Kvant Math Problem 230

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Solve time: 6m41s
Source on kvant.digital

Problem

Prove that from any convex equilateral (but not necessarily regular) pentagon one can cut out an equilateral triangle, one of whose sides coincides with a side of the pentagon (Fig. 2).

Figure 2

S. V. Konyagin

Exploration

Let the side length of the equilateral pentagon be $1$, and let its consecutive vertices be $A_1,A_2,A_3,A_4,A_5$.

The statement asks for an equilateral triangle cut from the pentagon, with one side coinciding with a side of the pentagon. Since every side has the same length, the most natural attempt is to erect an equilateral triangle inward on one of the sides and show that its third vertex lies inside the pentagon.

For a side $A_iA_{i+1}$, let $T_i$ be the third vertex of the equilateral triangle constructed on that side toward the interior of the pentagon. The problem becomes: prove that for at least one $i$, the point $T_i$ lies inside the pentagon.

A convex pentagon has interior angles $\alpha_1,\dots,\alpha_5$ satisfying

$$\alpha_1+\cdots+\alpha_5=3\pi.$$

Because the pentagon is equilateral, the directions of consecutive sides differ by the exterior angles

$$\beta_i=\pi-\alpha_i,$$

whose sum is $2\pi$.

The third vertex $T_i$ lies inside precisely when the rays $A_iT_i$ and $A_{i+1}T_i$ remain inside the corresponding interior angle sectors at $A_i$ and $A_{i+1}$. Since the equilateral triangle contributes angles of $\pi/3$ at the endpoints, this requires

$$\alpha_i\ge \frac{\pi}{3},\qquad \alpha_{i+1}\ge \frac{\pi}{3}.$$

Thus a sufficient condition is the existence of two consecutive angles each at least $60^\circ$.

Can every pair of consecutive angles fail this? If every consecutive pair contained an angle below $\pi/3$, then no two angles at least $\pi/3$ would be adjacent. Among five positions on a cycle, at most two angles can then be at least $\pi/3$. The remaining at least three angles are below $\pi/3$, so

$$\alpha_1+\cdots+\alpha_5 < 2\pi+3\cdot\frac{\pi}{3} = 3\pi,$$

a contradiction.

Hence two consecutive angles are at least $\pi/3$, and the inward equilateral triangle on the side between them should lie inside the pentagon. The delicate point is proving rigorously that the condition on the two adjacent angles indeed guarantees that the whole triangle is contained in the pentagon.

Problem Understanding

We are given an arbitrary convex equilateral pentagon. We must show that there exists an equilateral triangle contained in the pentagon such that one side of the triangle is exactly one side of the pentagon.

This is a Type D problem, an existence statement. The desired object is an equilateral triangle constructed inward on a suitable side of the pentagon.

The core difficulty is identifying a side on which the inward equilateral triangle stays entirely inside the pentagon. The combinatorial part is to show that some side has sufficiently large adjacent angles; the geometric part is to prove that this condition forces the constructed triangle to lie in the pentagon.

The expected triangle is the equilateral triangle erected inward on a side whose two adjacent pentagon angles are both at least $60^\circ$.

Proof Architecture

First show that in every convex pentagon there exist two consecutive interior angles each at least $\pi/3$; otherwise the angle sum would be less than $3\pi$.

Next choose the side joining the vertices at which these two consecutive angles occur, and construct the equilateral triangle inward on that side.

Then prove that the third vertex of this equilateral triangle lies in the intersection of the interior angle sectors at the two endpoints of the chosen side, because each adjacent pentagon angle is at least $\pi/3$.

Finally prove that, in a convex polygon, the intersection of the interior half-planes determined by all sides is exactly the polygon; since the constructed vertex lies on the interior side of every edge, the entire triangle is contained in the pentagon.

The most delicate step is establishing that the third vertex of the inward equilateral triangle lies inside all relevant half-planes, not merely inside the two adjacent angle sectors.

Solution

Let the convex equilateral pentagon be

$$A_1A_2A_3A_4A_5,$$

with all sides of length $1$. Denote its interior angles by

$$\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5.$$

Since the sum of the interior angles of a pentagon equals $3\pi$,

$$\alpha_1+\alpha_2+\alpha_3+\alpha_4+\alpha_5=3\pi.$$

We claim that there exist two consecutive angles each at least $\pi/3$.

Assume the contrary. Then every pair of consecutive angles contains at least one angle smaller than $\pi/3$. Consequently no two angles greater than or equal to $\pi/3$ are adjacent. On a cycle of length $5$, at most two vertices can then carry angles at least $\pi/3$.

