Question-15006

Question Number 15006 by ajfour last updated on 06/Jun/17

Commented by RasheedSoomro last updated on 07/Jun/17

Which app do you use to draw so

nice figures ?!

Commented by ajfour last updated on 06/Jun/17

s o l u t i o n t o Q . 14940

s i d e s o f △ A O B d i v i d e d i n 3 p a r t s

(s p e c i a l c a s e) :

[i m a g e g o t p o s t e d a s q u e s t i o n s e v e r a l

t i m e s; i t o o n e e d t o c o m p l a i n t h i s i s s u e . .]

Commented by ajfour last updated on 07/Jun/17

L e k h d i a g r a m

Commented by Tinkutara last updated on 07/Jun/17

Please tell me also Sir! Which app is

this ?

Commented by RasheedSoomro last updated on 07/Jun/17

TO WHOM WHO INTREST IN GEOMETRY

A free geometry book in PDF .

The book is treasure of theorems .

For the book and about the book

see the image below

Commented by RasheedSoomro last updated on 07/Jun/17

Commented by Tinkutara last updated on 07/Jun/17

I found PDF of this book on net but

this book does not contain any questions

(problems) . It only contains figures .

How to find its question ?

Commented by RasheedSoomro last updated on 07/Jun/17

Statements of theorems (data and what to

prove both) can be understand from figures!

Answered by ajfour last updated on 06/Jun/17

e q n . o f U T :

\frac{x_{a}}{a} - \frac{y_{b}}{b} = 1 (t r i a n g u l a r c o o r d i n a t e s)

e q n . o f S T :

\frac{x_{a}}{2 a} + \frac{y}{b} = 1

s o l v i n g t h e m t o g e t T \equiv (\frac{4 a}{3}, \frac{b}{3})

e q n . o f R S :

\frac{x_{a}}{a} + \frac{y_{b}}{2 b} = 1

s o l v i n g w i t h t h a t f o r S T t o g e t

S \equiv (\frac{2 a}{3}, \frac{2 b}{3})

{A r e a}_{△ r e d} = \frac{9 a p}{2} - 3 a p = \frac{3 a p}{2}

{A r e a}_{△ S a T} = \frac{1}{2} (2 a - a) (y_{S} - y_{T})

= \frac{1}{2} (a) (\frac{2 p}{3} - \frac{p}{3}) = \frac{a p}{6}

{A r e a}_{P Q R S T U} = {A r e a}_{△ r e d} - 3 {A r e a}_{S a T}

= \frac{3 a p}{2} - 3 (\frac{a p}{6}) = a p

t o t a l a r e a o f △ A O B = \frac{9 a p}{2}

s o, \frac{{A r e a}_{P Q R S T U}}{{A r e a}_{△ A O B}} = \frac{a p}{(9 a p / 2)} = \frac{2}{9} .

Commented by mrW1 last updated on 06/Jun/17

very nice sir! just go on .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/Jun/17

n i c e s o l u t i o n m r A j f o u r! t h a n k s .

i l o v e t h i s .

Commented by ajfour last updated on 06/Jun/17

t h a n k s s i r, s y m m e t r y i s a

b e a u t i f u l p r o p e r t y . .

Commented by mrW1 last updated on 07/Jun/17

mr ajfour :

you only need to replace 2 a with (n - 1) a

and 2 b with (n - 1) b, then you can

get the general solution for case n .

To make the thing a little bit easier,

you can put point A on the y - axis,

since this will not change the areas

as we know .

Commented by ajfour last updated on 07/Jun/17

A l l r i g h t S i r, s h a l l a t t e m p t . .

Commented by mrW1 last updated on 07/Jun/17

G reat job sir! F antastic!

Commented by ajfour last updated on 07/Jun/17

G o t t h e a n s w e r S i r,

f o r n ⩾ 2 a n d n \in N

\frac{{A r e a}_{P Q R S T}}{{A r e a}_{△ A O B}} = \frac{2 {(n^{2} - 3 n + 3)}^{2}}{n^{3} (2 n - 3)} .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

m r A j f o u r! p l e a s e p o s t y o u r s o l u t i o n h e r .