IMC 2017 · Day 2 · P10

For an arbitrary $\varepsilon > 0$ we will establish a lower bound for the sum of perimeters that would tend to $+\infty$ as $\varepsilon \to +0$; this solves the problem.

Rotate and scale the picture so that one of the sides of $K$ is the segment from $(0,0)$ to $(0,1)$, and stretch the picture horizontally in such a way that the projection of $K$ to the $x$ axis is $[0,1]$. Evidently, we may work with the lengths of the projections to the $x$ or $y$ axis instead of the perimeters and consider their sum, that is why we may make any affine transformation.

Let $f_i(a)$ be the length of intersection of the straight line $\{x = a\}$ with $T_i$ and put $f(a) = \sum_i f_i(a)$. Then $f$ is piece-wise increasing with possible downward gaps, $f(a) \le 1 - a$, and $$\int_0^1 f(x)\,dx \ge \frac12 - \varepsilon.$$ Let $d_1, \dots, d_N$ be the values of the gaps of $f$. Every gap is a sum of side-lengths of some of $T_i$ and every $T_i$ contributes to one of $d_j$, we therefore estimate the sum of the gaps of $f$.

In the points of differentiability of $f$ we have $f'(a) \ge f(a)/a$; this follows from $f'_i(a) \ge f_i(a)/a$ after summation. Indeed, if $f_i$ is zero this inequality holds trivially, and if not then $f'_i = 1$ and the inequality reads $f_i(a) \le a$, which is clear from the definition.

Choose an integer $m = \lfloor 1/(8\varepsilon) \rfloor$ (considering $\varepsilon$ sufficiently small). Then for all $k = 0, 1, \dots, [(m-1)/2]$ in the section of $K$ by the strip $k/m \le x \le (k+1)/m$ the area, covered by the small triangles $T_i$ is no smaller than $1/(2m) - \varepsilon \ge 1/(4m)$. Thus $$\int_{k/m}^{(k+1)/m} f'(x)\,dx \ge \int_{k/m}^{(k+1)/m} \frac{f(x)\,dx}{x} \ge \frac{m}{k+1} \int_{k/m}^{(k+1)/m} f(x)\,dx \ge \frac{m}{k+1} \cdot \frac{1}{4m} = \frac{1}{4(k+1)}.$$ Hence, $$\int_0^{1/2} f'(x)\,dx \ge \frac14 \left( \frac{1}{1} + \dots + \frac{1}{[(m-1)/2]} \right).$$ The right hand side tends to infinity as $\varepsilon \to +0$. On the other hand, the left hand side equals $$f(1/2) + \sum_{x_i < 1/2} d_i;$$ hence $\sum_i d_i$ also tends to infinity.