2006 Canadian MO Problems/Problem 2

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 $\frac{AA'}{MY} = \frac{AH}{BI} = \frac{\frac{AH}{DE}}{\frac{BI}{DE}} = \frac{AD}{BD}$ , 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.

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

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2006 Canadian MO Problems/Problem 2

Contents

Problem

Solution

Solution 2

See also