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Difference between revisions of "2006 Canadian MO Problems/Problem 2"

 
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==Problem==
 
==Problem==
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Let <math>ABC</math> be an acute angled triangle. Inscribe a rectangle <math>DEFG</math> in this triangle so that <math>D</math> is on <math>AB</math>, <math>E</math> on <math>AC</math>, and <math>F</math> and <math>G</math> on <math>BC</math>. Describe the locus of the intersections of the diagonals of all possible rectangles <math>DEFG</math>.
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Let <math>ABC</math> be an [[acute angle]]d [[triangle]]. Inscribe a [[rectangle]] <math>DEFG</math> in this triangle so that <math>D</math> is on <math>AB</math>, <math>E</math> on <math>AC</math>, and <math>F</math> and <math>G</math> on <math>BC</math>. Describe the [[locus]] of the [[intersection]]s of the [[diagonal]]s of all possible rectangles <math>DEFG</math>.
 
==Solution==
 
==Solution==
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The locus is the [[line segment]] which joins the [[midpoint]] of side <math>BC</math> to the midpoint of the [[altitude]] to side <math>BC</math> of the triangle.
  
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{{solution}}
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Let <math>r = \frac{AD}{AB}</math> and let <math>H</math> be the foot of the altitude from <math>A</math> to <math>BC</math>.  Then by [[similarity]], <math>\frac{AE}{AC} = \frac{GH}{BH} = \frac{FH}{CH} = r</math>. 
  
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*[[2006 Canadian MO]]
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Now, we use [[vector]] geometry: intersection <math>I</math> of the diagonals of <math>DEFG</math> is also the midpoint of diagonal <math>DF</math>, so
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 +
<math>I = \frac{1}{2}(D + F) = \frac{1}{2}((rA + (1 - r)B) + (rH + (1 - r)C)) = r \frac{A + H}{2} + (1 - r)\frac{B + C}{2}</math>,
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 +
and this point lies on the segment joining the midpoint <math>\frac{A + H}{2}</math> of segment <math>AH</math> and the midpoint <math>\frac{B + C}{2}</math> of segment <math>BC</math>, dividing this segment into the [[ratio]] <math>r : 1 - r</math>.
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==See also==
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*[[2006 Canadian MO Problems/Problem 1 | Previous problem]]
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*[[2006 Canadian MO Problems/Problem 3 | Next problem]]
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*[[2006 Canadian MO Problems]]
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[[Category:Olympiad Geometry Problems]]

Revision as of 16:18, 6 February 2007

Problem

Let $ABC$ be an acute angled triangle. Inscribe a rectangle $DEFG$ in this triangle so that $D$ is on $AB$, $E$ on $AC$, and $F$ and $G$ on $BC$. Describe the locus of the intersections of the diagonals of all possible rectangles $DEFG$.

Solution

The locus is the line segment which joins the midpoint of side $BC$ to the midpoint of the altitude to side $BC$ of the triangle.

Let $r = \frac{AD}{AB}$ and let $H$ be the foot of the altitude from $A$ to $BC$. Then by similarity, $\frac{AE}{AC} = \frac{GH}{BH} = \frac{FH}{CH} = r$.

Now, we use vector geometry: intersection $I$ of the diagonals of $DEFG$ is also the midpoint of diagonal $DF$, so

$I = \frac{1}{2}(D + F) = \frac{1}{2}((rA + (1 - r)B) + (rH + (1 - r)C)) = r \frac{A + H}{2} + (1 - r)\frac{B + C}{2}$,

and this point lies on the segment joining the midpoint $\frac{A + H}{2}$ of segment $AH$ and the midpoint $\frac{B + C}{2}$ of segment $BC$, dividing this segment into the ratio $r : 1 - r$.

See also

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