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, 1997

Geometry Difficulty 7.9 National Olympiad, round 2 Prove it Hong Kong

Assume the dimensions of an answer sheet to be $297 \text{ mm}$ by $210 \text{ mm}$ . Suppose that your pen leaks and makes some non-intersecting ink stains on the answer sheet. It turns out that the area of each ink stain does not exceed $1 \text{ mm}^2$ . Moreover, any line parallel to an edge of the answer sheet intersects at most one ink stain. Prove that the total area of the ink stains is at most $253.5 \text{ mm}^2$ . (You may assume a stain is a connected piece.)

Solution

Suppose there are $n$ ink stains, having areas $S_1, S_2, \dots, S_n$ (in mm²) respectively. Suppose the lengths of the projections of the ink stains on the top edge of the answer sheet are $x_1, x_2, \dots, x_n$ (in mm) and the lengths of the projections of the ink stains on the left edge of the answer sheet are $y_1, y_2, \dots, y_n$ (in mm) respectively. Since any line parallel to an edge of the answer sheet intersects at most one ink stain, we have $\sum_{j=1}^{n} x_j \le 210$ and $\sum_{j=1}^{n} y_j \le 297$ . By the condition $S_j \le 1$ and the AM-GM inequality, we have
$\sum_{j=1}^{n} S_j \le \sum_{j=1}^{n} \sqrt{S_j} \le \sum_{j=1}^{n} \sqrt{x_j y_j} \le \sum_{j=1}^{n} \frac{x_j + y_j}{2} \le \frac{1}{2}(210 + 297) = 253.5.$

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