From-a-point-A-on-the-circum-ference-of-a-circle-of-radius-r-a-perpendicular-AF-is-dropped-on-a-tangent-to-the-circle-at-P-Find-the-maximum-possible-area-of-APF-

Question Number 15736 by ajfour last updated on 13/Jun/17

F r o m a p o i n t A o n t h e c i r c u m -

f e r e n c e o f a c i r c l e o f r a d i u s r, a

p e r p e n d i c u l a r A F i s d r o p p e d o n

a t a n g e n t t o t h e c i r c l e a t P .

F i n d t h e m a x i m u m p o s s i b l e

a r e a o f Δ A P F .

Answered by mrW1 last updated on 13/Jun/17

let θ = ∠ AOP

O = center of circle

PF = rsin θ

AF = r - rcos θ

A_{Δ APF} = \frac{1}{2} \times rsin θ \times (r - rcos θ) = \frac{r^{2}}{2} \sin θ (1 - \cos θ)

= \frac{r^{2}}{2} f (θ)

with f (θ) = \sin θ (1 - \cos θ)

\frac{df}{d θ} = \cos θ (1 - \cos θ) + \sin^{2} θ = \cos θ - {2 \cos}^{2} θ + 1 = 0

\cos θ = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} = 1, - \frac{1}{2}

\Rightarrow θ = 0 ° (not what we need)

\Rightarrow θ = 120 °

\max . A_{Δ APF} = \frac{r^{2}}{2} \times \sin 120 \times (1 - \cos 120)

= \frac{r^{2}}{2} \times \frac{\sqrt{3}}{2} \times (1 + \frac{1}{2})

= \frac{3 \sqrt{3} r^{2}}{8}

Commented by mrW1 last updated on 13/Jun/17

Commented by ajfour last updated on 13/Jun/17

g r e a t s i r, i s t h e r e a l s o a w a y

o f g e o m e t r y t o a r r i v e a t t h e

r e s u l t w i t h o u t h a v i n g t o

d i f f e r e n t i a t e . .

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