Question-15440

Question Number 15440 by mrW1 last updated on 10/Jun/17

Commented by mrW1 last updated on 10/Jun/17

The sides of a convex quadrilateral

ABCD are elongated 2 times in one

direction, see diagram, to form a new

quadrilateral EFGH . What is the

area of the new quadrilateral if the

area of ABCD is 1 .

What is the result, if the sides will be

elongated n times ?

Commented by ajfour last updated on 10/Jun/17

n^{2} + n + 1

i s i t c o r r e c t s i r ?

Commented by mrW1 last updated on 10/Jun/17

no, this is not correct .

Commented by ajfour last updated on 10/Jun/17

Commented by ajfour last updated on 10/Jun/17

E x p r e s s i o n f o r A r e a o f o r i g i n a l

q u a d r i l a t e r a l :

l e t a r e a o f o r i g i n a l q u a d r i l a t e r a l

b e A_{0} . T h e n

2 A_{0} = \frac{1}{2} (a b \sin β + b c \sin γ + c d \sin δ

+ d a \sin α)

\Rightarrow 2 A_{0} = \frac{1}{2} Σ a b \sin β

o r Σ a b \sin β = 4 A_{0} .

T h i s i s u s e d i n n e x t f i g u r e

w b e r e i n t h e s o l u t i o n i s c o n t i n u e d \dots

Commented by ajfour last updated on 10/Jun/17

Commented by ajfour last updated on 10/Jun/17

l e t a r e a o f n e w q u a d r i l a t e r a l

b e S .

S = A_{0} + \frac{1}{2} [n a (n + 1) b \sin β

+ n b (n + 1) c \sin γ

+ n c (n + 1) d \sin δ

+ n d (n + 1) a \sin α]

\Rightarrow S = A_{0} + \frac{n (n + 1)}{2} Σ a b \sin β

i t h a s b e e n f o u n d i n p r e v i o u s

c o m m e n t t h a t Σ a b \sin β = 4 A_{0},

t h e r e f o r e

S = A_{0} + \frac{n (n + 1)}{2} (4 A_{0})

S = A_{0} (1 + 2 n^{2} + 2 n)

o r \frac{S}{A_{0}} = 2 n^{2} + 2 n + 1 .

Commented by mrW1 last updated on 10/Jun/17

Answer {2 n}^{2} + 2 n + 1 is correct!

Commented by ajfour last updated on 10/Jun/17

w i t h n = 2, \frac{S}{A_{0}} = 13 .

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 11/Jun/17

∡ A D C = α, ∡ A B C = β, ∡ D C B = θ, ∡ D A B = δ

A B = a, B C = b, C D = c, D A = d

S_{A B C D} = \frac{1}{2} d c . s i n α + \frac{1}{2} a b . s i n β = S_{1}

S_{A B C D} = \frac{1}{2} b c . s i n θ + \frac{1}{2} d a . s i n δ = S_{1}

S_{1} + \frac{1}{2} . 2 b . 3 c . s i n (π - θ) + \frac{1}{2} . 2 c . 3 d . s i n (π - α) +

+ \frac{1}{2} . 2 d . 3 a . s i n (π - δ) + \frac{1}{2} . 2 a . 3 b . s i n (π - β) = S

S_{1} + 6 \times (\frac{1}{2} b c . s i n θ + \frac{1}{2} d a s i n δ) +

+ 6 \times (\frac{1}{2} c d . s i n α + \frac{1}{2} a b s i n β) = S

\Rightarrow S_{1} + 6 S_{1} + 6 S_{1} = S \Rightarrow 13 S_{1} = S .

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 11/Jun/17

i n c a s e o f : n

S_{1} + 2 n (n + 1) S_{1} = S

\Rightarrow \frac{S}{S_{1}} = 2 n^{2} + 2 n + 1 .

Commented by mrW1 last updated on 11/Jun/17

{That}^{'} s right .

Answered by mrW1 last updated on 11/Jun/17

I have an other solution without using

trigonometry and analytic geometry :

A_{Δ BCF} = {nA}_{Δ ACB}

A_{Δ BGF} = (n + 1) A_{Δ BCF} = (n + 1) {nA}_{Δ ACB}

similarly

A_{Δ DEH} = (n + 1) {nA}_{Δ ACD}

\Rightarrow A_{Δ BGF} + A_{Δ DEH} = (n + 1) n (A_{Δ ACB} + A_{Δ ACD}) = (n + 1) {nA}_{ABCD}

similarly

\Rightarrow A_{Δ CHG} + A_{Δ AFE} = (n + 1) n (A_{Δ CDB} + A_{Δ ADB}) = (n + 1) {nA}_{ABCD}

A_{EFGH} = A_{Δ BGF} + A_{Δ DEH} + A_{Δ BGF} + A_{Δ DEH} + A_{ABCD}

= 2 (n + 1) {nA}_{ABCD} + A_{ABCD}

= [2 n (n + 1) + 1] A_{ABCD}

= 2 n (n + 1) + 1

= {2 n}^{2} + 2 n + 1

Commented by mrW1 last updated on 11/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 11/Jun/17

n i c e & s m a r t . b u t : n^{2} + 2 n + 1 (o r 2) ?

Commented by mrW1 last updated on 11/Jun/17

thank you! {2 n}^{2} + 2 n + 1 is correct .

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