In a triangle ABC

Question

In a triangle ΔABC, it is known that AB = 10, AC = 9, cos ∠BAC = 3/8. Points P, Q on sides AB, AC such that the area of ΔAPQ is half of the area of ΔABC. For all points P, Q satisfying the above, the smallest possible value of PQ is....

Answer 1

I will not solve this completely for you, but I will show you how to do it.

The area of ΔABC is (1/2)*10*9*sin ∠BAC.

The area of ΔPAQ is (1/2)*AP*AQ* sin ∠BAC

Equating areas gives AP*AQ=45 (remember the factor of 1/2!)

Now substitute AP=45/AQ into the law of cosines.as a nasty looking

Let y=length of PQ and x=length of AQ

y² = x² + (45/x)² - 2*45*(3/8)

Differentiate implicitly and after some simplification since y≠0, you will get

2x-(2*45²)/x³ = 0 and then x=3√5

I'm sorry this took so many tries and so long!