| [ QuizWit ] in KIDS 글 쓴 이(By): scheme (스킴이어요�p) 날 짜 (Date): 1994년06월11일(토) 02시14분05초 KDT 제 목(Title): [re] rational coordinates... Sorry, I cannot use hangul on this terminal. I think that both answers are NO. Suppose that you have an equilateral triangle whose coordinates are all rational. Then, by magnifying(multiplying) by the lcm of all denominators of coordinates of vertices, we get an equilateral triangle whose coordinates are all integral. Then, two times the area of this triangle is integer. (Why?) Let a be the length of an edge of this triangle. By Pytagoras' Theorem (Did I spell right?), the square of a is an integer. But the area A is A = sqrt(3)/4 * a^2 , which must be of form n/2 Hence, sqrt(3)/4 = n/(2*a^2), a rational number! Similar method works for 3 dimensional version. (I think.) !@#$%^&*()_+~!@#$%^&*()_+~ "The time has come," the Walrus said, "To talk of many things: 스킴이어요 ...... Of shoes-and ships-and sealing-wax -- !@#$%^&*()_+~!@#$%^&*()_+~ Of cabbages-and kings --" -Lewis Carroll |