# Difference between revisions of "2006 Romanian NMO Problems/Grade 8/Problem 1"

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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 consider a prism with 6 faces, 5 of which are circumscriptible quadrilaterals. Prove that all the faces of the prism are circumscriptible quadrilaterals. | ||

==Solution== | ==Solution== | ||

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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. | 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. | ||

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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. | 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. | ||

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==See also== | ==See also== | ||

*[[2006 Romanian NMO Problems]] | *[[2006 Romanian NMO Problems]] | ||

[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |

## Latest revision as of 22:22, 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

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.