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BP/PC=CQ/QA=AR/BR;show that AP^2+BQ^2+CR^2 iis least when P,Q,R bisect the sides

BP/PC=CQ/QA=AR/BR;show that AP^2+BQ^2+CR^2 iis least when P,Q,R bisect the sides

Grade:12

1 Answers

mycroft holmes
272 Points
7 years ago
Let the ratios be equal to t.
 
We know that  2AP^2 + BP^2 +CP^2 = AB^2+AC^2 .
 
where BP^2 = \frac{t^2}{(t+1)^2} BC^2 and CP^2 = \frac{1}{(t+1)^2} BC^2
 
Hence 2AP^2 = AB^2+AC^2 - BC^2 \left (\frac{1+t^2}{(t+1)^2} \right)
 
Adding three similar relations and simplifying we get
 
2(AP^2+BQ^2+CR^2)= \left(AB^2+AC^2 +BC^2 \right) \left (2 -\frac{1+t^2}{(t+1)^2} \right)
 
Hence LHS is minimized when the expression \frac{1+t^2}{(t+1)^2} is minimized i.e
 
when \frac{(t+1)^2}{1+t^2} is maximized  which is when \frac{2t}{1+t^2} is maximised.
 
\frac{2t}{1+t^2} = \frac{2}{t+\frac{1}{t}} \le 1 as by AM_GM Ineqt+\frac{1}{t} \ge 2 with equality when t=1 i.e. when BP=PC, CQ = QA, AR = BR etc
 
Hence AP^2+BQ^2+CR^2 is minimized when P,Q, R bisect BC, CA, AP resp.

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