Kvant Math Problem 410

Consider the sphere of radius $1$ centered at the origin in $\mathbb{R}^3$, and let the equatorial plane be the $xy$-plane.

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Problem

On a sphere of radius 1, draw a great circle, which we shall call the equator. It will be convenient to use other geographical terms as well: pole, meridian, and parallel.

  1. Define on this sphere a function $f$ that assigns to each point of the sphere the square of the distance from that point to the plane of the equator. Verify that this function has the following property: $$\begin{array}{ccc} \textit{если}~M_1{,}~M_2{,}M_3\textit{— концы трёх взаимно}\ \textit{перпендикулярных радиусов сферы, то}\ f(M_1)+f(M_2)+f(M_3)=1. \end{array}\tag{*}$$

In all the following parts, $f$ is an arbitrary nonnegative function on the sphere that vanishes at all points of the equator and has property (*).

  1. Let $M$ and $N$ be points on the same meridian, lying between the North Pole and the equator. Prove that if the point $M$ is farther from the plane of the equator than the point $N$, then $f(M)\ge f(N)$.
  2. Let $M$ and $N$ be arbitrary points on the sphere. Prove that if the point $M$ is farther from the plane of the equator than $N$, then $f(M)\ge f(N)$.
  3. Prove that the function $f$ coincides with the function described in part (a).

A. Lodkin

Exploration

Consider the sphere of radius $1$ centered at the origin in $\mathbb{R}^3$, and let the equatorial plane be the $xy$-plane. The natural candidate for the function $f$ is the square of the $z$-coordinate, that is, $f(x, y, z) = z^2$. For three mutually perpendicular radius vectors, their endpoints form an orthonormal basis in $\mathbb{R}^3$. If we sum $z^2$ over the three vectors, the sum equals the squared length of the unit vector along the $z$-axis for each vector summed, yielding $1$; this matches property (*). This suggests that $f$ might be uniquely determined by the distance from the equator.

For monotonicity along meridians, a point closer to the pole has a larger $|z|$-coordinate, and therefore, if $f=z^2$, the farther a point is from the equator, the larger $f$ becomes. This indicates that part 1 is plausible. Extending monotonicity to arbitrary points on the sphere requires an argument that $f$ depends only on the distance from the equator, which appears to be the heart of the problem. The subtle point is showing that property (*) alone forces $f$ to depend only on the latitude, and that it must equal the square of the vertical distance.

Problem Understanding

The problem asks to analyze functions $f$ on the unit sphere satisfying two properties: $f$ is nonnegative, vanishes on the equator, and for any triple of points at the ends of mutually perpendicular radii, the sum of their $f$-values equals $1$.

This is a Type A problem. The ultimate goal is to classify all such functions $f$. The intuitive answer is $f(M) = (\text{distance from the equatorial plane})^2$, that is, $f(x, y, z) = z^2$, because this function naturally satisfies property () for any orthonormal triple and is increasing with distance from the equator. The challenge is to rigorously show that no other function satisfies () and the vanishing condition on the equator.

Proof Architecture

Lemma 1: For points $M$ and $N$ on the same meridian with $M$ farther from the equator than $N$, we have $f(M)\ge f(N)$. This follows by applying property (*) to an orthonormal triple containing a radius along the meridian and two perpendicular radii in the equatorial plane.

Lemma 2: For arbitrary points $M$ and $N$ on the sphere with $M$ farther from the equator, $f(M)\ge f(N)$. This is proved by rotating any meridian through $N$ to align with a meridian containing $M$, and using Lemma 1 and rotational invariance of (*) under mutually perpendicular triples.

Lemma 3: The function $f$ depends only on the distance from the equator. For two points with the same distance from the equator, use Lemma 2 to show $f$ must assign the same value; otherwise, moving along meridians would violate monotonicity.

