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
Permanent Citation