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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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==Solution 2==
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We claim that the desired locus is the line segment from the midpoint <math>A'</math> of altitude <math>AD</math> to the midpoint of <math>BC</math>, <math>M</math>, not including both endpoints.
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A homothety about <math>A</math> maps the rectangle <math>DEFG</math> onto rectangle <math>BCHI</math> in the exterior of <math>ABC</math>. The scale factor of the homothety is <math>\frac{AD}{AB}</math>, which is also the scale factor of the mapping of the intersection of diagonals (the original we call <math>X</math> and the new we call <math>Y</math>. Hence <math>\frac{AX}{XY} = \frac{AD}{DB}</math>. But <math>\frac{AA'}{MY} = \frac{AH}{BI} = \frac{\frac{AH}{DG}}{\frac{BI}{DG}} = \frac{AD}{BD}</math>, and <math>MY // AH</math>, so <math>MYX</math> and <math>A'AX</math> are similar, and so <math>X</math> lies on <math>A'M</math>, as desired. Reversing the argument proves the other direction for a locus, and we are done.
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==See also==
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*[[2006 Canadian MO Problems]]
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{{CanadaMO box|year=2006|num-b=1|num-a=3}}
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[[Category:Olympiad Geometry Problems]]

Latest revision as of 20:00, 21 September 2014

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 $r : 1 - r$.

Solution 2

We claim that the desired locus is the line segment from the midpoint $A'$ of altitude $AD$ to the midpoint of $BC$, $M$, not including both endpoints.

A homothety about $A$ maps the rectangle $DEFG$ onto rectangle $BCHI$ in the exterior of $ABC$. The scale factor of the homothety is $\frac{AD}{AB}$, which is also the scale factor of the mapping of the intersection of diagonals (the original we call $X$ and the new we call $Y$. Hence $\frac{AX}{XY} = \frac{AD}{DB}$. But $\frac{AA'}{MY} = \frac{AH}{BI} = \frac{\frac{AH}{DG}}{\frac{BI}{DG}} = \frac{AD}{BD}$, and $MY // AH$, so $MYX$ and $A'AX$ are similar, and so $X$ lies on $A'M$, as desired. Reversing the argument proves the other direction for a locus, and we are done.

See also

2006 Canadian MO (Problems)
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 Followed by
Problem 3
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