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Question Number 49183 by ajfour last updated on 04/Dec/18

Commented by ajfour last updated on 04/Dec/18

$F i n d \frac{A r e a △ P Q R}{A r e a △ A B C}, i f c e n t r o i d G$ $o f △ A B C l i e s o n z a x i s a n d e a c h$ $v e r t e x o n s u r f a c e o f p a r a b o l o i d .$
Commented by mr W last updated on 04/Dec/18

$d o e s Δ A B C h a v e g i v e n s i d e l e n g t h e s$ $a, b, c ?$
Commented by ajfour last updated on 04/Dec/18

$y e s s i r!$
	Answered by mr W last updated on 05/Dec/18
	$z=kr^2 with k=1 lengthes of medians of ΔABC: AG=(2/3)m_a =g_a =((√(2(b^2 +c^2 )−a^2 ))/3) BG=(2/3)m_b =g_b =((√(2(c^2 +a^2 )−b^2 ))/3) CG=(2/3)m_c =g_c =((√(2(a^2 +b^2 )−c^2 ))/3) let z_G =h let z_A =z_G +AG sin α=h+g_a sin α let z_B =z_G +BG sin β=h+g_b sin β we have three unknowns: h,α,β. r_A =AG cos α=g_a cos α r_B =BG cos β=g_b cos β z_A =kr_A ^2 ⇒h+g_a sin α=kg_a ^2 cos^2 α ⇒sin^2 α+(1/(kg_a )) sin α+(h/(kg_a ^2 ))−1=0 ...(i) similarly ⇒sin^2 β+(1/(kg_b )) sin β+(h/(kg_b ^2 ))−1=0 ...(ii) ⇒sin^2 γ+(1/(kg_c )) sin γ+(h/(kg_c ^2 ))−1=0 z_(B′) =z_G −B′G sin β=h−(1/3)m_b sin β z_(B′) =(1/2)(z_A +z_C ) z_C =2z_(B′) −z_A =2h−(2/3)m_b sin β−h−(2/3)m_a sin α ⇒z_C =h−(2/3)(m_a sin α+m_b sin β) z_C =h+(2/3)m_c sin γ ⇒(2/3)m_c sin γ=−(2/3)(m_a sin α+m_b sin β) ⇒ sin γ=−(1/g_c )(g_a sin α+g_b sin β) ⇒k(g_a sin α+g_b sin β)^2 −(g_a sin α+g_b sin β)+h−kg_c ^2 =0 ...(iii) from (i) and (ii): k(g_a ^2 sin^2 α−g_b ^2 sin^2 β)+(g_a sin α−g_b sin β)−k(g_a ^2 −g_b ^2 )=0 (1+kg_a sin α+kg_b sin β)(g_a sin α−g_b sin β)=k(g_a ^2 −g_b ^2 ) from (i) and (iii): k[g_a ^2 sin^2 α−(g_a sin α+g_b sin β)^2 ]+[g_a sin α+(g_a sin α+g_b sin β)]−k(g_a ^2 −g_c ^2 )=0 (1−kg_b sin β)(2g_a sin α+g_b sin β)=k(g_a ^2 −g_c ^2 ) we can solve (nummerically) { ((((1/k)+g_a sin α+g_b sin β)(g_a sin α−g_b sin β)=g_a ^2 −g_b ^2 )),((((1/k)−g_b sin β)(2g_a sin α+g_b sin β)=g_a ^2 −g_c ^2 )) :} for sin α and sin β. note: there can be more than one solution! with sin α and sin β we get sin γ from g_a sin α+g_b sin β+g_c sin γ=0 let e_a ,e_b ,e_c =projection of AG,BG,CG e_a =g_a cos α=g_a (√(1−sin^2 α)) e_b =g_b cos β=g_b (√(1−sin^2 β)) e_c =g_c cos γ=g_c (√(1−sin^2 γ)) Δ_(PQR) =((3(√((e_a +e_b +e_c )(−e_a +e_b +e_c )(e_a −e_b +e_c )(e_a +e_b −e_c ))))/4) Δ_(ABC) =((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4) (Δ_(PQR) /Δ_(ABC) )=((3(√((e_a +e_b +e_c )(−e_a +e_b +e_c )(e_a −e_b +e_c )(e_a +e_b −e_c ))))/(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c))))$
	$z = {k r}^{2} w i t h k = 1$ $l e n g t h e s o f m e d i a n s o f Δ A B C :$ $A G = \frac{2}{3} m_{a} = g_{a} = \frac{\sqrt{2 (b^{2} + c^{2}) - a^{2}}}{3}$ $B G = \frac{2}{3} m_{b} = g_{b} = \frac{\sqrt{2 (c^{2} + a^{2}) - b^{2}}}{3}$ $C G = \frac{2}{3} m_{c} = g_{c} = \frac{\sqrt{2 (a^{2} + b^{2}) - c^{2}}}{3}$ $l e t z_{G} = h$ $l e t z_{A} = z_{G} + A G \sin α = h + g_{a} \sin α$ $l e t z_{B} = z_{G} + B G \sin β = h + g_{b} \sin β$ $w e h a v e t h r e e u n k n o w n s : h, α, β .