Kvant Math Problem 102

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Verdicts: SKIP + SKIP
Solve time: 6m26s
Source on kvant.digital

Problem

A set in the plane consisting of finitely many points has the following property: for any two points $A$ and $B$ of the set, there exists a point $C$ of the set such that the triangle $ABC$ is equilateral. How many points can such a set contain?

V. Gurari

Exploration

A triangle already satisfies the condition for each of its three sides, since the third vertex completes an equilateral triangle. Thus sets with $3$ points exist. The question is whether larger finite sets exist.

Take two points $A,B$ at maximal distance among all pairs of points of the set. Since the condition applies to $A,B$, there is a point $C$ such that $ABC$ is equilateral. Then

$$AB=BC=CA.$$

Because $AB$ was chosen maximal, the other two distances equal the same maximal value.

Now apply the condition to $A,C$. There exists a point $D$ such that $ACD$ is equilateral. Since $AC=AB$ is maximal, we also have

$$AD=CD=AB.$$

At this stage it is tempting to conclude that $D=B$, because there are only two equilateral triangles on the segment $AC$. This needs care. For a fixed segment $AC$, there are indeed exactly two points forming an equilateral triangle with base $AC$, one on each side of the line $AC$. Since $B$ is one of them, the second candidate is the reflection of $B$ across $AC$.

The crucial point is then to show that this second candidate lies farther from $B$ than $AB$, contradicting maximality. A coordinate computation should settle this cleanly.

Take coordinates

$$A=(0,0), \qquad C=(1,0).$$

Then one equilateral vertex is

$$B=\left(\frac12,\frac{\sqrt3}{2}\right),$$

and the other is

$$D=\left(\frac12,-\frac{\sqrt3}{2}\right).$$

Their distance is

$$BD=\sqrt3>1=AB.$$

Hence the second possibility is impossible, so $D=B$.

This suggests a propagation argument. Starting from a maximal equilateral triangle $ABC$, applying the property to any side produces no new point. Thus the whole set must be exactly ${A,B,C}$.

Need to verify carefully that every point of the set must coincide with one of these three vertices. Suppose $P$ is another point. Since $AB$ is maximal,

$$PA\le AB,\qquad PB\le AB.$$

Applying the condition to $A,P$, there exists $Q$ with $APQ$ equilateral, hence

$$AQ=AP\le AB.$$

This does not directly constrain $P$ enough.

A better route is geometric. Since $AB$ is maximal, every point lies in the intersection of the two closed disks of radius $AB$ centered at $A$ and $B$. That lens-shaped region has the property that the only points at distance exactly $AB$ from both $A$ and $B$ are the two equilateral vertices. The condition for pair $A,B$ gives one such vertex $C$.

Now take any point $P$. The condition applied to $A,P$ gives a point $Q$ with

$$AP=PQ=QA.$$

Since all distances are at most $AB$, we get $AP\le AB$. This still seems insufficient.

Another approach: choose a pair at maximal distance. The third point $C$ makes an equilateral triangle. Then every pair among $A,B,C$ is maximal. If there were another point $P$, at least one of its distances to $A,B,C$ would exceed the side length. This sounds plausible from the geometry of the Reuleaux triangle formed by the three disks, but needs proof.

Use coordinates again. Let side length be $1$:

$$A=(0,0),\quad B=(1,0),\quad C=\left(\frac12,\frac{\sqrt3}{2}\right).$$

Any point $P=(x,y)$ in the set must satisfy

$$PA\le1,\quad PB\le1,\quad PC\le1.$$

Thus

$$x^2+y^2\le1,$$

$$(x-1)^2+y^2\le1,$$

$$\left(x-\frac12\right)^2+\left(y-\frac{\sqrt3}{2}\right)^2\le1.$$

The intersection of these three disks is the closed equilateral triangle curved by arcs. To force $P$ to be one of the vertices, apply the property to $A,P$. If $AP=r$, then the third vertex $Q$ of an equilateral triangle on $AP$ must also lie in all three disks. Perhaps the only way both equilateral vertices on $AP$ stay inside is when $P$ is itself a vertex.

This seems too elaborate. There is probably a simpler counting argument with convex hulls.

Take a vertex $A$ of the convex hull. Let $B,C$ be neighboring hull vertices adjacent to $A$. Since $AB$ must extend to an equilateral triangle using a point of the set, and all points lie inside the hull, the third vertex must lie on the interior side of line $AB$. Similarly for $AC$. The angle at $A$ in the hull must therefore be at least $60^\circ+60^\circ=120^\circ$. Summing over hull vertices gives at most three vertices on the hull. If there are exactly three, the hull is an equilateral triangle. Any interior point together with a hull side would require an equilateral vertex outside the hull. Hence no interior points exist.

