otkaznik (otkaznik) wrote,
otkaznik
otkaznik

Старинный приятель напомнил, как лет 50 назад я помог ему сдать зачет, решив задачку. Я этот эпизод забыл начисто. Теперь вспомнив заинтресовался, сумею ли решить ее опять. Половину сумел, а остаток пришлось подсмотреть:

Bisect a given line segment AB

Geometric Construction with the Compass Alone

Solution

Construct a point C such that AC = 2·AB (Problem #1). Swing an arc of radius CA with center C. This will intersect the circle of radius AB centered at A at points D and E. Draw two arcs of radii DA and EA (which are both equal to AB) and centers at, respectively, D and E. Besides A, the two arcs will intersect at another point F which is the middle of the segment AB.

Indeed, ΔACE is similar to the ΔAFE (they are both isosceles and share ∠EAF. From here,AC/AE = AE/AF. But AC/AE = 2·AB/AB = 2. Therefore, AB = AE = 2·AF.

Tags: remembrance, былое и думы, ностальгическое
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