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Question Number 65491 by mr W last updated on 30/Jul/19

Commented by mr W last updated on 30/Jul/19

$[Q 65312 r e p o s t e d]$ $f i n d x = ?$
	Answered by mr W last updated on 30/Jul/19

	Commented by mr W last updated on 30/Jul/19

	$I d e v e l o p e d f o l l o w i n g s o l u t i o n a f t e r$ $s o m e t r i a l & e r r o r .$ $v e c t o r m e t h o d s h o u l d a l s o b e a g o o d$ $a p p r o a c h . a j f o u r s i r : c a n y o u t r y i t ?$ $t h i s i s m y s o l u t i o n :$ $l e t$ $O A = a, O B = b, O C = c, O D = d$ $O E = e, O F = f, O G = g, O H = h$ $S A = p, S E = q$ $u s i n g {M e n e l a u s}^{'} t h e o r e m (s e e b e l o w) :$ $\frac{q + 2}{3} \times \frac{g}{c} \times \frac{1}{p + 1} = 1 . . . (i)$ $\frac{q}{5} \times \frac{g}{c} \times \frac{2}{p} = 1 . . . (i i)$ $\frac{q}{5 + x} \times \frac{h}{d} \times \frac{3}{p} = 1 . . . (i i i)$ $\frac{q + 2}{3 + x} \times \frac{h}{d} \times \frac{2}{p + 1} = 1 . . . (i v)$ $(i) \div (i i) :$ $\frac{q + 2}{3} \times \frac{5}{q} \times \frac{1}{p + 1} \times \frac{p}{2} = 1 . . . (I)$ $(i i i) \div (i v) :$ $\frac{q}{5 + x} \times \frac{3 + x}{q + 2} \times \frac{3}{p} \times \frac{p + 1}{2} = 1 . . . (I I)$ $(I) \times (I I) :$ $\frac{5 \times 1 \times (3 + x) \times 3}{3 \times 2 \times (5 + x) \times 2} = 1$ $\frac{5 (3 + x)}{4 (5 + x)} = 1$ $\Rightarrow 15 + 5 x = 20 + 4 x$ $\Rightarrow x = 5$
	Commented by mr W last updated on 30/Jul/19

	Commented by MJS last updated on 31/Jul/19

	$what I did to find the solution was the$ $coordinates - method once more but I had$ $no time to type it . anyway I {didn}^{'} t expect$ $an answer from the original poster . . .$
	Commented by Tony Lin last updated on 31/Jul/19

	$a c t u a l l y t h e r e i s a s p e c i a l w a y t o s o l v e$ $t h i s q u e s t i o n c a l l e d ‘ ‘ c r o s s {r a t i o}^{″}$ $R (A, B; C, D) = R (E, F; G, H)$ $\frac{\frac{A C}{A D}}{\frac{B C}{B D}} = \frac{\frac{E G}{E H}}{\frac{F G}{F H}}$ $\frac{\frac{2}{3}}{\frac{1}{2}} = \frac{\frac{5}{5 + x}}{\frac{3}{3 + x}}$ $20 + 4 x = 15 + 5 x$ $x = 5$ $b u t i h o p e t h a t t h e r e a r e s o m e s i m p l e$ $w a y s t o s o l v e w i t h o u t a t h e o r e m t h a t$ $f e w p e o p l e h e a r d$ $t h a n k s f o r y o u r h e l p$
	Commented by Tony Lin last updated on 31/Jul/19

	Commented by Tony Lin last updated on 31/Jul/19

	Commented by Tony Lin last updated on 31/Jul/19
	$without loss of generality ⇒using coordinate-system hypothesis let O(0,0),A(0,−s),B(1,−s),C(2,−s) D(3,−s),E(0,−t),F((6/5),−t−(8/5)) G(3,−t−4),H(x,y) O,B,F are collinear ⇒t+(8/5)=(6/5)s O,C,G are collinear ⇒t+4=(3/2)s ⇒ { ((t+(8/5)=(6/5)s)),((t+4=(3/2)s)) :} ⇒(t,s)=(8,8) ∵H is the point of intersection of L_(OD) &L_(EG) ∴ { ((y=((−8)/3)x)),((y=((−4)/3)x−8)) :} ⇒(x,y)=(6,−16) GH=(√((6−3)^2 +[−16−(−12)]^2 ))=5 i only think of this idea now$
	$w i t h o u t l o s s o f g e n e r a l i t y$ $\Rightarrow u s i n g c o o r d i n a t e - s y s t e m h y p o t h e s i s$ $l e t O (0, 0), A (0, - s), B (1, - s), C (2, - s)$ $D (3, - s), E (0, - t), F (\frac{6}{5}, - t - \frac{8}{5})$ $G (3, - t - 4), H (x, y)$ $O, B, F a r e c o l l i n e a r$ $\Rightarrow t + \frac{8}{5} = \frac{6}{5} s$ $O, C, G a r e c o l l i n e a r$ $\Rightarrow t + 4 = \frac{3}{2} s$ $\Rightarrow {\begin{cases} t + \frac{8}{5} = \frac{6}{5} s \\ t + 4 = \frac{3}{2} s \end{cases}$ $\Rightarrow (t, s) = (8, 8)$ $∵ H i s t h e p o i n t o f i n t e r s e c t i o n o f L_{O D} & L_{E G}$ $∴ {\begin{cases} y = \frac{- 8}{3} x \\ y = \frac{- 4}{3} x - 8 \end{cases}$ $\Rightarrow (x, y) = (6, - 16)$ $G H = \sqrt{{(6 - 3)}^{2} + {[- 16 - (- 12)]}^{2}} = 5$ $i o n l y t h i n k o f t h i s i d e a n o w$
	Commented by MJS last updated on 31/Jul/19

	$. . . this is exactly what I did .$ $thank you both for above theorems$
	Commented by mr W last updated on 31/Jul/19

	$L i n s i r, t h a n k s f o r a l l t h e i n f o r m a t i o n!$
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