Tuesday, December 06, 2005
20:18
Okay, I'm not in this semester's Modern Geometry class, but I have become mildly obsessed with one of their homework problems. The statement is simple enough: Given that the circumcenter of a triangle is the same as the Gergonne point, prove that the triangle is equilateral. The circumcenter is the center of the circumcircle, which is the circle that circumscribes the triangle, i.e., intersects all three vertices. It is also the intersection point of the perpendicular bisectors of the sides. The Gergonne point is defined as the point of intersection of the lines connecting the vertices with the points of tangency for the incircle, where the incircle is the circle inscribed in the triangle. The approach I've been trying is to show that when the Gergonne point and the circumcenter are the same, they must also be the same as the incenter, which is the center of the incircle. But I haven't had success yet. Apparantly no one in the class, including the teacher, has been able to solve this.
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Okay, I'm probably over simplifying this. But if the Gergonne point is the circumcenter, then the distance from the Gergonne point to each vertex is r, the radius of the circle. Then the perpendicular distance from the Gergonne point to each of the sides must be the same (call it d), so each of the triangle's diagonals is (r+d). If the diagonals are all equal, it must be equilateral. What am I missing?
Wait... I haven't shown that those lengths of side r have to be at 60 degrees to one another. So I'd need to argue that if not, then one of the assumed properties would be contraindicated.
Um, where do you get that the perpendicular lines going through the circumcenter point must be equal?
:) I have a solution. There's a hint on the whiteboard in the math center. (Yes, my initial attempt assumed that the line through the center and vertex was the perpendicular bisector. My current solution doesn't). I'm sure there's a more elegant solution, though. Mine's a bit...bottom up.
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