Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 166520 by ajfour last updated on 21/Feb/22

Commented by ajfour last updated on 21/Feb/22

$I f t h e c i r c l e s h a v e r a d i i a, b, c$ $f i n d m a x i m u m s i d e l e n g t h o f$ $s u c h a n e q u i l a t e r a l t r i a n g l e .$
Commented by mr W last updated on 22/Feb/22

$t h e g e n e r a l c a s e s e e Q 16194 .$
	Answered by mr W last updated on 21/Feb/22

	Commented by mr W last updated on 21/Feb/22

	$t h i s i s t o f i n d t h e t r i a n g l e w i t h t h e$ $m a x i m u m a r e a, n o t t h e m a x i m u m$ $e q u i l a t e r a l t r i a n g l e .$ $\tan α = \frac{(1 + \cos β) b + (1 + \cos γ) c}{(1 + \sin β) b + (1 + \sin γ) c}$ $\cos (α + γ) = (\frac{b}{c} + 1) (\sin α - \cos α) - \frac{b}{c} \cos (α + β)$ $\tan β = \frac{(1 + \cos α) a + (1 + \cos γ) c}{(1 + \sin α) a - (1 + \sin γ) c}$ $\cos (β - γ) = (\frac{a}{c} - 1) \sin β - (\frac{a}{c} + 1) \cos β - \frac{a}{c} \cos (α + β)$ $\tan γ = \frac{(1 + \cos α) a - (1 + \cos β) b}{(1 + \sin α) a + (1 + \sin β) b}$ $s o l v e f o r α, β, γ .$
	Commented by mr W last updated on 21/Feb/22

	Commented by mr W last updated on 21/Feb/22

	Answered by mr W last updated on 22/Feb/22

	Commented by mr W last updated on 22/Feb/22

	$A (a, a)$ $B (- b, b)$ $C (c, - c)$ $s a y G (h, k)$ $e q n . o f G P :$ $\frac{y - k}{a - k} = \frac{x - h}{a - h}$ $(a - h) y = (a - k) x + a (k - h)$ $y = \frac{a - k}{a - h} x + \frac{a (k - h)}{a - h}$ $e q n . o f c i r c l e A :$ ${(x - a)}^{2} + {(y - a)}^{2} = a^{2}$ ${(x - a)}^{2} + {(\frac{a - k}{a - h} x + \frac{a (k - h)}{a - h} - a)}^{2} = a^{2}$ $[{(a - h)}^{2} + {(a - k)}^{2}] {(x - a)}^{2} = a^{2} {(a - h)}^{2}$ $x - a = \pm \frac{a (a - h)}{\sqrt{{(a - h)}^{2} + {(a - k)}^{2}}}$ $x_{P} = a + \frac{a}{\sqrt{1 + {(\frac{a - k}{a - h})}^{2}}}$ $y_{P} = a + \frac{a}{\sqrt{1 + {(\frac{a - h}{a - k})}^{2}}}$ $e q n . o f G B :$ $\frac{y - k}{b - k} = \frac{x - h}{- b - h}$ $- (b + h) y = (b - k) x - b (h + k)$ $e q n . o f c i r c l e B :$ ${(x + b)}^{2} + {(y - b)}^{2} = b^{2}$ $[{(b + h)}^{2} + {(b - k)}^{2}] {(x + b)}^{2} = b^{2} {(b + h)}^{2}$ $x + b = \pm \frac{b (b + h)}{\sqrt{{(b + h)}^{2} + {(b - k)}^{2}}}$ $x_{Q} = - b - \frac{b}{\sqrt{1 + {(\frac{b - k}{b + h})}^{2}}}$ $y_{Q} = b + \frac{b}{\sqrt{1 + {(\frac{b + h}{b - k})}^{2}}}$ $e q n . o f G C :$ $\frac{y - k}{- c - k} = \frac{x - h}{c - h}$ $(c - h) y = - (c + k) x + c (h + k)$ $e q n . o f c i r c l e C :$ ${(x - c)}^{2} + {(y + c)}^{2} = c^{2}$ $[{(c - h)}^{2} + {(c + k)}^{2}] {(x - c)}^{2} = c^{2} {(c - h)}^{2}$ $x - c = \pm \frac{c (c - h)}{\sqrt{{(c - h)}^{2} + {(c + k)}^{2}}}$ $x_{R} = c + \frac{c}{\sqrt{1 + {(\frac{c + k}{c - h})}^{2}}}$ $y_{R} = - c - \frac{c}{\sqrt{1 + {(\frac{c - h}{c + k})}^{2}}}$ $s^{2} = {(x_{P} - x_{Q})}^{2} + {(y_{P} - y_{Q})}^{2}$ $s^{2} = {(a + b + \frac{a}{\sqrt{1 + {(\frac{a - k}{a - h})}^{2}}} + \frac{b}{\sqrt{1 + {(\frac{b - k}{b + h})}^{2}}})}^{2} + {(a - b + \frac{a}{\sqrt{1 + {(\frac{a - h}{a - k})}^{2}}} - \frac{b}{\sqrt{1 + {(\frac{b + h}{b - k})}^{2}}})}^{2}$ $s^{2} = {(x_{P} - x_{R})}^{2} + {(y_{P} - y_{R})}^{2}$ $s^{2} = {(a - c + \frac{a}{\sqrt{1 + {(\frac{a - k}{a - h})}^{2}}} - \frac{c}{\sqrt{1 + {(\frac{c + k}{c - h})}^{2}}})}^{2} + {(a + c + \frac{a}{\sqrt{1 + {(\frac{a - h}{a - k})}^{2}}} + \frac{c}{\sqrt{1 + {(\frac{c - h}{c + k})}^{2}}})}^{2}$ $s^{2} = {(x_{Q} - x_{R})}^{2} + {(y_{Q} - y_{R})}^{2}$ $s^{2} = {(b + c + \frac{b}{\sqrt{1 + {(\frac{b - k}{b + h})}^{2}}} + \frac{c}{\sqrt{1 + {(\frac{c + k}{c - h})}^{2}}})}^{2} + {(b + c + \frac{b}{\sqrt{1 + {(\frac{b + h}{b - k})}^{2}}} + \frac{c}{\sqrt{1 + {(\frac{c - h}{c + k})}^{2}}})}^{2}$ $s o l v e f o r h, k, s$ $. . . .$
	Commented by ajfour last updated on 22/Feb/22
	$A(a,a) P(a+acos θ, a+asin θ) Q(a+acos θ−scos φ, a+asin θ−ssin φ) (a−b+acos θ−scos φ)^2 +(a−b+asin θ−ssin φ)^2 =b^2 ⇒ a^2 +2a{cos θ(a−b−scos φ) +sin θ(a−b−ssin φ)} +(a−b−scos φ)^2 +(a−b−ssin φ)^2 =b^2 (√((a−b−scos φ)^2 +(a−b−ssin φ)^2 )) =m mcos (θ−δ)=b^2 −a^2 −m^2 tan δ=((a−b−ssin φ)/(a−b−scos φ)) θ=δ+cos^(−1) (((b^2 −a^2 −m^2 )/m)) ..(i) M midpoint of PQ R{a+acos θ−((scos φ)/2)−((s(√3))/2)sin φ, a+asin θ−((ssin φ)/2)−((s(√3))/2)cos φ} And as (RC)^2 =c^2 ⇒ (a+c+acos θ−((scos φ)/2)−((s(√3))/2)sin φ)^2 +(a+c+asin θ−((ssin φ)/2)−((s(√3))/2)cos φ)^2 =c^2 using (i) we have θ=f(φ) now g(s,φ)=0 s_(max) can be found from g(s,φ) graph.$
	$A (a, a)$ $P (a + a \cos θ, a + a \sin θ)$ $Q (a + a \cos θ - s \cos ϕ, a + a \sin θ - s \sin ϕ)$ ${(a - b + a \cos θ - s \cos ϕ)}^{2}$ $+ {(a - b + a \sin θ - s \sin ϕ)}^{2} = b^{2}$ $\Rightarrow a^{2} + 2 a {\cos θ (a - b - s \cos ϕ)$ $+ \sin θ (a - b - s \sin ϕ)}$ $+ {(a - b - s \cos ϕ)}^{2} + {(a - b - s \sin ϕ)}^{2}$ $= b^{2}$ $\sqrt{{(a - b - s \cos ϕ)}^{2} + {(a - b - s \sin ϕ)}^{2}}$ $= m$ $m \cos (θ - δ) = b^{2} - a^{2} - m^{2}$ $\tan δ = \frac{a - b - s \sin ϕ}{a - b - s \cos ϕ}$ $θ = δ + \cos^{- 1} (\frac{b^{2} - a^{2} - m^{2}}{m}) . . (i)$ $M m i d p o i n t o f P Q$ $R {a + a \cos θ - \frac{s \cos ϕ}{2} - \frac{s \sqrt{3}}{2} \sin ϕ,$ $a + a \sin θ - \frac{s \sin ϕ}{2} - \frac{s \sqrt{3}}{2} \cos ϕ}$ $A n d a s {(R C)}^{2} = c^{2} \Rightarrow$ ${(a + c + a \cos θ - \frac{s \cos ϕ}{2} - \frac{s \sqrt{3}}{2} \sin ϕ)}^{2}$ $+ {(a + c + a \sin θ - \frac{s \sin ϕ}{2} - \frac{s \sqrt{3}}{2} \cos ϕ)}^{2}$ $= c^{2}$ $u s i n g (i) w e h a v e θ = f (ϕ)$ $n o w g (s, ϕ) = 0$ $s_{m a x} c a n b e f o u n d f r o m g (s, ϕ)$ $g r a p h .$
	Commented by Tawa11 last updated on 22/Feb/22

	$Great sirs$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum