ABC is a triangle P is a point on the plane of the triangle such that |AR(PAB)| = |AR(PAC)| = |AR(PBC)|. Find possible number of positions of P.
Solution:
The answer is Four
But how...???....think !!!...let me know if you want to know
Solution:
The answer is Four
But how...???....think !!!...let me know if you want to know
The first points is the centroid: resulting area = A/3
ReplyDeleteThen there are three points on each line containing the medians at a distance two times the length of the median from the respective vertices. resulting area = A.
Geometric Proofs (you may imagine/draw it):
@centroid: it is the "center of mass" of the triangle assuming uniform mass density.
@exterior points on the median: draw lines parallel to the edges through the opposite vertex. in this way you will get three parallelograms. The three points are the intersections of these parallel lines. The result can be shown using the fact that the diagonals bisect the parallelogram into equal parts.
One can even use vector expressions for area of a triangle (area = 0.5* (a x b); a,b are edge vectors of a triangle) to get these results.
Dr Prashanth
ReplyDeleteU r right in all we have 4 posiible positions for `p`