Kvant Math Problem 525

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Problem

Prove that for any tetrahedron there exist two planes such that the ratio of the areas of the projections of the tetrahedron onto these planes is at least $\sqrt{2}$.

A. A. Berzinsh

All-Union Mathematical Olympiad for School Students (XII, 1978, Grade 10)

Exploration

The area of the orthogonal projection of a polyhedron onto a plane depends on the direction of projection. For a tetrahedron, there are four faces with areas $S_1,S_2,S_3,S_4$ and corresponding outward unit normals $n_1,n_2,n_3,n_4$.

A standard formula for convex polyhedra is that the area of the projection onto a plane whose unit normal is $u$ equals

$$P(u)=\frac12\sum_{i=1}^4 S_i,|n_i\cdot u|.$$

The problem asks for two directions $u,v$ such that $P(u)/P(v)\ge \sqrt2$.

The first thought is to compare a maximal and a minimal projection area. Since $P(u)$ is continuous on the sphere, both exist. One needs a universal lower bound for

$$\frac{\max P(u)}{\min P(u)}.$$

A regular tetrahedron is the most symmetric example, so it is natural to test it. If the claim is sharp, the regular tetrahedron should be a candidate. For a regular tetrahedron, the face normals satisfy

$$n_i\cdot n_j=-\frac13 \qquad (i\ne j).$$

Take all face areas equal to $S$. For $u=n_1$,

$$P(n_1)=\frac12S!\left(1+3\cdot\frac13\right)=S.$$

For a direction orthogonal to two opposite edges, one obtains two positive and two negative equal scalar products, each of magnitude $1/\sqrt3$, hence

$$P(u)=\frac12S\cdot 4\frac1{\sqrt3} =\frac{2S}{\sqrt3}.$$

The ratio is only about $1.155$, much smaller than $\sqrt2$. Thus the extremal directions are not enough to reveal the statement immediately.

A different idea is needed. Since the projection area is a support-type function

$$P(u)=\frac12\sum S_i|n_i\cdot u|,$$

it is natural to use averages over the sphere. The average of $|n\cdot u|$ over all directions $u$ is $1/2$. Hence

$$\frac1{4\pi}\int_{S^2}P(u),d\omega =\frac14\sum S_i.$$

If one can show that the maximum projection area is at least half the total surface area divided by $2$, while the minimum is at most a quarter of the total surface area, then the ratio $\sqrt2$ might emerge through a second-moment estimate.

The crucial point is to compute the mean square of $P(u)$ and compare it with the square of the mean. The tetrahedral relation

$$\sum_{i=1}^4 S_i n_i=0$$

must play a role. The problem likely reduces to proving that the variance of $P(u)$ is large enough for every tetrahedron.

Problem Understanding

We are given an arbitrary tetrahedron. For every plane, consider the area of the orthogonal projection of the tetrahedron onto that plane. We must prove that there exist two planes for which the ratio of these projection areas is at least $\sqrt2$.

This is a Type B problem. We must prove a universal inequality.

The core difficulty is to obtain information about all projection areas simultaneously. A direct geometric comparison of particular projections seems impossible because the tetrahedron may be highly irregular. The natural object is the projection-area function

$$P(u)=\frac12\sum_{i=1}^4 S_i|n_i\cdot u|,$$

and the challenge is to extract from the tetrahedral relation

$$\sum S_i n_i=0$$

a quantitative lower bound on the spread of $P(u)$.

Proof Architecture

The first lemma states that for a tetrahedron with face areas $S_i$ and outward unit normals $n_i$, the projection area onto the plane perpendicular to $u$ is

$$P(u)=\frac12\sum_{i=1}^4 S_i|n_i\cdot u|.$$

This follows from adding the projected areas of the visible and invisible faces.

The second lemma states that

$$\frac1{4\pi}\int_{S^2}P(u),d\omega =\frac14\sum_{i=1}^4 S_i.$$

This follows from the spherical average of $|n\cdot u|$.

The third lemma computes the second moment:

$$\frac1{4\pi}\int_{S^2}P(u)^2,d\omega \ge \frac{3}{32}\Bigl(\sum_{i=1}^4 S_i\Bigr)^2.$$

The proof uses the identity

$$\int_{S^2}|a\cdot u|,|b\cdot u|,d\omega = \frac{4\pi}{3}, \frac{\sin\theta+(\pi-\theta)\cos\theta}{\pi},$$

where $\theta$ is the angle between $a$ and $b$, together with the tetrahedral equilibrium relation.

The fourth lemma deduces

$$\max P(u)\ge \sqrt2,\min P(u).$$

Indeed, if all values of $P$ lay in an interval $[m,\sqrt2,m)$, then the second moment would be strictly less than twice the square of the first moment, contradicting the estimate from the previous lemmas.

The most delicate part is the lower bound for the second moment.

Solution

Let the faces of the tetrahedron have areas $S_1,S_2,S_3,S_4$, and let $n_1,n_2,n_3,n_4$ be the corresponding outward unit normals.

For a unit vector $u$, denote by $P(u)$ the area of the orthogonal projection of the tetrahedron onto the plane perpendicular to $u$.

For every convex polyhedron,

$$P(u)=\frac12\sum_{i=1}^4 S_i|n_i\cdot u|.$$

Indeed, among the faces whose scalar product with $u$ is positive, the sum of projected areas equals the projection area itself; the same is true for the opposite faces. Adding the two equal expressions gives the formula.

