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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.  (As shown in fig. the circles  touch each other ).

$${Find}\:{maximum}\:{area}\:{of}\:{a}\:{triangle} \\ $$$${when}\:{its}\:{vertices}\:{lie}\:{on}\:{the} \\ $$$${cirumference}\:{of}\:{three}\:{circles}. \\ $$$$\left({As}\:{shown}\:{in}\:{fig}.\:{the}\:{circles}\right. \\ $$$$\left.{touch}\:{each}\:{other}\:\right). \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 31/May/18

i have deleted the post...yes i was wrong

$${i}\:{have}\:{deleted}\:{the}\:{post}...{yes}\:{i}\:{was}\:{wrong} \\ $$

Commented by MJS last updated on 31/May/18

anyway thank you for all the input, it′s  good to have many people with many  different points of view

$$\mathrm{anyway}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{all}\:\mathrm{the}\:\mathrm{input},\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{good}\:\mathrm{to}\:\mathrm{have}\:\mathrm{many}\:\mathrm{people}\:\mathrm{with}\:\mathrm{many} \\ $$$$\mathrm{different}\:\mathrm{points}\:\mathrm{of}\:\mathrm{view} \\ $$

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 ...

$${mrW}\:{Sir},\:{i}\:{m}\:{again}\:{wishing} \\ $$$${your}\:{cooperation}\:... \\ $$

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)  .

$${let}\:\:{DE}\:\:{be}\:{x}-{axis}\:\:{with}\:{D}\:{as}\:{origin}. \\ $$$${let}\:\angle{FDE}\:=\alpha \\ $$$${A}\equiv\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:\:,\:\:{B}\equiv\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right)\:\:,\:{C}\equiv\left({x}_{\mathrm{3}} ,{y}_{\mathrm{3}} \right) \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} ={p}^{\mathrm{2}} \\ $$$$\left({x}_{\mathrm{2}} −{p}−{q}\right)^{\mathrm{2}} +{y}_{\mathrm{2}} ^{\mathrm{2}} ={q}^{\mathrm{2}} \\ $$$$\left[{x}_{\mathrm{3}} −\left({p}+{r}\right)\mathrm{cos}\:\alpha\right]^{\mathrm{2}} +\left[{y}_{\mathrm{3}} −\left({p}+{r}\right)\mathrm{sin}\:\alpha\right]={r}^{\mathrm{2}} \\ $$$$\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{{x}_{\mathrm{1}} }&{{y}_{\mathrm{1}} }&{\mathrm{1}}\\{{x}_{\mathrm{2}} }&{{y}_{\mathrm{2}} }&{\mathrm{1}}\\{{x}_{\mathrm{3}} }&{{y}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}\:\:\:{with} \\ $$$$\frac{\partial\bigtriangleup}{\partial{x}_{\mathrm{1}} }\:=\mathrm{0}\:,\:\:\frac{\partial\bigtriangleup}{\partial{x}_{\mathrm{2}} }\:=\mathrm{0}\:,\:\frac{\partial\bigtriangleup}{\partial{x}_{\mathrm{3}} }\:=\mathrm{0} \\ $$$$\Rightarrow\:\:\begin{vmatrix}{\mathrm{1}}&{−\frac{{x}_{\mathrm{1}} }{{y}_{\mathrm{1}} }}&{\mathrm{1}}\\{{x}_{\mathrm{2}} }&{{y}_{\mathrm{2}} }&{\mathrm{1}}\\{{x}_{\mathrm{3}} }&{{y}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left({y}_{\mathrm{2}} −{y}_{\mathrm{3}} \right)+\frac{{x}_{\mathrm{1}} }{{y}_{\mathrm{1}} }\left({x}_{\mathrm{2}} −{x}_{\mathrm{3}} \right)+\left({x}_{\mathrm{2}} {y}_{\mathrm{3}} −{x}_{\mathrm{3}} {y}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\:{similarly}\:\:{two}\:{more}\:{equations} \\ $$$$\:{for}\:\:\:\:\:\:\frac{\partial\bigtriangleup}{\partial{x}_{\mathrm{2}} }\:=\mathrm{0}\:,\:\frac{\partial\bigtriangleup}{\partial{x}_{\mathrm{3}} }\:=\mathrm{0}\:. \\ $$$${using}\:{these}\:{six}\:{equations}\:{we}\:{find} \\ $$$$\left({x}_{{i}} \:,\:{y}_{{i}} \right)\:{and}\:{thus}\:\bigtriangleup_{{max}} \:. \\ $$

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

$${the}\:{triangle}\:{with}\:{the}\:{max}.\:{area}\:{should} \\ $$$${be}\:{that}\:{one}\:{which}\:{fulfills}: \\ $$$${tangent}\:{at}\:{A}\:{is}\:{parallel}\:{to}\:{BC} \\ $$$${tangent}\:{at}\:{B}\:{is}\:{parallel}\:{to}\:{CA} \\ $$$${tangent}\:{at}\:{C}\:{is}\:{parallel}\:{to}\:{AB} \\ $$

Answered by MJS last updated on 02/Jun/18

in A the tangent must be parallel to the line  M_q M_r , same goes for B & M_p M_r  and C & M_p M_q   I came to this conclusion by calculating it for  p=q=r and then reducing r. I tried to find the  formulas but couldn′t handle them. with given  values for p, q, r it′s easy to calculate and see  that changing one of the verticles A, B, C  (use △θ) by any small value will decrease the  area.  I′ll try to show...

$$\mathrm{in}\:{A}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{must}\:\mathrm{be}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line} \\ $$$${M}_{{q}} {M}_{{r}} ,\:\mathrm{same}\:\mathrm{goes}\:\mathrm{for}\:{B}\:\&\:{M}_{{p}} {M}_{{r}} \:\mathrm{and}\:{C}\:\&\:{M}_{{p}} {M}_{{q}} \\ $$$$\mathrm{I}\:\mathrm{came}\:\mathrm{to}\:\mathrm{this}\:\mathrm{conclusion}\:\mathrm{by}\:\mathrm{calculating}\:\mathrm{it}\:\mathrm{for} \\ $$$${p}={q}={r}\:\mathrm{and}\:\mathrm{then}\:\mathrm{reducing}\:{r}.\:\mathrm{I}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{formulas}\:\mathrm{but}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{handle}\:\mathrm{them}.\:\mathrm{with}\:\mathrm{given} \\ $$$$\mathrm{values}\:\mathrm{for}\:{p},\:{q},\:{r}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{and}\:\mathrm{see} \\ $$$$\mathrm{that}\:\mathrm{changing}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{verticles}\:{A},\:{B},\:{C} \\ $$$$\left(\mathrm{use}\:\bigtriangleup\theta\right)\:\mathrm{by}\:\mathrm{any}\:\mathrm{small}\:\mathrm{value}\:\mathrm{will}\:\mathrm{decrease}\:\mathrm{the} \\ $$$$\mathrm{area}. \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{try}\:\mathrm{to}\:\mathrm{show}... \\ $$

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.

$${I}\:{think}\:{your}\:{conclusion}\:{is}\:{not}\:{correct},\:{sir}. \\ $$$${The}\:{tangent}\:{at}\:{A}\:{should}\:{be}\:{parallel}\:{to}\:{the} \\ $$$${line}\:{BC},\:{not}\:{to}\:{the}\:{line}\:{M}_{{q}} {M}_{{r}} ,\:{etc}. \\ $$

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