2. From a square steel sheet with a side of 1 meter, a triangle is cut off from each of the four corners so that a regular octagon remains. Determine the mass of this octagon if the sheet thickness is 3 mm and the density of steel is 7.8 g/cm ${ }^{3}$. Give your answer in kilograms, rounding to the nearest whole number if necessary.
Answer: $46.8(\sqrt{2}-1) \approx 19$ kg.

Solution. A regular octagon must have equal angles and sides. Therefore, four equal triangles with angles $45^{\circ}, 45^{\circ}$, and $90^{\circ}$ are cut off. If the legs of this triangle are equal to $x$, then the hypotenuse is $x \sqrt{2}$ - this will be the side of the octagon. Therefore, the length of the side of the square is $2 x+x \sqrt{2}$, and we get the equation $2 x+x \sqrt{2}=100$. Hence, $x=\frac{100}{2+\sqrt{2}}=50(2-\sqrt{2})$ cm, and the area of the remaining octagon is $10000-4 \cdot \frac{x^{2}}{2}=10000-2 \cdot 2500 \cdot(2-\sqrt{2})^{2}=10000-5000 \cdot(6-4 \sqrt{2})=20000(\sqrt{2}-1)\left(\mathrm{cm}^{2}\right)$. Therefore, its volume is $20000(\sqrt{2}-1) \cdot 0.3=6000(\sqrt{2}-1)$ (cm ${ }^{3}$ ), and its mass is $6000(\sqrt{2}-1) \cdot 7.8$ $=46800(\sqrt{2}-1)$ g, which is $46.8(\sqrt{2}-1)$ kg.

Rounding to the nearest whole number can be done in different ways. One way: since $\sqrt{2}>1.4$, the mass is greater than 18.72. Since $\sqrt{2}<1.415$ (proven by squaring), the mass is less than 19.422. Therefore, the nearest whole number is 19 kg. The main requirement is that the proof must be based on strict estimates, not on approximate calculations without accuracy assessment. In the case of unjustified rounding, a score of 10 points was given. In the case of an incorrect answer (regardless of the reasons), the problem was scored 0 points.

Answer to variant 172: $43.2(\sqrt{2}-1) \approx 18$ kg.

Answer to variant 173: $31.2(\sqrt{2}-1) \approx 13$ kg.

Answer to variant 174: $64.8(\sqrt{2}-1) \approx 27$ kg.