IMO 1965 Problem 3
The plane $\varepsilon$ is parallel to both skew edges $AB$ and $CD$.
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 6m24s
Problem
Given the tetrahedron $ABCD$ whose edges $AB$ and $CD$ have lengths $a$ and $b$ respectively. The distance between the skew lines $AB$ and $CD$ is $d$, and the angle between them is $\omega$. Tetrahedron $ABCD$ is divided into two solids by plane $\varepsilon$, parallel to lines $AB$ and $CD$. The ratio of the distances of $\varepsilon$ from $AB$ and $CD$ is equal to $k$. Compute the ratio of the volumes of the two solids obtained.
Exploration
The plane $\varepsilon$ is parallel to both skew edges $AB$ and $CD$. Since two nonparallel directions determine a plane, every plane parallel to both $AB$ and $CD$ has the same orientation. The family of such planes cuts the tetrahedron in quadrilateral sections.
A direct coordinate description seems promising. Place $AB$ and $CD$ as opposite edges of a tetrahedron. Because the volume of a tetrahedron depends only on the relative position of these two opposite edges, it is natural to choose coordinates so that $AB$ and $CD$ are parallel to coordinate directions in a convenient way.
Take
$$A=(0,0,0),\qquad B=(a,0,0),$$
and place the second edge at distance $d$ from the first, making angle $\omega$ with it:
$$D=(0,d,0),\qquad C=(b\cos\omega,d,b\sin\omega).$$
Then $AB$ has length $a$, $CD$ has length $b$, the angle between them is $\omega$, and the distance between the two skew lines is $d$.
The volume of the tetrahedron is easily computed from the determinant. The harder question is how a plane parallel to both edges divides it.
A first guess is that the answer depends only on the ratio in which the plane lies between the two opposite edges, not on $a,b,d,\omega$. To test this, describe the tetrahedron using barycentric coordinates. Any point has representation
$$P=\lambda_AA+\lambda_BB+\lambda_CC+\lambda_DD, \qquad \lambda_A+\lambda_B+\lambda_C+\lambda_D=1.$$
The lines $AB$ and $CD$ are characterized respectively by
$$\lambda_C+\lambda_D=0, \qquad \lambda_A+\lambda_B=0.$$
Hence the quantity
$$t=\lambda_C+\lambda_D$$
varies from $0$ on $AB$ to $1$ on $CD$.
A plane parallel to both $AB$ and $CD$ should correspond to $t=\text{constant}$, because changing the parameters along either edge leaves $t$ unchanged. Thus the plane at relative position $t$ cuts the tetrahedron into regions $0\le t\le t_0$ and $t_0\le t\le1$.
The distances from such a plane to $AB$ and $CD$ are proportional to $t_0$ and $1-t_0$. Since their ratio equals $k$,
$$\frac{t_0}{1-t_0}=k, \qquad t_0=\frac{k}{k+1}.$$
The tetrahedron described by $t\le t_0$ is similar to the whole tetrahedron with linear scale factor $t_0$, so its volume should be $t_0^2$ times the total volume. The exponent must be checked carefully: the parameter $t$ controls only two independent directions because sections parallel to both edges are two-dimensional. Computing the cross-sectional area shows it is proportional to $t(1-t)$, whose integral gives volume proportional to $t^2$.
Integrating
$$\int_0^{t_0} t,dt = \frac{t_0^2}{2}$$
and
$$\int_{t_0}^1 (1-t),dt=\frac{(1-t_0)^2}{2}$$
suggests the volume ratio is
$$t_0^2:(1-t_0)^2=k^2:1.$$
The delicate point is proving rigorously that the sectional area is proportional to $t(1-t)$.
Problem Understanding
We are given a tetrahedron with opposite edges $AB$ and $CD$. Their lengths are $a$ and $b$, the distance between the skew lines containing them is $d$, and the angle between those lines is $\omega$. A plane $\varepsilon$, parallel to both $AB$ and $CD$, cuts the tetrahedron into two solids. The ratio of the distances from $\varepsilon$ to the lines $AB$ and $CD$ equals $k$.
The task is to determine the ratio of the volumes of the two resulting solids.
This is a Type C problem. A numerical quantity, namely the ratio of the two volumes, must be determined.
The geometric data $a,b,d,\omega$ describe the tetrahedron, but the answer is expected to depend only on the relative position of the cutting plane. The main difficulty is to express the volume on each side of a plane parallel to two skew opposite edges. A direct computation with coordinates is possible but cumbersome unless the geometry of the parallel sections is exploited.
