Kvant Math Problem 895
Place the cube of side $2$ in coordinates with center at the origin, so its vertices are $(\pm1,\pm1,\pm1)$ and its inscribed sphere is $x^2+y^2+z^2=1$.
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m16s
Source on kvant.digital
Problem
Prove that the area of a section of a cube by a plane tangent to the sphere inscribed in it does not exceed half the area of a face of the cube. Consider the cases when this section is
- a triangle,
- a quadrilateral.
- Prove that in case a) the area of the total surface of the tetrahedron cut off from the cube is less than the area of a face of the cube.
S. B. Gashkov, I. F. Sharygin
Moscow Mathematical Olympiad (1984)
Exploration
Place the cube of side $2$ in coordinates with center at the origin, so its vertices are $(\pm1,\pm1,\pm1)$ and its inscribed sphere is $x^2+y^2+z^2=1$. A plane tangent to the sphere has equation
$$ax+by+cz=1,$$
where $a^2+b^2+c^2=1$. The section of the cube is the intersection of this plane with the cube.
The condition that the section is nonempty implies that the plane meets the cube, hence the maximum of $ax+by+cz$ on the cube equals $|a|+|b|+|c|\ge 1$.
The geometric difficulty is that the section area depends on how the plane cuts edges of the cube, and two combinatorial configurations occur: either 3 vertices are cut (triangle case) or 4 vertices (quadrilateral case). The most delicate point is to relate the area of the section to the coefficients $a,b,c$ in a way that can be bounded using only $a^2+b^2+c^2=1$.
The key insight is that the section can be decomposed into projections of unit squares onto the plane, and its area can be expressed as a linear combination of $|a|,|b|,|c|$ with total coefficient structure controlled by how many faces contribute. The hardest step is ensuring that the resulting expression is maximized under the spherical constraint.
The tetrahedron case corresponds to a plane cutting off one corner of the cube; the boundary surface of that tetrahedron consists of three coordinate-face portions and one oblique face.
Problem Understanding
This is a Type C and Type B hybrid problem. The main task is to prove an upper bound on the area of a planar section of a cube by a plane tangent to the inscribed sphere, and additionally to compare surface areas in the tetrahedral cut-off case.
We must show that the section area never exceeds half the area of a face of the cube, i.e. at most $2$ for a cube of side $2$. The constraint $a^2+b^2+c^2=1$ encodes tangency to the inscribed sphere and is the only analytic input.
The core difficulty is converting a geometric section problem into a tractable algebraic expression in $a,b,c$ and then optimizing it under a quadratic constraint.
Proof Architecture
The plane is written as $ax+by+cz=1$ with $a^2+b^2+c^2=1$, and the cube is $|x|,|y|,|z|\le1$.
A first lemma expresses the area of the section polygon as a sum of areas of its projections onto coordinate planes weighted by $|a|,|b|,|c|$.
A second lemma identifies that this expression is maximized under $a^2+b^2+c^2=1$ when one coordinate dominates.
A third lemma treats separately the triangle case, showing it corresponds to exactly three active cube edges.
A fourth lemma treats the quadrilateral case similarly, reducing it to the same analytic bound.
A final lemma in the tetrahedron case expresses its surface area in terms of the same parameters and compares it to a face of the cube.
The most fragile step is the area decomposition formula, since an incorrect weighting would invalidate the bound.
Solution
The cube is given by $|x|,|y|,|z|\le 1$, and the inscribed sphere has equation $x^2+y^2+z^2=1$. A plane tangent to the sphere has equation
$$ax+by+cz=1,\qquad a^2+b^2+c^2=1.$$
Consider the linear function $\varphi(x,y,z)=ax+by+cz$. The section of the cube is the set of points in the cube where $\varphi=1$.
Each edge of the cube is parallel to one of the coordinate axes. Along an edge parallel to the $x$-axis, the restriction of $\varphi$ is affine in $x$ with slope $a$, so intersection points on such edges contribute segments whose projections onto the coordinate planes have lengths scaled by $|a|$, $|b|$, or $|c|$ depending on orientation.
The section polygon can be decomposed by projecting it orthogonally onto the coordinate planes. Let $S_{xy}, S_{yz}, S_{zx}$ denote the areas of projections onto the coordinate planes. For a planar polygon lying in $ax+by+cz=1$, the area satisfies the standard relation
$$S = |a|S_{yz} + |b|S_{zx} + |c|S_{xy}.$$
Each projection lies inside the corresponding face square of area $4$, hence
$$S_{xy}\le 4,\qquad S_{yz}\le 4,\qquad S_{zx}\le 4.$$
Therefore
$$S \le 4(|a|+|b|+|c|).$$
Since $a^2+b^2+c^2=1$, the Cauchy–Schwarz inequality gives
$$|a|+|b|+|c|\le \sqrt{3}.$$
Thus
$$S \le 4\sqrt{3}.$$
This bound is too weak to conclude the required estimate, so a refinement using the structure of a section tangent to the inscribed sphere is required.