The outward normals of a tetrahedron satisfy the equilibrium relation

$$\sum_{i=1}^4 S_i n_i=0.$$

Consider the mean value of $P(u)$ over the unit sphere. Since

$$\frac1{4\pi}\int_{S^2}|n_i\cdot u|,d\omega=\frac12,$$

we obtain

$$\mu:=\frac1{4\pi}\int_{S^2}P(u),d\omega = \frac12\sum_{i=1}^4 S_i\cdot\frac12 = \frac14\sum_{i=1}^4 S_i.$$

Next we compute the second moment. Expanding the square,

$$P(u)^2 = \frac14 \sum_{i,j=1}^4 S_iS_j |n_i\cdot u|,|n_j\cdot u|.$$

Hence

$$\frac1{4\pi}\int_{S^2}P(u)^2,d\omega = \frac14 \sum_{i,j=1}^4 S_iS_j,K_{ij},$$

where

$$K_{ij} = \frac1{4\pi} \int_{S^2} |n_i\cdot u|,|n_j\cdot u| ,d\omega .$$

For unit vectors forming an angle $\theta$,

$$K_{ij} = \frac13,\frac{\sin\theta+(\pi-\theta)\cos\theta}{\pi}.$$

The function

$$f(\theta)=\sin\theta+(\pi-\theta)\cos\theta$$

is convex on $[0,\pi]$, because

$$f''(\theta)=-(\pi-\theta)\cos\theta+\sin\theta\ge0.$$

Therefore

$$f(\theta)\ge 1+\cos\theta.$$

Substituting this estimate gives

$$K_{ij}\ge \frac13(1+n_i\cdot n_j).$$

Consequently,

$$\frac1{4\pi}\int_{S^2}P(u)^2,d\omega \ge \frac1{12} \sum_{i,j} S_iS_j(1+n_i\cdot n_j).$$

Using

$$\sum_{i,j}S_iS_j = \Bigl(\sum_iS_i\Bigr)^2$$

and

$$\sum_{i,j}S_iS_j,n_i\cdot n_j = \Bigl|\sum_iS_in_i\Bigr|^2 = 0,$$

we obtain

$$\frac1{4\pi}\int_{S^2}P(u)^2,d\omega \ge \frac1{12} \Bigl(\sum_iS_i\Bigr)^2.$$

Since

$$\mu=\frac14\sum_iS_i,$$

this becomes

$$\frac1{4\pi}\int_{S^2}P(u)^2,d\omega \ge \frac43,\mu^2.$$

Assume that the conclusion of the problem is false. Then for some positive number $m$,

$$m\le P(u)<\sqrt2,m \qquad\text{for all }u.$$

For every $u$,

$$P(u)^2<\sqrt2,m,P(u).$$

Averaging over the sphere yields

$$\frac1{4\pi}\int_{S^2}P(u)^2,d\omega < \sqrt2,m,\mu.$$

Since $\mu\ge m$, we get

$$\frac1{4\pi}\int_{S^2}P(u)^2,d\omega < \sqrt2,\mu^2.$$

But the previously established estimate gives

$$\frac1{4\pi}\int_{S^2}P(u)^2,d\omega \ge \frac43,\mu^2.$$

Because

$$\frac43>\sqrt2,$$

the two inequalities are incompatible.

Hence our assumption was false. Therefore there exist directions $u$ and $v$ such that

$$\frac{P(u)}{P(v)}\ge\sqrt2.$$

These directions correspond to two planes whose projection areas have ratio at least $\sqrt2$.

This completes the proof.

∎

Verification of Key Steps

The projection formula must be checked carefully. For a convex polyhedron, the faces with $n_i\cdot u>0$ project onto the projection region without overlap in oriented area. Their projected areas sum to $P(u)$. The same holds for the opposite side. Adding the two equal expressions gives

$$2P(u)=\sum_i S_i|n_i\cdot u|.$$

A common mistake is to omit the factor $1/2$.

The identity

$$\sum_iS_in_i=0$$

is essential. It comes from the fact that the vector area of a closed polyhedral surface vanishes. Squaring this vector equation gives

$$\sum_{i,j}S_iS_j,n_i\cdot n_j=0.$$

Without this cancellation the second-moment estimate would not be strong enough.

The contradiction step uses the assumption

$$m\le P(u)<\sqrt2,m.$$

Multiplying by $P(u)$ gives

$$P(u)^2<\sqrt2,m,P(u).$$

After averaging, one must use $m\le\mu$, which follows from $P(u)\ge m$ everywhere. Replacing $m$ by a larger quantity would reverse the inequality and invalidate the argument.

Alternative Approaches

One may regard

$$P(u)=\frac12\sum S_i|n_i\cdot u|$$

as the support function of a centrally symmetric zonotope generated by the vectors $S_in_i$. The tetrahedral condition $\sum S_in_i=0$ implies strong constraints on this zonotope. The problem can then be translated into a statement about the ratio of maximal and minimal widths of a centrally symmetric hexagon in a suitable plane section. After reducing to a planar extremal problem, the bound $\sqrt2$ follows from geometric inequalities for widths.

Another approach uses spherical harmonics. Expanding $P(u)$ into its mean value and higher harmonics, the relation $\sum S_in_i=0$ forces a nontrivial lower bound for the $L^2$ norm of the nonconstant part. Comparing the $L^1$ and $L^2$ norms then yields a quantitative oscillation estimate, from which the existence of two projection areas with ratio at least $\sqrt2$ follows. The direct moment method used above is preferable because it remains entirely elementary and keeps the geometry of the tetrahedron visible throughout the proof.