1979 IMO Problems/Problem 4
Problem
We consider a point in a plane
and a point
. Determine all the points
from
for which
is maximum.
Solution
Let be the orthogonal projection of the point
on the plane
. Then, the line
is the orthogonal projection of the line
on the plane
, and thus forms the least angle with the line
among all lines through the point
which lie in the plane
; hence,
, and equality holds if and only if the point
lies on the ray
(the only exception is when
; in this case,
, so the ray
is undefined, and equality holds for all points
in the plane
, since we always have
).
From , it follows that
; also, since the angles
and
are
, their half-angles
and
are < 90°, so that from
we can conclude that
. Equality holds, as in the above, if and only if the point
lies on the ray
(and, respectively, for all points
in the plane
if
).
On the other hand, we obviously have with equality if and only if
, i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base
, i. e. if and only if
, i. e. if and only if the point
lies on the sphere with center
and radius
.
Now, applying the Mollweide theorem in triangle , we get
(since
)
(since
),
and equality holds here if and only if equality holds in both of the inequalities and
that we have used, i. e. if and only if the point
lies both on the ray
(this condition should be ignored if
) and on the sphere with center
and radius
.
Hence, the point for which the ratio
is maximum is the point of intersection of the ray
with the sphere with center
and radius
(or, respectively, it can be any arbitrary point on the intersection of the plane
with the sphere with center
and radius
if
).
See Also
1979 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |