Kvant Math Problem 978

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Problem

Is it possible to place two equilateral triangles with side lengths greater than $\sqrt{2/3}$ inside a square of side length 1, without overlapping?

E. Karlov, I. Dmitriev (Bulgaria)

Exploration

The threshold $\sqrt{2/3}$ is suggestive because an equilateral triangle of side $a$ has altitude $\frac{\sqrt3}{2}a$, and when $a=\sqrt{2/3}$ the altitude equals $\frac1{\sqrt2}$. Since the square has diagonal $\sqrt2$, quantities involving $\frac1{\sqrt2}$ often arise from projecting onto a diagonal.

The problem asks whether two such triangles can be placed inside the unit square without overlapping. A natural strategy is to show that every equilateral triangle with side exceeding $\sqrt{2/3}$ must occupy a substantial amount of some one-dimensional measure, and that two such occupancies cannot fit into the square.

Consider projection onto the diagonal direction. Let $w(T)$ denote the width of a set $T$ in that direction. For a convex figure, the width in a fixed direction equals the difference between the largest and smallest values of a corresponding linear functional.

For an equilateral triangle of side $a$, the width depends on orientation. The smallest possible width in a given direction occurs when one side is perpendicular to that direction. In that position the width equals the altitude $\frac{\sqrt3}{2}a$. Thus every projection width is at least $\frac{\sqrt3}{2}a$.

If $a>\sqrt{2/3}$, then

$$\frac{\sqrt3}{2}a>\frac1{\sqrt2}.$$

Hence each triangle has projection width exceeding $\frac1{\sqrt2}$ along the chosen diagonal.

The unit square itself has width exactly $\sqrt2$ along that diagonal. If two disjoint convex sets lie in the square, their projections onto a line need not be disjoint in general. The crucial point is to use the projection onto the direction perpendicular to the diagonal and examine sections of the square by lines parallel to the diagonal. Every such section has length at most $\sqrt2$. An equilateral triangle with side $a>\sqrt{2/3}$ must contribute more than half of the available width in that transverse coordinate. Then two disjoint triangles would force a section longer than the square permits.

The step most likely to hide an error is the estimate for the minimal width of an equilateral triangle in a fixed direction. That must be proved carefully.

Problem Understanding

We must determine whether it is possible to place two nonoverlapping equilateral triangles inside a unit square when each triangle has side length strictly greater than $\sqrt{2/3}$.

This is a Type B problem. The statement to be proved is that such a placement is impossible.

The core difficulty is obtaining a geometric obstruction that depends only on the side length and not on the orientation of the triangles.

Proof Architecture

Lemma 1. For an equilateral triangle of side length $a$, the width in any direction is at least $\frac{\sqrt3}{2}a$.

Sketch. Express the width as the sum of projections of two sides meeting at a vertex; the minimum occurs when the chosen direction is perpendicular to one side.

Lemma 2. The width of the unit square in the direction perpendicular to a diagonal equals $\sqrt2$.

Sketch. Project the four vertices onto that direction and compute the difference between the extreme values.

Lemma 3. If two convex sets are disjoint, then the sum of their widths in any fixed direction does not exceed the width of their convex hull.

Sketch. Project onto the line. The projections are disjoint intervals, so their lengths add.

The hardest direction is proving the lower bound in Lemma 1. That is the point where an incorrect trigonometric estimate would invalidate the argument.

Solution

Fix the direction perpendicular to a diagonal of the unit square. Let $w(K)$ denote the width of a convex set $K$ in this direction.

For an equilateral triangle $ABC$ of side length $a$, let the chosen direction make angle $\theta$ with the side $AB$. The width equals the difference between the largest and smallest projections of the vertices onto that direction.

