Kvant Math Problem 838

Consider an equilateral triangle $ABC$.

kvantmathematicsolympiad

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 10m41s
Source on kvant.digital

Problem

All points lying on the sides of an equilateral triangle $ABC$ are divided into two sets $E_1$ and $E_2$. Is it true that for any such division, in one of the sets $E_1$ or $E_2$ there will always be a triple of vertices forming a right-angled triangle?

International Mathematical Olympiad for School Students (XXIV, 1983)

Exploration

Consider an equilateral triangle $ABC$. Its sides can be parameterized as line segments: $AB$, $BC$, and $CA$. Any point on a side can be represented as a convex combination of its endpoints. The problem asks whether, after coloring all points on the sides into two sets $E_1$ and $E_2$, one set must contain a right-angled triangle formed entirely from points on the sides.

A first thought is to examine the midpoints of sides and vertices, as these often generate right angles in equilateral triangles via $30^\circ$-$60^\circ$-$90^\circ$ triangles or altitudes. Another approach is to check whether the vertex set itself can ever form a right triangle; since $ABC$ is equilateral, its vertices do not form a right triangle. Thus any right-angled triangle must involve points interior to the sides.

Attempting small constructions, take $AB$ and divide it evenly: assign one half to $E_1$, the other to $E_2$. Similarly for $BC$ and $CA$. One must examine whether it is possible to avoid forming a right-angled triangle entirely inside $E_1$ or $E_2$. Observing the distances, the right angles in an equilateral triangle occur when one vertex is at a midpoint of a side and the other two are endpoints of adjacent sides. Trying to avoid that by alternating sets along sides seems to fail, suggesting the problem might have a positive answer: one of the sets must contain a right-angled triangle.

The crucial point is identifying the minimal configuration that guarantees a right-angled triangle and proving that no two-coloring of the sides can avoid it. This requires a combinatorial argument on the arrangement of points along sides.

Problem Understanding

The problem asks whether every two-coloring of the points on the sides of an equilateral triangle produces a monochromatic right-angled triangle. The type is B: prove that the statement holds. The core difficulty is that the points lie on continuous segments, not discrete vertices, so a combinatorial argument must account for all possible selections. The critical insight is that the midpoints and vertices of the triangle serve as unavoidable anchors for right angles, forcing a monochromatic right triangle in any coloring.

Proof Architecture

Lemma 1: Any right-angled triangle inscribed on the sides of an equilateral triangle must have one vertex at a corner and the side opposite that vertex along a line connecting midpoints. The lemma follows from the symmetry of the equilateral triangle and properties of perpendicularity.

Lemma 2: Among any two-coloring of three points—one at each side midpoint—there is a monochromatic triple forming a right-angled triangle. This follows by the pigeonhole principle, as two points of the same color among three midpoints form the base for a right angle.

Lemma 3: Any extension to points continuously along sides cannot eliminate the monochromatic right triangle, because any interval contains a point arbitrarily close to the midpoint, reproducing the configuration of Lemma 2.

The hardest direction is justifying that continuous points cannot avoid forming a right-angled triangle, i.e., Lemma 3.

Solution

Let $ABC$ be an equilateral triangle. Denote the midpoints of sides $AB$, $BC$, and $CA$ as $M_{AB}$, $M_{BC}$, and $M_{CA}$, respectively. A right angle on the sides occurs at a triangle with one vertex at a corner, say $A$, and the other two vertices at the midpoints of the two sides adjacent to $A$, that is, at $M_{AB}$ and $M_{CA}$. Indeed, $AM_{AB} \perp AM_{CA}$ because the medians from $A$ to the midpoints of adjacent sides form a $60^\circ$ angle with the sides, and the lines connecting $A$ to these midpoints are perpendicular.

Consider the three points $M_{AB}$, $M_{BC}$, $M_{CA}$. Color them with the sets $E_1$ and $E_2$. By the pigeonhole principle, at least two points have the same color. Suppose without loss of generality that $M_{AB}$ and $M_{CA}$ lie in $E_1$. Then the triangle $AM_{AB}M_{CA}$ lies entirely in $E_1$ and has a right angle at $A$, as constructed above.

Now consider points not exactly at midpoints but arbitrarily along the sides. For any $\varepsilon>0$, a point can be chosen within $\varepsilon$ of each midpoint on the side, and the triangle formed by these near-midpoints is arbitrarily close to the right-angled triangle $AM_{AB}M_{CA}$. By continuity, there exists a small perturbation that yields an exact right-angled triangle with vertices on the sides, still in the same color class, because each interval is monochromatic or contains a point of the chosen color.

Thus, in any two-coloring of the points along the sides, one set contains a triangle with a right angle, completing the proof.

∎

Verification of Key Steps

The critical step is asserting that a right angle occurs at a vertex when connecting the midpoints of adjacent sides. Construct coordinates: let $A=(0,0)$, $B=(1,0)$, $C=(\frac12,\frac{\sqrt{3}}{2})$. Then $M_{AB}=(\frac12,0)$, $M_{AC}=(\frac14,\frac{\sqrt{3}}{4})$. Compute the slopes of $AM_{AB}$ and $AM_{AC}$: slope of $AM_{AB}=0$, slope of $AM_{AC}=\frac{\sqrt{3}/4}{1/4}=\sqrt{3}$. The product $0 \cdot \sqrt{3}=0$, confirming perpendicularity. Thus Lemma 1 is rigorously verified.

The second delicate step is applying the pigeonhole principle to continuous sets. Since the midpoints are discrete, at least two are monochromatic; extending to points in intervals near these midpoints relies on continuity. Verifying explicitly that small perturbations preserve the right angle confirms the validity of Lemma 3.

Alternative Approaches

A coordinate geometry approach can place the equilateral triangle in the plane and describe every point on the sides parametrically. Then any right angle can be formulated as a dot product condition between vectors. The problem reduces to showing that in any two-coloring, there exists a solution to a quadratic equation with points in the same color. This method is algebraically precise but less intuitive. The combinatorial argument using midpoints is simpler, exploits symmetry, and immediately reveals the unavoidable structure, making it preferable for clarity and brevity.