610 Geometry and TrigonometryThe lemma is proved.Remark.A. Dang gave an alternative solution by observing that trianglesAH C and
QH Pare orthological, and then using the property of orthological triangles proved by
us in the introduction.
(Indian Team Selection Test for the International Mathematical Olympiad, 2005,
proposed by R. Gelca)586.Let−→a,−→
b,−→c,−→
d, and−→p denote vectors from a common origin to the vertices
A, B, C, Dof the tetrahedron, and to the pointPof concurrency of the four lines. Then
the vector equation for the altitude fromAis given by
−→ra=−→a +λ[(−→b +−→c +−→d)/ 3 −−→p].The position vector of the point corresponding toλ=3is−→a +−→
b +−→c +−→
d − 3 −→p,
which is the same for all four vertices of the tetrahedron. This shows that the altitudes
are concurrent.
For the converse, if the four altitudes are concurrent at a pointHwith position vector
−→
h, then the line through the centroid of the faceBCDand perpendicular to that face is
described by
−→r′
a=[(−→
b +−→c +−→
d)/ 3 ]+λ′(−→a −−→
h).This time the common point of the four lines will correspond, of course, toλ′=^13 , and
the problem is solved.
(proposed by M. Klamkin forMathematics Magazine)
587.The double of the area of triangleONQis equal to‖
−−→
ON×
−−→
OQ‖=
∥
∥∥
∥
(
1
3
−→
OA+
2
3
−→
OB
)
×
(
2
3
−−→
OD+
1
3
−→
OC
)∥
∥∥
∥.
Since−→
OAis parallel to−→
OCand−→
OBis parallel to−−→
OD, this is further equal to
∥
∥∥
∥2
9
(
−→
OA×
−−→
OD+
−→
OB×
−→
OC)
∥
∥∥
∥.
A similar computation shows that this is equal to|
−−→
OM×
−→
OP|, which is twice the area
of triangleOMP. Hence the conclusion.
588.The area of triangleAMNis equal to
1
2