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 |