"Avni Pllana" <avniu66@hotmail.com> wrote > Triangle ABC with sides AB = 6, BC = 10, AC = 9 is a > normal projection of an equilateral triangle. > Construct the original equilateral triangle. Hi Avni. The beautiful geometric solution, don't forget to show it to us. Algebraically, lifting one corner x, another y, and solving, I get the triangle side 10.5017 Since I will remember this problem, I tried a search for integer sides. 8,9,12 gives SQRT(145) 25, 43, 53 gives 55 If my calculations are correct. Thanks Hans.

"Avni Pllana" <avniu66@hotmail.com> wrote in message news:24847854.1169211645541.Java Mail.jakarta@nitrogen.mathforum.org... > > > > "Avni Pllana" <avniu66@hotmail.com> wrote > > > Triangle ABC with sides AB = 6, BC = 10, AC = 9 is > > a > > > normal projection of an equilateral triangle. > > > Construct the original equilateral triangle. > > > > Hi Avni. > > > > The beautiful geometric solution, don't forget to > > show it to us. > > > > Algebraically, lifting one corner x, another y, and > > solving, I get the triangle side > > 10.5017 > > > > Since I will remember this problem, I tried a search > > for integer sides. > > 8,9,12 gives SQRT(145) > > 25, 43, 53 gives 55 > > > > If my calculations are correct. > > > > Thanks Hans. > > > > > > > > Hi Hans, > > your results are correct and excellent. Many thanks for the contribution. > > To construct the original equilateral triangle we need only the radius of its circumscribed circle. We first construct the center of mass M of triangle ABC. We draw through M a line 'l' parallel to the side AB. Let C1 be the middle point of side AB. We have to construct a point D on the line 'l' to the right of M, such that MD : C1B = 1 : sqrt(3)/2. For this purpose we draw a line 'h' through A at 30° with respect to AB. Let E be the intersection point of 'h' with the perpendicular of AB through C1. Then we construct D such that MD = AE. Now MC and MD are two conjugate axes of the ellipse that is the orthogonal projection of the circumscribed circle of the original equilateral or regular triangle. Using Rytz's method > > http://de.wikipedia.org/wiki/Rytzsche_Achsenkonstruktion > > we obtain the major and minor semiaxes of the ellipse from conjugate axes MC and MD. The major semiaxis is equal to the radius of the circumscribed circle of the original equilateral triangle. > > Best regards, > Avni Hi, Avni. I was sure it was possible, and the M centered circumscribed ellipse is called the Steiner ellipse, but thats as far as I got. Best wishes Hans.