The reflection in the horizontal diameter of the big circle creates a second Yin-Yang pair of regions whose borderline supplies the necessary cut. The rest of the proof is an easy consequence of the Carpets theorem and is continued below. Obviously, the area of the circle is πR²/2 which is then the area of the left over annulus. It follows that the straight cut inclined 45° to the diameter, as shown, solves Dudeney's problem. The two pieces together add to πR²/4 which is exactly half of Yin's area. Just above the small semicircle, adjacent to the diameter is a circular sector of the big circle with central angle 45°. Part of the Yin (black) piece below the horizontal diameter of the big circle is a semicircle with area πR²/8, where R is assumed to be the radius of the big circle, so that the small semicircle is of radius R/2. Another (sixth) solution was suggested by the solution to the problem of dividing a circle into N (> 1) equal parts. Trigg added three more solutions (#3, 4, 5 below). In response to the column, several readers offered a simplification of Dudeney's proof for Solution 2. Gardner in one of his Scientific American columns and also included in two of his collections. He also gave two solutions: Solutions 1 and 2 below. Dudeney posed a question of bisecting the Yin-Yang symbol by the common Euclidean means: straightedge and compass. The symbol is composed of two regions of a circle separated by two semicircles of half the radius of the big circle. That represents the struggle, merger and co-existence of two opposites (could be hot/cold, male/female, sky/earth, moon/sun, etc.) In the diagram Yin (the negative aspect) is rendered in black, with Yang (the positive aspect) rendered in white. The flag of South Korea (and of Kingdom of Korea from 1883)Ĭontains the ancient yin-yang symbol ( Taijitu in Chinese, Tomoye in Japanese and Taegeuk in Korean)
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