ABC is a Triangle and P be a Fermat Point Inside it.draw three equilateral triangle based on the three sides i.e$\triangle ABA'$, $\triangle ACC'$, $\triangle BCB'$ respectively.Join $AB'$,$BC'$ and$CA'$ .Show that $ AB'$,$BC'$ and $CA'$ pass through a single piont i.e they are concurrent.
Rotation
Geometry
shortest distance
But try the problem first...
Regional Math Olympiad, India
Challenges and thrills of pre college mathematics
First hint
Rotation:
ABC is a Triangle . Let P Be any point join $AP,BP$ and $CP$. Now if we rotate the $\triangle ABP$ about the point at B $ 60 ^{\circ} $ anti clockwise we will get $\triangle BP'A'$.
Second Hint
SHORTEST DISTANCE:
Join the point P and P'.Now In the triangle BPP' we have
BP-BP'
$\angle PBP'=60 ^{\circ} $, SO $\triangle BPP' $ is a equilateral triangle. so $BP=BP'=PP'$
and also $AP'=AP$ (Length remain unchange after Rotation).
So from the point $A'$ to $C$ the path is $A'P'+PP'+PC$.This path will be Shortest distance if $A'P'+PP'+PC$ i.e A'C be a straight line. and also $AP+PB+AB=A'P'+PP'+PC$
the shortest path betwween two points is a straight line and so $ PA+PB+PC$ reaches its minimum if and only if the point $p$ and $P'$ lie on the line $A'C$
By symmetry it follows that $ P $ must also lie on the line $BC'$ and $AB'$.
So the point of intersection of these lines is a fermat point of a $\triangle ABC$.
Final Step
EQUILATERAL TRIANGLE :
Now the triangle $AA'B$ we have
$A'B=AB$ (length remain unchange due to rotation)
$\angle A'BA =60^{\circ}$. so the triangle $AA'B $ is a equilateral triangle .
similarly for the other two triangles $AC'C$ and $BB'C$