FoldPoints
Let a "foldpoint" of a triangle be defined as a place on the triangle to which
the three corners can all be folded simultaneously, such that no folds overlap.
What are the shape and area of the region surrounding all foldpoints for:

an equilateral triangle with side 1?

a 306090 triangle with hypotenuse 1?
If you find any extensions to this problem, let me know. For example:
 Are there any triangles for which no foldpoints can be found?
 For which other convex polygons can foldpoints be found?
 Others?
Source: Original, based on a puzzle in rec.puzzles. Used as Ken's
Puzzle of the Day 4/26/94.
Solutions were received from
Philippe Fondanaiche:
1) an equilateral triangle with side 1
Let call ABC the equilateral triangle with P, Q, R respectively at the middle
of the sides BC, AC and AB.
The region surrounding all foldpoints is defined by a kind of escutcheon
interior to the triangle ABC which is the common part of 3 circles:
C1 of center P and radius PB=PC=1/2, C2 of center Q and radius QA=QC=1/2, C3
of center R and radius RA=RB=1/2.
The area A1 of this region is equal to the area of the equilateral triangle
PQR + the area of the 3 halflentils alongside PQ,QR,PR . As the area of a
halflentil is equal to the area of a sector of angle 60° and radius 1/2 less
the area of PQR, that is to say: 1/6*area of a circle of radius 1/2  area
PQR, we have:
A1 = 1/2 * area of circle of radius 1/2  2 * area PQR = ( pi  sqrt(3) ) /8
= 0,17619273....
2) a 306090 triangle with hypotenuse 1
Let call ABC the triangle with :
 the right angle at B ,
 BC=1/2, AC=1 and AB=sqrt(3)/2,
 P,Q,R at the middle of the sides BC,AC and AB.
The region surrounding all foldpoints is defined by a kind of lentil which is
the common part of 2 circles: C1 of center P and radius PB=PC=1/4, C2 of
center R and radius RA=RB=sqrt(3)/4. C1 and C2 cuts AC at S. B and S are the
edges of this lentil.
The area A2 of this region is equal to: area of sector BPS area of triangle
BPS + area of sector BRS  area of trianlge RBS.
As area of sector BPS = 1/3 * pi/16 and area of sector BRS = 1/6 * 3*pi/16,
then:
A2 = 5*pi/96  sqrt(3) /16 = ( 5*pi  6*sqrt(3) ) / 96 = 0,055371441....
3) generalization
With any triangle (A,B and C not on the same line) having the points P,Q and R
as middles of the sides BC,AC and AB, the common part is defined by the
intersection of the circles of centers P,Q and R and radius PB, QC and RA. The
area of this region is always >0.
With convex polygons the common part is reduced to one point when the polygon
is regular (square,pentagon,hexagon,etc...). With any convex polygon,there is
no common part.
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