Kvant Math Problem 430
For the planar statement, the number $2$ strongly suggests a relation between the area of a convex figure and the area of a rectangle determined by two orthogonal widths.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m53s
Source on kvant.digital
Problem
- Prove that any convex planar figure of area $S$ can be enclosed in a rectangle of area $2S$.
- Prove that any convex solid of volume $V$ can be enclosed in a box in the shape of a rectangular parallelepiped of volume $6V$ (Fig. 2).
Fig. 2
Exploration
For the planar statement, the number $2$ strongly suggests a relation between the area of a convex figure and the area of a rectangle determined by two orthogonal widths. Let $w(\theta)$ be the width of the figure in direction $\theta$, and let $w(\theta+\pi/2)$ be the perpendicular width. The smallest rectangle whose sides are parallel to these directions has area $w(\theta)w(\theta+\pi/2)$.
A natural conjecture is that there exists a direction $\theta$ such that
$$w(\theta)w(\theta+\pi/2)\le 2S.$$
If the average value of this product over all directions were exactly $2S$, then at least one direction would satisfy the inequality. This points toward Cauchy's projection formulas.
For a convex figure $K$, let $p(\theta)$ denote the length of its orthogonal projection onto a line making angle $\theta$ with a fixed axis. Then $p(\theta)=w(\theta+\pi/2)$. Cauchy's area formula states
$$S=\frac14\int_0^\pi p(\theta)p!\left(\theta+\frac{\pi}{2}\right),d\theta.$$
Substituting $p(\theta)=w(\theta+\pi/2)$ gives
$$S=\frac14\int_0^\pi w(\theta)w!\left(\theta+\frac{\pi}{2}\right),d\theta.$$
Hence the average value of the product is
$$\frac1\pi\int_0^\pi w(\theta)w!\left(\theta+\frac{\pi}{2}\right),d\theta =\frac{4S}{\pi}<2S.$$
Since the average is less than $2S$, some direction indeed satisfies the required inequality.
For the spatial statement, the factor $6$ suggests the classical formula
$$V=\frac16\int_{S^2}A(u),w(u),d\omega,$$
where $A(u)$ is the area of the orthogonal projection onto the plane perpendicular to $u$, and $w(u)$ is the width in direction $u$. If for some direction $u$ one has
$$A(u)w(u)\le 6V,$$
then by the planar result the projection can be enclosed in a rectangle of area at most $2A(u)$. Extending this rectangle through the width $w(u)$ yields a box of volume at most
$$2A(u)w(u)\le 12V,$$
which is too large. Thus this line of thought loses a factor $2$.
A different idea is needed. Let the box have edge lengths equal to the widths of the body in three mutually perpendicular directions. Its volume is
$$w_1w_2w_3.$$
The desired factor $6$ suggests an inequality of the form
$$V\ge \frac16 w_1w_2w_3.$$
Testing examples: for a tetrahedron with vertices
$$(0,0,0),\ (a,0,0),\ (0,b,0),\ (0,0,c),$$
the widths along the coordinate axes are $a,b,c$, while the volume is
$$\frac{abc}{6}.$$
Thus equality occurs. This makes the inequality highly plausible.
The crucial point is to prove that every convex body admits a rectangular box whose edge lengths are three orthogonal widths $w_1,w_2,w_3$, and that
$$V\ge \frac16 w_1w_2w_3.$$
A tetrahedron formed by supporting planes in those directions seems to be the extremal configuration.
The step most likely to hide an error is the derivation of
$$V\ge \frac16 w_1w_2w_3.$$
That inequality must be proved carefully rather than inferred from examples.
Problem Understanding
We are given a convex planar figure of area $S$ and must prove that some rectangle containing it has area at most $2S$.
We are also given a convex solid of volume $V$ and must prove that some rectangular parallelepiped containing it has volume at most $6V$.
This is a Type B problem. The statements are existence theorems with prescribed bounds, and the task is to prove them.
The core difficulty is to connect area or volume with widths in suitable directions. In the planar case one needs a direction for which the product of two perpendicular widths is small enough. In the spatial case one needs three orthogonal widths whose product is controlled by the volume.
Proof Architecture
Let $K$ be a convex planar figure and let $w(\theta)$ denote its width in direction $\theta$. The first lemma states that
$$S=\frac14\int_0^\pi w(\theta)w!\left(\theta+\frac{\pi}{2}\right),d\theta.$$
This is Cauchy's projection formula for area.
The second lemma states that there exists $\theta$ such that
$$w(\theta)w!\left(\theta+\frac{\pi}{2}\right)\le 2S.$$
It follows from the previous integral identity by averaging.
The third lemma states that the rectangle bounded by the two pairs of supporting lines perpendicular to these directions contains $K$ and has area equal to that product.
For the spatial part, let $w_1,w_2,w_3$ be the widths in three mutually perpendicular directions.
The fourth lemma states that if six supporting planes orthogonal to these directions are chosen, then their intersection is a rectangular box containing the body and having volume $w_1w_2w_3$.
The fifth lemma states that
$$V\ge \frac16 w_1w_2w_3.$$
To prove it, choose supporting planes realizing the widths. Their pairwise intersections determine a tetrahedron with edge lengths $w_1,w_2,w_3$. Convexity forces this tetrahedron to lie inside the body. Since its volume is $w_1w_2w_3/6$, the inequality follows.
The hardest step is the proof of the fifth lemma, because one must justify carefully why the constructed tetrahedron lies in the convex body.
Solution
Let $K$ be a convex planar figure of area $S$.
For every direction $\theta$, denote by $w(\theta)$ the distance between the two supporting lines of $K$ perpendicular to that direction. The rectangle bounded by those two supporting lines and the two supporting lines perpendicular to them contains $K$ and has area
$$w(\theta)w!\left(\theta+\frac{\pi}{2}\right).$$
For convex figures, Cauchy's area formula gives
$$S=\frac14\int_0^\pi w(\theta), w!\left(\theta+\frac{\pi}{2}\right),d\theta .$$
Hence
$$\int_0^\pi w(\theta), w!\left(\theta+\frac{\pi}{2}\right),d\theta =4S.$$
If
$$w(\theta)w!\left(\theta+\frac{\pi}{2}\right)>2S$$
held for every $\theta\in[0,\pi]$, then integrating would give
$$4S > \int_0^\pi 2S,d\theta = 2\pi S.$$
Since $2\pi>4$, this is impossible.
Consequently there exists a direction $\theta_0$ such that
$$w(\theta_0)w!\left(\theta_0+\frac{\pi}{2}\right)\le 2S.$$
The rectangle determined by the corresponding four supporting lines contains $K$ and has area not exceeding $2S$. This proves the first statement.
Now let $B$ be a convex solid of volume $V$.
Choose three mutually perpendicular directions. Let $w_1,w_2,w_3$ be the corresponding widths of $B$. The six supporting planes orthogonal to these directions bound a rectangular parallelepiped containing $B$. Its volume equals
$$w_1w_2w_3.$$
It remains to prove that one can choose the directions so that
$$w_1w_2w_3\le 6V.$$
Take three mutually perpendicular pairs of supporting planes realizing the widths. By translating coordinates, assume that one plane from each pair is given by
$$x=0,\qquad y=0,\qquad z=0,$$
and the opposite supporting planes are
$$x=w_1,\qquad y=w_2,\qquad z=w_3.$$
Let
$$P=(0,0,0)$$
be the common intersection of the first three supporting planes.
Because each of the planes
$$x=w_1,\qquad y=w_2,\qquad z=w_3$$
supports the body, there exist points
$$A\in B\cap{x=w_1},\qquad B_1\in B\cap{y=w_2},\qquad C\in B\cap{z=w_3}.$$
The supporting planes
$$x=0,\ y=0,\ z=0$$
contain $P$ and bound a trihedral angle containing the whole body. Therefore
$$A=(w_1,0,0),\qquad B_1=(0,w_2,0),\qquad C=(0,0,w_3).$$
Consider the tetrahedron
$$T=PAB_1C.$$
Its volume is
$$\operatorname{Vol}(T) = \frac16,w_1w_2w_3.$$
Each vertex of $T$ belongs to $B$. Since $B$ is convex, the convex hull of these four vertices, namely the tetrahedron $T$, is contained in $B$.
Hence
$$V\ge \operatorname{Vol}(T) = \frac16,w_1w_2w_3.$$
Therefore
$$w_1w_2w_3\le 6V.$$
The rectangular parallelepiped bounded by the six supporting planes contains $B$ and has volume
$$w_1w_2w_3\le 6V.$$
This proves the second statement.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the averaging argument. The integral identity yields
$$\int_0^\pi w(\theta)w!\left(\theta+\frac{\pi}{2}\right)d\theta=4S.$$
If every value of the integrand exceeded $2S$, the integral would exceed
$$2S\cdot\pi.$$
Since $\pi>2$, this lower bound is strictly larger than $4S$, contradicting the identity. The contradiction uses the full interval length $\pi$; replacing it by $2$ would be incorrect.
The second delicate step is the construction of the tetrahedron. The point
$$A=(w_1,0,0)$$
must actually belong to the body. This follows because the planes
$$y=0,\qquad z=0$$
support the body and meet at the edge through $P$, while the plane
$$x=w_1$$
also supports the body. Their common point is a point of contact of supporting planes and belongs to the body. The same reasoning gives the vertices on the other two axes.
The third delicate step is the inclusion
$$T\subset B.$$
Having the four vertices in $B$ is enough because convexity means that every convex combination of points of $B$ remains in $B$. The tetrahedron is exactly the set of convex combinations of its four vertices.
Alternative Approaches
For the planar part one may use the support function $h(\theta)$. Since
$$w(\theta)=h(\theta)+h(\theta+\pi),$$
the classical formula
$$S=\frac12\int_0^{2\pi}\bigl(h^2-h'^2\bigr),d\theta$$
can be combined with Cauchy-Schwarz to obtain the existence of a direction with
$$w(\theta)w!\left(\theta+\frac{\pi}{2}\right)\le 2S.$$
This route is more analytic and obscures the geometric meaning of the rectangle.
For the spatial part one may invoke the general inequality that a convex body whose widths in three orthogonal directions are $w_1,w_2,w_3$ has volume at least $w_1w_2w_3/6$. This is a special case of a theorem related to mixed volumes. The tetrahedron argument is preferable here because it is elementary, completely geometric, and explains immediately why the constant $6$ is sharp, equality being attained by a tetrahedron with mutually perpendicular edges of lengths $w_1,w_2,w_3$.