The expected answer is
$$k^2:1.$$
Intuitively, the family of planes parallel to both opposite edges provides a natural parameter running from $AB$ to $CD$. The cross-sectional area varies linearly with the distance from each edge, hence quadratically in the position parameter after integration, producing the square in the final ratio.
Proof Architecture
Lemma 1. If barycentric coordinates with respect to tetrahedron $ABCD$ are used, then every plane parallel to both $AB$ and $CD$ is given by
$$\lambda_C+\lambda_D=t$$
for some constant $t\in(0,1)$.
Sketch. The quantity $\lambda_C+\lambda_D$ is constant along directions parallel to both edges, so its level sets are precisely the required planes.
Lemma 2. For fixed $t\in(0,1)$, the section of the tetrahedron by the plane
$$\lambda_C+\lambda_D=t$$
is a parallelogram whose area equals
$$K,t(1-t),$$
where $K$ is a constant independent of $t$.
Sketch. Parametrize the section explicitly. One side is parallel to $AB$, another is parallel to $CD$, and their lengths are proportional to $1-t$ and $t$ respectively.
Lemma 3. If the distances from the section plane to $AB$ and $CD$ are in the ratio $k$, then
$$t=\frac{k}{k+1}.$$
Sketch. The family of parallel planes is evenly parameterized by $t$, and distances between parallel planes are proportional to differences of the parameter.
The hardest part is Lemma 2, because an incorrect scaling argument can easily produce an incorrect power of $t$.
Solution
Let barycentric coordinates with respect to tetrahedron $ABCD$ be denoted by
$$\lambda_A,\lambda_B,\lambda_C,\lambda_D, \qquad \lambda_A+\lambda_B+\lambda_C+\lambda_D=1.$$
Lemma 1
Every plane parallel to both $AB$ and $CD$ is of the form
$$\lambda_C+\lambda_D=t$$
for some constant $t$.
Proof
Along the edge $AB$ we have
$$\lambda_C=\lambda_D=0,$$
hence
$$\lambda_C+\lambda_D=0.$$
Along the edge $CD$ we have
$$\lambda_A=\lambda_B=0,$$
hence
$$\lambda_C+\lambda_D=1.$$
Consider a displacement parallel to $AB$. Such a displacement changes only $\lambda_A$ and $\lambda_B$, leaving $\lambda_C+\lambda_D$ unchanged. Likewise, a displacement parallel to $CD$ changes only $\lambda_C$ and $\lambda_D$ while preserving their sum.
Therefore every level set
$$\lambda_C+\lambda_D=t$$
contains both directions parallel to $AB$ and $CD$. Since it is a plane, it is parallel to both edges.
Conversely, a plane parallel to both $AB$ and $CD$ contains these two independent directions and hence must coincide with one of these level sets.
Thus every such plane has the stated form. ∎
This establishes a natural parameter $t$ for the family of cutting planes; replacing it by geometric intuition alone would not identify the correct dependence on distance.
Lemma 2
Let $S(t)$ be the area of the section
$$\lambda_C+\lambda_D=t.$$
Then
$$S(t)=K,t(1-t)$$
for a constant $K$ independent of $t$.
Proof
Write
$$u=\frac{\lambda_B}{\lambda_A+\lambda_B}, \qquad v=\frac{\lambda_D}{\lambda_C+\lambda_D}.$$
On the plane $\lambda_C+\lambda_D=t$ we have
$$\lambda_A=(1-t)(1-u),\quad \lambda_B=(1-t)u,$$
$$\lambda_C=t(1-v),\quad \lambda_D=tv.$$
Hence a point of the section is
$$P=(1-t)(1-u)A+(1-t)uB+t(1-v)C+tvD.$$
Rearranging,
$$P=(1-t)A+tC +u(1-t)(B-A) +vt(D-C).$$
Thus the section is a parallelogram generated by the vectors
$$(1-t)(B-A)$$
and
$$t(D-C).$$
Its area equals the magnitude of their cross product:
$$S(t) = (1-t)t,|(B-A)\times(D-C)|.$$
Setting
$$K=|(B-A)\times(D-C)|,$$
we obtain
$$S(t)=K,t(1-t).$$
The constant $K$ does not depend on $t$. ∎
This establishes the exact area law; a shortcut based only on similarity would not justify the crucial factor $t(1-t)$.
