InterviewSolution
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InterviewSolution
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If O(vec0) is the circumcentre and O' the orthocentre of a triangle ABC, then prove that i. vec(OA)+vec(OB)+vec(OC)=vec(OO') ii. vec(O'A)+vec(O'B)+vec(O'C)=2vec(O'O) iii. vec(AO')+vec(O'B)+vec(O'C)=2 vec(AO)=vec(AP) where AP is the diameter through A of the circumcircle. |
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Answer» Solution :O is the CIRCUMCENTRE, which is the intersection of the right bisectiors of the sides of the triangle, and O' is the ORTHOCENTER, which is the point of intersection of ALTITUDES DRAWN from the vertices. Also, from geometry, we know that `""2OD=AO'` `therefore""2vec(OD)=VEC(AO')""(i)` i. To prove : `vec(OA)+vec(OB)+vec(OC)=vec(OO')` Now `vec(OB)+vec(OC)=2vec(OD)=vec(AO')` `rArr" "vec(OA)+vec(OB)+vec(OC)=vec(OA)+vec(AO')=vec(OO')""`[by (i)] ii. To prove : `vec(O'A)+vec(O'B)+vec(O'C)=2vec(OO')` `"""L.H.S."=2vec(DO)+2vec(O'D)""`[by (i)] `""2(vec(O'D)+vec(DO))=2vec(O'O)` iii. To prove : `vec(AO')+vec(O'B)+vec(O'C)=2vec(AO)=vec(AP)` `"""L.H.S."=2vec(AO')-vec(AO')+vec(O'B)+vec(O'C)` `""=2vec(AO')+(vec(O'A)+vec(O'B)+vec(O'C))` `""=2vec(AO')+2vec(O'O)=2vec(AO)` `""=vec(AP)` (where `AP` is the diameter through `A` of the circumcircle). 
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