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 13548 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/May/17

$i n t r i a n g l e A B C :$ $1) ∡ A B D = ∡ D B E = ∡ E B C$ $2) E H ⊥ A B, D G ⊥ B C$ $3) S_{A B C} = 1$ $. . . . . . . . . . . . . . . . . . . . . . .$ $S_{D F E} = ? (a r e a o f Δ D E F)$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/May/17

$s p e c i a l c a s e : ∡ A B C = 90^{°}, A B = B C$
Commented by mrW1 last updated on 21/May/17

$I {d o n}^{'} t t h i n k \frac{S_{D F E}}{S_{A B C}} = c o n s t a n t .$ $A c c o r d i n g t o t h e 3 c o n d i t i o n s w e$ $c a n n o t d e t e r m i n e S_{D F E} .$ $S e e p i c t u r e :$ $B o t h c a s e s h a v e t h e s a m e c o n d i t i o n s,$ $b u t S_{D F E} i s n o t t h e s a m e .$
Commented by mrW1 last updated on 21/May/17

Commented by mrW1 last updated on 21/May/17

$T h e r e w e r e a s o l u t i o n i f t h e c o n d i t i o n s$ $w e r e t h a t H a n d G a r e t h e m i d p o i n t$ $o f A B a n d B C, D a n d E a r e t h e 1 / 3$ $p o i n t s o f A C .$ $I n t h i s c a s e \frac{S_{Δ D E F}}{S_{Δ A B C}} = c o n s t a n t = \frac{1}{15}$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/May/17

