Let A=(0,0), B(1/2,sqrt(3)/2), C=(1,0), and P=(x,y)
The the equation becomes
x^2 - y^2 = (x-1/2)^2+ (y-sqrt(3)/2)^2 + (x-1)^2 + y^2
which, after expanding and completing the square on x becomes
(x-3/2)^2+(y-sqrt(3)/2)^2 = 1
Which implies P can be anywhere on the arc of the circle of radius 1 centered at (3/2,sqrt(3)/2). This circle contains B and C.
Furthermore the arc from B to C is 60 degrees. The arc external to the triangle is 300 degrees and central angle BPC = 150 degrees.
https://www.desmos.com/calculator/rsi6mqixax
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Posted by Jer
on 2024-12-08 17:41:05 |
Solution