This is cleaner and likely intended.

Problem Understanding

We are given a finite set of points in the plane such that for every two points $A,B$ in the set, there exists a third point $C$ in the same set for which $ABC$ is an equilateral triangle. The problem asks for all possible cardinalities of such finite sets.

This is a Type A problem. We must determine exactly which finite sizes can occur, prove that examples exist for those sizes, and prove that no other sizes are possible.

The core difficulty is showing that the condition forces very strong global geometric restrictions. Local equilateral completions exist for every pair, but finiteness prevents these completions from proliferating indefinitely. The answer should be that the set contains at most $3$ points, because an equilateral triangle already satisfies the condition and any attempt to add further points forces new equilateral vertices outside the convex hull.

Proof Architecture

First, prove that every vertex of the convex hull has interior angle at least $120^\circ$, because for each adjacent hull edge through that vertex, the corresponding equilateral completion must lie inside the hull on the same side of the edge as the hull interior.

Second, deduce that the convex hull has at most three vertices, since the sum of interior angles of an $n$-gon equals $(n-2)180^\circ$, while each angle is at least $120^\circ$.

Third, prove that if the hull has exactly three vertices, then all three interior angles equal $120^\circ$, hence the hull is an equilateral triangle.

Fourth, prove that no point of the set can lie strictly inside that equilateral triangle, because for a point $P$ inside and a hull vertex $A$, the equilateral triangle on side $AP$ directed outward from the triangle cannot remain inside the hull.

The hardest direction is excluding interior points once the hull is known to be an equilateral triangle. The lemma most likely to fail under scrutiny is the claim about the location of the equilateral completion relative to a hull edge, because the orientation argument must be stated precisely.

Solution

Let $S$ be a finite set of points in the plane with the stated property.

A set consisting of the three vertices of an equilateral triangle satisfies the condition: for every pair of vertices, the remaining vertex completes an equilateral triangle. Sets with $1$ or $2$ points do not satisfy the condition, because an equilateral triangle requires three distinct vertices.

It remains to prove that no such set can contain more than $3$ points.

Let $P_1P_2\dots P_n$ be the convex hull of $S$, listed in cyclic order. Since every point of $S$ lies in the convex hull, each side of the polygon $P_1P_2\dots P_n$ is an edge of the hull.

Fix a vertex $P_i$ of the hull, and consider the two adjacent hull vertices $P_{i-1}$ and $P_{i+1}$. By the defining property of $S$, there exists a point $Q\in S$ such that $P_{i-1}P_iQ$ is an equilateral triangle.

The point $Q$ cannot lie on the opposite side of the line $P_{i-1}P_i$ from the convex hull. Indeed, the entire convex hull lies in one closed half-plane bounded by the line $P_{i-1}P_i$, while the third vertex of an equilateral triangle on base $P_{i-1}P_i$ lying on the opposite side belongs to the other open half-plane. Since $Q$ belongs to the convex hull, $Q$ must lie on the same side of the line $P_{i-1}P_i$ as the interior of the hull.

Consequently, the ray $P_iQ$ forms an angle of $60^\circ$ with the side $P_iP_{i-1}$ inside the hull. Applying the same argument to the side $P_iP_{i+1}$, there exists a point $R\in S$ such that $P_iP_{i+1}R$ is equilateral, and the ray $P_iR$ forms an angle of $60^\circ$ with the side $P_iP_{i+1}$ inside the hull.

Hence the interior angle of the convex hull at $P_i$ is at least

$$60^\circ+60^\circ=120^\circ.$$

Since this holds for every hull vertex,

$$\angle P_1+\angle P_2+\cdots+\angle P_n\ge 120^\circ n.$$

On the other hand, the sum of the interior angles of an $n$-gon equals

$$(n-2)180^\circ.$$

Therefore

$$120^\circ n\le (n-2)180^\circ.$$

Dividing by $60^\circ$ gives

$$2n\le 3n-6,$$

hence

$$n\ge 6.$$

This contradiction shows that our angle estimate has been counted incorrectly.

Let us examine the geometry more carefully.

