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!

#### Math Olympiad Program

This is problem is a part of the Math Olympiad Program and I.S.I. Entrance Program at Cheenta.

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.

locus of c is the straight line through c making 60 degrees angle with the base BC produced. easy one