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Question Number 14306 by ajfour last updated on 30/May/17

Commented by prakash jain last updated on 30/May/17

$Are a, b and c sides of triangle ?$
Commented by ajfour last updated on 30/May/17

$y e s . s i d e s o f △ A B C .$
Commented by ajfour last updated on 30/May/17

$F i n d x, y, a n d z i n t e r m s o f$ $a, b, a n d c, (w h i c b a r e r e a l) .$
	Answered by ajfour last updated on 30/May/17

	$A s s u m e C a s t h e o r i g i n w i t b$ $s i d e a a l o n g x a x i s .$ $a l s o A (α, β) a n d M (h, k)$ $α^{2} + β^{2} = b^{2}$ ${(a - α)}^{2} + {(- β)}^{2} = c^{2}$ $s u b t r a c t i n g :$ $2 a α - a^{2} = b^{2} - c^{2}$ $α = \frac{a^{2} + b^{2} - c^{2}}{2 a}$ $β^{} = \pm \sqrt{b^{2} - α^{2}}$ $l e t m = \frac{k}{h}$ $s l o p e o f C M = m = \frac{k}{h}$ $s l o p e o f R C = - \frac{1}{s l o p e o f A M}$ $= - \frac{α - h}{β - k}$ $∠ b e t w e e n C M a n d R C = \frac{π}{6}$ $S o, \tan \frac{π}{6} = \frac{m + (\frac{α - h}{β - k})}{1 - (\frac{α - h}{β - k})} = \frac{1}{\sqrt{3}}$ $a s m = k / h$ $\frac{\sqrt{3} k}{h} + \sqrt{3} (\frac{α - h}{β - k}) = 1 - (\frac{α - h}{β - k})$ $o r,$ $\sqrt{3} k^{2} - k (α + β \sqrt{3})$ $= - h^{2} \sqrt{3} + α \sqrt{3} h - β h . . . (i)$ $A s i m i l a r s e c o n d c a s e,$ $s l o p e o f B Q = - \frac{1}{m} = - \frac{h}{k}$ $s l o p e o f B M = \frac{k}{h - a}$ $∠ b e t w e e n B M a n d B Q = \frac{π}{6}$ $S o, \frac{\frac{k}{h - a} + \frac{1}{m}}{1 - (\frac{k}{h - a}) \frac{1}{m}} = \frac{1}{\sqrt{3}}$ $w i t b m = \frac{k}{h}, t h i s l e a d s t o$ $\sqrt{3} k^{2} + a k = - h^{2} \sqrt{3} + a h \sqrt{3} . . . (i i)$ $(i i) - (i) g i v e s :$ $k (a + α + β \sqrt{3}) = h (a \sqrt{3} - α \sqrt{3} + β h)$ $m = \frac{k}{h} = \frac{(a - α) \sqrt{3} + β}{a + α + β \sqrt{3}}$ $r e p l a c i n g k = m h i n (i i)$ $\sqrt{3} (1 + m^{2}) h^{2} = (a \sqrt{3} - m) h$ $h {s h o u l d n}^{'} t b e z e r o, s o$ $h = \frac{a (\sqrt{3} - m)}{\sqrt{3} (1 + m^{2})} a n d k = m h$ $n o w,$ $y^{2} = h^{2} + k^{2} (s e e f i g u r e)$ $z^{2} = {(α - h)}^{2} + {(β - k)}^{2}$ $x^{2} = {(a - h)}^{2} + k^{2} .$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $i f a = 2, b = 3, a n d c = 4$ $w e h a v e α = \frac{a^{2} + b^{2} - c^{2}}{2 a} = - \frac{3}{4}$ $p o s i t i v e β = \sqrt{b^{2} - α^{2}} = \frac{\sqrt{135}}{4}$ $m = \frac{(a - α) \sqrt{3} + β}{a + α + β \sqrt{3}} = 1.222$ $h = \frac{a (\sqrt{3} - m)}{\sqrt{3} (1 + m^{2})} = 0.236$ $k = m h = 0.29$ $y = \sqrt{h^{2} + k^{2}} \approx 0.374$ $z = \sqrt{{(α - h)}^{2} + {(β - k)}^{2}} \approx 2.79$ $x = \sqrt{{(a - h)}^{2} + k^{2}} \approx 1.79$ $t h e n$ $x^{2} + y^{2} + x y = a^{2} = 4$ $y^{2} + z^{2} + y z = b^{2} = 9$ $z^{2} + x^{2} + x z = c^{2} = 16$ $i h a v e v e r i f i e d . . . .$
	Commented by ajfour last updated on 30/May/17

