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.
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.