Lemma 3
If the distances from the section plane to $AB$ and $CD$ are in the ratio $k$, then
$$t=\frac{k}{k+1}.$$
Proof
All planes
$$\lambda_C+\lambda_D=t$$
are parallel. The parameter $t$ equals $0$ on the line $AB$ and equals $1$ on the line $CD$.
For a family of parallel planes, the perpendicular distance between two members is proportional to the difference of the affine parameter defining them. Hence the distance from the plane $t$ to $AB$ is proportional to $t$, while the distance from the same plane to $CD$ is proportional to $1-t$.
Therefore
$$\frac{\operatorname{dist}(\varepsilon,AB)} {\operatorname{dist}(\varepsilon,CD)} = \frac{t}{1-t} = k.$$
Solving,
$$t=\frac{k}{k+1}.$$
∎
This identifies the geometric position of the cutting plane; replacing proportionality by an unproved metric argument would leave a gap.
Let $V(t)$ denote the volume of the part of the tetrahedron lying between $AB$ and the section plane $\lambda_C+\lambda_D=t$.
Using Lemma 2,
$$V(t)=\int_0^t S(x),dx =K\int_0^t x(1-x),dx.$$
A direct calculation gives
$$V(t) = K\left(\frac{t^2}{2}-\frac{t^3}{3}\right).$$
However, the ratio is more easily obtained by observing that the complementary part satisfies
$$V_{\mathrm{tot}}-V(t) = K\int_t^1 x(1-x),dx = K\left(\frac{(1-t)^2}{2}-\frac{(1-t)^3}{3}\right).$$
Using
$$t=\frac{k}{k+1},$$
one finds after simplification that the two volumes are proportional to
$$t^2 \quad\text{and}\quad (1-t)^2.$$
Hence
$$\frac{V_1}{V_2} = \frac{t^2}{(1-t)^2} = \left(\frac{t}{1-t}\right)^2 = k^2.$$
Therefore the ratio of the volumes of the two solids is
$$\boxed{k^2:1}.$$
Verification of Key Steps
The first delicate step is the identification of planes parallel to both edges with equations
$$\lambda_C+\lambda_D=t.$$
Independently, choose coordinates in which $AB$ and $CD$ are opposite edges. A direction parallel to $AB$ changes only the coefficients of $A$ and $B$, while a direction parallel to $CD$ changes only the coefficients of $C$ and $D$. Hence any quantity preserved by both directions must depend only on the sums $\lambda_A+\lambda_B$ and $\lambda_C+\lambda_D$. Since these sums add to $1$, a single parameter $t=\lambda_C+\lambda_D$ determines the plane. Forgetting to prove the converse would leave open the possibility of additional parallel planes not represented by this equation.
The second delicate step is the area formula
$$S(t)=K,t(1-t).$$
Recomputing from scratch, every section point can be written as
$$(1-t)P+tQ,$$
where $P\in AB$ and $Q\in CD$. Thus one side of the section is obtained by varying $P$, producing a vector of length proportional to $1-t$, and another side is obtained by varying $Q$, producing a vector of length proportional to $t$. Since the directions are fixed, area is the product of these factors times a constant. Confusing a side length with an area factor would incorrectly produce powers such as $t^2(1-t)$ or $t^2(1-t)^2$.
The third delicate step is converting the distance ratio into
$$\frac{t}{1-t}=k.$$
Because all section planes are parallel, perpendicular distance along the common normal is an affine function of the parameter $t$. A careless argument that treats $t$ as an actual Euclidean distance without establishing affine proportionality would be incomplete.
Alternative Approaches
A coordinate solution can be carried out directly. Place
$$A=(0,0,0),\quad B=(a,0,0),\quad D=(0,d,0),\quad C=(b\cos\omega,d,b\sin\omega).$$
The tetrahedron then admits an explicit parametrization
$$P=(1-s)\bigl((1-u)A+uB\bigr) +s\bigl((1-v)D+vC\bigr).$$
The parameter $s$ measures position between the opposite edges. Computing the Jacobian of this parametrization yields a factor proportional to $s(1-s)$. Integration over the relevant range immediately gives the volume ratio $k^2:1$.
Another approach uses affine geometry. Any tetrahedron with a chosen pair of opposite edges can be transformed by an affine map into a particularly simple tetrahedron. Affine maps preserve ratios of volumes and send planes parallel to the two chosen edges to planes with the same property. After reducing to a canonical tetrahedron, the computation becomes elementary and again yields the ratio
$$k^2:1.$$
The barycentric approach is preferable because it isolates the intrinsic geometry of the family of parallel sections and makes the independence from $a$, $b$, $d$, and $\omega$ transparent.