A cylindrical log with a diameter of $46 \mathrm{~cm}$ is to be sawn into planks that are $4 \mathrm{~cm}$ thick and $12 \mathrm{~cm}$ wide. Develop a plan to saw as many planks of the specified size as possible. (Do not consider the thickness of the saw blade. The length of the planks should be equal to the length of the log.)





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According to the last remark, the task is equivalent to placing as many $4 \times 12$ cm rectangles as possible within a circle of $46 \mathrm{~cm}$ diameter, without overlapping. Each rectangle must have one of its sides parallel to the diameter.

![](https://cdn.mathpix.com/cropped/2024_05_02_2557e4d745511d8477dag-1.jpg?height=512&width=683&top_left_y=313&top_left_x=710)

The diameter can fit the width of the rectangle 11 times, and since 11 is an odd number, we divide the circle into $4 \mathrm{~cm}$ wide strips so that the central strip is symmetric about a diameter. Thus, the shorter of the chords delimiting the strips are 2, 6, 10, 14, 18, 22 cm from the center, so their lengths, by the Pythagorean theorem, are 45.8, 44.4, 41.4, 36.4 (> $3 \cdot 12$), 28.6 (> $2 \cdot 12$), 13.4 (> 12), respectively. Therefore, 1 rectangle can be cut from each of the 2 outermost strips, 2 rectangles from each of the next strips, and 3 rectangles from each of the 7 central strips. Since the length of the second chord mentioned is more than 3 times the length of the rectangle plus twice the width: $44.4 > 3 \cdot 12 + 2 \cdot 4$ (and the rectangle must be 3 times as long as it is wide), 2 more rectangles can be placed crosswise at the end of the central 3 strips (or 1 at each end). Thus, we can draw $2 + 4 + 21 + 2 = 29$ rectangles, or cut 29 planks from the log.

We do not prove that more planks cannot be cut, but we note that the area of the circle is more than 34 times the area of the rectangle, but less than 35 times.

Sailer Kornél (Ózd, József A. Gymn., III. grade) Szengofszky Oszkár (Budapest, Berzsenyi D Gymn., III. grade)