	$x^{2} + y^{2} + x y = a^{2}$ $\Rightarrow {(y + \frac{x}{2})}^{2} + {(\frac{x \sqrt{3}}{2})}^{2} = a^{2}$ $y + \frac{x}{2} = a \sin θ; \frac{x \sqrt{3}}{2} = a \cos θ$ $f r o m o t h e r t w o e q u a t i o n s$ $z + \frac{y}{2} = b \sin ϕ; \frac{y \sqrt{3}}{2} = b \cos ϕ$ $x + \frac{z}{2} = c \sin ψ; \frac{z \sqrt{3}}{2} = c \cos ψ .$ $h e n c e t h e g e o m e t r i c a l i d e a . .$ $o f a p r i o r q u e s t i o n o f A l g e b r a .$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/May/17

	$o k m r A j f o u r! i t i s a m a z i n g .$ $i t i s a p e r f e c t m e t h o d . i l o v e i t .$ $t h a n k y o u s o m u c h .$
	Commented by ajfour last updated on 30/May/17

	$t h a n k s f o r t h e a p p r e c i a t i o n S i r .$ $i h a v e n o t h o w e v e r c o n s i d e r e d$ $t h e s p e c i a l c a s e s w h e n$ $h = 0, k = 0, a^{2} + b^{2} = c^{2}, . . .$
	Commented by prakash jain last updated on 30/May/17

	$For a = 2, b = 3, c = 4$ $I got 4 s o l u t i o n s e t .$ $x = d, y = e, z = f is solution$ $than x = - d, y = - e, z = - f is also$ $a solution .$
	Commented by ajfour last updated on 30/May/17

	$a n y i n t e g e r o n e s ?$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 31/May/17

	$h e l l o m r A j f o u r .$ $t h i s i s a n i c e w a y t o s o l v e t h i s Q . b u t :$ $1) i s p o i n t : M, u n i q u e ?$ $2) ∡ Q B P \overset{?}{=} \frac{π}{6}$ $3) i t h i n k i f M, b e a p o i n t o f i n t e r s e c t i o n$ $o f s o m e s e g m e n t s o f Δ A B C, s u c h t h a t$ $m i d a n s, b i s e c t s, . . . . w e c a n s o l v e t h i s$ $v e r y e a s y b y u s i n g r e l a t i o n s a b o u t$ $a t r i a n g l e s e g m e n t s .$ $4) a v e r y s p i c i a l c a s e i n a t r i a n g l e,$ $w h e n M, i s t h e i n t e r s e c t i o n o f m i d a n s$ $o f Δ A B C a n d : ∠ A M B = ∠ B M C = ∠ C M A .$
	Commented by prakash jain last updated on 30/May/17

	$No integer solution .$ $all 4 I wrote in question 14211 .$ $There are only 4 solutions for$ $a = 2, b = 3, c = 4$ $One of solutions matches with$ $yours . I followed simplification$ $to quadractic in one variable$ $approach .$
	Commented by mrW1 last updated on 31/May/17

	$M r . a j f o u r : G r e a t j o b d o n e!$
	Commented by Tawa11 last updated on 15/Jan/22

	$Great sir$
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