#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

mycroft holmes
272 Points
4 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 Ineq$t+\frac{1}{t} \ge 2$ 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.