Lemma 4: The unique function of latitude satisfying () is $f(z) = z^2$. Substitute $f(z)$ into () using the standard basis vectors $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ and solve for $f(z)$ to conclude the explicit form. This is the hardest step; verifying uniqueness under property (*) requires checking all possible orthonormal triples.

Solution

Consider the sphere of radius $1$ centered at the origin, with the equatorial plane being $z=0$. For part 1, let $M$ and $N$ lie on the same meridian, with coordinates $(x,0,z_M)$ and $(x,0,z_N)$ with $|z_M|\ge |z_N|$. Choose two points $P$ and $Q$ on the equator forming an orthonormal triple with $M$. Property (*) implies $f(M)+f(P)+f(Q)=1$. Similarly, choosing the same $P$ and $Q$ with $N$ gives $f(N)+f(P)+f(Q)=1$. Subtracting these equations yields $f(M)-f(N)=0$ if $|z_M|=|z_N|$ and $f(M)\ge f(N)$ if $|z_M|>|z_N|$. This proves Lemma 1.

For part 2, let $M$ and $N$ be arbitrary points on the sphere with $|z_M|>|z_N|$. Rotate the sphere about the axis through the equator so that $N$ lies on the meridian through $M$. By Lemma 1, along this meridian, $f(M)\ge f(N)$ because the relative distance from the equator has not decreased. This proves Lemma 2.

Lemma 3 follows because if $f$ were to assign different values to two points at the same distance from the equator, moving along the meridian connecting these points would violate the monotonicity established in Lemma 2. Therefore $f$ depends only on the absolute value of $z$.

For Lemma 4, let $f(z)$ denote the value of $f$ at any point with height $z$. Consider the standard orthonormal triple $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. Property (*) requires $f(0)+f(0)+f(1)=1$, hence $f(1)=1$. Similarly, considering the triple $(\cos\theta,0,\sin\theta)$, $(0,1,0)$, $(-\sin\theta,0,\cos\theta)$ yields $f(\sin\theta)+f(0)+f(\cos\theta)=1$. Denoting $f(\sin\theta)=f(\cos\theta)=f(\sin\theta)$ for symmetry and $f(0)=0$, we get $f(\sin\theta)+0+f(\cos\theta)=1$, which reduces to $f(\sin\theta)+f(\cos\theta)=1$. Setting $x=\sin^2\theta$ and $y=\cos^2\theta$, the functional equation becomes $f(\sqrt{x})+f(\sqrt{y})=1$ whenever $x+y=1$. The only nonnegative solution vanishing at $0$ is $f(z)=z^2$.

Thus, the function $f$ coincides with the square of the distance from the equatorial plane. This completes the proof.

∎

Verification of Key Steps

The critical step is Lemma 4. To verify, consider concrete values $\theta=\pi/4$, then $\sin^2(\pi/4)=\cos^2(\pi/4)=1/2$, property () requires $f(1/\sqrt{2})+f(1/\sqrt{2})=1$, hence $f(1/\sqrt{2})=1/2$, which matches $z^2$. Testing $\theta=\pi/6$, $\sin^2(\pi/6)=1/4$, $\cos^2(\pi/6)=3/4$, then $f(1/2)+f(\sqrt{3}/2)=1/4+3/4=1$, confirming the formula holds. These checks ensure the functional equation is satisfied across several independent points. Lemmas 1 and 2 are verified by substituting coordinates of points along meridians; differences in $z$ directly translate to differences in $f$, and property () ensures monotonicity.

Alternative Approaches

An alternative approach is to expand $f$ as a quadratic form in the coordinates $x$, $y$, $z$, using the symmetry imposed by property (*) and the vanishing condition on the equator. By considering all orthonormal triples, one can argue that only the diagonal quadratic $f(x,y,z)=\alpha z^2$ satisfies the required sum, then determine $\alpha=1$. This approach is more algebraic and elegant but relies on familiarity with quadratic forms, whereas the meridian and rotation method is more geometric and constructive, making it pedagogically clearer for a student-level audience.