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Question Number 15284 by ajfour last updated on 09/Jun/17

Commented by ajfour last updated on 09/Jun/17

$Q . 14940 (g e n e r a l p r o o f : s i d e s$ $o f Δ A B C d i v i d e d i n n e q u a l p a r t s)$
Commented by RasheedSoomro last updated on 09/Jun/17

$Sir you have told elsewhere that you$ $use L e k h d i a g r a m .$ $C an we draw scaled figures with it like$ $geogebra ?$
Commented by ajfour last updated on 10/Jun/17

$i t i s n o t h i n g c o m p a r e d t o g e o g e b r a,$ $i u s e t h e s e o t h e r a p p s c o z m y$ $o l d a n d s i m p l e d e v i c e d o n t s u p p o r t$ $g e o g e b r a a n d m a n y o t h e r s . .$
	Answered by ajfour last updated on 09/Jun/17

	$O r i g i n : c e n t r o i d o f Δ A B C o f$ $s i d e s A B = n c, B C = n a, A C = n b$ $a + b + c = 0 (v e c t o r s u m)$ $a \times a = b \times b = c \times c = 0$ $a \times b = b \times c = c \times a = 2 t_{0}$ $\forall t_{0} i s a r e a o f t r i a n g l e w i t b s i d e s$ $a, b, a n d c .$ $p o s i t i o n v e c t o r o f p o i n t B :$ ${\bar{r}}_{B} = (\frac{1}{2}) (\frac{2}{3}) (- n a) + (\frac{1}{3}) (n c)$ $= \frac{n}{3} (c - a) .$ $E q n o f l i n e I G :$ $\bar{r} = O I + λ (I G)$ $= {\bar{r}}_{B} + a + λ [(n - 2) a - c] . . . (i)$ $E q n o f l i n e E F :$ $\bar{r} = O E + μ (E F)$ $= {\bar{r}}_{B} - c + μ [c + (n - 1) a] . . . (i i)$ $P i s i n t e r s e c t i o n o f I G a n d E F$ $e q u a t i n g (i i) a n d (i) f o r p o i n t P :$ $a [1 + λ (n - 2) - μ (n - 1)] +$ $c [- λ + 1 - μ] = 0$ $\Rightarrow λ (n - 2) - μ (n - 1) = - 1 a n d$ $λ + μ = 1$ $o r λ (n - 1) + μ (n - 1) = n - 1$ $a d d i n g t o g e t λ = \frac{n - 2}{2 n - 3} = \frac{I P}{I G}$ $μ = 1 - \frac{(n - 2)}{2 n - 3} = \frac{n - 1}{2 n - 3} = \frac{E T}{E F}$ $E q n o f H T :$ $\bar{r} = O I - ρ (H I)$ $= \frac{n}{3} (c - a) + a - ρ [(n - 1) c + a]$ $E q n o f E F :$ $\bar{r} = \frac{n}{3} (c - a) - c + κ [(c + (n - 1) a]$ $e q u a t i n g t h e m f o r p o i n t T :$ $a [1 - ρ - κ (n - 1)] + c [- ρ (n - 1) + 1 - κ] = 0$ $\Rightarrow ρ + κ (n - 1) = 1 . . . . (i i i) a n d$ $ρ (n - 1) + κ = 1$ $o r ρ {(n - 1)}^{2} + κ (n - 1) = n - 1 . . (i v)$ $(i v) - (i i i) g i v e s :$ $ρ = \frac{n - 2}{{(n - 1)}^{2} - 1} = \frac{1}{n} = \frac{T I}{H I}$ $κ = 1 - ρ (n - 1) = 1 - \frac{(n - 1)}{n} = \frac{1}{n}$ $= \frac{E T}{E F} .$ $A r e a (Δ T I P) = (I P \times I T) / 2$ $= \frac{λ (I G) \times (- ρ) (H I)}{2}$ $= (\frac{1}{2 n}) (\frac{n - 2}{2 n - 3}) [(n - 2) a - c] \times [- (n - 1) c - a]$ $= \frac{(n - 2)}{2 n (2 n - 3)} [(n - 1) (n - 2) + 1] (c \times a)$ $= \frac{(n - 2) (n^{2} - 3 n + 3)}{n (2 n - 3)} t_{0}$ $A r e a (Δ I G H) = \frac{I G \times I H}{2}$ $= \frac{[(n - 2) a - c] \times [- a - (n - 1) c]}{2}$ $= \frac{[(n - 1) (n - 2) + 1]}{2} (c \times a)$ $= (n^{2} - 3 n + 3) t_{0} .$ $A r e a (c o n s i d e r e d h e x a g o n)$ $= A r e a (Δ I G H) - 3 A r e a (Δ T I P)$ $= (n^{2} - 3 n + 3) t_{0} - \frac{3 (n - 2) (n^{2} - 3 n + 3) t_{0}}{n (2 n - 3)}$ $= (n^{2} - 3 n + 3) [1 - \frac{3 (n - 2)}{n (2 n - 3)}] t_{0}$ $= (n^{2} - 3 n + 3) \frac{(2 n^{2} - 6 n + 6)}{n (2 n - 3)} t_{0}$ $= \frac{2 {(n^{2} - 3 n + 3)}^{2}}{n (2 n - 3)} t_{0} .$ $A r e a (Δ A B C) = \frac{n c \times n a}{2}$ $\frac{n^{2}}{2} (c \times a) = \frac{n^{2}}{2} (2 t_{0}) = n^{2} t_{0}$ $\frac{A r e a (h e x a g o n)}{A r e a (Δ A B C)} = \frac{2 {(n^{2} - 3 n + 3)}^{2}}{n^{3} (2 n - 3)} .$
	Commented by mrW1 last updated on 09/Jun/17

	$Very beautiful work sir! congratulation!$ $Since you have the points T and P,$ $I have expected, that you will calculate$ $the area of hexagon directly in this way :$ $A_{Hexagon} = 6 \times A_{Δ OTP}$ $Can you check if this will get the same$ $result, but the process will be shorter .$
	Commented by ajfour last updated on 09/Jun/17

	$y e s S i r, c o m p l e t e l y t r u e .$ $I c h e c k e d i t t h r o u g h, s e e m s t e d i o u s$ $a g a i n b u t t h e y f a c t o r i z e o u t .$ $T h a n k s f o r t h e g e n u i n e s u g g e s t i o n .$
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