t-3)^2.$$
Hence
$$\frac{V_1}{V_2} = \frac43+\frac{(t-3)^2}{6(t-1)}.$$
Since $t>1$, the denominator is positive, so
$$\frac{V_1}{V_2}\ge\frac43,$$
with equality exactly when $t=3$. This proof is shorter and avoids differentiation entirely.
Another approach uses the half-apex angle $\alpha$ directly. Since
$$R=h\tan\alpha, \qquad s=h\sec\alpha,$$
Lemma 1 yields
$$r=\frac{h\tan\alpha}{\tan\alpha+\sec\alpha}.$$
Substituting into the volume formulas reduces the ratio to a trigonometric expression in $\alpha$. The resulting optimization is equivalent but algebraically less transparent than the substitution
$$t=\sqrt{1+x^2}.$$