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

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==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.
  
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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

This problem needs a solution. If you have a solution for it, please help us out by adding it. 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 $O_1$ and $O_2$ and let them intersect at $X$ and $Y$. Draw the line $l_1$ through (the center of) $O_1$ perpendicular to the plane of the circle. We know that any point on that line is equidistant from any point on $O_1$ because for a point $P$ on the line and a point $Q$ on the circle, $PQ=\sqrt{PO_1^2+O_1Q^2}=\sqrt{PO_1^2+r_1^2}$, which does not depend on the point $Q$ chosen. Similarly, we draw the line $l_2$ through $O_2$. Since $X$ and $Y$ lie on both circles, any point on $l_1$ or $l_2$ is equidistant from $X$ and $Y$. However, the locus of all points equidistant from $X$ and $Y$ is the plane that perpendicularly bisects $\overline{XY}$. Therefore, $l_1$ and $l_2$ lie on one plane. Since our two circles are not coplanar, $l_1$ and $l_2$ must intersect at our desired point.

Let the cube have vertices $A$, $B$, $C$, $D$, $A'$, $B'$, $C'$, $D'$, where all sides but $A'B'C'D'$ are known to be cyclic quadrilaterals. First, we consider the circumcircles of quadrilaterals $ABCD$ and $ABB'A'$. By our lemma, there exists a point $O$ equidistant from $A$, $B$, $A'$, $B'$, $C$, $D$. Let the perpendicular from $O$ to the plane $AA'D'D$ intersect the plane at $O'$. By HL congruency, the triangles $OO'A$, $OO'A'$, and $OO'D$ are congruent. Since $O'A=O'A'=O'D$, O is the center of quadrilateral $AA'D'D$. By SAS congruency, $\triangle OO'D'$ is congruent to the aforementioned triangles, so $OA'=OD'$. Similarly, if we focus on quadrilateral $BCC'B'$, we get that $OB'=OC'$. Therefore, $OA'=OB'=OC'=OD'$. Let the perpendicular from $O$ to the plane $A'B'C'D'$ intersect the plane at $O''$. By HL congruency, the triangles $OO''A'$, $OO''B'$, $OO''C'$, and $OO''D'$ are congruent. Thus, $O''A'=O''B'=O''C'=O''D'$ and $A'B'C'D'$ is cyclic.

See also

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