Menu Close

Question-218428

Tinku Tara 0 Comments
Pin




Question Number 218428 by hardmath last updated on 09/Apr/25
Answered by A5T last updated on 09/Apr/25
Let F ∈ BC such that AF⊥BC AB=AE⇒BF=FE=((7+3)/2)=5 BF×BC=BA^2 ⇒5(10+x)=BA^2 ...(i) AB^2 −5^2 =AC^2 −(5+x)^2 ...(ii) AD^2 −2^2 =AE^2 −5^2 =AB^2 −5^2 ...(iii) ((AD)/3)=((AC)/x)⇒AD=((3AC)/x)...(iv) (iv)&(iii)⇒((9AC^2 )/x^2 )−4=AB^2 −5^2 ...(v) (ii)&(v)⇒((9(AB^2 −5^2 +(5+x)^2 ))/x^2 )−4=AB^2 −5^2 ...(vi) (i)&(vi)⇒((x^2 +27x+90)/x^2 )=x+5⇒x=5
$$\mathrm{Let}\:\mathrm{F}\:\in\:\mathrm{BC}\:\mathrm{such}\:\mathrm{that}\:\mathrm{AF}\bot\mathrm{BC} \\ $$$$\mathrm{AB}=\mathrm{AE}\Rightarrow\mathrm{BF}=\mathrm{FE}=\frac{\mathrm{7}+\mathrm{3}}{\mathrm{2}}=\mathrm{5} \\ $$$$\mathrm{BF}×\mathrm{BC}=\mathrm{BA}^{\mathrm{2}} \Rightarrow\mathrm{5}\left(\mathrm{10}+\mathrm{x}\right)=\mathrm{BA}^{\mathrm{2}} …\left(\mathrm{i}\right) \\ $$$$\mathrm{AB}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} =\mathrm{AC}^{\mathrm{2}} −\left(\mathrm{5}+\mathrm{x}\right)^{\mathrm{2}} …\left(\mathrm{ii}\right) \\ $$$$\mathrm{AD}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} =\mathrm{AE}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} =\mathrm{AB}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} …\left(\mathrm{iii}\right) \\ $$$$\frac{\mathrm{AD}}{\mathrm{3}}=\frac{\mathrm{AC}}{\mathrm{x}}\Rightarrow\mathrm{AD}=\frac{\mathrm{3AC}}{\mathrm{x}}…\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{iv}\right)\&\left(\mathrm{iii}\right)\Rightarrow\frac{\mathrm{9AC}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }−\mathrm{4}=\mathrm{AB}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} …\left(\mathrm{v}\right) \\ $$$$\left(\mathrm{ii}\right)\&\left(\mathrm{v}\right)\Rightarrow\frac{\mathrm{9}\left(\mathrm{AB}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} +\left(\mathrm{5}+\mathrm{x}\right)^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} }−\mathrm{4}=\mathrm{AB}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} …\left(\mathrm{vi}\right) \\ $$$$\left(\mathrm{i}\right)\&\left(\mathrm{vi}\right)\Rightarrow\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{27x}+\mathrm{90}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{x}+\mathrm{5}\Rightarrow\mathrm{x}=\mathrm{5} \\ $$
Answered by vnm last updated on 09/Apr/25
