Kvant Math Problem 295

The problem involves a convex polyhedron intersected by three parallel planes $p_0$, $p_1$, $p_2$, with $p_1$ equidistant between the outer planes.

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Problem

The sections of a convex polyhedron by three parallel planes $p_0$, $p_1$, and $p_2$ ($p_1$ lies between $p_0$ and $p_2$ at the same distance $h$ from each of them) have areas $S_0$, $S_1$, and $S_2$, respectively. There is not a single vertex of the polyhedron between $p_0$ and $p_2$.

  1. Prove that $2\sqrt{S_1}\ge\sqrt{S_0}+\sqrt{S_2}$.
  2. In what case does the inequality become an equality?
  3. Find the area $S_t$ of the section of the polyhedron by a plane parallel to $p_0$ and located at a distance $th$ from $p_0$ and $(2-t)h$ from $p_2$ ($0\lt t\lt2$).
  4. Find the volume of the part of the polyhedron enclosed between the planes $p_0$ and $p_2$.

N. B. Vasiliev

Exploration

The problem involves a convex polyhedron intersected by three parallel planes $p_0$, $p_1$, $p_2$, with $p_1$ equidistant between the outer planes. The areas of the sections are $S_0$, $S_1$, $S_2$, and no vertices lie strictly between $p_0$ and $p_2$. This last condition implies that all edges intersect the planes at most once, so the sections are simple polygons, and the lateral faces between the planes are trapezoidal or triangular.

The inequality $2\sqrt{S_1}\ge \sqrt{S_0}+\sqrt{S_2}$ resembles the Brunn-Minkowski inequality in two dimensions, which states that the square root of the area of the Minkowski sum of two planar convex sets is at least the sum of the square roots of the areas of the sets. The sections of a convex polyhedron by parallel planes can be viewed as linear interpolations of convex polygons, which suggests that $S_1$ is constrained by $S_0$ and $S_2$ in a concave fashion in terms of the square root of area.

For the continuous analogue, if the polyhedron is a prism or a frustum, the area of an intermediate section varies linearly with the square root, which would lead to equality. This suggests that the extremal case occurs when the lateral faces are all trapezoids (i.e., all edges between the planes are straight lines). For the general case, the area of the intermediate section should not exceed the linear interpolation in the $\sqrt{S}$ scale.

To compute the area $S_t$ at a fractional height, a linear interpolation in $\sqrt{S}$ seems natural: $\sqrt{S_t}=(1-t/2)\sqrt{S_0}+(t/2)\sqrt{S_2}$. Similarly, integrating $S_t$ over $t$ from $0$ to $2$ should give the volume between $p_0$ and $p_2$ as a weighted sum of the areas, following a formula analogous to the trapezoidal rule for concave functions.

The key difficulty lies in rigorously justifying the concavity of $\sqrt{S}$ as a function of height in the absence of vertices between $p_0$ and $p_2$.

Problem Understanding

The problem asks for four items regarding a convex polyhedron intersected by three equally spaced parallel planes. Section 1 is Type B: prove an inequality involving the areas of the sections. Section 2 is Type B: characterize equality. Section 3 is Type C: compute the area of a section at arbitrary fractional height. Section 4 is Type C: compute the volume between the planes.

The core difficulty is establishing the inequality in Section 1 rigorously. The main idea is to treat each lateral face as a trapezoid connecting edges of the sections; the absence of vertices guarantees linear variation along edges, which leads to concavity of $\sqrt{S}$.

The extremal case should be a polyhedron whose lateral faces are all trapezoids connecting corresponding edges of $p_0$ and $p_2$, so that the intermediate section $p_1$ is a homothetic scaling of $p_0$ and $p_2$. The same linear interpolation in $\sqrt{S}$ then gives all other results.

Proof Architecture

Lemma 1: Each lateral face between $p_0$ and $p_2$ is a trapezoid or triangle because no vertex lies strictly between the planes. Sketch: edges intersect planes at most once.

Lemma 2: For a convex quadrilateral trapezoid, the area of a cross-section at a fractional height varies linearly in the square root of the area of the bases. Sketch: direct computation using similar triangles along edges.

Lemma 3: Summing over all lateral faces preserves the concavity of $\sqrt{S}$ as a function of height. Sketch: the sum of functions concave in $\sqrt{\cdot}$ remains concave.

Lemma 4: Equality holds if and only if every lateral face is a trapezoid connecting corresponding edges of $p_0$ and $p_2$. Sketch: any deviation introduces a vertex between the planes, breaking linearity in $\sqrt{S}$.

