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Difference between revisions of "2006 Canadian MO Problems/Problem 2"
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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>.
 
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>.
 
==See also==
 
==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]]
 
 
*[[2006 Canadian MO Problems]]
 
*[[2006 Canadian MO Problems]]
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{{CanadaMO box|year=2006|before=First question|num-a=2}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 19:56, 7 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

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