Difference between revisions of "2006 Canadian MO Problems/Problem 2"
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==Problem== | ==Problem== | ||
− | Let <math>ABC</math> be an acute | + | 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== | ||
+ | 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. | ||
− | {{ | + | 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>. |
− | *[[2006 Canadian MO]] | + | 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 |
+ | |||
+ | <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>, | ||
+ | |||
+ | 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== | ||
+ | *[[2006 Canadian MO Problems/Problem 1 | Previous problem]] | ||
+ | *[[2006 Canadian MO Problems/Problem 3 | Next problem]] | ||
+ | *[[2006 Canadian MO Problems]] | ||
+ | |||
+ | [[Category:Olympiad Geometry Problems]] |
Revision as of 16:18, 6 February 2007
Problem
Let be an acute angled triangle. Inscribe a rectangle
in this triangle so that
is on
,
on
, and
and
on
. Describe the locus of the intersections of the diagonals of all possible rectangles
.
Solution
The locus is the line segment which joins the midpoint of side to the midpoint of the altitude to side
of the triangle.
Let and let
be the foot of the altitude from
to
. Then by similarity,
.
Now, we use vector geometry: intersection of the diagonals of
is also the midpoint of diagonal
, so
,
and this point lies on the segment joining the midpoint of segment
and the midpoint
of segment
, dividing this segment into the ratio
.