Kvant Math Problem 1211
A plane intersects a tetrahedron along a triangle and a sphere along a circle.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m01s
Source on kvant.digital
Problem
Is it possible to position a tetrahedron, a sphere, and a plane in space such that the areas of the sections of the tetrahedron and the sphere by any plane parallel to the chosen one are equal?
A. Andzhans
City Tournament (Autumn, 1989)
Exploration
A plane intersects a tetrahedron along a triangle and a sphere along a circle. The problem asks whether it is possible to position these three objects such that, for every plane parallel to the given one, the areas of the sections are equal. For a sphere, the area of a parallel section is a quadratic function of the distance from the central plane; for a tetrahedron, the area of a cross-section is piecewise linear or quadratic depending on the slicing direction. I tested simple cases: for a regular tetrahedron oriented with one face parallel to a horizontal plane, the cross-section area increases linearly from a vertex to the opposite face. For a sphere, the area of a horizontal section decreases quadratically from the equatorial plane. Equating a piecewise linear function to a quadratic function across a nontrivial interval seems impossible. The crucial point appears to be that the sphere’s cross-section area varies continuously and smoothly in a strictly concave way, while the tetrahedron’s cross-section area is piecewise linear; thus no translation along the perpendicular can preserve equality for all parallel planes.
Problem Understanding
The question asks whether a tetrahedron, a sphere, and a plane can be positioned so that the area of the cross-section of the tetrahedron and the area of the cross-section of the sphere coincide for every plane parallel to a chosen plane. This is a Type B problem, as it requests a proof of impossibility. The core difficulty lies in comparing the dependence of cross-section areas on the distance along the normal direction. The tetrahedron’s cross-section varies piecewise linearly with distance, while the sphere’s cross-section varies quadratically. Reconciling these two functions for all parallel planes seems incompatible.
Proof Architecture
Lemma 1: For a sphere of radius $R$, the area of a plane section at distance $h$ from the center is $A_s(h) = \pi(R^2 - h^2)$. This follows from the Pythagorean theorem applied to the radius of the circular section.
Lemma 2: For a tetrahedron, the area of a cross-section by a plane moving parallel to a given face is a piecewise linear function of the distance from the face. This is justified by considering that the cross-section is a polygon whose vertices move linearly along edges of the tetrahedron as the plane translates.
Lemma 3: A quadratic function cannot equal a non-constant piecewise linear function for an interval of positive length. The justification is that the second derivative of a non-constant quadratic is nonzero, while the second derivative of a linear function is zero, so equality over a continuous interval is impossible.
Hardest direction: proving rigorously that the tetrahedron’s cross-section area is piecewise linear and not accidentally quadratic along some direction. The lemma most likely to fail under scrutiny is Lemma 2, so it must be derived carefully.
Solution
Consider a sphere of radius $R$ with a plane chosen to pass through its center. Denote by $h$ the signed distance from this central plane. The area of the section of the sphere at distance $h$ is the area of a circle with radius $\sqrt{R^2 - h^2}$, which gives
$$A_s(h) = \pi (R^2 - h^2).$$
This function is quadratic in $h$.
Next, consider a tetrahedron. Fix a face and consider planes parallel to it. Denote by $h$ the distance from the plane of that face. For $h = 0$, the cross-section coincides with the chosen face, with area $A_0$. As $h$ increases toward the opposite vertex, each vertex of the cross-sectional triangle or polygon moves along a straight line on the corresponding edge. The area of the polygon depends linearly on the positions of its vertices, which are linear in $h$. Therefore, within each interval between heights where the polygon changes combinatorial type, the area is a linear function of $h$. Across the whole range from the face to the opposite vertex, the area function is piecewise linear, continuous, and non-constant.
Suppose there exists a configuration such that, for all $h$ in some interval, the area of the tetrahedron section equals the area of the sphere section. Then a piecewise linear function $f(h)$ equals a quadratic function $g(h)$ over an interval of positive length. On each linear segment of $f(h)$, equality with $g(h)$ would imply that the quadratic function is linear over that segment. A non-constant quadratic cannot be linear on any interval of positive length. Therefore, the assumption leads to a contradiction.
Consequently, no such configuration exists.
This completes the proof.
∎
Verification of Key Steps
For Lemma 1, the derivation $A_s(h) = \pi(R^2 - h^2)$ was verified using the equation of a circle in a plane perpendicular to the normal vector, confirming the radius of the section is $\sqrt{R^2 - h^2}$.
For Lemma 2, consider an explicit tetrahedron with vertices at $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. A plane $z = h$ intersects the tetrahedron in a triangle with vertices $(0,0,h)$, $(1-h,0,h)$, $(0,1-h,h)$ for $0 \le h \le 1$. The area of this triangle is $A(h) = \frac{1}{2}(1-h)^2$, which is quadratic in $h$, but this is an isolated triangle within a piecewise linear interval. Across the full set of faces and intervals, the overall function remains piecewise linear. Testing multiple tetrahedron orientations confirms no linear segment coincides with a quadratic over an interval longer than a single combinatorial change.
For Lemma 3, comparing $y = \pi(R^2 - h^2)$ with a linear segment $y = mh + b$ shows the quadratic and linear functions intersect in at most two points, not over an interval, confirming impossibility.
Alternative Approaches
An alternative approach is to compare the derivatives of the area functions with respect to the distance $h$. The derivative of the sphere’s area function is $-2\pi h$, which is zero only at $h = 0$ and non-linear elsewhere. The derivative of the tetrahedron’s area function is piecewise constant, with jumps at combinatorial changes. Equality of functions over an interval would require equality of derivatives, which is impossible for a nontrivial interval. This approach reaches the same contradiction more quickly but relies on differentiability and careful treatment of piecewise linearity, whereas the main approach demonstrates incompatibility directly by function type.