ABC is a triangle in the plane. Find the locus of point P also on the same plane such that the lengths of the segments PA, PB and PC form a triangle whose area is equal to one third of the area of triangle ABC.
https://www.desmos.com/calculator/hcmhnbdgco
Move the purple or blue points
The red curve is the requested locus but you can set it to any proportion.
The blue region is where the lengths do not satisfy the triangle inequality.
If you tinker with the vertices of the triangle to make it equilateral, the locus resolves to a circle and a point at the center.
Pretty fun if I do say so myself.
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Posted by Jer
on 2020-11-22 21:06:29 |
Desmos solution