Two equilateral triangles OPQ and ORS of respective sides a and b.

Case 1: a=b then OP=OQ=OR=OS=a=b=r, i.e. (P,Q,R,S) lie on the same circle, irrespective of triangles' orientation. Hence a^2+a.b+b^2=r^2+r.r+r^2=3.r^2

Case 2: a<>b Assume PQ and RS are not parallel. (P,Q,R,S) on the same circle implies that circle's center I is the intersection of PQ and RS bisectors, which is O by construction. Since OP=a<>b=OR we have a contradiction. Therefore PQ//RS.

PQ and RS share a common bisector as depicted on the following figure: http://bleaug.free.fr/8foxes/8foxes354.png

Since P,J,S belong to circle centered in I we have QPR=QJR=QSR=QIR/2=60°. Hence QR=r.sqrt(3)

PQ//RS implies QOR=120°, therefore in triangle OQR: QR^2=OQ^2+OR^2-2.OQ.OR.cos(QOR) which can be rewritten as: 3.r^2=a^2+b^2+a.b

to be complete, we should also consider the (sub-)case where PQ and RS lie on the same side w.r.t. O as depicted in: http://bleaug.free.fr/8foxes/8foxes354b.png

If a and b represent positive length values, formula becomes: 3.r^2=a^2+b^2-a.b

If we adopt the convention that (a.b) = scalar product (OQ.OS) including sign, the above formula holds in all cases.

Two equilateral triangles OPQ and ORS of respective sides a and b. then arc(PQ) + arc(RS) = 2 . 60 degree =120 degree but arc(PR) = arc(QS) . Then arc(PR) = arc(QS) = 120 degree. and PR = QS = r.sqrt(3) . In the triangle OPR angle(POR) = 120 degree. Then OP^2 + OR^2 + OP.OR = PR^2 and a^2 + b^2 + a.b = r.sqrt(3) όπερ έδει δείξαι. Philip Pydnaios , from Hellas

Two equilateral triangles OPQ and ORS of respective sides a and b.

ReplyDeleteCase 1: a=b

then OP=OQ=OR=OS=a=b=r, i.e. (P,Q,R,S) lie on the same circle, irrespective of triangles' orientation.

Hence a^2+a.b+b^2=r^2+r.r+r^2=3.r^2

Case 2: a<>b

Assume PQ and RS are not parallel. (P,Q,R,S) on the same circle implies that circle's center I is the intersection of PQ and RS bisectors, which is O by construction. Since OP=a<>b=OR we have a contradiction. Therefore PQ//RS.

PQ and RS share a common bisector as depicted on the following figure:

http://bleaug.free.fr/8foxes/8foxes354.png

Since P,J,S belong to circle centered in I we have QPR=QJR=QSR=QIR/2=60°. Hence QR=r.sqrt(3)

PQ//RS implies QOR=120°, therefore in triangle OQR:

QR^2=OQ^2+OR^2-2.OQ.OR.cos(QOR)

which can be rewritten as:

3.r^2=a^2+b^2+a.b

bleaug

to be complete, we should also consider the (sub-)case where PQ and RS lie on the same side w.r.t. O as depicted in:

ReplyDeletehttp://bleaug.free.fr/8foxes/8foxes354b.png

If a and b represent positive length values, formula becomes:

3.r^2=a^2+b^2-a.b

If we adopt the convention that (a.b) = scalar product (OQ.OS) including sign, the above formula holds in all cases.

bleaug

Two equilateral triangles OPQ and ORS of respective sides a and b.

ReplyDeletethen arc(PQ) + arc(RS) = 2 . 60 degree =120 degree

but arc(PR) = arc(QS) .

Then arc(PR) = arc(QS) = 120 degree.

and PR = QS = r.sqrt(3) .

In the triangle OPR angle(POR) = 120 degree.

Then OP^2 + OR^2 + OP.OR = PR^2

and a^2 + b^2 + a.b = r.sqrt(3) όπερ έδει δείξαι.

Philip Pydnaios , from Hellas