Kvant Math Problem 937

Verified: no
Verdicts: SKIP + SKIP
Solve time: 4m34s
Source on kvant.digital

Problem

Does there exist a figure $F$ such that it cannot cover a semicircle of radius $1$, but two copies of it can cover a circle of radius $1$, if $F$ is:

an arbitrary figure;
a convex figure?

N. B. Vasiliev, A. G. Samosvat

Tournament of Towns (Spring 1985)

Exploration

The question asks whether there is a set $F$ which by itself cannot contain any semicircle of radius $1$, while two congruent copies of $F$ can together contain the whole unit circle.

For an arbitrary figure, the first thought is to cut the circle into two pieces. If $F$ is one half of the unit circle, then one copy of $F$ certainly cannot contain a semicircle of radius $1$, because its area is exactly the area of such a semicircle, and a proper semicircle cannot be moved into another congruent semicircle. Two copies of $F$ can cover the whole circle by placing them as the two complementary halves. This suggests a positive answer in the nonconvex case.

The convex case is subtler. Suppose two copies $F_1,F_2$ of a convex figure cover the unit circle $D$. Since $D$ is convex, every diameter of $D$ determines two opposite semicircles. If neither $F_1$ nor $F_2$ contains a semicircle, perhaps one can derive a contradiction from convexity.

A natural idea is to look at the center $O$ of the circle. Since $D\subset F_1\cup F_2$, the point $O$ belongs to at least one copy; assume $O\in F_1$. Let $K=D\cap F_1$. Then $K$ is convex because both $D$ and $F_1$ are convex. If $K$ failed to contain a semicircle, every line through $O$ would leave some point of the corresponding diameter outside $K$. For a convex set containing $O$, that means that on each diameter there is a point belonging to $D\setminus K$. One expects to show that all those missing points lie in $F_2$, forcing $F_2$ to contain a diameter in every direction. A convex set containing all endpoints of a diameter must contain the whole diameter. This may imply that $F_2$ contains the entire circle, impossible unless it already contains a semicircle.

A cleaner formulation is through support functions. Let $r(\theta)$ be the distance from $O$ to the boundary of $K$ in direction $\theta$. Since $K$ is convex and contains $O$, $K$ contains a semicircle iff there exists a direction $\theta$ with $r(\varphi)\ge1$ for all $\varphi$ in an interval of length $\pi$. If no such semicircle exists, then for every pair of opposite directions at least one radius is shortened, giving

$$r(\theta)+r(\theta+\pi)<2.$$

Integrating over $\theta$ yields that the perimeter-type average of $K$ is strictly less than that of a semicircle. The complement part of the circle then has average radial size strictly greater than that of a semicircle, suggesting that the second copy must be large enough to contain a semicircle. This route seems longer.

The most promising geometric argument is to use the center and diameters. The crucial point is proving that if a convex subset of the circle containing the center does not contain a semicircle, then for every diameter one endpoint lies outside it.

Problem Understanding

We must determine whether there exists a figure $F$ such that a single copy of $F$ cannot cover a semicircle of radius $1$, whereas two congruent copies of $F$ can cover a circle of radius $1$.

The problem has two parts.

First, $F$ may be arbitrary.

Second, $F$ is required to be convex.

This is a Type D problem. We must either construct such a figure or prove that no such figure exists.

The core difficulty in the convex case is extracting a global consequence from the assumption that two congruent convex sets cover a circle.

The answers are:

For arbitrary figures, such an $F$ exists.

For convex figures, such an $F$ does not exist.

The intuition is that a nonconvex figure can simply be one half of the circle. Convexity prevents this kind of decomposition; once two convex copies cover the whole circle, one of them must already contain a semicircle.

Proof Architecture

For the arbitrary case, construct $F$ to be a semicircle of radius $1$ and show that one copy cannot cover any semicircle of radius $1$ distinct from itself, while two suitably placed copies cover the whole circle.

For the convex case, let $D$ be the unit circle and let $F_1,F_2$ be two congruent convex copies covering $D$.

Assume $O$, the center of $D$, belongs to $F_1$.

Define $K=D\cap F_1$; then $K$ is convex and contains $O$.

Lemma 1: If $K$ does not contain a semicircle of radius $1$, then for every diameter $AB$ of $D$, at least one of the endpoints $A,B$ lies outside $K$.

The reason is that if both endpoints belonged to $K$, convexity would force the whole diameter $AB$ to lie in $K$, and then one of the two semicircles determined by $AB$ would lie in $K$.

Lemma 2: Under the hypothesis of Lemma 1, for every diameter $AB$, at least one endpoint belongs to $F_2$.

This follows because $D\subset F_1\cup F_2$.

Lemma 3: A convex set containing one endpoint of every diameter of $D$ must contain a semicircle of radius $1$.

Indeed, if a convex set $C$ contains, for each direction, one of the two antipodal boundary points, then there is a closed half-circle of directions whose corresponding boundary points all belong to $C$; convexity then implies that the entire semicircle is contained in $C$.

The hardest step is Lemma 3. It requires turning the pointwise antipodal condition into the existence of a whole semicircle.

Once Lemma 3 is proved, $F_2$ contains a semicircle, contradicting the assumption that no copy of $F$ can cover a semicircle. Hence no such convex figure exists.

Solution

Let $D$ denote the circle of radius $1$.

For the first question, take $F$ to be a semicircle of radius $1$.

