Difference between revisions of "2006 Canadian MO Problems/Problem 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. | 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. | ||
− | 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}{ | + | 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. |
==See also== | ==See also== |
Latest revision as of 19:00, 21 September 2014
Contents
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 .
Solution 2
We claim that the desired locus is the line segment from the midpoint of altitude to the midpoint of , , not including both endpoints.
A homothety about maps the rectangle onto rectangle in the exterior of . The scale factor of the homothety is , which is also the scale factor of the mapping of the intersection of diagonals (the original we call and the new we call . Hence . But , and , so and are similar, and so lies on , 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 |