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Topic: Parallelogram Orthocenters
Replies: 7   Last Post: Oct 22, 2013 7:28 AM

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Avni Pllana

Posts: 531
Registered: 12/6/04
Re: Parallelogram Orthocenters
Posted: Oct 18, 2013 9:39 AM
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> > Let ABCD be a parallelogram such that AB=3DDC=3D10, and
> > AD=3DBC=3D6,
> > and angle(BAD)=3D60=C2=B0. Let H1,H2 be the orthocenters

> of
> > triangles ABD and CDB respectively. Determine the
> > distance between H1 and H2.
> >=20
> > Best regards,
> > Avni

>=20
> A pretty figure.
> Two normals divide the parallelogram sides into 3+7
> and 1+5 so that the orthocenters are the opposite
> corners of a rectangle 7x(3.sqrt3-2.2/sqrt3) =3D 7 x
> (5/sqrt3)
> whence H1H2 =3D sqrt(172/3) =3D 2.sqrt(43/3)
>=20
> Regards, Peter Scales.


Hi Peter,

I expected an analytical solution in terms of a, b and fi,

H1H2=3Dsqrt((b*sin(fi) - 2*(b-a*cos(fi))/sin(fi))^2 + (a-b*cos(fi))^2) . =
=20

In fact you also provided an analytical solution.=20

I want to add that in general case H1 i concyclic with C,D,B and the circum=
center O2 lies on the segment H1C. In the special case a =3D 10, b =3D 6, f=
i =3D 60=C2=B0, even O1 is concyclic with C,B,D, see attachment.

Best regards,
Avni



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