KVPY SX SB Exam 2012 Question Paper

Question : 17 of 120

Marks: +1, -0

Let H be the orthocentre o f an acute-angled triangle ABC and O be its circumcenter. Then

\ov {HA}+\ov {HB}+\ov {HC}=?

Solution:

G is centroid

G={A+B+C}/3

G={2O+H}/3

2O+H=3G

\ov {HA}+\ov {HB}+\ov {HC}

=\ov A-\ov H+\ov B-\ov H+\ov C-\ov H

=\ov A+\ov B+\ov C-3\ov H

=2\ov G-3\ov H

=2\ov O+\ov H-3\ov H

==2\ov O-2\ov H=2\ov {OH}

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