Question Number 36270 by ajfour last updated on 30/May/18
Commented by ajfour last updated on 30/May/18
$${Find}\:{maximum}\:{area}\:{of}\:{a}\:{triangle} \\ $$$${when}\:{its}\:{vertices}\:{lie}\:{on}\:{the} \\ $$$${cirumference}\:{of}\:{three}\:{circles}. \\ $$$$\left({As}\:{shown}\:{in}\:{fig}.\:{the}\:{circles}\right. \\ $$$$\left.{touch}\:{each}\:{other}\:\right). \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 31/May/18
$${i}\:{have}\:{deleted}\:{the}\:{post}…{yes}\:{i}\:{was}\:{wrong} \\ $$
Commented by MJS last updated on 31/May/18
$$\mathrm{anyway}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{all}\:\mathrm{the}\:\mathrm{input},\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{good}\:\mathrm{to}\:\mathrm{have}\:\mathrm{many}\:\mathrm{people}\:\mathrm{with}\:\mathrm{many} \\ $$$$\mathrm{different}\:\mathrm{points}\:\mathrm{of}\:\mathrm{view} \\ $$
Commented by ajfour last updated on 31/May/18
Commented by ajfour last updated on 31/May/18
$${mrW}\:{Sir},\:{i}\:{m}\:{again}\:{wishing} \\ $$$${your}\:{cooperation}\:… \\ $$
Commented by ajfour last updated on 31/May/18
![let DE be x-axis with D as origin. let ∠FDE =α A≡(x_1 ,y_1 ) , B≡(x_2 ,y_2 ) , C≡(x_3 ,y_3 ) x_1 ^2 +y_1 ^2 =p^2 (x_2 −p−q)^2 +y_2 ^2 =q^2 [x_3 −(p+r)cos α]^2 +[y_3 −(p+r)sin α]=r^2 △=(1/2) determinant ((x_1 ,y_1 ,1),(x_2 ,y_2 ,1),(x_3 ,y_3 ,1)) with (∂△/∂x_1 ) =0 , (∂△/∂x_2 ) =0 , (∂△/∂x_3 ) =0 ⇒ determinant ((1,(−(x_1 /y_1 )),1),(x_2 ,y_2 ,1),(x_3 ,y_3 ,1))=0 ⇒ (y_2 −y_3 )+(x_1 /y_1 )(x_2 −x_3 )+(x_2 y_3 −x_3 y_2 )=0 similarly two more equations for (∂△/∂x_2 ) =0 , (∂△/∂x_3 ) =0 . using these six equations we find (x_i , y_i ) and thus △_(max) .](https://www.tinkutara.com/question/Q36325.png)
$${let}\:\:{DE}\:\:{be}\:{x}-{axis}\:\:{with}\:{D}\:{as}\:{origin}. \\ $$$${let}\:\angle{FDE}\:=\alpha \\ $$$${A}\equiv\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:\:,\:\:{B}\equiv\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right)\:\:,\:{C}\equiv\left({x}_{\mathrm{3}} ,{y}_{\mathrm{3}} \right) \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} ={p}^{\mathrm{2}} \\ $$$$\left({x}_{\mathrm{2}} −{p}−{q}\right)^{\mathrm{2}} +{y}_{\mathrm{2}} ^{\mathrm{2}} ={q}^{\mathrm{2}} \\ $$$$\left[{x}_{\mathrm{3}} −\left({p}+{r}\right)\mathrm{cos}\:\alpha\right]^{\mathrm{2}} +\left[{y}_{\mathrm{3}} −\left({p}+{r}\right)\mathrm{sin}\:\alpha\right]={r}^{\mathrm{2}} \\ $$$$\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{{x}_{\mathrm{1}} }&{{y}_{\mathrm{1}} }&{\mathrm{1}}\\{{x}_{\mathrm{2}} }&{{y}_{\mathrm{2}} }&{\mathrm{1}}\\{{x}_{\mathrm{3}} }&{{y}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}\:\:\:{with} \\ $$$$\frac{\partial\bigtriangleup}{\partial{x}_{\mathrm{1}} }\:=\mathrm{0}\:,\:\:\frac{\partial\bigtriangleup}{\partial{x}_{\mathrm{2}} }\:=\mathrm{0}\:,\:\frac{\partial\bigtriangleup}{\partial{x}_{\mathrm{3}} }\:=\mathrm{0} \\ $$$$\Rightarrow\:\:\begin{vmatrix}{\mathrm{1}}&{−\frac{{x}_{\mathrm{1}} }{{y}_{\mathrm{1}} }}&{\mathrm{1}}\\{{x}_{\mathrm{2}} }&{{y}_{\mathrm{2}} }&{\mathrm{1}}\\{{x}_{\mathrm{3}} }&{{y}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left({y}_{\mathrm{2}} −{y}_{\mathrm{3}} \right)+\frac{{x}_{\mathrm{1}} }{{y}_{\mathrm{1}} }\left({x}_{\mathrm{2}} −{x}_{\mathrm{3}} \right)+\left({x}_{\mathrm{2}} {y}_{\mathrm{3}} −{x}_{\mathrm{3}} {y}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\:{similarly}\:\:{two}\:{more}\:{equations} \\ $$$$\:{for}\:\:\:\:\:\:\frac{\partial\bigtriangleup}{\partial{x}_{\mathrm{2}} }\:=\mathrm{0}\:,\:\frac{\partial\bigtriangleup}{\partial{x}_{\mathrm{3}} }\:=\mathrm{0}\:. \\ $$$${using}\:{these}\:{six}\:{equations}\:{we}\:{find} \\ $$$$\left({x}_{{i}} \:,\:{y}_{{i}} \right)\:{and}\:{thus}\:\bigtriangleup_{{max}} \:. \\ $$
Commented by MrW3 last updated on 11/Jun/18
Great!
Commented by MrW3 last updated on 12/Jun/18
$${the}\:{triangle}\:{with}\:{the}\:{max}.\:{area}\:{should} \\ $$$${be}\:{that}\:{one}\:{which}\:{fulfills}: \\ $$$${tangent}\:{at}\:{A}\:{is}\:{parallel}\:{to}\:{BC} \\ $$$${tangent}\:{at}\:{B}\:{is}\:{parallel}\:{to}\:{CA} \\ $$$${tangent}\:{at}\:{C}\:{is}\:{parallel}\:{to}\:{AB} \\ $$
Answered by MJS last updated on 02/Jun/18
$$\mathrm{in}\:{A}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{must}\:\mathrm{be}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line} \\ $$$${M}_{{q}} {M}_{{r}} ,\:\mathrm{same}\:\mathrm{goes}\:\mathrm{for}\:{B}\:\&\:{M}_{{p}} {M}_{{r}} \:\mathrm{and}\:{C}\:\&\:{M}_{{p}} {M}_{{q}} \\ $$$$\mathrm{I}\:\mathrm{came}\:\mathrm{to}\:\mathrm{this}\:\mathrm{conclusion}\:\mathrm{by}\:\mathrm{calculating}\:\mathrm{it}\:\mathrm{for} \\ $$$${p}={q}={r}\:\mathrm{and}\:\mathrm{then}\:\mathrm{reducing}\:{r}.\:\mathrm{I}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{formulas}\:\mathrm{but}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{handle}\:\mathrm{them}.\:\mathrm{with}\:\mathrm{given} \\ $$$$\mathrm{values}\:\mathrm{for}\:{p},\:{q},\:{r}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{and}\:\mathrm{see} \\ $$$$\mathrm{that}\:\mathrm{changing}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{verticles}\:{A},\:{B},\:{C} \\ $$$$\left(\mathrm{use}\:\bigtriangleup\theta\right)\:\mathrm{by}\:\mathrm{any}\:\mathrm{small}\:\mathrm{value}\:\mathrm{will}\:\mathrm{decrease}\:\mathrm{the} \\ $$$$\mathrm{area}. \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{try}\:\mathrm{to}\:\mathrm{show}… \\ $$
Commented by MrW3 last updated on 12/Jun/18
$${I}\:{think}\:{your}\:{conclusion}\:{is}\:{not}\:{correct},\:{sir}. \\ $$$${The}\:{tangent}\:{at}\:{A}\:{should}\:{be}\:{parallel}\:{to}\:{the} \\ $$$${line}\:{BC},\:{not}\:{to}\:{the}\:{line}\:{M}_{{q}} {M}_{{r}} ,\:{etc}. \\ $$