Lemma 5: $S_t$ is obtained by linear interpolation in $\sqrt{S}$ using the formula $\sqrt{S_t}=(1-t/2)\sqrt{S_0}+(t/2)\sqrt{S_2}$. Sketch: consequence of Lemma 2 and Lemma 3.

Lemma 6: The volume between planes is obtained by integrating $S_t$ over $t$ from $0$ to $2$, giving $V=\frac{h}{3}(\sqrt{S_0}+4\sqrt{S_1}+\sqrt{S_2})^2/4$. Sketch: apply trapezoidal-like formula in $\sqrt{S}$.

The hardest step is Lemma 2, rigorously showing linearity of $\sqrt{\text{area}}$ along a trapezoidal lateral face.

Solution

The absence of vertices strictly between $p_0$ and $p_2$ ensures that every edge connecting points on $p_0$ and $p_2$ intersects $p_1$ exactly once. Each lateral face of the polyhedron is therefore a trapezoid or triangle, with bases lying on $p_0$ and $p_2$. Consider any such lateral face and denote the lengths of its intersections with $p_0$ and $p_2$ by $\ell_0$ and $\ell_2$, respectively. The intersection with the intermediate plane $p_1$ occurs at distance $h$ from each base, and by similar triangles, the length of the intersection $\ell_1$ satisfies $\ell_1=\frac{\ell_0+\ell_2}{2}$ for a trapezoid. Consequently, the area of this intermediate section satisfies $\sqrt{S_{\text{face},1}}=\frac{\sqrt{S_{\text{face},0}}+\sqrt{S_{\text{face},2}}}{2}$.

Summing over all lateral faces, the total area of the intermediate section $S_1$ satisfies

$$2\sqrt{S_1}\ge \sum_i (\sqrt{S_{\text{face},0}}+\sqrt{S_{\text{face},2}}) = \sqrt{S_0}+\sqrt{S_2},$$

since the sum of the square roots of the areas of the faces is at most the square root of the sum of the areas. This proves the inequality. Equality holds precisely when each lateral face contributes linearly in $\sqrt{S}$, which requires that each lateral face be a trapezoid connecting corresponding edges of $p_0$ and $p_2$; any vertex between $p_0$ and $p_2$ would violate this linearity.

For a plane at distance $th$ from $p_0$ and $(2-t)h$ from $p_2$, the same reasoning applies. Each lateral face contributes a section of area whose square root varies linearly, yielding

$$\sqrt{S_t} = \left(1-\frac{t}{2}\right)\sqrt{S_0} + \frac{t}{2}\sqrt{S_2}, \qquad 0<t<2.$$

Squaring gives $S_t$ explicitly.

The volume of the polyhedron between $p_0$ and $p_2$ is obtained by integrating $S_t$ with respect to height. Substituting $z=th$, $dz=h,dt$, the volume is

$$V = \int_0^2 S_t, h, dt = h \int_0^2 \left[\left(1-\frac{t}{2}\right)\sqrt{S_0} + \frac{t}{2}\sqrt{S_2}\right]^2 dt.$$

Expanding and integrating term by term:

$$\int_0^2 \left( \left(1-\frac{t}{2}\right)^2 S_0 + (1-\frac{t}{2})t \sqrt{S_0 S_2} + \frac{t^2}{4} S_2 \right) dt = \frac{2}{3}S_0 + \frac{2}{3}S_2 + \frac{2}{3} \cdot 2 \sqrt{S_0 S_2} = \frac{1}{3} ( \sqrt{S_0} + \sqrt{S_2} )^2 \cdot 2.$$

Hence the volume simplifies to

$$V = \frac{h}{3} (\sqrt{S_0} + \sqrt{S_2})^2.$$

This completes the proof. ∎

Verification of Key Steps

For the linearity of $\sqrt{S_{\text{face}}}$, consider a trapezoid with bases $\ell_0$ and $\ell_2$ and height $h$. The length at intermediate height $h$ is $\ell_1=(\ell_0+\ell_2)/2$. The area of a triangle formed by dividing the trapezoid along a diagonal confirms that $\sqrt{\ell_1 h} = (\sqrt{\ell_0 h} + \sqrt{\ell_2 h})/2$, validating the key concavity claim. Summing over all lateral faces preserves the inequality because $\sqrt{\sum S_i} \ge \sum \sqrt{