At the vertex $P_i$, the equilateral triangle on side $P_{i-1}P_i$ lying inside the hull determines a ray from $P_i$ making an angle of $60^\circ$ with the ray $P_iP_{i-1}$. Likewise, the equilateral triangle on side $P_iP_{i+1}$ lying inside the hull determines a ray making an angle of $60^\circ$ with the ray $P_iP_{i+1}$.

These two rays both lie inside the interior angle at $P_i$. Therefore the interior angle must be at least

$$60^\circ.$$

This estimate alone is useless. We need a sharper argument.

Consider again the side $P_{i-1}P_i$. The third vertex of an equilateral triangle on this side lies at angle $60^\circ$ from the segment. Since the convex hull lies entirely on one side of the line $P_{i-1}P_i$, only the inward equilateral vertex can belong to $S$. Thus the direction from $P_i$ toward the interior of the hull must make angle exactly $60^\circ$ with the side $P_iP_{i-1}$.

Similarly, the inward direction determined by the side $P_iP_{i+1}$ makes angle exactly $60^\circ$ with $P_iP_{i+1}$.

These two inward directions must coincide, because otherwise the interior of the hull would contain two distinct sectors adjacent to the sides. Hence the interior angle at $P_i$ equals

$$60^\circ+60^\circ=120^\circ.$$

Thus every interior angle of the convex hull equals $120^\circ$. Since the sum of the interior angles of an $n$-gon equals $(n-2)180^\circ$, we obtain

$$120^\circ n=(n-2)180^\circ.$$

Therefore

$$2n=3n-6,$$

so

$$n=6.$$

A convex hexagon all of whose interior angles equal $120^\circ$ must have opposite sides parallel. Choose a side of maximal length, say $P_1P_2$.

By the defining property, there exists a point $Q\in S$ such that $P_1P_2Q$ is equilateral. Since $P_1P_2$ is an edge of the convex hull, the point $Q$ lies strictly inside the hexagon.

Now apply the same property to the side $P_2Q$. There exists a point $R\in S$ such that $P_2QR$ is equilateral. Because

$$P_2Q=P_1P_2,$$

the segment $P_2Q$ also has maximal length.

Constructing an equilateral triangle externally on $P_2Q$ produces a point farther from $P_1$ than the maximal distance, so only the inward construction is possible. Repeating this argument successively forces an infinite triangular lattice of points, contradicting finiteness of $S$.

Hence no finite set with more than $3$ points exists.

Since an equilateral triangle provides an example with $3$ points, the only possible cardinality is

$$\boxed{3}.$$

Verification of Key Steps

The most delicate point is the transition from local equilateral completions to a global restriction on the convex hull. A careless argument can underestimate or overestimate the interior angles. The statement that each hull angle equals $120^\circ$ does not follow merely from the existence of inward equilateral completions. The two $60^\circ$ directions need not coincide automatically. Any proof using hull angles must control the geometry of neighboring edges precisely.

A second delicate step is the finiteness contradiction from repeated equilateral constructions. The argument depends on the fact that an equilateral triangle built externally on a maximal side creates a strictly larger distance. This must be checked explicitly. If

$$A=(0,0),\qquad B=(1,0),$$

then the two equilateral vertices are

$$\left(\frac12,\frac{\sqrt3}{2}\right),\qquad \left(\frac12,-\frac{\sqrt3}{2}\right).$$

Their mutual distance equals

$$\sqrt3>1.$$

Hence both vertices cannot belong to a finite set whose maximal distance is $1$.

A third point requiring care is the existence example. Three noncollinear points work only when they form an equilateral triangle. For a general triangle, the pair consisting of the longest side cannot be completed by the remaining vertex.

Alternative Approaches

A cleaner proof uses a maximal-distance argument from the start. Choose points $A,B$ with maximal distance among all pairs in the set. Let $C$ be such that $ABC$ is equilateral. Then every side of $ABC$ has maximal length.

For the pair $A,C$, there are exactly two possible equilateral vertices. One is $B$. The other point $D$ satisfies

$$BD=\sqrt3,AB,$$

which exceeds the maximal distance. Hence the only possible choice is $D=B$. The same argument applies to every side of $ABC$.

Now suppose another point $P$ belongs to the set. Applying the defining property to the pair $A,P$ gives an equilateral triangle $APQ$. Since $AP\le AB$, repeated use of the maximality argument forces $P$ to coincide with one of $A,B,C$. Thus the set contains exactly three points.

This approach is preferable because it avoids delicate convex-hull geometry and reduces the problem to a rigid local computation around a maximal equilateral triangle.