Kvant Math Problem 721

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Verdicts: FAIL + FAIL
Solve time: 41m44s
Source on kvant.digital

Problem

Each side of a triangle is divided into 3 equal parts. The division points serve as the vertices of two triangles whose intersection is a hexagon. Find the area of this hexagon if the area of the given triangle is $S$.

A. Zolotykh, 10th-grade student (Moscow, Specialized Physics and Mathematics School No. 18 at Moscow State University)

Exploration

Let $ABC$ be the given triangle. Each side of the triangle is divided into three equal segments, producing trisection points that can be labeled consecutively along the perimeter. Selecting alternating trisection points around the boundary produces two smaller triangles whose intersection is a hexagon. The problem requires computing the area of this hexagon as a fraction of the area of the original triangle $ABC$. Affine transformations preserve ratios of areas and map lines to lines, so it suffices to perform computations for a conveniently chosen reference triangle and then generalize to any triangle by affine invariance.

Problem Understanding

The problem involves a geometric construction in which the intersection of two triangles defined by alternating trisection points forms a hexagon. The task is to determine the area of this hexagon relative to the area of the original triangle. By choosing a reference triangle with simple coordinates, the trisection points can be expressed explicitly, the two triangles can be described by linear inequalities, and the intersection hexagon can be computed precisely. The ratio of the hexagon area to the triangle area is preserved under affine transformations, allowing generalization.

Proof Architecture

The solution proceeds by selecting a right triangle $ABC$ with convenient integer coordinates. The trisection points on each side are computed, and the two triangles formed by alternating trisection points are defined explicitly. Each triangle is represented by linear inequalities corresponding to its sides. The intersection of these inequalities gives the hexagon, and its vertices are determined by solving the corresponding linear systems. The area of the hexagon is computed using the shoelace formula. Finally, the area ratio of the hexagon to the original triangle is obtained and generalized to any triangle using affine invariance.

Solution

Choose the reference triangle

$A=(0,0), \qquad B=(3,0), \qquad C=(0,3),$

which has area

$[ABC] = \frac{1}{2} \cdot 3 \cdot 3 = \frac{9}{2}.$

Divide each side into three equal parts. The trisection points are

$P=(1,0), \quad Q=(2,0),$

on $AB$,

$R=(2,1), \quad U=(1,2),$

on $BC$, and

$V=(0,2), \quad W=(0,1),$

on $CA$.

Form the two triangles using alternating trisection points:

$T_1 = \triangle PRV, \qquad T_2 = \triangle QUW.$

The sides of $T_1$ satisfy the equations

$PR: y = x-1, \qquad RV: x + 2y = 4, \qquad VP: 2x + y = 2.$

Choosing a point inside $T_1$, the interior is described by the inequalities

$y \ge x-1, \qquad x + 2y \le 4, \qquad 2x + y \ge 2.$

The sides of $T_2$ satisfy the equations

$QU: 2x + y = 4, \qquad UW: y = x + 1, \qquad WQ: x + 2y = 2.$

Using a point inside $T_2$, the interior is described by

$2x + y \le 4, \qquad y \le x + 1, \qquad x + 2y \ge 2.$

The intersection hexagon $H = T_1 \cap T_2$ is thus given by

$x-1 \le y \le x+1, \qquad 2 \le 2x + y \le 4, \qquad 2 \le x + 2y \le 4.$

The vertices of the hexagon are obtained by solving the systems of equations formed by the pairs of boundary lines:

Solving $2x + y = 2$ and $x + 2y = 2$ yields

$y = 2 - 2x \quad \Rightarrow \quad x + 2(2 - 2x) = 2 \quad \Rightarrow \quad x + 4 - 4x = 2 \quad \Rightarrow \quad -3x = -2 \quad \Rightarrow \quad x = \frac{2}{3}, \quad y = \frac{2}{3}.$

Solving $x + 2y = 2$ and $y = x - 1$ yields

$x + 2(x - 1) = 2 \quad \Rightarrow \quad 3x - 2 = 2 \quad \Rightarrow \quad x = \frac{4}{3}, \quad y = \frac{1}{3}.$

Solving $y = x - 1$ and $2x + y = 4$ yields

$2x + (x - 1) = 4 \quad \Rightarrow \quad 3x - 1 = 4 \quad \Rightarrow \quad x = \frac{5}{3}, \quad y = \frac{2}{3}.$

Solving $2x + y = 4$ and $y = x + 1$ yields

$2x + (x + 1) = 4 \quad \Rightarrow \quad 3x + 1 = 4 \quad \Rightarrow \quad x = 1, \quad y = 2.$

Solving $y = x + 1$ and $x + 2y = 4$ yields

$x + 2(x + 1) = 4 \quad \Rightarrow \quad 3x + 2 = 4 \quad \Rightarrow \quad x = \frac{2}{3}, \quad y = \frac{5}{3}.$

Solving $x + 2y = 4$ and $2x + y = 2$ yields

$2x + y = 2 \quad \Rightarrow \quad y = 2 - 2x$

$x + 2(2 - 2x) = 4 \quad \Rightarrow \quad x + 4 - 4x = 4 \quad \Rightarrow \quad -3x = 0 \quad \Rightarrow \quad x = 0, \quad y = 2.$

Thus the vertices of $H$ in cyclic order are

$\left(\frac{2}{3},\frac{2}{3}\right), \left(\frac{4}{3},\frac{1}{3}\right), \left(\frac{5}{3},\frac{2}{3}\right), (1,2), \left(\frac{2}{3},\frac{5}{3}\right), (0,2).$

Applying the shoelace formula, compute

$\sum x_i y_{i+1} = \frac{2}{3} \cdot \frac{1}{3} + \frac{4}{3} \cdot \frac{2}{3} + \frac{5}{3} \cdot 2 + 1 \cdot \frac{5}{3} + \frac{2}{3} \cdot 2 + 0 \cdot \frac{2}{3} = \frac{2}{9} + \frac{8}{9} + \frac{10}{3} + \frac{5}{3} + \frac{4}{3} + 0 = \frac{67}{9},$

$\sum y_i x_{i+1} = \frac{2}{3} \cdot \frac{4}{3} + \frac{1}{3} \cdot \frac{5}{3} + \frac{2}{3} \cdot 1 + 2 \cdot \frac{2}{3} + \frac{5}{3} \cdot 0 + 2 \cdot \frac{2}{3} = \frac{8}{9} + \frac{5}{9} + \frac{2}{3} + \frac{4}{3} + 0 + \frac{4}{3} = \frac{37}{9}.$

The area of the hexagon is therefore

$[H] = \frac{1}{2} \left( \frac{67}{9} - \frac{37}{9} \right) = \frac{1}{2} \cdot \frac{30}{9} = \frac{15}{9} = \frac{5}{3}.$

The area ratio relative to the reference triangle is

$\frac{[H]}{[ABC]} = \frac{5/3}{9/2} = \frac{10}{27}.$

Affine invariance guarantees that this ratio holds for any triangle. Therefore, if the original triangle has area $S$, the area of the hexagon is

$\boxed{\frac{10}{27} S}.$