Question-67770

Question Number 67770 by TawaTawa last updated on 31/Aug/19

Answered by MJS last updated on 31/Aug/19

AD+DE=AE=6 ⇒ AD=q; DE=6−q [⇒ q<6] AB=DE−AD=6−2q [⇒ q<3] A= ((0),(0) ) D= ((q),(0) ) E= ((6),(0) ) line AB: y=xtan 60° ⇔ y=(√3)x [1] B on this circle: x^2 +y^2 =(6−2q)^2 [2] [1] in [2] x=3−q ⇒ B= (((3−q)),(((√3)(3−q))) ) line BD: y=(√3)((q−3)/(2q−3))(x−q) C_(∈BD) = ((p),(((√3)((q−3)/(2q−3))(p−q))) ) ∣CD∣^2 =(((p−q)^2 )/((2q−3)^2 ))(7q^2 −30q+36) ∣BD^2 ∣=7q^2 −30q+36=∣CD∣^2 ⇒ (((p−q)^2 )/((2q−3)^2 ))=1 ⇒ p=3q−3 c=∣AB∣=6−2q b=∣AC∣=2(√(3q^2 −9q+9)) a=∣BC∣=2(√(7q^2 −30q+36)) radius of circumcircle of ΔABC R=((abc)/δ)=(2/q)(√((q^2 −3q+3)(7q^2 −30q+36))) center O= ((((1/q)(4q^2 −15q+18))),((−((√3)/q)(q−2)(2q−3))) ) ∣OE∣^2 =R^2 ⇒ q=(3/2) ⇒ A= ((0),(0) ) B= (((3/2)),(((3(√3))/2)) ) C= (((3/2)),((−((3(√3))/2))) ) D= (((3/2)),(0) ) E= ((6),(0) ) O= ((3),(0) ) S_(ABC) =6+3(√3)

