2006 Romanian NMO Problems/Grade 8/Problem 1
Problem
We consider a prism with 6 faces, 5 of which are circumscriptible quadrilaterals. Prove that all the faces of the prism are circumscriptible quadrilaterals.
Solution
We use the lemma that given two non-coplanar circles in space that intersect at two points, there exists a point P such that P is equidistant from any point on one circle and any point on the other circle.
Proof of lemma: Let the two circles be and
and let them intersect at
and
. Draw the line
through (the center of)
perpendicular to the plane of the circle. We know that any point on that line is equidistant from any point on
because for a point
on the line and a point
on the circle,
, which does not depend on the point
chosen. Similarly, we draw the line
through
. Since
and
lie on both circles, any point on
or
is equidistant from
and
. However, the locus of all points equidistant from
and
is the plane that perpendicularly bisects
. Therefore,
and
lie on one plane. Since our two circles are not coplanar,
and
must intersect at our desired point.
Let the cube have vertices ,
,
,
,
,
,
,
, where all sides but
are known to be cyclic quadrilaterals. First, we consider the circumcircles of quadrilaterals
and
. By our lemma, there exists a point
equidistant from
,
,
,
,
,
. Let the perpendicular from
to the plane
intersect the plane at
. By HL congruency, the triangles
,
, and
are congruent. Since
, O is the center of quadrilateral
. By SAS congruency,
is congruent to the aforementioned triangles, so
. Similarly, if we focus on quadrilateral
, we get that
. Therefore,
. Let the perpendicular from
to the plane
intersect the plane at
. By HL congruency, the triangles
,
,
, and
are congruent. Thus,
and
is cyclic.