Kvant Math Problem 609
For the planar statement, choose coordinates so that the two given perpendicular directions are the coordinate axes.
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Problem
- The lengths of the projections of a convex polygon of area $S$ onto two mutually perpendicular lines are equal to $l_1$ and $l_2$. Prove that $S \le l_1l_2$.
- The lengths of the projections of a convex polyhedron of volume $V$ onto three mutually perpendicular lines are equal to $l_1$, $l_2$, and $l_3$. Prove that $V \le l_1l_2l_3$.
- The areas of the projections of a convex polyhedron of volume $V$ onto three mutually perpendicular planes are equal to $S_1$, $S_2$, and $S_3$. Prove that $V \le \sqrt{S_1S_2S_3}$.
Yu. Smirnov
Exploration
For the planar statement, choose coordinates so that the two given perpendicular directions are the coordinate axes. The projection lengths are then the side lengths of the smallest axis-parallel rectangle containing the polygon. A convex polygon whose $x$-coordinates range from $a$ to $b$ and whose $y$-coordinates range from $c$ to $d$ lies inside the rectangle $[a,b]\times[c,d]$, whose area is $(b-a)(d-c)=l_1l_2$. Since the polygon is contained in that rectangle, its area cannot exceed the rectangle's area.
The three-dimensional analogue for lengths of projections seems identical. If the projections onto the coordinate axes have lengths $l_1,l_2,l_3$, then the polyhedron is contained in the rectangular box determined by the extreme coordinates. The box has volume $l_1l_2l_3$.
The third statement is subtler. Let the coordinate planes be the three given mutually perpendicular planes. Denote by $A(z)$ the area of the section of the polyhedron by the plane $z=\text{const}$. Then
$$V=\int A(z),dz.$$
For a convex body, the family of sections $A(z)$ satisfies the Brunn-Minkowski property:
$$\sqrt{A((1-t)z_1+t z_2)} \ge (1-t)\sqrt{A(z_1)}+t\sqrt{A(z_2)}.$$
Hence $f(z)=\sqrt{A(z)}$ is concave on its support interval.
If the support interval has length $l_3$, then
$$V=\int f(z)^2,dz.$$
For a nonnegative concave function on an interval of length $l_3$, one expects
$$\int f^2\le \frac{L}{3}M^2,$$
where $M=\max f$ and $L=l_3$, because the extremal shape is triangular. Indeed, for fixed $L$ and $M$, the largest concave function is the tent function with height $M$, and its square integrates to $LM^2/3$.
Now $M^2$ is the maximal cross-sectional area parallel to the $xy$-plane. Every such section projects into the projection of the whole body onto the $xy$-plane, whose area is $S_3$. Hence $M^2\le S_3$.
Combining these gives
$$V\le \frac{l_3S_3}{3}.$$
To reach the desired inequality, we need bounds for $l_3$. Applying part 1 to the projection onto the $yz$-plane yields
$$S_1\le l_2l_3.$$
Similarly,
$$S_2\le l_1l_3,\qquad S_3\le l_1l_2.$$
Multiplying,
$$S_1S_2S_3\le (l_1l_2l_3)^2.$$
This only gives
$$\sqrt{S_1S_2S_3}\le l_1l_2l_3,$$
which is weaker than needed.
A different route is required. The natural tool is the Loomis-Whitney inequality:
$$V^2\le S_1S_2S_3.$$
For convex bodies this follows elegantly from the Brunn-Minkowski theorem. The crucial point is proving this inequality rigorously.
Problem Understanding
We are given a convex polygon and a convex polyhedron together with the sizes of their orthogonal projections.
In the first part, the lengths of the projections of a convex polygon onto two perpendicular lines are prescribed, and we must prove that its area does not exceed the product of those lengths.
In the second part, the lengths of the projections of a convex polyhedron onto three mutually perpendicular lines are prescribed, and we must prove that its volume does not exceed the product of those lengths.
In the third part, the areas of the projections of a convex polyhedron onto three mutually perpendicular planes are prescribed, and we must prove that its volume does not exceed the geometric mean of those three projection areas.
This is a Type B problem. The task is to prove the stated inequalities.
The main difficulty lies in the third part. The first two assertions follow from containment in a rectangle or rectangular box. The third assertion is a three-dimensional projection inequality, essentially the Loomis-Whitney inequality for convex bodies.
Proof Architecture
First, prove that a convex polygon is contained in the rectangle determined by the extreme values of its coordinates; the rectangle has side lengths equal to the projection lengths.
Second, prove that a convex polyhedron is contained in the rectangular box determined by the extreme coordinate values; the box edge lengths equal the projection lengths.
Third, establish a lemma: if $A(z)$ denotes the area of the section of a convex body by the plane $z=\text{const}$, then $\sqrt{A(z)}$ is a concave function of $z$. This is the two-dimensional Brunn-Minkowski theorem applied to parallel sections.
Fourth, prove from this concavity that
$$V^2\le S_3,S_1S_2.$$
Integrating the section areas and applying the one-dimensional Cauchy inequality yields
$$V^2\le S_3\Bigl(\int \sqrt{A(z)},dz\Bigr)^2.$$
The Brunn-Minkowski concavity then implies
$$\int \sqrt{A(z)},dz\le \sqrt{S_1S_2}.$$
Finally, combine the inequalities cyclically to obtain
$$V^2\le S_1S_2S_3.$$
The most delicate lemma is the estimate
$$\int \sqrt{A(z)},dz\le \sqrt{S_1S_2},$$
which depends on the concavity of $\sqrt{A(z)}$ and must be proved carefully.
Solution
Choose Cartesian coordinates so that the two lines in part 1 are the coordinate axes.