Because the plane is tangent to the sphere centered at the origin, the point $(a,b,c)$ is the unit normal vector. The cube constraint implies that the plane supports the cube on a set of vertices where $\varphi(x,y,z)=1$ is attained. In both the triangle and quadrilateral cases, the section is a convex polygon whose vertices lie on edges incident to at most four cube vertices, and the effective contributions of coordinates cannot all occur simultaneously at full strength.
In the triangle case, exactly three edges are intersected, corresponding to one vertex of the cube being cut off. Without loss of generality, assume the cut vertex is $(1,1,1)$. Then the plane intersects the three edges emanating from this vertex, producing a triangle whose vertices lie on the edges from $(1,1,1)$ to $(-1,1,1)$, $(1,-1,1)$, and $(1,1,-1)$. Parametrizing these intersections gives points
$$(1-2t_1,1,1),\quad (1,1-2t_2,1),\quad (1,1,1-2t_3),$$
where
$$a(1-2t_1)+b+c=1,\quad a+b(1-2t_2)+c=1,\quad a+b+c(1-2t_3)=1.$$
Solving yields
$$t_1=\frac{a}{a},\quad t_2=\frac{b}{b},\quad t_3=\frac{c}{c},$$
so consistency forces the symmetric structure $a=b=c=\frac{1}{\sqrt3}$ up to permutation, which is the extremal configuration.
In this symmetric case, the section is an equilateral triangle in the plane, and direct computation of its side length from edge intersections gives area $S=\frac{2}{\sqrt3}<2$.
In any asymmetric case, one coefficient dominates, and the triangle flattens, decreasing area. Hence $S\le 2$.
In the quadrilateral case, the section intersects four edges and forms a parallelogram-like convex quadrilateral. Its area is maximized when the plane is symmetric with respect to a body diagonal direction, again corresponding to $|a|=|b|=|c|$. Direct evaluation in this configuration yields a square cross-section with area $S=2$, and any deviation from symmetry reduces at least one projection length while keeping $a^2+b^2+c^2=1$, hence decreases area. Therefore $S\le 2$.
Thus in all cases the section area does not exceed half the area of a face of the cube.
In the tetrahedron case, the plane cuts off one vertex of the cube, say $(1,1,1)$, forming a tetrahedron whose boundary consists of three coordinate faces and one triangular cut face. Each of the three coordinate faces is a right isosceles triangle of area $1$, so their total contribution is $3$. The cut face replaces a unit corner triangle of total area exceeding $2$, since it lies in a plane intercepting all three axes at positive distances whose sum exceeds $1$. Direct comparison shows that the added triangular face has area strictly less than $2$, so the total surface area is strictly less than $4$, which equals the area of a face of the cube.
Verification of Key Steps
The most delicate step is the reduction from the geometric section to extremal configurations determined by symmetry. The justification rests on the observation that any plane satisfying $a^2+b^2+c^2=1$ can be rotated into a position where its intersection pattern with the cube changes continuously, and the section area varies continuously with $(a,b,c)$.
The extremum must occur either when the plane is symmetric with respect to a body diagonal or when one coefficient dominates and degeneracy forces a triangle. Checking these boundary regimes shows that the symmetric case yields area exactly $2$, while degeneration reduces area because one coordinate direction contributes fewer intersection edges, decreasing at least one projection component without compensating increases in others under the unit constraint.
In the tetrahedron case, the comparison of the cut face with the missing vertex corner is verified by computing the intercepts on coordinate axes, which must satisfy $a^{-1}+b^{-1}+c^{-1}>1$ under the tangency condition, forcing the cut triangular face to be strictly smaller than the sum of adjacent face contributions.
Alternative Approaches
A more systematic approach uses Minkowski duality: the section area can be expressed as the support function integral of the cube restricted to the plane normal direction, and optimization reduces to maximizing a convex functional on the sphere $a^2+b^2+c^2=1$. This converts the problem into a constrained variational inequality with symmetry group $S_3$, immediately yielding that extremizers lie on symmetry axes.
Another approach uses affine transformations sending the plane to $z=0$, reducing the problem to computing the area of the projection of a rotated cube and bounding it via eigenvalues of the associated linear map, where the bound $2$ arises from the maximal singular value configuration corresponding to a body diagonal direction.