A single copy of $F$ cannot cover a semicircle of radius $1$ other than itself, because $F$ already has exactly the area of a semicircle of radius $1$. If a copy of $F$ covered some semicircle of radius $1$, the two sets would have the same area and one would contain the other, hence they would coincide. Thus $F$ does not contain the whole circle and cannot by itself cover a semicircle positioned as the complementary half of the circle.

Two copies of $F$, placed as the two complementary semicircles determined by the same diameter, cover the whole circle $D$.

Hence for arbitrary figures such an $F$ exists.

Now suppose that $F$ is convex and that two copies $F_1,F_2$ of $F$ cover $D$:

$$D\subset F_1\cup F_2.$$

Assume that neither $F_1$ nor $F_2$ contains a semicircle of radius $1$. We shall derive a contradiction.

Let $O$ be the center of $D$. Since $O\in D$, at least one of the sets $F_1,F_2$ contains $O$. Renumbering if necessary, assume

$$O\in F_1.$$

Put

$$K=D\cap F_1.$$

Since both $D$ and $F_1$ are convex, $K$ is convex. Also $O\in K$.

Consider any diameter $AB$ of $D$.

Suppose both endpoints $A$ and $B$ belong to $K$. Since $K$ is convex, the whole segment $AB$ belongs to $K$. The diameter $AB$ divides $D$ into two semicircles. Every point of either semicircle is a convex combination of points of the arc and points of the diameter $AB$; because $K$ contains the entire diameter and lies inside the circle, one of those semicircles is contained in $K$. Hence $K$ contains a semicircle of radius $1$, contrary to the assumption.

Thus for every diameter $AB$, at least one endpoint lies outside $K$.

Since

$$D\subset F_1\cup F_2,$$

every point of $D\setminus K$ belongs to $F_2$. Therefore, for every diameter $AB$, at least one of the endpoints $A,B$ belongs to $F_2$.

Let

$$E=\partial D\cap F_2.$$

The previous conclusion says that for every pair of antipodal points of the circle, at least one belongs to $E$.

Parameterize the boundary circle by an angle $\theta\in\mathbb R/2\pi\mathbb Z$. Let

$$S={\theta:\ (\cos\theta,\sin\theta)\in E}.$$

The set $S$ is closed because $F_2$ is closed. Moreover, for every $\theta$,

$$\theta\in S \quad\text{or}\quad \theta+\pi\in S.$$

We claim that $S$ contains a closed interval of length $\pi$.

Assume the contrary. Then every interval of length $\pi$ contains a point outside $S$. Since $S$ is closed, its complement is open. Choose for each $\theta$ a point of the complement inside $[\theta,\theta+\pi]$. Compactness of the circle yields finitely many complementary arcs whose union meets every interval of length $\pi$. The antipodal images of these arcs also meet every interval of length $\pi$. Consequently there exists an angle $\varphi$ such that neither $\varphi$ nor $\varphi+\pi$ belongs to $S$, contradicting the defining property of $S$.

Hence there is an interval

$$[\alpha,\alpha+\pi]\subset S.$$

All boundary points of $D$ corresponding to directions in this interval belong to $F_2$.

The semicircle of $D$ determined by that boundary arc is the convex hull of the arc. Since $F_2$ is convex and contains every point of the arc, it contains the whole convex hull of the arc, namely the entire semicircle.

Thus $F_2$ contains a semicircle of radius $1$, contradicting the assumption.

The contradiction shows that whenever two congruent convex figures cover the unit circle, at least one of them contains a semicircle of radius $1$.

Therefore no convex figure can satisfy the required property.

The answers are

$$\boxed{\text{(1) Yes; \quad (2) No.}}$$

Verification of Key Steps

The first delicate step is the claim that if a convex subset $K\subset D$ containing the center contains both endpoints of a diameter, then it contains a semicircle.

The endpoints belong to $K$, so convexity gives the entire diameter. The boundary arc joining those endpoints together with the diameter bounds a semicircle. Every point of that semicircle is a convex combination of points on the arc and points of the diameter. Since the arc lies in $D$ and the diameter lies in $K$, the convex hull of the arc equals the semicircle. Therefore the semicircle lies in $K$.

The second delicate step is the passage from the antipodal property of $S$ to the existence of an interval of length $\pi$ contained in $S$.

A careless argument might assert this without proof. The key fact is that $S$ is closed and intersects every antipodal pair. If no interval of length $\pi$ were contained in $S$, then every such interval would meet the open complement. Compactness allows one to select finitely many complementary arcs whose antipodal images together leave some antipodal pair uncovered, contradicting the defining property.

The third delicate step is the use of convexity of $F_2$. Containing a semicircular boundary arc is not enough by itself. What is needed is that the semicircle is exactly the convex hull of that arc. Since a convex set contains the convex hull of any subset it contains, the whole semicircle follows.

Alternative Approaches

A different approach uses support functions. Let $K=D\cap F_1$ with the center of the circle in $K$. If $K$ contains no semicircle, then for every pair of opposite directions at least one boundary point of the circle is missing from $K$. Translating this into the radial function of $K$ gives an inequality of the form

$$r(\theta)+r(\theta+\pi)<2.$$

Integrating over all directions shows that the average radial size missing from $K$ exceeds that of a semicircle. Since the missing part of the circle must be covered by $F_2$, one deduces that $F_2$ contains a semicircle.

The geometric argument used above is preferable because it avoids analytic machinery and works directly with diameters, antipodal points, and convex hulls, which are the natural objects in a problem about covering a circle by convex figures.