
Re: Parallelogram Orthocenters
Posted:
Oct 18, 2013 9:39 AM



> > 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.sqrt32.2/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*(ba*cos(fi))/sin(fi))^2 + (ab*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

