“An equilateral triangle ABC has side length 2. A square,PQRS, is such that P lies on AB, Q lies on BC, and R and S lie on AC as shown. The points P, Q, R, and S move so that P, Q and R always remain on the sides of the triangle and S moves from AC to AB through the interior of the triangle. If the points P, Q, R and S always form the vertices of a square, show that the path traced out by S is a straight line parallel to BC.”
The CEMC’s solution seems a little tricky. Here I want to discuss a method using analytic geometry and congruent triangles.
Place the equilateral triangle ABC on a coordinate plane such that and . Let the coordinates of be . The problem is actually asking for showing that whatever the value of is, the -coordinate of is constant.
Let the coordinates of other points be , , , , and .
Since the point O is the midpoint of SQ and the y-coordinate of Q is always 0, we just need to show the is constant, by the midpoint formula of analytic geometry. Since O is also the midpoint of PR, it’s equivalent to show is constant.
Plugging in the coordinates of P, Q to the equations of line AB and AC, we have the following equations.
(1)
(2)
Notice that , then we have the following connections between the coordinates of P and Q.
() => (3)
() => (4)
Plugging the equations (3)(4) to (1)(2), we have the following equations.
(5)
(6)
(5)+(6) shows that .
Therefore is constant. Done.
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