$${AD}+{DE}={AE}=\mathrm{6} \\ $$$$\Rightarrow\:{AD}={q};\:{DE}=\mathrm{6}−{q}\:\:\left[\Rightarrow\:{q}<\mathrm{6}\right] \\ $$$${AB}={DE}−{AD}=\mathrm{6}−\mathrm{2}{q}\:\:\left[\Rightarrow\:{q}<\mathrm{3}\right] \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{{q}}\\{\mathrm{0}}\end{pmatrix}\:\:{E}=\begin{pmatrix}{\mathrm{6}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{line}\:{AB}:\:{y}={x}\mathrm{tan}\:\mathrm{60}°\:\Leftrightarrow\:{y}=\sqrt{\mathrm{3}}{x}\:\:\left[\mathrm{1}\right] \\ $$$${B}\:\mathrm{on}\:\mathrm{this}\:\mathrm{circle}:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{6}−\mathrm{2}{q}\right)^{\mathrm{2}} \:\:\left[\mathrm{2}\right] \\ $$$$\left[\mathrm{1}\right]\:\mathrm{in}\:\left[\mathrm{2}\right] \\ $$$${x}=\mathrm{3}−{q} \\ $$$$\Rightarrow\:{B}=\begin{pmatrix}{\mathrm{3}−{q}}\\{\sqrt{\mathrm{3}}\left(\mathrm{3}−{q}\right)}\end{pmatrix} \\ $$$$\mathrm{line}\:{BD}:\:{y}=\sqrt{\mathrm{3}}\frac{{q}−\mathrm{3}}{\mathrm{2}{q}−\mathrm{3}}\left({x}−{q}\right) \\ $$$${C}_{\in{BD}} =\begin{pmatrix}{{p}}\\{\sqrt{\mathrm{3}}\frac{{q}−\mathrm{3}}{\mathrm{2}{q}−\mathrm{3}}\left({p}−{q}\right)}\end{pmatrix} \\ $$$$\mid{CD}\mid^{\mathrm{2}} =\frac{\left({p}−{q}\right)^{\mathrm{2}} }{\left(\mathrm{2}{q}−\mathrm{3}\right)^{\mathrm{2}} }\left(\mathrm{7}{q}^{\mathrm{2}} −\mathrm{30}{q}+\mathrm{36}\right) \\ $$$$\mid{BD}^{\mathrm{2}} \mid=\mathrm{7}{q}^{\mathrm{2}} −\mathrm{30}{q}+\mathrm{36}=\mid{CD}\mid^{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\left({p}−{q}\right)^{\mathrm{2}} }{\left(\mathrm{2}{q}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{1}\:\Rightarrow\:{p}=\mathrm{3}{q}−\mathrm{3} \\ $$$${c}=\mid{AB}\mid=\mathrm{6}−\mathrm{2}{q} \\ $$$${b}=\mid{AC}\mid=\mathrm{2}\sqrt{\mathrm{3}{q}^{\mathrm{2}} −\mathrm{9}{q}+\mathrm{9}} \\ $$$${a}=\mid{BC}\mid=\mathrm{2}\sqrt{\mathrm{7}{q}^{\mathrm{2}} −\mathrm{30}{q}+\mathrm{36}} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{circumcircle}\:\mathrm{of}\:\Delta{ABC} \\ $$$${R}=\frac{{abc}}{\delta}=\frac{\mathrm{2}}{{q}}\sqrt{\left({q}^{\mathrm{2}} −\mathrm{3}{q}+\mathrm{3}\right)\left(\mathrm{7}{q}^{\mathrm{2}} −\mathrm{30}{q}+\mathrm{36}\right)} \\ $$$$\mathrm{center} \\ $$$${O}=\begin{pmatrix}{\frac{\mathrm{1}}{{q}}\left(\mathrm{4}{q}^{\mathrm{2}} −\mathrm{15}{q}+\mathrm{18}\right)}\\{−\frac{\sqrt{\mathrm{3}}}{{q}}\left({q}−\mathrm{2}\right)\left(\mathrm{2}{q}−\mathrm{3}\right)}\end{pmatrix} \\ $$$$\mid{OE}\mid^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\:{q}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{2}}}\\{\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{2}}}\\{−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{pmatrix} \\ $$$${D}=\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix}\:\:{E}=\begin{pmatrix}{\mathrm{6}}\\{\mathrm{0}}\end{pmatrix}\:\:{O}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${S}_{{ABC}} =\mathrm{6}+\mathrm{3}\sqrt{\mathrm{3}} \\ $$

Commented by TawaTawa last updated on 31/Aug/19

Wow, God bless you sir. Thanks for your time.

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$

Commented by TawaTawa last updated on 31/Aug/19

Sir, i learnt the answer should be: ((9(√3))/4) = 3.8971 or ((4.5(√3))/2) = 3.8971

$$\mathrm{Sir},\:\mathrm{i}\:\mathrm{learnt}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}:\:\:\:\:\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{4}}\:\:=\:\:\mathrm{3}.\mathrm{8971}\:\:\:\:\:\:\:\: \\ $$$$\mathrm{or}\:\:\:\:\:\:\:\frac{\mathrm{4}.\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:=\:\:\mathrm{3}.\mathrm{8971} \\ $$

Commented by MJS last updated on 31/Aug/19

just draw the points and the circle of my solution, you can see I′m right

$$\mathrm{just}\:\mathrm{draw}\:\mathrm{the}\:\mathrm{points}\:\mathrm{and}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{my} \\ $$$$\mathrm{solution},\:\mathrm{you}\:\mathrm{can}\:\mathrm{see}\:\mathrm{I}'\mathrm{m}\:\mathrm{right} \\ $$

Commented by TawaTawa last updated on 01/Sep/19

$$\mathrm{I}\:\mathrm{will}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by TawaTawa last updated on 01/Sep/19

$$\mathrm{Sir}\:\:\:\:\mathrm{S}_{\mathrm{ABC}} \:\:\mathrm{or}\:\:\mathrm{S}\left(\mathrm{ABC}\right)\:\:\mathrm{means}\:\mathrm{what}\:\mathrm{sir}\:\:? \\ $$

Commented by TawaTawa last updated on 01/Sep/19