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Question Number 189973 by mr W last updated on 25/Mar/23

Commented by mr W last updated on 25/Mar/23

$f i n d t h e a r e a a n d s i d e l e n g t h s o f t h e$ $s h a d e d t r i a n g l e .$
Commented by MJS_new last updated on 25/Mar/23

$\frac{1}{7} of the area of Δ A B C$ $I^{'} ll show my work later$
Commented by MJS_new last updated on 25/Mar/23

$btw to a given triangle with sides a, b, c$ $there are 2 possible ‘ ‘ {inner}^{″} triangles with$ $sides o_{k}, p_{k}, q_{k} :$ $o_{1} = \frac{\sqrt{6 a^{2} - 2 b^{2} + 3 c^{2}}}{7}; o_{2} = \frac{\sqrt{6 a^{2} + 3 b^{2} - 2 c^{2}}}{7}$ $[p_{k}, q_{k} by cycling a, b, c]$ ${it}^{'} s easy to solve these for a, b, c with given$ $o, p, q :$ $a_{1} = \sqrt{6 o^{2} + 3 p^{2} - 2 q^{2}}; a_{2} = \sqrt{6 o^{2} - 2 p^{2} + 3 q^{2}}$ $[again b_{k}, c_{k} by cycling o, p, q]$ $all these triangles share the same mass center$
Commented by mr W last updated on 25/Mar/23

$t h a n k s s i r!$
	Answered by ajfour last updated on 25/Mar/23
	$B origin. vectors a, c, p r_C =a r_A =c A ′=P and so on.. r_Q =2p r_R =a+2(2p−c)=c+((p−c)/2) ⇒ 2a+7p=5c 2Area_(sh) =∣p×{a+2(2p−c)}∣ ⇒ 98A_(sb) ^� =(5c−2a)×{7a+4(5c−2a)−14c} =(5c−2a)(6c−a) =7∣a×c∣ ⇒ A_(sh) =((∣a×c∣)/(14))=(△_(ABC) /7) ★ 2△_(ABC) =∣a×c∣$
	$B o r i g i n .$ $v e c t o r s a, c, p$ $r_{C} = a$ $r_{A} = c$ $A^{'} = P a n d s o o n . .$ $r_{Q} = 2 p$ $r_{R} = a + 2 (2 p - c) = c + \frac{p - c}{2}$ $\Rightarrow 2 a + 7 p = 5 c$ $2 {A r e a}_{s h} =∣ p \times {a + 2 (2 p - c)} ∣$ $\Rightarrow 98 {\bar{A}}_{s b}$ $= (5 c - 2 a) \times {7 a + 4 (5 c - 2 a) - 14 c}$ $= (5 c - 2 a) (6 c - a)$ $= 7 ∣ a \times c ∣$ $\Rightarrow A_{s h} = \frac{∣ a \times c ∣}{14} = \frac{△_{A B C}}{7}$ $★ 2 △_{A B C} =∣ a \times c ∣$
	Commented by mr W last updated on 25/Mar/23

	$t h a n k s s i r!$
	Answered by mr W last updated on 25/Mar/23

	Commented by mr W last updated on 25/Mar/23

	$w e s e e a l l s e v e n s m a l l t r i a n g l e s h a v e$ $t h e s a m e a r e a, t h e r e f o r e$ $Δ_{A^{'} B^{'} C^{'}} = \frac{Δ_{A B C}}{7} .$ $s a y {A C}^{'} = C^{'} A^{'} = p, {B A}^{'} = A^{'} B^{'} = q, {C B}^{'} = B^{'} C^{'} = r$ $\cos β = - \frac{4 q^{2} + r^{2} - a^{2}}{4 q r} = \frac{q^{2} + r^{2} - p^{2}}{2 q r}$ $- 2 p^{2} + 6 q^{2} + 3 r^{2} = a^{2} . . . (i)$ $s i m i l a r l y$ $3 p^{2} - 2 q^{2} + 6 r^{2} = b^{2} . . . (i i)$ $6 p^{2} + 3 q^{2} - 2 r^{2} = c^{2} . . . (i i i)$ $\Rightarrow p^{2} = \frac{- 2 a^{2} + 3 b^{2} + 6 c^{2}}{49} \Rightarrow p = \frac{\sqrt{- 2 a^{2} + 3 b^{2} + 6 c^{2}}}{7}$ $\Rightarrow q^{2} = \frac{6 a^{2} - 2 b^{2} + 3 c^{2}}{49} \Rightarrow q = \frac{\sqrt{6 a^{2} - 2 b^{2} + 3 c^{2}}}{7}$ $\Rightarrow r^{2} = \frac{3 a^{2} + 6 b^{2} - 2 c^{2}}{49} \Rightarrow r = \frac{\sqrt{3 a^{2} + 6 b^{2} - 2 c^{2}}}{7}$
	Answered by manxsol last updated on 26/Mar/23

	$2 A = a c s i n x = b c s i n y = b a s i n z$ $2 A_{1} = 2 c b s i n y = 2 (2 A) = 4 A$ $2 A_{2} = 2 a c s i n x = 2 (2 A) = 4 A$ $2 A_{3} = 2 b a s i n z = 2 (2 A) = 4 A$ $\frac{Δ b l u e}{Δ t o t a l} = \frac{A}{7 A} = \frac{1}{7}$
	Commented by manxsol last updated on 26/Mar/23

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