2002 Pan African MO Problems/Problem 2
Problem
is a right triangle with . and are moving on and respectively such that . Show that there is a fixed point through which the perpendicular bisector of always passes.
Solution
We can consider two cases: one where the legs have equal length and one where the legs don’t have equal length.
Case 1:
Since and , by the Segment Addition Postulate, . This, is a 45-45-90 triangle.
Let be the midpoint of . By SSS Congruency, . Thus, , so .
Now extend to , and let be the point of intersection. By SAS Similarity, , so and . Thus, .
By HL Congruency, , so . Therefore, is a perpendicular bissector of . Thus, all perpendicular bissectors of where are also a perpendicular bisector of , so there is a point where the perpendicular bissector of passes through.
Case 2:
WLOG, let . Additionally, let and , so . Let be points where , making . Thus, , so is on . Since the midpoint of is on , the y-coordinate of must equal .
Let be points where , making . The midpoint of is , and the slope of is . Thus, the equation of the line perpendicular to is . Since the y-coordinate of equals , substituting and solving for results in . The coordinates of are ; now we need to prove that this point is on any perpendicular bissector of .
Let be any point of at a given time, and let the coordinates of be . That means the coordinates of are . The midpoint of is , and the slope of is . Thus, the equation of the line perpendicular to is . Plugging in for means that , so point is on the line.
Thus, there is a fixed point through which the perpendicular bisector of always passes.
See Also
2002 Pan African MO (Problems) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All Pan African MO Problems and Solutions |