Let $ ABCD$ be a parallelogram with no angle equal to $ 60^{\textrm{o}}$. Find all pairs of points $ E, F$, in the plane of $ ABCD$, such that triangles $ AEB$ and $ BFC$ are isosceles, of basis $ AB$, respectively $ BC$, and triangle $ DEF$ is equilateral. Valentin Vornicu
Problem
Source: Romanian TST 5 2007, Problem 1
Tags: geometry, parallelogram, geometric transformation, rotation, reflection, analytic geometry, perpendicular bisector
07.06.2007 10:48
07.06.2007 12:52
nice solution valenitn,you can also note the following similar problem in valentin`s solution:
13.06.2007 05:31
in the contest i solved it using computational methods. take $B(0,0)$, $A(x,y)$, $C(1,0)$, and make a mix of complex numbers (with $z=x+iy$) and analytic geometry. the equations are pretty easy, and they reflect the construction in the solution above.
24.06.2008 18:42
Valentin Vornicu wrote: On the other hand, when triangle $ DEF$ is equilateral, a rotation of angle $ \pm 60^{\textrm{o}}$ will rotate the perpendicular bisector of $ AB$, meeting the perpendicular bisector of $ BC$ in exactly two points, be them $ F_{ + }, F_{ - }$, originating from points $ E_{ + }, E_{ - }$. But the triangles $ DE_{ + }F_{ + }$ and $ DE_{ - }F_{ - }$ are clearly equilateral, and this accounts for the two cases proven in the paragraph above, therefore no other cases, with isosceles triangles $ AEB$ and $ BFC$ may exist anymore. This is unreadable! What are you rotating when you mention a rotation of angle $ \pm 60^{\textrm{o}}$? What's the origin of rotation? What's $ F_{+},F_{-}$? It would be a surprise if anyone can understand this just by reading what you wrote. By the way, I also just solved this problem using only coordinate method. It was absolutely easy and required no brainwork. There's no need to use a nuclear bomb to kill an ant, you know? Moreover, sometimes the ant might not die.