If-one-vertex-of-the-triangle-having-maximum-area-that-can-be-inscribed-in-the-circle-z-i-5-is-3-3i-then-find-other-vertices-of-triangle-

Question Number 32184 by rahul 19 last updated on 21/Mar/18

\boldsymbol I f o n e v e r t e x o f t h e t r i a n g l e h a v i n g

m a x i m u m a r e a t h a t c a n b e i n s c r i b e d

i n t h e c i r c l e ∣ \boldsymbol z - \boldsymbol i ∣= 5 i s 3 - 3 \boldsymbol i, t h e n

f i n d o t h e r v e r t i c e s o f t r i a n g l e .

Answered by MJS last updated on 21/Mar/18

the triangle with maximum area

is equilateral

the given circle has {it}^{'} s center

at z = i, so we move everything

to : ∣ z ∣= 5 and the vertex to

A^{'} = 3 - 4 i = {5 e}^{i (\tan^{- 1} \frac{3}{4} - \frac{π}{2})}

the other vertices are

B^{'} = {5 e}^{i (\tan^{- 1} \frac{3}{4} + \frac{π}{6})} = (2 \sqrt{3} - \frac{3}{2}) + (\frac{3 \sqrt{3}}{2} + 2) i

C^{'} = {5 e}^{i (\tan^{- 1} \frac{3}{4} - \frac{7 π}{6})} = (- 2 \sqrt{3} - \frac{3}{2}) + (- \frac{3 \sqrt{3}}{3} + 2) i

now we move them back :

A = A^{'} + i = 3 - 3 i

B = B^{'} + i = (2 \sqrt{3} - \frac{3}{2}) + (\frac{3 \sqrt{3}}{2} + 3) i

C = C^{'} + i = (- 2 \sqrt{3} - \frac{3}{2}) + (- \frac{3 \sqrt{3}}{2} + 3) i

Commented by rahul 19 last updated on 21/Mar/18

t h a n k u s i r!

t h a n k u s i r!