Problem: ABC is an isosceles right angled triangle with AB = BC = 1. If P is any point on BC, then min{max{(area APQ), (area PQBR), area PRC)}} is

(A) 2/9

(B) 1/9

(C) 1/3

(D) none of these

Sol. (A). Let B ≡ (0, 0), P ≡ (h, k)

⇒ area of APQ = $\large \frac{1}{2}h^2$

area of PQBR = hk

area of PRC = $\large \frac{1}{2}k^2$

Clearly max{area APQ, area PQBR, area PRC}

= Area APQ $\large = \frac{1}{2}h^2 [for \frac{h}{k}\ge 2 ] \ge \frac{2}{9}$

= Area PQBR for h k $\large [for \frac{1}{2} \le \frac{h}{k} < 2 ] > \frac{2}{9}$

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