A Circle Defined from an Equilateral Triangle via Pythagoras

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Consider an equilateral triangle and the set of points such that . Then is either 30° or 150°, depending on whether is inside or outside , and so the locus of is a circle through and .

Contributed by: Dominic C. Milioto (September 2019)
Open content licensed under CC BY-NC-SA


Details

Use the "rotate" slider to rotate the triangle by . This gives:

Snapshot 1: with

Snapshot 2: the point is outside the triangle with

Snapshot 3: the point is at the top vertex of the triangle and is undefined

Proof [1]:

Suppose is inside the circle.

Rotate by so that falls along . Call the rotated triangle ; since and have equal lengths, . Also, since and have equal lengths and , is equilateral and and have equal lengths. Then:

.

Since and is equilateral, .

When is outside , a similar argument gives .

Reference

[1] AoPS Online. (Sep 10, 2019) artofproblemsolving.com/community/c3t48f3h1903996_geometry.


Snapshots



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