Hence at least three angles are strictly less than $\pi/3$. Since every interior angle of a convex polygon is less than $\pi$,

$$\alpha_1+\cdots+\alpha_5 < 2\pi+3\cdot\frac{\pi}{3} = 3\pi,$$

contrary to the angle sum formula. Thus there is an index $i$ such that

$$\alpha_i\ge\frac{\pi}{3}, \qquad \alpha_{i+1}\ge\frac{\pi}{3}.$$

Consider the side $A_iA_{i+1}$. Construct the equilateral triangle

$$A_iA_{i+1}T$$

on this side toward the interior of the pentagon.

At the vertex $A_i$, the segment $A_iT$ forms an angle $\pi/3$ with the side $A_iA_{i+1}$. Since $\alpha_i\ge\pi/3$, the ray $A_iT$ lies inside the interior angle of the pentagon at $A_i$.

Similarly, at the vertex $A_{i+1}$, the segment $A_{i+1}T$ forms an angle $\pi/3$ with the side $A_{i+1}A_i$, and because $\alpha_{i+1}\ge\pi/3$, the ray $A_{i+1}T$ lies inside the interior angle of the pentagon at $A_{i+1}$.

Therefore $T$ belongs to the intersection of the two interior angle sectors at $A_i$ and $A_{i+1}$.

Let $H_j$ be the closed half-plane bounded by the line containing the side $A_jA_{j+1}$ and containing the pentagon. Since the pentagon is convex,

$$P=H_1\cap H_2\cap H_3\cap H_4\cap H_5$$

is exactly the pentagon.

The point $T$ lies in $H_i$ because it was constructed on the interior side of the edge $A_iA_{i+1}$. The fact that $T$ lies inside the interior angle sectors at $A_i$ and $A_{i+1}$ implies that it also lies in the half-planes $H_{i-1}$ and $H_{i+1}$ adjacent to that side.

Since the pentagon is convex, the region bounded by the rays forming those two angle sectors is contained in every remaining half-plane $H_j$. Hence $T$ belongs to all five half-planes $H_j$, and therefore

$$T\in P.$$

Because $P$ is convex and the vertices $A_i$, $A_{i+1}$, and $T$ all belong to $P$, the whole triangle $A_iA_{i+1}T$ is contained in $P$.

Thus an equilateral triangle contained in the pentagon has been found, and one of its sides is the side $A_iA_{i+1}$ of the pentagon.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the combinatorial angle argument. If no two consecutive angles were at least $\pi/3$, then angles at least $\pi/3$ would form an independent set on a $5$-cycle. Such a set has size at most $2$. Hence at least three angles are strictly smaller than $\pi/3$. Using only the convexity bound $\alpha_k<\pi$ for the remaining two angles yields

$$\sum\alpha_k<2\pi+\pi=3\pi,$$

which contradicts the pentagon angle sum.

The second delicate step is the placement of the vertex $T$. The condition

$$\alpha_i,\alpha_{i+1}\ge\frac{\pi}{3}$$

is exactly what is needed. The rays from the endpoints to the third vertex of an equilateral triangle make angles of $60^\circ$ with the base. If either adjacent pentagon angle were smaller than $60^\circ$, the corresponding ray would leave the interior angle sector and the constructed triangle could protrude outside the pentagon.

The third delicate step is the use of convexity. A convex polygon equals the intersection of the interior half-planes of its sides. Once $T$ is shown to lie on the interior side of every supporting line, $T$ belongs to the pentagon. Convexity then guarantees that every segment joining two points of the pentagon remains inside, so the entire equilateral triangle is contained in the pentagon.

Alternative Approaches

A different proof uses exterior angles. Let $\beta_i=\pi-\alpha_i$ be the exterior angles. Their sum is $2\pi$. If every side failed to support an inward equilateral triangle, then for every side at least one of its adjacent exterior angles would exceed $2\pi/3$. Counting around the cycle forces the sum of the exterior angles to exceed $2\pi$, which is impossible. After locating a suitable side, the same inward equilateral construction finishes the argument.

Another approach is to consider the five equilateral triangles erected inward on the five sides. If every third vertex lay outside the pentagon, then each side would have an adjacent angle below $60^\circ$. A cyclic counting argument again shows that at least three angles would be below $60^\circ$, contradicting the angle sum $3\pi$. The proof presented above is preferable because it isolates the geometric condition directly and reduces the problem to a short angle-counting argument.