$i n A \overset{Δ}{B} \overset{}{C} : d r a w b i s e c t o f ∡ B .$ $l e t : B D = d_{b} (b i s e c t o f ∡ B)$ $S_{A B C} = S_{A B D} + S_{B D C}$ $\frac{1}{2} a c . s i n B = \frac{1}{2} c . d_{b} . s i n \frac{B}{2} + \frac{1}{2} a . d_{b} . s i n \frac{B}{2}$ $\Rightarrow d_{b} (a + c) s i n \frac{B}{2} = 2 a c . s i n \frac{B}{2} . c o s \frac{B}{2}$ $\Rightarrow d_{b} = \frac{2 a c}{a + c} . c o s \frac{B}{2}$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/May/17
$in trianvle:AB^Δ C^ ,draw trisect of∡B. now:∡ABE=∡EBF=∡FBC=((∡B)/3) let:BE=d_1 ,BF=d_2 from the commenet above,in AB^Δ F, BE is bisect of ∡ABF,so we have: d_1 =((2ad_2 )/(a+d_2 ))cos(B/3),similarly:d_2 =((2cd_1 )/(c+d_1 ))cos(B/3) { ((ad_1 +d_1 d_2 =2ad_2 .cos(B/3))),((cd_2 +d_1 d_2 =2cd_1 .cos(B/3))) :}from this system: { ((d_1 =((ac)/(c+2a.cos(B/3)))(4cos^2 (B/3)−1))),((d_2 =((ac)/(a+2c.cos(B/3)))(4cos^2 (B/3)−1))) :}$
$i n t r i a n v l e : A \overset{Δ}{B} \overset{}{C}, d r a w t r i s e c t o f ∡ B .$ $n o w : ∡ A B E = ∡ E B F = ∡ F B C = \frac{∡ B}{3}$ $l e t : B E = d_{1}, B F = d_{2}$ $f r o m t h e c o m m e n e t a b o v e, i n A \overset{Δ}{B F},$ $B E i s b i s e c t o f ∡ A B F, s o w e h a v e :$ $d_{1} = \frac{2 {a d}_{2}}{a + d_{2}} c o s \frac{B}{3}, s i m i l a r l y : d_{2} = \frac{2 {c d}_{1}}{c + d_{1}} c o s \frac{B}{3}$ ${\begin{cases} {a d}_{1} + d_{1} d_{2} = 2 {a d}_{2} . c o s \frac{B}{3} \\ {c d}_{2} + d_{1} d_{2} = 2 {c d}_{1} . c o s \frac{B}{3} \end{cases} f r o m t h i s s y s t e m :$ ${\begin{cases} d_{1} = \frac{a c}{c + 2 a . c o s \frac{B}{3}} (4 {c o s}^{2} \frac{B}{3} - 1) \\ d_{2} = \frac{a c}{a + 2 c . c o s \frac{B}{3}} (4 {c o s}^{2} \frac{B}{3} - 1) \end{cases}$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/May/17
$if ∡B=90,a=c,from the commonet above ,we have: ((B/3)=30°) S_(ABC) =(1/2)ac=1⇒ac=2⇒a=c=(√2),b=2 4cos^2 (B/3)−1=4(((√3)/2))^2 −1=3−1=2 ⇒ { ((d_1 =((ac)/(c+2a.cos(B/3)))(4cos^2 (B/3)−1)=(((√(2.))(√2))/((√2)+2(√2).((√3)/2))).2=(√6)−(√2))),((d_1 =((2ac)/(c+a(√3))),d_2 =((2ac)/(a+c(√3))))) :} ⇒d_1 =d_2 =(√6)−(√2) AD^2 =AB^2 +d_1 ^2 −2AB.d_1 .cos(B/3)⇒ AD^2 =c^2 +((4a^2 c^2 )/((c+a(√3))^2 ))−2c((2ac)/(c+a(√3))).((√3)/2)= =(c^2 /((c+a(√3))^2 ))(c^2 +3a^2 +2ac(√3)+4a^2 −6a^2 −2ac(√3))= =((c^2 (c^2 +a^2 ))/((c+a(√3))^2 ))=((c^2 b^2 )/((c+a(√3))^2 ))⇒ ⇒ { ((AD=((bc)/(c+a(√3)))=(2/(1+(√3)))=(√3)−1)),((CE=((ba)/(a+c(√3)))=AD=(√3)−1.)) :} DE=b−((bc)/(c+a(√3)))−((ba)/(a+c(√3)))= =(b/((c+a(√3))(a+c(√3))))(ac+c^2 (√3)+a^2 (√3)+3ac−ac−c^2 (√3)−ac−a^2 (√3))= ⇒DE=((2abc)/(4ac+b^2 (√3)))=((2.(√2).2.(√2))/(4(√2).(√2)+4(√3)))=(8/(4(2+(√3))))=2(2−(√3))=((√3)−1)^2 ⇒or:DE=AC−(AD+CE)=2−2((√3)−1)= =4−2(√3)=((√3)−1)^2 DH=HE=((DE)/(√2))=((((√3)−1)^2 )/(√2)) ⇒S_(DHE) =(1/2)DH.HE=(1/2).((((√3)−1)^2 )/(√2)).((((√3)−1)^2 )/(√2))= =(1/4)(4−2(√3))^2 =(2−(√3))^2 .■$
$i f ∡ B = 90, a = c, f r o m t h e c o m m o n e t$ $a b o v e, w e h a v e : (\frac{B}{3} = 30 °)$ $S_{A B C} = \frac{1}{2} a c = 1 \Rightarrow a c = 2 \Rightarrow a = c = \sqrt{2}, b = 2$ $4 {c o s}^{2} \frac{B}{3} - 1 = 4 {(\frac{\sqrt{3}}{2})}^{2} - 1 = 3 - 1 = 2$ $\Rightarrow {\begin{cases} d_{1} = \frac{a c}{c + 2 a . c o s \frac{B}{3}} (4 {c o s}^{2} \frac{B}{3} - 1) = \frac{\sqrt{2 .} \sqrt{2}}{\sqrt{2} + 2 \sqrt{2} . \frac{\sqrt{3}}{2}} . 2 = \sqrt{6} - \sqrt{2} \\ d_{1} = \frac{2 a c}{c + a \sqrt{3}}, d_{2} = \frac{2 a c}{a + c \sqrt{3}} \end{cases}$ $\Rightarrow d_{1} = d_{2} = \sqrt{6} - \sqrt{2}$ ${A D}^{2} = {A B}^{2} + d_{1}^{2} - 2 A B . d_{1} . c o s \frac{B}{3} \Rightarrow$ ${A D}^{2} = c^{2} + \frac{4 a^{2} c^{2}}{{(c + a \sqrt{3})}^{2}} - 2 c \frac{2 a c}{c + a \sqrt{3}} . \frac{\sqrt{3}}{2} =$ $= \frac{c^{2}}{{(c + a \sqrt{3})}^{2}} (c^{2} + 3 a^{2} + 2 a c \sqrt{3} + 4 a^{2} - 6 a^{2} - 2 a c \sqrt{3}) =$ $= \frac{c^{2} (c^{2} + a^{2})}{{(c + a \sqrt{3})}^{2}} = \frac{c^{2} b^{2}}{{(c + a \sqrt{3})}^{2}} \Rightarrow$ $\Rightarrow {\begin{cases} A D = \frac{b c}{c + a \sqrt{3}} = \frac{2}{1 + \sqrt{3}} = \sqrt{3} - 1 \\ C E = \frac{b a}{a + c \sqrt{3}} = A D = \sqrt{3} - 1 . \end{cases}$ $D E = b - \frac{b c}{c + a \sqrt{3}} - \frac{b a}{a + c \sqrt{3}} =$ $= \frac{b}{(c + a \sqrt{3}) (a + c \sqrt{3})} (a c + c^{2} \sqrt{3} + a^{2} \sqrt{3} + 3 a c - a c - c^{2} \sqrt{3} - a c - a^{2} \sqrt{3}) =$ $\Rightarrow D E = \frac{2 a b c}{4 a c + b^{2} \sqrt{3}} = \frac{2 . \sqrt{2} . 2 . \sqrt{2}}{4 \sqrt{2} . \sqrt{2} + 4 \sqrt{3}} = \frac{8}{4 (2 + \sqrt{3})} = 2 (2 - \sqrt{3}) = {(\sqrt{3} - 1)}^{2}$ $\Rightarrow o r : D E = A C - (A D + C E) = 2 - 2 (\sqrt{3} - 1) =$ $= 4 - 2 \sqrt{3} = {(\sqrt{3} - 1)}^{2}$ $D H = H E = \frac{D E}{\sqrt{2}} = \frac{{(\sqrt{3} - 1)}^{2}}{\sqrt{2}}$ $\Rightarrow S_{D H E} = \frac{1}{2} D H . H E = \frac{1}{2} . \frac{{(\sqrt{3} - 1)}^{2}}{\sqrt{2}} . \frac{{(\sqrt{3} - 1)}^{2}}{\sqrt{2}} =$ $= \frac{1}{4} {(4 - 2 \sqrt{3})}^{2} = {(2 - \sqrt{3})}^{2} . ◼$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17

	$S_{D F E} = p . q^{2} . r^{2}$ $p = \frac{c o s A . c o s C}{2 s i n B}$ $q = \frac{a b c}{(a + 2 c . c o s \frac{B}{3}) (c + 2 a . c o s \frac{B}{3})} = \frac{a b c}{2 {(a + c)}^{2} . c o s \frac{B}{3} + a c}$ $r = 4 {c o s}^{2} \frac{B}{3} - 1 = \frac{s i n B}{s i n \frac{B}{3}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum