2005 Indonesia MO Problems/Problem 4
Problem
Let be a point in triangle
such that
,
,
. The centers of circumcircles of triangles
are
, respectively. Prove that the area of
is greater than the area of
.
Solution
From the above constraints, we can let be
,
be
,
be
, and
be
, where
are positive. Note that the centers of circumcircles are on the perpendicular bisectors of
,
, or
.
Because is a right triangle, the circumcenter of
is the midpoint of
, so the coordinates of
are
. Additionally, the midpoint of
is
and the slope of line
on the coordinate grid is
. Therefore, the equation of the perpendicular bisector of
is
.
Note that is on the perpendicular bisector of
, so the y-coordinate of
is
. Therefore, the x-coordinate of
is
. Thus, the length of
is
.
Now we'll calculate the area of and
. By using a triangle area formula,
Since
is a 30-60-90 triangle,
. Therefore,
Now what's left is to show that
. Since
,
. Additionally, by using the AM-GM Inequality,
Therefore,
.
See Also
2005 Indonesia MO (Problems) | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 5 |
All Indonesia MO Problems and Solutions |