BC=a ∠ABC=(π/4)+ϕ ∠ACB=(π/4)−ϕ AB=asin((π/4)−ϕ)=(a/( (√2)))(cosϕ−sinϕ) AC=asin((π/4)+ϕ)=(a/( (√2)))(cosϕ+sinϕ) BE=2BAcos(∠ABC)= 2(a/( (√2)))(cosϕ−sinϕ)∙cos((π/4)+ϕ)=a(cosϕ−sinϕ)^2 =3+7=10 CE=BC−BE=asin2ϕ=x ((EC)/(sin(∠EAC)))=((AC)/(sin∠AEC))=((AC)/(sin(∠ABC))) ((asin2ϕ)/(sin(∠EAC)))=a, sin(∠EAC)=sin2ϕ ∠BAD=∠EAC=2ϕ ∠BDA=π−(∠ABD+∠BAD)=((3π)/4)−3ϕ ((BD)/(sin(∠BAD)))=((AB)/(sin(∠BDA))) (3/(sin2ϕ))=(((a/( (√2)))(cosϕ−sinϕ))/(sin(((3π)/4)−3ϕ)))=((a(cosϕ−sinϕ))/(cos3ϕ+sin3ϕ)) { ((a(cosϕ−sinϕ)^2 =10)),(((3/(sin2ϕ))=((a(cosϕ−sinϕ))/(cos3ϕ+sin3ϕ)))) :} a=((10)/((cosϕ−sinϕ)^2 )) 3(cos3ϕ+sin3ϕ)=((10sin2ϕ)/(cosϕ−sinϕ)) 3(cos3ϕcosϕ+sin3ϕcosϕ−cos3ϕsinϕ−sin3ϕsinϕ)=10sin2ϕ 3(cos4ϕ+sin2ϕ)=10sin2ϕ 3(1−2sin^2 2ϕ)=7sin2ϕ 6sin^2 2ϕ+7sin2ϕ−3=0 sin2ϕ=((−7+(√(49+72)))/(12))=(1/3) a=((10)/(1−sin2ϕ))=15 x=15−10=5
$$ \\ $$$${BC}={a} \\ $$$$\angle{ABC}=\frac{\pi}{\mathrm{4}}+\varphi \\ $$$$\angle{ACB}=\frac{\pi}{\mathrm{4}}−\varphi \\ $$$${AB}={a}\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−\varphi\right)=\frac{{a}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{cos}\varphi−\mathrm{sin}\varphi\right) \\ $$$${AC}={a}\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}+\varphi\right)=\frac{{a}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{cos}\varphi+\mathrm{sin}\varphi\right) \\ $$$${BE}=\mathrm{2}{BA}\mathrm{cos}\left(\angle{ABC}\right)= \\ $$$$\mathrm{2}\frac{{a}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{cos}\varphi−\mathrm{sin}\varphi\right)\centerdot\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}+\varphi\right)={a}\left(\mathrm{cos}\varphi−\mathrm{sin}\varphi\right)^{\mathrm{2}} =\mathrm{3}+\mathrm{7}=\mathrm{10} \\ $$$${CE}={BC}−{BE}={a}\mathrm{sin2}\varphi={x} \\ $$$$\frac{{EC}}{\mathrm{sin}\left(\angle{EAC}\right)}=\frac{{AC}}{\mathrm{sin}\angle{AEC}}=\frac{{AC}}{\mathrm{sin}\left(\angle{ABC}\right)} \\ $$$$\frac{{a}\mathrm{sin2}\varphi}{\mathrm{sin}\left(\angle{EAC}\right)}={a},\:\:\:\mathrm{sin}\left(\angle{EAC}\right)=\mathrm{sin2}\varphi \\ $$$$\angle{BAD}=\angle{EAC}=\mathrm{2}\varphi \\ $$$$\angle{BDA}=\pi−\left(\angle{ABD}+\angle{BAD}\right)=\frac{\mathrm{3}\pi}{\mathrm{4}}−\mathrm{3}\varphi \\ $$$$\frac{{BD}}{\mathrm{sin}\left(\angle{BAD}\right)}=\frac{{AB}}{\mathrm{sin}\left(\angle{BDA}\right)} \\ $$$$\frac{\mathrm{3}}{\mathrm{sin2}\varphi}=\frac{\frac{{a}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{cos}\varphi−\mathrm{sin}\varphi\right)}{\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}−\mathrm{3}\varphi\right)}=\frac{{a}\left(\mathrm{cos}\varphi−\mathrm{sin}\varphi\right)}{\mathrm{cos3}\varphi+\mathrm{sin3}\varphi} \\ $$$$\begin{cases}{{a}\left(\mathrm{cos}\varphi−\mathrm{sin}\varphi\right)^{\mathrm{2}} =\mathrm{10}}\\{\frac{\mathrm{3}}{\mathrm{sin2}\varphi}=\frac{{a}\left(\mathrm{cos}\varphi−\mathrm{sin}\varphi\right)}{\mathrm{cos3}\varphi+\mathrm{sin3}\varphi}}\end{cases} \\ $$$${a}=\frac{\mathrm{10}}{\left(\mathrm{cos}\varphi−\mathrm{sin}\varphi\right)^{\mathrm{2}} } \\ $$$$\mathrm{3}\left(\mathrm{cos3}\varphi+\mathrm{sin3}\varphi\right)=\frac{\mathrm{10sin2}\varphi}{\mathrm{cos}\varphi−\mathrm{sin}\varphi} \\ $$$$\mathrm{3}\left(\mathrm{cos3}\varphi\mathrm{cos}\varphi+\mathrm{sin3}\varphi\mathrm{cos}\varphi−\mathrm{cos3}\varphi\mathrm{sin}\varphi−\mathrm{sin3}\varphi\mathrm{sin}\varphi\right)=\mathrm{10sin2}\varphi \\ $$$$\mathrm{3}\left(\mathrm{cos4}\varphi+\mathrm{sin2}\varphi\right)=\mathrm{10sin2}\varphi \\ $$$$\mathrm{3}\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{2}\varphi\right)=\mathrm{7sin2}\varphi \\ $$$$\mathrm{6sin}^{\mathrm{2}} \mathrm{2}\varphi+\mathrm{7sin2}\varphi−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{sin2}\varphi=\frac{−\mathrm{7}+\sqrt{\mathrm{49}+\mathrm{72}}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}=\frac{\mathrm{10}}{\mathrm{1}−\mathrm{sin2}\varphi}=\mathrm{15} \\ $$$${x}=\mathrm{15}−\mathrm{10}=\mathrm{5} \\ $$
Commented by hardmath last updated on 09/Apr/25
thank you dear professor cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$
Answered by mr W last updated on 10/Apr/25
Commented by mr W last updated on 10/Apr/25
ΔABG≡ΔAEC BF=FE=((3+7)/2)=5 h^2 =BF×FC=5(5+x) AG=(√(h^2 +(5+x)^2 ))=(√(x^2 +15x+50)) AD=(√(h^2 +2^2 ))=(√(5x+29)) AB is bisector of ∠GAD, ⇒((GA)/(GB))=((DA)/(DB)) ((√(x^2 +15x+50))/x)=((√(5x+29))/3) ((x^2 +15x+50)/x^2 )=((5x+29)/9) ⇒x^3 +4x^2 −27x−90=0 ⇒(x+3)(x+6)(x−5)=0 x=−3, −6 <0 ⇒rejected ⇒x=5 >0 ✓
$$\Delta{ABG}\equiv\Delta{AEC} \\ $$$${BF}={FE}=\frac{\mathrm{3}+\mathrm{7}}{\mathrm{2}}=\mathrm{5} \\ $$$${h}^{\mathrm{2}} ={BF}×{FC}=\mathrm{5}\left(\mathrm{5}+{x}\right) \\ $$$${AG}=\sqrt{{h}^{\mathrm{2}} +\left(\mathrm{5}+{x}\right)^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{50}} \\ $$$${AD}=\sqrt{{h}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\sqrt{\mathrm{5}{x}+\mathrm{29}} \\ $$$${AB}\:{is}\:{bisector}\:{of}\:\angle{GAD}, \\ $$$$\Rightarrow\frac{{GA}}{{GB}}=\frac{{DA}}{{DB}} \\ $$$$\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{50}}}{{x}}=\frac{\sqrt{\mathrm{5}{x}+\mathrm{29}}}{\mathrm{3}} \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{50}}{{x}^{\mathrm{2}} }=\frac{\mathrm{5}{x}+\mathrm{29}}{\mathrm{9}} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{27}{x}−\mathrm{90}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{3}\right)\left({x}+\mathrm{6}\right)\left({x}−\mathrm{5}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{3},\:−\mathrm{6}\:<\mathrm{0}\:\Rightarrow{rejected} \\ $$$$\Rightarrow{x}=\mathrm{5}\:>\mathrm{0}\:\checkmark \\ $$
Commented by hardmath last updated on 10/Apr/25
thankyou dearprofessor
$$\mathrm{thankyou}\:\mathrm{dearprofessor} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *