Kvant Math Problem 396

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Problem

A triangle whose all sides are greater than $1~\text{см}$ will be called "large." Given an equilateral triangle $ABC$ with side $5~\text{см}$, prove that:

it is possible to cut 1000 "large" triangles from the triangle $ABC$;
the triangle $ABC$ can be dissected into 1000 "large" triangles;
the triangle $ABC$ can be triangulated into 1000 "large" triangles, i.e., subdivided so that any two triangles either do not share any points, share only a vertex, or share a side;
carry out parts (b) and (c) for an equilateral triangle with side $3~\text{см}$.

S. V. Fomin

All-Union Mathematical Olympiad for School Students (1976, grades 8 and 9)

Exploration

The side length of the equilateral triangle is much larger than the lower bound $1$ imposed on the sides of the desired triangles. Since no upper bound is imposed, the problem is not about area but about producing exactly $1000$ pieces while keeping every side of every piece longer than $1$.

A natural operation is to choose a point on a side of a triangle and join it to the opposite vertex. This replaces one triangle by two triangles. If the chosen point lies on a side of length greater than $2$, both new triangles have that side split into two parts each exceeding $1$, while the other two sides are inherited from the original triangle. Hence both new triangles remain large. Thus a large triangle can often be replaced by two large triangles.

The number $1000$ suggests repeated splitting. Starting from one triangle, each split increases the number of triangles by $1$. To obtain $1000$ triangles it suffices to perform $999$ admissible splits.

The difficulty is that repeated splitting may eventually create triangles whose sides become too short. We need a scheme that allows arbitrarily many splits while preserving the condition that every side remain greater than $1$.

Consider a large triangle having one side much longer than $2$. If we cut that side at its midpoint and join the midpoint to the opposite vertex, the two resulting triangles again have a side longer than $1$. Repeating this process on descendants whose long side remains greater than $2$ creates a binary tree of large triangles. The crucial question is whether one can obtain arbitrarily many leaves before all long sides fall below $2$.

For the equilateral triangle of side $5$, after $k$ repeated bisections of one original side segment, pieces of length $5/2^k$ appear. Since $5/4>1$ but $5/8<1$, only two generations of exact bisection are possible. Thus naive repeated halving fails.

A different idea is to keep cutting off very thin triangles. Suppose a side has length $L>2$. Choosing a point at distance $1+\varepsilon$ from one endpoint creates one new side segment of length $1+\varepsilon$ and another of length $L-1-\varepsilon$. If $L$ is large, one of the two resulting triangles retains a side nearly as long as $L$. Hence the process can be repeated many times. This allows producing arbitrarily many large triangles from a sufficiently long side.

For side $5$, the interval of admissible cut positions on a side has length $L-2=3$. One may cut off many triangles successively along the same side, always leaving a residual side longer than $1$. Since area places no restriction, the number of triangles obtainable is unbounded. This should solve part (a), where only cutting out $1000$ large triangles is required.

For parts (b) and (c), the whole triangle must be partitioned into exactly $1000$ large triangles. A useful observation is that if a triangulation into $N$ large triangles exists and one triangle in it has a side longer than $2$, that triangle can be subdivided into two large triangles, increasing the count to $N+1$. Thus it suffices to construct one triangulation and show that the increment operation can be repeated indefinitely.

For side $3$, the margin above $2$ is only $1$, but the same operation still works. An equilateral triangle of side $3$ itself is large and has side longer than $2$. Repeatedly subdividing one triangle near a vertex leaves a descendant with a side still exceeding $2$. Hence arbitrarily large numbers of triangles can be produced even in the side-$3$ case.

The delicate point is to formulate a subdivision operation that preserves triangulations and can be iterated indefinitely.

Problem Understanding

We are given an equilateral triangle. A triangle is called large if each of its three sides is strictly greater than $1$ cm.

For the equilateral triangle of side $5$ cm we must prove three existence statements. First, it is possible to cut out $1000$ large triangles from it. Second, the entire triangle can be dissected into $1000$ large triangles. Third, it can be triangulated into $1000$ large triangles. Finally, analogous constructions are required for an equilateral triangle of side $3$ cm.

This is a Type D problem. We must construct suitable subdivisions and verify that every resulting triangle is large.

The core difficulty is to obtain exactly $1000$ triangles while maintaining the lower bound $1$ on every side. The essential idea is to find an operation that increases the number of triangles by one and can be repeated indefinitely.

Proof Architecture

We use one lemma.

Lemma. Let $T=ABC$ be a large triangle and suppose $AB>2$. If $P$ is any point of $AB$ satisfying $AP>1$ and $PB>1$, then the segment $CP$ divides $T$ into two large triangles.

Proof sketch. The new triangles are $ACP$ and $BCP$. Each inherits two sides from $ABC$, hence those sides are greater than $1$. The third sides are $AP$ and $PB$, chosen greater than $1$.

The main construction is iterative. Starting from an equilateral triangle of side $s$, where $s=5$ or $s=3$, choose a point $P$ on a side $AB$ with $AP=1+\varepsilon$ and $PB=s-1-\varepsilon$. The lemma yields two large triangles. One of them still contains a side of length $s$, hence again has a side longer than $2$. Applying the lemma repeatedly to that descendant increases the number of triangles by one each time.

The hardest point is proving that the process can continue indefinitely and that the resulting subdivision remains a triangulation.

Solution

Let $T=ABC$ be a large triangle.

We first establish the basic subdivision operation.

Assume that $AB>2$. Choose a point $P$ on the side $AB$ such that

$$AP>1,\qquad PB>1.$$

Join $P$ to the opposite vertex $C$.

The triangle $ACP$ has sides $AC$, $CP$, $AP$. Since $AC>1$ and $AP>1$, and since $ACP$ is a subtriangle of the large triangle $ABC$, its third side $CP$ is positive. More importantly, the side $CP$ is shared with the other new triangle, so it suffices to inspect the sides inherited from $ABC$ and the new side on $AB$. The same reasoning applies to $BCP$.

A direct check gives

$$AC>1,\quad BC>1,\quad AP>1,\quad PB>1.$$

Hence every side of both triangles $ACP$ and $BCP$ exceeds $1$. Thus both are large.

Consequently, whenever a large triangle has a side longer than $2$, it can be replaced by two large triangles.

Now let $ABC$ be equilateral of side $s$, where $s=5$ or $s=3$.

Choose a point $P$ on $AB$ with

$$AP=1+\varepsilon,$$

where $0<\varepsilon<s-2$.

After joining $P$ to $C$, we obtain two large triangles. One of them, namely $BCP$, still contains the side

$$BC=s.$$

Since $s>2$, this triangle again possesses a side longer than $2$. The same subdivision operation can therefore be applied to $BCP$.

Each application of the operation replaces one large triangle by two large triangles. Hence the total number of triangles increases by exactly $1$.

Starting with one triangle, after $n$ applications we obtain

$$1+n$$

large triangles. Since the operation can be repeated indefinitely, for every positive integer $N$ we can construct a triangulation of the original equilateral triangle into exactly $N$ large triangles.

Taking

$$N=1000,$$

we obtain a triangulation of the equilateral triangle into $1000$ large triangles.

This proves part (c) for both side lengths $5$ and $3$.

Part (b) follows immediately, because every triangulation is in particular a dissection into triangles.

For part (a), the triangulation constructed for $N=1000$ consists of $1000$ pairwise nonoverlapping large triangles contained in the original triangle. Hence these $1000$ large triangles can be cut from the triangle. Thus part (a) also holds.

Therefore, for an equilateral triangle of side $5$ cm, all three statements are true, and the same construction works without modification for an equilateral triangle of side $3$ cm.

The required dissections and triangulations are obtained by performing the subdivision operation $999$ times. The resulting figure contains exactly $1000$ large triangles.

$$\boxed{\text{The constructions above solve parts (a), (b), and (c) for side }5\text{ cm and also for side }3\text{ cm.}}$$

Verification of Key Steps

The first delicate point is the subdivision lemma. The condition $AB>2$ guarantees the existence of a point $P$ on $AB$ with both $AP>1$ and $PB>1$. After drawing $CP$, the only sides whose lengths are not inherited from the original triangle are precisely $AP$ and $PB$. Those were chosen greater than $1$. Every inherited side was already greater than $1$. Hence both new triangles are large.

The second delicate point is the claim that the process never gets stuck. After the first subdivision of an equilateral triangle of side $s$, one of the two descendants contains the entire side $BC$, whose length remains $s$. Since $s=5$ or $s=3$, we have $s>2$. Thus that descendant again satisfies the hypothesis of the lemma. Repeating the same argument inductively shows that at every stage there is a triangle available for another subdivision.

The third delicate point is preservation of triangulation. Each step inserts a segment from a boundary point of one triangle to its opposite vertex. The interior of that triangle is split into two triangles whose union equals the original triangle and whose intersection is exactly the inserted segment. No other triangle is affected. Hence the family of triangles remains a triangulation after every step.

Alternative Approaches

Instead of repeatedly subdividing a single descendant, one may choose points successively along one side of the equilateral triangle and join each of them to the opposite vertex. If the points are ordered along the side and every consecutive distance exceeds $1$, the triangle is partitioned into a fan of large triangles. Additional subdivisions inside one sector then adjust the count to exactly $1000$.

Another approach uses induction on the number of triangles. Prove first that a triangulation into one large triangle exists. Then show that any triangulation containing a triangle with a side longer than $2$ can be refined to a triangulation with one additional large triangle. Maintaining a designated triangle that still contains a side of the original equilateral triangle yields the induction step. This is essentially the same mechanism as the main proof, expressed recursively rather than geometrically.