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Difference between revisions of "1959 IMO Problems/Problem 6"

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== Problem ==
 
== Problem ==
  
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Two [[plane]]s, <math>\displaystyle P </math> and <math>\displaystyle Q </math>, [[intersect]] along the [[line]] <math>\displaystyle p </math>.  The point <math>\displaystyle A </math> is in the plane <math>\displaystyle P </math>, and the point <math>\displaystyle {C} </math> is in the plane <math>\displaystyle Q </math>; neither of these points lies on the straight line <math>\displaystyle p </math>.  Construct an [[isosceles trapezoid]] <math>\displaystyle ABCD </math> (with <math>\displaystyle AB </math> [[parallel]] to <math>\displaystyle DC </math>) in which a [[circle]] can be constructed, and with [[vertex | vertices]] <math>\displaystyle B </math> and <math>\displaystyle D </math> lying in the planes <math>\displaystyle P </math> and <math>\displaystyle Q </math>, respectively.
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Two [[plane]]s, <math>P </math> and <math>Q </math>, [[intersect]] along the [[line]] <math>p </math>.  The point <math>A </math> is in the plane <math>P </math>, and the point <math>{C} </math> is in the plane <math>Q </math>; neither of these points lies on the straight line <math>p </math>.  Construct an [[isosceles trapezoid]] <math>ABCD </math> (with <math>AB </math> [[parallel]] to <math>DC </math>) in which a [[circle]] can be inscribed, and with [[vertex | vertices]] <math>B </math> and <math>D </math> lying in the planes <math>P </math> and <math>Q </math>, respectively.
  
 
== Solution ==
 
== Solution ==
  
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We first observe that we must have both lines <math>\displaystyle AB </math> (which we shall denote <math> \displaystyle a </math>) and <math>\displaystyle DC </math> (which we shall denote <math> \displaystyle c </math>) parallel to <math>\displaystyle p </math>, since if one of them is not, then neither can be and they must both intersect <math>\displaystyle p </math> (since they are both coplanar with <math>\displaystyle p </math>), making them [[skew lines | skew]].
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We first observe that we must have both lines <math>AB </math> (which we shall denote <math>a </math>) and <math>DC </math> (which we shall denote <math>c </math>) parallel to <math>p </math>, since if one of them is not, then neither can be and they must both intersect <math>p </math> (since they are both coplanar with <math>p </math>), making them [[skew lines | skew]].
  
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Now we note since a circle can be inscribed in the trapezoid, we must have <math>\displaystyle AB + DC = AD + BC</math>, and since the trapezoid is isosceles, this implies that each of the trapezoid's legs has length equal to the [[arithmetic mean | average]] of the lengths of the bases.
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Now we note since a circle can be inscribed in the trapezoid, we must have <math>AB + DC = AD + BC</math>, and since the trapezoid is isosceles, this implies that each of the trapezoid's legs has length equal to the [[arithmetic mean|average]] of the lengths of the bases.
  
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We can find this average by dropping [[perpendicular]] <math>\displaystyle AA' </math> to <math>\displaystyle c </math> such that <math>\displaystyle A' </math> is on <math>\displaystyle c </math>.  The average will be <math>\displaystyle A'C </math>, which is one of the sides of the [[rectangle]] with sides on <math>\displaystyle a </math> and <math>\displaystyle c </math> with vertices at <math>\displaystyle A </math> and <math>\displaystyle {C} </math>.
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We can find this average by dropping [[perpendicular]] <math>AA' </math> to <math>c </math> such that <math>A' </math> is on <math>c </math>.  The average will be <math>A'C </math>, which is one of the sides of the [[rectangle]] with sides on <math>a </math> and <math>c </math> with vertices at <math>A </math> and <math>{C} </math>.
  
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We now draw a circle with center <math>\displaystyle {C} </math> that contains <math> \displaystyle A' </math>.  The intersections of this circle with <math>\displaystyle a </math> are the two possible values of <math>\displaystyle B </math>, from either of which it is trivial to determine the corresponding location for <math>\displaystyle D </math>.  It is worth noting that the intersection points may concur (in which case there is only one distinct possibility (a square)), or they may not occur at all.  Q.E.D.
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We now draw a circle with center <math>{C} </math> that contains <math>A' </math>.  The intersections of this circle with <math>a </math> are the two possible values of <math>B </math>, from either of which it is trivial to determine the corresponding location for <math>D </math>.  It is worth noting that the intersection points may concur (in which case there is only one distinct possibility (a square)), or they may not occur at all.  Q.E.D.
  
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
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== Resources ==
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== See Also ==
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* [[1959 IMO Problems/Problem 5 | Previous problem]]
 
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* [[1959 IMO Problems]]
 
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=341533#p341533 Discussion on AoPS/MathLinks]
 
  
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[[Category:Olympiad Geometry Problems]]
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{{IMO box|year=1959|num-b=5|after=Last question}}
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[[Category: Olympiad Geometry Problems]]
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[[Category: 3D Geometry Problems]]
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[[Category: Geometric Construction Problems]]

Latest revision as of 15:25, 29 July 2024

Problem

Two planes, $P$ and $Q$, intersect along the line $p$. The point $A$ is in the plane $P$, and the point ${C}$ is in the plane $Q$; neither of these points lies on the straight line $p$. Construct an isosceles trapezoid $ABCD$ (with $AB$ parallel to $DC$) in which a circle can be inscribed, and with vertices $B$ and $D$ lying in the planes $P$ and $Q$, respectively.

Solution

We first observe that we must have both lines $AB$ (which we shall denote $a$) and $DC$ (which we shall denote $c$) parallel to $p$, since if one of them is not, then neither can be and they must both intersect $p$ (since they are both coplanar with $p$), making them skew.

Now we note since a circle can be inscribed in the trapezoid, we must have $AB + DC = AD + BC$, and since the trapezoid is isosceles, this implies that each of the trapezoid's legs has length equal to the average of the lengths of the bases.

We can find this average by dropping perpendicular $AA'$ to $c$ such that $A'$ is on $c$. The average will be $A'C$, which is one of the sides of the rectangle with sides on $a$ and $c$ with vertices at $A$ and ${C}$.

We now draw a circle with center ${C}$ that contains $A'$. The intersections of this circle with $a$ are the two possible values of $B$, from either of which it is trivial to determine the corresponding location for $D$. It is worth noting that the intersection points may concur (in which case there is only one distinct possibility (a square)), or they may not occur at all. Q.E.D.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1959 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last question
All IMO Problems and Solutions
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