
RE: Triangle Problem
Posted:
May 4, 2006 3:23 AM


> Original Message > From: approve@support1.mathforum.org > [mailto:approve@support1.mathforum.org] On Behalf Of Philippe 92 > Sent: Wednesday, May 03, 2006 11:54 PM > To: geometrypuzzles@moderators.isc.org; geometry > puzzles@support1.mathforum.org > Subject: Re: Triangle Problem > Avni Pllana wrote : > > Let ABC be a triangle such that CA < 2*BC and BC < 2*CA . > > Construct points D and E on the sides BC and CA respectively, such that > > BD = DE = EA .
and Philippe replied: > Just a hint, as I won't solve this for you. > Consider to construct a triangle A'B'C' similar to ABC, from any > arbitrary choosen distance d = B'D' = D'E' = E'A', > with D' on B'C' and E' on A'C'. > > This is done by constructing d on AC, d on BC, and move BC to B'C' so > that D'E' becomes = d, and AB'C' remains similar to ABC.
No! This is a false, nonruler and compass construction. The main error is the step "and move BC to B'C' so that D'E' becomes = d".
Here is a way to solve the problem: Let x = BD = DE = EA (an unknown quantity), so that BC = ax, EC = bx. By the cosine rule on DCE we have DE^2 = CD^2 + CE^2  2.CD.CE. cosC = CD^2 + CE^2  2.CD.CE.(a^2 + b^2  c^2)/(2ab) so that
x^2 = (ax)^2 + (bx)^2 2(ax)(bx).(a^2 + b^2  c^2)/(2ab).
This is a quadratic in x and, as is well known, its roots are ruler and compass constructible. That is all.
One more remark: Only one of the roots is positive. If you interpret negative roots as going in the opposite direction, you have a second solution to the problem, but this time externally to the given triangle.
All the best, Michael Lambrou.

