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Learn MoreABC 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...

Source

Suggested Reading

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 $BB'C$

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