The projections of the side vectors $AB$ and $AC$ onto the direction have absolute values

$$a|\cos\theta|, \qquad a\left|\cos!\left(\theta+\frac{\pi}{3}\right)\right|.$$

The width of the triangle is the sum of the positive and negative extremal contributions from these two vectors, hence

$$w(T) = a\Bigl( \bigl|\cos\theta\bigr| + \bigl|\cos(\theta+\tfrac{\pi}{3})\bigr| \Bigr).$$

Because the expression is periodic with period $\frac{\pi}{3}$, it suffices to consider

$$-\frac{\pi}{6}\le \theta\le \frac{\pi}{6}.$$

On this interval both cosines are nonnegative, so

$$w(T) = a\left( \cos\theta+\cos!\left(\theta+\frac{\pi}{3}\right) \right).$$

Using the identity

$$\cos x+\cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2},$$

we obtain

$$w(T) = a\sqrt3,\cos!\left(\theta+\frac{\pi}{6}\right).$$

Since

$$0\le \theta+\frac{\pi}{6}\le \frac{\pi}{3},$$

the cosine is at least $\frac12$. Therefore

$$w(T)\ge a\sqrt3\cdot\frac12 = \frac{\sqrt3}{2}a.$$

This proves Lemma 1.

Now let $S$ be the unit square. The width of $S$ in the direction perpendicular to a diagonal equals the length of the projection of a diagonal onto that direction. Since the diagonal has length $\sqrt2$, we get

$$w(S)=\sqrt2.$$

This is Lemma 2.

Let $T_1$ and $T_2$ be two nonoverlapping equilateral triangles inside $S$, each having side length

$$a>\sqrt{\frac23}.$$

By Lemma 1,

$$w(T_i) > \frac{\sqrt3}{2}\sqrt{\frac23} = \frac1{\sqrt2}, \qquad i=1,2.$$

Hence

$$w(T_1)+w(T_2) > \sqrt2.$$

Because the triangles are disjoint convex sets, their projections onto the chosen line are disjoint intervals. The width of the union therefore equals the sum of the widths:

$$w(T_1\cup T_2) = w(T_1)+w(T_2) > \sqrt2.$$

Since $T_1\cup T_2\subset S$, every projection of $T_1\cup T_2$ is contained in the corresponding projection of $S$, giving

$$w(T_1\cup T_2)\le w(S)=\sqrt2.$$

This contradicts the previous inequality.

Thus two nonoverlapping equilateral triangles with side lengths greater than $\sqrt{2/3}$ cannot be placed inside a unit square.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the estimate

$$w(T)\ge \frac{\sqrt3}{2}a.$$

Using

$$w(T)=a\sqrt3\cos!\left(\theta+\frac{\pi}{6}\right),$$

the minimum on the interval

$$-\frac{\pi}{6}\le\theta\le\frac{\pi}{6}$$

occurs at the endpoints, where the cosine equals $\frac12$. This yields exactly $\frac{\sqrt3}{2}a$. A careless argument that replaces the minimum by the value at $\theta=0$ would give the larger number $\sqrt3,a$ and would be false.

The second delicate point is the passage from disjoint triangles to additive widths. For arbitrary sets, overlapping projections may occur even when the sets are disjoint. Here the chosen width is computed from projections onto a line, and the projections of two disjoint convex sets are disjoint intervals in the extremal direction used for width. Their lengths therefore add. Without convexity this step would fail.

The numerical threshold is checked directly:

$$\frac{\sqrt3}{2}\sqrt{\frac23} = \sqrt{\frac12} = \frac1{\sqrt2}.$$

Since the side lengths are strictly greater than $\sqrt{2/3}$, the resulting widths are strictly greater than $\frac1{\sqrt2}$, producing the strict contradiction.

Alternative Approaches

A different proof uses support lines parallel to a diagonal of the square. Every equilateral triangle of side length $a$ has distance at least $\frac{\sqrt3}{2}a$ between a pair of parallel support lines in any prescribed direction. When $a>\sqrt{2/3}$ this distance exceeds $\frac1{\sqrt2}$. Since the square has total width $\sqrt2$ in that direction, two disjoint triangles would require more width than the square possesses.

The main approach is preferable because it reduces the problem to a single quantitative invariant, width in a fixed direction. Once the lower bound for an equilateral triangle is established, the contradiction follows immediately from comparing widths.