2005 Indonesia MO Problems/Problem 4
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 .
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, .
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