Kvant
Miscellany

Construction of a regular 17-gon

Rough automatic draft, low quality: expect mistakes and awkward wording. It will be replaced by a better translation. Original →

38

Find the distance from the center of the heavy chain to the axis of rotation.

F134. A heavy plate moves with a speed vv in a direction perpendicular to its plane. At an angle α\alpha to the direction of motion of the plate, a light ball moves at a speed uu. Determine the magnitude and direction of the ball’s velocity vector after its elastic collision with the plate.

a) Distance when the ball speed is directed towards and away from the plate.

  1. How does the answer change if the plate moves in a direction making an angle β\beta with its plane?

F135. A mine lying on the ground explodes from detonation. The mine fragments begin to move symmetrically in all directions with the same speeds vv. The sizes of all osokols are the same. What fraction of the sedges will fall outside the circle of radius RR centered at the point of explosion?

G. L. Kstkin

F136. A parallel plate capacitor has a capacitance of CC. Charge +q+q was placed on one of the capacitor plates, and charge +4q+4q was placed on the other. Determine the potential difference between the plates of the capacitor.

V. E. Belonukkin

F137 *. Infrared radiation of the defining wavelength is absorbed by methane (CH4\text{CH}_4). Under normal conditions, a 1 cm thick layer of pure methane absorbs 98% of the radiation energy. How many times will such radiation be weakened when passing vertically through the Earth’s atmosphere?

When calculating, take the weight content of methane in the atmosphere equal to 1,4⋅10−61,4 \cdot 10^{-6}.

S. M. Kozel

Answers to the crossword (see page 31)

Horizontally: 1. VTSNIIOT-PUSK. 6. Obscurantist. 7. Telephone 8. Joint. 11. Egor. 18. Sax. 14. Akbin. 17. Ataman. 20. Ogly. 21. Stopkop. 22. Presentation.

Constructing a correct 17-gon

Richmond gave a simple construction of the seventeenth triangle P0P1…P16P_0P_1 \dots P_{16} (see figure).

Let’s connect point P0P_0 with point JJ, lying on radius OBOB at a distance 1/4OB1/4OB from the center. On the diameter passing through the point P0P_0, we choose points EE and FF so that ⇒OJE\Rightarrow OJE is equal to four times the angle OJP0OJP_0, and ⇒FJE\Rightarrow FJE is equal to 45∘45^\circ. Let a circle built on FP0FP_0 as a diameter intersect

OBOB at point KK, and let a circle with center EE and radius EKEK intersect OP0OP_0 at points N3N_3 (between OO and P0P_0) and N5N_5. Let us restore the perpendiculars to OP0OP_0 at these two points until they intersect with the original circle at points P3P_3 and P5P_5. Then the arc P3P5P_3P_5 (and the equal arc P1P3P_1P_3) is equal to the 2/172/17 circle. The proof uses several times the fact that the roots of the equation

x2+2xctg,2α−1=0x^2+2x\text{ctg}, 2\alpha-1=0 are equal to tgα\alpha and - ctgα\alpha.

(G. S. M. Coxter, Introduction to Geometry, M., “Hayka”, 1966, p. 49.)