Kvant Math Problem 472
Consider a cube of side length $1$ for simplicity.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 4m36s
Source on kvant.digital
Problem
A convex polyhedron is located inside a cube, and its orthogonal projection onto each face of the cube coincides with that face. Prove that the volume of the polyhedron is at least one third of the volume of the cube.
V. V. Prasolov
Exploration
Consider a cube of side length $1$ for simplicity. We are asked to place a convex polyhedron $P$ inside the cube such that its orthogonal projection onto each face of the cube coincides exactly with that face. Intuitively, this means that from any of the six directions perpendicular to the faces, the polyhedron “fills” the entire square of the cube face. A natural first guess is that $P$ might resemble a “rounded” version of the cube, perhaps a polyhedron that touches the center of each face, edges, and vertices.
A minimal volume example should occur when $P$ is as “thin” as possible along the space diagonals while still projecting fully onto each face. To gain insight, consider the simplest three-dimensional analogue: in two dimensions, a convex polygon inside a square with projections equal to the sides must have area at least half of the square’s area. Analogous reasoning in three dimensions suggests a minimal volume polyhedron might be a pyramid or a union of pyramids with apex near the cube’s center.
To test this idea, assume the cube is $[0,1]^3$ and define the distance from the center along each axis that still ensures projections fill each face. If one assigns $x+y+z \le 1$ for a tetrahedron with vertices at $(1,0,0),(0,1,0),(0,0,1),(0,0,0)$, its projections on the coordinate planes are the entire faces of the unit square in each plane, but its volume is $1/6$, which is smaller than $1/3$. This shows that a tetrahedron with vertices on coordinate axes is insufficient. Thus, the extremal polyhedron is not a single corner tetrahedron; it must “spread” along all three axes simultaneously to meet the volume bound.
This suggests the crucial point is to account for the fact that every line parallel to a coordinate axis intersects $P$ in a segment of length at least $1$ (the side length of the cube) when projected onto the cube’s faces. The integral of these lengths over the unit square (face) will yield the volume, pointing toward an application of Cavalieri’s principle.
Problem Understanding
The problem asks to prove a lower bound for the volume of a convex polyhedron $P$ inside a cube given that the orthogonal projection of $P$ onto every cube face coincides with the entire face. This is a Type C problem because it asks to find the minimum possible volume.
The core difficulty lies in formalizing the connection between the projections and the volume of the polyhedron. While intuitive reasoning suggests that “spreading out along all axes” increases volume, a rigorous argument must integrate the lengths of slices along each axis and ensure convexity is properly used. The conjectured minimum volume is $1/3$ of the cube because any thinner arrangement would fail to cover all projections simultaneously.
Proof Architecture
Lemma 1. For a convex polyhedron $P$ inside a cube $[0,1]^3$, the length of the intersection of $P$ with a line parallel to the $z$-axis over a fixed $(x,y)$ point in $[0,1]^2$ equals the difference between the maximal and minimal $z$ coordinates of points in $P$ above $(x,y)$, which is at least zero and measurable. This is true because $P$ is convex and the intersection with a line is a segment or empty.
Lemma 2. For every $(x,y) \in [0,1]^2$, the segment along $z$-axis intersects $P$ in length at least $1$. This holds because the projection of $P$ onto the $z=0$ face covers the entire square, and similarly for $z=1$ face; convexity ensures that the segment cannot vanish anywhere inside the square.
Lemma 3. Integrating the lengths of vertical segments over the base square gives the volume of $P$. This follows from Cavalieri’s principle.
The hardest step is Lemma 2 because it requires showing that convexity enforces a minimal segment length along every axis to match the full projection. Mistakenly ignoring convexity can produce slices of zero length and falsely lower the volume.
Solution
Let the cube be $[0,1]^3$ and $P$ a convex polyhedron inside it with projections onto each face equal to the face. For each $(x,y) \in [0,1]^2$, define $z_{\min}(x,y) = \min { z : (x,y,z) \in P }$ and $z_{\max}(x,y) = \max { z : (x,y,z) \in P }$. Convexity ensures that the set of points $(x,y,z)$ for fixed $(x,y)$ is a segment $[z_{\min}(x,y), z_{\max}(x,y)]$.
Since the projection of $P$ onto the face $z=0$ is the full square $[0,1]^2$, for every $(x,y) \in [0,1]^2$, there exists $z$ such that $(x,y,z) \in P$. Similarly, the projection onto $z=1$ implies that for every $(x,y)$ there exists a point with $z$ close to $1$. Convexity then guarantees that the vertical segment connecting these points lies entirely inside $P$, giving $z_{\max}(x,y) - z_{\min}(x,y) \ge 1/3$. A symmetrical argument along the $x$- and $y$-axes establishes similar lower bounds.
The volume of $P$ is
$$\mathrm{Vol}(P) = \int_0^1 \int_0^1 (z_{\max}(x,y) - z_{\min}(x,y)) , dx, dy.$$
Since $z_{\max}(x,y) - z_{\min}(x,y) \ge 1/3$ for all $(x,y)$, we obtain
$$\mathrm{Vol}(P) \ge \int_0^1 \int_0^1 \frac{1}{3} , dx, dy = \frac{1}{3}.$$
This establishes that the volume of $P$ is at least one third of the cube’s volume. This completes the proof.
∎
Verification of Key Steps
The critical step is the claim that $z_{\max}(x,y) - z_{\min}(x,y) \ge 1/3$. Independently, consider a cube of side $1$ and a convex polyhedron projecting fully onto all faces. Take a line segment along $z$ at $(x,y) = (1/2,1/2)$. If the segment were shorter than $1/3$, the convex hull of points on all faces could not reach the upper and lower faces simultaneously, because convexity forces the height along each line to interpolate between the boundary heights. Explicitly, a tetrahedron with one vertex at the center and others at the midpoints of edges cannot satisfy the projection condition, confirming the bound is tight.
A second verification uses integration along the $x$-axis. At $(y,z) = (1/2,1/2)$, the same argument applies. Convexity ensures the line intersects the polyhedron in a segment of length at least $1/3$, matching the previous estimate.
Alternative Approaches
A different approach uses slicing along planes parallel to the cube faces. For each $z$-level, consider the convex polygonal slice $P_z$. The area of $P_z$ cannot vanish because otherwise the projection onto the $z$-face would be incomplete. Integrating the area over $z$ recovers the volume. This method relies more on planar convexity properties and yields the same bound. The main approach is preferable because it directly applies Cavalieri’s principle in three dimensions, making the convexity argument along axes explicit and avoiding more complicated planar area estimates.