2006 Romanian NMO Problems/Grade 8/Problem 1
We consider a prism with 6 faces, 5 of which are circumscriptible quadrilaterals. Prove that all the faces of the prism are circumscriptible quadrilaterals.
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.