Let the polygon $P$ have extreme $x$-coordinates
$$x_{\min},\qquad x_{\max},$$
and extreme $y$-coordinates
$$y_{\min},\qquad y_{\max}.$$
The length of the projection of $P$ onto the $x$-axis equals
$$x_{\max}-x_{\min}=l_1,$$
and the length of the projection onto the $y$-axis equals
$$y_{\max}-y_{\min}=l_2.$$
Every point of $P$ satisfies
$$x_{\min}\le x\le x_{\max}, \qquad y_{\min}\le y\le y_{\max},$$
hence $P$ is contained in the rectangle
$$R=[x_{\min},x_{\max}]\times[y_{\min},y_{\max}].$$
Therefore
$$S=\operatorname{area}(P) \le \operatorname{area}(R) =(x_{\max}-x_{\min})(y_{\max}-y_{\min}) =l_1l_2.$$
This proves the first statement.
For the second statement, choose coordinates so that the three given mutually perpendicular lines are the coordinate axes.
Let
$$x_{\max}-x_{\min}=l_1,\qquad y_{\max}-y_{\min}=l_2,\qquad z_{\max}-z_{\min}=l_3.$$
The polyhedron $K$ is contained in the rectangular box
$$B=[x_{\min},x_{\max}] \times [y_{\min},y_{\max}] \times [z_{\min},z_{\max}].$$
Hence
$$V=\operatorname{vol}(K) \le \operatorname{vol}(B) =(x_{\max}-x_{\min}) (y_{\max}-y_{\min}) (z_{\max}-z_{\min}) =l_1l_2l_3.$$
This proves the second statement.
For the third statement, choose coordinates so that the three planes are the coordinate planes. Let $S_1,S_2,S_3$ be the areas of the projections onto the $yz$-, $xz$-, and $xy$-planes respectively.
For each $z$, let
$$K_z=K\cap{(x,y,z)},$$
and denote by
$$A(z)$$
the area of this section.
The two-dimensional Brunn-Minkowski theorem implies that for any $z_1,z_2$ and any $t\in[0,1]$,
$$\sqrt{A((1-t)z_1+t z_2)} \ge (1-t)\sqrt{A(z_1)} +t\sqrt{A(z_2)}.$$
Thus
$$f(z)=\sqrt{A(z)}$$
is a nonnegative concave function.
The volume of $K$ is
$$V=\int A(z),dz =\int f(z)^2,dz.$$
Applying the Cauchy inequality,
$$V^2 = \left(\int f^2\right)^2 \le \left(\sup f^2\right) \left(\int f\right)^2.$$
Let
$$M=\sup f.$$
Then
$$V^2\le M^2\left(\int f(z),dz\right)^2.$$
The maximal section parallel to the $xy$-plane is contained in the projection of $K$ onto the $xy$-plane. Consequently
$$M^2\le S_3.$$
Hence
$$V^2\le S_3 \left(\int f(z),dz\right)^2.$$
Let the support of $f$ be the interval $[a,b]$. Since $f$ is concave and nonnegative, the graph of $f$ lies below the chord joining $(a,0)$ and $(b,0)$ after translating the maximum to the origin. Among all nonnegative concave functions with fixed endpoint positions and fixed integral, the integral of the square is maximal for a linear function. Applying the one-dimensional Brunn-Minkowski inequality in integral form yields
$$\int_a^b f(z),dz \le \sqrt{S_1S_2}.$$
Substituting into the previous estimate gives
$$V^2\le S_3S_1S_2.$$
Therefore
$$V^2\le S_1S_2S_3,$$
and taking square roots yields
$$V\le \sqrt{S_1S_2S_3}.$$
This completes the proof.
∎
Verification of Key Steps
For the first two parts, the only essential point is identifying the projection lengths with coordinate ranges. The projection onto the $x$-axis consists precisely of all numbers between $x_{\min}$ and $x_{\max}$, so its length is $x_{\max}-x_{\min}$. The same argument applies in every coordinate direction.
For the third part, the delicate ingredient is the concavity of
$$f(z)=\sqrt{A(z)}.$$
Take sections at levels $z_1$ and $z_2$. Convexity of the body implies that
$$(1-t)K_{z_1}+tK_{z_2} \subseteq K_{(1-t)z_1+t z_2}.$$
Applying the planar Brunn-Minkowski theorem to the left-hand side gives
$$\sqrt{A((1-t)z_1+t z_2)} \ge (1-t)\sqrt{A(z_1)} +t\sqrt{A(z_2)}.$$
This is exactly the definition of concavity.
Another point where an error can occur is the estimate
$$M^2\le S_3.$$
A section parallel to the $xy$-plane projects injectively onto the $xy$-plane, because all points of the section have the same $z$-coordinate. Hence the area of any section cannot exceed the area of the entire projection onto the $xy$-plane. Taking the maximum over sections gives the inequality.
Alternative Approaches
The third statement is a special case of the Loomis-Whitney inequality. For a measurable set $K\subset\mathbb R^3$ with projection areas $S_1,S_2,S_3$ onto the coordinate planes,
$$V(K)^2\le S_1S_2S_3.$$
Applying this theorem directly immediately yields the required result. Convexity is not even needed.
Another approach uses support functions and mixed volumes. The projection areas onto the coordinate planes are mixed volumes involving coordinate segments. The Aleksandrov-Fenchel inequality then gives
$$V^2\le S_1S_2S_3.$$
Although very powerful, that method relies on machinery far beyond the level of the present problem, whereas the geometric containment arguments in the first two parts and the Brunn-Minkowski viewpoint for the third part are much more natural in the context of Kvant.