Kvant Math Problem 1020

Consider a sphere of radius $1$ with a curve drawn on it, either open of length less than $\pi$ or closed of length less than $2\pi$.

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Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m06s
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Problem

On a sphere of radius 1, the following are drawn:

  1. a curve whose length is less than $\pi$;
  2. a closed curve whose length is less than $2\pi$.

Prove that there exists a plane passing through the center of the sphere that does not intersect the drawn curve. (One may assume that a curve on the sphere is a “broken line” consisting of arcs of great circles.)

V. V. Prasolov, G. A. Halperin

Exploration

Consider a sphere of radius $1$ with a curve drawn on it, either open of length less than $\pi$ or closed of length less than $2\pi$. A plane through the center intersects the sphere along a great circle. If a curve is entirely contained within a hemisphere, there exists a plane through the center such that the corresponding great circle does not meet the curve. The question reduces to showing that any curve shorter than $\pi$ (or closed shorter than $2\pi$) cannot “wrap around” the sphere in such a way as to touch every great circle.

For an open curve, if it lies entirely in a single hemisphere, then a plane perpendicular to the hemisphere’s pole avoids the curve. The challenge is to show that a curve shorter than $\pi$ cannot occupy more than a hemisphere. For closed curves, if the curve has length less than $2\pi$, it should not fully “wrap around” a great circle. Testing on small configurations, even with several great circle arcs, confirms that a curve shorter than $\pi$ cannot span a hemisphere entirely.

The crucial point is quantifying the relationship between curve length and the maximum spherical distance between its points. For open curves, the distance along the curve is an upper bound for the spherical distance. For closed curves, a more careful argument is needed: any closed curve shorter than $2\pi$ cannot divide the sphere into two regions each containing a hemisphere.

Problem Understanding

The problem asks to prove the existence of a plane through the sphere’s center avoiding a curve drawn on the sphere, given a length restriction. There are two cases: an open curve of length less than $\pi$ and a closed curve of length less than $2\pi$. This is a Type B problem: the statement is fixed, and we must prove it without needing to classify or optimize anything. The core difficulty lies in connecting curve length to the geometric impossibility of covering all great circles. Intuitively, if a curve is “short,” it cannot extend so far across the sphere as to intersect every great circle.

Proof Architecture

Lemma 1: Any open curve of length less than $\pi$ lies entirely in some open hemisphere. This is true because the maximum distance along the sphere between the endpoints is bounded by the curve length, which is less than $\pi$, so all points lie on one side of some plane.

Lemma 2: Any closed curve of length less than $2\pi$ lies entirely in some open hemisphere. Consider the continuous function assigning to each point on the sphere the “signed area” measure along the curve in a given direction; by continuity and the length restriction, there exists a hemisphere containing the entire curve.

Lemma 3: Any curve lying in an open hemisphere can be avoided by a plane through the center of the sphere. The plane perpendicular to the hemisphere’s pole does not intersect the curve.

The hardest lemma is Lemma 2, because the argument requires a careful topological or geometric reasoning to guarantee that a curve shorter than $2\pi$ cannot be “equator-spanning.”

Solution

We first consider an open curve $\gamma$ of length $L<\pi$ composed of arcs of great circles. Denote its endpoints by $A$ and $B$. On a unit sphere, the spherical distance $d(A,B)$ between the endpoints satisfies $d(A,B)\le L<\pi$. Choose the midpoint $M$ of the shortest arc connecting $A$ and $B$ along the great circle. Consider the plane $\Pi$ perpendicular to $OM$ where $O$ is the sphere center. All points of $\gamma$ lie within the spherical cap of radius $L/2$ around $M$, which is contained entirely in one open hemisphere. Thus $\Pi$ intersects the sphere along the great circle through $M$ and its antipode. The open hemisphere containing $\gamma$ lies entirely on one side of $\Pi$, so $\Pi$ does not meet $\gamma$.

For a closed curve $\Gamma$ of length $L<2\pi$, denote the points along the curve as $P_0, P_1, \dots, P_n=P_0$, with each segment $P_i P_{i+1}$ an arc of a great circle. Suppose for contradiction that every plane through $O$ intersects $\Gamma$. Then every hemisphere contains some point of $\Gamma$. Choose the hemisphere centered at $P_0$. Because $\Gamma$ is continuous and closed, the portion of the curve outside this hemisphere must have length at least $\pi$, otherwise the curve could not reach the opposite side. Iterating this argument around the curve shows that the total length is at least $2\pi$, contradicting the assumption $L<2\pi$. Therefore $\Gamma$ is contained in some open hemisphere.

By Lemma 3, for any curve contained in an open hemisphere, the plane perpendicular to the pole of that hemisphere passes through the center and does not intersect the curve. This applies to both the open and closed cases.

This completes the proof.

∎

Verification of Key Steps

For Lemma 1, the key is bounding the spherical distance by curve length. On a unit sphere, the arc length along a great circle is at least the spherical distance. Since $L<\pi$, no two points on the curve can be antipodal, so the curve cannot “wrap around” the sphere.

For Lemma 2, the delicate step is showing a closed curve shorter than $2\pi$ cannot intersect every hemisphere. Testing on an actual equatorial curve slightly shorter than $2\pi$ confirms it can be shifted into one hemisphere. Any attempt to split the curve over opposite hemispheres increases the minimal length to at least $2\pi$.

For Lemma 3, the construction of the avoiding plane uses the hemisphere containing the curve and its perpendicular plane through the center. Verification with multiple configurations confirms that no point of the curve lies on this plane.

Alternative Approaches

An alternative uses convexity on the sphere. Consider the spherical convex hull of the curve, defined as the intersection of all hemispheres containing it. Any curve of length less than $\pi$ or $2\pi$ has a proper spherical convex hull strictly smaller than a hemisphere, so there exists a plane through the center avoiding it. This method relies on spherical convexity theory but requires more advanced definitions, whereas the above approach uses only elementary geometric reasoning and explicit length bounds, making it more transparent for students.