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 ==Solution==   ==Solution== 
 {{solution}}   {{solution}} 
−  We use the lemma that given two noncoplanar 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 <math>O_1</math> and <math>O_2</math> and let them intersect at <math>X</math> and <math>Y</math>. Draw the line <math>l_1</math> through (the center of) <math>O_1</math> perpendicular to the plane of the circle. We know that any point on that line is equidistant from any point on <math>O_1</math> because for a point <math>P</math> on the line and a point <math>Q</math> on the circle, <math>PQ=\sqrt{PO_1^2+O_1Q^2}=\sqrt{PO_1^2+r_1^2}</math>, which does not depend on the point <math>Q</math> chosen. Similarly, we draw the line <math>l_2</math> through <math>O_2</math>. Since <math>X</math> and <math>Y</math> lie on both circles, any point on <math>l_1</math> or <math>l_2</math> is equidistant from <math>X</math> and <math>Y</math>. However, the locus of all points equidistant from <math>X</math> and <math>Y</math> is the plane that perpendicularly bisects <math>\overline{XY}</math>. Therefore, <math>l_1</math> and <math>l_2</math> lie on one plane. Since our two circles are not coplanar, <math>l_1</math> and <math>l_2</math> must intersect at our desired point.
 
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−  Let the cube have vertices <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>A'</math>, <math>B'</math>, <math>C'</math>, <math>D'</math>, where all sides but <math>A'B'C'D'</math> are known to be cyclic quadrilaterals. First, we consider the circumcircles of quadrilaterals <math>ABCD</math> and <math>ABB'A'</math>. By our lemma, there exists a point <math>O</math> equidistant from <math>A</math>, <math>B</math>, <math>A'</math>, <math>B'</math>, <math>C</math>, <math>D</math>. Let the perpendicular from <math>O</math> to the plane <math>AA'D'D</math> intersect the plane at <math>O'</math>. By HL congruency, the triangles <math>OO'A</math>, <math>OO'A'</math>, and <math>OO'D</math> are congruent. Since <math>O'A=O'A'=O'D</math>, O is the center of quadrilateral <math>AA'D'D</math>. By SAS congruency, <math>\triangle OO'D'</math> is congruent to the aforementioned triangles, so <math>OA'=OD'</math>. Similarly, if we focus on quadrilateral <math>BCC'B'</math>, we get that <math>OB'=OC'</math>. Therefore, <math>OA'=OB'=OC'=OD'</math>. Let the perpendicular from <math>O</math> to the plane <math>A'B'C'D'</math> intersect the plane at <math>O''</math>. By HL congruency, the triangles <math>OO''A'</math>, <math>OO''B'</math>, <math>OO''C'</math>, and <math>OO''D'</math> are congruent. Thus, <math>O''A'=O''B'=O''C'=O''D'</math> and <math>A'B'C'D'</math> is cyclic.
 
 ==See also==   ==See also== 
 *[[2006 Romanian NMO Problems]]   *[[2006 Romanian NMO Problems]] 
 [[Category:Olympiad Geometry Problems]]   [[Category:Olympiad Geometry Problems]] 
Revision as of 22:21, 15 August 2012
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
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See also