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
Valentin Vornicu
07.06.2007 10:48
Take $E$ and $F$ such that triangles $AEB$ and $BFC$ are equilateral, with interiors disjunct
with $ABCD$. Then $AD = BC = CF = BF$, $AE = BE = AB = CD$, and, denoting $\alpha = \angle BAD$,
$\angle DAE = \angle FBE = \angle FCD = \alpha+60^{\textrm{o}}$, therefore triangles $DAB$, $FBE$ and $FCD$ are
congruent (SAS), whence $DE = EF = FD$, i.e. triangle $DEF$ is equilateral. Similarly, for $E$ and $F$ such that
triangles $AEB$ and $BFC$ are equilateral, with interiors intersecting $ABCD$.
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.
BaBaK Ghalebi
07.06.2007 12:52
nice solution valenitn,you can also note the following similar problem in valentin`s solution:
for a fixed line $\ell$ and a fixed point $A$ denote $A'$ as the reflection of $A$ respect to $\ell$,assume point $B$ as an arbitary point on $\ell$,now the locus of points $M$ in this plane s.t $\triangle MAB$ is equilateral is two lines $d,d'$ passing thruogh $A'$ such that the angle between $\ell,d$ and also the angle between $\ell,d'$ is equal to $60^{\textrm{o}}$...
maky
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.
Carlez Tevos
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.