Let I be the incenter of ABC. AI, BI, CI intersect the circumcircle of ABC again at A', B', C' respectively. Show that the area of A'B'C' >= the area of ABC
No, truly there is no java script here but what I am attempting to describe is what I believe is the mechanics of the solution that I feel fell somewhat short in my prior comment.
If the premise of the statement that Area ABC = Area A'B'C' is true at some point of time then let the triangles be congruent, or rather, equilateral triangles.
Accepting that, let all 6 vertices be located on the circumference of a circle with centre at "O" for the extent of this discussion and ass defined in the problem body text. I just make one stipulation. Point "B" is located perpendicularly below "O" and is fixed.
Now "I" is defined as the incentre of ABC so that it initially has a coincidence with "O" with AB, BC and AC being tangent to the incircle at "I".
ObservationsMovement of the "I" allows for one of two situations:
1. Along the IB 'axis' forms varying isosceles triangles
whereas
2. veering left or right off that path creates scalene triangles.
In both cases the height and area of the initially formed triangle are reduced.
Considerations 1. Three points are generated under the rule (the A'B'C' triangle).
2. The point B' appears somewhere on the other side of point "O".
3. As "I" approaches "B" the angle ABC widens to eventually become obtuse; in the meantime the other two angles become more acute.
4. Let "I'" be the yet undefined incentre of triangle A'B'C'. As "I" approaches "B" then so does "I'" but with a difference. ABC begins to close in on "I" at the point "B" to become just a point whereas A'B'C' converges upon "I'" as a straight line from A' to, I'm thinking "B"/"I".
Although I have again failed in my proof, I think I've adequately explained the behaviour of the two incentres which I believe are at the root of a solution.
Praneeth, enjoyed having time to look into this, hope someone gives a more analytical dialogue.
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Posted by brianjn
on 2007-11-09 06:14:05 |