 Try this beautiful problem. We will give you hints!

ABC is an equilateral triangle with vertex A fixed and B moving in a given straight line. Find the locus of C

##### What is the big idea ?

Think about what does not change.

1. Point A is fixed.
2. The line on which B moves is fixed.
3. The fact that ABC is always equilateral is fixed.

##### What changes?

1. The position of C will change (as B moves).
2. Length of the sides of the triangle will change (it needs to stay equilateral).

##### How to get started?

Carefully draw several positions of B (and corresponding positions of C). This will give you an idea about what is the possible path traced out by C

# Guess the path! Drop a perpendicular from A on the line L. Suppose the feet of the perpendicular is P.

Make AB such that $\angle BAP = 30^o$

Reflect BAP about AP. Let the image of B under reflection be C.

Draw a line through C that makes an angle of $60^0$ with L. We will call this Red line.

Claim: This Red Line is part of the locus of point C.

Pick another point $B_1$ on L.

Joint $AB_1$

Make angle $60^0$ with $AB_1$ and draw a line. Let it hit the red line at $C_1$

Show that $AB_1 C_1$ is equilateral.

Notice that $AB_1 C C_1$ is cyclic. Why? After all $\angle B_1 A C_1 = 60^0$ and $\angle B_1 C C_1 = 120^o$. Hence opposite angles add up to $180^o$.

Any quadrilateral in which opposite angles add up to $180^o$ is cyclic.

This immediately proves $AB_1 C_1$ is equilateral. Hence red line is part of the locus.

Find the other part of the locus by interchanging the positions of B and C n first step.