Difference between revisions of "1960 IMO Problems/Problem 5"
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
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Latest revision as of 23:28, 18 July 2016
Problem
Consider the cube (with face
directly above face
).
a) Find the locus of the midpoints of the segments , where
is any point of
and
is any point of
;
b) Find the locus of points which lie on the segment
of part a) with
.
Solution
Let ,
,
,
,
,
,
, and
. Then there exist real
and
in the closed interval
such that
and
.
The midpoint of has coordinates
. Let
and
be the
- and
-coordinates of the midpoint of
, respectively. We then have that
and
, so
and
. The region of points that satisfy these inequalities is the closed square with vertices at
,
,
, and
. For every point
in this region, there exist unique points
and
such that
is the midpoint of
.
If and
, then
has coordinates
. Let
and
be the
- and
- coordinates of
. We then have that
and
, and
and
. The region of points that satisfy these inequalities is the closed rectangle with vertices at
,
,
, and
. For every point
in this region, there exist unique points
and
such that
and
.
See Also
1960 IMO (Problems) | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 • 7 | Followed by Problem 6 |