$ $r_{A} = A G \cos α = g_{a} \cos α$ $r_{B} = B G \cos β = g_{b} \cos β$ $z_{A} = {k r}_{A}^{2}$ $\Rightarrow h + g_{a} \sin α = {k g}_{a}^{2} \cos^{2} α$ $\Rightarrow \sin^{2} α + \frac{1}{{k g}_{a}} \sin α + \frac{h}{{k g}_{a}^{2}} - 1 = 0 . . . (i)$ $s i m i l a r l y$ $\Rightarrow \sin^{2} β + \frac{1}{{k g}_{b}} \sin β + \frac{h}{{k g}_{b}^{2}} - 1 = 0 . . . (i i)$ $\Rightarrow \sin^{2} γ + \frac{1}{{k g}_{c}} \sin γ + \frac{h}{{k g}_{c}^{2}} - 1 = 0$ $z_{B^{'}} = z_{G} - B^{'} G \sin β = h - \frac{1}{3} m_{b} \sin β$ $z_{B^{'}} = \frac{1}{2} (z_{A} + z_{C})$ $z_{C} = 2 z_{B^{'}} - z_{A} = 2 h - \frac{2}{3} m_{b} \sin β - h - \frac{2}{3} m_{a} \sin α$ $\Rightarrow z_{C} = h - \frac{2}{3} (m_{a} \sin α + m_{b} \sin β)$ $z_{C} = h + \frac{2}{3} m_{c} \sin γ$ $\Rightarrow \frac{2}{3} m_{c} \sin γ = - \frac{2}{3} (m_{a} \sin α + m_{b} \sin β)$ $\Rightarrow \sin γ = - \frac{1}{g_{c}} (g_{a} \sin α + g_{b} \sin β)$ $\Rightarrow k {(g_{a} \sin α + g_{b} \sin β)}^{2} - (g_{a} \sin α + g_{b} \sin β) + h - {k g}_{c}^{2} = 0 . . . (i i i)$ $f r o m (i) a n d (i i) :$ $k (g_{a}^{2} \sin^{2} α - g_{b}^{2} \sin^{2} β) + (g_{a} \sin α - g_{b} \sin β) - k (g_{a}^{2} - g_{b}^{2}) = 0$ $(1 + {k g}_{a} \sin α + {k g}_{b} \sin β) (g_{a} \sin α - g_{b} \sin β) = k (g_{a}^{2} - g_{b}^{2})$ $f r o m (i) a n d (i i i) :$ $k [g_{a}^{2} \sin^{2} α - {(g_{a} \sin α + g_{b} \sin β)}^{2}] + [g_{a} \sin α + (g_{a} \sin α + g_{b} \sin β)] - k (g_{a}^{2} - g_{c}^{2}) = 0$ $(1 - {k g}_{b} \sin β) (2 g_{a} \sin α + g_{b} \sin β) = k (g_{a}^{2} - g_{c}^{2})$ $w e c a n s o l v e (n u m m e r i c a l l y)$ ${\begin{cases} (\frac{1}{k} + g_{a} \sin α + g_{b} \sin β) (g_{a} \sin α - g_{b} \sin β) = g_{a}^{2} - g_{b}^{2} \\ (\frac{1}{k} - g_{b} \sin β) (2 g_{a} \sin α + g_{b} \sin β) = g_{a}^{2} - g_{c}^{2} \end{cases}$ $f o r \sin α a n d \sin β .$ $n o t e : t h e r e c a n b e m o r e t h a n o n e$ $s o l u t i o n!$ $w i t h \sin α a n d \sin β w e g e t \sin γ f r o m$ $g_{a} \sin α + g_{b} \sin β + g_{c} \sin γ = 0$ $l e t e_{a}, e_{b}, e_{c} = p r o j e c t i o n o f A G, B G, C G$ $e_{a} = g_{a} \cos α = g_{a} \sqrt{1 - \sin^{2} α}$ $e_{b} = g_{b} \cos β = g_{b} \sqrt{1 - \sin^{2} β}$ $e_{c} = g_{c} \cos γ = g_{c} \sqrt{1 - \sin^{2} γ}$ $Δ_{P Q R} = \frac{3 \sqrt{(e_{a} + e_{b} + e_{c}) (- e_{a} + e_{b} + e_{c}) (e_{a} - e_{b} + e_{c}) (e_{a} + e_{b} - e_{c})}}{4}$ $Δ_{A B C} = \frac{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}{4}$ $\frac{Δ_{P Q R}}{Δ_{A B C}} = \frac{3 \sqrt{(e_{a} + e_{b} + e_{c}) (- e_{a} + e_{b} + e_{c}) (e_{a} - e_{b} + e_{c}) (e_{a} + e_{b} - e_{c})}}{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}$
	Commented by mr W last updated on 04/Dec/18

	Commented by ajfour last updated on 06/Dec/18

	$T h a n k s s i r! a n d s o r r y f o r s u c h$ $n o t s o i n t e r e s t i n g q u e s t i o n . .$
	Commented by mr W last updated on 06/Dec/18

	$b u t i t i s a n i n t e r e s t i n g q u e s t i o n . w h a t$ $i s t h e r e a s o n f o r t h e r e s t r i c t i o n t h a t$ $t h e c e n t r o i d s h o u l d l i e o n z - a x i x ?$ $w i t h t h i s r e s t r i c t i o n n o t e v e r y t r i a n g l e$ $c a n b e p l a c e d i n s i d e t h e c u p, f o r$ $e x a m p l e a t r i a n g l e w i t h s i d e s 3, 4, 5 .$
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